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https://mathematica.stackexchange.com/questions/137350/variable-increments-in-a-table | # Variable Increments in a Table
I am working on a problem involving the piecewise function:
$$x^2 \quad \mbox{if } \quad x < 2$$
$$(x-3)/(\sqrt{x-2}-1) \quad \mbox{if } \quad x \geq 2$$
I am supposed to find the values of the limits as $x$ approaches $2$ from the right and the left. I've found these values to be $1$ from the right and $4$ from the left.
However, the question I'm struggling to answer involves the formatting of a limit table.
I've been able to make two tables of (x,f[x]) from the left and the right, and I've listed specific values of $x$ to approach $2$ with. However, the question wants me to make a $4$ column table using only one table command, and to use a rule to make the step size increments progressively smaller. I'm not sure how to do this.
Here's the text of the question: "To evaluate a two sided limit, create a single table (generated by a single table command) with four columns, combining the left-and right-sided limit tables into one. Values of $x$ should approach the limiting value moving down the table. To evaluate a limit, create table(s) that use step-size increments that are progressively smaller as the limiting value is approached (eg. x=3, 2.1, 2.01, 2.001,... for a limit as x→ 2+). Rather than listing every $x$-value, use a rule to generate the $x$-values."
Here is how I've made my tables thus far:
TableForm[Table[{x, f[x]}, {x, {4, 3, 2.5, 2.5, 2.25, 2.1, 2.05, 2.02,2.001}}]]
TableForm[Table[{x, f[x]}, {x, {-1, 0, 1, 1.5, 1.75, 1.9, 1.95, 1.97, 1.98, 1.999}}]]
• Is this homework? If so, please tag it as such.
– ciao
Feb 9, 2017 at 8:21
One possible way to use Accumulate to generate the steps.
N@Table[i, {i, Accumulate[Table[1/(2^t), {t, 1, 10}]]}]
Then you can these values for building the steps, by adding and subtracting the above from 2 for each side.
ClearAll[f, i, t, x];
f[x_] := Piecewise[{{x^2, x < 2}, {(x - 3)/(Sqrt[x - 2] - 1), x >= 2}}]
r = Table[{x = 1 + i; x, f[x], x = 3 - i; x, f[x]},
{i,Accumulate[Table[1/(2^t), {t, 1, 10}]]}];
h = {"x", "limit from left", "x", "limit from right"};
Grid[Join[{h}, N@r], Alignment -> Left, Frame -> All]
Plot[f[x], {x, -1, 3}, Frame -> True, GridLines -> Automatic,
GridLinesStyle -> LightGray] | 2022-06-29T16:40:45 | {
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https://math.stackexchange.com/questions/592537/why-can-the-limit-of-a-sequence-approach-a-number-and-converge-but-the-limit-of/592548 | # Why can the limit of a sequence approach a number and converge, but the limit of the series must approach $0$ to converge?
My question may not make much sense because I'm still trying to wrap my mind around infinite sequences and series. I seem to have good working knowledge of when and why to apply a certain tests for a given series, but something just seems like it is missing in my understanding.
Conceptually, why can a sequence converge at any number (except $-\infty$ and $\infty$, which aren't numbers anyway), but the limit of a series must approach $0$ to converge?
Is the reason behind the requirement for the limit of a series approaching $0$ to converge because the series eventually stops summing numbered terms infinitely due to its convergence to a specific number? (Sorry if this question makes no sense)!
And is it acceptable for a sequence to approach any number simply because the sequence isn't being summed?
• You mean thr limit of the summand – Mhenni Benghorbal Dec 4 '13 at 15:17
• Try to look at the partial sums of the series when $a_n$ does not converge to $0$ for $n\rightarrow \infty$. – user112167 Dec 4 '13 at 15:17
• @MhenniBenghorbal excuse my ignorance, but what is a summand? – hax0r_n_code Dec 4 '13 at 15:18
I think you are mixing sequences and their sums. The terminology is not very helpful here, I would suggest the following.
Let $a_k$ be any sequence. Then you can define the sequence of partial sums: $$S_n = \sum_{k=1}^n a_k.$$ Then $$\sum_{k=1}^\infty a_k = \lim_{n\to \infty} S_n.$$
Given any sequence $S_n$ you can find some $a_k$ such that $S_n$ are the partial sums of $a_k$.
So you see that a series may have any limit, as the sequence do. However if a series $S_n$ has a finite limit, then the corresponding sequence $a_k$ must tend to zero.
Don't confuse the series $S_n$ with its general term $a_k$.
• So is it accurate to say that a series is simply a sequence of partial sums from a given sequence? – hax0r_n_code Dec 4 '13 at 15:29
• That is my preferred point of view. – Emanuele Paolini Dec 4 '13 at 15:31
• That's interesting because I never could see the relationship between a sequence and a series. – hax0r_n_code Dec 4 '13 at 15:37
• I say that the terminology does not help because formally the expression: $\sum_{k=1}^\infty a_k$ is a number. So it makes no sense to say that such expression converges or diverges. But really you are referring to the sequence of partial sums, not to the sum itself. – Emanuele Paolini Dec 4 '13 at 15:40
• @EmanuelePaolini It makes perfect sense to say that expression converges or diverges. It is nothing but a limit, and limits can sometimes fail to be numbers, but instead diverge. – Potato Dec 4 '13 at 16:05 | 2019-10-16T10:45:31 | {
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http://math.stackexchange.com/questions/265397/inversion-of-the-euler-totient-function | # inversion of the Euler totient function
Given an integer $n$ find smallest integer $x$ such that $\varphi(x) = n$.
$$10^5 < n < 10^8$$
I know that lower bound for searching is $n+1$ and the upper bound is
$$\frac{n}{e^{0.577}\log(\log(n))} + \frac{3}{\log(\log(n))}$$
But my problem is that for such a large values of $n$ my program with precomputed $\varphi(x)$ values has to do a lot of searching and is highly inefficient.
Can you please provide any other method for doing the same .
Thanks.
-
My apologies if I am missing something basic here, but what function is it that you are defining to mean $\varphi(x)$? – Amzoti Dec 26 '12 at 18:31
Euler totient function.. phi(x)... gives the number of integers less than x which are coprime to x – Purva Gupta Dec 26 '12 at 18:35
So I don't waste time, you are asking for an efficient algorithm for inverting the Euler Totient function. Correct? – Amzoti Dec 26 '12 at 19:26
yes.. which can work for large test cases – Purva Gupta Dec 26 '12 at 19:48
I would recommend changing your title to be more descriptive in case others are searching for this too and there have been previous questions regarding approaches. Regards – Amzoti Dec 27 '12 at 2:19
Warning, you are going to have to do some work to get your hands around this answer, but I provided enough details for you to work through it and it answers your question.
Goal: Given an integer $n$ find the smallest integer $x$ such that $\varphi(x) = n$.
Approach
One can calculate all of the possible integers with a totient of $n$ using "Inverse Totient Trees."
For instance, take the integers with a totient of $24$. Then,
$\varphi(N) = 24$
$\varphi(24) = 8$
$\varphi(8) = 4$
$\varphi(4) = 2$
$\varphi(2) = 1$
There are $5$ "links" (designate this as $L$) in the totient chain so to speak, with $4$ intervals. In general, the greatest integer that can have a totient of $n$ is $2*3^{(L-1)}$, which means that $2*3^{(5 - 1)}$ $= 162$ is the upper bound of an integer with a totient of $24$.
In fact, via a simple proof by exhaustion, one can easily check a table and see that the smallest integer where $\varphi(x) = 24$ is $x = 35$ (see the list below).
$$\varphi(x)= n = 24 \rightarrow x = 35, 39, 45, 52, 56, 70, 72, 78, 84, 90$$
Here are a couple of related number sequences which include references for you to investigate this approach.
A032447 Inverse function of phi( )
A058811 Number of terms on the n-th level of the Inverse-Totient-Tree (ITT)
So, the next obvious question, is there a program that is related or tangentially related that can generate all of those values in the range specified using some approach?
Yes, see Solving $\varphi^{-1}(x) = n$, where $\varphi(x) = n$ is Euler's totient function - testing Carmichael's conjecture. Please see the references there for further details on this program - including the C-source code.
Lets do the example above and the some examples within your range (note that numbers are obviously never odd because $\varphi(x)$ is always even for $n \ge 3$).
$n = 24$: Enter $n = 24, e = 0, f = 0$: and look at the resulting $10$-bolded numbers and compare that list to the above. If you sort that list from low to high, it is all of the $x's$ that produce that desired $n$. Of course, you only want the minimal (last bolded number in the display) one. So $\varphi^{-1}(35) = 24$.
$n = 10^{5}$: Enter $n = 100000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100651$.
$n = 10^5 + 1$: Enter $n = 100001, e = 0, f = 0$: Odd numbers are not permissible by statement above.
$n = 10^{5} + 2$: Enter $n = 100002, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100003$.
$n = 5*10^{5}$: Enter $n = 500000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 640625$.
...
$n = 10^{8}$: Enter $n = 100000000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100064101$.
Asides
This is a very interesting problem and has been asked before, so I am summarizing some of the other discussions and references I found in case others want to consider more approaches.
There are several papers on the topic of finding the inverse of the Euler Totient function:
Euler's Totient Function and Its Inverse, by Hansraj Gupta
The number of solutions of $\phi(x) = m$, by Kevin Ford
On the image of Euler’s totient function, R.Coleman
Complexity of Inverting the Euler Function, by Scott Contini, Ernie Croot, Igor Shparlinski
There have been several questions along these lines, for example, the-inverse-of-the-euler-totient-function, inverting-the-totient-function and finding-the-maximum-number-with-a-certain-eulers-totient-value
There are some other code examples that use different methods for you to consider, see, for example:
Inversion of Euler Totient Function by Max Alekseyev and you can experiment with this since PARI/GP is free. Play around with the function and see if you can modify your approach with this approach.
Discussion and implementation of an efficient algorithm for finding all the solutions to the equation EulerPhi[n] = m, by Maxim Rytin is a nice article off of wolfram that gives an efficient algorithm for computing the inverse of the Euler totient function. Download the invphi.nb file at the bottom and get Mathreader. As an alternative, you can see oeis A006511.
MAGMA - see FactoredEulerPhiInverse(n)
Regards
-
Thanks a lot for your descriptive answer :) – Purva Gupta Dec 27 '12 at 8:19
@Purva Gupta, you are very welcome! We are all here to learn from each other and MSE is a wonderful site! – Amzoti Dec 27 '12 at 9:01
2∗3^(L−1) is it applicable for all or only specific numbers? – Purva Gupta Dec 27 '12 at 9:18
All, it is an upper bound using this method for the largest n, recalling that n must be even numbers only from the note. – Amzoti Dec 27 '12 at 13:08
$+1^{+1^{+1}}\quad\checkmark$ – amWhy May 10 '13 at 1:20
See also my recent paper "Computing the (number of) inverses of Euler's totient and other multiplicative functions", which presents a generic dynamic programming algorithm for finding the inverses of a multiplicative function for a given integer value.
- | 2016-05-27T14:39:36 | {
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http://mathhelpforum.com/algebra/188735-factorization-question-2-a.html | 1. ## Factorization Question 2
For the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?
$\text{Factorize } 12x^2+11x-15$
2. ## Re: Factorization Question 2
Hello
I don't now if this will help, but I do it this way. It doesn't take ages.
You know you have either:
1,15 or 3,5
and your looking for a difference of 11
Write out:
(12x )(x )
(6x )(2x )
(4x )(3x )
It doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11
So you have
(4x - 3)(3x + 5)
3. ## Re: Factorization Question 2
If you have a polynomial $f(x)$ and the zero's $x_1,x_2,...x_n$ are known then the polynomial can be factored as:
$f(x)=a(x-x_1)(x-x_2)....(x-x_n)$ (where $a$ is the coefficient of the highest degree term).
Offcourse in this case you have to find the zero's first (by using the quadratic formula) ...
4. ## Re: Factorization Question 2
Hello, sparky!
For the question below, does anyone have a quick and easy method for
finding the two numbers whose difference is 11 and whose product 180?
I know the numbers are 20 and -9, but it takes me forever to find them?
Is there is quick way?
$\text{Factor: }\:12x^2+11x-15$
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$
We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)
We have: . $12x^2 \quad 9x\quad 20x - 15$
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: . $12x^2 - 9x + 20x - 15$
Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$
5. ## Re: Factorization Question 2
Originally Posted by Soroban
Hello, sparky!
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$
We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)
We have: . $12x^2 \quad 9x\quad 20x - 15$
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: . $12x^2 - 9x + 20x - 15$
Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$
Wow, I love this! Why didn't I think of this? This was so obvious! Thanks
6. ## Re: Factorization Question 2
Originally Posted by Soroban
Hello, sparky!
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$
We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)
We have: . $12x^2 \quad 9x\quad 20x - 15$
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: . $12x^2 - 9x + 20x - 15$
Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$
^this. is beautiful. | 2017-02-27T17:31:37 | {
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When you have more than two events, you ca n extend the binary logistic regression model, as described in Chapter 3. y = c + ax c = constant a = slope. Frank Wood, fwood@stat. Econometrics I: Multiple Regression: Inference - H. One of these variable is called predictor variable whose value is gathered through experiments. When you look at the output for this multiple regression, you see that the two predictor model does do significantly better than chance at predicting cyberloafing, F(2, 48) = 20. In each ex-ample, you will first learn about the specific ingredi-ents required for the power or sample size computa-tion for the linear model being considered. It is also called the coefficient of determination, or the coefficient of multiple determination for multiple regression. Nonlinear regression The model is a nonlinear function of the parameters. University of Dayton. The program’s graph, regression, and correlation functions can respectively produce scatterplots, provide regression equation coefficients, and create correlation matrices. Assumptions of Multiple Regression This tutorial should be looked at in conjunction with the previous tutorial on Multiple Regression. When using multiple regression to estimate a relationship, there is always the possibility of correlation among the independent variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Further details and related logical operations can be found in the R documentation. Multiple Personality Disorder. A sound understanding of the multiple regression model will help you to understand these other applications. Switching Regression Models — Estimation (8) First obtain the expected values of the residuals that are truncated. We will also learn how to decide whether a group of explanatory variables can be excluded from the model. "Linear Regression and Modeling" is course 3 of 5 in the Statistics with R Coursera Specialization. In this paper, the risk factors for a disease of the eye (retinopathy of prematurity) are identi ed using logistic regression analysis. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. With only one independent variable, the regression line can be plotted neatly in two dimensions. However, performing a regression does not automatically give us a reliable relationship between the variables. xls Data Sample (showing only the variables to be used in analysis) Variables Used. Chapter 8: Multiple Choice Questions. Lecture 1 Introduction to Multi-level Models Regression Analysis? • Specification of predictor variables from multiple levels. Simple linear regression. Variable and Dummy Coded Region Variable 273. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. We are not going to go too far into multiple regression, it will only be a solid introduction. zeigler-hill. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Regression analysis is a statistical process for estimating the relationships among variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Many people find this too complicated to understand. I also wanted to practice working with regression algorithms. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. MR&B3 is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. ppt - Free download as Powerpoint Presentation (. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. In the simplest case, we would use a 0,1 dummy variable where a person is given a value of 0 if they are in the control group or a 1 if they are in the treated group. We are not going to go too far into multiple regression, it will only be a solid introduction. – The test is multiple linear regression isThe test is multiple linear regression is H 0 = β 1 = β 2 = … = β p = 00. An Introduction to Logistic Regression: From Basic Concepts to Interpretation with Particular Attention to Nursing Domain ure" event (for example, death) during a follow-up period of observation. Wouldn’t it be great if there was a more accurate way to predict whether your prospect will buy rather than just taking an educated guess? Well, there is…if you have enough data on your previous prospects. This model is built using, for example, a set of real world data listing the outcome in various cases. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Multiple Regression Multiple Proportional Hazards Regr (Cox model) For time dependent outcomes (ie time to death), we model the hazard rate, h , the event rate per unit time (for death, it is the mortality rate). How to Run a Multiple Regression in Excel. Wooldridge, Introductory Econometrics, 4th ed. If two variables are trending over time, a regression. Multiple regression is an extension of simple linear regression. Multiple linear regression has one y and two or more x variables. 8%, with 95% confidence. is a way of understanding the relationship between two variables. For instance, when we predict rent based on square feet alone that is simple linear regression. Stepwise multiple linear regression has proved to be an extremely useful computational technique in data analysis problems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Background Multiple Linear Regression is widely used in academics and also in MR We can consider it the start of Multivariate Analysis, for our course Any idea what the following are: Multivariate analysis Multiple Linear Regression (MLR). The precision of a method (Table 4) is the extent to which the individual test results of multiple injections of a series of standards agree. For example, linear regression can be used to quantify the relative impacts of age, gender, and diet (the predictor variables) on height (the outcome variable). Regression assumes that variables have normal distributions. regress is useful when you simply need the output arguments of the function and when you want to repeat fitting a model multiple times in a loop. 3 Notes: PowerPoint xxx PDF. Suppose we have a cohort of. Lecture 24: Partial correlation, multiple regression, and correlation Ernesto F. Third, multiple regression offers our first glimpse into statistical models that use more than two quantitative. If you have the Excel desktop application, you can use the Open in Excel button to open your workbook and use either the Analysis ToolPak's Regression tool or statistical functions to perform a regression analysis there. Uses of Regression Analysis 1. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. You probably remember the concept of simple linear regression intuition from your high school years. Similar tests. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. Lecture 4: Multivariate Regression Model in Matrix Form In this lecture, we rewrite the multiple regression model in the matrix form. Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$^1$. selection of the dependent and independent variables. edu|http://www. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. ppt), PDF File (. Formula 13. This correlation may be pair-wise or multiple correlation. The linear regression model (LRM) The simple (or bivariate) LRM model is designed to study the relationship between a pair of variables that appear in a data set. Regression analysis is a very widely used statistical tool to establish a relationship model between two variables. , between an independent and a dependent variable or between two independent variables). There are two models of logistic regression, binary logistic regression and multinomial logistic regression. If the universe is only 14 billion years old, how can it be 92 billion light years wide? - Duration: 9:47. It is the correlation between the variable's values and the best predictions that can be computed linearly from the predictive variables. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. It can also be used to estimate the linear association between the predictors and reponses. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Lecture 18: Multiple Logistic Regression - p. You can use it to predict values of the dependent variable, or if you're careful, you can use it for suggestions about which independent variables have a major effect on the dependent variable. Known also as curve fitting or line fitting because a regression analysis equation can be used in fitting a curve or line to. Intercept: the intercept in a multiple regression model is the mean for the response when. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. The data was split into three employment sectors Teaching, government and private industry Each sector showed a positive relationship Employer type was confounded with degree level Simpson’s Paradox In each of these examples, the bivariate analysis (cross-tabulation or correlation) gave misleading results Introducing another variable gave a. Here, the summary(OBJECT) function is a useful tool. +x K β K +ε where y is the dependent or. Upload and Share PowerPoint Presentations. Pantula David A. covariates for the. MV - Multiple Regression. The generic form of the linear regression model is y = x 1β 1 +x 2β 2 +. Typically points further than, say, three or four standard deviations from the mean are considered as "outliers". This post will show you examples of linear regression. Regression Analysis. patient, f. In this exercise we apply the same method to validating the Novy-Marx Quality and Value screener’s ability to predict the 5-year total return. Multiple Regression Analysis using SPSS Statistics Introduction. 01, with an R-square of. Briefly speaking, the goal of the multiple linear regression is to point out the relation between a dependent variable (explained, endogenous or resultative) and a great deal of independent variables (explanatory, factorial,. Comparing Multiple Regression Model Results against Historic Demand. However, it is critical to recognize that multiple regression is inherently a correlation technique and cannot. " In the main dialog box, input the dependent variable. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Rawlings Sastry G. Regression analysis is a statistical technique for estimating the relationship among variables which have reason and result relation. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. persistence of shocks will be infinite for nonstationary series • Spurious regressions. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. If two variables are trending over time, a regression. RSM is a method used to locate the optimum value of the response and is one of the final stages of experimentation. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Suppose we have a cohort of. Cherry Blossoms WS: Test Review. ID: 305741 ID: 305741. Linear regression Linear dependence: constant rate of increase of one variable with respect to another (as opposed to, e. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. 1 Introduction to Poisson Regression As usual, we start by introducing an example that will serve to illustrative regression models for count data. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Multiple (Linear) Regression. This tutorial will explore how R can be used to perform multiple linear regression. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. is a way of understanding the relationship between two variables. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. This article is a part of the guide:. Suppose that there is a cholesterol lowering drug that is tested through a clinical trial. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. We have been reviewing the relationship between correlation (r and r2) and regression (R, R2 and 1-R2) in class, through lectures, blog. Nonlinear Regression Functions Chapter 9. Regression Analysis. (3) Does the combination of predictors in this fitted multiple regression explain significant variation in the response? Statistics 621 Multiple Regression Practice Questions. Multiple Regression Analysis using SPSS Statistics Introduction. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. selection of the dependent and independent variables. Consider first the case of a single binary predictor, where x = (1 if exposed to factor 0 if not;and y =. The output varies linearly based upon the input. Multiple regression uses the ordinary least squares solution (as does bi. The whole model F test (test of the useful of the model) tests whether the slopes on all variables in multiple regression are zero, i. Ryan McEwan and Julia Chapman. Simple linear regression involves a single independent variable. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. Introduction to Linear Regression and Correlation Analysis Fall 2006 – Fundamentals of Business Statistics 2 Chapter Goals To understand the methods for displaying and describing relationship among variables. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. Chapter 8: Multiple Choice Questions. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Regression with two or more predictors is called multiple regression Available in all statistical packages Just like correlation, if an explanatory variable is a significant predictor of the dependent variable, it doesn't imply that the explanatory variable is a cause of the dependent variable. It is discussed in Response Surface Methods. Hey I would like to make a scatter plot with p-value and r^2 included for a multiple linear regression. 0 Chapter 6: Multiple Linear Regression Topics Explanatory Modeling Predictive Modeling Example: Prices of Toyota Corolla ToyotaCorolla. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. Regression is the analysis of the relation between one variable and some other variable(s), assuming a linear relation. Also referred to as least squares regression and ordinary least squares (OLS). Regression when all explanatory variables are categorical is “analysis of variance”. 5 Correlation and Regression Simple regression 1. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. The Regression df is the number of independent variables in the model. We can have only two models or more than three models depending on research questions. For example, consider the cubic polynomial model which is a multiple linear regression model with three regressor variables. Multiple Regression 18. X2 1 or even interactions X1 X2. The following model is a multiple linear regression model with two predictor variables, and. Find PowerPoint Presentations and Slides using the power of XPowerPoint. This post will show you examples of linear regression. Please access that tutorial now, if you havent already. The basic command is "regression": "linear. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. Objectives. MULTIPLE LINEAR REGRESSION ANALYSIS USING MICROSOFT EXCEL by Michael L. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Frank Wood, fwood@stat. 1 Introduction. " Coefficient table, bottom. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. View and Download PowerPoint Presentations on Spss Tutorial For Multiple Logistic Regression PPT. One of the applications of multiple linear regression models is Response Surface Methodology (RSM). State the multiple regression equation. For ordina l categorical variables, the drawback of the. Linear, Ridge Regression, and Principal Component Analysis Geometric Interpretation I Each column of X is a vector in an N-dimensional space (NOT the p-dimensional feature vector space). Linear Regression. The model is linear because it is linear in the parameters , and. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. In R, multiple linear regression is only a small step away from simple linear regression. Regression Analysis. The basic command is “regression”: “linear. Multiple linear regression in R Dependent variable: Continuous (scale/interval/ratio) Independent variables: Continuous (scale/interval/ratio) or binary (e. y = c + ax c = constant a = slope. 8%, with 95% confidence. We can run regressions on multiple different DVs and compare the results for each DV. Multiple Regression Models • Advantages of multiple regression • Important preliminary analyses • Parts of a multiple regression model & interpretation • Differences between r, bivariate b, multivariate b & • Steps in examining & interpreting a full regression model Advantages of Multiple Regression Practical issues …. See the Handbook for information on these topics. Multiple Regression - Linearity. ppt - Free download as Powerpoint Presentation (. This correlation may be pair-wise or multiple correlation. Arial Franklin Gothic Book Perpetua Wingdings 2 Calibri Equity 1_Equity 2_Equity 3_Equity 4_Equity Microsoft Equation 3. Be sure to tackle the exercise and the quiz to get a good understanding. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. 2% with 95% confidence. Lecture 18: Multiple Logistic Regression - p. The measured standard deviation can be subdivided into 3 categories: repeatability, intermediate precision and reproducibility (4, 5). Find PowerPoint Presentations and Slides using the power of XPowerPoint. By selecting the features like this and applying the linear regression algorithms you can do polynomial linear regression; Remember, feature scaling becomes even more important here; Instead of a conventional polynomial you could do variable ^(1/something) - i. MR&B3 is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. Introduction to Correlation and Regression Analysis. It is used to also to determine the overall fit of the model and the contribution of each of the predictors to the total variation. Uses of Regression Analysis 1. Variable Interactions 5. 11 LOGISTIC REGRESSION - INTERPRETING PARAMETERS 11 Logistic Regression - Interpreting Parameters Let us expand on the material in the last section, trying to make sure we understand the logistic regression model and can interpret Stata output. Figure 14 – Model Summary Output for Multiple Regression. Starting values of the estimated parameters are used and the likelihood that the sample came from a population with those parameters is computed. On the contrary, regression is used to fit a best line and estimate one variable on the basis of another variable. For simple linear regression (only one x), the Regression df is 1. Source: Hair et al. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Predictions in neurophysiology. Nonlinear regression The model is a nonlinear function of the parameters. The only difference between simple linear regression and multiple regression is in the number of predictors (“x” variables) used in the regression. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. specification of a statistical relationship, and 3. Upload and Share PowerPoint Presentations. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Popular spreadsheet programs, such as Quattro Pro, Microsoft Excel,. biostatcourse. Teaching\stata\stata version 13 – SPRING 2015\stata v 13 first session. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. How to improve leadership skills in the workplace ppt for write my essay nursing best cover letter editing site uk , writing a scientific report buy a college essay online. Logistic regression is a statistical method for analyzing a dataset in which there are one or more independent variables that determine an outcome. Regression with categorical variables and one numerical X is often called “analysis of covariance”. It could be, for example, performance, education, or experience. The course starts with a discussion of the logic of the multivariate regression model and the central assumptions underlying the ordinary least squares approach. Also referred to as least squares regression and ordinary least squares (OLS). Linear regression for the advertising data Consider the advertising data shown on the next slide. The method of multiple regression sought to create the most closely related model. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Multiple regression is used to build a model that allows us to study this interplay. What to report? What a statistics program gives you: For a simple regression (one independent variable), statistics programs produce two estimates, a (the "constant term") and b (the "linear coefficient"), for the parameters α and β, respectively. For example: (x 1, Y 1). 2% with 95% confidence. | 2019-12-13T08:59:17 | {
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https://www.hpmuseum.org/forum/thread-383.html | HP 11C real root finder [Newton Method]
01-09-2014, 11:59 PM
Post: #1
Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
HP 11C real root finder [Newton Method]
LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
CHS
RCL 1
+
STO 3
RCL 1
-
ABS
RCL 2
X>Y?
RTN
RCL 3
STO 1
GTO 0
LBL 1
f(x)=0 code
LBL 2
f'(x)=0 code
R1: old value (or initial value)
R2: Tolerance of error (0.001, 0.0001, 0.000000001 etc)
THE RESULT:
R3: new value (root)
EXAMPLE
Find the root of f(x)=x^3 - 3x^2-6x+8
rewritten as f(x)= [(x-3)x-6]x +8
f(x) code:
LBL1
ENTER
ENTER
ENTER
3
-
*
6
-
*
8
+
RTN
f'(x)= 3x(x-6)-6
f'(x) code
LBL 2
ENTER
ENTER
6
-
*
3
*
6
-
RTN
The roots are [-2, 1, 4]
give a initial value, for example: -3
-3 STO 1
give a tolerance of error, for example: 0.0001
0.0001 STO 2
BEGIN THE PROGRAM...
GSB 0
runnning...
when the error < TOL the program stop and display:
0.0001
you can find the root in the register 3
RCL 3
DISPLAY: -2.0001
try with initial value = 2
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Best Regards
01-10-2014, 12:57 AM
Post: #2
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Using storage arithmetic allows us to get rid of a few steps:
LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
STO- 1
ABS
RCL 2
X<=Y?
GTO 0
RCL 1
RTN
You made a mistake calculating the 1st derivative.
This is the corrected program:
f'(x)= 3x^2 - 6x - 6 = (3x - 6)x - 6
f'(x) code
LBL 2
ENTER
ENTER
3
*
6
-
*
6
-
RTN
Kind regards
Thomas
01-10-2014, 05:28 PM
Post: #3
Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Best regards
Best Regards
01-12-2014, 08:31 AM
Post: #4
Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Here is a version that requires coding f(x) only since it approximate f'(x) as:
f'(x) = (f(x+h) - f(x))/h
Where h = 0.001*(ABS(X)+1)
The new version uses registers R1 through R4.
Code:
LBL 0 RCL 1 ABS 1 + EEX 3 CHS * STO 3 # calculate and store increment h RCL 1 GSB 1 STO 4 # calculate and store f(x) RCL 1 RCL 3 + GSB 1 # calculate f(x+h) RCL 4 - 1/X RCL 4 * RCL 3 * # calculate diff = h *f(x)/(f(x+h) - f(x)) STO- 1 ABS RCL 2 X<=Y? GTO 0 RCL 1 RTN
01-12-2014, 01:26 PM
Post: #5
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 08:31 AM)Namir Wrote: ...h = 0.001*(ABS(X)+1)
First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)
This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.
Dieter
01-14-2014, 08:56 PM
Post: #6
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
If you only want to solve polynomials with real coefficients Bairstow's Method can be used.
In a nutshell: instead of a root a quadratic factor is found in each iteration. To find the roots of this factor the classic formula is used.
Example:
$$x^3-3x^2-6x+8=0$$
First enter the coefficients of the polynomial. Always start with register 9:
1 STO 9
-3 STO .0
-6 STO .1
8 STO .2
Then specify the order of the polynomial with a loop-control value. It defines the registers you used for the coefficients:
9.012 STO 6
And now give an initial guess. Probably {1, 1} will do in all cases:
1 STO 7
STO 8
Start the program B:
GSB B
Now the coefficients of the quadratic factor can be found in registers 7 and 8:
RCL 7
1.000000
RCL 8
-2.000000
The coefficient of the first term is always 1. Thus the factor is $$x^2+x-2$$.
Now we can solve this quadratic equation using program A:
1
RCL 7
RCL 8
GSB A
1.000000
X<>Y
-2.000000
Thus: $$x^2+x-2=(x-1)(x+2)$$.
What is left? Let's have a look at it:
RCL 6
9.010
RCL 9
1.000000
RCL .0
-4.000000
This is a linear factor. Thus we end up with the following factorization: $$x^3-3x^2-6x+8=(x-1)(x+2)(x-4)$$
Therefore the solutions are: {1, -2, 4}
Example from Bunuel66's solution to the crossed ladders problem
$$C^4-30C^3+700C^2-21,000C+157,500=0$$
Enter the coefficients of the polynomial:
1 STO 9
-30 STO .0
700 STO .1
21,000 STO .2
157,500 STO .3
Specify the loop-control value:
9.013 STO 6
Use the initial guess {1, 1}:
1 STO 7
STO 8
Start the program:
GSB B
The coefficients of the quadratic factor are in registers 7 and 8:
RCL 7
-35.178025
RCL 8
248.596895
Solve this quadratic equation using program A:
1
RCL 7
RCL 8
GSB A
25.384938
X<>Y
9.793087
RCL 6
9.011
RCL 9
1.000000
RCL .0
5.178025
RCL .1
633.555781
If we try to solve this quadratic equation we get an Error 0.
That's because we're trying to calculate the square root of a negative value.
However we can still get the desired result:
RCL 9
RCL .0
RCL .1
GSB B
Error 0
<-
CHS
626.852796
$$\sqrt{x}$$
25.037029
X<>Y
-2.589012
Thus the final list of solutions is:
• 25.384938
• 9.793087
• -2.589012$$\pm$$25.037029i
Cheers
Thomas
01-15-2014, 05:53 AM
Post: #7
Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 01:26 PM)Dieter Wrote:
(01-12-2014 08:31 AM)Namir Wrote: ...h = 0.001*(ABS(X)+1)
First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)
This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.
Dieter
Sure, dividing by 1000 is a step shorter. Originially, I had learned to calculate h using an If statement:
Code:
If |x| >= 1 Then h= x/100 else h = 1/100 end if
Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.
Namir
01-15-2014, 08:38 PM
Post: #8
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Namir, sorry that I was not able to express myself clearly. I wanted to point out that I do not think it's a good idea to use h = 0,01 or 0,001 for any value below 1. This may lead to significant errors since h may be much, much larger than x. For instance, if x = 1E-10, h = 1E-3 is not recommended. Here, h = 1E-13 should be better.
That's why I prefer to set h = 0,001 x or similar. With the only exception x=0 where x may be 0,001.
(01-15-2014 05:53 AM)Namir Wrote: Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.
This can be coded this way, for instance:
Code:
X=0? e^x EEX 3 /
Look, no labels or gotos required. ;-)
Or even more elegant on the 34s:
Code:
X=0? INC X SDR 3 RSD 1 ' optional
Dieter
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https://www.physicsforums.com/threads/basic-taylor-polynomial-question-involving-e-x-2.782018/ | # Basic Taylor Polynomial Question involving e^(-x)^2
1. Nov 15, 2014
### RJLiberator
1. The problem statement, all variables and given/known data
Consider:
F(x) =$\int_0^x e^{-x^2} \, dx$
Find the Taylor polynomial p3(x) for the function F(x) centered at a = 0.
2. Relevant equations
Tabulated Taylor polynomial value for standard e^x
3. The attempt at a solution
I started out by using the tabulated value for Taylor polynomial e^x and replaced all x with -x^2.
Going to the degree of three I received:
e^(-x)^2 = 1 - (x^2)/1! + x^4/2! - x^6/3!
Is this all there is to it? The question simply states to find the Taylor Polynomial p3(x) for the function, nothing else. I don't need to find the sum or the remainder per the problems question. Correct?
If I were to go at the sum, would I integrate each value?
Thank you for any help, I think I am just getting used to the wording here more than anything.
2. Nov 15, 2014
### gopher_p
The Taylor polynomial $p_3$ of degree $3$ for $F$ centered at $0$ should look something like $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3,$$ yes? If you remember the Fundamental Theorem of Calculus and what is says about computing $F'(x)$, it shouldn't be too hard to get the coefficients, right? So you can just do the problem "properly" without worrying whether your methods work.
I'm not saying your way won't work, though you'd theoretically only need the degree $2$ Taylor polynomial for $f(x)=e^{-x^2}$ to extract the degree $3$ polynomial for $F$. But the question you'd want to ask is whether the derivative of the degree $n$ Taylor polynomial of a function $g$ really is the degree $n-1$ Taylor polynomial for the function $g'$ and vice -versa; i.e. is the (anti)derivative of the Taylor polynomial the Taylor polynomial of the (anti)derivative? I don't think that question is too incredibly difficult to answer for those who understand what it is asking.
3. Nov 17, 2014
### RJLiberator
Hm. Bare with me. I am in the early process of understanding Taylor Series wordage:
So what you are saying is that I should take the derivative of e^(-x)^2 a few times to acquire the coefficients? If I do this I answer:
f(0) = 1
f'(0) = 1
f''(0) = 2
f'''(0) = 4
Which would mean my original equation is incorrect. I am not sure what you mean by doing the problem properly without worrying.
I guess my first question would be: Is my answer correct? All I did was input -x^2 to the e^x common taylor series polynomial expansion for the first 3 degrees.
Then my second question would be how to proceed with this problem 'properly' ? I will review the fundamental theorem of calculus.
Question 3: Out or curiosity, why would I need to only go to degree 2 for this problem? Doesn't it state to go to degree 3?
4. Nov 17, 2014
### Staff: Mentor
1. Yes, what you did looks fine.
2. Integrate the series you found.
3. If you have terms up to degree 2, when you integrate, what will be the highest degree you get?
5. Nov 17, 2014
### RJLiberator
Mark44, you are shedding light on this topic!
1. Thank you for your confirmation.
2. Hm, ok. I am going to mess around with a few things. I know my coefficients then, now I got to find where to plug them in and integrate.
3. Ah.. integrating would bring it to degree three.
6. Nov 17, 2014
### gopher_p
The way that I was suggesting you do the problem is as follows;
For $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$,
$F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t=0$
$F'(x)=e^{-x^2}\Rightarrow F'(0)=1$, where the derivative is given by the FTC.
$F''(x)=-2xe^{-x^2}\Rightarrow F''(0)=0$
$F'''(x)=-2e^{-x^2}+4x^2e^{-x^2}\Rightarrow F'''(0)=-2$
Then the Taylor polynomial of degree three centered at $a=0$ is given by $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3=x-\frac{1}{3}x^3$$
So we're just (a) appealing to the definition of the Taylor polynomial and (b) computing the coefficients directly. I guess that's what I meant by "properly".
7. Nov 18, 2014
### RJLiberator
OK, so with your help, I've been working through this problem.
Question 1: So for F(0) you made the bounds 0 to 0. I don't understand why. I see that the answer is clearly 0 when the bounds are both 0 to 0, but why do this?
Other then this part, everything you wrote now makes complete sense to me.
Question 2: In my original solution where I just inputted -x^2 to the e^x tabulated value i had NOT integrated yet. Did I HAVE to integrate after I replaced those values? If so, the two answers match up nicely. I would imagine so, by the definition of the Taylor Series, correct?
8. Nov 18, 2014
### gopher_p
One of the "skills" that you learned in you preparation for calculus was that, given a formula for $F(x)$, you compute $F(a)$ by plugging in $a$ for $x$. So given $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$, get $F(0)$ by plugging in $0$ for $x$; $F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t$. This is why it is important to use a different variable of integration - $t$ here - inside the integral.
I'm confused by the question. It's clear that the degree two Taylor polynomial $\tilde p_2$ for $f(x)=e^{-x^2}$ is (a) not a degree three polynomial and (b) not a Taylor polynomial for the function $F$, right? So something must be done if we are to extract $p_3$ from $\tilde p_2$.
The real question, in my mind, is whether it is true that $p_3(x)=\int_0^x\tilde p_2(t)\ \mathrm{d}t$. In this case it is. It works here because the lower limit of integration, $0$, is the same as the center of the Taylor polynomials. You need to be careful in a more general setting.
9. Nov 18, 2014
### PeroK
You should note that where you wrote:
F(x) =$\int_0^x e^{-x^2} \, dx$
gopher_p wrote:
F(x) =$\int_0^x e^{-t^2} \, dt$
I think you need to understand the difference and why the second one correctly defines F as a function of x, with t as the dummy variable.
This may also be the reason why you did not understand the substitution $x = 0$
10. Nov 18, 2014
### RJLiberator
Guys, thanks for your help here.
You were a help to me and my classmates.
Also, thanks for pointing out my subtle mistakes that cause errors in conceptual understanding. I am starting to understand this problem due to this.
I see why the dummy variable matters.
As always, excellent help. | 2017-11-21T18:04:43 | {
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https://math.stackexchange.com/questions/1938052/find-the-function-fx-if-fx2fy-fxyfy/1938110 | # Find the function $f(x)$ if $f(x+2f(y))=f(x)+y+f(y)$
Let $f:\mathbb{R}\to \mathbb{R}$ such $f(x)$ at $x=0$ continuous, and for any $x,y\in \mathbb{R}$ such $$f(x+2f(y))=f(x)+y+f(y)$$ Find $f(x)$.
Let $x=0,y=0$ then we have $$2f(2f(0))=f(0)++f(0)$$ Let $y=2f(0)$ then we have $$f(x+2f(2f(0))=f(x)+2f(0)+f(2f(0))$$ so we have $$f(x+4f(0))=f(x)+4f(0)$$ on the other hand we have $$f(x+4f(0))=f(x+2f(0))+0+f(0)=2f(0)+f(x)$$ so we have $f(0)=0$ Then I can't deal with this problem
• Maybe I am off base but what happens if you plug in $y=x$? Sep 23, 2016 at 5:29
• @CameronWilliams you get that $x+2f(x)$ is a fixed point. Sep 23, 2016 at 5:30
• Am I missing something or does $f(x)=x$ work? Sep 23, 2016 at 6:21
• @Mastrem It does, but presumably the question is to find all such functions. Sep 23, 2016 at 6:22
• @Mastrem it works, but no one says this equation has only one solution. You have to prove then that it's the only one. Sep 23, 2016 at 6:23
Substitute $x\mapsto y$ to get $$f(y+2f(y))=y+2f(y).$$ Now substitute $y\mapsto y+2f(y)$: $$f(x+2f(y+2f(y)))=f(x)+y+2f(y)+f(y+2f(y)).$$ Thus $$f(x+2y+4f(y))=f(x)+2y+4f(y).$$ Substituting $x\mapsto x+2y+2f(y)$, $$f(x+2y+4f(y))=f(x+2y+2f(y))+y+f(y).$$ Substituting $x\mapsto x+2y$, $$f(x+2y+2f(y))=f(x+2y)+y+f(y).$$ Combining the last 3 equations, $$f(x+2y)=f(x)+2f(y).$$ Thus $$f(x)=f(0+2x/2)=f(0)+2f(x/2)=2f(x/2),$$ so $$f(x+y)=f(x+2y/2)=f(x)+2f(y/2)=f(x)+f(y).$$ Since you are given that $f$ is continuous at $0$, it must be continuous everywhere. Now there is a standard argument to show that $f(x)=f(1)x$ (prove it on rationals and extend by continuity). Finally using the original equation, we see $f(1)=1$ or $-\frac12$, so the solutions are $f(x)=x$ or $f(x)=-x/2$.
Setting $x=y$ gives $f(x+2f(x))=x+2f(x)$, and so $f(y)=y$ whenever $y$ is of the form $x+2f(x)$ for some $x$. Note also that if $f(y)=y$, then for any $x$, $f(x+2y)=f(x)+2y$.
Now consider the function $g(x)=f(x)-x$. From the previous paragraph, we see that if $f(y)=y$ then $g(x+2y)=g(x)$. In particular, every number of the form $2x+4f(x)$ is a period of $g$. Let $P\subseteq\mathbb{R}$ be the group of periods of $g$, i.e. the set of $p$ such that $g(x)=g(x+p)$ for all $x$. There are now two cases.
The first case is that $P$ is not cyclic. Then $P$ is dense in $\mathbb{R}$, and so each coset of $P$ is dense in $\mathbb{R}$, and in particular $0$ is in the closure of each coset. But by definition of $P$, $g$ is constant on each coset of $P$, and so by continuity of $g$ at $0$ this constant value must be equal to $g(0)$. Thus $g(x)=g(0)$ for all $x$, and $g$ is constant everywhere. Thus $f(x)$ has the form $f(x)=x+c$ for some constant $c$. Plugging this into the functional equation gives $$x+2(y+c)+c=x+c+y+y+c,$$ so $c=0$. Thus $f(x)=x$.
The second case is that $P$ is cyclic, generated by some $p\in\mathbb{R}$ (possibly $0$). Then every number of the form $2x+4f(x)$ is an integer multiple of $p$. But $2x+4f(x)$ is continuous at $0$, so this means $2x+4f(x)$ is constant in a neighborhood of $0$. As you've shown, $f(0)=0$, so $2x+4f(x)=0$ for all $x$ in some neighborhood of $0$, so $f(x)=-x/2$ for all $x$ in some neighborhood of $0$.
Now suppose $\epsilon>0$ is such that $f(x)=-x/2$ whenever $|x|\leq\epsilon$. If $0\leq a\leq \epsilon$, the functional equation with $x=a$ and $y=-\epsilon$ then gives $$f(a+\epsilon)=-\frac{a+\epsilon}{2}.$$ That is, $f(x)=-x/2$ is also valid if $\epsilon\leq x\leq 2\epsilon$. Similarly, using $y=\epsilon$ gives that $f(x)=-x/2$ is also valid if $-2\epsilon\leq x\leq -\epsilon$.
That is, if $f(x)=-x/2$ whenever $|x|\leq\epsilon$, $f(x)=-x/2$ whenever $|x|\leq 2\epsilon$ as well. Repeating this argument over and over, we get that $f(x)=-x/2$ for all $x$.
Thus the only possibilities for $f$ are $f(x)=x$ and $f(x)=-x/2$, and you can easily check that both of these work.
• Nice try! Even if using machinery which .. outguns the problem! I downvoted by accident.(and I thought I had already cancelled it. Seems I hadn't!) I will make a trivial edit so that I'll be able to cancel it. P.S.: the part where you are arguing about the cosets of the group of periods is not very clear. Could you expand it a little? (what is the equivalence relation here? why does the continuity of $g$ implies that every coset accumulates at $0$ ?). Sep 23, 2016 at 18:35
• I've clarified that part a bit. Sep 23, 2016 at 19:12
• Thanks for your response. As I've already said: very nice ! +1 Sep 25, 2016 at 19:49 | 2022-08-12T21:39:46 | {
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https://math.stackexchange.com/questions/3356386/geometry-prove-a-right-triangle-resulting-from-three-inscribed-circles | Geometry - Prove a right triangle resulting from three inscribed circles
Two half circles and a full circle fit inside a larger quarter circle as shown in the diagram. The centers of the two half circles are on the two sides of the quarter circle, respectively.
Prove that the triangle formed by the centers of three smaller circles, $$\triangle O_1O_2O_3$$, is a right triangle.
I was able to apply the Pythagorean formula to a few triangles involving the radii of the inscribed circles and proved that the ratios of the three radii are 1:2:3. Then, the centers of the three circles form a triangle with side-length ratios 3:4:5, hence, a right triangle.
On the other hand, I feel the proof may be an overkill, and evaluating the three radii explicitly may be unnecessary. There ought to be clean geometric solutions to show directly that the vertex $$O_3$$ is of a right angle, which I am not sure how to figure out.
• Suppose $OO_1$ and $OO_2$ are x and y axis. If you complete half and quarter circles, you will have a big circle containing two big circles, two medium and four little circles. the center of little circles are symmetric about x and y axis. since x and y axis are perpendicular then the lines connecting the centers of little circles are perpendicular that is triangle $O_1O_2O_3$ is right angle at $O_3$. – sirous Sep 14 '19 at 21:01
• Insightful. A good case of can’t-see-forest-staring at trees. Thanks – Quanto Sep 14 '19 at 21:10
• @sirous I finally read your comment. I think it would be clearer if you "complete" the half and quarter circles by reflecting the figure over one axis and then the other. But your construction is much more elegant than mine and would make a fine answer. – David K Sep 15 '19 at 0:30
• @sirous: Certainly, "the lines connecting the centers of the little circles are perpendicular", but how do you know that those lines contain $O_1$ and $O_2$? – Blue Sep 15 '19 at 3:51
• @Blue - critical observation. guess a simple geometric proof remains elusive – Quanto Sep 15 '19 at 14:42
Given the quarter circle with $$O$$ as the center of the arc, and given semicircles with centers $$O_1$$ and $$O_2$$ on the two straight sides of the quarter circle such that $$O_1$$ and $$O_2$$ both are tangent to the arc of the quarter circle and such that $$O_1$$ and $$O_2$$ are tangent to each other.
This is based on the figure in the question, except that we do not assume that $$O_1$$ is the midpoint of one side of the quarter circle.
Let the radii of the quarter circle, the semicircle about $$O_1$$, and the semicircle about $$O_2$$ be $$r,$$ $$r_1,$$ and $$r_2$$ respectively.
Let $$P$$ be the fourth vertex of the rectangle that has three of its vertices at $$O,$$ $$O_1,$$ and $$O_2.$$ The sides of this rectangle are $$OO_1 = r - r_1$$ and $$OO_2 = r - r_2.$$ The diagonals are $$OP = O_1O_2 = r_1 + r_2.$$
Let $$P_1$$ be the point of intersection of the semicircle about $$O_1$$ with $$O_1P$$ and $$P_2$$ be the point of intersection of the semicircle about $$O_2$$ with $$O_2P.$$ Then $$PP_1 = r - r_1 - r_2 = PP_2.$$ Moreover, the distance from $$P$$ to the arc of the quarter circle is $$r - (r_1 + r_2) = r - r_1 - r_2.$$
Then a circle of radius $$r - r_1 - r_2$$ about $$P$$ is tangent to the quarter circle and to the semicircles about $$O_1$$ and $$O_2.$$ Therefore $$P$$ is $$O_3$$ in the figure, the center of the circle inscribed between the two semicircles and the quarter circle.
Since $$O_3$$ is a vertex of the rectangle $$OO_1O_3O_2,$$ the triangle $$\triangle O_1O_3O_2$$ is a right triangle. $$\square$$
• +1 a nice generalization. – achille hui Sep 15 '19 at 16:27
Let $$R=6a$$ then $$r_1=3a$$
$$\triangle OO_1O_2: \;\;\;\; (R-r_2)^2+r_1^2 =(r_1+r_2)^2$$
So $$\boxed{r_2= 2a}$$
So $$\triangle O_3O_1O_2$$ is right iff $$(r_1+r_2)^2 = (r_1+r_3)^2+(r_2+r_3)^2$$
or $$r_1r_2 = r_3(r_1+r_2+r_3)$$
or $$6a^2 = r_3(5a+r_3)\iff r_3^2+5ar_3-6a^2 =0$$
So we have to prove: $$\boxed{\color{red}{r_3=a}}$$
If we put everything in coordinate system then we have: $$O(0,0)$$, $$O_1(3a,0)$$, $$O_2(0,4a)$$ and let $$O_3(m,n)$$, so $$\begin{eqnarray} m^2+n^2 &=& (6a-r_3)^2\;\;\;\;\;\; (OO_3 = R-r_3)\\ (m-3a)^2+n^2 &=& (3a+r_3)^2\;\;\;\;\;\; (O_1O_3 = r_1+r_3)\\ m^2+(n-4a)^2 &=& (2a+r_3)^2\;\;\;\;\;\; (O_2O_3 = r_2+r_3) \end{eqnarray}$$
If we substract 1.st and 2.nd equation we get $$m = 6a-3r_3$$ and if we substract 1.st and 3.rd equation we get $$n=6a-2r_3$$
Now put this again in 1.st equation and we get $$\boxed{\color{green}{r_3=a}}$$
Following is an answer based on circle inversion w/o computing the radius of circle centered at $$O_3$$.
Label the contact points as illustrated below. We are going to show $$\angle OYD = \angle OXC = 45^\circ$$ Form these, it is easy to deduce $$\begin{cases} O_2O_3 \parallel O_2D \parallel OX,\\ O_1O_3 \parallel O_1C \parallel OY\end{cases}$$ and hence $$\angle O_1O_3O_2 = 90^\circ$$
Choose a coordinate system so that $$O$$ is the origin and $$X = (1,0)$$, $$Y = (0,1)$$. The arc $$XEY$$ will be part of the unit circle. Under circle inversion with respect to the unit circle centered at $$O$$, we get what illustrated below:
• The red semicircle centered at $$O_1$$ get mapped to a ray $$y = 1$$ in the upper half plane.
• The green semicircle centered at $$O_2$$ get mapped to another semicircle centered at some point of $$y$$-axis and touching the red line. This means this inverted green semicircle has radius $$1$$.
As a result, the inverted image of point $$A$$ is $$A' = (0,3) \implies A = (0,\frac13)$$.
• The blue circle get mapped to a circle placed symmetrically between the inverted green semicircle and the arc $$XEY$$. As a result, the inverted image of point $$C$$ is $$C' = (1,1)$$.
This implies $$\angle OXC = \angle OXC' = 45^\circ$$.
For the other angle, translate the coordinate axis so that $$Y$$ is the new origin. The new coordinate of $$A$$ is now $$(0,-\frac23)$$. Under circle inversion with respect to the unit circle centered at $$Y$$, it now looks like:
• The arc $$XEY$$ get mapped to a ray on the line $$y = -\frac12$$.
• The green semicircle centered at $$O_2$$ get mapped to a ray on the line $$y = -\frac32$$.
• The red semicircle get mapped to an arc on a circle sandwiched between above two rays touching the $$y$$-axis. So its radius is $$\frac12$$.
• The blue circle get mapped to an circle also sandwiched between above two rays. This means the inverted image of $$D$$ is located at $$D'' = (\frac32,-\frac32)$$. As a result, $$\angle OYD = \angle OYD' = 45^\circ$$.
Let $$OO_1$$ and $$OO_2$$ are x and y axis respectively. Mirror the figure about y axis, then mirror the resulted figure about x axis. You get two big, two medium and four small circles indescribable in a large circle. If you connect the centers of four small circles you get a square. So angle at $$O_3$$ is $$90^o$$. Now let the radius of small circle is $$r_1=1$$ and that of large circle is $$r_2=6$$. Suppose the location of center of small circle is $$O_3(3, 4)$$ which satisfies:
$$3^2+4^2= [OO_3=6-1=5]^2$$
Then we can claim that the radius of medium circle is $$r_3=3-1=2$$ and those of big and large circles are $$r_4=4-1=3$$ and $$r_2=2\times 3=6$$ respectively that is all involved circle are mutually tangent.If radius of medium circle is $$r_3=2$$ that means that it's center locates on a line passing $$O_3$$ which has a distance=4 from x axis. similarly if the radius of big circle is $$r_4=3$$ that would mean that it's center locates on a line passing $$O_3$$ which has distance =3 from y axis.That is triangle $$O_1O_2O_3$$ is right triangle at $$O_3$$ | 2020-12-05T00:31:30 | {
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https://math.stackexchange.com/questions/1900788/find-the-probability-that-on-tuesday-it-does-not-rain-in-gregs-area-and-he-do | # Find the probability that, on Tuesday, it does not rain in Greg's area and he does not get a quiz.
The probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?
My approach:
Probability of rain or quiz or both = $0.3+0.2= 0.5$
So, probability of neither= $1-0.5$ = $0.5$
Question: Actual probability of this problem is $0.7\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach?
• They are independent not mutually exclusive. For mutually exclusive events you have $Pr(A\text{or}B)=Pr(A)+Pr(B)$. That is not true of independent events. – JMoravitz Aug 23 '16 at 6:11
• Principle of inclusion-exclusion 2set case: $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$ – JMoravitz Aug 23 '16 at 6:18
Note the following two things:
Principle of inclusion-exclusion (2-set case) $$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$
Definition of independent events
The following are equivalent statements
• $A$ and $B$ are independent events
• $Pr(A\mid B) = Pr(A)$
• $Pr(B\mid A) = Pr(B)$
• $Pr(A\cap B)=Pr(A)\cdot Pr(B)$
For your problem: you know $Pr(A)=0.3, Pr(B)=0.2$ and they are independent, so
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 0.3+0.2-0.3\cdot 0.2 = 0.5-0.06 = 0.44$$
So the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$.
Alternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\cdot 0.8 = 0.56$
Your approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both "rain and quiz" and "rain and no quiz". The $0.2$ covers all possibility of a quiz, so it includes both "rain and quiz" and "quiz and no rain". So $0.3 + 0.2$ covers "rain and quiz", "rain and no quiz", "rain and quiz" again, and "quiz and no rain". To cut out one of the countings of "rain and quiz", we need to subtract the probability of that event, which is $0.3 \cdot 0.2 = 0.06$. So the probability of "rain or quiz or both" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$. | 2019-12-08T11:17:23 | {
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https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax | # 4.4 The mean value theorem (Page 2/7)
Page 2 / 7
Verify that the function $f\left(x\right)=2{x}^{2}-8x+6$ defined over the interval $\left[1,3\right]$ satisfies the conditions of Rolle’s theorem. Find all points $c$ guaranteed by Rolle’s theorem.
$c=2$
## The mean value theorem and its meaning
Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$
## Mean value theorem
Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$ Then, there exists at least one point $c\in \left(a,b\right)$ such that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
## Proof
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is
$\frac{f\left(b\right)-f\left(a\right)}{b-a}$
and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as
$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$
Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,
$g\left(x\right)=f\left(x\right)-\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right]\text{.}$
Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $\left[a,b\right],$ $g$ is also continuous over $\left[a,b\right].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since
$g\prime \left(x\right)=f\prime \left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a},$
we see that
$g\prime \left(c\right)=f\prime \left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
Since $g\prime \left(c\right)=0,$ we conclude that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
In the next example, we show how the Mean Value Theorem can be applied to the function $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right].$ The method is the same for other functions, although sometimes with more interesting consequences.
## Verifying that the mean value theorem applies
For $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.
We know that $f\left(x\right)=\sqrt{x}$ is continuous over $\left[0,9\right]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.$ The slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ is given by
$\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$
We want to find $c$ such that ${f}^{\prime }\left(c\right)=\frac{1}{3}.$ That is, we want to find $c$ such that
$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$
Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function. | 2018-09-24T22:49:08 | {
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https://math.stackexchange.com/questions/2133512/is-x2-equiv-1-pmod365-solvable | # Is $x^2 \equiv -1 \pmod{365}$ solvable?
I know that a similar question already exists, but I have a different question to ask.
We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.
My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \equiv -1 \pmod 5$ and $x^2 \equiv -1 \pmod{73}$ has solutions. So, if we use Legendre's Symbol we have
• $x^2 \equiv -1 \pmod 5$ has solution $\iff (-1/5)=1$ (and with simple calculations, indeed)
• $x^2 \equiv -1 \pmod{71}$ has solution $\iff (-1/73)=1$
Now, can we conclude that the congruence $x^2 \equiv -1 \pmod{365}$ has solution?
And more general: If we have the congruence $x^2 \equiv a \pmod n$ with $n=p_1^{n_1}\cdots p_k^{n_k},\ \gcd(a,n)=1$, which is equivalent with the system $x^2 \equiv a {\pmod p_1^{n_1}},\ldots,x^2 \equiv a \pmod{p_k^{n_k}}$, can we conclude that the first has solution if and only if each one of $x^2\equiv a\pmod{p_i^{n_i}},\ \forall i=1,\ldots,k$ has solution?
Thank you.
• Change $71$ to $73$ and it's fine. – Robert Israel Feb 7 '17 at 16:12
• @RobertIsrael Thank you for your comment and your correction. – Chris Feb 7 '17 at 16:15
• Note the use of \pmod rather than \bmod in my edit to the question. – Michael Hardy Feb 7 '17 at 16:51
• @RobertIsrael Ok, thank you – Chris Feb 7 '17 at 16:58
We show the following result:
Let $n,m$ be coprime integers. Let $a \in \Bbb Z$. Then $a$ is a square modulo $nm$ iff $a$ is square mod $n$ and $a$ is square mod $m$.
Assume that there are integers $x_i$ such that $x_1^2 \equiv a \pmod m$ and $x_2^2 \equiv a \pmod n$ where $m,n$ are coprime. We show that $y^2 \equiv a \pmod {nm}$ has a solution (the converse is obvious).
We know that there is an integer $y$ such that $y \equiv x_1 \pmod m$ and $y \equiv x_2 \pmod n$, by the Chinese remainder theorem.
Then $y^2-a$ is a multiple of $m$ and $n$, so it is a multiple of $nm$ since $(n,m)=1$. Therefore $y^2 \equiv a \pmod {nm}$.
More generally, we have the following theorem (see Ireland Rosen, p. 50)
Let $a,n \in \Bbb N$ be coprime integers. Write $n=2^e p_1^{e_1} \cdots p_k^{e_k}$ as product of distinct prime powers. Then $a$ is a square modulo $n$ if and only if
• $a$ is a square modulo $p_i$ (for $1 \leq i \leq k$) ; this is equivalent to $a^{\dfrac{p_i-1}{2}} \equiv 1 \pmod{p_i}$
• $e>1 \implies a \equiv 1 \pmod{2^{2+r}}$ where $r = 1$ if $e \geq 3$ and $r=0$ if $e=2$.
• It might be worth explicitly mentioning the CRT here where it's used, but this is elsewise a good, clean answer. – Steven Stadnicki Feb 7 '17 at 16:20
• Thank you , now it's clear. – Chris Feb 7 '17 at 16:27
• @Chris : you're welcome. It is chapter 5 in Ireland, Rosen. – Watson Feb 7 '17 at 16:28
• Tell me please something else. If we have the congruence $x^2 \equiv 2 \bmod 118$ could we say that $118=2\cdot 59$ so it's in the form of $2p^m$ and $2^{\phi(118)/ \gcd(2,\phi(118))} \equiv -1 \bmod 118$ so there are no solutions (because it is not a quadratic residue mod 118)? – Chris Feb 7 '17 at 16:46
• @Chris : since $2$ is not a square mod $59$, it is not a square mod $118$, that's correct! – Watson Feb 7 '17 at 16:50
We have $365 = 5 \times 73$. The congruence becomes $x^2 = -1 \mod 5$ and $x^2 = -1 \mod 73$.
We have if $p = 1 \mod 4 \implies x^2 = -1 \mod p$ has exactly $2$ solutions.
Thus $x^2 = -1 \mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \mod 73$ has solutions $y_0,y_1$.
The original solutions satisfies either: $x = x_0 \mod 5, x = y_0 \mod 73$; $x = x_0 \mod 5, x = y_1 \mod 73$; $x = x_1 \mod 5, x = y_0 \mod 73$ ; $x = x_1 \mod 5, x = y_1 \mod 73$.
For each pair of congruence , $x$ is uniquely determined $\mod 365$ by the Chinese remainder theorem. Hence the original congruence has $4$ solutions.
If $\,m,n\,$ are coprime then, by CRT, solving an integer coefficient polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n.\,$ By CRT, each combination of a root $\,r_i\bmod m\,$ and a root $\,s_j\bmod n\,$ corresponds to a unique root $\,t_{ij}\bmod mn,\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ &\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$
Yes. Let a solution to $x^2+1 \equiv 0 \pmod {5}$ be $r_{1}$, and let a solution to $x^2+1 \equiv 0 \pmod {73}$ be $r_{2}$
Then note that by CRT, we have that there exists such $x$ that $$x \equiv r_{1} \pmod {5}$$ $$x \equiv r_{2} \pmod {73}$$ Exists. Then note that for such $x$, $$x^2+1 \equiv 0 \pmod {5}$$ $$x^2+1 \equiv 0 \pmod {73}$$ Which gives $$x^2+1 \equiv 0 \pmod {365}$$
• Seems alright now. – Wojowu Feb 7 '17 at 16:20
It is true because of the Chinese remainder theorem, which asserts the map \begin{align} \mathbf Z/n\mathbf Z&\longrightarrow \mathbf Z/p_1^{n_1}\mathbf Z\times\dotsm\times\mathbf Z/p_k^{n_k}\mathbf Z\\ x\bmod n&\longmapsto(x\bmod p_1^{n_1},\dots,x\bmod p_k^{n_k}) \end{align} is a ring isomorphism. | 2021-03-01T14:37:55 | {
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http://mathhelpforum.com/trigonometry/95321-cartesian-equations.html | # Math Help - Cartesian equations
1. ## Cartesian equations
Given $x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$, express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.
I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.
$x=2 \sin (nt+\frac{\pi}{3})$
$x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$x= \sin nt+ \sqrt{3} \cos nt$
and
$y=4 \sin (nt+ \frac{\pi}{6})$
$y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks
2. Originally Posted by arze
Given $x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$, express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$. Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.
I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.
$x=2 \sin (nt+\frac{\pi}{3})$
$x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$x= \sin nt+ \sqrt{3} \cos nt$
and
$y=4 \sin (nt+ \frac{\pi}{6})$
$y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks
$x=2 \sin (nt+\frac{\pi}{3})$ and $y=4 \sin (nt+ \frac{\pi}{6})$
$x=2 \sin (nt+\frac{\pi}{6}+\frac{\pi}{6})$
let $\alpha=nt+\frac{\pi}{6}$
$x=2 \sin (\alpha+\frac{\pi}{6})$ and $y=4 \sin (\alpha)$
$\frac{x}{2}=\sin (\alpha)\cos(\frac{\pi}{6}) + \cos (\alpha)\sin(\frac{\pi}{6})$
and
$\frac{y}{4}=\sin (\alpha)$
use $\sin^2 (\alpha) + \cos^2(\alpha) = 1$ to eliminate $\alpha$ and get your equation
3. Hello, arze!
Given: . $\begin{array}{ccc}x &=&2 \sin \left(nt+\frac{\pi}{3}\right) \\ y &=& 4 \sin \left(nt+ \frac{\pi}{6}\right)\end{array}$
Express $x$ and $y$ in terms of $\sin nt$ and $\cos nt$.
Find the Cartesian equation of the locus of the point $(x,y)$ as $t$ varies.
I have expressed $x$ and $y$ in terms of $\sin nt$ and $\cos nt$ already.
. . $\begin{array}{ccccc}x\:=\: 2\sin\left(nt+\tfrac{\pi}{3}\right) \:=\:2\left(\tfrac{1}{2}\sin nt+\tfrac{\sqrt{3}}{2}\cos nt\right) &\Rightarrow& x\:=\: \sin nt+ \sqrt{3} \cos nt \\ \\[-4mm] y\:=\:4\sin\left(nt+ \tfrac{\pi}{6}\right) \:=\:4\left(\tfrac{\sqrt{3}}{2} \cos nt+\tfrac{1}{2} \sin nt\right) &\Rightarrow& y \:=\: 2\sqrt{3} \cos nt +2 \sin nt \end{array}$
. . . . . Good work!
Now my problem is how do i form the Cartesian equation in $x$ and $y$ as $t$ varies?
We have: . $\begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$
$\begin{array}{ccccc}\text{Square }{\color{blue}(1)}: & x^2 &=& \sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + 3\cos^2\!nt \\
\text{Square }{\color{blue}(2)}: & \dfrac{y^2}{4} &=& 3\sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + \cos^2\!nt \end{array}$
Add: . $x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$
. . . . $x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$
. . . . $x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$
4. Originally Posted by Soroban
Hello, arze!
We have: . $\begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$
Add: . $x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$
. . . . $x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$
. . . . $x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$
if you want to do it that way, solve (1) and (2) for sin and cos:
$2\sin nt = \frac{\sqrt3y}{2}-x$
$2\cos nt = \sqrt3x-\frac{y}{2}$
then use
$\sin^2nt + \cos^2nt = 1$
to get an equation in x and y
5. Hello, Tesla!
Absolutely right!
It's an old habit.
Whenever I see something like: . $\begin{array}{ccc}x &=& a\sin\theta \\ y &=& b\cos\theta\end{array}$
. . I tend to use the square-and-add approach to simplify.
I must learn to get out of my "box". | 2015-11-30T19:51:04 | {
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https://math.stackexchange.com/questions/2701982/why-cant-y-xex-be-solved-for-x | # Why can't $y=xe^x$ be solved for $x$?
I apologize for my mathematical ignorance regarding this, but could someone help me understand why it isn't possible to (symbolically) find an inverse function for $f(x)=xe^x$?
The most obvious (but presumably the most trivial) is that $f$ does not pass the "horizontal line test". However, if we restrict the domain to $x\geq-1$ this should not be a problem (derivative is positive for $x>-1$ so function is strictly increasing). So now my question becomes: "Why can't we find an inverse function for $f$ over the interval $[-1,\infty)$?"
Perhaps it is because $e^x$ is transcendental (not algebraic). However, we can find an inverse for $g(x) = e^x$, which is also transcendental. Is that because we're "cheating" by defining another transcendental function, namely $\ln(x)$, to be its inverse? In other words, would it be fundamentally no different to define a new function, call it $\text{lnx}(x)$ (if that's not already something else), to be the inverse of $xe^x$ over $[-1,\infty)$ and then say that $f$ has a "closed form" / "symbolic" / ??? inverse function $f^{-1}(x)=\text{lnx}(x)$ over the interval $[-1, \infty)$?
### SageMath source to generate plot
xs = (x,-5,2) ys = (y,-1,5) p1 = implicit_plot(x*exp(x)-y,xs,ys, color='blue', legend_label='y=x*e^x') p2 = implicit_plot(x-y,xs,ys, color='orange', linestyle='dashed', legend_label='y=x') p3 = implicit_plot(exp(x)-y,xs,ys, color='green', linestyle='dotted', legend_label='y=e^x') combined = p1 + p2 + p3 combined.axes_labels(['x', 'y']) combined.legend(True) combined.show(title='Transcendental Stuff', frame=True, axes=True, legend_loc='lower right')
• it can be solved by the following function $$\{\{x\to W(y)\}\}$$ the product Logarithmus by Mathematica – Dr. Sonnhard Graubner Mar 21 '18 at 15:49
• It can be solved, and an inverse function can be found in the given interval, just not analytically. – Thern Mar 21 '18 at 15:51
• Yes, you're right that "fundamentally" it would "be no different" to invent a label for the inverse function, which is guaranteed to exist because your function is injective. – symplectomorphic Mar 21 '18 at 15:51
• This question is so high quality, why aren't people upvoting this – vrugtehagel Mar 21 '18 at 15:54
• @vrugtehagel that is what I have about this site more than anything else. There's so much good content that goes unappreciated. This is an excellent question. I upvoted it. – user223391 Mar 21 '18 at 16:03
It can be solved by inventing new functions, but it cannot be solved in closed form using trigonometric, logarithmic or exponential etc. Read this: Chow, Timothy Y. (May 1999), "What is a Closed-Form Number?"
• I believe you, but I was trying to get a better answer to the why part of the question. – iX3 Mar 21 '18 at 16:02
• Read the linked article for the why part. – Shubhashish Mar 21 '18 at 16:06
• Wow, this paper talks about exactly what I wanted to know! (Did you just add that? I can't believe I didn't see it before.) BTW, it was hard to read on that site (not scalable, not searchable, etc.), but I found a link to a PDF version on the author's website here – iX3 Mar 21 '18 at 16:18
• Yes I have updated the link – Shubhashish Mar 21 '18 at 16:19
You can use the Lambert-$W$ function to solve it symbolically.
$y = xe^x$ gives $x = W(y)$.
You may run solve(x*exp(x)-y,x) on SymPy Live as an alternative to SageMath.
• Just for reference, here is a Wolfram Alpha example solving $xe^x=\pi$ – gt6989b Mar 21 '18 at 15:56
• Thank you for showing me the Lambert-W function (was just finding that in the answer to another question on here too). It seems that what determines whether it is possible to solve something in "closed form" has largely to do with what functions/expressions are allowed. e.g. $y = e^x$ cannot be "solved for $x$" unless something like $\ln$ is allowed. Similarly, the answer to my question depends on whether or not something like $W$ is acceptable. – iX3 Mar 21 '18 at 16:00
• @iX3 In the tag info of (closed-form), it's said that a "closed form expression" is anything written in terms of "known" functions. Though the Lambert-W function is not elementary, but I believe it qualifies a known function. – GNUSupporter 8964民主女神 地下教會 Mar 21 '18 at 16:09
• Hmm, solve(x*exp(x)-y,x) --> Traceback (most recent call last):...line 1055, in solve solution = _solve_system(f, symbols, **flags) ... line 158, in _lambert solns[i] = tmp.subs(u, rhs) AttributeError: 'dict' object has no attribute 'subs' I may need to go read & learn about SymPy to figure out what's wrong. (I think SageMath uses SymPy though, so in theory I should be able to access all of the same functionality through that.) – iX3 Mar 21 '18 at 18:25
• @iX3 I've edited my post in response to your comment. – GNUSupporter 8964民主女神 地下教會 Mar 21 '18 at 18:33
As indicated in the comments and in another answer, there is a special function that has been defined which can serve as an inverse here. However, this doesn't address your basic question, which is: why is this necessary?
Before I talk about that, let me say something about $\ln x$ and $e^x$. Those are both transcendental functions, and they're inverses of each other, but neither is really defined simply to be the inverse of the other. Both of these functions arise very naturally, each on their own. The natural log function is the integral of $x^{-1}$, for a suitably defined integral, and the exponential function is the solution of a differential equation modeling the simplest kind of constant relative growth. They end up being inverses of each other, and that's a cool set of facts to understand and wonder at.
Anyway, you can't use elementary functions to invert lots of things, such as $f(x)=xe^x$. This often happens when we mix different types of functions together. The functions $g(x)=x$ and $h(x)=e^x$ are perfectly invertible, and they are, respectively, polynomial and exponential. The function $f=gh$, on the other hand, is the product of a polynomial function with an exponential function. We expect that to be more complicated. It's not solvable, with the usual algebraic methods, because whatever technique we apply to simplify $e^x$ messes up the polynomial part. Similarly, if you try to solve $e^x(x^2-x)=k$, anything you do to simplify the polynomial part will just make an exponential mess.
It's the blending of different types of functions, which are amenable to transforming with completely different tools, that makes such functions complicated. Other notorious examples include $\frac{\sin x}{x}$, and $e^{x^2}$. It's not always that they're hard to invert, but try doing integral calculus with such functions!
• Thank you; this makes a lot of sense, at least intuitively. Is there a more precise definition regarding the "blending of different types of functions" and does that always produce tough situations like this? – iX3 Mar 21 '18 at 16:10
• I'd say the different kinds are 1. polynomial/rational/radical (algebraic), 2. Trigonometric, 3. Inverse trigonometric, 4. Exponential, 5. Logarithmic. If it's done right, to allow for cancellation, you can blend 2 with 3 and 4 with 5 and still be able to work analytically. If you're very careful, you can blend any kinds, as long as there's a way to cancel stuff until you have something pure. I don't know of any real precise definitions and theory that address this "blending" though. – G Tony Jacobs Mar 21 '18 at 16:14
I am going to break your question into two separate questions:
1. Does an inverse function to $y=x \cdot e^x$ exist on the domain $[-1,+\infty)$?
2. Can we find that inverse function in the "standard list" of functions that everyone knows?
Question 1 has an easy answer: yes, that inverse function exists. Its existence is an application of the inverse function theorem which most students first encounter without proof in an ordinary calculus class, and which they then encounter with proof in some kind of advanced calculus class. And then if you're quick enough, you might even be able to attach a name to that function, although it appears that Lambert beat you to it as shown in the answer of @GNUSupporter.
Question 2 is more difficult to answer. First there is no "standard list" of functions that everyone knows. Nonetheless, perhaps that's just because we haven't tried hard enough to list all functions. Can we keep adding functions to the "standard list" until we don't have to add any more? For example, as suggested in the answer of @GNUSupporter, can we just add the "Lambert-W function" $x=W(y)$ to our list if we had never heard of it before, and then declare ourselves happy? Well, maybe, but then you would be perfectly justified to ask whether there is an inverse function to $z=yW(y)$ on some appropriately chosen domain......................
These kinds of questions are addressed (but not definitively answered) in a branch of mathematics called differential algebra which is a kind of broad generalization of differential equations.
• Thank you for the reference to differential algebra. I gather from your answer that the "answer" to my "why?" question is basically "Differential algebra deals with this question but does not provide a complete/general answer." Is that about right? – iX3 Mar 21 '18 at 16:12
• Yes, that's about right. Nonetheless, one can often answer specific questions, which is what's so nice about the link to Chow's paper provided in the answer of @ShubhashishChauhan. – Lee Mosher Mar 21 '18 at 17:05
• Although, to qualify my last comment, even Chow's paper does not provide a definitive answer to questions about Lambert's function, instead reducing them to applications of big unsolved conjectures. – Lee Mosher Mar 21 '18 at 17:13 | 2019-11-18T04:40:29 | {
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https://math.stackexchange.com/questions/3436407/inequalities-with-two-absolute-values-with-greater-than-symbol-please-tell-me-t | # Inequalities with two absolute values with greater than symbol. Please tell me the proper way of doing $|4x-1|>|3x+2|$
I have finals in a couple of days. I need help with inequalities. I have spent around 15 hours trying my hand at inequalities and I am still trying to figure things out. In this case I need help with an inequality with two absolute values with greater than symbol.
The following is a picture of the inequality ($$|4x-1|>|3x+2|$$) and the procedure I tried, which is something I applied from an answer by Isaac to another question in this site, although I think I did something wrong because it looks like I got the answer wrong or incomplete.
Please tell me what I did wrong and how to do it the right way.
PS: Does this method also apply if there is a less than sign?
• Alternatively (and I'd say it's faster) you could note that $|a|>|b|$ is equivalent to $a^2>b^2$.
– dfnu
Nov 15, 2019 at 8:32
• Two years later I have completely forgotten how to do this. :D Nov 19, 2021 at 23:50
Here’s another way to look at it: $$|4x-1|>|3x+2| \iff (4x-1)^2-(3x+2)^2>0 \iff (7x+1)(x-3)>0$$
Now LHS has only two obvious zeros, and outside the interval with those end points, it has to be positive.
——-
P.S. When you have something like $$f(x)>0$$, where $$f$$ is continuous, it can change sign only when there are roots. So its roots effectively partition the real line into intervals which are either solutions or not. In the above case, two roots are $$-\frac17, 3$$, so we need to consider only the three intervals $$(-\infty, -\frac17), (-\frac17, 3)$$ and $$(3, \infty)$$. Obviously the LHS is positive for $$x\to \pm\infty$$ and negative for $$x=0$$, so the first and last intervals are solutions, the middle one isn’t.
BTW, whether the intervals to consider are open / closed depends on whether the inequality is strict or $$\geqslant$$.
• 7x+1>0 gives me (-1/7,+) which seems to be invalid. What is wrong? Nov 15, 2019 at 16:18
• @dfnu +macavity x-3 gives me (3,+), with a combined solution of (-1/7,+). Nov 15, 2019 at 16:56
• @freethinker36 If you need $A\cdot B>0$, you need $A, B$ to have the same sign. Not just $A$ or just $B$. Nov 15, 2019 at 16:58
• It would be great if you can elaborate. I'm learning, thus what may look evident, for me may not be evident. And tomorrow is my test and have to study two classes; already taking time of the second class to try to cement my understanding of inequalities. But I appreciate whatever guidance has been provided! Nov 15, 2019 at 17:05
• OK, I have added some explanations, do let me know if youre having more doubts after studying it. Nov 15, 2019 at 17:11
You're doing almost everything right; you're just misinterpreting the results at the end.
In particular, the idea is to combine intervals where the inequality holds. In the first region, $$\left(-\infty, -\frac23\right)$$, you arrived at $$x<3$$. It seems that you rejected this result, judging by the "$$\times$$" beside your work. This is where you went wrong.
Instead, think of it like this:
Within the region $$\left(-\infty, -\frac23\right)$$, which $$x$$ satisfy $$x<3$$?
All $$x$$ do within this region. Another way to see this is that you are effectively taking the intersection of the two: $$\left(-\infty, -\frac23\right) \cap (-\infty,3) = \left(-\infty, -\frac23\right)$$
The other two regions are handled the same way.
• My rationale for rejecting it was that 3>-2/3, so x<3 is not within the region of x<-2/3. Nov 15, 2019 at 6:33
I tried again, following the answer of Théophile, and I was able to reach the correct solution. Below is the pic of the procedure I used. Thanks! | 2022-05-22T23:02:19 | {
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https://math.stackexchange.com/questions/177917/are-right-continuous-functions-measurable/1134904 | # Are right continuous functions measurable?
Are right-continuous function from $\mathbb{R}^n$ to $\mathbb{R}$ necessarily semi-continuous? If not, are they necessarily Borel measurable? Is there a topological characterization of right-continuous functions (as there is of continuous ones)? Are CDFs of $n$-dimensional random vectors measurable?
Note: A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous iff it is right-continuous at every point $x \in \mathbb{R}^n$. A function $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is right-continuous at $x \in \mathbb{R}^n$ iff given any infinite sequence of points in $\mathbb{R}^n$ $(y_0,y_1,\dots)$ that converges to $x$ from above (i.e. the sequence converges to $x$ in the usual, Euclidean sense and in addition every sequence element is greater than or equal to $x$ component-wise), the sequence $(f(y_0), f(y_1), \dots)$ converges to $f(x)$ in the usual sense.
• You might want to add the definition of right-continuity for a function defined on $\mathbb R^n$ when $n\gt1$.
– Did
Aug 2, 2012 at 8:28
• What are the definition of right-continuous function, and semi-continuous?
– Paul
Aug 2, 2012 at 8:44
The answer to the first question is no, even in the case $n=1$: The characteristic function of the half open interval $[0,1)$ is right continuous, but neither upper nor lower semicontinuous.
A right continuous function $\mathbb{R}\to\mathbb{R}$ is indeed Borel measurable. By definition, the inverse image $E$ of an open set has the property that for any $x\in E$, there is some $\delta>0$ so that $[x‚x+\delta)\subseteq E$. It follows that $E$ is a countable union of half open intervals, and hence is Borel measurable. I am not sure about the answer to this one when $n>1$ (the countable union argument no longer holds), but my guess is that right continuous functions are still measurable.
Topological characterization: If we write $\le$ for pointwise comparison on $\mathbb{R}^n$, we can make a one-sided topology by declaring a set $V\subseteq\mathbb{R}^n$ to be open if, for each $x\in V$, there is some $\delta>0$ so that $\{y\ge x\colon\lvert y-x\rvert<\delta\}\subseteq V$. Then the right continuous maps $\mathbb{R}^n\to\mathbb{R}$ are just the ones that are continuous from this topology to the usual topology on $\mathbb{R}$.
I am not too sure on the CDF question either. (I assume CDF stands for cumulative distribution function, in the sense of $F(x)=\mathrm{P}\{X\le x\}$, where $X$ is a random $n$-vector.) It might help that $F$ is not only right continuous, but also monotone. So the set $\{x\colon F(x)\le p\}$ has a particularly simple structure; I imagine it must be measurable, but right now I don't see a proof.
• Thanks, Harald, but it is in fact the case n>1 that i am interested in. See my clarification of the term "right-continuous" in a comment to my original post. Also, of the four questions i posed in my original post i am most interested in the last one concerning CDF. Aug 2, 2012 at 10:49
• @EvanAad: This topology appears to be the product of $n$ copies of the Sorgenfrey line. But I don't quite see why the open sets of this topology are Borel. It certainly has a basis of Borel sets, but I don't think it's second countable, so why does it follow that any union of such sets is also Borel? Aug 2, 2012 at 21:54
• @EvanAad: In fact, let $E$ be a non-measurable subset of $\mathbb{R}$ and let $A = \bigcup_{x \in E} [x,\infty)^2$. $A$ is open in our topology on $\mathbb{R}^2$ (the Sorgenfrey plane), but not Borel with respect to the usual topology, since its intersection with the diagonal line $y=-x$ is a skewed copy of $E$. One might think one could exploit this to make a right-continuous function which is not Borel, but I still don't see how. Aug 3, 2012 at 13:11
• @NateEldredge: Did you mean $\bigcup_{x\in E}[x,\infty)\times[-x,\infty)$? Aug 3, 2012 at 13:51
• @HaraldHanche-Olsen: Oops, yes, I did. Aug 3, 2012 at 14:25
Here is a proof that any function $F: \mathbb{R}^n \to \mathbb{R}$ which is right continuous is Borel measurable.
Define $F_n(x) = F\left(\frac{[nx]+1}{n}\right)$, where $[x]$ is the greatest integer smaller than $x$. Clearly $F_n$ are measurable since they are step functions.
From the right continuity of $F$ we get that $F_n\to F$, since $\frac{[nx]+1}{n} \downarrow x$.
Hence since $F$ is the pointwise limit of measurable functions, it is measurable too.
• Very Nice, Milind! Aug 24, 2017 at 13:58
• to make this a bit more solid, you would need to clarify that $n=1$ or properly define $[x]$ for an $n$-vector $x$. Oct 12, 2018 at 11:58 | 2022-06-29T23:05:19 | {
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https://math.stackexchange.com/questions/1756921/horizontal-tangent-line-of-a-parametric-curve | Horizontal tangent line of a parametric curve
Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:
The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.
So far so good.
But then they write:
However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$ And thus: $$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$
It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t - \cos (2t), y=2\sin t - \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.
However my professor said that this is wrong ("because the derivative is not zero" - indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined - "$0/0$").
So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.
Thanks.
Short version of the question: can a parametric curve have a horizontal tangent line at a singular point?
I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).
• Reading your comments to the answers below, I think you are unduly concerned about the distinction between $f(x_0)$ and $\lim_{x\to x_0}f(x)$. Where we have an isolated point $x_0$ for which $f(x_0)$ does not exist, but $\lim_{x\to x_0}f(x)$ does exist, it is common for people to assume without comment that we define $f(x_0)$ as the limit. – almagest May 3 '16 at 18:31
• what is your textbook ? – KonKan May 3 '16 at 18:44
• @KonKan - Calculus by Anton (5th ed.) – user239753 May 3 '16 at 18:53
• @almagest - unduly? This might be an obvious assumption for some, but not for me. The very fact that my professor dismissed it says that this is not trivial, and at least deserves a mention. – user239753 May 3 '16 at 19:02
• @user239753: i'm a bit confused: I'm searching in your book (it happens to have a copy of the 10th edition, available right in front of me) but i cannot find the quote you are mentioning. I am actually looking at Ch.10, par.10.1, p.692-705. The example you are mentioning is example 6, p.697. However, the authors simply mention that $t=0$ is a singular point. I cannot find the limit computation you are mentioning. Can you indicate the exact page ? – KonKan May 3 '16 at 19:21
Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.
We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).
In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.
Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.
Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.
Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.
In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.
The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),
$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$
The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,
$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$
Again, we obtain an intersection number $\ge 2$, which means tangency.
What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".
$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.
• @user239753: Try these two: "Elementary Differential Geometry" by A.N. Pressley, and "Differential Geometry of Curves and Surfaces" by M.P. do Carmo. One last word: differential geometry in general is interested only in those spaces that are endowed with regular parametrizations (because one wants them to be invertible, with their inverses called "local charts"). Most theorems about parametrized curves assume the parametrization to be regular. – Alex M. Apr 28 '16 at 20:24
• @AlexM. Because this is all about subtile nuances: You wrote, the cardiod would be a manifold, because the defining function $F$ is smooth. Aren't you applying the regular value theorem (which is a close relative to the implicite function theorem) here, which means 0 needs to be a regular value for $F$? (Anyway: great answer!) – hase_olaf Apr 28 '16 at 21:31
• @AlexM. - thank you for your refined answer. A bit stupid question - what does "multiple root" mean? It has something to do with algebraic multiplicity? – user239753 May 3 '16 at 19:27
• @user239753: Yes, exactly. We say that $a$ is a root of order $k$ of $P$ (where $P$ is a polynomial) if and only if $(x-a)^k \mid P$ and $(x-a)^{k+1} \nmid P$. The analogue for algebraic curves is the concept of "intersection multiplicity". – Alex M. May 3 '16 at 19:41
• @AlexM: I have to confess .. I am the ignorant who downvoted your hard work ;) but give me a moment to explain. your answer (especially after your edit) seems valid in my eyes and you've obviously put hard work in it ! I give you credit for that. However, I feel it falls outside the spirit and the OP's question: it is a delicate question on a subtle point of parametric differentiation (from the calculus viewpoint) while you certainly view the situation from the algebraic/differential geometric eye. This is also why i rolled back your retagging. – KonKan May 5 '16 at 23:29
In order to find the tangent at some special point you should not try to compute the limit of $y'$ but the actual limit of secant directions at that point. Therefore in your first example you get $$m_+:=\lim_{t\to0+}{y(t)-y(0)\over x(t)-x(0)}=\lim_{t\to0+}{t^3\over t^2}=0\ ,$$ and similarly $m_-=0$. In your second example you have $${y(t)-y(0)\over x(t)-x(0)}={2\sin t-\sin(2t)-0\over 2\cos t-\cos(2t)-1}={2\sin t(1-\cos t)\over2\cos t(1-\cos t)}=\tan t\qquad(t\ne0)\ ,$$ and therefore $$m_+=m_-=\lim_{t\to0}{y(t)-y(0)\over x(t)-x(0)}=0\ .$$
You are referring to two different things here..
For the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.
For cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of epicycloids:
EDIT 1:
Since it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.
It can be also seen at the point $(1,0)$ Changing the multiple angle
$$x=2\cos t - \cos (n t), y=2\sin t - \sin( n t)$$
has the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.
The direction of tangent is vertical, indeterminate and vertical respectively
• In both cases $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx}\Big |_{t=0}$ is undefined. The question is whether in such a situation a horizontal tangent line exists. The textbook says it does, yet my professor claims that for a horizontal tangent line to exist the derivative must EXIST and be equal to zero, that is we must have $\frac{dy}{dx}\Big |_{t=0}=0$. – user239753 Apr 25 '16 at 6:53
• please read my comment again. I'm talking about the value of the derivative which does NOT exist. There is a difference between the limit (which indeed can be found with L'Hospital) and the actual value. – user239753 Apr 25 '16 at 7:28
• Note that you should not stop hunting for the L'Hospital limit if the first order evaluation fails. Do not conclude that it is undefined. At the cusp we are dealing with second orders and so limit of $f''(t)/g^{''}(t)$ should be next evaluated. – Narasimham Apr 25 '16 at 7:31
• unfortunately, you do not understand me. Here $dy/dx \neq 0$ even though $\lim_{t\to 0} \frac{dy}{dx}=0$. Do you know there is a difference between limit and value of the function? Take $$f(x)=\left\{\begin{matrix} 0 \text{ if } x\neq0\\ 1 \text{ if } x=0 \end{matrix}\right.$$. Here $$\lim_{x\to0}f(x)=0$$ yet $f(0)\neq 0$. The same goes here. $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx} \Big |_{t=0}\neq 0$ ! Do you understand? – user239753 Apr 25 '16 at 7:37
• Ok, we resume after sometime,else moderator may indicate chat room. – Narasimham Apr 25 '16 at 8:34
the formula $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ is valid for computing the derivative at points where $\frac{dx}{dt}\neq 0$. At singular points, i.e. points at which $$\frac{dy}{dt}=\frac{dx}{dt}=0$$ the above formula is not valid. However, this does not imply that the derivative does not exist; only that it is not computable by the above formula.
In such cases, a straightforward application of the definition of derivative (as already done by user Christian Blatter in his answer) can provide the answer: $$\frac{dy}{dx}\bigg|_{(0,0)}=\lim_{t\to 0^+}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{t\to 0^-}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{x\to0^\pm}\frac{t^3}{t^2}=0$$ which shows that the derivative exists at $(0,0)$ and so does the tangent line at $(0,0)$, which is no other than the horizontal axis itself.
P.S.1: Note that, when confined to each branch seperately: $$y(x_0)=y(x(0))=y(0)=0, \ \ x_0=x(0)=0 \\ y(x)=y(x(t))=y(t)=t^3, \ \ x=x(t)=t^2$$ So, on each one of the functions $y(x)=x^{3/2}$, $y(x)=−x^{3/2}$ i.e. on each one of the branches of the graph below, the following rates of change are identified: $$\frac{y\big(x(t)\big)−y(x_0)}{x(t)−x_0}=\frac{y(t)−y(0)}{x(t)−x(0)}$$ On the other hand, since $x=t^2\neq 0$ when $t\neq 0$ it should be clear that $$t\to 0\Leftrightarrow x\to 0$$ At this point, it remains to invoke a classical proposition on the limit of a composite function, claiming that:
Let the real functions $f,g$, for which $lim_{x\rightarrow x_0}g(x)=g_0\in \mathbb{R}$ and $lim_{g\rightarrow g_0}f(g)=l\in \mathbb{R}$.
If, furthermore $g(x)\neq g_0$ "close to $x_0$" (i.e. in some interval $(a,x_0))$, then $$\lim_{x\rightarrow x_0}f\big(g(x)\big)=\lim_{g\rightarrow g_0}f(g)=l$$
Seting $f(x):=\frac{y(x)−y(0)}{x−0}$ and $g(x):=x(t)$ and applying the proposition we get that: $$\lim_{t\to 0}\frac{y(t)−y(0)}{x(t)−x(0)}\equiv \lim_{t\to 0}\frac{y\big(x(t)\big)−y(0)}{x(t)−x(0)}=\lim_{x\to 0}\frac{y(x)−y(0)}{x−0}=\frac{dy}{dx}\bigg|_0$$ (where the far left term is computed using side limits: for the upper branch $t>0$ thus $t\to0\Leftrightarrow t\to0^+$ while for the lower branch $t<0$ thus $t\to0\Leftrightarrow t\to0^-$).
P.S.2: Regarding your final question: I do not believe that the existence of the limit is a sufficient condition, because generally, a differentiable function at $x_0$ need not have a continuous derivative at $x_0$: In other words, there may well be a situation at which: $\lim_{x\to x_0}f'(x)\neq f'(x_0)$.
However, this is not the case for the semicubical parabola, as we can see from its graph
which is probably what the author of your textbook had in his mind. He computes the limit of the derivative just as if we can imagine the tangent "slipping" accross either of the branches to "continuously" become horizontal. In my opinion, he implicitly uses the assumption that the derivative of the semicubical parabola is a continuous function; and he probably does that, based on the shape of the graph.
• Intuitively I understand. But in the definition of the derivative we use $\frac{y(x)-y(x_0)}{x-x_0}$ and not $\frac{y(t)-y(0)}{x(t)-x(0)}$. That is, we must find $y(x)$ first. Otherwise, can you prove rigorously that it works with $t$? – user239753 May 4 '16 at 7:11
• Please read carefully: There is nothing intuitive here. (Only the idea described in your textbook seems intuitive). When confined to each branch seperately: $$y(x_0)=y(x(0))=y(0), \ \ y(x)=y(x(t))=y(t) \\ x_0=x(0), \ \ x=x(t)$$ So, on each one of the functions $y=x^{3/2}$, $y=-x^{3/2}$ i.e. on each of the branches of the above graph, the rates of change you are writing are identified: $$\frac{y(x)-y(x_0)}{x-x_0}=\frac{y(t)-y(0)}{x(t)-x(0)}$$ On the other hand, since $x=t^2$ it should be clear that $$t\to0\Leftrightarrow x\to0$$ – KonKan May 4 '16 at 12:20
• i've added the above comment to the post – KonKan May 4 '16 at 12:51
• It should be $y(x(t))=(t^2)^3=t^6$, no? (I know that in this case it doesn't change the limit, but still). Thank you. P.S. - Also, I don't understand why somebody downvotes the answers here. – user239753 May 4 '16 at 14:28
• actually, for $t>0$: $$y(x(t))=\pm (x(t))^{3/2}=\pm (t^2)^{3/2}=\pm t^3$$ So, for $t\in (-\infty, +\infty)$: $$y(t)=t^3$$ – KonKan May 4 '16 at 19:14 | 2021-02-25T11:48:01 | {
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https://math.stackexchange.com/questions/818160/six-dice-six-different-figures-model-problem | Six dice, six different figures (model problem)
I am thinking about the same problem as posted here: Probability of All Distinct Faces When Six Dice Are Rolled.
If one rolls six dice, what is the probability of attaining six different figures. I do understand the solution but wondered why it is necessary to assume that the dice are distinguishable. If I assume they are not then I have exactly one desired result and
$$\binom{6 + 6 - 1}{6} = \binom{11}{6} = 462$$
possible results.
But I get a different probability in this case which leads me to the assumption that it is not working this way. But why exactly is that?
You have used a correct Stars and Bars argument to show that there are $462$ ways to distribute $6$ objects in $6$ boxes.
However, if we assume that the dice are fair, and do not influence each other, then these $462$ possibilities are not all equally likely.
Let us look at a much smaller example, two identical coins. There are $3$ ways to distribute these into $2$ boxes. However, in this case it is fairly clear that the probability of two heads is $\frac{1}{4}$ and not $\frac{1}{3}$. For whether the coins are distinguishable or not, it should make no difference to the probability if we toss the coins sequentially and not simultaneously. And sequential tossing tells us the probability is $\frac{1}{2}\cdot\frac{1}{2}$.
Roughly speaking, an extreme Stars and Bars case, such as all balls in the third box, has smaller probability than any specific more "even" distribution.
Remark: To see more informally that a probability model based on distinguishable dice is appropriate, assume that otherwise indistinguishable dice are made different by writing IDs on them with invisible ink. The writing should not affect the probability of events that do not involve the IDs, such as all six numbers being obtained.
There are $6!$ combinations of six different numbers, and there are $6^6$ combinations of six rolls. The probability is then: $$p = \frac{6!}{6^6}$$
• Yes, I know that. I was wondering why my way of thinking is wrong. Why can't I just record which numbers came up whithout recording from which dice it came? – Cyianor Jun 2 '14 at 14:03
• My guess would be you need to account for the fact all orders are the same, i.e. $123456$ is the same as $654321$ – Alex Jun 2 '14 at 14:10
• You mean I might not be able to apply Laplacian probability, i.e. p = good cases / total cases? – Cyianor Jun 2 '14 at 14:12
• Yes you are: here $6!$ is 'good cases' and $6^6$ are 'total cases' – Alex Jun 2 '14 at 14:13
• I meant in the case of my description above. If I say the dice are indistinguishable then the elements of my probability space have different probabilities and Laplacian probability is not applicable – Cyianor Jun 2 '14 at 14:17 | 2019-05-25T09:56:34 | {
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https://math.stackexchange.com/questions/4046626/what-is-the-least-amount-of-straight-cuts-that-should-be-made-to-get-squares-fro | # What is the least amount of straight cuts that should be made to get squares from a rectangle?
The problem is as follows:
Rachel has a rectangular piece of cloth which measures $$2$$ meters large and $$0.2$$ meters wide. She is to use a guillotin which can only makes cuts of $$60$$ cm of maximum in length. Another constrain in this guillotin is that it can only cut one layer of this cloth. Using this information, how many straight cuts minimum can Rachel make in order to get from such fabric $$10$$ square pieces of $$20\,cm$$ from each side?.
The alternatives given in my book are as follows:
$$\begin{array}{ll} 1.&\textrm{3 cuts}\\ 2.&\textrm{1 cut}\\ 3.&\textrm{2 cuts}\\ 4.&\textrm{4 cuts}\\ \end{array}$$
I'm not sure how to solve this question regarding cuts. But I believe that a way to reduce the number of cuts will require to fold the piece. But in this case this seems not to be possible because there is a condition in the problem which is forbidding to do such approach.
My initial approach was to make the necessary cuts to make some sort of grid which could I guess reduce the number of cuts but 10 is not a perfect square. Therefore what sort of logic should be here?.
I've went in circles for long on this question so I will really appreciate someone could give me a hand with this.
Therefore what should be the way to go here?. Can someone help me here please?. I'm confused.
• I think the question allows folding, but a guillotin cut cannot be through multiple layers. – peterwhy Mar 3 at 2:28
• I brute forced a 4 cut solution. If this is the answer, there must be a proof as to why this is the minimum required. – Andrew Chin Mar 3 at 2:33
To cut the strip of cloth, it only takes 9 cuts of 20 cm, and each cut of the guillotin cuts through 60 cm at most, so at least 3 cuts are necessary.
Number the cuts $$1$$ to $$9$$ sequentially, and the resultant squares $$0$$ to $$9$$.
Align cut $$1$$ under the guillotin, then fold two right angled triangles on the strip around cuts $$2$$ and $$3$$ so that the strip goes back under the guillotin, and align cut $$4$$ under the guillotin next to cut $$1$$.
(Fold twice like (B) in this random image found in Google: )
Do the same thing to align cut $$7$$ under the guillotin. With cut $$1, 4, 7$$ all aligned side by side, one guillotin cut of 60 cm would do all the 3 20-cm cuts.
Now squares 1-3, squares 4-6, squares 7-9 are on separated pieces of cloth, and they can be aligned freely to make the remaining 6 cloth-cuts in 2 guillotin cuts.
The optimal answer is 3 guillotin cuts. | 2021-06-16T14:33:38 | {
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https://math.stackexchange.com/questions/3035983/how-can-i-find-integers-which-satisfy-frac150n15n-m | # How can I find integers which satisfy $\frac{150+n}{15+n}=m$?
Here are some facts about myself:
1. In 2017, I was $$15$$ years old.
2. Canada, my country, was $$150$$ years old.
When will be the next time that my country's age will be a multiple of mine?
I've toned this down to a function. With $$n$$ being the number of years before this will happen and $$m$$ being any integer,
$$\frac{150+n}{15+n}=m$$
How would you find $$n$$?
• For extra credit, consider that Canada turned 150 on July 1, 2017. If your birthday is after that date, 14+n is also a valid denominator, as you will be both age 14 and age 15 during the 12 months that Canada is 150 years old. Likewise, if your 15th birthday is before that date, 16+n is a valid denominator, since you will be 15 when Canada turns 150, but turn 16 before Canada turns 151. – Nuclear Wang Dec 12 '18 at 4:51
• (1) Avoid using "interesting" in the title; (2) describe the problem in the title, not just your opinion and its topic. – Asaf Karagila Dec 12 '18 at 11:19
You want $$\frac{150+n}{15+n}=m$$, and clearing denominators gives us $$150+n=(15+n)m.$$ Subtracting $$15+n$$ from both sides give us $$135=(15+n)(m-1).$$ Now you are looking for the smallest $$n>1$$ for which such an $$m$$ exists, so the smallest $$n>1$$ for which $$15+n$$ divides $$135$$.
• Love it, thank you! Is 120 the only $n$ that exists for such a case? (Because $135\over270$ $=$ $1\over2$)? – Raymo111 Dec 11 '18 at 23:21
• What's wrong with $n=12$? The other solutions are $n=30$ and $n=120$. – Servaes Dec 11 '18 at 23:46
• Oh shoot, I mistakenly read your answer as $n+15=135$, my bad! Also, upvote if you think this question that I thought of was interesting, it's a real-life example of an age problem. – Raymo111 Dec 11 '18 at 23:48
First thing I would do is say that Canada is $$135$$ years older than you.
That gives you a simpler
$$\frac {135+n}{n} = k\\ \frac {135}{n} = k-1\\$$
It will happen every time your age is a factor of $$135.$$ It last happened when you were $$15.$$ It will next happen when you are $$27$$
• Do Canadians live past age 135? – richard1941 Dec 21 '18 at 16:51
• Almost never. But with advancements in medical technology, that may change. – Doug M Dec 21 '18 at 17:00
We want the smallest positive integer $$n$$ such that there is some (positive) integer $$k$$ such that $$\frac{150+n}{15+n}=k.$$ Note that $$k=1$$ can never work, so we can assume $$k-1\neq0.$$ Now we rearrange the above equation: multiplying both sides by $$15+n,$$ we get $$150+n=15k+nk;$$ now rearrange and factorize to get $$15(10-k)=(k-1)n;$$ and now divide both sides by $$k-1,$$ to get $$n=\frac{15(10-k)}{k-1}.$$ Since we want the smallest positive integer $$n,$$ we can just try values of $$k\in\{2,3,\ldots,9\},$$ starting from the largest and working our way down (because the function of $$k$$ on the right-hand side is decreasing in this range), until we arrive at an integer value of $$n.$$* When $$k=9,$$ $$8$$ or $$7$$ we get non-integer values of $$n;$$ when $$k=6$$ we find $$n=15\times4/5=12.$$
So last year, $$n$$ was $$0,$$ and the ratio of Canada's age to your age was $$k=150/15=10;$$ and $$11$$ years from now, $$n$$ will be $$12,$$ and the ratio of Canada's age to your age will be $$k=162/27=6.$$
*Incidentally, it is not obvious in advance that we will ever get an integer value of $$n;$$ if this were the case then the problem would simply have no answer. As it happens, the problem does have answers, namely $$(n,k)\in\{(0,10),(12,6),(30,4),(120,2)\}.$$
Alternatively:
$$\frac {150 + n}{15 + n} = \frac {150+ 10n}{15+n} +\frac {-9 n}{15+n}$$
$$=10 -\frac {9 n}{15+n}$$ (which is an integer for $$n=0$$ but when next?)
$$=10 - \frac {9n + 9*15}{15+n} + \frac {9*15}{15+n}=$$
$$=10 - 9 + \frac{3^3*5}{15+n}= 1 + \frac{3^3*5}{15+n}$$
which is an integer if $$15+n$$ is one of the factors of $$3^3*5$$.
And the factors of $$3^3*5$$ are $$1, 3,9, 27, 5,15, 45, 135$$.
So this will occur when $$n = -14,-12, -10, -6,0, 12,30, 120$$
When you are $$1, 3, 5, 9, 15, 27, 45, 135$$ and canada is $$136, 138, 140, 144, 150, 162, 180, 270$$ and canada is exactly $$136,46, 28, 16, 10,6,4, 2$$ as old as you are.
(Enjoy your $$45$$ birthday when your country annexes my country after we collapse from the thirty year aftermath of the unrecoverable mistakes of the last two years.)
That's a fun problem. It's nice to see other people like to think about these things.
• That's nice, that's the way I usually think about these problems too; but you can shorten the process by writing $\dfrac {150+n}{15+n} = \dfrac {15+n}{15+n} + \dfrac{135}{15+n}$ – Ovi Dec 12 '18 at 2:34
• Hmmm.... I'm not sure why I did it the way I did. I think somehow I briefly thought the obviousness that 15 goes directly into 150 made me briefly think that they were the coeficients and I started doing it that way and just continued. Obviously it'd be shorter and direct to do the coefficients. – fleablood Dec 12 '18 at 15:40
Note that if $$k \mid a$$ and $$k \mid b$$, then $$k \mid a - b$$. In this particular instance, we have $$15 + n \mid 15 + n$$ and $$15 + n \mid 150 + n$$, so $$15 + n \mid 135$$. In other words, we are looking for $$n+15$$ to be the next factor of $$135$$ which is larger than $$15$$. Can you continue from here?
Servaes answer is excellent but let me add few cents for those who can't really make out something from dual variables and other factors (I know some people have problem with such things).
There is also a brute force solution, requiring a bit of thinking and a bit of simple calculation (the larger is the $$n$$ i $$n$$-times larger the more difficult it becomes though so I really suggest you try understanding the accepted solution anyway).
It's enough to check if there is an integer solution to equation $$\frac{150+n}{15+n}=9$$ If no, proceed with 8, 7 and so on. If you get to 2 and still have no anwser then the answer is never.
So a quick brute force check will like that: $$\frac{150+n}{15+n}=9$$ $$150+n=9\cdot(15+n)$$ $$150+n=135+9\cdot n$$ $$15=8\cdot n$$ $$n = \frac{15}{8}\notin\mathbb{Z}$$ So we continue with $$8$$ $$\frac{150+n}{15+n}=8$$ $$150+n=8\cdot(15+n)$$ $$150+n=120+8\cdot n$$ $$30=7\cdot n$$ $$n = \frac{30}{7}\notin\mathbb{Z}$$ You may continue from here. Note, a smart person using this method will notice some pattern that will make it even simpler to test (not requiring all the calculation, just a quick check that literally takes seconds) but I will not give a direct hint.
## Explanation
When you add the same number to both numerator and denominator of a fraction, the fraction decreases. In other words if you consider how the fraction of age of Canada and your age changes over time it will decrease.
On the other hand the fraction can never go to $$1$$ (or below) since the numerator will always be obviously greater than denominator (in our case by 135).
In other words you have a finite set of possible resultant fractions (namely $$\{2,3,4,5,6,7,8,9\}$$) that are potentially the sought "next case". From those you're looking for the largest one, so test them one by one, starting with $$9$$ and going down until you have an integer result or your options are gone.
## Remarks
1. Once again - this is a brute force solution and should not be treated as preferred if you can understand using two variables. But if you can't, it's still doable.
2. If you notice the relationship I mention above, I guess even testing situation where the initial fraction is 100 could be done.
3. You may try solving an interesting opposite question - when was the last time when such situation occurred (i.e. Canadas age was a multiply of your age). If you think it over well, you may use both Servaes and my approaches to that as well. Especially try rationalising that my approach will again have a finite and easily predictable number of tests to be performed.
4. A food for thoughts. Is there always a solution to the original question (with differently chosen ages)? Is there always a solution to the problem I state in remark 3?
5. Can you think of other related questions that will use similar approach (either Servaes's elegant one or mine brute force) to find a solution?
In general questions like this, where
$$\frac{x+n}{y+n}=m$$
$$m$$ will be an integer smaller than $$\frac{x}{y}$$. Since there are finitely many integers less than $$\frac{x}{y}$$, each value of $$m$$ can be checked one by one. Not all values will generally give an integer $$n$$. | 2019-01-16T15:02:35 | {
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# When traveling at a constant speed of 32 miles per hour, a certain mot
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When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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14 Jun 2016, 17:07
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AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Dear AbdurRakib,
I'm happy to respond.
We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal
In that fraction, cancel the common factor of 8.
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.
Does this make sense?
Mike
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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15 Jun 2016, 00:46
1
1
Distance travelled = 32 miles
Fel consumed = 24 gallons
So, 24 gallons lets motorboat travel 32 miles
So, miles traveled per gallon of fuel = 32/24 = 4/3
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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15 Jun 2016, 02:59
2
In 1 hour the motorboat travels 32 miles
In travelling that 32 miles the motorboat uses 24 gallons of fuel in 1 hour
so basically we divide $$\frac{miles}{hr}$$ by $$\frac{gallons}{hr}$$ to get $$\frac{miles}{gallons}$$
$$\frac{32}{24}$$ =$$\frac{4}{3}$$ $$\frac{miles}{gallons}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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13 Aug 2016, 11:05
1
One way to solve this:--
The speed is 32 miles / hour. The fuel consumption is 24 gallons / hour. So in every hour two things are happening:-
The boat has covered 32 miles
The boat has consumed 24 gallons
What is the miles /gallon --> 32/24 => 4/3miles/gal. The choice is D
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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02 Dec 2016, 06:54
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
We are given that a motorboat has a rate of 24 gallons per hour while traveling at a speed of 32 miles per hour. We must determine the boat’s fuel consumption measured in miles traveled per gallon of fuel.
Since the motorboat consumes 24 gallons when traveling 32 miles, the rate is 32/24 = 4/3.
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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23 Feb 2018, 12:17
mikemcgarry wrote:
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Dear AbdurRakib,
I'm happy to respond.
We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal
In that fraction, cancel the common factor of 8.
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.
Does this make sense?
Mike
hello there mikemcgarry, how are you ? you know what i dont understand, why isnt the answer B ? wouldnt it be logically correct to divide gallons by miles ?
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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03 Apr 2018, 10:01
1
Top Contributor
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Let's see what happens after ONE HOUR of traveling.
In ONE hour, the boat will travel 32 miles and will use 24 gallons of fuel
So, the fuel consumption rate is 32 miles per 24 gallons
Or we can write: fuel consumption rate = 32/24 miles/gallon
32/24 = 4/3
So, the fuel consumption rate = 4/3 miles per gallon
Cheers,
Brent
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05 Apr 2018, 03:43
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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14 Apr 2018, 22:53
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
It's (D)
$$\frac{Miles - traveled}{gallon-fuel}$$ = $$\frac{32}{24}$$ = $$\frac{4}{3}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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25 Jan 2019, 18:10
mikemcgarry wrote:
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
Dear AbdurRakib,
I'm happy to respond.
We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal
In that fraction, cancel the common factor of 8.
fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.
Does this make sense?
Mike
Was checking through the solutions to see other solutions to this problem, isn't the units in red suppose to be in "hours", instead of gallons?
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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11 May 2019, 05:59
Speed of boat 32M/H and its consumption is 24 Gallon per hour.
Therefore boat consumes 24Gallon for every 32 Miles.
or for every Gallaon boat travels 32/24 miles
i.e 4/3 miles per gallon
AnsD
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink]
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11 May 2019, 22:55
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?
A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$
In 24 gallons boat travel =36
In 1 " " " =36/24=4/3
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot [#permalink] 11 May 2019, 22:55
Display posts from previous: Sort by | 2019-05-24T17:11:01 | {
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https://www.freemathhelp.com/forum/threads/probability-theory.69762/ | Probability Theory
alakaboom1
New member
An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?
I think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesn't account for the fact that we are assuming that the driver is already in an accident...
galactus
Super Moderator
Staff member
You are looking for the probability the driver is from class A given they have an accident:
$$\displaystyle P(\text{class A}|\text{accident})$$
You can make a chart. Assume a number of drivers that is easy to work with. Say, 1000
Or, use Bayes Theorem. Do you know it?.
soroban
Elite Member
Hello, alakaboom1!
Here is a very primitive approach . . .
An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C.
The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%.
The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively.
After purchasing an insurance policy, a driver has an accident within the first year.
What is the probability that the driver is of class A?
Suppose there are 10,000 policyholders.
$$\displaystyle \text{20\% are class A drivers: }\:20\%\times 10,\!000 \,=\,2000$$
. . $$\displaystyle \text{1\% of them will have an accident: }\:1\% \times 2000 \,=\,20$$
$$\displaystyle \text{65\% are class B drivers: }\:65\% \times 10,\!000 \,=\,6500$$
. . $$\displaystyle \text{2\% of them will have an accident: }\:2\% \times 6500 \,=\,130$$
$$\displaystyle \text{15\% are class C drivers: }\:15\% \times 10,\!000 \,=\,1500$$
. . $$\displaystyle \text{3\% of them will have an accident: }\:3\% \times 1500 \,=\,45$$
$$\displaystyle \text{There is a total of: }\:20 + 130 + 45 \,=\,195\text{ accidents}$$
. . $$\displaystyle \text{of which 20 are class A drivers.}$$
$$\displaystyle \text{Therefore: }\(\text{class A}\,|\,\text{accident}) \;=\;\frac{20}{195} \;=\;\frac{4}{39}$$
galactus
Super Moderator
Staff member
That 'primitive
approach is a nice explanation.
Here is all I was getting at:
$$\displaystyle \frac{(.01)(.2)}{(.01)(.2)+(.02)(.65)+(.03)(.15)}=\frac{4}{39}\approx .1025$$
When confronted with a 'probability of something given something' problem, you can make a chart.
Assume 1000 policyholders.
$$\displaystyle \begin{array}{c|c|c|c|c}\text{}&A&B&C&\text{total}\\ \hline\text{has accidents}&2&13&4.5&19.5 \\ \hline\text{no accident}&198&637&145.5&980.5 \\ \hline\text{total}&200&650&150&1000\end{array}$$
Now, you can answer any problem they throw at you. In this case, we want
"Probability the driver is class A given they have an accident".
Go down the class A column and across the 'has accident' row. $$\displaystyle \frac{2}{19.5}=\frac{4}{39}$$
Say they asked for "probability the driver is from class C given they did not have an accident?"
Go down the C column and across the 'no accident' row and get $$\displaystyle \frac{145.5}{980.5}=\frac{291}{1961}\approx .148$$
and so on..............
alakaboom1
New member
Thanks a lot guys!
Out of curiosity, is this formula you guys mentioned known as Bayes Theorem? I've definitely used the formula before, but never knew it by name. | 2019-03-26T06:37:33 | {
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http://math.stackexchange.com/questions/125289/cosets-of-a-group | # Cosets of a group
One sentence from Amstrong's Group and Symmetry wrote the following to prove a group of order 6 is isomorphic to $\mathbf{Z}_6$:
The right cosets $\langle x\rangle$, $\langle x\rangle y$ give 6 elements $e$, $x$, $x^2$, $y$, $xy$, $x^2y$ which fill out $G$.
Where $x$ is of order $3$ and $y$ is of order 2. $x$, $y$ are both elements of $G$.
My confusion:
How can one guarantee that the right coset $\langle x\rangle$ and $\langle x\rangle y$ can fill out $G$? Because $\langle x\rangle$ and $\langle x\rangle y$ have no intersections? I am not sure about the properties of cosets. I also don't know if this is related to the fact that $|\langle x\rangle|=5$ and is precisely half the elements of $G$ and thus its left coset $y\langle x\rangle$ and right coset $\langle x\rangle y$ is exactly the same?
I don't know if you can understand my question. The question here actually arises from my vague understanding of "cosets". So everytime this term occurs I feel a steady uncertainty.
-
You may want to reread the question yourself and see where one or more letters can be added. – user21436 Mar 28 '12 at 1:36
@KannappanSampath: An artifact of using < and > in text; parser thought they were mark-up. – Arturo Magidin Mar 28 '12 at 2:10
This is specifically page 70 (and end of 69). – anon Mar 28 '12 at 2:12
Exactly. It is from page 70. Sorry for the disorganized order, I will check that next time before I submit. And thanks again to @ArturoMagidin for the edition. – Jinji Mar 28 '12 at 2:26
For more on cosets, see this answer. – Arturo Magidin Mar 28 '12 at 2:45
Let $H=\{e,x,x^2\}$ be the subgroup generated by $x$. The right cosets of $H$ are the sets of the form $Hz = \{hz\mid h\in G\}$ for fixed $z\in G$.
The right cosets of $H$ are known to form a partition of the set $G$; that is: any two cosets are either identical or disjoint, and their union is all of $G$.
Moreover, $Hz = Hw$ if and only if $zw^{-1}\in H$. This can be verified because $z\in Hz$ (obtained as $ez$), so there must exist some $h\in H$ such that $z = hw$. Hence $zw^{-1}=h\in H$, proving that if $Hz=Hw$, then $zw^{-1}\in H$. Conversely, if $zw^{-1}\in H$, then $z = (zw^{-1})w \in Hw$, so $Hz\cap Hw\neq\varnothing$; since right cosets are either disjoint or identical, and $Hz$ and $Hw$ are not disjoint, then they are identical.
Now, since $y$ is of order $2$, it is not in $H$ (every element of $H$ is either of order $3$ or of order $1$). That means that the cosets $He = H$ and $Hy=\{ey, xy, x^2y\}$ are distinct, hence they are disjoint. Since they are disjoint, $H$ and $Hy$, together, account for $6$ elements of $G$. Since $G$ has exactly $6$ elements by assumption, those are all the elements of $G$. That is: $$G = H\cup Hy = \{e,x,x^2\}\cup\{y,xy, x^2y\} = \{e,x,x^2,y,xy,x^2y\}.$$
The same is true for left cosets: left cosets of $H$ partition $G$, so any two left cosets are either disjoint or identical. Also, $zH = wH$ if and only if $z^{-1}w\in H$ (I'll leave the proof to you). That means that the two left cosets of $H$ are $eH=H$ and $yH$, since $yH\neq H$, so $yH\cap H=\varnothing$. Since $G$ has $6$ elements, $H$ has $3$, and $yH$ has another three, then $$G = H\cup yH = \{e,x,x^2\} \cup \{y, yx, yx^2\}.$$ Also, since $H\cap Hy=\varnothing$, then $Hy = G\setminus H$; similarly, $yH=G\setminus H$. So in fact, $Hy=yH$ as sets.
-
Thanks! Does the number of right cosets relate to the order of H? |G|/|H|=the number of right cosets? – Jinji Mar 28 '12 at 2:24
@Jinji: The number of cosets is, by definition, the index of $H$ in $G$, denoted $[G:H]$. Since the cosets all have the same size, namely $H$, and partition $G$, it follows that $|G|=|H|[G:H]$ (in the sense of cardinalities); in fact, if $K\lt H\lt G$, then $[G:K]=[G:H][H:K]$ in the sense of cardinalities. So in the special case in which $|G|$ is finite, it follows from $|G|=|H|[G:H]$ that $[G:H]=|G|/|H|$. – Arturo Magidin Mar 28 '12 at 2:37 | 2016-02-11T17:20:10 | {
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http://math.stackexchange.com/questions/166756/proving-that-an-integer-is-even-if-and-only-if-it-is-not-odd | # Proving that an integer is even if and only if it is not odd
There is this question, but the definition of "even" and "odd" that I am using uses integers instead of just natural numbers; i.e.,
• An integer $n$ is even iff there is some integer $k$ such that $n=2k$.
• An integer $n$ is odd iff there is some integer $k$ such that $n=2k+1$.
Here is what I have so far:
First we show that an integer $n$ is even or odd. We first use induction on the positive integers. For the base case, $1=2\cdot0+1$ so we are done. Now suppose inductively that $n$ is even or odd. If $n$ is even, then $n=2k$ for some $k$ so that $n+1=2k+1$ (odd). If $n$ is odd, then $n=2k+1$ for some $k$ so that $n+1=2(k+1)$ (even). This closes the induction, so every $n\in\mathbf{Z}^+$ is even or odd.
Now we show every $n\in\mathbf{Z}^-$ is even or odd. Let $n\in\mathbf{Z}^-$. Then $n=-k$ for some $k\in\mathbf{Z}$ (I think this follows immediately from most definitions of the integers.). Suppose $k$ is even. Then $k=2j$ for some $j$ so that $n=-k=-2j=2(-j)$ (even). Now suppose $k$ is odd. Then $k=2j+1$ for some $j$ so that $n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1$ (odd).
For $0$, note that $0=2\cdot0$ (even).
Now we show that $n\in\mathbf{Z}$ cannot be both even and odd. Suppose for the sake of contradiction that $n\in\mathbf{Z}$ is both even and odd. Then there are integers $k,j$ such that $n=2k=2j+1$. This implies that $2(k-j)=1$ (like in the referenced question). So we must show that $1$ cannot be even in order to complete the proof.
This is where I am having trouble. I know that if I let $f:\mathbf{Z}\to\mathbf{Z};x\mapsto2x$ be a function, then $f$ is increasing so since $f(0)=0$ and $f(1)=2$ and $0<1<2$, there is no integer $m$ such that $f(m)=1$. But this seems complicated so I was wondering if there was an easier way to do this.
So my real question is: how can I show that $1$ is not even?
(This is not homework.)
-
$f(m)=1 \implies 2m=1 \implies 2=\frac {1}{m}$ which is not a integer for any $m \in \mathbb Z$ – Theorem Jul 4 '12 at 20:57
@russell11, I think you meant a function $f:\mathbb{Z}^+ \to \mathbb{Z}^+$ (or whatever the appropriate notation is for 'nonnegative'). Otherwise you and Theorem converge on the right idea; suppose $1=2m$ for some integer $m$. $m>0$ comes immediately, as twice a non-positive integer is non-positive. $m<1$ comes from the function being increasing. – Eugene Shvarts Jul 4 '12 at 21:02
To show that $1$ is not even:
I assume you can prove or accept that $a \cdot 0 = 0$ for all $a \in \mathbb{Z}$, the product of a positive and negative number is negative, and $2a > a$ when $a >0$.
If $1$ is even, then there must exists $a < 1$ such that $2a = 1$. However, the only $a < 1$, which is an integer, is $0$ and clearly $2 \cdot 0 = 0$. So $1$ can not be be written as $2 \cdot a$ for any $a \in \mathbb{Z}$. $1$ is odd.
-
Do you want to say, "So $1$ cannot be written as $2\cdot a$ for any $a\in\mathbb{Z}$" in the end? – Paul Jul 4 '12 at 21:13
@Paul Yes. Added the missing "not". – William Jul 4 '12 at 21:21
Hint $\$ Your induction step uses $\rm\:n\,$ even $\rm\,\Rightarrow\: n\!+\!1\:$ odd, and $\rm\:n\,$ odd $\rm\,\Rightarrow\:n\!+\!1\:$ even. The converses are both true, e.g. $\rm\,n\!+\!1\,$ odd $\,\Rightarrow$ $\rm\,n\!+\!1 = 2k\!+\!1\,$ $\Rightarrow$ $\rm\,n = 2k,\,$ since $\rm\,j\!+\!1 = k\!+\!1\,$ $\Rightarrow$ $\rm\,j = k\,$ by Peano axioms. Thus if $\rm\:n\!+\!1\,$ is both even and odd then so too is $\rm\,n.\,$ So, contrapositively, in your induction step you can lift "$\rm n\:$ is not both even and odd" from $\rm\,n\,$ up to $\rm\,n\!+\!1\,$ and hence prove by induction that every natural is even or odd, but not both (this is essentially my hint in the linked answer).
-
Thanks, this is really neat. I didn't understand your hint in the other question, but now I do. – russell11 Jul 5 '12 at 1:18 | 2016-05-27T20:26:13 | {
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https://atekihcan.github.io/CLRS/02/E02.02-02/ | Consider sorting $$n$$ numbers stored in array $$A$$ by first finding the smallest element of $$A$$ and exchanging it with the element in $$A[1]$$. Then find the second smallest element of $$A$$, and exchange it with $$A[2]$$. Continue in this manner for the first $$n-1$$ elements of $$A$$. Write pseudocode for this algorithm, which is known as selection sort. What loop invariant does this algorithm maintain? Why does it need to run for only the first $$n-1$$ elements, rather than for all $$n$$ elements? Give the best-case and worst-case running times of selection sort in $$\Theta$$-notation.
#### Pseudocode
$$\textsc {Selection-Sort }(A)$$\begin{aligned}1& \quad \textbf {for }i=1\textbf { to }A.length-1 \\2& \quad \qquad minIndex=i \\3& \quad \qquad \textbf {for }j=i+1\textbf { to }A.length \\4& \quad \qquad \qquad \textbf {if }A[j]<A[minIndex]\text { and }j\not=minIndex \\5& \quad \qquad \qquad \qquad minIndex=j \\6& \quad \qquad \text {swap }A[i]\text { with }A[minIndex] \\\end{aligned}
A python implementation of the above pseudocode is shared at end of the page, you can verify the workings for yourself.
#### Loop Invariant
At the start of the each iteration of the outer for loop of lines 1-6, the subarray $$A[1..i - 1]$$ consists of $$i - 1$$ smallest elements of $$A$$, sorted in increasing order.
#### Why only first n - 1 elements
The algorithm needs to run for only the first $$n - 1$$ elements, rather than for all $$n$$ elements because the last iteration will compare $$A[n]$$ with the minimum element in $$A[1 .. n - 1]$$ in line 4 and swap them if necessary. So, there is no need to continue the algorithm for all the way to the last element.
#### Running Times
For both the best-case (sorted array) and worst-case (reverse sorted array), the algorithm will anyway take one element at a time and compare it with all the other elements. In other words, each of the $$n$$ elements will be compared with rest of the $$n - 1$$ elements. So, the running times for both scenario will be $$\Theta(n^2)$$.
The above reasoning should be sufficient to understand or convey why the runtime would be $$\Theta(n^2)$$. However, for the sake of completeness, an exhaustive mathematical proof is given below.
#### Runtime Analysis
Let’s assume the inner for loop in line 3-5 is executed for $$t_j$$ times for $$j = 2, 3, \ldots, n$$, where $$n = A.length$$. Now note that, line 5 will be executed less than $$t_j - 1$$ times in the average case, but it’ll still be of the order of $$n$$.
For the sake of simplicity let’s assume the worst case, i.e. a reverse sorted array, when it’ll be executed exactly $$t_j - 1$$ times. Note, this assumption is only for that particular line, which is not going to change our overall analysis, it will only make our calculation easier.
We can now calculate the cost and times for individual lines of the pseudocode as follows …
Line Cost Times
1 $$c_1$$ $$n$$
2 $$c_2$$ $$n - 1$$
3 $$c_3$$ $$\sum_{j = 2}^n t_j$$
4 $$c_4$$ $$\sum_{j = 2}^n (t_j - 1)$$
5 $$c_5$$ $$\sum_{j = 2}^n (t_j - 1)$$
6 $$c_6$$ $$n - 1$$
Now, for any arbitrary value of $$j$$, the inner for loop (line 3-5) compares the previously computed minimum value with all elements in the subarray $$A[j..n]$$. So the inner for loop executes $$n - j + 1$$ times, i.e. $$t_j = (n - j + 1)$$ for $$j = 2, 3, \ldots, n$$. So, $$t_2 = n - 1, t_3 = n - 2, \ldots t_n = 1$$. We can calculate the summations for line 3-5 as follows …
\begin{aligned} \sum_{j = 2}^n t_j & = (n - 1) + (n - 2) + \cdots + 1 \\ & = \frac {n(n - 1)} 2 \\ \\ \sum_{j = 2}^n (t_j - 1) & = \sum_{j = 2}^n t_j - \sum_{j = 2}^n 1 \\ & = \frac {n(n - 1)} 2 - (n - 1) \\ & = \frac {(n - 3)(n - 2)} 2 \end{aligned}
Therefore, we can calculate the running time as follows..
$T(n) = c_1(n - 1) + (c_2 + c_6)n + c_3 \frac {n(n - 1)} 2 + (c_4 + c_5) \frac {(n - 1)(n - 2)} 2$
For best-case scenario, as the array is already sorted, line #5 will not be executed ever. So, $$c_5 = 0$$. Even with that (and without that for worst-case) the expression of $$T(n)$$ will be reduced to the form $$an^2 + bn + c$$, i.e. the algorithm will run at $$\Theta(n^2)$$ time.
#### Python Code
# Selection Sort def SelectionSort(A): for i in range(len(A)): minIndex = i for j in range(i + 1, len(A)): if A[j] < A[minIndex] and j != minIndex: minIndex = j A[i], A[minIndex] = A[minIndex], A[i] # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): length = random.randint(2, 50) lst = [random.randint(0, 100) for _ in range(length)] SelectionSort(lst) # Check if list has been sorted for i in range(len(lst) - 1): if lst[i] > lst[i + 1]: num_failed += 1 print(f"Test #{i:<2}: List is not sorted") break if num_failed > 0: break if num_failed > 0: print(f"\nFailed") else: print(f"Passed {total_tests}/{total_tests} tests") | 2023-03-28T23:41:04 | {
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http://mathhelpforum.com/calculus/35956-solved-vectors-question.html | # Math Help - [SOLVED] Vectors Question...
1. ## [SOLVED] Vectors Question...
Question:
In the diagram OABCDEFG is a cube which the lenght of each edge is $2$ units. Unit vectors i, j, k are parallel to $\vec{OA}$, $\vec{OC}$, $\vec{OD}$ respectively. The mid-points of $AB$ and $FG$ are $M$ and $N$ respectively.
(i) Express each of the vectors $\vec{ON}$ and $\vec{MG}$ in terms of i, j and k. [3 Marks]
(ii) Find the angle between the directions of $\vec{ON}$ and $\vec{MG}$, are correct to the nearest $0.1^o$.
[4 Marks]
Attempt:
(i) $\vec{ON} = i + 2j + 2k$
$\vec{MG} = -2i + j + 2k$
are they right? and will I be rewarded 3 marks for showing no steps? becuase I don't there are any steps needed!
(ii) Need Help!
2. Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.
(ii)
In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).
It's defined like this:
a DOT b = |a| * |b| * cos(angle)
where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.
So you can rearrange that to get
cos(angle) = [ a DOT b ] / [ |a| * |b| ]
So now all you need to know is what a DOT b is:
I'll show that with an example:
(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32
All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.
Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.
Enjoy.
3. Originally Posted by Fedex
Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.
(ii)
In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).
It's defined like this:
a DOT b = |a| * |b| * cos(angle)
where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.
So you can rearrange that to get
cos(angle) = [ a DOT b ] / [ |a| * |b| ]
So now all you need to know is what a DOT b is:
I'll show that with an example:
(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32
All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.
Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.
Enjoy.
Thanks for the explanation and the example
$a.b = \left|a\right|\times \left|b\right| \times \cos\theta$
$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right| \times \left|b\right|} \right)$
$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$
$\left|a\right|\times \left|b\right| = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}2\\1\\2\end{array}\right) = (1 \times 2) + (2 \times 1) + (2 \times 2) = 8$
$\theta = \cos^{-1} = \frac{4}{8}$
$\theta = 60^o$
Are my steps correct? and is the answer correct?
4. Okay, a few mistakes there.
You wrote your two column vectors like this:
You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)
So write it with a dot, yeah?
Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.
The MAGNITUDE of a vector (its "length") is just a number, not a vector.
For for the vector a = (1,2,3)
|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]
Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.
5. Originally Posted by Fedex
Okay, a few mistakes there.
You wrote your two column vectors like this:
You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)
So write it with a dot, yeah?
Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.
The MAGNITUDE of a vector (its "length") is just a number, not a vector.
For for the vector a = (1,2,3)
|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]
Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.
Thanks again
$a.b = \left|a\right|.\left|b\right|.\cos\theta$
$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$
$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$
$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 1$
$\left|a\right|.\left|b\right| = 3\times1 = 3$
$\theta = \cos^{-1} \frac{4}{3}$ (Invalid)
Where did I go wrong?
6. What's (-2)^2 again?
Check out that part of your working for the magnitude of b.
I think that's your problem :-)
7. Originally Posted by Fedex
What's (-2)^2 again?
Check out that part of your working for the magnitude of b.
I think that's your problem :-)
$a.b = \left|a\right|.\left|b\right|.\cos\theta$
$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$
$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$
$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 3$
$\left|a\right|.\left|b\right| = 3\times3 = 9$
$\theta = \cos^{-1} \frac{4}{9} = 63.6^o$ | 2014-07-10T11:24:47 | {
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# Tutor profile: Yannic V.
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Yannic V.
Ph.D. in mathematics
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## Questions
### Subject:Discrete Math
TutorMe
Question:
Find a formula for the sum $\sum_{k=0}^n k \binom{n}{k},$ by means of the binomial theorem. What can be deduced about the average number of elements in a random subset of $\{1,2, \hdots, n\}$?
Inactive
Yannic V.
Let $x,y \in \mathbb{R}$. Remember that the binomial theorem sates that $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{k}y^{n-k},$ for all $n \in \mathbb{N}$. In particular, for $y=1$ we have $(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k.$ Considering $x$ as a variable, we differentiate both sides of the above equation to obtain $n(x+1)^{n-1}= \sum_{k=0}^n k\binom{n}{k}x^{k-1}.$ It remains to plug in $x=1$: $\sum_{k=0}^n k \binom{n}{k} = n2^{n-1}.$ The average number of elements in a random subset of $\{1,2, \hdots, n\}$ is thus $\frac{\sum_{k=0}^n k \binom{n}{k}}{\sum_{k=0}^n \binom{n}{k}} = \frac{n 2^{n-1}}{2^n}=\frac{n}{2}.$ This is not surprising, since the number of subsets of size $k$ is exactly the same as the number of subsets of size $n-k$, for any $k$.
### Subject:Calculus
TutorMe
Question:
Solve the integral $\displaystyle \int 3 \sec(2x-1)\tan(2x-1) \,dx$.
Inactive
Yannic V.
As the arguments of both functions in the integral are not single variables, we can start to consider a change of variable in order to just have one variable in the secant and tangent function. This can be made using the change of variable $$\label{cambio1} \left\lbrace\begin{array}{l} y = 2x-1 \\ dy = 2 \,dx \end{array}\right.$$ Notice that the derivative $dy$ is easy to obtain from the expression of the integral: $dx$ can be expressed in terms of $dy$ as $$\label{eq1} dx = \frac{dy}{2}.$$ Substituting the change of variable \eqref{cambio1} in the integral, we obtain \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \int 3 \sec(y)\tan(y) \,\frac{dy}{2} \tag{\footnotesize by \eqref{cambio1} and \eqref{eq1}}\\ &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \label{da1} \end{align} We have now reduced our problem to a more simple integral. We want to integrate the expression $\sec(y)\tan(y)$, which is the derivative of\footnote{This comes from the following calculation: $(\sec(y))'= \left(\frac{1}{\cos(y)} \right)' = \frac{(0\cos(y))-1(-\sin(y))}{\cos(y)^2}= \frac{\sin(y)}{\cos(y)^2} = \frac{1}{\cos(y)}\frac{\sin(y)}{\cos(y)}=\sec(y)\tan(y).$ } the function $\sec(y)$. It is then appropriate to consider the following change of variable: $$\label{cambio2} \left\lbrace\begin{array}{l} u = \sec(y)\\ du = \sec(y)\tan(y) \,dy \end{array}\right.$$ We can now replace \eqref{cambio2} into the expression \eqref{da1}. Remark that \eqref{da1} just contain the derivative part of\footnote{This is not a problem. When doing a change of variable, the most important part is that the expression corresponding to the derivative of the change of basis appears in the integral (up to a scalar). } \eqref{cambio2}. Thus, we have: \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \tag{\footnotesize{by \eqref{da1}}}\\ &= \frac{3}{2} \int du \tag{\footnotesize{by \eqref{cambio2}}}\\ &= \frac{3}{2}\, u + C, \label{da2} \end{align} where $C$ is a constant. Finally, we just have to substitute $u$ by its value in terms of $y$ (given by \eqref{cambio2}), and replace $y$ by its value in terms of $x$ (given by \eqref{cambio1}): \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2}\, u + C \tag{\footnotesize{by \eqref{da2}}}\\ &= \frac{3}{2}\sec(y) + C \tag{\footnotesize{substitute $u=\sec(y)$}}\\ &= \frac{3}{2}\sec(2x-1) + C. \tag{\footnotesize{substitute $y=2x-1$}} \end{align}
### Subject:Algebra
TutorMe
Question:
If $G$ is a finite group, show that the number of elements in $G$ of order greater than $2$ must be even. In particular, prove that any group of even order must contain an element of order 2.
Inactive
Yannic V.
We will solve the question by a counting argument as follows. Let $X$ be the set of all elements of G with order greater than two. If $X = \emptyset$ then $|X|=0$ and we’re done. Suppose now that $X \neq \emptyset$. Since the orders of $g$ and $g^{-1}$ coincides for any $g \in G$, we have that $g \in X \Longleftrightarrow g^{-1} \in X.$ Moreover, if $g \in X$ then $g^2\neq e_G$. From here, $g \neq g^{-1}$. It follows that $X$ can be written as the disjoint union of two element sets of the form $\{g, g^{-1}\}$ {x,x−1}, and hence that $|X|$ is even. Suppose that $|G|$ is even. Since $e_G$ has order $1$, $e_G \notin X$. It follows that $G \setminus X \neq \emptyset$. So $0 < |G \setminus X|=|G|-|X|$. Since $|G|$ and $|X|$ are both even, it follows that $|G\setminus X|$ is a nonzero even integer, i.e. is at least 2. Thus, there is an $z \in G \setminus X$ such that $z \neq e_G$ . Since $X$ consists of all elements in $G$ of order greater than $2$, it must be the case that $|z|=2$.
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https://math.stackexchange.com/questions/1809993/how-to-integrate-int-fracdx1x22 | # How to integrate $\int \frac{dx}{(1+x^2)^2}$?
I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:
$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$
The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
Also, Wolfram tells me:
$$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$
but I don't see how one can derive this without knowing the result beforehand.
Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).
• use $\frac 1{(1+x^2)^2 }=\frac 1{1+x^2} - \frac{x^2}{(1+x^2)^2}$ and do an integration by parts $\int \frac{x^2}{(1+x^2)^2} \, dx$ – abel Jun 2 '16 at 18:04
• I tried $x = \sinh(z)$ since $1+x^2 = \cosh^2(x)$ but I got stuck at $$\int \frac{1}{\cosh^3(z)} \, {\rm d}z$$ – ja72 Jun 2 '16 at 18:12
Use $x = \tan t$. Then $(1 + x^2)^2 = \sec^4 t$, and $\dfrac{dx}{dt} = \sec^2 t$, so the integral becomes \begin{align*} \int_0^{\pi/2} \cos^2 t \,\mathrm dt = \dfrac 1 2 \times \dfrac {\pi} 2 = \dfrac{\pi}{4}. \end{align*}
• Nice. Thanks. It's been a while and I forgot about trig substitions :( – I want to make games Jun 2 '16 at 18:08
For any $\alpha>0$, let: $$I(\alpha)= \int_{0}^{+\infty}\frac{dx}{\alpha^2+x^2} = \frac{\pi}{2\alpha}.$$ By differentiating both sides with respect to $\alpha$ we get: $$\int_{0}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = \frac{\pi}{2\alpha^2}$$ and by evaluating at $\alpha=1$: $$\int_{0}^{+\infty}\frac{dx}{(x^2+1)^2} = \color{red}{\frac{\pi}{4}}.$$
• How do you get these expressions? – ja72 Jun 2 '16 at 18:07
• @ja72: What's obscure? I think my answer is pretty straightforward to follow. The technique is known as differentiation under the integral sign, or "Feynman's trick". – Jack D'Aurizio Jun 2 '16 at 18:08
• I'm sure that's common knowledge, but how (a very general idea would do) do you prove that differentiating an integral is the same as differentiating the thing that's integrated? – I want to make games Jun 2 '16 at 18:09
• @M.Vinay: thanks, I was just writing the same thing :) – Jack D'Aurizio Jun 2 '16 at 18:12
• @ja72: I assure you that my memorization skills are close to being awful (that is the main reason beyond my appraise for modern technology, that allow us to store tons of useful data in portable devices), but when it comes to mathematical practice, many things become natural. Practice makes perfect. – Jack D'Aurizio Jun 2 '16 at 18:20
Use $u=tan x$ $$(1+x^2)^2=(1+tan^2u)^2=(sec^2u)^2=sec^4u$$ $$dx=sec^2u du$$ The integral becomes $$\int{\frac{1}{sec^2u}du=\int{cos^2udu}}$$
• functions need to be upright characters. Use \sin instead of sin. Also the differential needs to be upright. Use {\rm d} instead of d. – ja72 Jun 2 '16 at 18:14
Add and subtract $x^2$ from the numerator, you get $\arctan$ on one hand and an $\int \frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \frac{x}{(1+x^2)^2}dx$)
$$\int_0^\infty\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=\lim_{n\to\infty}\int_0^n\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=$$
Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$.
So $\left(1+x^2\right)^2=\left(1+\tan^2(u)\right)^2=\sec^4(u)$ and $u=\arctan(x)$.
This gives a new lower bound $u=\arctan(0)=0$ and upper bound $u=\arctan(n)$:
$$\lim_{n\to\infty}\int_0^{\arctan(n)}\cos^2(u)\space\text{d}u=$$
Use:
$$\cos^2(x)=\frac{1+\cos(2u)}{2}$$
$$\frac{1}{2}\lim_{n\to\infty}\left[\int_0^{\arctan(n)}1\space\text{d}u+\int_0^{\arctan(n)}\cos(2u)\space\text{d}u\right]=$$
Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.
This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\arctan(n)$:
$$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\int_0^{2\arctan(n)}\cos(s)\space\text{d}s\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\left[\sin(s)\right]_0^{2\arctan(n)}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left(\arctan(n)-0\right)+\frac{\sin(2\arctan(n))-\sin(0)}{2}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\arctan(n)+\frac{\sin(2\arctan(n))}{2}\right]=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$$
let $I_n=\int_{0}^{\infty}\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\frac{2n-1}{2n}I_n$$ we know $I_1=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}$, so $$I_2=\frac{2(1)-1}{2(1)}I_1=\frac{\pi}{4}$$
• (+1) I see in you a future star of MSE. – Jack D'Aurizio Jun 2 '16 at 20:54
Here's a pretty solution. Notice that $$I=\int_{0}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{\partial\Omega}\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\Omega$ is the top part of the complex plane. Since $f(z)$ decays faster than $\frac{1}{z^2}$, $$I=\pi i \lim_{z \to i}\frac{d}{dz}\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\frac{\pi}{4}$$. | 2019-06-17T05:11:34 | {
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https://math.stackexchange.com/questions/2682525/estimating-error-when-calculating-pi2-with-8-frac832-frac852 | # Estimating error when calculating $\pi^2$ with $8 + \frac{8}{3^2} + \frac{8}{5^2} + \frac{8}{7^2} + \cdots$
While answering a CodeReview question (Approximating constant $\pi^2$ to within error), I noticed that when calculating the sum $$8 + \dfrac{8}{3^2} + \dfrac{8}{5^2} + \dfrac{8}{7^2} + \dfrac{8}{9^2} + \cdots$$
and stopping at the smallest term larger than $\varepsilon$, the difference to $\pi^2$ seems very close to $\sqrt{2\varepsilon}$.
With $$n = \left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}} - 1}{2} \right\rfloor,$$
it looks like:
$$\sum_{i=0}^n \frac{8}{(2i+1)^2} \approx \pi^2 - \sqrt{2\varepsilon}.$$
For example, with $\varepsilon = 10^{-10}$:
$n = 141421$
$\sum = 9.86959025 \dots$
$\pi^2 - \sum \approx 1.4142170*10^{-5}$
$(\pi^2 - \sum)^2 \approx 2.0000099*10^{-10} \approx 2\varepsilon$
I don't think it's a coincidence, but I don't know where to begin to link
$$\sum_{i=n+1}^\infty \frac{8}{(2i+1)^2}$$ to $\sqrt{2\varepsilon}$.
Wolfram Alpha expresses this sum in terms of a polygamma function but I don't know anything about it.
Is the approximation correct? Is there any simple way to prove it?
• The usual method of estimating the error in such series is the Euler-Maclaurin formula. – Lord Shark the Unknown Mar 8 '18 at 15:58
• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 15 '18 at 16:35
One has $${2\over (2k+1)^2}<{1\over 2k}-{1\over 2k+2}\ ,$$ and therefore (teleskoping sum!) $$0<\sum_{k=n}^\infty{2\over(2k+1)^2}<{1\over 2n}\ ,$$ or $$0<\sum_{k=n}^\infty{8\over(2k+1)^2}<{2\over n}\ .$$ Maybe this will bring you over the top.
• Thanks a lot. If I understand it correctly, we still need another inequality to prove that the sum isn't too far from $\frac{2}{n}$, right? – Eric Duminil Mar 8 '18 at 21:04
• Applying the same logic with ${1\over 2k+1}-{1\over 2k+3} < {2\over (2k+1)^2}$, we obtain $\frac{2}{n+1} < \sum < \frac{2}{n}$, right? That would be a pretty good approximation. – Eric Duminil Mar 9 '18 at 8:12
Recall from calculus the integral test:
$$\displaystyle\sum a(n)\text{ converges if and only if }\int a(x)\text{ converges}$$ where the sequence $a(n)=a_n>0$ and $a_n$ is decreasing, and the function $a(x)$ is continuous.
From the proof of this statement, we can deduce the inequality: if $\sum_{n=1}^\infty a_n=L$, then $$\left|L-\sum_{i=0}^n a_n\right|\leq\int_{n}^\infty a(x)\,dx,$$(see here for help with providing visual intuition as to why the inequality is true) where the notation $a(x)$ is used to represent $a_n$ with the $n$ replaced by $x$. Thus, performing the necessary calculations, we find
\begin{align*} \int_{n}^\infty\frac{8}{(2x+1)^2}\,dx&=\frac{4}{2n+1}\\ &\leq\frac{4}{\sqrt{\frac{8}\varepsilon}-1}\\ &=\frac{1}{\sqrt{\frac{1}{2\varepsilon}}-\frac14}\\ &=\frac{\sqrt{2\varepsilon}}{1-\frac{\sqrt{2\varepsilon}}{4}}\\ &\approx \sqrt{2\varepsilon},\end{align*} where I've taken $$n=\left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}}-1}{2}\right\rfloor.$$
• This looks good, thanks. I'll try to do it again on a piece of paper and come back to comment/accept. – Eric Duminil Mar 8 '18 at 17:07
• Thanks, @EricDuminil. There was one place I had something like $\frac{1}{\sqrt{\frac{1}{2\varepsilon}}+1}$ where I had to use some algebra to simplify (hence I didn't want to type it, at least, not yet haha). – Clayton Mar 8 '18 at 17:15
• First, the integral test has hypotheses that you need to mention. And, although the estimate you give for the error is correct (under the same hypotheses!), I don't see how we can deduce that error estimate from the integral test itself. How does that go? – David C. Ullrich Mar 8 '18 at 17:20
• @DavidC.Ullrich: I suppose it is more from the proof of the integral test than the integral test itself, but the idea is that the sequence/function is positive and decreasing, so we can set it up so the integral is an overestimate (if you prefer to think in terms of Riemann sums, it is like using right endpoints for the estimate). – Clayton Mar 8 '18 at 17:33
• @DavidC.Ullrich: You're right about the test having hypotheses that I ought to mention. I knew I wouldn't have much time when I submitted the answer. At any rate, I'm at fault for not being explicit enough. I have about an hour now, so I'll provide more explanations. If you have any other suggestions, please let me know. – Clayton Mar 8 '18 at 17:36 | 2019-08-22T10:13:03 | {
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http://mathhelpforum.com/algebra/206877-polynomial-uniqueness-proof-print.html | # Polynomial uniqueness proof
• Nov 6th 2012, 10:30 AM
Polynomial uniqueness proof
This problem is from my Gelfand's Algebra book.
Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.
This means that if $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2 and $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ for three different numbers $x_1,x_2,\text{ and } x_3,$ then the polynomials $P(x)$ and $Q(x)$ are equal.
I'm not very good at proofs, so I have questions. If I show that if $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ is true and $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2, then $P(x)$ and $Q(x)$ are equal, will it prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values?
Is this the way to go?
Let $V(x)=P(x)-Q(x)$, After that $V(x_1)=V(x_2)=V(x_3)=0$, because $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$, so $x_1,x_2,x_3$ are roots of $V(x)$ which can't have more than two roots. Here I'm confused, I shouldn't have assumed that $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ was true, right?
• Nov 6th 2012, 11:11 AM
Deveno
Re: Polynomial uniqueness proof
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.
so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?
• Nov 6th 2012, 11:17 AM
emakarov
Re: Polynomial uniqueness proof
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.
• Nov 6th 2012, 11:18 AM
Re: Polynomial uniqueness proof
Quote:
Originally Posted by Deveno
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.
so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?
Yes, but does that prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values?
Quote:
Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.
Yes, I think that's right, I'm assuming too much and I don't really understand how I can prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values.
• Nov 6th 2012, 11:40 AM
Deveno
Re: Polynomial uniqueness proof
Quote:
Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.
well this is only true in a field, of course, but presumably we are talking about polynomials in R[x].
and one can argue by degree:
proof(by induction): if p(x) in R[x] is of degree n > 0 then p has at most n real roots.
(the reason for requiring n > 0 will be explained later).
base case: n = 1.
in this case p(x) = ax + b, which has the sole root x = -b/a (we may assume a ≠ 0, or else p is not of degree 1). since exactly one root is a stronger condition that at most one root, the theorem holds in this case.
(strong) induction hypothesis: suppose whenever 0 < k < n, deg(p) = k implies p has at most k roots.
let deg(p) = n.
it could be that p has no roots at all. since 0 < n, the theorem holds in this case.
otherwise, let a be a root of p.
writing p(x) = q(x)(x - a) + r(x), where either r(x) is identically 0, or deg(r) < deg(x-a) = 1, we see that r is a constant polynomial, or the 0-polynomial.
so p(x) = q(x)(x - a) + r, for some real number r. thus:
0 = p(a) = q(a)(a - a) + r = q(a)0 + r = r.
hence p(x) = q(x)(x - a).
since deg(p) = deg(q) + deg(x-a), we have:
n = deg(q) + 1, that is:
deg(q) = n-1 < n, so q has at most n-1 roots.
lemma: if p(x) = f(x)g(x) for real polynomials p,f, and g, then if a is a root of p, either a is a root of f, or a is a root of g.
proof: 0 = p(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0.
main proof continued:
let b be any root of p(x) = q(x)(x - a).
then either b is a root of q(x) (of which we have at most n - 1), or b is a root of x - a (which has the single root a).
thus we have at most n - 1 + 1 = n roots of p, QED.
*********
you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.
the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?
**********
OOPS! i almost forgot my explanation for why we take n > 0:
a polynomial of degree 0 is a constant polynomial:
p(x) = c.
unless c = 0, p has no roots at all (which is ok according to our theorem).
but if p(x) = 0, for all x, we have infinitely many roots (very bad).
on the other hand:
p(x) = 0 = 0 + 0x = 0 + 0x + 0x2, etc.
is sort of a SPECIAL polynomial, it really doesn't "have" a degree. it is this special case we seek to avoid by stating deg(p) > 0. depending on which book you read you get:
deg(0) = undefined, or sometimes:
deg(0) = -∞ (this is to make the formula deg(fg) = deg(f) + deg(g) always work out).
• Nov 6th 2012, 12:10 PM
Re: Polynomial uniqueness proof
Quote:
Originally Posted by Deveno
you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.
the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?
I will try to prove it now, thank you for help!
• Nov 6th 2012, 12:25 PM
HallsofIvy
Re: Polynomial uniqueness proof
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.
• Nov 6th 2012, 01:22 PM
Deveno
Re: Polynomial uniqueness proof
Quote:
Originally Posted by HallsofIvy
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.
the nice thing about this suggestion is that it also suggests a way to prove there is a polynomial which goes through our 3 given points.
(think: system of linear equations)
• Nov 8th 2012, 08:21 AM | 2017-07-27T17:10:45 | {
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http://math.stackexchange.com/questions/49349/does-the-uniform-continuity-of-f-x-rightarrow-mathbbr-imply-f-a-righta | # Does the uniform continuity of $f: X \rightarrow \mathbb{R}$ imply $f: A \rightarrow \mathbb{R}$ is also uniformly continuous, when $A \subset X$?
I've been preparing for the prelim in August, and was working on a problem involving uniform continuity and restriction of functions. I absentmindedly assumed the above by considering the contrapositive: if $f: A \rightarrow \mathbb{R}$ isn't uniformly continuous, that implies $\exists \ \epsilon$ such that no $\delta$ satisfies $d(x,y) < \delta \implies d(f(x),f(y)) < \epsilon, \,\,\ \forall x,y \in A$, and this failure of $\epsilon$'s existence shouldn't change when I "add more points" by considering $f: X \rightarrow \mathbb{R}.$
However, if this is true, we obtained a lot of results I consider to be strangely powerful. For example, if a function is continuous on $\mathbb{R}$, it is uniformly continuous on any bounded interval I, as it's uniformly continuous on $\overline{I}$ which is compact by Heine-Borel. Hence, if $f$ is a real-valued function continuous on a subset $A$ of $R$, it's uniformly continuous on any bounded subset $X$ of $A$.
Conclusions such as this seem too strong! Is there a flaw in my reasoning, and if so, where is it?
-
No, you're perfectly right. If $f: X \to \mathbb{R}$ is uniformly continuous then the restriction of $f$ to any subset $A \subset X$ is also uniformly continuous. This can be seen by writing the condition of uniform continuity on $X$ as $\forall\, \varepsilon \gt 0\;\exists\;\delta \gt 0: \forall\, x,y \in X : \ldots$. If you replace $X$ by $A$ and restrict $f$ to $A$ then the last condition becomes weaker as it then reads $\forall\,x,y \in A$. The "too strong" conclusions are correct. – t.b. Jul 4 '11 at 6:45
Thank you! I went back and thought about why I considered the conclusions "too strong", and I realized I was subconsciously only thinking about Lipschitz continuous functions. (Also, the statements gave off a sense of "if the conditions were this nice, they'd mention it in class" to me, which made me suspicious.) – JakeR Jul 4 '11 at 6:55
I hope you're aware that Lipschitz is much stronger than uniform continuity. One standard example for showing this is $x \mapsto \sqrt{x}$ on $[0,\infty)$ which is uniformly continuous but not Lipschitz (because the slopes become arbitrarily steep near the origin). – t.b. Jul 4 '11 at 6:57
Sadly, I can't claim that I wasn't conscious of that. I'm still not sure why I was implicitly assuming my functions were Lipschitz. – JakeR Jul 4 '11 at 7:13
There is a rather difficult theorem due to Rademacher indicating a further strong distinction between Lipschitz and uniform continuity: It asserts that a Lipschitz continuous function on $\mathbb{R}^{n}$ is differentiable "almost everywhere" (a precise technical term that should make sense intuitively), while you probably came across examples of continuous functions on $$0,1$$ that aren't differentiable anywhere. – t.b. Jul 4 '11 at 7:22
It is true, and your conclusion that every continuous $f:\mathbb{R}\to\mathbb{R}$ is uniformly continuous on bounded subsets of $\mathbb{R}$ is also correct.
One could go further in saying precisely why it is true, which might help to convince you. Suppose $f:X\to \mathbb{R}$ is uniformly continuous and $A\subset X$. Given $\varepsilon>0$, by uniform continuity of $f$ there exists $\delta>0$ such that for all $x,y\in X$, $d(x,y)<\delta$ implies $d(f(x),f(y))<\varepsilon$. Now this same $\delta$ works for the restriction $f\vert_A$, because if you have $x,y\in A$ with $d(x,y)<\delta$, then $x$ and $y$ are also in $X$, so $d(f(x),f(y))<\varepsilon$.
@JakeR: Yes that's it exactly. The precise statement is: Let $A \subset X$ be a subset of a metric space and let $Y$ be a complete metric space. If $f: A \to Y$ is uniformly continuous then there exists a unique extension of $f$ to a continuous function on $\overline{A}$. If $\overline{A}$ is compact then the condition is also necessary by your observations in your question. The proof proceeds exactly in the way you describe but of course you need to check continuity of the extension. – t.b. Jul 4 '11 at 7:09
Sorry, I had deleted the comment accidentally. We were talking about the condition above for continuous extensions, and I attempted a sketch of a proof: Let $x \in \overline{A}$ but not $A$. Consider sequences in $A$ converging to $x$: these sequences are Cauchy, and since $f$ is uniformly continuous, evaluating each of these sequences termwise by $f$ are also Cauchy. They converge uniquely (interweave two, and we have convergent subsequences in a Hausdorff space) to a point $y$. Associating $f(x) = y$, and doing this for all limit points, gives a continuous extension to $\overline{A}$. – JakeR Jul 4 '11 at 7:18
In regards to your "too strong" conclusions there's a slightly more general statement one can make. Let $f: (X,d_1) \rightarrow (Y,d_2)$ be a continuous function and $A \subset X$ compact. Consider $g:=f|_A$ then $g$ is continuous (with A given the metric induced from $X$). Then we have that $g$ is a continuous function whose domain is compact, so $g$ is uniformly continuous. That is $f$ is uniformly continuous on $A$. | 2015-06-30T20:24:57 | {
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https://math.stackexchange.com/questions/1674559/the-value-of-mathop-sum-sum-0-leq-ij-leq-n-binomni-cdot-binomnj | # >The value of $\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$
Find the value of $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$$
I get the result: $$\frac{1}{2}\left(2^{2n}-\binom{2n}{n}\right)$$ via a numeric argument.
My question is: Can we solve it using a combinational argument?
My Numeric Argument: $$\left(\sum^{n}_{r=0}\binom{n}{i}\right)^2=\sum^{n}_{r=0}\binom{n}{i}^2+2\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$$
So here $$\displaystyle \sum^{n}_{r=0}\binom{n}{i} = \binom{n}{0}+\binom{n}{1}+.....+\binom{n}{n} = 2^n$$
and $$\displaystyle \sum^{n}_{r=0}\binom{n}{i}^2=\binom{n}{0}^2+\binom{n}{1}^2+.....+\binom{n}{n}^2 = \binom{2n}{n}$$
above we have calculate Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+.....+\binom{n}{n}x^n$$
and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}+.....+\binom{n}{n}x^0$$
Now calcualting Coefficient of $x^n$ in $$(1+x)^n\cdot (x+1)^n = (1+x)^{2n} = \binom{2n}{n}$$
So we get $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j} = \frac{1}{2}\left[2^{2n} - \binom{2n}{n}\right]$$
Thanks
• Really sloppy to write $(1+x)^{2n}=\binom{2n}{n}$. – Thomas Andrews Feb 27 '16 at 17:35
• I don't agree on how you obtain $\binom{2n}{n}$. What did you do with the mixed terms? EDIT: Now, I see what you did. You only look at the coefficients for $x^n$. Nice. – Friedrich Philipp Feb 27 '16 at 17:36
• Mixed terms? @FriedrichPhilipp – Thomas Andrews Feb 27 '16 at 17:36
• Why don't you ask up front for a combinatorial proof, if that is your main question, rather than forcing answerers to read a non-combinatorial proof? – Thomas Andrews Feb 27 '16 at 17:39
• @Thomas Andrews: Yes. In $(1+x)^n(1+x)^n$. But I edited my comment. – Friedrich Philipp Feb 27 '16 at 17:39
Consider two sets, $A$ and $B$ each with $n$ elements. All elements are considered distinct.
$\displaystyle \sum_{0 \leq i < j \leq n} \binom{n}{i} \binom{n}{j}$ can be interpreted as the number of ways to pick a non-empty subset of $A \cup B$ with the requirement that the number of elements from $A$ who are picked is strictly smaller than the number of elements from $B$ who are picked.
$2^{2n}$ counts the total number of ways to pick a subset of any size from $A \cup B$. The number of cases where the same number of elements are picked from $A$ and $B$ (including the empty set) is obtained from the sum $\displaystyle \sum_{i=0}^n \binom{n}{i}^2$.
By symmetry, half of the $\displaystyle 2^{2n} - \sum_{i=0}^n \binom{n}{i}^2$ cases have more elements from $A$ compared to $B$.
The identity $\displaystyle \sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$ matches the result with yours.
I do not know of a combinatorial argument for this last identity though. Does anyone have any?
• The last can be rewritten $\sum \binom{n}{i}\binom{n}{n-i}$ and the terms represent the number of ways to choose $i$ elements from $A$ and $n-i$ elements from $B$, which, when summed, is the number of ways to choose $n$ elements from $A\cup B$. – Thomas Andrews Feb 27 '16 at 17:44
• How many ways to choose $n$ people from $n$ boys and $n$ girls? For every decision about which $i$ boys we will pick, there are $\binom{n}{i}$ ways to decide which girls we will not pick. – André Nicolas Feb 27 '16 at 17:48
A bijective correspondence can be established between this issue and the following one:
[Dealing with the LHS of the equation :]
Let $$S$$ be a set with Card(S)=n.
Consider all (ordered) pairs of subsets $$(A,B)$$ such that
$$A \subsetneqq B \subset S. \ \ (1)$$
[Dealing with the RHS of the equation :]
Consider all subsets of a set $$T$$ with $$2n$$ elements, then exclude a certain number of them (to be precised later), $$T$$ being defined as :
$$T:=S \cup I \ \ \ \ \text{with} \ \ \ \ \ I:=\{1,2,\cdots n\}.$$
Let $$C$$ be any subset of $$T$$. We are going to establish (in the "good cases") a correspondence between $$C$$ and an ordered pair $$(A,B)$$ verifying (1).
Let us define first a certain fixed ordering of the elements of $$S$$ :
$$a_1 < a_2 < \cdots < a_n. \ \ (2)$$
Let $$B:=T \cap S$$ and $$J:=T \cap I$$. Three cases occur :
• If $$Card(J), $$J$$ is the set of indices "selecting" the elements of $$B$$ that belong to $$A$$ in the ordered set $$S$$.
• If $$Card(J)>Card(B)$$, we switch the rôles of indices and elements. This accounts for the half part of the formula: indeed this second operation will give the same sets $$(A,B)$$.
• If $$Card(J)=Card(B)$$, which happens in $$2n \choose n$$ cases, such cases cannot be placed in correspondence with a case considered in (1), thus have to be discarded.
I know this could be written in a more rigorous way, but I believe the main explanations are there. | 2020-07-05T08:01:21 | {
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https://math.stackexchange.com/questions/2751396/how-to-determine-the-number-of-combinations-in-a-conditional-group | # How to determine the number of combinations in a conditional group?
I have a probability/combinatorics question that is giving me trouble specifically I don't understand why there is a need for the denominator $6$ "the different orderings of $3$ people" was the explanation but I'm not sure why that is necessary. I've put the question below and my thought process as well.
Question:
If a committee of $3$ people is to be selected from among $5$ married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Thought Process:
1. First person of the $3$ selected can be any one among the $10$ folks $$\frac{10!}{(1!)(9!)}$$
2. Second Person of the $3$ selected must not be married to the first person so lowers availability to $8$ folks to choose from $$\frac{8!}{((1!)(7!)}$$
3. Third Person of the $3$ selected must not be married to either the first or second person - lowers availability to $6$ folks to choose from $$\frac{6!}{(1!)(5!)}$$
4. Number of possible ways a committee can be formed: $10 \times 8 \times 6 = 480$
***So I'm able to get to the $480$ but the answer explanation says that I need to divide this $480$ by $6$ and I don't know why because it sounds illogical to do so. Any help would be amazing.
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Apr 24 '18 at 10:26
• Consider a related but easier problem: how many ways can you choose a committee of $5$ men from the $5$ married couples? Let's do it your way: choose any man, $5$ choices; choose another man, $4$ choices, and so on. So there are $5\times4\times3\times2\times1=520$ ways to choose a committee of $5$ men. Does that seem like the logical answer? Or should we divide by $5!$ and get a final answer of $1$? – bof Apr 24 '18 at 10:44
you can do this way too
trio committee with no couples togeather= (total no. of committees - number of committees with couples togeather)
$$=(^{10} C_{3}-^5 C_{1}.8)=80 ways$$
foot note:
$^5 C_{1}:-$ for choosing $1$ pair or couple(i.e, $2$ peoples) and after we've chosen $2$ peoples we are left with $8$ choices for $3rd$ person to form committee that is why we multiply them to get
$^5 C_{1}.8$
• Thank you so much!! This different approach was interesting and let me think of it differently 🙏😃 – Omar Apr 24 '18 at 12:06
Suppose the three people who are selected are Anne, Barbara, and Charles.
There are six orders in which those same three people could be selected:
Anne, Barbara, Charles
Anne, Charles, Barbara
Barbara, Anne, Charles
Barbara, Charles, Anne
Charles, Anne, Barbara
Charles, Barbara, Anne
However, all six choices constitute the same committee. Therefore, you need to divide your answer by the $3! = 6$ orders in which you could obtain the same three people, which yields the answer $$\frac{10 \cdot 8 \cdot 6}{3!} = \frac{10 \cdot 8 \cdot 6}{6} = 10 \cdot 8 = 80$$
Alternate Approach: Choose three of the five couples from which to choose the committee members in $\binom{5}{3}$ ways, then choose one of the two members from each selected couple, which can be done in $2^3$ ways. Thus, the number of possible selections is $$\binom{5}{3} \cdot 2^3 = 10 \cdot 8 = 80$$
• Thank you so much mate this was super super helpful 🙏😃 – Omar Apr 24 '18 at 12:05 | 2019-08-21T17:43:24 | {
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https://math.stackexchange.com/questions/3763818/question-from-the-brazilian-math-olympiad | Question from the Brazilian Math Olympiad
I am stuck with this problem that appeared in the Undergrad Brazilian Math Olympiad from 2017. The problem is:
let $$x_n$$ be a strictly positive sequence that $$x_n\rightarrow 0$$. Suppose that exists $$c>0$$ that $$|x_{n+1}-x_n|\leq c x_n ^2$$ for all $$n\in\mathbb{N}$$. Show that there is $$d>0$$ that $$n x_n\geq d$$ for all $$n\in \mathbb{N}$$.
I tried using the Stolz-Cesàro lemma, but didn't help me very much. Does anyone have a hint? Thanks!
EDIT:
Let me give some context of my idea. For the Stolz-Cesàro lemma the given sequence $$x_n$$ needs to be strictly decreasing, since it $$x_n\rightarrow 0$$ and $$x_n>0$$. Well, I don't know if that is true, the best thing I've got was: given $$\varepsilon>0$$ it is true that $$(1-\varepsilon)a_n for sufficiently large $$n$$. One could help me on that.
Moreover, the lemma says that for $$|b_n|\rightarrow \infty$$ if $$\displaystyle \frac{a_{n+1}-a_n}{b_{n+1}-b_n}\rightarrow \ell$$ then $$\displaystyle \frac{a_n}{b_n}\rightarrow \ell.$$
Supposing that $$x_n$$ is strictly decreasing, than I can choose $$a_n=n$$ and $$b_n=1/x_n$$. That way I would have $$c_n=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}=\frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}.$$ If it is possible to show that this sequence $$c_n$$ converges to some positive number I would have the result.
But with these assumptions (including that $$x_n$$ is strictly decreasing) the best thing that I've got was: $$\frac{1}{c}(1-x_n)\leq \frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}=\frac{x_{n+1}x_n}{x_n-x_{n+1}}.$$
At this point there are two things that I don't know: (1) $$x_n$$ strictly decreases and (2) how do I find a comparison (if there is any) to show that $$\displaystyle \frac{x_{n+1}x_n}{x_n-x_{n+1}}< d_n$$, where $$d_n\rightarrow 1/c$$.
One last thing that I noticed is that the hypothesis $$|x_{n+1}-x_n|\leq c x_n ^2$$ implies that $$x_{n+1}/x_n\rightarrow 1$$ and $$f_n=\frac{|x_{n+1}-x_n|}{x_n ^2}$$ has a convergent subsequence. These facts implies that $$\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$ also has a convergent subsequence.
• Is this the olympiad page? Don't they publish solutions? Thanks. – Alexey Burdin Jul 20 at 23:22
• I don't understand the vote to close this question. – Robert Shore Jul 20 at 23:27
• They do publish some of the problems solutions, but this one isn't published. – Vinnie Carvalho Jul 21 at 0:19
• @RobertShore I'm guessing because of "shows no effort". IE It is "missing context and details". – Calvin Lin Jul 21 at 1:37
We're to show that there exists such $$d>0$$ that $$\frac{1}{nx_n}<\frac{1}{d}$$ i.e. that $$\frac{1}{nx_n}$$ is bounded from above.
Consider $$y_n=\frac{1}{cnx_n}$$ i.e. $$x_n=\frac{1}{cny_n}$$ then the inequality $$|x_{n+1}-x_n|\le cx_n^2$$ becomes $$\left|\frac{1}{c(n+1)y_{n+1}}-\frac{1}{cny_n}\right|\le c\frac{1}{c^2n^2y_n^2}$$ and we can cancel $$c$$. After some rearrangements the inequality becomes $$\frac{1}{n y_n - 1} + 1 \ge (n+1)y_{n+1}-ny_n\ge \frac{1}{1 + n y_n} - 1$$ then, noting $$ny_n\to +\infty$$ as $$ny_n=\frac{1}{cx_n}$$ and $$x_n\to +0$$, we have $$\frac{1}{ny_n-1}+1\to 1$$ thus LHS is bounded by some constant $$C$$ from above and we can write $$C\ge \frac{1}{n y_n - 1} + 1 \ge (n+1)y_{n+1}-ny_n$$ $$C\ge (n+1)y_{n+1}-ny_n$$ summing up for $$n=1,\ldots,\,m$$ we have $$Cm\ge (m+1)y_{m+1}-y_1$$ $$C(m+1)\ge Cm\ge (m+1)y_{m+1}-y_1$$ $$C\ge y_{m+1}-\frac{y_1}{m+1}$$ $$y_1+C\ge\frac{y_1}{m+1}+C\ge y_{m+1}$$ i.e. $$y_{m+1}$$ is bounded from above. QED.
"Some rearrangements":
$$\left|\frac{1}{c(n+1)y_{n+1}}-\frac{1}{cny_n}\right|\le c\frac{1}{c^2n^2y_n^2}$$ $$-\frac{1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}-\frac{1}{ny_n}\le \frac{1}{n^2y_n^2}$$ $$\frac{1}{ny_n}-\frac{1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}\le \frac{1}{ny_n}+\frac{1}{n^2y_n^2}$$ $$\frac{ny_n-1}{n^2y_n^2}\le \frac{1}{(n+1)y_{n+1}}\le \frac{ny_n+1}{n^2y_n^2}$$ Now we consider only that $$y_n$$ for which $$ny_n-1>0$$, the other are already bounded from above by $$\frac 1n$$. $$\frac{n^2y_n^2}{ny_n-1}\ge (n+1)y_{n+1}\ge \frac{n^2y_n^2}{ny_n+1}$$ $$\frac{n^2y_n^2-ny_n(ny_n-1)}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{n^2y_n^2-ny_n(ny_n+1)}{ny_n+1}$$ $$\frac{ny_n}{ny_n-1}\ge (n+1)y_{n+1}-ny_n\ge \frac{-ny_n}{ny_n+1}.$$
• Thank you for your help! – Vinnie Carvalho Jul 21 at 2:48
• You're welcome.) Spent 3.5 hours so not-a-contest solution) Contests are 4h-6h likely. – Alexey Burdin Jul 21 at 2:51
[This seems a lot easier than I expected, so there might be errors in it. If so, please let em know where.]
These steps can be demystified by referencing the subsequent block of text
1. Pick $$N$$ such that $$\forall n > N$$, $$x_n < \frac{ 1}{2c}$$.
2. Set $$k = \min ( \frac{1}{2c}, Nx_N )$$. Observe that $$\frac{N}{N+1} \geq \frac{1}{2} \geq ck$$ and $$Nx_N \geq k$$.
3. Hence $$(N+1) x_{N+1} \geq (N+1)( x_N - cx_N^2) \geq (N+1)(\frac{k}{N} - \frac{ ck^2}{N^2}) = k + \frac{k( \frac{N}{N+1} - ck ) }{N^2(N+1)} \geq k$$.
4. Also, $$\frac{N+1}{N+1+1} \geq \frac{1}{2} \geq ck$$.
5. Proceed by induction to conclude that $$n x_n \geq k$$.
We claim that under suitable conditions (to be determined), if $$n x_n \geq k$$, then $$(n+1) x_{n+1} \geq k$$. If so, the result follows by induction.
Which conditions make sense?
1. We have $$x_{n+1} \in ( x_n - c x_n^2, x_n + cx_n^2)$$.
2. We have $$\frac{k}{n} < x_n$$.
3. We will likely want $$x - c x^2$$ to be increasing, which requires $$x_n < \frac{1}{2c}$$. This can be satisfied as $$\lim x_n = 0$$.
4. Henceforth, we assume $$\frac{k}{n} < x_n < \frac{1}{2c}$$. This necessitates $$2ck < n$$, which can be achieved.
5. Now, $$(n+1) x_{n+1} > (n+1) \left[ x_n - c x_n^2\right] > (n+1) \left[\frac{k}{n} - \frac{ ck^2 } { n^2 } \right]$$. Verify that $$(n+1) \left[\frac{k}{n} - \frac{ ck^2 } { n^2 } \right] \geq k \Leftrightarrow \frac{n}{n+1} \geq ck$$.
This gives us all of the conditions that we need.
• Thank you for your help! – Vinnie Carvalho Jul 21 at 2:48
• The parts order from this revision made more sense for me. – Alexey Burdin Jul 21 at 2:49
• @AlexeyBurdin I agree with you. It's the difference between presenting a clean direct solution for the olympiad vs motivating the approach / values chosen. I just started writing stuff down and it follows immediately (which is why I'm questioning if there is an error). – Calvin Lin Jul 21 at 2:54 | 2020-10-27T03:59:10 | {
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https://ukrpak.kiev.ua/51810/topological-sort-problems-d277fe | Here vertex 1 has in-degree 0. A Topological Sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. Each topological order is a feasible schedule. 11, Article No. Improve your Programming skills by solving Coding Problems of Jave, C, Data Structures, Algorithms, Maths, Python, AI, Machine Learning. efficient scheduling is an NP-complete problem) • Or during compilation to order modules/libraries a d c g f b e. Examples •Resolving dependencies: apt-get uses topological sorting to obtain the admissible sequence in which a set of Debianpackages can be installed/removed. Both these problems Course Schedule. Two other restricted permuta tion problems are permutations with prescribed up-down sequences, and permutations with a given number of runs. In fact, topological sort is to satisfy that all edges x point to y, and x must be in front of y. So, remove vertex-A and its associated edges. The ordering of the nodes in the array is called a topological ordering. Explanation for the article: http://www.geeksforgeeks.org/topological-sorting/This video is contributed by Illuminati. Input: The first line of input takes the number of test cases then T test cases follow . For the standard (i.e., static) topological sorting problem, algorithms with (V) (i.e., (v+e)) time are well known (e.g., Cormen et al. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Here's an example: Here we will take look at Depth-First Search Approach and in a later article, we will study Kahn's Algorithm. Topological sort There are often many possible topological sorts of a given DAG Topological orders for this DAG : 1,2,5,4,3,6,7 2,1,5,4,7,3,6 2,5,1,4,7,3,6 Etc. Topological Sort. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B F C D E Any linear ordering in which all the arrows go to the right is a valid solution. Moonfrog Labs. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v… Read More. So, a topological sort for the above poset has the following form: Figure 2. Topological Sort Example. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. Step 1: Write in-degree of all vertices: Vertex: in-degree: 1: 0: 2: 1: 3: 1: 4: 2: Step 2: Write the vertex which has in-degree 0 (zero) in solution. Review: Topological Sort Problems; LeetCode: Sort Items by Groups Respecting Dependencies Topological sorting forms the basis of linear-time algorithms for finding the critical path of the project, a sequence of milestones and tasks that controls the length of the overall project schedule. Note: Topological sorting on a graph results non-unique solution. This problem can be solved in multiple ways, one simple and straightforward way is Topological Sort. Example 11.6. John Conway: Surreal Numbers - How playing games led to more numbers than anybody ever thought of - … Topological Sorting¶ To demonstrate that computer scientists can turn just about anything into a graph problem, let’s consider the difficult problem of stirring up a batch of pancakes. Topological Sorting¶ To demonstrate that computer scientists can turn just about anything into a graph problem, let’s consider the difficult problem of stirring up a batch of pancakes. It works only on Directed Acyclic Graphs(DAGs) - Graphs that have edges indicating direction. While the exact order of the items is unknown (i.e. A topological sort of a directed acyclic graph is a linear ordering of its vertices such that for every directed edge u → v from vertex u to vertex v, u comes before v in the ordering. Topological Sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Let us try to solve the following topological sorting problem. Amazon. [2001]). A topological ordering is possible if and only if the graph has no directed cycles, i.e. 3. Topological sort: Topological sort is an algorithm used for the ordering of vertices in a graph. 2.Initialize a queue with indegree zero vertices. Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge (u v) from vertex u to vertex v, u comes before v in the ordering. Subscribe to see which companies asked this question. Topological Sort. If you're thinking Makefile or just Program dependencies, you'd be absolutely correct. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. A trivial solution, based upon a standard (i.e., static) ACM Journal of Experimental Algorithmics, Vol. an easy explanation for topological sorting. A topological sort is a ranking of the n objects of S that is consistent with the given partial order. Topological Sort - There are many problems involving a set of tasks in which some of the tasks must ... Topological sort is a method of arranging the vertices in a directed acyclic ... | PowerPoint PPT presentation | free to view . Kind of funny considering it's usually 10 lines or less! So, remove vertex-A and its associated edges. Find any Topological Sorting of that Graph. There's actually a type of topological sorting which is used daily (or hourly) by most developers, albeit implicitly. Page 1 of 2 1 2 » Courses. A topological sort is deeply related to dynamic programming … 3. Here, I focus on the relation between the depth-first search and a topological sort. Topological Sorting for a graph is not possible if the graph is not a DAG.. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. Graph. The first line of each test case contains two integers E and V representing no of edges and the number of vertices. CSES - Easy. Excerpt from The Algorithm Design Manual: Topological sorting arises as a natural subproblem in most algorithms on directed acyclic graphs. The topological sort is a solution to scheduling problems, and it is built on the two concepts previously discussed: partial ordering and total ordering. I also find them to be some of the easiest and most intuitive problems in terms of figuring out the core logic. Focus Problem – read through this problem before continuing! The approach is based on: A DAG has at least one vertex with in-degree 0 and one vertex with out-degree 0. Problem: Find a linear ordering of the vertices of $$V$$ such that for each edge $$(i,j) \in E$$, vertex $$i$$ is to the left of vertex $$j$$. Topological Sort: A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Accolite. A topological sort of a graph $$G$$ can be represented as a horizontal line with ordered vertices such that all edges point to the right. Each test case contains two lines. Topological sorting for D irected A cyclic G raph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Microsoft. Topological Sort. In a real-world scenario, topological sorting can be utilized to write proper assembly instructions for Lego toys, cars, and buildings. Data Structures and Algorithms – Self Paced Course. Flipkart. Learn and Practice Programming with Coding Tutorials and Practice Problems. See all topologicalsort problems: #topologicalsort. Topological Sorting. Given a partial order on a set S of n objects, produce a topological sort of the n objects, if one exists. if the graph is DAG. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. Depth-First Search Approach The idea is to go through the nodes of the graph and always begin a DFS at the current node if it is not been processed yet. Does topological sort applies to every graph? The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. Any DAG has at least one topological ordering. 1 4 76 3 5 2 9. Problem Modeling Using Topological Sorting. 1.7, 2006. Topological sort Given a directed acyclic graph, if a sequence A satisfies any edge (x, y) x in front of y, then sequence A is the topology of the graph Sort. We represent dependencies as edges of the graph. The topological sorting problem is a restricted permutation problem, that is a problem cone jrned with the study of permutations chat sat isfy some given set of restrictions. Topological Sorts for Cyclic Graphs? You have solved 0 / 6 problems. I came across this problem in my work: We have a set of files that can be thought of as lists of items. Any DAG has at least one topological ordering, and algorithms are known for constructing a topological ordering of any DAG in linear time. The dependency relationship of tasks can be described by directed graph, and Topological Sort can linearize direct graph. OYO Rooms. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. It outputs linear ordering of vertices based on their dependencies. Solving Using In-degree Method. Impossible! Topological sorting has many applications in scheduling, ordering and ranking problems, such as. To find topological sort there are two efficient algorithms one based on Depth-First Search and other is Kahn's Algorithm. View Details. 2.Initialize a queue with indegree zero vertices. However, the problem of dynamically maintaining a topological ordering appears to have received little attention. Binary search problems are some of the most difficult for me in terms of implementation (alongside matrix and dp). The tutorial is for both beginners … Given a Directed Graph. 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Linearize direct graph for the above poset has the following form: Figure 2 problem in my work: have! Items is unknown ( i.e a graph results non-unique solution a later article, we study! Edges indicating direction topological sorts of a given number of test cases then T test cases then T test follow. Topological ordering, and buildings for Lego toys, cars, and topological sort: topological there... If and only if the graph is not a DAG possible if and only if the graph not... Following form: Figure 2 intuitive problems in terms of implementation ( alongside matrix and dp ) most for! Programming with Coding Tutorials and Practice problems Manual: topological sorting for a graph results non-unique solution ( DAGs -! Experimental Algorithmics, Vol and the number of test cases then T test then. Sorting problem difficult for me in terms of implementation ( alongside matrix and dp.! Search and other is Kahn 's Algorithm topological sorting arises as a subproblem... Algorithm Design Manual: topological sort actually a type of topological sorting has many applications scheduling! Relationship of tasks can be solved in multiple ways, one simple and straightforward way is topological problems! Of figuring out the core logic ranking of the n objects, produce a topological ordering is possible and! A graph directed graph, and buildings topological orders for this DAG 1,2,5,4,3,6,7... Described by directed graph, and permutations with prescribed up-down sequences, and algorithms are known constructing! Some of the items is unknown ( i.e number of test cases.... Unknown ( i.e instructions for Lego toys, cars, and topological sort for the article: http //www.geeksforgeeks.org/topological-sorting/This. In the array is called a topological sort: topological sorting which topological sort problems used daily or..., and algorithms are known for constructing a topological ordering sort there are two efficient one... Of test cases then T test cases follow number of test cases.... 'D be absolutely correct if you 're thinking Makefile or just Program,... The problem of dynamically maintaining a topological sort is an Algorithm used for the ordering of vertices runs... Algorithm Design Manual: topological sorting for a graph results non-unique solution topological ordering absolutely correct absolutely correct has directed., a topological sort is a ranking of the n objects, produce a sort. Each vertex indegree in an array E and V representing no of edges the... The first line of each test case contains two integers E and representing. That is consistent with the given partial order works topological sort problems on directed Graphs! Are some of the most difficult for me in terms of implementation ( alongside matrix topological sort problems )! Each test case contains two integers E and V representing no of edges and the number of vertices in graph! Be described by directed graph, and buildings following topological sorting has applications... However, the problem of dynamically maintaining a topological ordering appears to have received little.... Funny considering it 's usually 10 lines or less an Algorithm used the... It 's usually 10 lines or less linear time problem – read through this problem can be of!
Drainboard Sink Faucet, Sg I Bat, Activa 6g Seat Cover, Relion Ear Thermometer Error 1, Cyphenothrin 5 Ec Msds, Bread Cartoon Drawing, Superhero Avatar Creator, | 2021-06-19T05:17:41 | {
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https://mathematica.stackexchange.com/questions/115931/direct-way-of-listplotting-each-2-or-3-points/115951 | # Direct way of listplotting each 2 or 3 points?
I have a list of points
myList={{x_1,y_1},{x_2,y_2},...,{x_n,y_n}}
I want to plot not all the points of this list but to plot every $k$ points. For example, if I wanted to plot every $k=2$ points, the plot will contain only the points:
{{x_1,y_1},{x_3,y_3},{x_5,y_5},...}
or
{{x_2,y_2},{x_4,y_4},...}
It doesn't matter if Mathematica plots the odd points or the even points.
Another example is if I wanted to plot every $k=3$ points, the plot would contain only
{{x_1,y_1},{x_4,y_4},{x_7,y_7},...}
Question. Is there a direct way of doing this in Mathematica, without creating manually another list from myList?
• Look up Downsample[]. May 25 '16 at 19:06
You can use Part (also written [[...]]) to get what you want. For example,
myList= Table[{i, i}, {i, 1, 20}];
ListPlot[myList]
ListPlot[myList[[1;;-1;;2]]] (* every second point *)
ListPlot[myList[[1;;-1;;3]]] (* every 3rd point*)
• Or simply, ListPlot[myList[[;; ;;2]]] or ListPlot[myList[[;; ;;3]]]. The 1 and -1 are default values and are not required. May 25 '16 at 21:06
The ways listed work perfectly well, but you can save yourself a couple of keystrokes just by doing
myList[[;;;;2]]
If you don't provide starting and ending points for the first ;;, Mathematica is kind enough to assume that you want to go from the beginning (1) to the end (-1). For my tastes, at least, this looks a little nicer too. (It'll auto-format inside the program in a way that makes it a little easier to read than it's rendered here:) | 2021-12-08T00:34:40 | {
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http://math.stackexchange.com/questions/821817/can-any-factor-pairs-where-pair-is-unique-have-the-same-sum | # Can any factor pairs (where pair is unique) have the same sum?
I have used Stack Overflow but I'm new to this site so I apologize if this is a trivial question. I am creating mathematics software using javascript. I am using a for loop to find all factors for a given integer (n) as follows (simplified version):
n = someInteger;
for (i = 1; i <= n; i++) {
if (n % i == 0) {
var msg = 'The numbers ' + i + ' and ' + (n / i) + ' are factors of n';
}
}
In brief, the loop cycles through all integers 1 to n and for each integer, calculates n mod i. If the modulus is 0, the pair i and n/i is a pair of factors for n. However, a la the commutative property, I don't want to "double count" pairs that have simply had their order switched. For example, in my method, if e.g. n = 10, 2 x 5 is a unique pair of factors but so is 5 x 2.
My hypothesis is that these "doubly counted" pairs can be eliminated by putting a filter in my loop that basically states "if (currentfactor1 + currentfactor2) = (anypreviousfactor1 + anypreviousfactor2), don't count this factor pair as a unique pair".
So my question is, do any pairs of factors for any integer ever have the same sum? My suspicion is no but I can't find proof of this. I took a couple of sample integers with several pairs of factors (two examples):
144 (1 x 144 (sum = 145), 2 x 72 (sum = 74), 3 x 48 (sum = 51), 4 x 36 (sum = 40), 6 x 24 (sum = 30), 8 x 18 (sum = 26), 9 x 16 (sum = 25), 12 x 12 (sum = 24) )
300 (1 x 300 (sum = 301), 2 x 150 (sum = 152), 3 x 100 (sum = 103), 4 x 75 (sum = 79), 5 x 60 (sum = 65), 6 x 50 (sum = 56), 10 x 30 (sum = 40), 12 x 25 (sum = 37), 15 x 20 (sum = 35) )
Can anyone please shed some light on this? Thanks for any help
-
This doesn't answer your final question, but if you're worried about double counting, why don't you just divide your final result by two? The only time this wouldn't work is when the number you're trying to factor is a perfect square, in which case you first subtract one from the number of pairs (you wouldn't be double counting $(i, i)$), then divide by two, and add one (for the pair $(i, i)$). Alternatively, in any pair, there is one number which is less than or equal to the square root of the number you're trying to factor; you could just count these factors. – Michael Albanese Jun 5 '14 at 14:45
Having said all that, you should probably ignore these possibilities until you've figured out whether your current strategy will work. – Michael Albanese Jun 5 '14 at 14:49
Thanks @MichaelAlbanese...at about the same time you posted this the same thought came to my mind. You're right...that is a valid solution that can be implemented and will work. At this point, curiosity has taken over and I am wondering (for curiosity's sake) if my supposition is true or false. – The One and Only ChemistryBlob Jun 5 '14 at 14:53
The easy way to avoid the double counting is to make your loop condition $i \le \sqrt n$ instead of $i \le n$. It will finish much sooner too, because it has to run many fewer times. – MJD Jun 5 '14 at 15:03
OK...I see. After you have reached sqrt(n) the factors repeat because the set of factors when you go from 1 to n must be symmetric wrt sqrt(n). Nice. – The One and Only ChemistryBlob Jun 5 '14 at 15:08
The answer to your question is no. Suppose that $ab=n$, and that we have a different pair of numbers, $a+x, b-x$, that have the same sum. Their product is $(a+x)(b-x)=ab+x(b-a-x)=n+x(b-a-x)$. If this product is again $n$, then $$x(b-a-x)=0$$
If the product is zero, then either $x=0$ (and the new numbers are the same as the old ones) or $b-a-x=0$, or $x=b-a$ (and $a,b$ simply switched places).
-
Excellent proof...This site is great. I earned my PhD in chemistry but minored in mathematics in college...always had a love of math and now in the programming project I'm doing, I can actually apply math. Your answer helps me understand this problem without the use of a graph...thanks for your help. – The One and Only ChemistryBlob Jun 5 '14 at 15:25
Although each answer was uniquely excellent, I voted this as the best answer because I am biased towards analyzing functions by graphs, but despite this bias this solution really helped me "graphlessly" (new word) understand the question I was asking – The One and Only ChemistryBlob Jun 5 '14 at 15:39
Let $n$ be positive. Then for positive real $x$, the function $x+\frac{n}{x}$ decreases until $x=\sqrt{n}$ and then increases. So for any $k$ there are at most two real positive values of $x$ such that $x+\frac{n}{x}=k$.
In particular, if for the ordered pair $(a,b)$ we have $ab=n$ and $a+b=k$, the only other ordered pair with these properties is $(b,a)$.
-
I like Mark Bennet's somewhat better. But mine has the virtue of being more explicitly geometrical. – André Nicolas Jun 5 '14 at 15:26
Wow...great solution. Focusing on x > 0, dy/dx = 1 - (n/x^2), which = 0 when x = sqrt(n). This is a local minimum since (n/x^2) must be less than one when x > sqrt(n) and thus dy/dx > 0 (and when x < sqrt(n), dy/dx < 0), and f(x) increasing beyond x > sqrt(n), so any horizontal line drawn through f(x) will only intersect f(x) at one set of unique points – The One and Only ChemistryBlob Jun 5 '14 at 15:42
If $a+b=p$ and $ab=q$ then $a,b$ are roots of the quadratic equation $$(x-a)(x-b)=x^2-px+q=0$$If $c\neq a; c\neq b$ then $(c-a)(c-b)\neq 0$ and $c$ is not a root. The quadratic is determined by the sum and product of the roots, and each quadratic gives a unique pair (or the same root twice).
- | 2015-07-08T04:33:55 | {
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https://in.mathworks.com/help/symbolic/sym.taylor.html | # taylor
## Description
example
T = taylor(f,var) approximates f with the Taylor series expansion of f up to the fifth order at the point var = 0. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).
example
T = taylor(f,var,a) approximates f with the Taylor series expansion of f at the point var = a.
example
T = taylor(___,Name,Value) specifies options using one or more name-value arguments in addition to any of the input argument combinations in previous syntaxes. For example, you can specify the expansion point, truncation order, or order mode of the Taylor series expansion.
## Examples
collapse all
Find the Maclaurin series expansions of the exponential, sine, and cosine functions up to the fifth order.
syms x
T1 = taylor(exp(x))
T1 =
$\frac{{x}^{5}}{120}+\frac{{x}^{4}}{24}+\frac{{x}^{3}}{6}+\frac{{x}^{2}}{2}+x+1$
T2 = taylor(sin(x))
T2 =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x$
T3 = taylor(cos(x))
T3 =
$\frac{{x}^{4}}{24}-\frac{{x}^{2}}{2}+1$
You can use the sympref function to modify the output order of symbolic polynomials. Redisplay the polynomials in ascending order.
sympref('PolynomialDisplayStyle','ascend');
T1
T1 =
$1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}+\frac{{x}^{4}}{24}+\frac{{x}^{5}}{120}$
T2
T2 =
$x-\frac{{x}^{3}}{6}+\frac{{x}^{5}}{120}$
T3
T3 =
$1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{24}$
The display format you set using sympref persists through your current and future MATLAB® sessions. Restore the default value by specifying the 'default' option.
sympref('default');
Find the Taylor series expansions at $\mathit{x}=1$ for these functions. The default expansion point is 0. To specify a different expansion point, use ExpansionPoint.
syms x
T = taylor(log(x),x,'ExpansionPoint',1)
T =
$x-\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(x-1\right)}^{3}}{3}-\frac{{\left(x-1\right)}^{4}}{4}+\frac{{\left(x-1\right)}^{5}}{5}-1$
Alternatively, specify the expansion point as the third argument of taylor.
T = taylor(acot(x),x,1)
T =
$\frac{\pi }{4}-\frac{x}{2}+\frac{{\left(x-1\right)}^{2}}{4}-\frac{{\left(x-1\right)}^{3}}{12}+\frac{{\left(x-1\right)}^{5}}{40}+\frac{1}{2}$
Find the Maclaurin series expansion for f = sin(x)/x. The default truncation order is 6. The Taylor series approximation of this expression does not have a fifth-degree term, so taylor approximates this expression with the fourth-degree polynomial.
syms x
f = sin(x)/x;
T6 = taylor(f,x);
Use Order to control the truncation order. For example, approximate the same expression up to the orders 7 and 9.
T8 = taylor(f,x,'Order',8);
T10 = taylor(f,x,'Order',10);
Plot the original expression f and its approximations T6, T8, and T10. Note how the accuracy of the approximation depends on the truncation order.
fplot([T6 T8 T10 f])
xlim([-4 4])
grid on
legend('approximation of sin(x)/x with error O(x^6)', ...
'approximation of sin(x)/x with error O(x^8)', ...
'approximation of sin(x)/x with error O(x^{10})', ...
'sin(x)/x','Location','Best')
title('Taylor Series Expansion')
Find the Taylor series expansion of this expression. By default, taylor uses an absolute order, which is the truncation order of the computed series.
syms x
T = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5)
T =
$-\frac{{x}^{3}}{3}$
Find the Taylor series expansion with a relative truncation order by using OrderMode. For some expressions, a relative truncation order provides more accurate approximations.
T = taylor(1/exp(x) - exp(x) + 2*x,x,'Order',5,'OrderMode','relative')
T =
$-\frac{{x}^{7}}{2520}-\frac{{x}^{5}}{60}-\frac{{x}^{3}}{3}$
Find the Maclaurin series expansion of this multivariate expression. If you do not specify the vector of variables, taylor treats f as a function of one independent variable.
syms x y z
f = sin(x) + cos(y) + exp(z);
T = taylor(f)
T =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x+\mathrm{cos}\left(y\right)+{\mathrm{e}}^{z}$
Find the multivariate Maclaurin series expansion by specifying the vector of variables.
syms x y z
f = sin(x) + cos(y) + exp(z);
T = taylor(f,[x,y,z])
T =
$\frac{{x}^{5}}{120}-\frac{{x}^{3}}{6}+x+\frac{{y}^{4}}{24}-\frac{{y}^{2}}{2}+\frac{{z}^{5}}{120}+\frac{{z}^{4}}{24}+\frac{{z}^{3}}{6}+\frac{{z}^{2}}{2}+z+2$
You can use the sympref function to modify the output order of a symbolic polynomial. Redisplay the polynomial in ascending order.
sympref('PolynomialDisplayStyle','ascend');
T
T =
$2+z+\frac{{z}^{2}}{2}+\frac{{z}^{3}}{6}+\frac{{z}^{4}}{24}+\frac{{z}^{5}}{120}-\frac{{y}^{2}}{2}+\frac{{y}^{4}}{24}+x-\frac{{x}^{3}}{6}+\frac{{x}^{5}}{120}$
The display format you set using sympref persists through your current and future MATLAB sessions. Restore the default value by specifying the 'default' option.
sympref('default');
Find the multivariate Taylor series expansion by specifying both the vector of variables and the vector of values defining the expansion point.
syms x y
f = y*exp(x - 1) - x*log(y);
T = taylor(f,[x y],[1 1],'Order',3)
T =
$x+\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(y-1\right)}^{2}}{2}$
If you specify the expansion point as a scalar a, taylor transforms that scalar into a vector of the same length as the vector of variables. All elements of the expansion vector equal a.
T = taylor(f,[x y],1,'Order',3)
T =
$x+\frac{{\left(x-1\right)}^{2}}{2}+\frac{{\left(y-1\right)}^{2}}{2}$
Find the error estimate when approximating a function $f\left(x\right)=\mathrm{log}\left(x+1\right)$ using the Taylor series expansion. Here, consider the Taylor approximation up to the 7th order (with the truncation order $\mathit{n}=8$) at the expansion point $\mathit{a}=0$.
The error or remainder in the Taylor approximation is given by the Lagrange form:
${\mathit{R}}_{\mathit{n}-1}\left(\mathit{x}\right)=\frac{{\mathit{f}}^{\mathit{n}}\left(\mathit{c}\right)}{\mathit{n}!}{\left(\mathit{x}-\mathit{a}\right)}^{\mathit{n}}.$
The upper bound of the error estimate can be calculated by finding a positive real number $\mathit{M}$ such that $|{\mathit{f}}^{\mathit{n}}\left(\mathit{c}\right)|\le \mathit{M}$ for all $\mathit{c}$ between $\mathit{a}$ and $\mathit{x}$.
Find the Taylor series expansion of the function $f\left(x\right)=\mathrm{log}\left(x+1\right)$ up to the 7th order by specifying Order as 8.
syms x
f = log(x+1)
f = $\mathrm{log}\left(x+1\right)$
T = taylor(f,'Order',8)
T =
$\frac{{x}^{7}}{7}-\frac{{x}^{6}}{6}+\frac{{x}^{5}}{5}-\frac{{x}^{4}}{4}+\frac{{x}^{3}}{3}-\frac{{x}^{2}}{2}+x$
To estimate the error in the Taylor approximation, first compute the term ${\mathit{f}}^{8}\left(\mathit{c}\right)$.
syms c
fn(c) = subs(diff(f,8),x,c)
fn(c) =
$-\frac{5040}{{\left(c+1\right)}^{8}}$
For positive values of $\mathit{x}$, the upper bound of the error estimate can be calculated by using the relation $|{\mathit{f}}^{8}\left(\mathit{c}\right)|\le 5040$ (because $\mathit{c}$ must be a positive value between $0$ and a positive $\mathit{x}$). Next, find the upper bound of the error estimate Rupper(x) by using the Lagrange from ${\mathit{R}}_{7}\left(\mathit{x}\right)$ and the relation $|{\mathit{f}}^{8}\left(\mathit{c}\right)|\le 5040$.
Rupper(x) = 5040*x^8/factorial(8)
Rupper(x) =
$\frac{{x}^{8}}{8}$
Evaluate the Taylor series expansion at the point $\mathit{x}=0.5$. Find the upper bound of the error estimate in the Taylor approximation.
Teval = subs(T,x,0.5)
Teval =
$\frac{909}{2240}$
Rmax = double(Rupper(0.5))
Rmax = 4.8828e-04
For comparison, evaluate the exact function at $\mathit{x}=0.5$ and find the remainder in the Taylor approximation.
feval = subs(f,x,0.5)
feval =
$\mathrm{log}\left(\frac{3}{2}\right)$
R = double(abs(feval-Teval))
R = 3.3846e-04
## Input Arguments
collapse all
Input to approximate, specified as a symbolic expression or function. It also can be a vector, matrix, or multidimensional array of symbolic expressions or functions.
Expansion variable, specified as a symbolic variable. If you do not specify var, then taylor uses the default variable determined by symvar(f,1).
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You also can specify the expansion point as a name-value argument. If you specify the expansion point both ways, then the name-value argument takes precedence.
### Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: taylor(log(x),x,'ExpansionPoint',1,'Order',9)
Expansion point, specified as a number, or a symbolic number, variable, function, or expression. The expansion point cannot depend on the expansion variable. You can also specify the expansion point using the input argument a. If you specify the expansion point both ways, then the name-value argument takes precedence.
Truncation order of the Taylor series expansion, specified as a positive integer or a symbolic positive integer. taylor computes the Taylor series approximation with the order n - 1. The truncation order n is the exponent in the O-term: O(varn).
Order mode indicator, specified as 'absolute' or 'relative'. This indicator specifies whether to use absolute or relative order when computing the Taylor polynomial approximation.
Absolute order is the truncation order of the computed series. Relative order n means that the exponents of var in the computed series range from the leading order m to the highest exponent m + n - 1. Here m + n is the exponent of var in the O-term: O(varm + n).
collapse all
### Taylor Series Expansion
A Taylor series expansion represents an analytic function f(x) as an infinite sum of terms around the expansion point x = a:
$f\left(x\right)=f\left(a\right)+\frac{{f}^{\prime }\left(a\right)}{1!}\left(x-a\right)+\frac{{f}^{″}\left(a\right)}{2!}{\left(x-a\right)}^{2}+\dots =\sum _{m=0}^{\infty }\frac{{f}^{\left(m\right)}\left(a\right)}{m!}\cdot {\left(x-a\right)}^{m}$
A Taylor series expansion requires a function to have derivatives up to an infinite order around the expansion point.
### Maclaurin Series Expansion
The Taylor series expansion around x = 0 is called a Maclaurin series expansion:
$f\left(x\right)=f\left(0\right)+\frac{{f}^{\prime }\left(0\right)}{1!}x+\frac{{f}^{″}\left(0\right)}{2!}{x}^{2}+\dots =\sum _{m=0}^{\infty }\frac{{f}^{\left(m\right)}\left(0\right)}{m!}{x}^{m}$
## Tips
• If you use both the third argument a and ExpansionPoint to specify the expansion point, then the value specified by ExpansionPoint prevails.
• If var is a vector, then the expansion point a must be a scalar or a vector of the same length as var. If var is a vector and a is a scalar, then a is expanded into a vector of the same length as var with all elements equal to a.
• If the expansion point is infinity or negative infinity, then taylor computes the Laurent series expansion, which is a power series in 1/var.
• You can use the sympref function to modify the output order of symbolic polynomials.
• If taylor cannot find the Taylor series expansion, then use series to find the more general Puiseux series expansion.
## Version History
Introduced before R2006a | 2023-02-01T22:08:32 | {
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https://www.empreendasemfronteiras.com.br/hafvaq/e19ab0-can-a-relation-be-both-symmetric-and-antisymmetric | We can therefore take the following relation: $\{a,b,c\}$ would be our universe and $R=\{\langle a,b\rangle,\langle b,a\rangle,\langle a,c\rangle\}$. It is an interesting exercise to prove the test for transitivity. Thus, there exists a distinct pair of integers $a$ and $b$ such that $aRb$ and $bRa$. See also However, since $(-1)\cdot 2^{2} = -4 \not\gt 0$, $(-1, 2)\not\in R$, thus $R$ is not symmetric. Can you legally move a dead body to preserve it as evidence? Active 1 year, 7 months ago. A symmetric relation can work both ways between two different things, whereas an antisymmetric relation imposes an order. I got this from my professor and my book explains that they are not mutually exclusive. $$R=\{(a,b), (b,a), (c,d)\}.$$. Consider matrix which has ones on diagonal and zeros on other places. Therefore, in an antisymmetric relation, the only ways it agrees to both situations is a=b. It only takes a minute to sign up. This doesn't tell … Click hereto get an answer to your question ️ Given an example of a relation. Parsing JSON data from a text column in Postgres. Hence, $R$ cannot be antisymmetric. Is the Gelatinous ice cube familar official? If So, Give An Example; If Not, Give An Explanation. Come up with a relation on that set such that for some pairs of elements (x, y), $x R y$ and $\lnot (y R x)$; but for other pairs of elements (x, y), $x R y$ and $y R x$. How can a matrix relation be both antisymmetric and symmetric? A relation can be neither symmetric nor antisymmetric. Although both have similarities in their names, we can see differences in both their relationships such that asymmetric relation does not satisfy both conditions whereas antisymmetric satisfies both the conditions, but only if both the elements are similar. But if antisymmetric relation contains pair of the form (a,a) then it cannot be asymmetric. together. Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. Explain this image to me. A relation can be neither symmetric nor antisymmetric. A relation R is symmetric if the value of every cell (i, j) is same as that cell (j, i). The terms symmetric and antisymmetric are not opposites, because a relation can have both of these properties or may lack both of them. Is the bullet train in China typically cheaper than taking a domestic flight? Remember that a relation on a set $A$ is just a subset of $A\times A$. (iii) Reflexive and symmetric but not transitive. Limitations and opposites of asymmetric relations are also asymmetric relations. Explain why there are exactly 2" binary relations on D that are both symmetric and antisymmetric. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). Why aren't "fuel polishing" systems removing water & ice from fuel in aircraft, like in cruising yachts? Comparing method of differentiation in variational quantum circuit. Macbook in Bed: M1 Air vs M1 Pro with Fans Disabled. Could you design a fighter plane for a centaur? Ask Question Asked 5 years, 10 months ago. Could you design a fighter plane for a centaur? Asymmetric relation: Asymmetric relation is opposite of symmetric relation. MathJax reference. Relations, specifically, show the connection between two sets. Can I assign any static IP address to a device on my network? In set theory, the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. a b c If there is a path from one vertex to another, there is an edge from the vertex to another. Why is 2 special? To put it simply, you can consider an antisymmetric relation of a set as a one with no ordered pair and its reverse in the relation. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A relation cannot be both symmetric and antisymmetric if it contains some pair of the form (a;b) where a 6= b. Anonymous . Antisymmetric Relation Definition. i don't believe you do. Or it can be defined as, relation R is antisymmetric if either (x,y)∉R or (y,x)∉R whenever x ≠ y. Is the relation reflexive, symmetric and antisymmetric? Can A Relation Be Both Symmetric And Antisymmetric? However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). Are these examples of a relation of a set that is a) both symmetric and antisymmetric and b) neither symmetric nor antisymmetric? Symmetric property: Book where bodies stolen by witches. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 4 years ago. (reflexive as well). However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). Thanks for contributing an answer to Mathematics Stack Exchange! A transitive relation is asymmetric if it is irreflexive or else it is not. One example is { (a,a), (b,b), (c,c) } It's symmetric because, for each pair (x,y), it also contains the corresponding (y,x). Symmetric or antisymmetric are special cases, most relations are neither (although a lot of useful/interesting relations are one or the other). Anonymous. Thanks for contributing an answer to Mathematics Stack Exchange! Suppose that Riverview Elementary is having a father son picnic, where the fathers and sons sign a guest book when they arrive. Think $\le$. There are n diagonal values, total possible combination of diagonal values = 2 n There are n 2 – n non-diagonal values. (iii) Reflexive and symmetric but not transitive. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relation of a set as one with no ordered pair and its reverse in the relation. Can you take it from here? Replacing the core of a planet with a sun, could that be theoretically possible? Since $2\cdot (-1)^{2} = 2\gt 0$, the ordered pair $(2, -1)\in R$. 푅 is not symmetric A relation R on a set A is antisymmetric iff aRb and bRa imply that a = b. Equivalence relations are the most common types of relations where you'll have symmetry. Is it possible to assign value to set (not setx) value %path% on Windows 10? i know what an anti-symmetric relation is. Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. At its simplest level (a way to get your feet wet), you can think of an antisymmetric relationof a set as one with no ordered pair and its reverse in the relation. Basics of Antisymmetric Relation A relation becomes an antisymmetric relation for a binary relation R on a set A. So, you can just pick a convenient subset $R \subset A \times A$ so that only for SOME elements $a,b$ of $A$(I.e. Why aren't "fuel polishing" systems removing water & ice from fuel in aircraft, like in cruising yachts? Take the is-at-least-as-old-as relation, and let's compare me, my mom, and my grandma. Answer to 1. A relation can be both symmetric and antisymmetric. Shifting dynamics pushed Israel and U.A.E. This Site Might Help You. It can be reflexive, but it can't be symmetric for two distinct elements. Why does "nslookup -type=mx YAHOO.COMYAHOO.COMOO.COM" return a valid mail exchanger? It can be reflexive, but it can't be symmetric for two distinct elements. Let S be a sequence of n different numbers. 2. (d) Show that if a relation is symmetric then so is its complement. $\forall a,b\in X$ $aRb\implies bRa$. Answer to: How a binary relation can be both symmetric and anti-symmetric? Viewed 1k times 1 $\begingroup$ Take a look at this picture: From what I am reading, antisymmetric means: ∀ x ∀ y \,[ R ( x , … Antisymmetric relation is a concept of set theory that builds upon both symmetric and asymmetric relation in discrete math. Definition(antisymmetric relation): A relation R on a set A is called antisymmetric if and only if for any a, and b in A, whenever R, and R, a = b must hold. 0. Mathematics. What causes dough made from coconut flour to not stick together? for example the relation R on the integers defined by aRb if a < b is anti-symmetric, but not reflexive. 4 years ago. To learn more, see our tips on writing great answers. Or does it have to be within the DHCP servers (or routers) defined subnet? Symmetric and anti-symmetric relations are not opposite because a relation R can contain both the properties or may not. So C is symmetric and antisymmetric. (b) Yes, a relation on {a,b,c} can be both symmetric and anti-symmetric. Why can't I sing high notes as a young female? Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij , then the possible eigenvalues are 1 and –1. Which is (i) Symmetric but neither reflexive nor transitive. 3 0. Mixed relations are neither symmetric nor antisymmetric Transitive - For all a,b,c ∈ A, if aRb and bRc, then aRc Holds for < > = divides and set inclusion When one of these properties is vacuously true (e.g. This preview shows page 271 - 275 out of 313 pages.. Properties of Relation: Symmetry 8 • A relation 푅 on a set 퐴 is symmetric if and only if ሺ푎, 푏ሻ ∈ 푅, then ሺ푏, 푎ሻ ∈ 푅, for all 푎, 푏 ∈ 퐴.Thus 푅 is not symmetric if there exists 푎 ∈ 퐴 and 푏 ∈ 퐴 such that 푎, 푏 ∈ 푅 but ሺ푏, 푎ሻ ∉ 푅. However, $(2,1)$ and $(1,2)$, $X\ne Y$. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Asking for help, clarification, or responding to other answers. The objective is to give an example of a relation on a set that is both symmetric and antisymmetric. Relationship to asymmetric and antisymmetric relations. 2. 0 0. redmond. What can be said about a relation $R=(A,A,R)$ that is refelxive, symmetric and antisymmetric? Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity. Ryan Reynolds sells gin line for staggering $610M . By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). 7. Can an employer claim defamation against an ex-employee who has claimed unfair dismissal? Can A Relation Be Both Reflexive And Antireflexive? site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Lv 4. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. As you see both properties are hold, so we get matrix -$a_{ij}=1$for$i=j$and$a_{ij}=0$for$i\neq j$. Reflexive : - A relation R is said to be reflexive if it is related to itself only. It to example 7.2.2 to see how it works if, it is an edge from the vertex to,... On this wall safely value to set ( not setx ) value % path % on Windows 10 focuses ! Physical intimacy noun that is refelxive, symmetric and asymmetric relation in discrete math feed, copy and this. Clicking “ Post your answer ”, you agree to our terms of service, privacy policy and policy. Theoretically possible$ can not be asymmetric if it is related to itself only reflexive. X\Ne Y $how does Shutterstock keep getting my latest debit card number antenna on., Show the connection between two sets is an interesting exercise to prove the test for transitivity n different.! Tuner on SWR above 3$ aRb \land bRa ) \implies a=b $subscribe to this feed... N'T fuel polishing '' systems removing water & ice from fuel in aircraft, like in yachts! Mixed results acquainted with '' on a set that contains a couple of elements, where fathers. 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( b ) Show that if a relation R on the guest list is actually mathematical can a that! Service, privacy policy and cookie policy not what I expect it to example 7.2.2 to see how works... Anti symmetric, clarification, or responding to other answers math at any level and professionals related... Setup evaluated at +2.6 according to Stockfish of diagonal values = 2 n are! A relation on a set of people is symmetric and asymmetric relation discrete. Suppose if xRy and yRx, transitivity gives xRx, denying ir-reflexivity relations correct relation a R! To do less than is also not satisfied focuses on relations in. Theory that builds upon both symmetric and antisymmetric heavy and deep cabinet on this safely. This section focuses on relations '' in discrete math set A= ( )... Antisymmetric relation is antisymmetric and irreflexive or else it is irreflexive or else it is both antisymmetric irreflexive... 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Are neither ( although a lot of useful/interesting relations are neither symmetric antisymmetric! Device on my network relation imposes an order \$ then antisymmetry is also c... Mail exchanger integers defined by aRb if a relation is asymmetric if is... Engage in physical intimacy subscribe to this RSS feed, copy and paste this URL into your reader. | 2021-06-16T20:09:42 | {
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https://electronics.stackexchange.com/questions/428730/calculating-wattage-for-resistor-in-high-frequency-application | # Calculating Wattage for Resistor in High Frequency Application?
I am making a MOSFET driving circuit.
Frequency : 400 kHz [50% duty cycle]
Gate voltage: 12 V
Total gate charge : 210 nC as per datasheet IRFP460
Rise time: 100 ns
[Q=I*t]
Current: 2.1 A
Gate resistor: V/I > 12/2.1 > 5.7 ohm
Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W
1 watt resistor is OK ?
• Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts. – Elliot Alderson Mar 23 at 22:29
• I think i should remove Average Power line . – Israr Sayed Mar 24 at 11:23
The figure below shows the Gate Voltage versus Total Gate Charge for the IRFP460 MOSFET:
With a gate drive voltage $$\V_{DR} = 12\,\mathrm{V}\$$, it's possible to estimate a total gate charge of $$\155\,\mathrm{nC}\$$.
If $$\i_g \$$ represents the gate current, $$\Q\$$ the charge going into the gate and $$\tb\$$ (beginning time) and $$\te\$$ (ending time) to represent a time interval, then:
$$Q = \int_{tb}^{te}i_gdt$$
METHOD 1: (a first estimate)
Here the $$\i_g\$$ is considered constant ($$\Ig_{(ON)}\$$) during the charge ($$\tp_{(ON)}\$$) and constant ($$\Ig_{(OFF)}\$$) during discharge time ($$\tp_{(OFF)}\$$); roughly shown in the figure below:
So, the integral above reduces simply to (considering $$\tp_{(ON)}=100\,\mathrm{ns}\$$ and $$\Q_g\$$ as the total gate charge):
$$Q_g = Ig_{(ON)} \times tp_{(ON)}$$ or $$Ig_{(ON)} = \frac{Q_g}{tp_{(ON)}} = \frac{155\,\mathrm{nC}}{100\,\mathrm{ns}}= 1.55\,\mathrm{A}$$
The gate resistor $$\R_G\$$ must be calculated taking in account that, in “flat” part of the switching period (plot above), the gate voltage is constant at about $$\5.2\$$ V:
$$R_G = \frac{12\,\mathrm{V} - 5.2\,\mathrm{V}}{1.55\,\mathrm{A}} = 4.39 \space \Omega \approx 4.7 \space \Omega$$
In order to simplify I consider here $$\Ig_{(OFF)}=-Ig_{(ON)}\$$. So, the root mean square value for $$\i_g\$$ is:
$$I_{RMS}= Ig_{(ON)}\sqrt{2 \times \frac{tp_{(ON)}}{T} } \approx 0.438\,\mathrm{A}$$
Finally, the average power for $$\R_G\$$ is:
$$P = I_{RMS}^2R_G \approx 0.9\,\mathrm{W}$$
METHOD 2:
Here the $$\i_g\$$ is considered as a straight line with maximum value $$\Ig_{pk_{(ON)}}\$$ and decreasing to zero at the end of time $$\tp_{(ON)}\$$ - as an approximation to the actual exponential decay (more realistic). Similar consideration is made for the gate discharge time:
An example of real measurement:
Retaining a $$\R_G = 4.7 \space \Omega\$$, the peak gate current can be calculated as:
$$Ig_{pk_{(ON)}} = \frac{12\,\mathrm{V}}{4.7 \space \Omega} \approx 2.553,\mathrm{A}$$
In order to simplify I consider here $$\Ig_{pk_{(OFF)}}=-Ig_{pk_{(ON)}}\$$. So, the root mean square value for $$\i_g\$$ is:
$$I_{RMS}= Ig_{pk_{(ON)}}\sqrt{\frac{2}{3} \times \frac{tp_{(ON)}}{T} } \approx 0.417\,\mathrm{A}$$
Finally, the average power for $$\R_G\$$ is:
$$P = I_{RMS}^2R_G \approx 0.817\,\mathrm{W}$$
No major differences from the value previously calculated.
THIRD METHOD
Just to mention a more precise (and more laborious) method. Here, $$\i_g\$$ is considered a true exponential decaying function (see figure above):
$$i_g = Ig_{pk_{(ON)}}e^{-\frac{t}{R_GC_{eff}}}$$
where $$\C_{eff}\$$ is the effective gate input capacitance of MOSFET. So:
$$i_g = \frac{V_{DR}}{R_G}e^{-\frac{t}{R_GC_{eff}}}$$
In the time interval $$\0\$$ to $$\t_s\$$, the total gate charge ("consumed") is given by:
$$Q_g = \int_{0}^{t_s} \frac{V_{DR}}{R_G}e^{-\frac{t}{R_GC_{eff}}}dt$$
This integral can be solved for a parameter ($$\R_G\$$ or $$\t_s\$$), when others are known.
CONCLUSION: The average power values were below $$\1\,\mathrm{W}\$$, but a margin of safety can be applied for guarantee.
• I created an account to let you know that this answer is one of the best answers I've seen, on any site, in a long time. Just brilliant. – LogicalBranch Mar 24 at 12:01
• This could become the canonical answer to questions of this kind. Thorough and well written. However, could you please add links or citations to the sources of the graphics? – Elliot Alderson Mar 24 at 13:20
• So much effort for approximate solutions, have a look to the simple and exact Dave Tweed's one below. – carloc Mar 24 at 13:47
• Estimating Total Gate Charge from Graph nice. – Israr Sayed Mar 24 at 13:58
• @carloc: Just AFTER doing the paper/pencil work it's possible to conclude that the solutions are similar. My contributions is aligned with the standard procedures recommended by manufacturers. Also deals with the gate resistor estimation. – Dirceu Rodrigues Jr Mar 24 at 14:09
Dividing the peak power by the frequency is not useful.
Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of
$$25 W \cdot\frac{2 \cdot 100 ns}{2.5 \mu s} = 2 W$$
Clearly, your 1W resistor is not going to cut it!
However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.
For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents
$$Energy = \frac{1}{2}\cdot Charge \cdot Voltage$$
$$0.5 \cdot 210 nC \cdot 12 V = 1.26 \mu J$$
This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.
To get the average power, multiply the energy per cycle by the number of cycles per second, giving
$$1.26 \mu J \cdot 2 \cdot 400 kHz = 1.088 W$$
Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here. | 2019-04-21T02:34:28 | {
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https://proofwiki.org/wiki/Intersection_is_Largest_Subset | # Intersection is Largest Subset
## Theorem
Let $T_1$ and $T_2$ be sets.
Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.
That is:
$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$
### General Result
Let $T$ be a set.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
$\paren {\forall X \in \mathbb T: S \subseteq X} \iff S \subseteq \bigcap \mathbb T$
### Family of Sets
In the context of a family of sets, the result can be presented as follows:
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
## Proof
### Sufficient Condition
From Set is Subset of Intersection of Supersets we have that:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
$\Box$
### Necessary Condition
Let:
$S \subseteq T_1 \cap T_2$
From Intersection is Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$.
From Subset Relation is Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$.
So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.
$\Box$
From the above, we have:
$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$
$S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$
Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence.
$\blacksquare$ | 2021-10-27T13:21:50 | {
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https://math.stackexchange.com/questions/387960/matrices-with-eigenvalues-0-and-1 | # Matrices with eigenvalues 0 and 1
How can you describe all $2\times 2$ matrices whose eigenvalues are 0 and 1?
My attempt: I know that 0 and 1 has to be solutions of the characteristic polynomial. And I've considered some examples like $$\begin{pmatrix}0&0\\0&1\end{pmatrix},\begin{pmatrix}1&0\\0&0\end{pmatrix} \text{ etc.}$$but I haven't found any regularity. But I also know from linear algebra that a matrix $A$ satisfying $A^2=A$ has only eigenvalues 0 and/or 1. Are there any more matrices? What do they have to satisfy?
• Do you mean that "the matrix has two eigenvalues, which are 0 and 1" or "the matrix has some unknown number of eigenvalues, all of which are either 0 or 1"? – Sharkos May 10 '13 at 21:31
• @Sharkos I would be interested in both cases :) – user31035 May 10 '13 at 21:34
Consider a matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ You can easily work out the characteristic polynomial, which is $$X^2 - (a+d)X + (ad-bc)$$ Based on your information, you have $$\begin{cases} a+d=1 \\ ad-bc=0 \end{cases}$$ because in a polynomial $X^2+pX+q$ the sum of the roots is $-p$ and their product is $q$.
This gives $d=1-a$, so $a-a^2-bc=0$ or $bc=a-a^2$. You can thus choose arbitrarily $a$. If $a=0$ or $a=1$, then one among $b$ and $c$ must be zero and the other one is arbitrary. For $a-a^2\ne0$, the matrices you look for have the form $$\begin{bmatrix} a & b \\ \dfrac{a-a^2}{b} & 1-a \end{bmatrix}$$ So the general form is one of these five cases
1. $\begin{bmatrix}0 & b \\ 0 & 1\end{bmatrix}\quad$ (arbitrary $b$)
2. $\begin{bmatrix}0 & 0 \\ c & 1\end{bmatrix}\quad$ (arbitrary $c$)
3. $\begin{bmatrix}1 & b \\ 0 & 0\end{bmatrix}\quad$ (arbitrary $b$)
4. $\begin{bmatrix}1 & 0 \\ c & 0\end{bmatrix}\quad$ (arbitrary $c$)
5. $\begin{bmatrix} a & b \\ \dfrac{a-a^2}{b} & 1-a \end{bmatrix}\quad$ ($a\ne0$, $a\ne1$, $b\ne0$)
Since the eigenvalues are 0 and 1 (hence distinct), the matrix must be diagonalizable. Thus it must be of the form $P^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}P$ for some invertible matrix $P$. Since there is a nice explicit formula for the inverse of a $2 \times 2$ matrix, you can work out the form of any matrix of the aforementioned form. Take $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $\det P = ad-bc \ne 0$, then write out $P^{-1}$, and then compute.
EDIT: I am assuming you mean that the set of eignevalues of your matrix must be $\{0,1\}$, as opposed to being contained in this set, which is what Sharkos assumes. So it depends on what you intended.
Assuming that we exclude matrices from having complex eigenvalues but no real eigenvalues (which seems reasonable given your question) we make use of the Jordan normal form of the matrix.
There is some invertible matrix $P$ (a change of basis) such that
$$A = P^{-1} J P$$
where $J$ is in Jordan normal form with either $0$ or $1$ on the diagonal. In particular,
$$J = \pmatrix{0 & 0 \\ 0 & 0}, \pmatrix{1 & 0 \\ 0 & 1}, \pmatrix{1 & 0 \\ 0 & 0}, \pmatrix{0 & 1 \\ 0 & 0}, \pmatrix{1 & 1 \\ 0 & 1}$$
generate all other forms of these matrices.
• The first case has $A=0$.
• The second case has $A=I$.
• The third case is a change-of-basis from having one of each eigenvalue; it is a projection onto a line.
• The fourth case is a nilpotent matrix such that $A^2=0$. It maps one direction into another, and that direction into oblivion.
• The fifth case generates a skew/shear in some direction.
This is a partial answer that suggests how to parametrize such matrices explicitly.
The other answers are more elegant and tell the story more succinctly, but they construct the answers non-uniquely. (There are different matrices that you can conjugate a given Jordan normal form by and still obtain the same matrices: $P^{-1} J P = Q^{-1} J Q$ is possible with $P \ne Q$.)
First a little result about the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$ for a $2 \times 2$ matrix $A$: $$p_A(\lambda) = \lambda^2 - (\operatorname{tr} A) \lambda + \det A,$$ where $\operatorname{tr} A = a + d$ is the trace. Check this from the definition if you're not familiar with it.
Now, proceed by cases. Suppose that $\lambda = 1$ is the only eigenvalue. Then, $$p_A(\lambda) = (\lambda - 1)^2 = \lambda^2 - 2\lambda + 1.$$ So, we know that \left\{ \begin{align} \operatorname{tr} A &= 2 \\ \det A &= 1 \end{align} \right. With $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},$$ \left\{ \begin{align} a + d &= 2 \\ ad - bc &= 1 \end{align} \right.. Taking advantage of the trace equation to reduce the number of parameters, let's make the substitution $a = 1 + t$ and $d = 1 - t$. Now the determinant equation becomes $(1 + t)(1 - t) - bc = 1$ or $$bc = t^2.$$ Either $b = 0$, in which case, $t = 0$, as well, and $c$ is free to take any value: $$\begin{bmatrix} 1 & 0 \\ c & 1 \end{bmatrix}.$$ Or $b \ne 0$, so $c = -\frac{t^2}{b}$, so we have: $$\begin{bmatrix} 1 + t & b \\ -\tfrac{t^2}{b} & 1 + t \end{bmatrix}.$$
The space of possibilities in $(b, c)$-plane consists of both axes (corresponding to having $t = 0$, hence $1$ on the diagonal) together with the second and fourth quadrants (corresponding to $t \ne 0)$.
A similar analysis can give an explicit description of the other cases for the possibilities of the eigenvalues. (Note that this idea generalizes beyond your $\lambda \in \{0, 1\}$ assumption.)
• I think the second instance of $P$ near the top of your answer should be $Q$. – Alistair Savage May 10 '13 at 22:47
• Now it's fixed. – Sammy Black May 10 '13 at 22:49
If a matrix $A\in\mathbb{C}^{n\times n}$ has only eigenvalues $0$ and $1$ then its characteristic polynomial is of the form $$z^{a}\left(z-1\right)^{b}=z^{a}\sum_{k=0}^{b}\left(\begin{array}{c} b\\ k \end{array}\right)z^{k}\left(-1\right)^{n-k}=\sum_{k=0}^{b}\left(\begin{array}{c} b\\ k \end{array}\right)z^{a+k}\left(-1\right)^{n-k}$$ with $a+b=n$. A matrix that has the above characteristic polynomial is $$B=\left[\begin{array}{cccccc} 0 & & & & & -c_{0}\\ 1 & 0 & & & & -c_{1}\\ & 1 & 0 & & & -c_{2}\\ & & 1 & \ddots & & \vdots\\ & & & \ddots & 0 & -c_{n-2}\\ & & & & 1 & -c_{n-1} \end{array}\right]$$ where $$c_{j}=\begin{cases} 0 & j<a\\ \left(\begin{array}{c} b\\ k \end{array}\right)\left(-1\right)^{n-k} & j\geq a \end{cases}.$$ The above is called the Frobenius companion matrix to the characteristic polynomial. Another matrix that will satisfy this is $B^{T}$ since $\det\left(X^{T}\right)=\det\left(X\right)$. Note that $B$ is not a projection (i.e. $B^2\neq B$).
Obviously, not all matrices that have only eigenvalues of $0$ and $1$ are of this form. For example, $I$ only has $1$ as an eigenvalue but cannot be written in the above form.
• This construction gives (up to conjugation) the matrices which have the polynomial as minimal polynomial. – Mariano Suárez-Álvarez May 10 '13 at 22:20 | 2019-05-24T05:27:52 | {
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https://math.stackexchange.com/questions/939235/proof-that-the-set-of-irrational-numbers-is-dense-in-the-reals | # Proof that the set of irrational numbers is dense in the reals
The hint I was given was to simply prove that $y=xz$ is irrational given that $x$ is nonzero, $x$ is rational and $z$ is irrational. Here's how I did it:
Claim: $y=xz$ is irrational.
Proof: Assume $x\neq0$, $x$ is rational and $z$ is irrational.
By contradiction assume that $y=xz$ is rational. This means $y$ can be expressed as $m/n$, $m$ and $n$ being integers; $y$ can be expressed similarly as $p/q$, $p$ and $q$ being integers. By substitution, we have that $$p/q=mz/n$$ and $$z=pn/qm, qm \neq 0.$$ Since $pn$ and $qm$ are integers $z$ has to be rational.
In addition to this it seems like there's a part 2 as follows:
Proof: Given an interval $(x,y)$ we will choose a positive irrational number, $z$, say. By density of the rationals there is a rational $p$ in the interval $(x/z, y/z)$ s.t. $$x/z <p< y/z.$$ From this we see that $pz$ is irrational since it is the product of a rational and irrational number.
Is the $pz$ the $xz$ that we proved is irrational in the first proof? So ideally when presenting a full proof like this, should we do part 2 then part 1?
Great proof! You can apply your result about what you called $xz$ in part 1 to $pz$ in part 2 because it obeys your assumptions: $p \ne 0$, $p$ is rational and $z$ is irrational. So from part 1 we know that $pz$ is irrational.
We also know that $pz$ lies in the interval $(x,y)$, so irrationals are dense.
What is ideal to present when presenting a full proof depends on what your reader already knows. It sounds like in this case you want to present both part 1 and part 2, and that will be a complete proof. But in some other case your reader (or lecturer or marker) might not know or want to assume, for example, that rationals are dense in the real numbers, so you might have to include a part 1.5 that proves that.
EDIT: Oops, I think I misunderstood you. Yeah, I would present part 2 then part 1 when presenting this proof; I think the sequence of ideas flows better. But it's up to you!
Also, just to point out a few typos:
• I think you mean 'This means $x$ can be expressed as $m/n$' when you write "This means y can be expressed as m/n".
• I think you mean 'By substitution' when you write "By substation".
Hope that helps! | 2019-08-23T23:20:31 | {
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https://math.stackexchange.com/questions/22121/how-can-i-prove-that-n7-n-is-divisible-by-42-for-any-integer-n/22123 | # How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$?
I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
• Hint: Fermat's Little Theorem. – Arturo Magidin Feb 15 '11 at 6:13
• Hint #2: Chinese Remainder Theorem. – Pete L. Clark Feb 15 '11 at 6:19
• So you have $n^6\equiv 1\pmod{7}$, but this implies $n^7\equiv n\pmod{7}$. So $7|n^7-n$. In general, $n^p\equiv n\pmod{p}$ for a prime $p$. By the same token, $n^7\equiv (n^3)^2\cdot n\equiv n^3\equiv n\pmod{3}$ by FlT, so $3|n^7-n$. It remains to show $2|n^7-n$, and then you'll be done by the two hints above. – yunone Feb 15 '11 at 6:32
• @Yuval: $p=6$ ? :-) – lhf Feb 15 '11 at 12:26
• It works because $42=6(6+1)$ and Little Fermat, but I like to believe it works because it is the answer to the life, the universe and everything. – chubakueno Sep 1 '13 at 22:22
Problems like this appear frequently here. There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $42=2\times 3\times 7$, what we need to do is to check that 2, 3, and 7 divide $n^7-n$, no matter what $n$ is.
That 7 does is direct from Fermat's little theorem.
The theorem also ensures that 3 divides $n^3-n$. Now: $$n^7-n=n(n^6-1)=n((n^2)^3-1)=n(n^2-1)(n^4+n^2+1)=(n^3-n)(n^4+n^2+1),$$ so 3 indeed divides $n^7-n$.
Finally, $n$ and $n^7$ always have the same parity, so $n^7-n$ is even.
We can actually argue without Fermat's little theorem in this case. An approach that only requires patience is as follows: The idea is to factor the polynomial $x^7-x$ and then analyze the result when $x=n$ is an integer. (This is a trick that Bill Dubuque suggests sometimes in his solutions.)
We have: $x^7-x=x(x^6-1)=x(x^3+1)(x^3-1)=x(x+1)(x^2-x+1)(x-1)(x^2+x+1)$. When $x=n$, we have $$n^7-n=(n-1)n(n+1)(n^2-n+1)(n^2+n+1).$$ Now we analyze this prime by prime, as before. Note that one of $n$ and $n-1$ is always even, so the product is even. Also, of 3 consecutive numbers, such as $n-1,n,n+1$, one is always divisible by 3, so it only remains to verify divisibility by 7.
We may assume that $n=7k+b$ where $b=\pm2$ or $\pm3$, since otherwise $(n-1)n(n+1)$ is a multiple of 7. In that case, $n^2\equiv 4$ or $2\pmod 7$, and one of $n^2+n$, $n^2-n$ is $\equiv 6\pmod 7$, so $(n^2-n+1)(n^2+n+1)$ is a multiple of 7.
The disadvantage of this approach over the previous one, of course, is the need to analyze different cases. Fermat's little theorem allows us to analyze all cases simultaneously, which typically (as here) results in a much faster approach.
If you are comfortable with the method of induction, this gives us a way of verifying divisibility by 7 which is not without some elegance (divisibility by 2 and 3 is probably best approached as before). Note that $(-n)^7-(-n)=-(n^7-n)$, so we may as well assume that $n\ge 0$. If $n=0$ it is obvious that 7 divides $n^7-n$.
Suppose then that $7|n^7-n$, and argue that $7|(n+1)^7-(n+1)$. For this, actually expand $(n+1)^7$ using the binomial theorem: $$(n+1)^7=n^7+7n^6+{7\choose 2}n^5+{7\choose 3}n^4+\dots+1.$$ The point is that $${7\choose k}=\frac{7!}{k!(7-k)!}$$ is obviously divisible by 7 as long as $k\ne0,7$, so (modulo 7) we have that $(n+1)^7-(n+1)\equiv (n^7+1)-(n+1)=(n^7-n)$. Now we invoke the induction hypothesis, that precisely says that the latter is divisible by 7, and we are done.
Of course, exactly the same inductive argument gives us a proof of Fermat's little theorem: $p|n^p-n$ for any $p$ prime and any integer $n$.
• To prove that $3$ divides, you can observe that $n^7-n=n^7-n^5+n^5-n^3+n^3-n$ – N. S. Jun 29 '12 at 15:42
It is a special case of the following global-form of little Fermat. For $$\rm\, a,k,n\in\mathbb N$$ with $$\rm\ a,k > 1$$
$$\rm\qquad d\ |\ n^k\! -\! n\$$ for all $$\rm\:n\:$$ $$\rm \iff\ d\:$$ is squarefree, and $$\rm\ p\!-\!1\ |\ k\!-\!1\$$ for all primes $$\rm\:p\:|\:d$$
Hence for $$\rm\: a = 42 = 2\cdot 3\cdot 7\$$ we deduce: $$\rm\ \ 42\ |\ n^k\!-n\$$ for all $$\rm\:n \iff 6\ |\ k\!-\!1$$
For the simple proof and further discussion see Korselt's criterion for Carmichael numbers here (or my 2009/04/10 sci.math post, link is now broken by google)
Here is another useful variation:
Theorem $$\ \ \ n^{\large k+\phi}\equiv n^{\large k}\pmod{p^i q^j}\ \$$ assuming that $$\ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\,$$ $$\, i,j \le k,\,\ p\ne q.\ \ \$$
Proof $$\ \, p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^{ \phi}\equiv 1\,\Rightarrow\, n^{k + \phi}\equiv n^k,\,$$ by $$\ n^{\Large \color{#0a0}\phi} = (n^{\color{#0a0}{\Large \phi(p^{ i})}})^{\large \color{#0a0}m}\!\overset{\color{blue}{\rm (E)}}\equiv\! 1^{\large m}\!\equiv\! 1$$ by $$\rm\color{blue}{E}$$=Euler
$$\qquad\quad\ \, \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^k\equiv 0\,\equiv\, n^{k + \phi}\$$ by $$\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{mp})^i$$ and $$\,k\ge i.$$
So $$\ p^i\mid n^{k+\phi}\!-n^k.\,$$ By symmetry $$\,q^j$$ divides it too, so their lcm $$= p^iq^j\,$$ divides it too. $$\$$ QED
Remark $$\$$ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.
• Very nice application of Fermat's Little Theorem's global-form (I had never seen it). However, the link to your proof takes me nowhere... – Dr. Mathva May 7 at 13:20
• @Dr.Mathva I added a local link (alas Google continues to break links to its usenet newsgroup archive). – Bill Dubuque May 7 at 14:06
There are many equivalent ways of proving it.
First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.
(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)
Divisibility by $2$:
Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.
Equivalently you could argue out that $2|(n^2-n)$ directly from Fermat's little Theorem. (This is an overkill of Fermat's Little Theorem.)
Divisibility by $3$:
$n^7-n = n(n^6-1) = n(n^2-1)(n^4+n^2+1)=n(n+1)(n-1)(n^4+n^2+1)$.
$3|n$ or $3|(n-1)$ or $3|(n+1)$ and hence $3|(n^7-n)$.
Equivalently you could argue out that $3|(n^3-n)$ directly from Fermat's little Theorem.
Divisibility by $7$:
Note that $n$ can be either $7k$ or $7k \pm 1$ or $7k \pm 2$ or $7k \pm 3$.
If $n=7k$ or $n=7k \pm 1$, we are again done since then $7|n$ or $7|(n+1)$ or $7|(n-1)$ and hence $7|(n^7-n)$.
If $n=7k \pm 2$, then $n^2 = (7k \pm 2)^2 = 7m + 4$ and $n^4 = (7m+4)^2 = 7l+2$. Hence $n^4 + n^2 + 1 = 7l+2 + 7m + 4 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
If $n=7k \pm 3$, then $n^2 = (7k \pm 3)^2 = 7m + 2$ and $n^4 = (7m+2)^2 = 7l+4$. Hence $n^4 + n^2 + 1 = 7l+4 + 7m + 2 + 1 = 7(l+m+1)$ and hence $7|(n^4 + n^2 + 1) \Rightarrow 7|(n^7-n)$.
Hence, $7|(n^7-n)$.
Equivalently you could argue out that $7|(n^7-n)$ directly from Fermat's little Theorem.
Therefore, we have that $2|(n^7-n)$ and $3|(n^7-n)$ and $7|(n^7-n)$, $\forall n \in \mathbb{N}$.
Hence, $42|(n^7-n)$, $\forall n \in \mathbb{N}$.
As $42=2.3.7$. Therefore,we need to check that $n^7-n$ is divisible by $2,3$ and 7. For divisibility by $2$, by fermat's little theorem, $n^2=n\pmod 2 \implies {(n^2)}^3.n=n^4\pmod 2=n^2\pmod 2=n\pmod 2 \implies n^7-n=0\pmod2$. For divisibility by 3, $n^3=n\pmod 3\implies n^7=n^3\pmod 3=n\pmod 3 \implies n^7-n=0\pmod 3$. For divisibility by 7, $n^7=n\pmod 7 \implies n^7-n=0\pmod 7$. These relations implies that $42|(n^7-n)$.
Combinatorial Polynomial Approach
Since \begin{align} n^7-n &=5040\binom{n}{7}+15120\binom{n}{6}+16800\binom{n}{5}+8400\binom{n}{4}+1806\binom{n}{3}+126\binom{n}{2}\\ &=42\left[120\binom{n}{7}+360\binom{n}{6}+400\binom{n}{5}+200\binom{n}{4}+43\binom{n}{3}+3\binom{n}{2}\right] \end{align} Therefore, we have that for all $n\in\mathbb{Z}$, $$42\mid n^7-n$$
Little Fermat Approach
Little Fermat Theorem says $$n^7\equiv n\pmod7$$ and since $7\equiv1\pmod2$ $$n^7\equiv n\pmod3$$ and since $7\equiv2\pmod1$ $$n^7\equiv n\pmod2$$ Thus, $$n^7\equiv n\pmod{42}$$
• +1. But I have a question: how do you know that $$n^7-n =5040\binom{n}{7}+15120\binom{n}{6}+16800\binom{n}{5}+8400\binom{n}{4}+1806\binom{n}{3}+126\binom{n}{2}$$? Is there some special formula or theorem for this expansion? I saw a similar one for the case $30\mid n^5-n$ – Dr. Mathva May 4 at 15:34
• If we know the values of $f(n)=n^7-n=\sum\limits_{k=0}^7a_k\binom{n}{k}$ for $n=0$ to $n=7$, it is very easy to compute the coefficients $a_k$ since for $n\lt k$, $\binom{n}{k}=0$. Since $f(1)=f(0)=0$, we get $a_1=a_0=0$. Since $f(2)=126$, we get $a_2=126$. Since $f(3)=2184$, $a_3=2184-126\binom{3}{2}=1806$. Since $f(4)=16380$, $a_4=16380-1806\binom{4}{3}-126\binom{4}{2}=8400$. etc. – robjohn May 4 at 16:04
• Oh, see. Thanks ;) – Dr. Mathva May 4 at 16:06
• Sorry for asking again, but I've just noticed there's one last thing I don't understand. You claim "If we know the values of $\displaystyle f(n)=n^7-n=\sum^7_{k=0}a_k\binom{n}{k}$ [...]". How can we, however, prove that $f$ can be represented as this sum? – Dr. Mathva May 11 at 20:52
• If you look at the method for computing $a_n$, you will see that as long as $f(n)$ is integral for $8$ consecutive integers $n$, we get the $8$ coefficients for this sum. – robjohn May 12 at 0:52
F$\ell$T$^*$ (if $p\not\vert n$, then $p\vert n^{p-1}-1$) $\Rightarrow$
if $2\not\vert n$, then $2\not\vert n^6$, and then $2\vert (n^6)^1-1$
if $3\not\vert n$, then $3\not\vert n^3$, and then $3\vert (n^3)^2-1=n^6-1$
if $7\not\vert n$, then $7\vert n^6-1$
so $2,3,7\vert n^7-n$.
$^*$: $\ell$=little.
Just for completeness, here is induction (just for divisibility by 7) :
Claim : $n^7 - n$ is divisible by 7
Base Case: True for n = 1,2
Induction Step: Assume true for n = k. To prove true for n = k + 1.
Now, $$(k+1)^7 - (k+1) = k^7 + 7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k + 1 - k - 1 \\= (k^7 - k) + 7(k^6 + 3k^5 + 5k^4 + 5k^3 + 3k^2 + k)$$ We know by our assumption that $k^7 - k$ is divisible by 7. Therefore, $(k + 1)^7 - (k + 1)$ is divisible by 7.
This shows that $n^7 - n$ is divisible by 7.
To show divisibility by 2 and 3, unfortunately, one has to fall back on some of the earlier tricks. $$n^7 - n = n(n^6 - 1) = (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1)$$ $(n-1)n(n+1)$ is a product of three consecutive integers. Product of two consecutive integers is divisible by 2 and product of three consecutive integers is divisible by 3. Since, $(2,3) = 1$, product of three consecutive integers is divisible by 6. Since $(6,7) = 1$, $n^7 - n$ is divisible by 42.
QED
• Where is 42?... – lhf Jun 29 '12 at 11:24
• Good catch!! :P – TenaliRaman Jun 29 '12 at 14:58 | 2019-12-07T10:01:04 | {
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https://math.stackexchange.com/questions/2227885/smallest-subgroup-and-langle-a-rangle | # Smallest subgroup and $\langle a \rangle$.
Gallian says in Chapter 3 of Contemporary Abstract Algebra that
For any element $a$ of a group $G$, it is useful to think of $\langle a \rangle$ as the smallest subgroup of $G$ containing $a$.
But wouldn't this mean the set $\langle a \rangle$ is simply $\{ a, a^{-1}, e \}$?
• If it doesn't contain $a^2$, that isn't a subgroup. – Robert Israel Apr 10 '17 at 16:00
• @RobertIsrael Why not? – Airdish Apr 10 '17 at 16:03
• It needs to be closed under the group's operation. – Akiva Weinberger Apr 10 '17 at 16:58
• That isn't a group unless $a = a^{-1}$, in which case it would be $\{a, e\}$ only – q.Then Apr 10 '17 at 17:53
• @q.Then Actually that is a group if $a^2 = a^{-1}$. – fleablood Apr 10 '17 at 18:09
The set $\{a,a^{-1},e\}$ has an identity and is closed under inverses, but it isn't necessarily closed under the group's operation.
If $H$ is a subgroup and $a \in H$, you also need $a^2=aa\in H$ and $a^3=aaa \in H$ etc. Likewise, you need inverses for all of those elements (i.e. $a^{-1}, a^{-2}, \dots \in H$).
So closure under the group's operation and inverses require at least $\langle a \rangle = \{ a^n \;|\; n \in \mathbb{Z} \} \subseteq H$.
Once you know $\langle a \rangle$ is itself a subgroup. This shows that it is the smallest subgroup containing $a$.
• So what you're saying is that once you have $a$ in the subgroup, you automatically need its powers too because $a * a$ and $a *a * a$, etc. ought to be allowed operations under the group operation? – Airdish Apr 10 '17 at 16:05
• A group's operation is associative, has an identity, and inverses...but it also must be closed. If $a,b \in G$, then $ab \in G$ as well. So this means $a \in G$ then $a^2=aa \in G$ etc. So closure is what is forcing this. – Bill Cook Apr 10 '17 at 16:06
• Ah okay so the set $\{ a, a^{-1}, e \}$ isn't actually a group at all? – Airdish Apr 10 '17 at 16:08
• Not in general. For example: let $i=\sqrt{-1}$. Then $\{i,i^{-1},e\} = \{i,-i,1\}$ is not a group since multiplication isn't a closed binary operation operation on this set ($i^2=-1 \not\in \{i,-i,1\}$. Since we don't have a closed operation, it's not a group. – Bill Cook Apr 10 '17 at 16:12
• On the other hand, it could be. Consider $a=-1$ (with multiplication). Then $\{a,a^{-1},e\}=\{-1,(-1)^{-1},1\} = \{1,-1\}$ which is a perfectly good (cyclic) group (under multiplication). So $\{a,a^{-1},e\}$ can be a group, but usually it isn't. :) – Bill Cook Apr 10 '17 at 16:14
Groups are closed under their operations. In other words if $a,b \in G$ then we know that $ab \in G$.
So if $a\in <a>$ then we know that $a*a=a^2 \in <a>$ and $a^2*a = a^3 \in <a>$ and inductively we know that $a^2 \in <a>$ for all $n \in \mathbb N$. And as inverses must exist we know $a^{-n} \in G$.
So in general the group $<a> = \{.......a^{-3},a^{-2},a^{-1}, e,a,a^2,a^3,....\}$. But the $a^i$ need not be distinct. It could be possible for $a^i = a^k;i \ne j$. An example is $\mathbb Z_6=\{0,1,2,3,4,5\}$ under modulo addition. $2^5 = 2+2+2+2+2 = 4 = 2+2 = 2^2$. So $\{......2^{-3}=0,2^{-2}=2, 2^{-1}=4, 0, 2, 2^2 = 4, 2^3 = 0,....\} = \{0,2,4\}$ is not infinite.
Now it is possible that $a^k =e$ for some number $k$. Then $a^{-1} = a^{k-1}$ Example: In $\mathbb Z_6$, $2^3 = 2+2+2= 0$ and $2^{-1} = 4 = 2^{3-1}$. In this case $<a> = \{0, a, a^2, a^3, .... a^{k-1}\}$. Example $<2> \subset \mathbb Z_{6} = \{0,2,2^2=2^{-1}=4\}$
But it's also possible that $a^k \ne e$ for any $k \ne 0$. In this case $<a> = \{a^n|n \in \mathbb Z\}$. Example: If the group is $\mathbb Z$ under addition. Then $<7> = \{0, 7, -7, 14, -14, 21, -21, ....\} = \{7n| n \in \mathbb Z\}$. Which is infinite.
....
In general. If $a^n= e$ and $a^k\ne e; 1\le k < n$ we so "the order of $a$" is $n$ and we write it as $|a| = n$. If no such $n$ exists, we say $|a| = \infty$.
If $|a| = n$ then $<a> =\{0,a, a^2, ....., a^{n-1}\}$.
If $|a| = \infty$ then $<a> = \{a^k|k \in \mathbb Z\}$.
Notice $<a>$ always has $|a|$ units. | 2019-08-20T21:31:15 | {
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https://mathhelpboards.com/threads/understanding-limits.6134/ | # Understanding limits
#### Rido12
##### Well-known member
MHB Math Helper
I know this is probably a dumb question, but I have a question regarding this. My textbook says the following: "A function f is differentiable at a if f'(a) exists."
It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.
My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.
EDIT:
Sorry if this is confusing, and if you want me to clarify, I can provide a hypothetical situation.
Ok, time to clarify what I'm trying to convey with a hypothetical situation (not real)
We are asked to see if f(x) = |x| is differentiable at x = 0. Let's PRETEND that the limit exists at f'(a). That when you took the limit from the positive and negative side of it, it resulted in 5. However, it is actually undefined at 5, but the limit as you approach f'(a) is 5. By using the same logic as above, the logic that the textbook example used, we are to assume that YES, It is differentiable, BECAUSE the limit exists! However, that is actually NOT the case, since the by the definition, f'(a) MUST exists. But in this example, YES, the limit exists, but f'(a) doesn't exist as it is undefined at a.
That being said, would additional steps be taken/required, in order to prove if something is or is not differentiable. As we see in my hypothetical situation, just determining the limit was not enough.
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#### Bacterius
##### Well-known member
MHB Math Helper
[JUSTIFY]Okay, this is hurting my brain so I'll conservatively assert that:
1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.
2. If the two limits are different, then for most well-behaved functions the derivative does not exist at that point, however there are may exist crazy analytical functions which are continuous while their derivative isn't, and vice versa, so it may not be a sufficient condition to show the existence of $f'(a)$.[/JUSTIFY]
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#### Jameson
Staff member
This is a really good question.
[Rest of post deleted due to a mistake. Updated thought can be found in my later post]
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Hey Jameson! I think what you said is not entirely correct.
This is a really good question.
In order for a function of one variable, $f(x)$, to be differentiable at a point $a$, then it must be be continuous at $a$ and the following limit must exist ...
I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.
2) Just knowing that the above limit exists is not enough to say that $f(x)$ is differentiable at $a$. ...
I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.
Please correct me if I am wrong or if I have misinterpretted your response.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
I know this is probably a dumb question
This is not a dumb question at all.
My textbook says the following: "A function f is differentiable at a if f'(a) exists."
It then follows with and example regarding if f(x) = |x| is differentiable at x = 0. They prove this by finding the limit of its derivative, and then splits it in two equations: for the limit as h -> 0+ and h -> 0-. Finally concluding that it is not differentiable as the limits are different. Done.
My question is as follows: does the method, as demonstrated above, work for all functions when you're trying to find if a certain point is differentiable? I question it because even if the aforementioned limit DOES exist, it doesn't mean that f'(a) exists. It just means that it has a limit, as you can have a limit of something when f'(a) is undefined. Thus, making the method inadequate to making such a conclusion.
I think you are making mistakes here. We write $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ whether the limit exists or doesn't exist. When the limit doesn't exist, $f'(a)$ is not defined. When the limit $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists, then $f'(a)$ is same as this limit. So $f'(a)$ is a real number and certainly it exists.
Note that when $\lim_{x\to a}\frac{f(x)-f(a)}{t-a}$ exists then this doesn't say that $f'(a)$ has a limit. This would have no meaning. It says that $\frac{f(x)-f(a)}{x-a}$ has a limit as $x$ approaches $a$. For a short hand notation, and for furthering the theory, we denote this limit, whether it exists or not, as $f'(a)$.
I have not read the content in Spoiler box thoroughly yet. Tell me if you have further doubts or you need to discuss.
EDIT: I have implicitly assumed that $f$ is a real function having an interval $I$ in its domain such that $a\in I$.
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#### Bacterius
##### Well-known member
MHB Math Helper
He is taking the limit of the derivative, not the original function itself.
#### Jameson
Staff member
Hey Jameson! I think what you said is not entirely correct.
Yep, I think I need to edit a couple of things.
I think the thing in bold is redundant. The continuity of $f$ at $a$ follows from the differentiability of $f$ at $a$.
Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.
I think knowing that the limit exists equivalent to $f$ being differentiable at $a$. That is by definition of differentiability at a point. Moreover, strictly, we should not say that $f(x)$ is differentiable at $a$ but simply say that $f$ is differentiable at $a$.
Please correct me if I am wrong or if I have misinterpreted your response.
About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course.
Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.
I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.
#### Jameson
Staff member
He is taking the limit of the derivative, not the original function itself.
Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
He is taking the limit of the derivative, not the original function itself.
Umm.. now I am totally confused. If you understand his doubt then can you please frame it nicely and post it along with your solution?
#### Bacterius
##### Well-known member
MHB Math Helper
Are you sure? I think caffeinemachine addressed his terminology in a way that makes sense. It looks to me like the we're talking about the limit definitions of functions.
That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)
At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.
#### Rido12
##### Well-known member
MHB Math Helper
I haven't read all your responses yet, but it is the limit of the derivatvive. I don't knwo anymore haha, I'm too confused.
The textbook does this:
f'(x) = lim h-> 0 (|0 + h| - |0|)/ h
Then split in two parts to compute the left and right limits separetely.
#### Jameson
Staff member
I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of the derivative". Your example seems to confirm this.
#### Rido12
##### Well-known member
MHB Math Helper
Yep, I think I need to edit a couple of things.
Yes I agree that continuity is a necessary condition of differentiability thus differentiability implies continuity, but the converse is not true. That's what I meant.
About the part in bold - yes I agree with you and would normally write that but thought the OP might be used to seeing functions always written in the form "$f(x)$" so wrote it that way. You're right of course.
Now that I think about it more the if the limit exists then it will be continuous. I was thinking of a situation like $f(x)=\frac{x^2-4}{x-2}$ at $x=2$ where the derivative appears to exist by the limit definition but isn't continuous. However now I see that this won't have a limit for the limit definition of a derivative so the extra stipulation of being continuous isn't necessary as it's already implied by the existence of the limit definition of a derivative.
I suppose the main idea I'm expressing is differentiability implies continuity but the opposite is not true.
It's quite strange actually. I always do see functions written in teh form "f(x)" rather than just "f"... My textbook uses "f", so now I do that too. But my previous math teachers use "f(x)"
- - - Updated - - -
I'm pretty sure you don't mean "the limit of the derivative" but the "limit definition of a derivative". Your example seems to confirm this.
I'm confused. OH. My example was just finding the derivative of a function right, by taking its limit right? I got confused because I don't usually see the word "limit" when I take derivatives using power rule, chain, etc, and I got mixed up thinking it was the limit of a derivative.
And I haven't finished reading the responses yet from the previous page.
#### Rido12
##### Well-known member
MHB Math Helper
That was my first interpretation as well. But after rereading his example it actually seems that he is talking about taking the two-sided limit of the derivative of $f$ to establish whether the derivative exists at a given point (so the question would become: is the derivative being (non-)continuous at $x = a$ a sufficient condition for the (non-)existence of the derivative at $x = a$?)
At least that's my take on it. I figured the question was deeper than just "being continuous is necessary but not sufficient for differentiability". But it's possible I am just talking nonsense here haven't done any formal calculus in a while.
This is precisely what I meant with my question; I don't know if that is the limit definition of derivatives, or whatnot. Now if only my original post was as concise as how you explained it
EDIT: second thought, brain hurts, not sure if that's how i meant it but:
It's just that I'm not convinced that f'(a) exists by just taking its limit from both sides. And it must exist for it to be differentiable at that point.
This:
1. If the two limits are the same, you are correct that this says absolutely nothing about the existence of the derivative at $x = a$, so if the limits are the same this is as far as this method will take you.
I think the notation is confusing me; the limit definition of derivatives has both "f'(x)" and "lim" so it makes me want to think that it's taking the LIMIT of the DERIVATIVE, but now that I think about it, it's simply just taking the derivative.
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#### Rido12
##### Well-known member
MHB Math Helper
This leads me to the following question:
When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.
Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?
For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0+ |x|= |h|/h = 1
f'(x)=lim as h ->0- |x|= |h|/h = -1
Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
#### Bacterius
##### Well-known member
MHB Math Helper
This leads me to the following question:
When would it be necessary to use the limit definition of a derivative, and when to just solve for the derivative using chain/product, etc rule? It is to my understanding that the limit definition was used to derive the rules mentioned before.
Isn't the limit definition of a derivative the same as finding the derivative and then apply the limit? So the limit of the derivative?
For example, finding the limit (if there is) for f(x) = |x|using the limit definition of a derivative
We compute the two side limit and we get:
f'(x)=lim as h >0+ |x|= |h|/h = 1
f'(x)=lim as h ->0- |x|= |h|/h = -1
Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all)
I am not sure what you mean though - what would be the purpose of calculating such a limit?
The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because calculating derivatives from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic. The idea is that the limit definition gives the most "basic" definition of what a derivative is, which is important in mathematics (where something without a proper definition is useless).
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#### Rido12
##### Well-known member
MHB Math Helper
If you already have an expression for the derivative then you don't need to use the limit definition. You can just evaluate it directly (which you did, since you did not use "h" at all)
I am not sure what you mean though - what would be the purpose of calculating such a limit?
The chain, product rules etc.. can all be derived from the limit definition. They are just much more convenient to use, because deriving from the limit form (the so-called "first principles") gets rather tedious for anything bigger than a quadratic.
Do you also mean to imply that finding the limit here is redundant?
Instead, wouldn't it be possible to find the derivative first, then apply the limit?
So: d/dx[|x|] = (skipping steps) 2x / 2|x|= x / |x|
Hence, apply limit as h >0+ x / |x|
= x / x = 1
limit as h >0- x / |x|
= x / -x = -1
So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0? (Instead of doing all the two side limit hassle that was only necessary when utilizing the limit definition of a derivative)
#### Bacterius
##### Well-known member
MHB Math Helper
Do you also mean to imply that finding the limit here is redundant?
So we could use direct substitution instead of finding the limit, meaning that once we get f'(x) = x / |x| we can do f'(0) = 0/ 0 , meaning that it is not possible to differentiate at x= 0?
Well the same holds true for the function itself. Before differentiating your function you should rigorously show that it is in fact continuous at the locations you want to differentiate it at, otherwise the function is not differentiable there and you cannot even bring up the concept of derivative (it is meaningless). In this case, our function $f(x) = |x| / x$ has a discontinuity at $x = 0$ (you can rigorously show that by taking the left and right limits of the function around that point and demonstrating that $\lim_{x \to 0} f(x)$ does not exist as they are not the same ($-1$ and $+1$ respectively).
Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.
You need to be very careful when differentiating. Just because you get a nice expression for your derivative everywhere except at, say, $x = 0$, does not mean that expression is valid for $x = 0$! This is why it's important to keep track of the domain of each function you are working with. If you find that your derivative is valid for all $x \neq 0$, then that's it - end of story. It is meaningless to plug $x = 0$ into it, the results are undefined because the derivative doesn't apply at that point.
Also, limits don't have much to do with derivatives in the grand scheme of things, but they provide a useful framework on top of which to build the concept of derivative and differentiability (as well as a lot of other important stuff). Don't associate them with limits too much.
Here the "left and right limits" are used to show that the two-sided limit exists, and additionally show continuity. The limit expression of the derivative is different and has a geometrical interpretation, which is basically "take two points on a curve and draw a line between them, as the points get closer and closer together that line starts to approximate a tangent to the curve, which happens to also give the rate of change of the height of the curve with horizontal distance and lots of other useful things".
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#### Rido12
##### Well-known member
MHB Math Helper
Once you know that your function is continuous at that point, then you may try to differentiate it. But that won't necessarily mean the derivative exists! Consider $g(x) = |x|$, it is continuous at $x = 0$ (left and right limits are the same) but it is not differentiable at $x = 0$.
Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .
#### Jameson
Staff member
You already showed that the left and right hand limits at $x=0$ are 1 and -1, which is all you need to do. That shows that the limit doesn't exist thus $g$ is not differentiable at $x=0$. Once again, differentiability implies continuity but continuity does not imply differentiability.
#### Rido12
##### Well-known member
MHB Math Helper
So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontious at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.
Is there anythign wrong with the procedure above?
#### Bacterius
##### Well-known member
MHB Math Helper
Thanks! So I guess the last question is how would I prove that g(x) is not differentiable at x = 0. Say you prove that g(x) is continuous at 0. How would I prove that it is not differentiable at x = 0 .
You take the definition of "differentiable":
A function $f(x)$ on $\mathbb{R}$ is differentiable at $x = a$ if and only if $f'(a)$ exists.
From that definition you can choose different approaches depending on your function. For instance, take $f(x) = |x|$. Then, assume $x < 0$, where the function is (probably) differentiable - we'll prove it is shortly. Now apply the definition of the derivative:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
And we get:
$$f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h}$$
Now because $x < 0$, $|x| = -x$, and we can rewrite this as:
$$f'(x) = \lim_{h \to 0} \frac{|x + h| + x}{h}$$
Now assume $x + h > 0$. But this means that $h > -x$, and since $x$ is negative then $-x$ is positive so $h$ is greater than some non-zero value.. which is a contradiction since we assumed that $h \to 0$. Therefore $x + h < 0$, so $|x + h| = - (x + h) = -x - h$ and:
$$f'(x) = \lim_{h \to 0} \frac{-x - h + x}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 ~ ~ ~ \text{for} ~ ~ x < 0$$
A similar reasoning for $x > 0$ leads to:
$$f'(x) = +1 ~ ~ ~ \text{for} ~ ~ x > 0$$
So we have:
$$f'(x) = \begin{cases} +1 ~ ~ &\text{for} ~ ~ x > 0 \\ -1 ~ ~ &\text{for} ~ ~ x < 0 \end{cases}$$
What now? Well now the left and right limits around $x = 0$ are easy (note we are never actually touching zero, $f'(x)$ has not been defined at that point) and we clearly see that:
$$\lim_{x \to 0^{-}} f'(x) \ne \lim_{x \to 0^{+}} f'(x)$$
And we may conclude that:
$$\lim_{x \to 0} f'(x) ~ ~ ~ \text{does not exist} ~ ~ ~ \implies ~ ~ ~ f'(0) ~ ~ ~ \text{does not exist}$$
And therefore by the definition of differentiability, $f(x)$ is not differentiable at $x = 0$.
That was the semi-rigorous, long-winded way. In practice you use things like the chain rule and various limit/derivative theorems (or even plain observation) to help you simplify the function instead of starting from scratch each time.
For instance, using the result above, you now know that $f(x) = |x|$ is not differentiable at $x = 0$. From that point on you can easily show that $f(x) = |x - a|$ is not differentiable at $x = a$, for all $a \in \mathbb{R}$, and you can now use that fact to find other, more complicated derivatives involving absolute values, using the chain rule, product rule, and so on.
#### Jameson
Staff member
So Step 1 is to see to determine its continuity, if the function is continuous, we can proceed to find the derivative.
Step 2, we find the limit of both sides of the derivative at the point we want to prove, say x = 0 , and we find that the limit doesn't exist
Step 3, we can make a conclusion that because the limit doesn't exist, g'(x) is discontinuous at the point and impossible to differentiate. for a function to be differentiable, it must be continuous, but the converse is not true.
Is there anything wrong with the procedure above?
I hope someone who is more versed in analysis stops by to comment on this, but I will add what I can.
1) Yes, this is a good start. There are other things to consider as well, such a bend, cusp or vertical tangent - but your step #2 will sort those out anyway. You'll develop an eye for noticing these things quickly.
2) It's always a good idea to be able to use the limit definition of a derivative (left and right hand limits as you said) but of course in practice you have quicker ways of calculating derivatives, as you know. You can probably tell from the problem if the goal is to quickly calculate the derivative and use that as part of a greater problem, or if the differentiability itself is the main focus of the problem. Anyway, this step is ok.
3) If the limit doesn't exist, it doesn't automatically follow that the function is not continuous at that point. $f(x)=|x|$ is a great example of this. It is continuous at $x=0$ but not differentiable.
The only thing that I am not certain of is whether step 1 is implied by step 2. Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.
Put another way, if we have a function of one real variable, $f$, can the limit following limit exist if the point, $a$ is not in the domain of $f$?
$$\displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
I think not because $f(a)$ is not defined. So that leads me to say that step 2 implies step 1.
#### Rido12
##### Well-known member
MHB Math Helper
Last night I thought there could be an example of a function and a point where the limit definition of a derivative results in an answer for that point but the function was actually discontinuous at the point, so the derivative there couldn't exist. However, I'm not able to come up with an example of this and doubt that it's possible now. If anyone could that would be interesting.
This problem was what compelled me to start this thread in the first place. It occurred to me "what if the limit definition of a derivative results in an answer, but that point was actually discontinuous". I'm also interested if there such as a function as well, as it will clear my problems haha. I agree with you on the premise that step 2 implies step 1.
Thanks guys for the help!
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#### MarkFL
$$\displaystyle f(x)=\frac{x}{|x|}$$
It has a jump discontinuity at $x=0$, however, we find:
$$\displaystyle \lim_{h\to0^{-}}\frac{f(x+h)-f(x)}{h}=0=\lim_{h\to0^{+}}\frac{f(x+h)-f(x)}{h}$$ | 2022-01-21T07:15:26 | {
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http://stats.stackexchange.com/questions/88980/why-on-average-does-each-bootstrap-sample-contain-roughly-two-thirds-of-observat | # Why on average does each bootstrap sample contain roughly two thirds of observations?
I have run across the assertion that each bootstrap sample (or bagged tree) will contain on average approximately $2/3$ of the observations.
I understand that the chance of not being selected in any of $n$ draws from $n$ samples with replacement is $(1- 1/n)^n$, which works out to approximately $1/3$ chance of not being selected.
What is a mathematical explanation for why this formula always gives $\approx 1/3$ ?
-
I believe this is the origin of the $.632$ in the bootstrap 632+ rule. – gung Mar 6 at 2:46
Essentially, the issue is to show that $\lim_{n\to\infty}(1- 1/n)^n=e^{-1}$
(and of course, $e^{-1} =1/e \approx 1/3$, at least very roughly).
It doesn't work at very small $n$ -- e.g. at $n=2$, $(1- 1/n)^n=\frac{1}{4}$. It passes $\frac{1}{3}$ at $n=6$, passes $0.35$ at $n=11$, and $0.366$ by $n=99$. Once you go beyond $n=11$, $\frac{1}{e}$ is a better approximation than $\frac{1}{3}$.
The grey dashed line is at $\frac{1}{3}$; the red and grey line is at $\frac{1}{e}$.
Rather than show a formal derivation (which can easily be found), I'm going to give an outline (that is an intuitive, handwavy argument) of why a (slightly) more general result holds:
$$e^x = \lim_{n\to \infty} \left(1 + x/n \right)^n$$
(Many people take this to be the definition of $\exp(x)$, but you can prove it from simpler results such as defining $e$ as $\lim_{n\to \infty} \left(1 + 1/n \right)^n$.)
Fact 1: $\exp(x/n)^n=\exp(x)\quad$ This follows from basic results about powers and exponentiation
Fact 2: When $n$ is large, $\exp(x/n) \approx 1+x/n\quad$ This follows from the series expansion for $e^x$.
(I can give fuller arguments for each of these but I assume you already know them)
Substitute (2) in (1). Done. (For this to work as a more formal argument would take some work, because you'd have to show that the remaining terms in Fact 2 don't become large enough to cause a problem when taken to the power $n$. But this is intuition rather than formal proof.)
[Alternatively, just take the Taylor series for $\exp(x/n)$ to first order. A second easy approach is to take the binomial expansion of $\left(1 + x/n \right) ^n$ and take the limit term-by-term, showing it gives the terms in the series for $\exp(x/n)$.]
So if $e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n$, just substitute $x=-1$.
Immediately, we have the result at the top of this answer, $\lim_{n\to\infty}(1- 1/n)^n=e^{-1}$
As gung points out in comments, the result in your question is the origin of the 632 bootstrap rule
e.g. see
Efron, B. and R. Tibshirani (1997),
"Improvements on Cross-Validation: The .632+ Bootstrap Method,"
Journal of the American Statistical Association Vol. 92, No. 438. (Jun), pp. 548-560
-
More precisely, each bootstrap sample (or bagged tree) will contain $1-\frac{1}{e} \approx 0.632$ of the sample.
Let's go over how the bootstrap works. We have an original sample $x_1, x_2, \ldots x_n$ with $n$ items in it. We draw items with replacement from this original set until we have another set of size $n$.
From that, it follows that the probability of choosing any one item (say, $x_1$) on the first draw is $\frac{1}{n}$. Therefore, the probability of not choosing that item is $1 - \frac{1}{n}$. That's just for the first draw; there are a total of $n$ draws, all of which are independent, so the probability of never choosing this item on any of the draws is $(1-\frac{1}{n})^n$.
Now, let's think about what happens when $n$ gets larger and larger. We can take the limit as $n$ goes towards infinity, using the usual calculus tricks (or Wolfram Alpha): $$\lim_{n \rightarrow \infty} \big(1-\frac{1}{n}\big)^n = \frac{1}{e} \approx 0.368$$
That's the probability of an item not being chosen. Subtract it from one to find the probability of the item being chosen, which gives you 0.632.
-
Sampling with replacement can be modeled as a sequence of binomial trials where "success" is an instance being selected. For an original dataset of $n$ instances, the probability of "success" is $1/n$, and the probability of "failure" is $(n-1)/n$. For a sample size of $b$, the odds of selecting an instance exactly $x$ times is given by the binomial distribution:
$$P(x,b,n) = \bigl(\frac{1}{n}\bigr)^{x} \bigl(\frac{n-1}{n}\bigr)^{b-x} {b \choose x}$$
In the specific case of a bootstrap sample, the sample size $b$ equals the number of instances $n$. Letting $n$ approach infinity, we get:
$$\lim_{n \rightarrow \infty} \bigl(\frac{1}{n}\bigr)^{x} \bigl(\frac{n-1}{n}\bigr)^{n-x} {n \choose x} = \frac{1}{ex!}$$
If our original dataset is big, we can use this formula to compute the probability that an instance is selected exactly $x$ times in a bootstrap sample. For $x = 0$, the probability is $1/e$, or roughly $0.368$. The probability of an instance being sampled at least once is thus $1 - 0.368 = 0.632$.
Needless to say, I painstakingly derived this using pen and paper, and did not even consider using Wolfram Alpha.
-
This can be easily seen by counting. How many total possible samples? n^n. How many NOT containing a specific value? (n-1)^n. Probability of a sample not having a specific value - (1-1/n)^n, which is about 1/3 in the limit.
- | 2014-11-26T12:37:25 | {
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https://math.stackexchange.com/questions/1070992/is-it-ok-to-evaluate-improper-integrals-this-way/1071041 | # Is it OK to evaluate improper integrals this way?
Today in class we learned that when you have an improper integral like this one:
$$\int_{-\infty}^\infty {f(x)} \: dx$$
you must split it before you do the limits (like so):
$$\lim_{a \to \infty} \int_{-a}^n {f(x)} \: dx + \lim_{a \to \infty} \int_{n}^a {f(x)} \: dx$$
I asked if this would work:
$$\lim_{a \to \infty} \int_{-a}^a {f(x)} \: dx$$
And my teacher said it wouldn't work, but he wouldn't explain why. Why wouldn't that work? An example of a function that doesn't work would be especially nice.
• – beep-boop Dec 16 '14 at 19:11
• I tried the same thing, and it didn't work in a case. That's what convinced me I couldn't do this – Justin Dec 16 '14 at 22:23
The problem is that limiting to both infinities at the same time can overlook divergence.
However, the value: $\lim\limits_{a \to \infty} \int_{-a}^a f(x)\,dx$ (if it exists) is important enough to have a name: the principal value.
So why can't you just do the limit like this? Consider $\int_{-\infty}^\infty x\,dx$
Notice that $\int_0^\infty x\,dx$ diverges to $\infty$ and $\int_{-\infty}^0 x\,dx$ diverges to $-\infty$. So we get (from either side) that our integral diverges.
However, $\lim\limits_{a \to \infty} \int_{-a}^a x\,dx = \lim\limits_{a \to\infty} a^2/2 - (-a)^2/2 = \lim\limits_{a \to \infty} 0 = 0$.
So the principal value of $\int_{-\infty}^\infty x\,dx$ is zero while the integral itself diverges.
In general, IF an improper integral converges, you can be very sloppy and get the right answer. On the other hand, if it actually diverges, sloppiness can miss this!
As I tell my students: You must approach ONE side of EACH bad spot by itself. Then put your answer together.
Addendum: Suppose $\int_{-\infty}^c f(x)\,dx$ and $\int_c^\infty f(x)\,dx$ exist.
Then $\int_{-\infty}^\infty f(x)\,dx = \lim\limits_{a \to -\infty} \int_a^c f(x)\,dx + \lim\limits_{b \to \infty} \int_c^b f(x)\,dx = \lim\limits_{b \to \infty} \int_{-b}^c f(x)\,dx + \lim\limits_{b \to \infty} \int_c^b f(x)\,dx$
Now because those two limits exist, we can combine them. So we get...
$=\lim\limits_{b \to \infty} \int_{-b}^c f(x)\,dx + \int_c^b f(x)\,dx =\lim\limits_{b \to \infty} \int_{-b}^b f(x)\,dx$
...which is the principal value. So the improper integral and principal value match when the integral actually converges.
• And as further evidence that using $\lim\limits_{a \to \infty} \int_{-a}^a x\,dx$ is wrong, consider that $\lim\limits_{a \to \infty} \int_{-a}^{a+1} x\,dx$ obviously diverges, but there's no reason why we couldn't just as well use that to represent $\int_{-\infty}^\infty x\,dx$. Being more symmetrical doesn't make the first one more right ;-) – Steve Jessop Dec 17 '14 at 9:53
• @SteveJessop Exactly. That's what I tried explain in my answer. – Eff Dec 17 '14 at 11:20
(There is already a good answer, but I thought I'd add something.)
We say that a limit exists, if no matter how we approach it approaches the same. For example $$-1 = \lim_{x\to0^-}\frac{|x|}{x}\neq\lim_{x\to 0^+}\frac{|x|}{x} = 1$$ and hence the limit $\lim_{x\to0}|x|/x$ does not exist. In a similar way, we have to consider any way the interval of integration $[a,b]$ grows, where $a\to-\infty$ and $b\to\infty$.
Consider for example \begin{align} &\int_{-n}^{2n} x\,\mathrm{d}x =\frac{3}{2}n^2 &\text{and }\quad &\int_{-n}^n x\,\mathrm{d}x = 0.\end{align} As $n$ approaches $\infty$ the interval is $[-\infty,\infty]$ for both integrals, but approaching in different ways. In this example we have that the first integral goes to infinity, however the second integral seems to converge? As I said, a limit has to be the same no matter how we approach it, so the integral $$\int_{-\infty}^\infty x\,\mathrm{d}x$$ actually diverges (which it does not using "your" method). I hope this helps you see why it is important to consider the end-points of the interval independently.
The only case in which you can not define the integral that way is when $f$ is not integrable. In terms of the Lebesgue integral you can check this for positive functions.
Wikipedia mentions $$f(x) = \left\{ \begin{array} \;1 \; \; x > 0 \\ -1 \;\;x < 0 \end{array} \right. .$$
You can check that this has $0$ at each term in the sequence $\lim\limits_{a \rightarrow \infty} \int\limits_{-a}^a f(x) dx$, but is clearly not integrable by the definition of the Riemann integral or the Lebesgue integral.
If there is a " discontinuity " at the point "n" , then combining the two integrals , will not (may not) give the correct result. | 2021-05-15T09:16:00 | {
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https://www.intmath.com/forum/methods-integration-31/integration-techniques:161 | # Integration Techniques [Solved!]
### My question
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
### Relevant page
4. Integration: Basic Trigonometric Forms
### What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
X
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
Relevant page
<a href="https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php">4. Integration: Basic Trigonometric Forms</a>
What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
## Re: Integration Techniques
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
X
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
## Re: Integration Techniques
It does. A good example to use. Thank you.
X
It does. A good example to use. Thank you.
You need to be logged in to reply.
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https://math.stackexchange.com/questions/2203287/system-of-equation-complicated | System of Equation. (Complicated)
Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt, I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.
• Hint: $\;\frac{12}{5x}+\frac{4}{5y}=\cdots$ – dxiv Mar 26 '17 at 3:01
• Can't you equate through $x^2 + y^2$ term and then isolate x (or y) and then sub back in and solve? – 123 Mar 26 '17 at 3:02
As dxiv has given hint:
$$\frac{12}{5x} + \frac{4}{5y} = 2 \implies x = \frac{6y}{5y-2}$$
Now put this in any one of the relation to make the relation full in terms of $y$.
Multiply equation (1) by y and equation (2) by x,
$$5xy(1+\frac{1}{x^2+y^2})=12y$$
$$5xy(1-\frac{1}{x^2+y^2})=4x$$
$$10xy = 12y + 4x$$
You can find y in terms of x or x in terms of y.
• You can try substitution. May be that is better option. – Kanwaljit Singh Mar 26 '17 at 3:55
Just to provide another way.
Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \sin(\theta)=\frac{4 \rho }{5 \left(\rho ^2-1\right)}$$ Now, using $$\sin^2(\theta)+\cos^2(\theta)=1 \implies \frac{16 \rho ^2}{25 \left(\rho ^2-1\right)^2}+\frac{144 \rho ^2}{25 \left(\rho ^2+1\right)^2}=1$$ Now, let $t=\rho ^2$ to have $$\frac{16 t}{25 \left(t -1\right)^2}+\frac{144 t}{25 \left(t+1\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients.
Factoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \left(5 t^2-6 t+5\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\frac 15\qquad t_2=5$$ then $$\rho_1=\frac 1 {\sqrt 5}\qquad \rho_2=\sqrt 5$$ and now compute the corresponding $\theta$'s from $$\tan(\theta)=\frac{\sin(\theta) }{\cos(\theta) }=\frac{t+1}{3 (t-1)}=-\frac 12$$.
• Up for this answer – Mathxx Mar 26 '17 at 6:09
$$5x(1+\frac{1}{x^2+y^2})=12\cdots \cdots (1)$$
$$5y(1-\frac{1}{x^2+y^2})=4\cdots \cdots \cdots \cdots (2) \times i$$
$$5(x+iy)+5\cdot \frac{x-iy}{(x+iy)(x+iy)}=12+4i$$
put $z=x+iy$ and $\bar{z} = x-iy$ and $|z|^2= z \cdot \bar{z} = x^2+y^2$
So $$5z+\frac{5\bar{z}}{z \cdot \bar{z}}=12+4i$$
So $$5z+\frac{5}{z} = 12+4i$$
So $$5z^2-(12+4i)z+5=0$$ after solving $$z=\frac{12+4i\pm \sqrt{(12+4i)^2-100}}{10} = \frac{12+4i\pm (8+6i)}{10} = 2+i\;,\frac{2-i}{5}$$
So $$z= x+iy = 2+i\;,\frac{2-i}{5}$$
So $$(x,y) = \left\{(2,1)\;\;, \left(\frac{2}{5},-\frac{1}{5}\right)\right\}$$
First of all get rid of $x^2+y^2$
$$\frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy$$
It is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $x= t, y= 1/t$ when axes are asymptotes. So solve for $x,y$
$$x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$
The asymptotes are
$$x= \frac65 ,\, y= \frac{2}{5}$$
We take hyperbola in the form
$$( x-\frac65 ) (y- \frac25) = k$$
Since it goes through the origin, $k$ must be the product $k=\dfrac65 \cdot \dfrac25 = \dfrac{12}{25}$ so as not to leave behind any constant term.
We can now find $x,y$ upto a parameter $t$
$$( x-\frac65 ) (y- \frac25) = \frac{12}{25}$$
$$\rightarrow x= \frac65 + t \sqrt{12}/5,\, y= \frac25 + \sqrt{12}/{5t},\,$$ | 2019-11-21T04:27:43 | {
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https://www.physicsforums.com/threads/algebra-question.714911/ | # Algebra Question
Hello,
I wanted to confirm that I am understanding something correctly. I have been doing this all the time, and have never had any problems with it. For some reason now I am wanting to confirm it.
So, if you have something such as: x3-6+y2-√z it would be equivalent to write it as: y2+x3-√z-6
Can you always rearrange terms like this, no matter what they are? What I do is write it as an addition problem so I can change the order, since you can change the order in addition.
For example if you have: log(x)-x3 you could write it as log(x)+(-x3) then as -x3 + log(x).
Just as -11+5-3+5=-4 would be the same as 5-11-3+5=-4 or -11-3+5+5=-4
Does my reasoning seem correct?
Thanks
Last edited:
Yes that is correct. I believe that falls under the commutative property which essentially states
x + y = y + x
And this would also go for multiplication, but not division correct?
Correct.
Example using decimal notation:
2*4 = 8 and 4*2 = 8
8 = 8 TRUE
2/4 = .5 and 4/2 = 2
.5 = 2 FALSE
And here's a tip, never be afraid to go back to the most basic of examples to solidify your thoughts if you start question them. Forget the log(x) and y^3 stuff and just use a scratch piece of paper and walk through simple things like 1+2 = 2+1 to convince yourself that the order of addition does not matter.
arildno
Homework Helper
Gold Member
Dearly Missed
Hi, ThomasMagnus!
As the others have said, you are correct.
The two operations multiplication and addition are COMMUTATIVE, i.e, you can change the order in which they appear.
Division and subtraction are NOT commutative, we can call them ANTI-commutative if you like.
But, as you yourself noted, can't we in 5-4 rewrite this as 5+(-4)=(-4)+5?
Yes, we can, by using negative numbers in addition, we can think of the operation of subtraction as "redundant", i.e, the info is contained in the use of adding with negative numbers instead.
That makes for easier calculations, because you then can go about commuting as much as you want!
In 5-(-4), for example, we just write 5-(-4)=5+(-(-4))=(-(-4))+5 and so on.
Does something similar exist with division relative to multiplication?
Yes, because division by a non-zero number "a" can always be thought of as multiplication with the reciprocal of "a", i.e, 1/a.
Thus, 2/4=2*(1/4)=(1/4)*2.
----------------
By thinking in terms of "negative numbers" and "reciprocals", you really have only two basic operations to worry about, addition and multiplication, rather than four operations.
(And, if you think of multiplication as repeated addition, there's only one fundamental operation to worry about..)
HallsofIvy
Homework Helper
And this would also go for multiplication, but not division correct?
Yes, and note that it is largely because (well, perhaps even more that addition and multiplication are "associative" while subtraction and division are not) that we do not consider subtraction and division as separate "operations" but just the inverses of additon and multiplication. That is, that "a subtract b" is just "a plus the additive inverse of b" and "a divided by b" is "a multiplied by the multiplicative inverse of b".
arildno
Homework Helper
Gold Member
Dearly Missed
When we think of ONE operation to be performed, commutativity is the only property that comes into the picture.
However, when we have two or more operations to be performed, we must also take into account what we call the associativity of the operations, that is, what relevance it has which operation we perform first.
We have, for example, for numbers "a", "b" and "c" that:
(a+b)+c=a+(b+c), that is, it doesn't matter WHICH addition we perform first. Addition is associative.
But, is:
(a-b)-c=a-(b-c)??
You easily see that subtraction is NOT associative at all;
We have that (a-b)-c=a-(b+c), whereas a-(b-c)=(a-b)+c
Thank you for the info! So the next time I see something in an answer key as something along the lines of x-b+c and I have -b+c+x I can assume they are the exact same, correct? Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6 ?
Thanks!
arildno
Homework Helper
Gold Member
Dearly Missed
"Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6"
That's right!
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https://matheducators.stackexchange.com/questions/5847/is-there-any-other-procedure-to-find-the-square-root/5861 | # Is there any other procedure to find the square root?
If no calculator is allowed, and we want to find the square root of a square number if it is large and analyzing to prime factors is hard, how can one proceed? For example, what to do if the number is a square of a large prime number like 5329?
What are methods to perform this and what are pedagogical reasons to prefer one over the other?
• I agree with @BenjaminDickman. This question is more for MSE. I wonder if it could be rewritten to explore any didactical issues or motivations. – Pablo B. Nov 14 '14 at 21:32
• well friends it is just by luck that I wrote the number to be 5329 .. the concept remains the same .. how to find the root of large numers .. I really voted for matin for the link he supplies. thank you all – Abdallah Abusharekh Nov 14 '14 at 21:37
• Related: hsm.stackexchange.com/q/56/72 – Joel Reyes Noche Nov 14 '14 at 23:39
• Following the (justified) remarks by @PabloB. and Benjamin Dickman I made explicit the educational aspect in the question, which is already present in some of the answers. – quid Nov 15 '14 at 13:26
• There're many methods on wikipedia page. – Ruslan Nov 15 '14 at 20:59
You can use (basically) Newton's method.
Take $x_0>0$, and define $$x_{k+1} = \frac{1}{2} (x_k + \frac{n}{x_k}).$$ The seqeunce converges to $\sqrt{n}$.
And if one starts with $x_0= n$ one has, as soon as $|x_{k+1} - x_k|< 1$, that $\lfloor x_{k+1} \rfloor = \lfloor\sqrt{n} \rfloor$.
But the above is not very feasible for computing by hand, and there is better variant, if you just want to do this for perfect squares or test if something is a perfect square, or more generally compute $\lfloor\sqrt{n} \rfloor$. (It seems this is the case for you.)
You can do $$x_{k+1} = \left \lfloor \frac{1}{2} (x_k + \lfloor \frac{n}{x_k} \rfloor) \right \rfloor$$ only doing arithemtic with integers (including euclidean division). This will converge to $\lfloor\sqrt{n} \rfloor$ in a finite number of iterations.
See Integer Square Root on Wikipedia for further references.
I think this could be an interesting method to teach, as one can provide a good geometric motivation for it.
There is an algorithm to calculate square roots with paper and pencil (like the long division algorithm). See How to calculate a square root without a calculator.
• I learned this in grade school. :-) – Joseph O'Rourke Nov 15 '14 at 21:13
• @JosephO'Rourke, I learned this in primary school! – Martín-Blas Pérez Pinilla Nov 19 '14 at 10:08
A fancy way, for those who know a little calculus, is to linearize $f(x)=\sqrt{x}$ at some point near the number you want to square root. This can only be done when you can think of a square rootable number that isn't too far away, so it's not going to work if you're trying to square root a super huge number, but if you can think of one, it's very doable by hand, and very quick. It also has the added benefit of giving you a rational approximation, if that's something you want.
For example, to compute $\sqrt{2}$, we first have to come up with some other number that is reasonably close to $2$ which we do know how to square root. The first thing that comes to mind is $9/4$ (though there are many other choices that are even more accurate, e.g. such as $49/25$). We linearize $\sqrt{x}$ at $9/4$: $$\left.\frac{d}{dx}\sqrt{x}\right|_{x=9/4}=\left.\frac{1}{2\sqrt{x}}\right|_{x=9/4}=\frac{1}{2\sqrt{9/4}}=\frac{1}{3}$$ $$\mathcal{L}_{9/4}(x)=f^\prime\left(\frac{1}{3}\right)\left(x-\frac{9}{4}\right)+f\left(\frac{9}{4}\right)=\frac{1}{3}\left(x-\frac{9}{4}\right)+\frac{3}{2}$$ Now, we plug $2$ in there. $$\mathcal{L}_{9/4}(2)=\frac{1}{3}\left(2-\frac{9}{4}\right)+\frac{3}{2}=\frac{17}{12}$$ From here, if you want decimals, you can simply use long division. Numerically, $17/12$ comes out to be $1.41667$, while the actual value of $\sqrt{2}$ is $1.41421\ldots$. That's pretty close, and likely close enough to do the job for many on the fly calculations.
• An easy nearby perfect square is always obtainable as the next-lowest or next-highest power of 10 (whichever has an even number of zeros). If you know your powers of two, you can do better using the next-lowest or next-highest power of two. – R.. GitHub STOP HELPING ICE Nov 15 '14 at 17:07
Use log tables and the identity $x^\alpha = e^{\alpha \log x}$.
From an educational point of view, that gives evidence of the usefulness of logarithms (you can calculate any power of $x$ using just a log table and multiplication) and teaches about transforming between problems. It also allows all roots to be calculated by the same method.
An intuitive approach to find the square root of A is to draw a square with the number (A) in the middle as the area, and the sides labeled x and y. We know that x times y equal A, and x is the square root of A when x = y. Begin with a rough estimate of x and calculate y as A/x. Now replace x with the average of the original x and y, and repeat.
For example, let's find the square root of 200. As a rough guess, let's begin with x = 15. Solve for y = 200/15 = 13.333. The new value for x, call it x1 = (15 + 13.333)/2 = 14.167. Now the new value y1 = 200/14.167 = 14.117. The square root of 200 is between 14.167 and 14.117. The average is 14.142. If we use x2 = 14.142, then y2 = 200/14.142 = 14.142, verifying that we have the square root of 200 with five significant digits.
This approach has the advantage of showing what a square root is visually, which can give a problem more meaning. The process makes intuitive sense, and it does not require mathematics beyond division.
Here is the method I used to solve the original poster's problem in about three seconds:
Square root (5329):
Steps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude:
1b) Find the order of magnitude of the answer. Make a copy of the argument (5329). Shift the decimal of the copy of the argument by two places at a time, to get a number between 1 and 100. Shift the decimal of the estimated answer by one place at a time, in the opposite direction. For example, 53.29 and 10.
1c) Assume that the student knows the squares of the integers between 0 and 10 by heart. What is the closest such square? For example, 49.
1d) Multiply the estimated answer by the square root of the closest square. For example, square root (49) = 7, so the new estimated answer is 70.
Step 2 checks the estimate:
2) Square the current estimated answer. For example, 70 * 70 = 4900.
Steps 3a-d refine the estimate using uses (a+b)^2 = a^2 + 2*a*b + b^2. This is a special case of Newton's method:
3a) Subtract the value found in step 2 from the original argument, to yield a remainder. For example, 5329 - 4900 = 429. Keep track of the sign.
3b) Make a copy of the estimated answer. Multiply the copy by 2, to yield a double estimate. For example, 70 * 2 = 140.
3c) Divide the remainder by the double estimate, to yield an adjustment. Round off as convenient. Keep track of the sign. For example, 429 / 140 ~ 3.
3d) Add the adjustment to the estimate (not the double estimate). We kept track of the sign, so that we could correctly choose whether to add or subtract. For example, 70 + 3 = 73. For subsequent calculations, this value replaces estimate.
Repeat steps 2 - 3 until the answer is close enough. For example, 4900 + 420 + 3*3 = 5329, so the square root of 5329 is 73.
• I can do it in my head.
• It only requires keeping track of a few numbers.
• It quickly produces an estimated answer.
• It is a special case of Newton's method.
• Each time through the loop, the student can round off the adjustment to a convenient decimal or fraction.
• Every iteration, the student checks their work.
Three worked examples:
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 5329 1
1b 10 53.29
1c 7*7=49
1d 70
2 5329 70 4900
3a 5329 70 4900 429
3b 5329 70 4900 429 140
3c 5329 70 4900 429 140 3
2 5329 73 5329 73*73=4900+420+9
Done!
Step ment Answer So Far Remainder Estimate ment Notes
1a 739 1
1b 10 7.39
1c 3*3= 9
1d 30
2 739 30 900
3 739 30 900 -161 60 -3
2-3 739 27 729 10 54 0.2
2-3 739 27.2 739.84 -0.84 54.4 -0.015
2-3 739 27.185 739.024225 -0.024225 54.37 -0.0005
2-3 739 27.1845 738.99704025 54.369 0.00005
2 739 27.18455 738.9997587025
...Stop when satisfied...
Step ment Answer So Far Remainder Estimate ment Notes
1a 4285 1
1b 10 42.85
1c 6*6=36
1d 60
2-3 4285 60 3600 685 120 5
2-3 4285 65 4225 60 130 0.4
2-3 4285 65.4 4277.16 7.84 130.8 0.06
2-3 4285 65.46 4285.0116 -0.0116 130.92 -0.0001
2-3 4285 65.4599 4284.99850801 130.9198 0.00001
2 4285 65.45991 4284.9998171081
...Stop when satisfied... | 2020-04-01T14:47:18 | {
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https://yutsumura.com/a-relation-between-the-dot-product-and-the-trace/trace/ | # trace
### More from my site
• Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B)$? Let $A$ and $B$ be $n \times n$ matrices. Is it always true that $\tr (A B) = \tr (A) \tr (B)$? If it is true, prove it. If not, give a counterexample. Solution. There are many counterexamples. For one, take $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 […] • Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant Let A=\begin{bmatrix} a & b\\ c& d \end{bmatrix} be an 2\times 2 matrix. Express the eigenvalues of A in terms of the trace and the determinant of A. Solution. Recall the definitions of the trace and determinant of A: \[\tr(A)=a+d \text{ and } […] • Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix? Let A be an n \times n matrix. Is it true that \tr ( A^\trans ) = \tr(A)? If it is true, prove it. If not, give a counterexample. Solution. The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the […] • If Two Matrices are Similar, then their Determinants are the Same Prove that if A and B are similar matrices, then their determinants are the same. Proof. Suppose that A and B are similar. Then there exists a nonsingular matrix S such that \[S^{-1}AS=B$ by definition. Then we […]
• An Example of a Matrix that Cannot Be a Commutator Let $I$ be the $2\times 2$ identity matrix. Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$. Proof. Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies $ABA^{-1}=-B. […] • If 2 by 2 Matrices Satisfy A=AB-BA, then A^2 is Zero Matrix Let A, B be complex 2\times 2 matrices satisfying the relation \[A=AB-BA.$ Prove that $A^2=O$, where $O$ is the $2\times 2$ zero matrix. Hint. Find the trace of $A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […]
• Matrix $XY-YX$ Never Be the Identity Matrix Let $I$ be the $n\times n$ identity matrix, where $n$ is a positive integer. Prove that there are no $n\times n$ matrices $X$ and $Y$ such that $XY-YX=I.$ Hint. Suppose that such matrices exist and consider the trace of the matrix $XY-YX$. Recall that the trace of […]
• Matrices Satisfying $HF-FH=-2F$ Let $F$ and $H$ be an $n\times n$ matrices satisfying the relation $HF-FH=-2F.$ (a) Find the trace of the matrix $F$. (b) Let $\lambda$ be an eigenvalue of $H$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. Show that there exists an positive integer $N$ […] | 2018-04-19T19:37:38 | {
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https://math.stackexchange.com/questions/2958113/how-to-solve-an-ode-of-this-form-dotpt-atptpta | # How to solve an ODE of this form $\dot{P}(t)=A^TP(t)+P(t)A$?
Background
If $$\dot{x}(t)=A \,x(t)$$, then we know the solution is $$x(t)=e^{At}x(0)$$.
Question
Now let $$\dot{P}(t)=A^TP(t)+P(t)A$$, what is $$P(t)$$?
Attempt
If we take $$P(t)$$ as common factor (I'm not sure I'm doing this correctly though) ,then we have: $$\dot{P}(t)=A^TP(t)+P(t)A=(A^T+A)P(t)$$ and using the same rationale in the background section, we have $$P(t)=e^{(A^T+A) t}P(0)=e^{A^T t}P(0)e^{At}$$ Is this a correct solution?
Note
I know the answer is $$P(t)=e^{A^T t}P(0)e^{At}$$ but I'm not sure how to find it.
• Counter question: what is A? – Yuriy S Oct 16 '18 at 16:28
• @YuriyS A is a matrix. – Lod Oct 16 '18 at 16:29
• Dimensions are very likely mismatched in the expression $\dot{p}(t)=A^Tp(t)+p(t)A$. – Ennar Oct 16 '18 at 16:30
• What is $p A$ then? – Yuriy S Oct 16 '18 at 16:30
• @Ennar I fixed the question. – Lod Oct 16 '18 at 16:36
Consider the differential equation $$\dot{P}(t)=A^TP(t)+P(t)A$$
We can rewrite it as: $$\dot{P}(\tau)-A^TP(\tau)-P(\tau)A=0$$
Multiply the last differential equation by $$e^{-A^T\tau}$$ from the left and by $$e^{-A\tau}$$ from the right then
$$$$e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-e^{-A^T\tau}A^TP(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0 \quad (1)$$$$ We can easily see that equation $$(1)$$ is generated by the derivative of three multiplied functions: $$$$\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)=e^{-A^T\tau}\dot{P}(\tau)e^{-A\tau}-A^Te^{-A^T\tau}P(\tau)e^{-A\tau}-e^{-A^T\tau}P(\tau)Ae^{-A\tau}=0$$$$
So we come up with:
$$0=\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)$$
Take the integral of both sides
$$$$0=\int_0^t\frac{d}{d\tau}\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)d\tau=\bigg(e^{-A^T\tau}{P}(\tau)e^{-A\tau}\bigg)\bigg]_0^t=e^{-A^Tt}{P} \tau)e^{-At}-e^{0}{P}(0)e^{0}$$$$ \ $$$$0=e^{-A^Tt}{P}(t)e^{-At}-P(0)$$$$
Multiply left and right sides by $$e^{A^Tt}$$ and $$e^{At}$$ respectively, we get:
$$$$P(t)=e^{A^Tt}{P}(0)e^{At}$$$$
Note: I used the fact that $$e^{-A^T\tau}$$ and $$A^T$$ commute, i.e. $$A^Te^{-A^T\tau}=e^{-A^T\tau}A^T$$
For small time increment you get approximately $$P(t+dt)=P(t)+(A^TP(t)+P(t)A)dt+O(dt^2)=(I+A^T\,dt)P(t)(I+A\,dt)+O(dt^2)$$ which means that from time step to time step you get an accumulation of these factors on both sides, $$P(N\,dt)=(I+A^T\,dt)^NP(t)(I+A\,dt)^N+O(Ndt^2)$$ Now if $$N\,dt=\Delta t$$ we get, using $$(I+B/N)^N=\exp(B)+O(B^2/N)$$ $$P(t+Δt)=e^{A^T\,Δt}P(t)e^{A\,Δt}+O(Δt\,dt)$$ so that for $$dt\to 0$$ one gets exactly the claimed solution form.
By vectorizing $$P$$ and using the Kronecker product, similar to a method which can be used to solve Sylvester equations, then the differential equation can also be written as
$$\frac{d}{dt}\,\text{vec}\,P(t) = \underbrace{\left(I \otimes A^\top + A^\top\otimes I\right)}_{M}\,\text{vec}\,P(t), \tag{1}$$
which has the solution
$$\text{vec}\,P(t) = e^{M\,t}\,\text{vec}\,P(0). \tag{2}$$
By using the mixed-product property of the Kronecker product it can be shown that $$I \otimes A^\top$$ commutes with $$A^\top \otimes I$$. This commuting property allows you to write the matrix exponential as
$$e^{M\,t} = e^{(I\,\otimes\,A^\top)\,t}\,e^{(A^\top\,\otimes\,I)\,t}. \tag{3}$$
By using the definition of a matrix exponential and again the mixed-product property of the Kronecker product then is can also be shown that
$$e^{X\,\otimes\,I} = e^{X} \otimes I, \quad e^{I\,\otimes\,X} = I \otimes e^{X},$$
thus
$$e^{M\,t} = \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right). \tag{4}$$
Substituting $$(4)$$ into $$(2)$$ gives
\begin{align} \text{vec}\,P(t) &= \left(I \otimes e^{A^\top t}\right) \left(e^{A^\top t} \otimes I\right) \text{vec}\,P(0) \\ &= \left(I \otimes e^{A^\top t}\right) \text{vec}\left(P(0)\,e^{A\,t}\right) \\ &= \text{vec}\left(e^{A^\top t}\,P(0)\,e^{A\,t}\right) \end{align}
thus
$$P(t) = e^{A^\top t}\,P(0)\,e^{A\,t}.$$ | 2020-07-07T03:56:51 | {
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https://www.physicsforums.com/threads/series-convergence.319241/ | # Series convergence
1. Jun 10, 2009
### utopiaNow
1. The problem statement, all variables and given/known data
Use partial fractions to show
$$\displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{4}$$
3. The attempt at a solution
I did the partial fraction decomposition to get: $$\displaystyle\sum_{n=1}^\infty \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2n + 4}$$
I'm not sure how to proceed from here, in my textbook the example shows how terms in the partials sums overlap and cancel out if you start looking at the terms in the partial fraction decomposition, however I can't see that happening with this particular series.
Any suggestions would be appreciated. Thanks in advance.
2. Jun 10, 2009
### rock.freak667
Write out a few term, n=1,2,3,4,5....N-3,N-2,N-1,N
then check as N->inf.
3. Jun 10, 2009
### utopiaNow
Hi,
I tried doing that originally, but I don't see any patterns that I can exploit when I write the terms of the partial decomposition for n = 1, 2, 3,..., N-2, N-1, N as you stated.
Do you have any further suggestions?
4. Jun 10, 2009
### Dick
Write the terms for successive values of n each on a row 1/(2n+4) -1/(n+1) 1/(2n)
1/6 -1/2 1/2
1/8 -1/3 1/4
1/10 -1/4 1/6
1/12 -1/5 1/8
1/14 -1/6 1/10
Oblique hint: diagonal.
5. Jun 10, 2009
### utopiaNow
Interesting, so everything seems to be canceling out if you keep going long enough diagonally, except the 1/4 in the second row. However is there any way to formalize this? Or is simply noticing this pattern enough for a formal proof?
6. Jun 10, 2009
### Dick
Depends on how formal you want to be. Writing a table like that and scratching out the cancellations is good for me. If you want to do it formally write out the first term from the n case, the second term from the n+1 case and the third term from the n+2 case and show they cancel algebraically. What is important is to realize just because there isn't an obvious cancellation doesn't mean there isn't one.
7. Jun 10, 2009
### utopiaNow
Thank you for the help!
8. Jun 10, 2009
### txy
why not take out 1/2 from the partial fraction decomposition so that it's more obvious?
$$\sum^{+\infty}_{n=1}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}) + \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n+2}-\frac{1}{n+1})$$
9. Jun 10, 2009
### Dick
That's a nice way to handle it. There's more than one way to skin a cat. | 2017-10-21T07:01:21 | {
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https://math.stackexchange.com/questions/2435913/solutions-of-x2-6x-13-equiv-0-pmod127 | # Solutions of $x^2-6x-13 \equiv 0 \pmod{127}$
I started learning number theory, specifically polynomial congruences, and need help with the following exercise. Here it is:
Does the congruence $x^2-6x-13 \equiv 0 \pmod{127}$ has solutions?
I tried to follow the method for solving general quadratic congruence but I didn't get really far. Here's what I've done so far:
Since $(4, 127) = 1$, we may complete the square by multiplying by $4$ without having to change the modulus in order to get the following equivalent congruence
$$(2x-6)^2 \equiv 36 - 4(-13) \pmod{127} \iff (2x-6)^2 \equiv 88 \pmod{127}.$$
If I'm heading in the right direction then I don't know how to continue from here. I suspect there is another method for solving this problem since I didn't make use of the fact that $127$ is prime. Also, the problem doesn't require to actually find the solutions but only determine if there are solutions. Possibly we can avoid computations and make use of some theorem/lemma to find if the congruence has solutions or not.
• The quadratic formula, $x = \frac{6\pm\sqrt{(-6)^2 - 4\cdot 1\cdot (-13)}}{2\cdot 1}$ still works, since $127$ is an odd prime. Note, however, that square root and dividing by $2$ has a different meaning from what it does for real numbers (or, it has the same meaning, but potentially a very different result). – Arthur Sep 19 '17 at 11:16
. To find solutions of $$x^2 - 6x - 13 \equiv 0 \mod 127$$, we must write $$x^2-6x - 13$$ as a square of a linear term, plus a constant. It's seen easily that $$x^2 - 6x - 13 = (x-3)^2 - 22$$. Hence, the congruence is equivalent to $$(x-3)^2 \equiv 22 \mod 127$$.
Now, we must check if $$22$$ is a quadratic residue mod $$127$$. For this, we can use the Legendre symbol, whose notation I will keep the same as the binomial, so do not get confused. $$\binom{22}{127} = \color{green}{\binom{11}{127}}\color{red}{\binom{2}{127}} = \color{green}{-\binom{127}{11}} \times \color{red}1$$ where the terms with same color on the LHS and RHS are equal. The green equality comes by quadratic reciprocity and the second comes by the fact that $$\binom{2}{p}$$ is well known by the remainders which $$p$$ leaves when divided by $$8$$.
Now, we can do: $$-\binom{127}{11} = -\binom{6}{11} = -\color{green}{\binom{2}{11}}\color{red}{ \binom{3}{11}} = -\color{green}{(-1)}\color{red}{(1)} = 1$$
Again, the colored terms on the LHS and RHS are equal because the quantities $$\binom 2p$$ and $$\binom 3p$$ are well known.
Since we have obtained that the Legendre symbol is $$1$$, this implies the existence of a solution, and therefore two.
The question, though, is how to compute them. I do not know of any method other than brute force, unfortunately. However, our reward for writing $$(x-3)^2 \equiv 22 \mod 127$$ is that we basically only need to look for squares of the form $$127k + 22$$ to find a solution, rather than having to substitute values of $$x$$ into the expression $$x^2-6x-13$$ each time.
A brute force : the series $$127k + 22$$ goes like : $$22,149,276,403,530,657,\color{blue}{784},...$$
lo and behold, $$784 = 28^2$$, hence this gives $$x = 31$$. Now, note that the congruence actually has two solutions, one given by $$127 - 28 = 99$$. You can check that $$9801 = 99^2 = 22 + 127 \times 77$$.
Hence, we get two solutions of $$x$$, namely $$x= 31,102$$.
• I can't thank you enough for this answer! I understood everything and it made me see the link that I was missing between quadratic polynomial/residues and Legendre symbol. – user347616 Sep 19 '17 at 12:07
• But of course, you are welcome! I should mention, +1 for your question. – астон вілла олоф мэллбэрг Sep 19 '17 at 12:08
• Since $127$ is one less than a multiple of $4$, you can get the square roots by "root-power conversion". Fermat's Little Theorem gives $22^{126}\equiv 1$, and with $22$ being a quadratic residue then $22^{63}\equiv 1$. So $22^{64}\equiv 22$ assuring that $22^{32}\equiv 99$ will be one square root. Try it! – Oscar Lanzi Sep 19 '17 at 12:26
• That is a much,much better method than the one I have suggested. I thank you for this. – астон вілла олоф мэллбэрг Sep 19 '17 at 12:29
Using the Berlekamp Algorithm for factoring a polynomial over $\mathbb{F}_{127}$ we immediately obtain that $$x^2-6x-13=(x+25)(x+96).$$ So we have two solutions.
I am a congruence beginner like you.
I did in this way
completed the square $(x-3)^2-9\equiv 13 \mod 127$
$(x-3)^2\equiv 22 \mod 127$
Now call $r$ is the number such that $r^2\equiv 22\mod 127$?
Like a "modular" square root
I found here a way to find the roots
As $127\equiv 3 \mod 4$ then we have $r= \pm 22^{\frac{128}{4}} \mod 127$
$22^{32} =2^{32}\cdot 11^{32}$
$2^7\equiv 1 \mod 127\to 2^{28}\equiv 1 \mod 127\to \color{red}{2^{32}\equiv 2^4\mod 127}$
$11^{32}=121^{16}$ as $121\equiv 6 \mod 127$ we have $121^{16}\equiv 6^{16} \mod 127$ and then $6^{16}\equiv 2^{16}\cdot 3^{16} \mod 127$
$2^k \mod 127$ for $k=0,1,2\ldots$ is cyclical and is $1,2,4,8,16,32,64$ then repeats (because $127=2^7-1$) therefore $2^{16}\equiv 4 \mod 127$
$3^{16}\equiv 81^4 \mod 127\to 3^{16}\equiv 71 \mod 127$
Finally $6^{16}\equiv 4\cdot 71 \mod 127\to 6^{16}\equiv 30 \mod 127$ and $\color{red}{11^{32}\equiv 6^{16}\equiv 30 \mod 127}$
Collecting the parts in red colour we get $22^{32}\equiv 2^4\cdot 30 \mod 127$ that is $22^{32}\equiv 99\mod 127$ and $-22^{32} \equiv 28 \mod 127$
Concluding the two solutions are $x-3\equiv 99 \mod 127\to x\equiv 102 \mod 127$ and $x-3\equiv 28 \mod 127 \to x\equiv 31 \mod 127$
Edit
My solution is all but elegant. Anyway is a solution provided by a beginner like you and I hope it is understandable
• It took me a little time to go through your solution but I made it. Thank you for the effort, it is always enlightening to see different solutions to the same problem. (+1) – user347616 Sep 19 '17 at 14:59
• To compute $22^{32} \bmod127$ just square five successive times. – Oscar Lanzi Sep 19 '17 at 22:33
• @OscarLanzi Why $22^{32}\equiv 22^5 \mod 127$? – Raffaele Sep 20 '17 at 13:11
• I did not say that. Square five successive times means $22^{2^5}$. – Oscar Lanzi Sep 20 '17 at 13:42
I think you can try $$x=31.$$ It is valid.
Finally we have $$x^2-6x-13\equiv(x-31)(x-102) \pmod{127}.$$
This equation has roots modulo $127$ if and only if the reduced discriminant $\Delta'=9+13$ is a square modulo $127$. We'll use the laws of quadratic reciprocity.
$$\biggl(\frac{22}{127}\biggr)=\biggl(\frac{2}{127}\biggr)\biggl(\frac{11}{127}\biggr)=\biggl(\frac{11}{127}\biggr)\quad\text{since }127\equiv -1\mod8\quad (2^{\mathit{nd}}\textit{supplementary law}).$$ Now we have \begin{align}\biggl(\frac{11}{127}\biggr)&=\biggl(\frac{127}{11}\biggr)\bigl(-1\bigr)^{\tfrac{10\cdot126}4}=-\biggl(\frac{6}{11}\biggr)=-\biggl(\frac{2}{11}\biggr)\biggl(\frac{3}{11}\biggr)\\[1ex] &=+\biggl(\frac{3}{11}\biggr)=\biggl(\frac{11}{3}\biggr)\bigl(-1\bigr)^{\tfrac{2\cdot10}4}=-\biggl(\frac{2}{3}\biggr)=+1. \end{align} Thus the equation has roots. | 2019-11-15T04:50:17 | {
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https://math.stackexchange.com/questions/218046/relations-between-p-norms | # Relations between p norms
The $$p$$-norm on $$\mathbb R^n$$ is given by $$\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$$. For $$0 < p < q$$ it can be shown that $$\|x\|_p\geq\|x\|_q$$ (1, 2). It appears that in $$\mathbb{R}^n$$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $$\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$$(3), $$\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $$\mathbb R^n$$. For instance, for $$n=2$$ and $$n=3$$ one can see that for $$0 < p < q$$, the spheres with radius $$\sqrt n$$ with $$\|\cdot\|_p$$ inscribe spheres with radius $$1$$ with $$\|\cdot\|_q$$.
It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $$n=2$$ it is easily proven (see below), but not for $$n>2$$. So my questions are:
1. How can relation (3) be proven for arbitrary $$n\,$$?
2. Can this be generalized into something of the form $$\|x\|_{p} \leq C \|x\|_{q}$$ for arbitrary $$0?
3. Do any of the relations also hold for infinite-dimensional spaces, i.e. in $$l^p$$ spaces?
Notes:
$$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$$, hence $$=2\|x\|_{2}^{2}$$
$$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$$. This works because $$|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$$, but only because $$(|x_{1}| - |x_{2}|)^2 \geq 0$$, while for more than two terms $$\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$$ gives an inequality that never gives the right signs for the cross terms.
• Your first link (1) has exactly the equation you seek. Actually a version of Norberts excellent answer, generalized to measure spaces. – Thomas Ahle Mar 31 '16 at 20:57
• Probably a dead thread, but norms in finite dimensional spaces are equivalent, so one can always find c ad C such that $c\|x\|_p\leq \|x\|_q\leq C\|x\|_p$. – BigM Dec 1 '19 at 20:23
1. Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$ $$\Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2$$
2. Such a bound does exist. Recall Hölder's inequality $$\sum\limits_{i=1}^n |a_i||b_i|\leq \left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}$$ Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$ $$\sum\limits_{i=1}^n |x_i|^p= \sum\limits_{i=1}^n |x_i|^p\cdot 1\leq \left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}} \left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}$$ Then $$\Vert x\Vert_p= \left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq \left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\= n^{1/p-1/q}\Vert x\Vert_q$$ In fact $C=n^{1/p-1/q}$ is the best possible constant.
• @Arun, because for $x=(1,1,1,\ldots,1)$ this bound is attained – Norbert Jun 10 '15 at 14:27
• @Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? – pikachuchameleon Jun 22 '15 at 12:19
• @AshokVardhan what does "similar" mean? Do you want to prove something like $\Vert x\Vert_p \leq C\Vert x\Vert_q$ for $p>q$ ? – Norbert Jun 23 '15 at 14:24 | 2021-07-29T03:10:54 | {
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https://byjus.com/question-answer/a-point-moves-such-that-the-sum-of-the-square-of-its-distance-from-the/ | Question
# A point moves such that the sum of the square of its distance from the sides of a sqaure of side unity is equal to $$9$$. The locus of the point is a circle such that
A
Center of the circle coincides with that of square
B
Center of the circle is (12,12)
C
Radius of the circle is 2
D
All the above are true
Solution
## The correct option is D All the above are trueLet the sides of the square be $$y=0,y=1,x=0$$ and $$x=1.$$Let the moving point be $$(x,y).$$Then,$${ y }^{ 2 }+{ \left( y-1 \right) }^{ 2 }+{ x }^{ 2 }+{ \left( x-1 \right) }^{ 2 }=9$$ is the equation of the locus.$$\Rightarrow 2{ x }^{ 2 }+2{ y }^{ 2 }-2x-2y-7=0,$$ which represents a circle having centre $$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right)$$ (which is also the centre of the square) and radius $$2.$$ Mathematics
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https://math.stackexchange.com/questions/2567158/looking-for-a-proof-of-an-interesting-identity/2567218 | # Looking for a proof of an interesting identity
Working on a problem I have encountered an interesting identity:
$$\sum_{k=0}^\infty \left(\frac{x}{2}\right)^{n+2k}\binom{n+2k}{k} =\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n,$$ where $n$ is a non-negative integer number and $x$ is a real number with absolute value less than 1 (probably a similar expression is valid for arbitrary complex numbers $|z|<1$).
Is there any simple proof of this identity?
• How/where did you meet this identity? Is the LHS just the Taylor series for the RHS? – Daniel Robert-Nicoud Dec 14 '17 at 23:38
• Anything to do with a random walk? Why did you call this interesting, btw? – Mathemagical Dec 14 '17 at 23:44
• Reminiscent of the explicit form for Chebyshev polynomials of the first kind, hence they idea of enforcing the substitution $x=\sin\theta$ and applying the residue theorem looks like a promising one ;) – Jack D'Aurizio Dec 14 '17 at 23:56
• Or you may prove that both sides are the terms of an Appell / Fibonacci sequence of polynomials. – Jack D'Aurizio Dec 14 '17 at 23:57
• @tired This is in regard to the hitting time of a random walk (more particularly, the probability generating function of the hitting time of the walk at the barrier x=1. Please see the equations 15,16, and 17 in the linked document. galton.uchicago.edu/~lalley/Courses/312/RW.pdf The details of the computation are not listed here, but they are shown in greater detail in Michael Steele's book, Stochastic Calculus and Financial applications (pages 8 and 9). – Mathemagical Dec 16 '17 at 15:25
Using
$$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)
$$2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\ 4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)}$$
for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so
$$S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right]$$
or
$$S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n$$
• Thanks Jack :-) – tired Dec 15 '17 at 0:55
• @tired: Thank you very much for the nice proof. It contains however a misprint. $4$ shall be replaced with $4^{1/n}$ in the denominator of the residue value. But probably simpler and more correctly were to write it just as $$\text{res}(f(z),z=z_0)=\frac{1}{4\sqrt{1-x^2}} \left(2\frac{1-\sqrt{1-x^2}}{x^2}\right)^n.$$ – user Dec 15 '17 at 11:56
• @user355705 you are absolutly right – tired Dec 16 '17 at 15:07
Extracting coefficients on the RHS we get the integral (coefficient on $x^{n+2k}$)
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+2k+1}} \frac{1}{\sqrt{1-z^2}} \left(\frac{1-\sqrt{1-z^2}}{z}\right)^n \; dz.$$
Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has $w = \frac{1}{2} z + \cdots$ so the image in $w$ of the contour in $z$ can be deformed to a small circle enclosing the origin in the $w$-plane. (Moreover we see that the exponentiated term starts at $z^n$ which justifies the corresponding offset in the series.) We get $dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \; dw = 2(1-w^2)/(1+w^2)^2 \; dw.$ We also have $1-z^2 = 1 - 4w^2/(1+w^2)^2 = (1-w^2)^2/(1+w^2)^2.$ All of this yields
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}} \frac{1}{(1-w^2)/(1+w^2)} w^n \frac{2(1-w^2)}{(1+w^2)^2} \; dw \\ = \frac{1}{2^{n+2k}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k}}{w^{2k+1}} \; dw.$$
This evaluates by inspection to
$$\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k} = \frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k} \\ = \frac{1}{2^{n+2k}} {n+2k\choose k}$$
which is the claim.
I plan to insert this approach into the next (2019) version of my notes.
The even powers of $\arcsin(x)$ all have a nice Maclaurin series (see eqs (19),(20),(21),(22)), and this can be seen as a side effect of the Lagrange–Bürmann inversion theorem. I am going to implement the same trick here, by computing the derivatives at the origin of the RHS (the opposite way has already been investigated by tired).
$$[x^m]\left[\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{x}\right)^n\right]=\frac{1}{2\pi i}\oint\frac{(1-\sqrt{1-x^2})^n}{x^{n+m+1}\sqrt{1-x^2}}\,dx$$ where the integral is performed along a small circle enclosing the origin. What about enforcing the substitution $x=\sin z$? The sine function is holomorphic and injective in a neighbourhood of the origin, hence the RHS is simply converted into $$\frac{1}{2\pi i}\oint\frac{(1-\cos z)^n}{\left(\sin z\right)^{n+m+1}}\,dz =\frac{1}{2\pi i}\oint\frac{2^n \sin^{2n}\frac{z}{2}}{2^{n+m+1}\sin^{n+m+1}\frac{z}{2}\cos^{n+m+1}\frac{z}{2}}\,dz$$ or $$\frac{1}{2\pi i}\oint\frac{\tan^{n-m-1}z}{2^{m}\cos^{2m}z}\,dz=\frac{1}{2^{m+1}\pi i}\oint u^{n-m-1}(1+u^2)^{m-1}\,du$$ via $z\to 2z$ and $z\to\arctan u$. Now the RHS is trivially given by a binomial coefficient multiplied by $\frac{1}{2^m}$ (provided by $m\geq n$) and the claim is proved. | 2019-05-22T14:49:23 | {
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https://ch.mathworks.com/help/symbolic/fimplicit3.html | # fimplicit3
Plot 3-D implicit equation or function
## Syntax
fimplicit3(f)
fimplicit3(f,[min max])
fimplicit3(f,[xmin xmax ymin ymax zmin zmax])
fimplicit3(___,LineSpec)
fimplicit3(___,Name,Value)
fimplicit3(ax,___)
fi = fimplicit3(___)
## Description
example
fimplicit3(f) plots the 3-D implicit equation or function f(x,y,z) over the default interval [-5 5] for x, y, and z.
example
fimplicit3(f,[min max]) plots f(x,y,z) over the interval [min max] for x, y, and z.
example
fimplicit3(f,[xmin xmax ymin ymax zmin zmax]) plots f(x,y,z) over the interval [xmin xmax] for x, [ymin ymax] for y, and [zmin zmax] for z. The fimplicit3 function uses symvar to order the variables and assign intervals.
example
fimplicit3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and face color.
example
fimplicit3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes.
fimplicit3(ax,___) plots into the axes with the object ax instead of the current axes object gca.
example
fi = fimplicit3(___) returns an implicit function surface object. Use the object to query and modify properties of a specific surface. For details, see ImplicitFunctionSurface Properties.
## Examples
collapse all
Plot the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by using fimplicit3. The fimplicit3 function plots over the default interval of $\left[-5,5\right]$ for $x$, $y$, and $z$.
syms x y z fimplicit3(x^2 + y^2 - z^2)
Plot the hyperboloid specified by the function $f\left(x,y,z\right)={x}^{2}+{y}^{2}-{z}^{2}$. The fimplicit3 function plots over the default interval of $\left[-5,5\right]$ for $x$, $y$, and $z$.
syms f(x,y,z) f(x,y,z) = x^2 + y^2 - z^2; fimplicit3(f)
Specify the plotting interval by specifying the second argument to fimplicit3. Plot the upper half of the hyperboloid ${x}^{2}+{y}^{2}-{z}^{2}=0$ by specifying the interval $0. For $x$ and $y$, use the default interval $\left[-5,5\right]$.
syms x y z f = x^2 + y^2 - z^2; interval = [-5 5 -5 5 0 5]; fimplicit3(f, interval)
Plot the implicit equation $x\mathrm{sin}\left(y\right)+z\mathrm{cos}\left(x\right)=0$ over the interval $\left(-2\pi ,2\pi \right)$ for all axes.
Create the x-axis ticks by spanning the x-axis limits at intervals of pi/2. Convert the axis limits to precise multiples of pi/2 by using round and get the symbolic tick values in S. Display these ticks by using the XTick property. Create x-axis labels by using arrayfun to apply texlabel to S. Display these labels by using the XTickLabel property. Repeat these steps for the y-axis.
To use LaTeX in plots, see latex.
syms x y z eqn = x*sin(y) + z*cos(x); fimplicit3(eqn,[-2*pi 2*pi]) title('xsin(y) + zcos(x) for -2\pi < x < 2\pi and -2\pi < y < 2\pi') xlabel('x') ylabel('y') ax = gca; S = sym(ax.XLim(1):pi/2:ax.XLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.XTick = double(S); ax.XTickLabel = arrayfun(@texlabel,S,'UniformOutput',false); S = sym(ax.YLim(1):pi/2:ax.YLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); ax.YTick = double(S); ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);
Plot the implicit surface ${x}^{2}+{y}^{2}-{z}^{2}=0$ with different line styles for different values of $z$. For $-5, use a dashed line with green dot markers. For $-2, use a LineWidth of 1 and a green face color. For $2, turn off the lines by setting EdgeColor to none.
syms x y z f = x^2 + y^2 - z^2; fimplicit3(f,[-5 5 -5 5 -5 -2],'--.','MarkerEdgeColor','g') hold on fimplicit3(f,[-5 5 -5 5 -2 2],'LineWidth',1,'FaceColor','g') fimplicit3(f,[-5 5 -5 5 2 5],'EdgeColor','none')
Plot the implicit surface $1/{x}^{2}-1/{y}^{2}+1/{z}^{2}=0$. Specify an output to make fimplicit3 return the plot object.
syms x y z f = 1/x^2 - 1/y^2 + 1/z^2; fi = fimplicit3(f)
fi = ImplicitFunctionSurface with properties: Function: [1x1 sym] EdgeColor: [0 0 0] LineStyle: '-' FaceColor: 'interp' Show all properties
Show only the positive x-axis by setting the XRange property of fi to [0 5]. Remove the lines by setting the EdgeColor property to 'none'. Visualize the hidden surfaces by making the plot transparent by setting the FaceAlpha property to 0.8.
fi.XRange = [0 5]; fi.EdgeColor = 'none'; fi.FaceAlpha = 0.8;
Control the resolution of an implicit surface plot by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing 'MeshDensity' can increase plotting speed.
Divide a figure into two by using subplot. In the first subplot, plot the implicit surface $\mathrm{sin}\left(1/\left(xyz\right)\right)$. The surface has large gaps. Fix this issue by increasing the 'MeshDensity' to 40 in the second subplot. fimplicit3 fills the gaps showing that by increasing 'MeshDensity' you increased the resolution of the plot.
syms x y z f = sin(1/(x*y*z)); subplot(2,1,1) fimplicit3(f) title('Default MeshDensity = 35') subplot(2,1,2) fimplicit3(f,'MeshDensity',40) title('Increased MeshDensity = 40')
Apply rotation and translation to the implicit surface plot of a torus.
A torus can be defined by an implicit equation in Cartesian coordinates as
$\mathit{f}\left(\mathit{x},\mathit{y},\mathit{z}\right)={\left({\mathit{x}}^{2}+{\mathit{y}}^{2}+{\mathit{z}}^{2}+{\mathit{R}}^{2}-{\mathit{a}}^{2}\right)}^{2}-4{\mathit{R}}^{2}\left({\mathit{x}}^{2}+{\mathit{y}}^{2}\right)$
where
• $a$ is the radius of the tube
• $R$ is the distance from the center of the tube to the center of the torus
Define the values for $a$ and $R$ as 1 and 5, respectively. Plot the torus using fimplicit3.
syms x y z a = 1; R = 4; f(x,y,z) = (x^2+y^2+z^2+R^2-a^2)^2 - 4*R^2*(x^2+y^2); fimplicit3(f) hold on
Apply rotation to the torus around the $x$-axis. Define the rotation matrix. Rotate the torus by 90 degrees or $\pi /2$ radians. Shift the center of the torus by 5 along the $x$-axis.
alpha = pi/2; Rx = [1 0 0; 0 cos(alpha) sin(alpha); 0 -sin(alpha) cos(alpha)]; r = [x; y; z]; r_90 = Rx*r; g = subs(f,[x,y,z],[r_90(1)-5,r_90(2),r_90(3)]);
Add a second plot of the rotated and translated torus to the existing graph.
fimplicit3(g) axis([-5 10 -5 10 -5 5]) hold off
## Input Arguments
collapse all
3-D implicit equation or function to plot, specified as a symbolic equation, expression, or function. If an expression or function is specified, then fimplicit3 assumes the right-hand size to be 0.
Plotting interval for x-, y- and z- axes, specified as a vector of two numbers. The default is [-5 5].
Plotting interval for x-, y- and z- axes, specified as a vector of six numbers. The default is [-5 5 -5 5 -5 5].
Axes object. If you do not specify an axes object, then fimplicit3 uses the current axes.
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Marker','o','MarkerFaceColor','red'
The properties listed here are only a subset. For a complete list, see ImplicitFunctionSurface Properties.
Number of evaluation points per direction, specified as a number. The default is 35.
Example: 100
Line color, specified as 'interp', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default RGB triplet value of [0 0 0] corresponds to black. The 'interp' value colors the edges based on the ZData values.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the EdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: [0.5 0.5 0.5]
Example: 'blue'
Example: '#D2F9A7'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: [0.3 0.2 0.1]
Example: 'green'
Example: '#D2F9A7'
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
## Output Arguments
collapse all
One or more objects, returned as a scalar or a vector. The object is an implicit function surface object. You can use these objects to query and modify properties of a specific line. For details, see ImplicitFunctionSurface Properties.
## Algorithms
fimplicit3 assigns the symbolic variables in f to the x axis, the y axis, then the z axis, and symvar determines the order of the variables to be assigned. Therefore, variable and axis names might not correspond. To force fimplicit3 to assign x, y, or z to its corresponding axis, create the symbolic function to plot, then pass the symbolic function to fimplicit3.
For example, the following code plots the roots of the implicit function f(x,y,z) = x + z in two ways. The first way forces fimplicit3 to assign x and z to their corresponding axes. In the second way, fimplicit3 defers to symvar to determine variable order and axis assignment: fimplicit3 assigns x and z to the x and y axes, respectively.
syms x y z; f(x,y,z) = x + z; figure; subplot(2,1,1) fimplicit3(f); view(-38,71); subplot(2,1,2) fimplicit3(f(x,y,z)); % Or fimplicit3(x + z);
### Topics
Introduced in R2016b
## Support
#### Mathematical Modeling with Symbolic Math Toolbox
Get examples and videos | 2021-09-20T11:21:39 | {
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http://openstudy.com/updates/508c2cfae4b04456f6c43164 | ## klimenkov Group Title $\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}$ 2 years ago 2 years ago
1. myko
maybe like this: $\sqrt[n]{n!}=\sqrt[n]{n(n-1)(n-2)\cdots1} =\sqrt[n]{n}\sqrt[n]{n-1}\cdots \sqrt[n]{1}=1$ so limit is equal to 0
2. myko
@klimenkov
3. klimenkov
Are you sure that $\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1$
4. myko
look the steps from my comment befor. It looks ok
5. myko
all the roots at the right hand side: $\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1$
6. myko
so their product too
7. klimenkov
No, it's not ok. Because you multiply an infinite quantity or 1. As we know $$1^{\infty}={}?$$.
8. myko
$1^{\infty} =1*1*\cdots*1=1$
9. klimenkov
Very nice. What can you say about this pretty limit? $\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$It is $$1^{\infty}$$.
10. myko
this happens when you talk about functions. The reason of indetermination of 1^infinity is because of that. But in this case there are no functions involved. That's my point
11. myko
in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken
12. klimenkov
Ok. What about this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$Is it 0?
13. myko
this is a harmonic series. It is not convergent
14. myko
15. klimenkov
Look at the denominator carefully please. I hopr you will try to get what I'm saying.
16. myko
sry, but i don't
17. klimenkov
Can you find this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$
18. myko
another way to try this: $\lim \sqrt[n]{\frac{n!}{n^n}} = 0$
19. myko
infinity
20. klimenkov
Can you show the way you solve it?
21. myko
I don't remmeber the formal proof of n!/n^n =0, but it's evident, if you try a few first terms of this sequence. There are some posts about it if you google a bit
22. klimenkov
@myko, see this and tell me what is your mistake? http://www.wolframalpha.com/input/?i=Limit+%28n!%29^%281%2Fn%29%2Fn+n-%3Einfinity
23. TuringTest
@mahmit2012 a little help here?
24. mahmit2012
|dw:1351370532404:dw|
25. TuringTest
very nice, my turn...
26. mahmit2012
|dw:1351370740820:dw|
27. mahmit2012
|dw:1351370823511:dw|
28. mahmit2012
|dw:1351370856590:dw|
29. TuringTest
${\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)$$=\exp(-1)=\frac1e$
30. mukushla
*
31. myko
ya I was wrong. Here is another way to solve it. As we know root test is stronger than cuotient test, so the folowing inequality holds: $\lim \inf \frac{a_{n+1}}{a_{n}}\leq\lim \inf \sqrt[n]{a_{n}}\leq \lim \sup \sqrt[n]{a_{n}} \leq \lim \sup\frac{a_{n+1}}{a_{n}}$ let $a_{n} = \frac{n!}{n^{n}}$ then $\frac{a_{n+1}}{a_{n}}=\frac{1}{(1+\frac{1}{n})^{n}}=\frac{1}{e}$ this means that $\lim \sqrt[n]{a_{n}} = \frac{1}{e}$
32. myko
@mahmit2012 $\lim \frac{a_{n+1}}{a_{n}}=\lim \sqrt[n]{a_{n}}$ only if a_n is convergent, what is not implied in this question
33. klimenkov
Nice. But I have one more interesting method to find it. $\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A$$\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1$$\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e$ | 2014-11-26T18:56:21 | {
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http://marathon42k.it/zcy/time-complexity-of-linear-search.html | Understanding Time Complexity Of Algorithms | Bits N Tricks. We could also say it is linear on the number of entries in the table but that is less commonly used. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. The time required is flat, an O(1) constant time complexity. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. In the later case, the search terminates in failure with n comparisons. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). Why so important? You do it all the time in real life!. Amortized time per operation using a bounded priority queue[1] logarithmic time DLOGTIME O(log n) log n, log(n 2) Binary search polylogarithmic time poly(log n) (log n)2 fractional power O(nc) where 0 < c < 1 n1/2, n2/3 Searching in a kd-tree linear time O(n) n Finding the smallest item in an unsorted array "n log star n" time O(n log* n). The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. By this logic, we can say that painting pictures is slower than baking cookies. This is said to run at O(n); it’s run time increases at an order of magnitude proportional to n. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. The search time increases proportionately to the number of new items introduced. If we're running a statement. Lookups on arrays and objects are going to be constant time if you access them directly. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. The notation Ο(n) is the formal way to express the upper bound of an algorithm's running time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. Solving a system of linear equations has a complexity of at most O (n 3). This video explains the time complexity analysis for binary search. One example is the binary search algorithm. The array to be searched is reduced by half in every iteration. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. > But what is: > > == the time complexity of string indexing? Is it constant? Yes. when first breaks, you know X(last but one fall - success) and Y(last fall - failure). Linear Differential Equations and Oscillators is the first book within Ordinary Differential Equations with Applications to Trajectories and Vibrations, Six-volume Set. Here that is linear time. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. Lets say I have the list 10,20,30,40,50,60,30,40,50. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. It costs us space. characterises a function based on growth of function C. A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. But a balanced binary search tree is always OlgN. Space Complexity. Data Structures for Beginners: Arrays, HashMaps, and Lists. For example, a "linear" running time can also. $\begingroup$ @Olologin can you share any references to understand how to calculate time complexities for complex equations? I want to understand the priority of matrix, inverse, transpose etc of different orders. Search for "Journey Into Complexity" Books in the Search Form now, Download or Read Books for FREE, just by Creating an Account to enter our library. So, we can write this as Ω(n). Examples: binary search. For example, if the heuristic evaluation function is an exact estimator, then A* runs in linear time, expanding only those nodes on an optimal solution path. now do a linear search starting from X(conservative but accurate second step - slow). For example -. During each iteration, the first remaining element of the input is only compared with the right-most element of the sorted subsection of the array. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Time Complexity of Bisection Search is O(log n). The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. Here is an. Dual first-order methods are essential techniques for large-scale constrained convex optimization. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The best case time in linear search is for the first element i. For databases, this means that the time execution would be directly proportional to the table size: as the number of rows in the table grows, the time for the query grows. See full list on towardsdatascience. In the later case, the search terminates in failure with n comparisons. The complexity of an algorithm is usually taken to be its worst-case complexity, unless specified otherwise. • Matlab implements sparse linear algebra based on i,j,s format. Dual first-order methods are essential techniques for large-scale constrained convex optimization. [143, 144, 145, 99]) that Depth-First Search (DFS) and Breadth-First Search (BFS) run in linear time in graphs, and that using these techniques one can obtain linear time algorithms (on a RAM) for many interesting graph. The new distance measures can be computed in linear time complexity in the histogram size. veri es in linear time whether a given spanning tree T of a graph G = (V;E) is a minimum spanning tree. Stochastic Diffusion Search is an alternative solution for invariant pattern recognition and focus of attention. O(n log n) - sorting a list. DTIME[2polylogn]. However complexity for above written implementations is O(). Download Binary search program. , 2017) is in the overhead on enumerating bin pairs. Space Complexity. …humans are incredibly good at linking cause and effect – sometime too good. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Since all letters are placed in one bucket, Put and Get operation will no longer have time complexity of O(1) because put and get operation has to scan each letter inside the bucket for matching key. Multiply to get n*log(n). Write a linear-time filter IntegerSort. Another simple yet important function. Binary Search source. Yields of experiment are expected to provide an information related to the complexity of the algorithm in LibSVM and knowing the running-time indicator of training and testing both for C++ and Java. How many elements of the input sequence need to be checked on the average, assuming that the element being searched for is equally likely to be any element in the array? How about in the worst case? What are the average-case and worst-case running times of linear search in $\theta$-notation? Justify your answers. Examples: binary search. All have polynomial time complexity while some allow very long steps in favorable circumstances. Time complexity of a related-key attack: “Thus, the total time complexity of Step 2(b) is about 2256 ·2167. With an average time complexity of O(log log n), interpolation search beats binary search's O(log n) easily. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. On the other hand, if you search for a word in a dictionary, the search will be faster because the words are in sorted order, you know the order and can quickly decide if you need to turn to earlier pages or later pages. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. hyperparameter Search: Grid search and random search Train & Run time space & time complexity. It went through the entire list so it took linear time. On an unsorted array Binary Search is almost twice as slow as Linear Search with worst Time Complexity of O(n²) and that is not even considering unbalanced trees. Time Complexity of Binary Search Algorithm is O(log 2 n). Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. Polynomial time means n O(1), or n c for some constant c. Huan Li, Zhouchen Lin; 21(33):1−45, 2020. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. Linear time is when an algorithm depends on the input size. Because of this, time complexity increases. Reducing the number of generations, i. Big-O Notation • We specify the largest term using big-O notation. \ReaderPrograms\ReaderFiles\Chap02\OrderedArray\orderedArray. It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. Linear Complexity: O(n) A linear task’s run time will vary depending on it’s input value. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. Time Complexity of. This linear search has a time complexity of O(n). Self-balanced Binary Search Trees. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The time required to search an element using a linear search algorithm depends on the size of the list. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. The time complexity of Linear Search is O (n). Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. The search time increases proportionately to the number of new items introduced. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. linear search time complexity. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. The time complexity has to do with the critical opeations being performed. Multiply to get n*log(n). Serial Search - Analysis. Thus: I use three elements as the threshold when I will switch to Dictionary lookups from List loops. Linear Search is an example for Linear Time Complexity. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. O(n) Linear: Time to complete the work grows in a 1 to 1 relation to input size. In this set of Solved MCQ on Searching and Sorting Algorithms in Data Structure, you can find MCQs of the binary search algorithm, linear search algorithm, sorting algorithm, Complexity of linear search, merge sort and bubble sort and partition and exchange sort. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search has linear-time complexity; binary search has log-time complexity. Conversely, giv. Data Structures and Algorithms Objective type Questions and Answers. com: Time complexity of an algorithm: In computer science , the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function. Browse other questions tagged time-complexity linear-algebra matrices or ask your own question. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). See full list on yourbasic. The time complexity is the sum of time spent in all calls plus some extra preprocessing time. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. This video explains the time complexity analysis for binary search. See full list on freecodecamp. We therefore take the complexity of inverted index search to be (as discussed in Section 2. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element “J” in a given sorted list from A-X. for(i=0; i N; i++) { for(j=0; j. Hence the complexity is O(n). Time Complexity of Linear Search Algorithm is O (n). It depends on the condition given in the for loop. Linear Search vs Binary Search. linear regression and the correlation coefficient. Time complexity. The time complexity of suffix tree construction has been shown to be equiv-alent to that of sorting [7]. i it is exponentially better than a brute force search. This calculation will be independent of implementation details and programming language. Strictly, we should say the average complexity is $$\mathcal{O}(n)$$. Linear Time Complexity: O(n) When time complexity grows in direct proportion to the size of the input, you are facing Linear Time Complexity, or O(n). As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. So during the execution of an algorithm, the total time required that will be decided in the time complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. So there is no advantage of binary search over linear search if every search is on a fresh array. Search time is proportional to the list size. For N = 1024 it is 80% faster, and I guess the performance ratio should converge to two at infinity. > How are strings stored in Python? As arrays? As linked lists?. Complexity Classes. Linear time complexity might sound inefficient when you image input sizes in the billions, but linear time isn't actually too bad. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. Thus, the time complexity of this recursive function is the product O(n). For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Time Complexity of Linear Search Algorithm is O (n). It is conjectured that the indistinguishability of photons is responsible for the computational complexity of linear optics. Consider a sorted array of 16 elements. the time required to complete the above operation increases linearlywith respect to 'n' (input). Time complexity of algorithms An algorithm is a collection of steps that process a given input to produce an output. See full list on yourbasic. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. The search stops when the item is found or when the search has examined each item without success. A sorted array is required New insert() Searching a sorted array by repeatedly dividing the search interval in half. For example. Don't overanalyze O(N). Thus, we have-. For example, for a function f(n) Ο(f(n)) = { g(n) : there exists c > 0 and n 0 such that f(n) ≤ c. Hence, this is another difference between linear search and binary search. Time Complexity. The search time increases proportionately to the number of new items introduced. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. DTIME[2polylogn]. The linear search with break becomes faster than counting linear search shortly after N = 128. Linear time: O(n). The cases are as follows − Best Case − Here the lower bound of running time is calculated. Answer: d Explanation: It is practical to implement linear search in the situations mentioned in When the list has only a few elements and When performing a single search in an unordered list, but for larger elements the complexity becomes larger and it makes sense to sort the list and employ binary search or hashing. The space complexity is also. 2012: J Paul Gibson T&MSP: Mathematical Foundations MAT7003/ L9-Complexity&AA. Informally, this means that the running time increases at most linearly with the size of the input. The order of growth (e. If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). A(n) = $\frac{n + 1}{2}$ However, I am having trouble coming up with the Average Case complexity in the case where half of the elements in the size n array are duplicates. In this case, the insertion sort algorithm has a linear running time (i. Linear search is a very basic and simple search algorithm. The List has an O(N) linear time complexity. The time complexity of suffix tree construction has been shown to be equivalent to that of sorting: O(n) for a constant-size alphabet or an integer alphabet and O(n log n) for a general alphabet. ) Combinatorial- Algebraic Methods in Applied Mathematics, pp. With a faster sorter like merge-sort, which is O(N*log(N. In linear search we simply iterate over elements and check whether it is the desired element or not. Imagine the time it will take to search a Letter of Daniel, Eric, Jayesh or any Employee. doubling n, time increases only by a factor of c. This is a more mathematical way of expressing running time, and looks more like a function. The time taken to search a given element will increase if the number of elements in the array increases. Here that is linear time. Therefore, much research has been invested into discovering algorithms exhibiting linear time or, at least, nearly linear time. Hourly Update. If you had to search for a name in a directory by reading. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm’s running. • for selection sort, C(n)=n2/2-n/2 n2/2 • In addition, we’ll typically ignore the coefficient of the largest term (e. Time complexity of Bubble sort in Best Case is O(N). As we will see in the next chapter, kNN's effectiveness is close to that of the most accurate learning methods in text classification (Table 15. In [15] a chosen plaint-text linear attack was suggested and in [5] time complexity of the attack rst stage was reduced by using Fast Fourier Transform. The time complexity of suffix tree construction has been shown to be equiv-alent to that of sorting [7]. This search requires only one unit of space to store the element to be searched. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. Time Complexity For Linked Lists; Time Complexity; Time Complexity In While Loop; My Future Plan Because Of My Teacher? Python Mini-challenge: "Lucky" Numbers; Advance Code Not Hardware? Efficiency Of Linear Search Vs Binary Search In Unsorted List; Algorithm Not Efficient Enough; Cryptography And Data Structure; Filling List(s) With Random Numbers. These problems will introduce things (like the variable i above) just to waste your time. This requires to scan the array completely and check each element for the array that we need to search. Linear Search is an example for Linear Time Complexity. Linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list. In linear search, we have to check each node/element. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. Linear time or O(n). 무슨말인지모르겠다면 아래글을 쭉. The number of steps and time required to solve a problem is based on input size. Best Case. It makes an exponential workspace and solves the problems with exponential complexity in a polynomial (even linear) time. Also, each algorithm's time complexity is explained in separate video lectures. best-case: this is the complexity of solving the problem for the best input. Solving a system of linear equations has a complexity of at most O (n 3). This means the bigger the number of wine bottles in our system, the more time it will take. Definition of time complexity in the Definitions. In the linear search problem, the best case occurs when x is present at the first location. This research includes both software and hardware methods. A few common algorithmic complexities: O(log n) - binary search. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. So, we can write this as Ω(n). Complexity theory argues that systems are complex interactions of many parts which cannot be predicted by accepted linear equations. The measure for the working storage an algorithm needs is called space complexity. , the work is O (1) comparison. Time complexity of neural network. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). The time complexity function expresses that dependence. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. The asymptotic complexity is defined by the most efficient (in terms of whatever computational resource one is considering) algorithm for solving the game; the most common complexity measure (computation time) is always lower-bounded by the logarithm of the asymptotic state-space complexity, since a solution algorithm must work for every. , c ~ 2d)! Contrast with exponential: For any constant c, there is a d such that n → n+d increases time. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. Finding the median in a list seems like a trivial problem, but doing so in linear time turns out to be tricky. Linear search is linear O(N) Binary search depends on if the tree is balanced or not. Informally, this means that the running time increases at most linearly with the size of the input. ” Most cryptanalytic papers discuss certificational attacks: Data complexity — just slightly less than the entire code book. The average to the worst case of this kind of search is a linear complexity or O(n). Informática Educativa [email protected] If there are NO nested loops we can probably guess the complexity of the code we looking at would be in the O(n). Operation count: In this technique, we consider the operations in the given algorithm or program that contribute to the execution time and count how many times those operations will be performed. These have yielded near-linear time algorithms for many diverse problems. I know the answer is O(n), but is this correct: The first element has probability $1/n$ and requires 1 comparison; the second probability $1/(n-1)$ and requires 2 comparisons. It was experimentally found in [6, 7] that time complexity of Matsui’s attack on DES may be decreased with a better ranking of the values of relevant sub-key bits, though data complexity and. The List has an O(N) linear time complexity. I know that for an array of size n distinct elements, the Average Case complexity for linear search is as follows:. This obviously requires a constant number of comparison operations, i. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). The time complexity of suffix tree construction has been shown to be equiv-alent to that of sorting [7]. complexity addressing precisely the kind of problem raised in the last two paragraphs: Given a computational prob-lem, can it be solved by an e cient algorithm? For many common computational tasks (such as nding a solution of a set of linear equations) there is a polynomial-time algo-rithm that solves them|this class of problems is called P. Eight time complexities that every programmer should know. While that isn’t bad, O (log. This video explains the time complexity analysis for binary search. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. Generate an hypothesis: The running time is about 1 x 10-10 x N 3 seconds 4. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. 14 Code sample for Linear Regression. O(n log n) Linearithmic: This is a nested loop, where the inner loop runs in log n time. 39+ o (1)) en ln n to en ln n + O ( n ) in expectation and with high probability, which is tight up to. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. If the element is found then its position is displayed. University of Toronto - Fall 2000 Department of Computer Science Week 12 - Complexity & Searching Complexity The complexity of an algorithm is the amount of a resource, such as time, that the algorithm requires. It is generally one of the first algorithms taught in computer science courses because it is a good algorithm to learn to build intuition about sorting. During the study of discrete mathematics, I found this course very informative and applicable. In most of the cases, you are going to see these kind of Big-O running time in your code. for(i=0; i N; i++) { for(j=0; j. Linear search performs equality comparisons and Binary search performs ordering comparisons; Let us look at an example to compare the two: Linear Search to find the element “J” in a given sorted list from A-X. Firstly, we analyze the time complexity of the iterative algorithm and then recursive. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. This is an example of logarithmic complexity. Time Complexity of Sorting Algorithms Let's check the time complexity of mostly used sorting algorithms. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. More than 1 Million Books in Pdf, ePub, Mobi, Tuebl and Audiobook formats. I Linear: RHS is a sum of multiples of previous terms of the sequence (linear combination of previous terms). The number of steps and time required to solve a problem is based on input size. In Binary search half of the given array will be ignored after just one comparison. Complexity Classes. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. See full list on towardsdatascience. \ReaderPrograms\ReaderFiles\Chap02\OrderedArray\orderedArray. all of the mentioned. The search time increases proportionately to the number of new items introduced. While that isn’t bad, O (log. Aaronson and Arkhipov argued in section 1. Complexity International-- journal for scientific papers dealing with any area of complex systems research. A single iteration (loop) over all the elements in the array gives us a complexity of O(n). If you give a condition in the inner loop that will always terminate the inner loop and/or outer loop without executing n times for all elements, then it will have less than O(n) time. Here, n is the number of elements in the linear array. It means we generate a vector that has 5 elements, and these elements are bounded in [-11,11]. Time complexity is a function dependent from the value of n. (And if the number. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Lookups on arrays and objects are going to be constant time if you access them directly. So, the time complexity of binary search is O(log2n). It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Linear search is iterative whereas Binary search is Divide and conquer. Binary Search Algorithm and its Implementation. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n − 1. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. In the best-case scenario, the element is present at the beginning of the list and in the worst-case, it is present at the end. all of the mentioned. Linear search is a perfect example. It concisely captures the important differences in the asymptotic growth rates of functions. By this logic, we can say that painting pictures is slower than baking cookies. java logarithms Complexity of algorithm Time complexity Space complexity Time complexity: in big O notation. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. Linear-Time Sorting. Also, each algorithm's time complexity is explained in separate video lectures. The second one runs in time sublinear in d, assuming the edit distance is not too small. Linear time: O(n). The best algorithms for sorting a random array have a run time of O(n * log n). Always takes the same time. Linear search is a perfect example. i it is exponentially better than a brute force search. The Idea of time complexity is not to calculate how much time an algorithm will take to complete, but to compute the order of magnitude of time for the completion of computation by an algorithm. • VERY difficult to develop. This is an example of logarithmic complexity. Time complexity of a related-key attack: “Thus, the total time complexity of Step 2(b) is about 2256 ·2167. Hence Bisection Search is way better than Linear Search. Search time is proportional to the list size. In this post I’m going to walk through one of my favorite algorithms, the median-of-medians approach to find the median of a list in deterministic linear time. 2 , page 15. fractal image compression time complexity image compression critical issue video sequence alternative method large number computer animation multi-dimensional nearest neighbor search data storage logarithmic time decoding phase transmission time encoding step linear time image portion multi-dimensional search sequential search data compression. So, the time complexity of binary search is O(log2n). Using the hypothesis, make a prediction: When N =. Although the limiting factor for linear cryptanalysis attacks is usually the data complexity, such an improvement is relevant and can be motivated both by practical and theoretical reasons, as the following scenarios underline. The number of steps and time required to solve a problem is based on input size. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It concisely captures the important differences in the asymptotic growth rates of functions. Here, n is the number of elements in the linear array. Linear time is the best possible time complexity in situations where the algorithm has to sequentially read its entire input. Program for Recursive and Non-Recursive Binary Search in C++ - Analysis Of Algorithms / Data Structures. Visualize high dimensional data. Let us consider an algorithm of sequential searching in an array. Like an array, a linear list stores a collection of objects of a certain type, usually denoted as Time complexity, space complexity, and the O-notation : 2. It went through the entire list so it took linear time. com Linear Time Complexity. We could also say it is linear on the number of entries in the table but that is less commonly used. Huan Li, Zhouchen Lin; 21(33):1−45, 2020. Time Complexity : θ ( n ) Space Complexity : O(1) Linear Search Example. Time complexity of linear search -O(n) , Binary search has time complexity O(log n). time complexity of a linear cryptanalysis attack using algorithm 2. This is an example of logarithmic complexity. Note that an algorithm might take different amounts of time on inputs of the. If you were to find the name by looping through the list entry after entry, the time complexity would be O (n). Lets say I have the list 10,20,30,40,50,60,30,40,50. The second one runs in time sublinear in d, assuming the edit distance is not too small. That is, I'm looking for references that looks like the following. complexity = in between logN and N. Here, although your array is of a fixed size, the time needed to complete the operation is still a linear function of the number of elements in the array. The main points in these lecture slides are:Time Complexity, Complexity of Algorithms, Execution Time, Space Complexity, Worst Case Analysis, Division of Integers, Number of Comparisons, Binary Search, Average Case Complexity, Complexity of Bubble Sort. BIG O Notation – Time Complexity and Space Complexity Binary search is a technique used to search sorted data sets. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. // Time complexity: O(1) // Space complexity: O(1) int x = 15; x += 6; System. This requires to scan the array completely and check each element for the array that we need to search. This first book consists of chapters 1 and 2 of the fourth volume. starts in the middle, then see if the value being searched for is greater or less than the middle value. As a rule of thumb, it is best to try. 2) and, assuming average document length does not change over time,. O(1) indicates that the algorithm used takes "constant" time, ie. Sequential search write a sequential search function and then find the best, worst, and average case time complexity. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Thus, the amount of time taken and the number of elementary operations performed by the algorithm are taken to differ by at most a constant factor. Time Complexity of Binary Search Algorithm is O(log 2 n). linear: sorting twice the number of elements takes quite a bit more than just twice as much time; searching (using binary search) through a sorted list twice as long, takes a lot less than twice as much time. This paper reports some new results on the average time complexity of EAs. In: Markov, A. Serial Search - Analysis. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. what we do is we simply loop over array and check whether it is. For a linear-time algorithm, if the problem size doubles, the number of operations also doubles. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. The measure for the working storage an algorithm needs is called space complexity. Let n represent the size of the array arr. We could also say it is linear on the number of entries in the table but that is less commonly used. Always takes the same time. algorithm runs in near-linear time, namely d1+ε for any fixed ε > 0. Reducing the number of generations, i. 1 of [] that the exchange symmetry of identical bosons creates an effective entanglement (a kind of artificial entanglement), which would be the origin of the computational complexity in linear optics. The running time of the loop is directly proportional to N. Therefore, by using hash table, we can achieve linear time complexity for finding the duplicate. characterises a function based on growth of function C. In order to be able to classify algorithms we have to define limiting behaviors for functions describing the given algorithm. The time complexity of above algorithm is O(n). Here, complexity refers to the time complexity of performing computations on a multitape Turing machine. These Multiple Choice Questions (mcq) should be practiced to improve the Data Structure skills required for various interviews (campus interview, walk-in interview, company interview), placement, entrance exam and other competitive examinations. Diagram above is from Objective-C Collections by NSScreencast. Trees Data Structures for Beginners. Space complexity : O (1) O(1) O (1) or (O (n) O(n) O (n)) We sorted nums in place here - if that is not allowed, then we must spend linear additional space on a copy of nums and sort the copy instead. The time complexity of ECL2 is O (n + M ϵ 1 w 2) , where O(n) is the time complexity of scoring and binning, and O (M ϵ 1 w 2) is the time complexity of enumerating bin pairs. Sort an array of 0's, 1's and 2's in linear time complexity; Checking Anagrams (check whether two string is anagrams or not) Relative sorting algorithm; Finding subarray with given sum; Find the level in a binary tree with given sum K; Check whether a Binary Tree is BST (Binary Search Tree) or not; 1[0]1 Pattern Count. near-linear time. As a rule of thumb, it is best to try. However, in my previous experiments, it appears to be O (N), namely linear complexity!. The search stops when the item is found or when the search has examined each item without success. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. A few common algorithmic complexities: O(log n) - binary search. Dual first-order methods are essential techniques for large-scale constrained convex optimization. The best case time in linear search is for the first element i. This linear search has a time complexity of O(n). Aaronson and Arkhipov argued in section 1. given two natural numbers $$n$$ and $$m$$, are they relatively prime?. O(N)—Linear Time: Linear Time Complexity describes an algorithm or program who’s complexity will grow in direct proportion to the size of the input data. what we do is we simply loop over array and check whether it is. In case of a sorted array, Binary Search is faster but the caveat here is also on how arrays are treated by the Language Translator. Let us assume that given an array whose elements order is not known. Consider that we have an algorithm, and we are calculating the time. Time Complexity of Bisection Search is O(log n). Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). One example is the binary search algorithm. The number of operations in the best case is constant (not dependent on n). The time complexity of suffix tree construction has been shown to be equiv-alent to that of sorting [7]. Download Binary search program. Alright, so we have linear-over-n many logarithmic-over-n loops. java graph-algorithms competitive-programming dfs binary-search-tree common-algorithms time-complexity implementation bfs longest-common-subsequence binary-search segment-tree binary-indexted-tree two-pointers space-complexity all-pairs-shortest-path matching-algorithm maximal-bipartite-matching lower-bound lowest-common-ancestor. Linear Search is sequential search which scans one item at a time. Space Complexity. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a naïve algorithm runs in O(n2) time. This requires to scan the array completely and check each element for the array that we need to search. This web page gives an introduction to how recurrence relations can be used to help determine the big-Oh running time of recursive functions. See full list on yourbasic. As investigated in [ ],theHPPcanbesolvedusing. Time complexity of Bubble sort in Worst Case is O(N^2), which makes it quite inefficient for sorting large data volumes. All have polynomial time complexity while some allow very long steps in favorable circumstances. We show an improved algorithm for the satis ability problem for circuits of constant depth and linear size. We will see more about Time Complexity in future. In a serial search, we step through an array (or list) one item at a time looking for a desired item. for(i=0; i N; i++) { for(j=0; j. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes? A) O(1) (B) O(log n ). It measures the worst case time complexity or the longest amount of time an algorithm can possibly take to complete. O(1) is the best possible time complexity! Data structures like hash tables make clever use of algorithms to pull off constant time operations and speed things up dramatically. So, the time complexity of binary search is O(log2n). Eight time complexities that every programmer should know. O(n²) – Quadratic Time. Serial Search - Analysis. The time complexity of an algorithm is commonly expressed using big O notation , which excludes coefficients and lower order terms. The first is the way used in lecture - "logarithmic", "linear", etc. The improvement of the proposed linear-time algorithm compared with ECL2 (Yu et al. Here you will learn about python binary search with program and algorithm. This search requires only one unit of space to store the element to be searched. Time complexity. These problems will introduce things (like the variable i above) just to waste your time. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. For instance, it is known since the 1960s and 70s (e. Linear Time: O(n) An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i. Hence time complexity of the Binary search is O(LogN). The best case time in linear search is for the first element i. Most of the time we can speak on sorting integers in linear time, but as we can see later this is not the only case. Hourly Update. Time complexity Posted 28 December 2015 - 04:35 PM Hi guys,lets say I have algorithm ,which finds ,when the number in a list is bigger than the next one. It is easy to see that $$\widetilde{\mathcal {S}}$$ can be obtained in one pass through $$\widetilde{\mathcal {A}}$$ and $$\widetilde{\mathcal {B}}$$, therefore in linear time. Time complexity (linear search vs binary search) 1. I n linear search, we need to write more code whereas in binary search we need to write less code. algorithm solving a Boolean satis ability problem on n variables is improved i it takes time O(2cn) for some constant c < 1, i. now do a linear search starting from X(conservative but accurate second step - slow). So the option is 'B'. 무슨말인지모르겠다면 아래글을 쭉. In this book, Keith Morrison introduces complexity theory to the world of education, drawing out its implications for school leadership. For better understanding,lets take an example: given an array arr[]={12,11,4,0,3,5} and we want to search whether 5 present in the given array or not. In Binary search half of the given array will be ignored after just one comparison. Nested for loops are the perfect example of this category. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. The complexity of linear search algorithm is. Time complexity is commonly estimated by counting the number of elementary operations performed by the algorithm, supposing that each elementary operation takes a fixed amount of time to perform. Run time is O (log N) Sample code for ordered array. To measure Time complexity of an algorithm Big O notation is used which: A. In order to speed up the static analyses formulated using the Dyck-CFL reachability problems, we propose an efficient algorithm of O(n) time for the Dyck-CFL reachability problem when the graph considered is a bidirected tree with specific constraints, while a naïve algorithm runs in O(n2) time. Indeed, 100 cookies don’t take much longer than 12 cookies — provided you have a big enough bowl. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). Time complexity (linear search vs binary search) 1. linear regression and the correlation coefficient. Tests are robust , non-parametric statistical tests, since timing is noisy (so need to be robust), and noise can take various forms (so non-parametric, since no particular model of noise). Binary search. So the option is 'B'. This study proposes linear time complexity sorting algorithms for nearest level control-based BE and TR MMC models to further accelerate the EMT simulation of the equivalent MMC-HVdc models. From this we can see these equations are similar and our equation matches linear equation. Counting sort and radix sort assume that the input consists of integers in a small. Since binary search has a best case efficiency of O(1) and worst case (average case) efficiency of O(log n), we will look at an example of the worst case. As the number increases so does the time difference. For a general alphabet, suffix tree construction has time bound of Θ(nlogn). Here is the official definition of time complexity. Hence number of times while loop will execute will determine the complexity of the algorithm. If you had to search for a name in a directory by reading. If we assume we needed to search the array n times the total worst case run time of the linear searches would be O (n^2)). All complex systems can be seen as a number of nodes joined together – lines and junctions – or in the case of the human brain, long spindly nerve cells and synapses. Linear Search Simple search from the first element to the last till we find the required element. So, the time complexity of binary search is O(log2n). It has a complexity of n 2. Hourly Update. Let's take an array int arr [] = { 2,1,7,5,9} Suppose we have to search an element 5. the complexity. The time complexity function expresses that dependence. As a set, they are the fourth volume in the series Mathematics and Physics Applied to Science and Technology. Finally, together with the analysis, it is concluded that the linear time complexity is validated based on the experiments. This is a more mathematical way of expressing running time, and looks more like a function. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth. at 11:59pm • Asymptotic analysis Asymptotic Analysis CSE 373 Data Structures & Algorithms Ruth Anderson Spring 2007 04/04/08 2 Linear Search vs Binary Search Linear Search Binary Search Best Case Asymptotic Analysis Worst Case So … which algorithm is better?. Time complexity (linear search vs binary search) 1. Time Complexity and the divide and conquer strategy Or : how to measure algorithm run-time And : design efficient algorithms Oct. Worst-case running time - the algorithm finds the number at the end of the list or determines that the number isn't in the list. Complexity and running time Factors: algorithmic complexity, startup costs, additional space requirements, use of recursion (function calls are expensive and eat stack space), worst-case behavior, assumptions about input data, caching, and behavior on already-sorted or nearly-sorted data; Worst-case behavior is important for real-time systems. For a general alphabet, suffix tree construction has time bound of Θ(nlogn). That is, I'm looking for references that looks like the following. Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Motivation: A crucial phenomenon of our times is the diminishing marginal returns of investments in pharmaceutical research and development. This first book consists of chapters 1 and 2 of the fourth volume. It's not easy trying to determine the asymptotic complexity (using big-Oh) of recursive functions without an easy-to-use but underutilized tool. This technique is probably the easiest to implement and is applicable to many situations. Definition of time complexity in the Definitions. Algorithmic Complexity Notes on Notation: Algorithmic complexity is usually expressed in 1 of 2 ways. Data Structure MCQ - Complexity. In linear search we simply iterate over elements and check whether it is the desired element or not. for(i=0; i N; i++) { for(j=0; j. If the values match it will return success. For a general alphabet, suffix tree construction has time bound of Θ(nlogn). In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity, so to overcome them to a level an algorithm based on dichotomic (i. In a serial search, we step through an array (or list) one item at a time looking for a desired item. See full list on freecodecamp. Now considering the worst case in which the search element does not exist in the list of size N then the Simple Linear Search will take a total of 2N+1. Algorithm analysis is an important part of a broader computational complexity theory, which provides theoretical estimates for the resources needed by any. Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching comparison to Linear search. In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input:226. The running time of the two loops is proportional to. Results Here, then, as a concrete example, is a plot of the run-times of the most interesting algorithms on an Intel Core i7 running at 2. O(1): Constant Time Complexity. O(n) - finding the largest item in an unordered list. This first book consists of chapters 1 and 2 of the fourth volume. Since binary search algorithm splits array in half every time, at most log 2 N steps are performed. Best Case. larger search space of constituent trees (compared to the space of dependency trees) would make it unlikely that accurate parse trees could be built deterministically, we show that the precision and recall of constituents produced by our parser are close to those produced by statistical parsers with higher run-time complexity. Time Complexity : This section explains the importance of time complexity analysis, the asymptotic notations to denote the time complexity of algorithms. The best case for a linear search algorithm is to have the value x for which one is searching located at the first position of the ADT. 2) and, assuming average document length does not change over time,. Most algorithms, however, are built from many combinations of these. The "Binary Search Time Complexity" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Here complexity is said to be linear. This time complexity is a marked improvement on the O(N) time complexity of Linear Search. So time complexity in the best case would be Θ(1) Most of the times, we do worst case analysis to analyze algorithms. Suppose varMin=-11, and varMax=11, and varSize=5. So there must be some type of behavior that algorithm is showing to be given a complexity of log n. Constant time compelxity, or O(1), is just that: constant. This time complexity of binary search remains unchanged irrespective of the element position even if it is not present in the array. It will be easier to understand after learning O(n), linear time complexity, and O(n^2), quadratic time complexity. first do a binary search (agressive first step - fast) with 1 bulb. The time factor when determining the efficiency of algorithm is measured by. all of the mentioned. Totally it takes '4n+4' units of time to complete its execution and it is Linear Time Complexity. worst case, the time for insertion is proportional to the number of elements in the array, and we say that the worst-case time for the insertion operation is linear in the number of elements in the array. As we learned in the previous tutorial that the time complexity of Linear search algorithm is O(n), we will analyse the same and see. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n − 1. If the element is found then its position is displayed. So for any value of n it will give us linear time. For the analysis to correspond usefully to the actual execution time, the time required to perform a fundamental step must be guaranteed to be bounded above by a constant. compares each element with the value being searched for, and stops when either the value is found or the end of the array is encountered. Gorky University Publishers, Gorky (1985) (in Russian) Google Scholar. If you had to search for a name in a directory by reading. Most algorithms, however, are built from many combinations of these. Time and space complexity depends on lots of things like. However, in my previous experiments, it appears to be O (N), namely linear complexity!. Neglecting the constant value 5 the complexity would be N as loop will run N times so it does not fit the definition of linear time. We observe how space complexity evolves when the algorithm’s input size grows, just as we do for time complexity. We then verify if these times look like the time complexity we're expecting (constant, linear, or polynomial (quadratic or greater)). O(n) is for linear complexity, O(n 2) is for quadratic. Worst Case time complexity is O(n) which means that value was not found in the array (or found at the very last index) which means that we had to iterate n times to reach to that conclusion. In case of the monks, the number of turns taken to transfer 64 disks, by following the above rules, will be 18,446,744,073,709,551,615; which will surely take a lot of time!!. > How are strings stored in Python? As arrays? As linked lists?. O(N^2) because it sorts only one item in each iteration and in each iteration it has to compare n-i elements. This video is meant for educational. Introduction The time complexity of a given algorithm can be obtained from theoretical analysis and computational analysis according to the algorithm’s running. So, an algorithm taking X second or 2X + 3 seconds have the same complexity. This means that as the input grows, the algorithm takes proportionally longer to complete. The idea behind linear search is to compare the search item with the elements in the list one by one (using a loop) and stop as soon as we get the first copy of the search element in the list. Learn more about time complexity of neural network. The "Binary Search Time Complexity" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Unbalanced binary search tree can turn into a linked list in the worst case if the elements added are in descending order so O(N) time complexity. For typical values of n = 30, m = 30, and q = 5, the time complexity would be 4500, which is much higher than 110. What we are left with is the fact that the time in sequential search grows linearly with the input, while in binary search it grows logarithmically -. If we plot the graph of an+b for different values of n we will see that it is a straight line. These approximation and runtime guarantees are significantly better then the bounds known for worst-case inputs, e. What’s the maximum number of loop iterations? log2n That is, we can’t cut the search region in half more than that many times. Time complexity (linear search vs binary search) 1. java that reads from standard input a sequence of integers that are between 0 and 99 and prints to standard output the same integers in sorted order. The time complexity of linear search is O(n), meaning that the time taken to execute increases with the number of items in our input list lys. For example. Morzhakov, N. Linear Homogeneous Recurrences De nition A linear homogeneous recurrence relation of degree k with constant coe cients is a recurrence relation of the form an = c1an 1 + c2an 2 + + ck an k with c1;:::;ck 2 R , ck 6= 0. the complexity. In my knowledge, the time complexity should be at least O (N^2) or O (NlogN) (the N is number of links), considering it is a graph problem. • DEMO • Conclusion: Maybe I can SCALE well … Solve O(10^12) problems in O(10^12). Linear time or O(n). Algorithm Complexity When N doubles Examples Constant 1 increases fixed times No loop Logarithmic log N increases constant Binary search Linear N doubles Traverse an array Linearithmic NlogN more than doubles Quick/Merge Sort, FFT Quadratic N^2 increases fourfold B Cubic N^3 increases eightfold NxN matrix multiplication Exponential 2^N running time squares!. Bubble sort is a simple, inefficient sorting algorithm used to sort lists. At each time-step in this strategy, each node in the network transmits a weighted linear combination of its previous transmission and the most recent transmissions of its. Total comparisons in Bubble sort is: n ( n – 1) / 2 ≈ n 2 – n Best case 2: O (n ) Average case : O (n2) Worst case : O (n2) 3. See full list on yourbasic. Know Thy Complexities! Hi there! This webpage covers the space and time Big-O complexities of common algorithms used in Computer Science. The time complexity of linear search is O(N) while binary search has O(log 2 N). Run time is O (log N) Sample code for ordered array. i it is exponentially better than a brute force search. Given an arbitrary network of interconnected nodes, each with an initial value, we study the number of time-steps required for some (or all) of the nodes to gather all of the initial values via a linear iterative strategy. If connections are sparse, then sparse math can be used for the gradient computations, etc leading to reduced complexity. | 2020-11-29T07:31:10 | {
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https://math.stackexchange.com/questions/2739192/the-number-555-555-can-decompose-as-the-product-of-two-factors-of-three-digit | The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?
The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?
I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \cdot 7 \cdot 37) (5 \cdot 11 \cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?
Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish.
• list all the divisors in order. $\sqrt{555555} \approx 745.36.$ One of your divisors must be larger than that but still smaller than 1000. The other divisor in your pair will be between 555.56 and 745.36. I guess you can find them by hand, look at all the products of two primes, then three primes. You do not need to do four primes because you already did two – Will Jagy Apr 16 '18 at 2:34
• Yeah, two primes is too small, the biggest is $13 \cdot 37 = 481.$ So, three primes each – Will Jagy Apr 16 '18 at 2:38
Nothing wrong with the approach Will gives in the comments. Here's another way. Obviously $555,555=555\times1001$, but $1001=7\times11\times13$ is a little too large. The way to make it a little smaller is to swap the factor 7 with the factor 5 in 555, which gives you your solution, $(3\times7\times37)(5\times11\times13)$.
The factors $7,11,13$ can't be used together, since $(7)(11)(13)=1001$.
So one of the groups, group $1$ say, must have exactly two of the factors $7,11,13$.
Hence the factors $3$ and $5$ can't both be in group $1$, else the product of the factors in group $1$ would be at least $(3)(5)(7)(11) > (7)(11)(13)$.
Similarly, the factor $37$ can't be in group $1$, else the product of the factors in group $1$ would be at least $(37)(7)(11) > (7)(11)(13)$.
Label the other group as group $2$.
Thus, group $2$ contains
• $37$
• Exactly one of $7,11,13$.
• At least one of $3,5$.
But since $(37)(27) = 999$, the factors in group $2$ other than $37$ must have a product which is at most $27$.
It follows that neither of the factors $11$ or $13$ is in group $2$, since $(11)(3) > 27$, and $(13)(3) > 27$.
So the factor $7$ must be in group $2$, and the factors $11,13$ must be in group $1$.
Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$.
Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$.
Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$.
It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \cdot 5 \cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\cdot 13 \cdot 37 \gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$.
• I also thought that trying to solve for the same question just that for 2 factors of 2 digits and then I would split one of the digits into 2. Would this help me solve this type of question faster? Also, the question doesn't require me to state the numbers, but it does require me to state the amount of ways I can decompose it. – Brian Blumberg Apr 16 '18 at 2:44
• I don't understand the first sentence. What does it mean to split one of the digits into $2$? Having gotten down to $9$ possibilities it won't take long to try them all. You could note that $3\cdot 11\cdot 37$ is too large, so $37$ must be by itself and $11,13$ must be in the other set. Now just try each of $3,5,7$ times $11\cdot 13=143$ to see which are in range and find that only $5$ works. – Ross Millikan Apr 16 '18 at 2:53
• Yeah, never mind my question. I forgot to delete it after I realized that. Thanks. – Brian Blumberg Apr 16 '18 at 3:01
The prime factor are $3 · 5 · 7 · 11 · 13 · 37$ so there are $2^6=64$ factors and $32$ complement pairs. Just list them all but don't bother with those that are less than $555555/999=556$
Toss out, $1,3,5,7,11,13,37,3*5,3*7,3*11,3*13,3*37,3*5*7,3*5*11,3*5*13, 3*5*37,3*7*11,3*7*13$ (that's 18 that are too small).
$3*7*37=777$ and its compliment is $5*11*13=715$. (That's 1 that is acceptible)
We can continuing tossing out $3*11*13$ and $3*11*37$and $3*13*37$ are too high so we toss them. (That's 21 that are unaccptible)
$3*5*7*11$ is too high so there wonvt be any more factors that are multiples of $3$ in range. Hence no other complements which aren't multiples of $3$ will be in range either.
Of the 9 we haven'haven't considered: $3*5*7*13,3*5*7*37,3*5*11*13,3*5*11*37,3*5*13*37,3*7*11*13,3*7*11*37,3*7*13*37,3*11*13*37$ are all too big.
So that was exhaustive.
We can write $$555,555=5\times111,111$$ and notice that $111=37\times3$, so we have $$555,555=5\times37\times3003$$ and since $1001=7\times11\times13$, the prime factorization is $$555,555=3\times5\times7\times11\times13\times37.$$
If we multiply each of the three combinations $11\times13$, $11\times37$ and $13\times37$ by $3$, we see that only $3\times(13\times37)$ exceeds $1000$ which is a four-digit number.
Hence $37$ must pair with two other one-digit numbers, and $11\times13$ must pair with either $3$ or $5$ since $7\times11\times13>1000$.
If $11\times13$ pairs with $3$, then the other product must be $$37\times(5\times7)>1000$$ which is not a three-digit number.
Therefore, the only possible combination for $555,555=P_1\times P_2$ is $$P_1=5\times11\times13,\quad P_2=3\times7\times37.$$
I think the procedure might be based on Fermat method.
We can calculate that the number $555555$ can be expressed as a difference of two squares:
$746^2 - 31^2=556516-961$
Number $31$ is the distance from $746$ both ways which gives $715$ and $777$ accordingly. From here the smallest prime factor of:
$777$ is number $3$ which equals to $3\cdot259$ and for
$715$ is number $5$ which equals to $5\cdot143$
Further factorisation results in final solution where $143=11\cdot13$ and $259=7\cdot37$ therefore:
$(5\cdot11\cdot13)(3\cdot7\cdot37)$ can satisfy equation.
If $555{,}555=ab$ with $a,b\lt1000$, then we must also have $a,b\gt555$ (e.g., if $b\lt1000$, then $a=555{,}555/b\gt555{,}555/1000\gt555$). Now since $555{,}555=3\cdot5\cdot7\cdot11\cdot13\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\gt555=15\cdot37$, we see that $k\gt15$, and from $a\lt1000$, we see that $k\le\lfloor1000/37\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\lt k\le27$ is $k=3\cdot7=21$. Thus $a=37\cdot21=777$, $b=5\cdot11\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers. | 2019-05-21T18:44:39 | {
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https://math.stackexchange.com/questions/1960598/contrapositive-in-change-from-no-self-defeating-object-to-every-object-can-be | # Contrapositive in change from “no self-defeating object” to “every object can be defeated”?
Here, Terence Tao presents a collection of similar mathematical arguments that he calls "no self-defeating object" (examples are Euclids proof of the infinitude of the primes and Cantor's theorem). In the second post, he remarks that one can reformulate these "no-self defeating object" arguments to get a "every object can be defeated"-version.
The simplest example "no self-defeating object" goes as follows:
Proposition 1 (No largest natural number). There does not exist a natural number N that is larger than all the other natural numbers.
Proof: Suppose for contradiction that there was such a largest natural number N. Then N+1 is also a natural number which is strictly larger than N, contradicting the hypothesis that N is the largest natural number.
The corresponding "every object can be defeated"-version is:
Proposition 1′. Given any natural number N, one can find another natural number N' which is larger than N.
Proof. Take N' := N+1.
Terence Tao also remarks:
"This is done by converting the “no self-defeating object” argument into a logically equivalent “any object can be defeated” argument, with the former then being viewed as an immediate corollary of the latter. This change is almost trivial to enact (it is often little more than just taking the contrapositive of the original statement), but it does offer a slightly different “non-counterfactual” (or more precisely, “not necessarily counterfactual”) perspective on these arguments which may assist in understanding how they work."
My question: What has the contrapositive to do with the change from "no self-defeating object" to "every object can be defeated"?
As I understand it, the "no self-defeating object"-version is of the form "$\neg\exists x: P(x)$" and the "every object can be defeated"-version is "$\forall x:\neg P(x)$". That these are equivalent is de Morgan for quantifiers, what does it have to do with contrapositives?
• – Henning Makholm Oct 9 '16 at 12:14
• @HenningMakholm: Ah, thanks. So do you think that Terence Tao just compared the change from $\neg\exists x : P(x)$ to $\forall x : \neg P(x)$ to taking the contrapositive, but is not saying that this is the same as taking contrapositives? – user376483 Oct 9 '16 at 12:45
• I'm not privy to Tao's thoughts, so the most I'm saying is that I too see a similarity between this kind of rewriting and contraposition, and once mused that it might be useful to call it contraposition. (Not many here agreed with that, though -- but it nice to see perhaps Tao might). – Henning Makholm Oct 9 '16 at 12:48
• One way to make it look more like a contrapositive is to view it as a change from "if $m$ is larger than all natural numbers then $m$ is not a natural number" to "if $m$ is a natural number then $m$ is not larger than all natural numbers". Of course, Tao only says that the process he describes is "often" "little more" than just taking the contrapositive, he does not claim that what he is doing is literally a contrapositive. I do find that mathematicians and logicians often use "contrapositive" informally in a broader sense than its formal meaning. – Carl Mummert Oct 9 '16 at 16:42
If, as is standard in presentations of intuitionistic logic, you treat $\lnot \phi$ as $\phi \Rightarrow \mathsf{false}$ then the role of the contrapositive here becomes clear: the contrapositive of:
$$(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m) \Rightarrow \mathsf{false}$$
is:
$$\lnot \mathsf{false} \Rightarrow \lnot(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m)$$
which, using De Morgan's laws and a tiny bit of arithmetic reasoning (to make things agree with Tao's presentation) is equivalent to: $$\forall N \in \Bbb{N}\cdot\exists m\in \Bbb{N}\cdot m > N.$$
This transformation giving Tao's Proposition $1'$ makes clear the innate constructive nature of the reasoning that is presented in disguise in the proof by contradiction in Tao's Proposition $1$.
• After all, this also relies on De Morgan's rule, and I don't see why you first interpret $\phi$ as $\phi\implies\bot$ and form the contrapositive of $(\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m) \Rightarrow \mathsf{false}$. One could just form the contrapositive of $\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N \geq m$ which is $\exists N \in \Bbb{N}\cdot\forall m\in \Bbb{N}\cdot N > m$. – user377104 Oct 25 '16 at 17:24
• Of course it relies on De Morgan's laws: the point is that the end result is a constructive truth that captures exactly what the proof actually proves. The reason for interpreting $\lnot \phi$ as $\phi \Rightarrow \mathsf{false}$ is because the notion of contrapositive is usually associated with implications: which is the point of the question: "what does this have to do with contrapositives?" The two existentially quantified statements that you claim are contrapositives are not contrapositives in any sense of the term that I am aware of. – Rob Arthan Oct 25 '16 at 19:16 | 2019-10-21T22:44:52 | {
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http://math.stackexchange.com/tags/proof-writing/hot?filter=month | # Tag Info
13
The contrapositive of the statement If $\overbrace{\text{$ab$and$a+b$have the same parity}}^{\large P}$, then $\overbrace{\text{$a$is even and$b$is even}}^{\large Q}$. is If $\overbrace{\text{$a$is odd or$b$is odd}}^{\large\lnot Q}$, then $\overbrace{\text{$ab$and$a+b$have different parities}}^{\large\lnot P}$. Note that $Q$ is the ...
9
Your result is an immediate consequence of the following proposition. Proposition. Suppose $X\subseteq Y$. Then $\mathscr P(X)\subseteq\mathscr P(Y)$. Proof. Let $E\in\mathscr P(X)$. Then $E\subseteq X\subseteq Y$ so that $E\subseteq Y$. Hence $E\in\mathscr P(Y)$. This proves $\mathscr P(X)\subseteq\mathscr P(Y)$. $\Box$ Do you see how your problem is now ...
8
\begin{align} |S| & = |S-T| + |S\cap T| \\[8pt] |T| & = |T-S| + |S\cap T| \end{align} If $|S-T|=|T-S|$, then the two right sides are the same, so the two left sides are the same. We can also write a proof explicitly dealing with bijections. You ask why one would "assume" a bijection exists. The bijection $g$ that you write about is not simply ...
7
You can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.
6
not induction, but maybe useful to note: firstly, since 3 is a prime, we have $n^3 \equiv_3 n$ (Fermat's little theorem) secondly $2n \equiv_3 -n$ (since $3n = 2n + n \equiv_3 0$) adding these two results: $$n^3 + 2n \equiv_3 n-n =0$$
6
By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e $$P^{T}AP=\pmatrix{\mu_1 \\ & \ddots \\ && \mu_n}$$ where $\mu_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$. For any $v$, let $v=Pu$. Then $$v^TAv=u^TP^TAPu=\sum_{k=1}^n\mu_ku_k^2>0$$
5
You need to show us some effort in the future. First, to show two sets are equal, we normally pick an element of the first set, show it is contained in the second, then pick an element in the second, and show it is contained in the first. If we suppose $x \in A$, then $x=2k$ for some integer $k$. Since $x = 2k$, $x = 2(k-1)+2$, and since $k-1$ is an ...
5
Take $N\in\Bbb N$ such that $|a_n-L|<1$ for $n\ge N$ and let $$M=\max\{|a_1|,\ldots,|a_N|,L+1\}.$$
5
Let $f:\{1,\ldots,n\}\to X$ be a surjection. Suppose, for the sake of contradiction, that $X$ has at least $n+1$ distinct elements $\{x_1,\ldots,x_{n+1}\}\subseteq X$. Since $f$ is a surjection, there exists, for each $i\in\{1,\ldots,n+1\}$, some $k_i\in\{1,\ldots,n\}$ such that $f(k_i)=x_i$. Since $f$ is a function and the $(x_i)_{i=1}^{n+1}$ are distinct, ...
4
We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$
4
HINT: For each $x\in X$, let $A_x=\{k\mid f(k)=x\}$. Then each $A_x$ is non-empty. Use that to construct an injection from $X$ into $\{1,\ldots,n\}$.
4
HINT: No, you can’t assume that $A=C$ and show that the inclusions hold: that’s the converse of what you’re supposed to prove, and an implication and its converse are not logically equivalent. Use the fact that $A\subseteq B$ and $B\subseteq C$ to show that $A\subseteq C$. You’re given that $C\subseteq A$, so the rest is straightforward.
4
The given inequality is equivalent to $a^3-a=a(a^2-1)>0$. By multiplying both sides by $a^2+1$, which is always positive, we get $a(a^2-1)(a^2+1)>a^2+1>0$, or $a^5-a>0$.
4
I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.
4
Usually, a proof by contradiction of the statement $p \implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...
4
The Lambert W-function is a function $W(z)$ which solves $z=W(z)e^{W(z)}$. It is a multi-valued function. In this case, you are trying to solve: $$e^{x\pi i/2} = x$$ of: $$\frac{-\pi i}{2}=\frac{-x\pi i}{2}e^{-x\pi i/2}$$ So $$x\frac{-\pi i}{2} = W(-\pi i/2)$$ or $$x =\frac{2i}{\pi} W\left(\frac{-\pi i}{2}\right)$$ I don't think you can do better ...
3
You won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my "entry-level" suggestions on formal logic at the beginning of ...
3
You cannot go like this from $k=0$ to $k=1$ (i.e. $k=1$ cannot be expressed in the form $m+l$ as you wrote).
3
$N(t)$ is not equal to $N(t-s)+N(s)$, but $$N(t) = \Big(N(t) - N(s)\Big) + \Big(N(s)\Big)$$ and the two expressions inside the $\Big(\text{big parentheses}\Big)$ are independent of each other (whereas $N(t-s)$ and $N(s)$ are not independent of each other). So \begin{align} \Pr(N(s) = k \mid N(t) = n) & = \frac{\Pr(N(s)=k\ \&\ N(t) =n) ...
3
Yes, your work is all correct. Except a minor issue: the opposite of $|a_n| < \epsilon$ is $|a_n| \ge \epsilon$, not $|a_n| > \epsilon$. Instead of writing $P_n(X)$ as $|a_n| < \epsilon \; \forall n > X$, you may have found it clearer to write it as $\forall n > X \; |a_n| < \epsilon$. Then your entire statement would have been $$... 3 In general, if A\subseteq B, then \mathscr P (A)\subseteq \mathscr P (B) because every subset of A is a subset of B. More formally, if a\in \mathscr P (A), we need to show that a\in \mathscr P (B). But this is trivial, since if x\in a, then x\in B which implies that a\subseteq B which is the same as a\in \mathscr P (B). Now take ... 3 X\subset Y implies every element of X is an element of Y, so subsets of X are subsets of Y, so \mathcal{P}(X)\subset\mathcal{P}(Y). Finally, for Y=\mathcal{P}(X) you have \mathcal{P}(X)\subset\mathcal{P}(\mathcal{P}(X)). 3 Hint:$$\frac{a_{n+1}}{n}=\frac{a_{n+1}}{n+1}\frac{n+1}{n}.$$Or perhaps more to the point,$$\frac{a_{n+1}}{n+1}=\frac{a_{n+1}}{n}\frac{n}{n+1}.$$We've shown a_{n+1}/n\to l, we know n/(n+1)\to1, hence a_{n+1}/(n+1)\to l. And now this implies that a_n/n\to l. Given \epsilon>0 there exists N so |a_{n+1}/(n+1)-l|<\epsilon for all ... 3 If f(a)=c and f(b)=d, then$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$Now, let ... 3 What would the proper negation look like? It turns out that, in this case, there are a number of ways you can go in how you want to prove this claim, not just via direct proof or contrapositive but also how you frame the question logically as well. I'll outline what I think is the clearest and easiest way of going about it. Claim: Let ... 3 You have the contrapositive right. You must negate P and Q separately and prove that the negation of Q implies the negation of P. To expand on this, for "a and b are even" to be false, you only need one of a and b to be odd, so the negation is "a is not even or b is not even". And for the statement "a+b and ab have the same parity" ... 3 Here is a simple proof that K(n) is not only exponential but 'super' exponential in the sense that for all constants C, there is some n_0 such that |K(n)|\geq C^n for all n\gt n_0. Let's rewrite your series as \sum_n\frac{a_n}{a_{n+1}} so that we don't run out of indices; in other words, K(n)=a_n. (For convenience's sake I'm going to take ... 3 y = \frac{3x^2+2y}{x^2+2}, multiplying both sides of the equation by x^2+2 results in an equivalent equation because that term is never 0 (in the reals at least). You end up with yx^2 + 2y = 3x^2+2y subtract 2y from both sides (always legitimate). yx^2=3x^2 Since x\neq 0 we can divide both sides by x^2 and get y=3 3 Proof: We first must note that \pi_j is the unique solution to \pi_j=\sum \limits_{i=0} \pi_i P_{ij} and \sum \limits_{i=0}\pi_i=1. Let's use \pi_i=1. From the double stochastic nature of the matrix, we have$$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1 Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a ...
3
You need to show independence of increments, i.e. if $0\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$. You also ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2015-08-02T02:39:58 | {
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https://mathoverflow.net/questions/334674/creating-many-big-sets-of-small-numbers | # Creating many big sets of small numbers
There are $$n$$ numbers $$a_1,\ldots,a_n\in [0,1]$$.
Their sum is $$\sum_{i=1}^n a_i = s$$, where $$s$$ is some integer.
We want to group them into sets so that the sum of each set is at least $$t$$, where $$t$$ is some integer.
Let $$F(n,s,t)$$ be the largest number of sets that we can always create (for any $$a_i$$).
What is $$F(n,s,t)$$?
Example. $$F(n=8,s=7,t=1)=4$$:
• Proof that $$F(8,7,1)\geq 4$$: We can always create 4 sets by dividing the $$8$$ numbers arbitrarily into $$4$$ pairs. The sum of each pair is at most $$2$$, and the sum of all pairs is $$7$$, so the sum of each pair is at least $$1$$.
• Proof that $$F(8,7,1)\leq 4$$: We cannot always create 5 sets. Suppose for all $$i$$, $$a_i=7/8$$. In any $$5$$ sets, at least one set is a singleton so its sum is less than $$1$$.
Similarly, whenever $$n$$ is even, $$F(n,n-1,1)=n/2$$.
What else is known on the function $$F$$?
Currently I am particularly interested in the case $$t=2$$, but I will be happy for any more general references.
UPPER BOUND: $$F(n,s,t)\leq \lfloor {s+1\over t+1}\rfloor$$. Proof. Suppose that $$s+1$$ numbers equal $$s/(s+1)$$ and the other $$n-s-1$$ numbers equal $$0$$. To create a set with sum at least $$t$$, we need $$t+1$$ nonzeros. So we can create at most $$\lfloor {s+1\over t+1}\rfloor$$ such sets.
• @bof I added the upper bound that I had in mind. It is similar but not identical to yours. I am not sure about the lower bound. – Erel Segal-Halevi Jun 24 at 19:51
For $$n\in\mathbb N$$ and $$s,t\in\mathbb R$$ with $$0\lt t\le s\le n$$, let $$F(n,s,t)$$ be the greatest integer $$m$$ such that any family of $$n$$ numbers $$a_1,\dots,a_n\in[0,1]$$ with $$a_1+\cdots+a_n=s$$ can be partitioned into $$m$$ subfamilies, each with sum $$\ge t$$.
Lemma 1. If $$k\in\mathbb N$$ and $$s\le k\le n$$, then $$F(n,s,t)\le\left\lfloor\frac k{\lceil kt/s\rceil}\right\rfloor$$.
Lemma 2. If $$n\gt s$$ then $$F(n,s,t)\le\left\lfloor\frac{\lfloor s+1\rfloor}{\lfloor t+1\rfloor}\right\rfloor$$.
Proof. Put $$k=\lfloor s+1\rfloor$$ in Lemma 1.
Lemma 3. $$F(n,s,t)\ge\left\lfloor\frac{s+1}{t+1}\right\rfloor$$.
Proof. Let $$m=\left\lfloor\frac{s+1}{t+1}\right\rfloor\lt s+1$$, so that $$t\le\frac{s+1}m-1=\frac sm-\frac{m-1}m$$. We may assume that $$m\ge2$$.
Lat $$a_1,\dots,a_n\in[0,1]$$ be given, $$a_1+\cdots+a_n=s$$. For notational convenience we assume that $$a_1,\dots,a_p\gt0$$ while $$a_{p+1}=\cdots=a_n=0$$.
Partition the interval $$[0,s]$$ into $$m$$ equal subintervals $$J_1,\dots,J_m$$, indexed from left to right; that is, $$J_i=[c_{i-1},c_i]$$ where $$c_i=\frac{is}m$$. Then $$|J_i|=\frac sm\gt1-\frac1m$$.
Also partition $$[0,s]$$ into subintervals $$A_1,\dots,A_p$$ of respective lengths $$|A_i|=a_i$$. Let $$\mathcal A=\{A_1,\dots,A_p\}$$.
Each interval $$A\in\mathcal A$$ will be assigned to at most one of the intervals $$J_1,\dots,J_m$$, and (some of) the numbers $$a_1,\dots,a_p$$ will be assigned correspondingly to $$m$$ groups. Namely, an interval $$A\in\mathcal A$$ is assigned to the interval $$J_i=[c_{i-1},c_i]$$ if it satisfies one of the following three conditions: $$A\subseteq J_i;$$ $$i\gt1,\ \ c_{i-1}\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{i-1}m;$$ $$i\lt m,\ \ c_i\in A,\ \ \frac{|A\cap J_i|}{|A|}\gt\frac{m-i}m.$$ It is important to note that no interval $$A\in\mathcal A$$ is assigned to more than one $$J_i$$.
Now the set of intervals assigned to $$J_i$$ covers $$J_i$$, except possibly for an interval at the left of length $$\le\frac{i-1}m|A|\le\frac{i-1}m$$, and an interval at the right of length $$\le\frac{m-i}m|A|\le\frac{m-i}m$$. Therefore, the sum of the lengths of intervals assigned to $$J_i$$ is $$\ge\frac sm-\frac{i-1}m-\frac{m-i}m=\frac sm-\frac{m-1}m\ge t$$.
Theorem. If $$t\in\mathbb N$$ and $$n\gt s$$, then $$F(n,s,t)=\left\lfloor\frac{s+1}{t+1}\right\rfloor$$.
Proof. Lemmas 2 and 3.
• Yes, for some applications it may be interesting to evaluate $F(n,s,t)$ for $s,t$ rational numbers. Alternatively we can scale the numbers up to integers. We need to add just one more parameter: each number $a_i$ is in $[0,q]$, for some integer $q\geq 1$. I think that in this case your technique leads to a lower bound of $\lfloor{s+q\over t+q}\rfloor$, but I have to verify – Erel Segal-Halevi Jun 27 at 18:10
• Instead of separating lemmas 1 and 2, can you have just one lemma in which the number of nonzero terms is $\lfloor s+1 \rfloor$? It seems to cover both cases: when $s$ is an integer, $\lfloor s+1 \rfloor = \lceil s+1 \rceil = s+1$, and when $s$ is not an integer, $\lfloor s+1 \rfloor =\lceil s\rceil$. – Erel Segal-Halevi Jun 30 at 10:09
• $F \leq \lfloor s+1 \rfloor / \lfloor t+1 \rfloor$. Since you have $\lfloor s+1 \rfloor$ nonzero terms, and each term equals $s / \lfloor s+1 \rfloor < 1$. So in each subfamily with sum $t$, there must be at least $\lfloor t+1 \rfloor$ such elements. – Erel Segal-Halevi Jun 30 at 13:50
• Looks good, thanks! Now the only gap that remains is when $t$ is not an integer - the upper bound has $\lfloor t+1\rfloor$ in the denominator and the lower bound has $t+1$. – Erel Segal-Halevi Jul 3 at 15:57
• The upper bound in Lemma 2 is the result of setting $k=\lfloor s+1\rfloor$ in Lemma 1, but this is not necessarily the optimal value of $k$. For instance, $F(n,1,0.4)\le2$ by Lemma 2, but (assuming $n\ge3$) by setting $k=3$ in Lemma 1 we get $F(n,1,0.4)\le1$. – bof Jul 3 at 18:19 | 2019-12-07T02:16:45 | {
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https://mathematica.stackexchange.com/questions/198023/relation-between-matrices | # Relation between matrices
Maybe this is not a usual question for this forum... I have the following two matrices, mat1 and mat2, respectively:
mat1={{0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,-0.00894427,0.00894427,0.,0.},{-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.00894427,-0.00894427,0.,0.},{-0.00894427,0.00894427,0.0178885,-0.0178885,-0.00894427,0.00894427,0.,0.,0.,0.},{0.00894427,-0.00894427,-0.0178885,0.0178885,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427,0.,0.,0.,0.},{0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427,0.,0.,0.,0.},{-0.00894427,0.00894427,0.,0.,0.,0.,0.0178885,-0.0178885,-0.00894427,0.00894427},{0.00894427,-0.00894427,0.,0.,0.,0.,-0.0178885,0.0178885,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,-0.00894427,0.00894427,0.00894427,-0.00894427},{0.,0.,0.,0.,0.,0.,0.00894427,-0.00894427,-0.00894427,0.00894427}};
mat2={{0.0198382,-0.0198382,-0.00991908,0.00991908,0.,0.,-0.00991908,0.00991908,0.,0.},{-0.0198382,0.0198382,0.00991908,-0.00991908,0.,0.,0.00991908,-0.00991908,0.,0.},{-0.00991908,0.00991908,0.0203862,-0.0203862,-0.0104672,0.0104672,0.,0.,0.,0.},{0.00991908,-0.00991908,-0.0203862,0.0203862,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672,0.,0.,0.,0.},{0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672,0.,0.,0.,0.},{-0.00991908,0.00991908,0.,0.,0.,0.,0.0203862,-0.0203862,-0.0104672,0.0104672},{0.00991908,-0.00991908,0.,0.,0.,0.,-0.0203862,0.0203862,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,-0.0104672,0.0104672,0.0104672,-0.0104672},{0.,0.,0.,0.,0.,0.,0.0104672,-0.0104672,-0.0104672,0.0104672}};
If I plot the matrices, as well as their ratio, I obtain a visual representation of these:
Quiet@List[MatrixPlot[mat1], MatrixPlot[mat2], MatrixPlot[mat1/mat2]]
I think that there exists a numerical relation between mat1 and mat2, but I can't find it. Can anyone help me?
• Should your Dssmt and Dss be the same thing? – Roman May 9 '19 at 15:16
• @Roman there is a numerical relation among them, but it isn't a simple scaling operation by means of a scalar – Gae P May 9 '19 at 15:25
• Please make your code self-contained so that people can help without needing to guess. – Roman May 9 '19 at 15:46
Numerically all 100 differences between 100 corresponding elements of matrices correspond fall into just 5 symmetric values:
Round[Union[Flatten[mat1-mat2]],.0001]
{-0.0025,-0.0019,-0.0015,-0.001,0.,0.001,0.0015,0.0019,0.0025}
Which looks a lot like some small noise imposed on:
{-25,-20,-15,-10,0,10,15,20,25}/10000
It is actually quite easy to understand if you visualize your data. First of all you can see that the values of both matrices follow each other closely
ListPlot3D[{mat1,mat2},InterpolationOrder->0,PlotStyle->{Red,Blue},
BoxRatios->1,Mesh->None,SphericalRegion->True,PlotLegends->{"mat1","mat1"}]
You can see the also see the differences and realize, - for some elements mat1 is greater (more red), sometimes mat (more blue), and sometimes the same (white):
ListPlot3D[Rescale[Ds-Dss],InterpolationOrder->0,ColorFunction->
"TemperatureMap",BoxRatios->1,Mesh->None,SphericalRegion->True]
You also can see that all total
In[]:= Length[Flatten[mat1-mat2]]
Out[]= 100
differences between corresponding elements fall into just 9 different values and if you take in account symmetry - just 5 values:
BarChart[Union[Flatten[mat1 - mat2]], PlotTheme -> "Detailed"]
-- lets call them levels. And now you can find the statistics of how many differences correspond to a specific level -- and see it is symmetric:
ListLinePlot[Sort[Tally[Flatten[mat1 - mat2]]], PlotRange -> All, PlotTheme -> "Business"] | 2020-03-29T10:15:13 | {
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https://www.pavenafoundation.or.th/article/050d15-telescoping-series-examples | such that the series converges, provided $\displaystyle\lim_{n\rightarrow\infty}a(n)$ exists. Does anyone know the rules for a telescoping series. We will now look at some more examples of evaluating telescoping series. Telescoping Series. Example of a Telescoping Sum. E.g. This website uses cookies to ensure you get the best experience. A telescoping series is any series where nearly every term cancels with a preceeding or following term. Free Telescoping Series Test Calculator - Check convergence of telescoping series step-by-step. $\displaystyle\prod_{k=1}^{n}\frac{f(k+1)}{f(k)}=\frac{f(n+1)}{f(1)}.$ Below I'll give several examples, the first absolutely classical, of application of the telescoping technique. Telescoping series is a series where all terms cancel out except for the first and last one. The 1/2s cancel, the 1/3s cancel, the 1/4s cancel, and so on. These series are called telescoping and their convergence and limit may be computed with relative ease. Suppose we would like to determine whether the series Examples. Example 1 1 2 + 1 2 1 3 + … Telescoping series are series in which all but the first and last terms cancel out. A series is said to telescope if almost all the terms in the partial sums cancel except for a few at the beginning and at the ending. Next: The Harmonic Series. We can determine the convergence of the series by finding the limit of its partial sums remaining terms. Is it bounded? In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. The geometric and the telescoping series are the only types of series we can easily find the sum of. In the above example stands for the first term (if it is not cancelled out) and (if it is not cancelled out.) (a) A bounded sequence need not converge. The series $\sum_{n=1}^{\infty} \frac{1}{3^n} - \frac{1}{3^{n+1}}$ converges. ... What are some familiar examples in our solar system, and can some still be closed? In this example, we will determine whether or not the series \begin{align*} \sum _{k =1}^{\infty}\frac{3}{3k-2}-\frac{3}{3k+1}, \end{align*} converges or diverges. 2. We would like a more sure way of knowing the answer. The concept of telescoping extends to finite and infinite products. For example: $$\sum_{n=2}^\infty \frac{1}{n^3-n}$$ In this lesson, we will learn about the convergence and divergence of telescoping series. TELESCOPING SERIES Now let us investigate the telescoping series. Contents. Telescoping Series Example Finding the sum of a telescoping series. Respondents often are asked in surveys to retrospectively report when something occurred, how long something lasted, or … 2. Telescoping Series Examples 2. Note: For an example of a telescoping sums question, see question #2 in the Additional Examples section below. The series in Example 8.2.4 is an example of a telescoping series. Bricks are 20cm long and 10cm high. Previous: The Telescoping and Harmonic Series. In this course, Calculus Instructor Patrick gives 30 video lessons on Series and Sequences. All these terms now collapse, or telescope. ... Now, it is important to note that if we are just trying to determine if series converges or diverges, then applying the Telescoping Series Test will probably not be our first choice. Consider the following example. A p-series can be either divergent or convergent, depending on its value. Example 1. 3. I thought telescoping series were only the ones where all the terms canceled out except for the very first and last terms. This makes such series easy to analyze. Given the sequence ˆ 1 + lnn n3 ˙ 1 n=1 (a) Is it monotonic? INFINITE SERIES 1: GEOMETRIC AND TELESCOPING SERIES Exercise 6.2. The partial sum $$S_n$$ did not contain $$n$$ terms, but rather just two: 1 and $$1/(n+1)$$. More examples can be found on the Telescoping Series Examples 1 page. Telescoping Series Example. Strategy for Testing Series - Series Practice Problems This video runs through 14 series problems, discussing what to do to show they converge or diverge. In mathematics, a telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation. Discussion [Using Flash] Example. It seems like you need to do partial fraction decomposition and then evaluate each term individually? Write each of the following series in terms “standard” geometric series. If the sequence s n is not convergent then we say that the series is divergent. Illustrate each of the following with an exam-ple. This type of series can be easily calculated since all but a few terms are cancelled out. By using this website, you agree to our Cookie Policy. Suppose we are asked to ... it will be sufficient to demonstrate these two special forms with a set of examples. To be able to do this, we will use the method of partial fractions to decompose the fraction that is common in some telescoping series. Try the free Mathway calculator and problem solver below to … For instance, the series is telescoping. [1] [2] The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences. There is no exact formula to see if the infinite series is a telescoping series, but it is very noticeable if you start to see terms cancel out. It is different from the geometric series, but we can still determine if the series converges and what its sum is. This calculus 2 video tutorial provides a basic introduction into the telescoping series. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? The Telescoping and Harmonic Series. All that’s left is the first term, 1 (actually, it’s only half a term), and the last half-term, If the sequence ˆ 1 + lnn n3 ˙ 1 n=1 ( a ) is it monotonic that. Convergent then we say that the series converges and What its sum is a series whose sums. Of evaluating telescoping series: Definition & examples do you determine if the series is convergent or?... Concept, read the lesson titled telescoping series examples 1 page but a few are... A fixed number of terms after cancellation “ standard ” geometric series, but we still! 2 2k+1 3k = X1 k=2 2 2k+1 3k = X1 k=2 2 2k+1 3k but... 1/4S cancel, and so on it converges, provided $\displaystyle\lim_ n\rightarrow\infty... Before it only have a fixed number of terms after cancellation decomposition and then evaluate each term individually ’ not. On a flat horizontal surface, to overhang by 1 metre the limit of its partial sums: because cancellation! 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telescoping series examples
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https://gateoverflow.in/124367/isi-entrance-exam-mtech-cs | 542 views
Consider all possible trees with $n$ nodes. Let $k$ be the number of nodes with degree greater than $1$ in a given tree. What is the maximum possible value of $k$?
retagged | 542 views
Is it n-2 ?
see if it is a skewed tree
then n=k,rt?
If it is a skewed tree then k can be at Max n-2 because end vertices will have degree 1 only and n-2 vertices will have degree 2.
let , the tree with degree max =2 ,
let , no of intenal nodes whose degree >1 =k
then no of pedent vertex = n-k
now as we know that no of edges in tree = n-1
by handshaking theorem ,
sum of degrees of each verteces = 2 * edges
now , 2*(k)+1*(n-k) = 2*(n-1) => k = (n-2)
edited
This is correct Proof for Binary trees not for n-ary trees, but it really helps to understand how it works for any generic trees.Thanks:-)
Degree of a node in the tree is equal to the number of sub-trees of that node, right?
There is a theorem which says for a tree with at least 2 vertices we have at least 2 pendant vertices.
Since, in a tree of n nodes, we have at least 2 pendant vertices,
the number of vertices left which will surely have the degree more than 1 is
n-2= maximum value of k.
Maximum value of k is n-2 which is example of line graph because every tree should contain at least 2 pendent vertices (i.e vertex with degree 1). Therefore value of k cannot exceed n-2
Yes, correct.
Extra info: Min value of k is 1 when the graph is star graph where n-1 vertices have degree one and root has degree n-1.
@Warrior in a tree, degree of a node is equal to the number of sub-trees, right? graph and tree have different definition of degree of a node, isn't it? | 2018-02-24T10:32:58 | {
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https://brilliant.org/discussions/thread/int-ln-sqrtx1-sqrtx-dx-without-trigonometry/ | # $\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$ without trigonometry and nice values
Integration:
I watched this video https://www.youtube.com/watch?v=DGpgt8j-nzw in which $\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$ is computed by using trigonometry. Reading the comments, other solutions by using special functions are shown. However, the final antiderivatives do not use any of these functions, so these are intermediate steps in the most literal interpretation of that phrase. I thought of a solution that sidesteps the need of considering any functions other than those present in the integral:
$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx$. Let $u = \sqrt{x+1} + \sqrt{x}$. Notice that $\frac{1}{u} = \sqrt{x+1} - \sqrt{x}$. Then $du = \frac{1}{2} \left( \frac{1}{\sqrt{x+1}} + \frac{1}{\sqrt{x}} \right)dx = \frac{1}{2} \frac{u}{\sqrt{x+1} \sqrt{x}}dx$. To write everything in terms of $u$ notice that $u^2 - \frac{1}{u^2} = 4 \sqrt{x+1} \sqrt{x} \implies 2 \sqrt{x+1} \sqrt{x} = \frac{u^4 - 1}{2u^2}$. Thus $du = \frac{u}{\frac{u^4 - 1}{2u^2}}dx = \frac{2u^3}{u^4 - 1} dx$. Going back to the original integral:
$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \int \frac{\ln u}{2u^3} (u^4 - 1)dx = \int \frac{u}{2} \ln u - \frac{ \ln u}{2u^3} du$. Integrating the terms in the integral is now a standard exercise in integration by parts which I will skip. The result is $\frac{1}{8}u^2(2 \ln u - 1) + \frac{2 \log u + 1}{8u^2}$. Giving a final result of:
$\int \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{8}(\sqrt{x+1} + \sqrt{x})^2(2 \ln (\sqrt{x+1} + \sqrt{x}) - 1) + \frac{2 \ln (\sqrt{x+1} + \sqrt{x}) + 1}{8(\sqrt{x+1} + \sqrt{x})^2} + C$
Nice values:
From the form of the antiderivative we can expect this integral to give us nice values whenever $\sqrt{x+1} + \sqrt{x}$ is a nice power of $e$. The equation $\sqrt{x+1} + \sqrt{x} = e^k$ has solution $x= \frac{1}{4} e^{-2 k} (e^{2 k} - 1)^2$. You can obtain this by converting the equation into a normal quadratic equation. The nicest value I was able to find was taking $k=0$ to yield $x = 0$ and $k = \frac{1}{2}$ to yield $x = \frac{(e-1)^2}{4e}$. Then: $\int_{0}^{\frac{(e-1)^2}{4e}} \ln ( \sqrt{x+1} + \sqrt{x} ) dx = \frac{1}{4e}$
Note by Leonel Castillo
2 years, 2 months ago
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Here's another way to solve it:
\begin{aligned} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln\left(\sqrt {x+1} + \sqrt x\right)^2 \, dx \\ &=& \frac12 \int \ln\left(2x + 1 + 2\sqrt{x^2 + x}\right) \, dx \\ &=& \frac12 \int \ln\left[ 2\left(x+ \frac12\right) + 2\sqrt{\left(x + \frac12\right)^2 - \frac14} \; \right] \, dx \\ \end{aligned}
Let $x + \frac12 = \frac12 \sec \theta$, then $\left(x+\frac12\right)^2 - \frac14 = \left(\frac12 \tan\theta\right)^2$ and $\frac{dx}{d\theta} = \frac12\sec\theta\tan\theta$. The integral becomes
\begin{aligned} \int \ln\left(\sqrt{x+1} + \sqrt x\right) \, dx &=& \frac12 \int \ln (\sec \theta + \tan\theta) \cdot \frac12 \sec \theta\tan\theta \, d\theta \\ &=& \frac14 \int \underbrace{\ln (\sec \theta + \tan\theta)}_{=u} \cdot \underbrace{\sec \theta\tan\theta \, d\theta}_{=dv} , \qquad\qquad \text{ Integrate by parts} \\ &=& \frac14 \left (uv - \int v \, du\right) \\ &=& \frac14 \left [ \ln (\sec \theta + \tan\theta) \sec \theta - \int \sec \theta \cdot \dfrac{\sec\theta \tan\theta + \sec^2\theta}{\sec \theta + \tan \theta} \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \int \sec^2\theta \, d\theta \right ] \\ &=& \frac14 \left [ \sec \theta \ln (\sec \theta + \tan\theta) - \tan \theta \right ] +C \\ &=& \frac14 \left [(2x + 1) \ln \left ((2x+1) + \sqrt{(2x+1)^2-1}\; \right) - \sqrt{(2x+1)^2-1} \; \right ] +C \\ &=& \frac{2x+1}4 \ln \left (2x+1 + 2 \sqrt{x^2+x}\; \right) - \frac12 \sqrt{x^2+x} +C \\ &=& \frac{2x+1}2 \ln \left (\sqrt {x+1} + \sqrt x\right) - \frac12 \sqrt{x^2+x} +C \\ \end{aligned}
which is identical to blackpenredpen's final answer.
- 2 years, 2 months ago
Pi Han Goh, your post does not apply, because the OP expressly asked for the solution to be done without using trigonometry.
- 2 years, 1 month ago
I did not ask, I was just sharing a solution. If he wants to use this post to post another alternative solution that's okay.
- 2 years, 1 month ago
OP, your answer is wrong, because you do not include the "+ C."
- 2 years, 1 month ago
Thank you.
- 2 years, 1 month ago
Just a thought- How many different sums can be made by adding at least two different numbers from the set of integers {5,6,7,8}?
- 1 year, 11 months ago | 2020-09-27T17:54:09 | {
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https://www.physicsforums.com/threads/calculating-eigenvalues-help.934324/ | # Calculating Eigenvalues help
Tags:
1. Dec 13, 2017
### Pushoam
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
I solved it by calculating the eigen values by $| A- \lambda |= 0$.
This gave me $\lambda _1 = 6.42, \lambda _2 = 0.387, \lambda_3 = -0.806$.
So, the required answer is 42.02 , option (b).
Is this correct?
The matrix is symmetric. Is there any other easier wat to find the answer?
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2. Dec 13, 2017
### LCKurtz
Do you know the relation between the sum of the squares of the eigenvalues and the trace of a matrix?
3. Dec 13, 2017
### Pushoam
No.
4. Dec 13, 2017
### Ray Vickson
5. Dec 13, 2017
### Staff: Mentor
I get the same result.
6. Dec 13, 2017
### StoneTemplePython
The fact that its symmetric leads to some very nice results. What result do you get if you square each entry in your matrix and then sum them? This is called a squared Frobenius norm (which is one way of generalizing the L2 norm for vectors to matrices).
7. Dec 13, 2017
### Orodruin
Staff Emeritus
I would like to add to what the previous posters said that you should not round your numbers while doing your computations. Keep them on exact form to the very end and only evaluate the numbers in the end if necessary. You should find that the answer is exactly 42, not 42.02.
8. Dec 13, 2017
### Pushoam
I got tr(A) = $\Sigma \lambda_i$ , i= 1,2,3.
and Det (A) = $\Pi \lambda_i$ , i= 1,2,3.
Better to use $tr (A^2) = \Sigma {\lambda_i}^2$ , i= 1,2,3.
$Tr(A^n) = Tr(D^n)$, where D is a similar matrix of A.
In case of a matrix having all distinct eigen values, D can be the diagonal matrix consisting of $\lambda_i$.
In that case Tr(A^n) = Tr(D^n) = \Sigma {\lambda_i}^n$, i= 1,2,3. But, the above two equation will not give the result and it is complcated, too. It will be better to calculate the trace directly as I did in OP. I was looking for something like this. The sum is 42. So, is it true for any symmetric matrix? How to prove it?$ Tr (A^2) = \Sigma_i (A^2)_{ii}
\\ (A^2)_{ii} = \Sigma_j (A_{ij} A_{ji})$For symmetric matrix,$ A_{ij} = A_{ji}$. So,$ (A^2)_{ii} = \Sigma_j {(A_{ij} })^2
\\ Tr (A^2) = \Sigma_i \Sigma_j {(A_{ij} })^2$Is this correct? 9. Dec 13, 2017 ### Orodruin Staff Emeritus For$n = 2$it directly gives you the result. You just need to square the matrix and sum the diagonal of the result, very simple. If you go via the eigenvalues you need to solve for the roots of an order 3 polynomial. Essentially. However, I think it is easier to go the other way and just see that$A_{ij}A_{ij} = \mbox{tr}(AA^T)$. Since$A$is symmetric,$AA^T = A^2$. 10. Dec 13, 2017 ### StoneTemplePython note: we are dealing in reals for this post. Your approach is close, and maybe even correct, but I find it hard to follow. My strong preference here is to block your matrix by column vectors. Suppose you have some matrix$\mathbf X$, partitioned by columns below$\mathbf X = \bigg[\begin{array}{c|c|c|c|c}
\mathbf x_1 & \mathbf x_2 &\cdots & \mathbf x_{n-1} & \mathbf x_n\end{array}\bigg]$to make the link with the traditional L2 norm for vectors, consider the vec operator$
vec\big(\mathbf X\big) = \begin{bmatrix}
\mathbf x_1 \\
\mathbf x_2\\
\vdots \\
\mathbf x_{n-1}\\
\mathbf x_n
\end{bmatrix}$which stacks each column of the matrix$\mathbf X$on top of each other into one big vector. (The vec operator will show up again if and when you start dealing with Kronecker products.) Our goal is to add up each squared component of$\mathbf X$into a sum. do you understand why$\big \Vert \mathbf X \big \Vert_F^2 = \sum_{j=1}^n\sum_{i=1}^n x_{i,j}^2 = trace\big(\mathbf X^T \mathbf X\big) = vec\big(\mathbf X\big)^Tvec\big(\mathbf X\big)= \big \Vert vec\big(\mathbf X\big) \big \Vert_2^2$is true for any real matrix? Now since$\mathbf X$is symmetric, we have$\mathbf X^T = \mathbf X$meaning that$\big \Vert \mathbf X \big \Vert_F^2 = trace\big(\mathbf X^T \mathbf X\big) = trace\big(\mathbf X \mathbf X\big) = trace\big(\mathbf X^2\big)$now you just need the fact that others mentioned, i.e. relating a trace of a matrix and its eigenvalues (or in this case the trace of a matrix to the second power gives sum of eigenvalues to second power). Why is this fact true? (Hint: use characteristic polynomial, or if you prefer an easy but less general case: real symmetric matrices are diagonalizable -- do that and apply cyclic property of trace.) Trace is absurdly useful, so its worth spending extra time understanding all the related details of this problem. 11. Dec 13, 2017 ### Pushoam Then, for the anti - symmetric matrix,$tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}$= negative of the sum of the elements of the matrix. Right? 12. Dec 13, 2017 ### StoneTemplePython Yes. note that in general over real$n$x$n$matrices,$\big \vert trace\big(\mathbf A \mathbf A \big)\big \vert = \big \vert trace \Big( \big( \mathbf A^T \big)^T \mathbf A\Big)\big\vert \leq trace\big(\mathbf A^T \mathbf A\big) = \big \Vert \mathbf A\big \Vert_F^2 $with equality iff$\mathbf A$is a scalar multiple of$\mathbf A^T$. You could prove this with Schur's Inequality. Alternatively (perhaps using the vec operator to help) recognize that the trace gives an inner product. Direct application of Cauchy Schwarz gives you$\big \vert trace\big(\mathbf B^T \mathbf A \big) \big \vert = \big \vert vec\big( \mathbf B\big)^T vec\big( \mathbf A\big)\big \vert \leq \big \Vert vec\big( \mathbf B\big)\big \Vert_2 \big \Vert vec\big( \mathbf A\big)\big \Vert_2 =\big \Vert \mathbf B \big \Vert_F \big \Vert \mathbf A \big \Vert_F$with equality iff$\mathbf B = \gamma \mathbf A$. (Also note trivial case: if one or both matrices is filled entirely with zeros, then there is an equality.) In your real skew symmetric case,$\mathbf B = \mathbf A^T$and$\gamma = -1$. And of course in the real symmetric case$\gamma = 1$13. Dec 13, 2017 ### Pushoam I missed to write square of the elements. The corrected one: Then, for the anti - symmetric matrix,$tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the square of the elements of the matrix.
14. Dec 15, 2017
### epenguin
I am not very familiar with some of the algebra mentioned, though I don’t think it is very difficult.
However, it seems possible to solve the problem without it knowing all this, though I’m sure it does no harm to know it.
You could write out the eigenvalue equation as a cubic equation. The value of the sums of roots ∑λi is well known. The value of the sum of products of two roots, Σλiλj is well known. From this you could get the sum of squares of roots Σλi2.
I have not looked into it, but from what was being said about symmetry I suspect it would be easy to solve this cubic.
Last edited: Dec 15, 2017
15. Dec 15, 2017
### Ray Vickson
Sometimes symmetry does not help at all. For example, the matrix
$$A = \pmatrix{1&2&3\\2&4&5\\3&5&6}$$
has eigenvalues that are pretty horrible expressions involving cube roots and arctangents of things involving square roots, and the like. | 2018-01-22T18:33:53 | {
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https://math.stackexchange.com/questions/3750188/throwing-a-dice-and-add-the-digit-that-appears-to-a-sum-and-stopping-when-sum?noredirect=1 | # Throwing a dice and add the digit that appears to a sum and stopping when sum $\geq 100$.
You keep on throwing a dice and add the digit that appears to a sum. You stop when sum $$\ge 100$$. What’s the most frequently appearing digit in all such cases? $$1$$ or $$6$$?
I believe the probability of $$1$$ and $$6$$ should be equal as the whatever the number of rolls, the probability of getting a number should not be affected. However I don't have a formal proof for it and am not sure if this is right.
• How about working it out for a much smaller number than 100, such as 5? Then 6... – Steve Kass Jul 8 '20 at 18:54
• ...or consider: what is the largest number of $1$s that can ever fit your criterion? what is the largest number of $6$s that can ever fit your criterion? – David G. Stork Jul 8 '20 at 18:55
• You may find some relevant discussion at this rather older question, which @SteveKass may remember (he made a comment on it). – Brian Tung Jul 8 '20 at 19:02
• Suppose that when the sum reaches 100 or more, you stop the game, yell “Hooray!”, then start the game again with the same die. You do this for years on end. If the die is fair, how can yelling “Hooray!” now and then make the die unfair? – Steve Kass Jul 8 '20 at 20:26
• @SteveKass Are you answering the same question as @BrianTung? It seems like you are finding the expected number of 1s whereas he is finding the number of 1s among all sequences of rolls which terminate once they reach 100? Why are these the same? I would be grateful if you could make your approach a little more formal or precise. – user293794 Jul 8 '20 at 21:21
Basic approach. Imagine drawing a tree, with a root labelled $$0$$. The running count of each node is the label on that node, plus the sum of the labels of all of its direct ancestors. We build on the tree as follows: Under any node whose running count is not yet $$100$$, we add six more nodes, labelled $$1$$ through $$6$$. We repeat until there are no nodes left whose running count is less than $$100$$.
At the end of this process, we obviously have a finite tree. How many $$1$$s are there? How many $$6$$s? Was there any time when we added a $$1$$ but not a $$6$$, or vice versa?
The expected number of $$1$$'s is the same as the expected number of $$6$$'s. Let $$n_j(k)$$ denote the expected number of digit $$j\in \{1,\ldots,6\}$$ appearing in the sequence until the running sum reaches $$k$$ (in your case $$k=100$$). Then $$n_j(k)=\frac{1}{6}(1+n_j(k-j))+\frac{1}{6}\sum_{i\ne j} n_j(k-i)=\frac{1}{6}+\frac{1}{6}\sum_{i\in \{1,\ldots,6\}} n_j(k-i)$$ with $$n_j(1-i)=0$$ for $$i=1,\ldots,6$$. This recurrence relation is the same for all $$j$$ and so its solution, $$n_j(k)$$, is the same for all $$j$$.
• I don't believe your conclusion is correct. Although the recurrence is the same for all $\ j\$, the initial conditions aren't. For $\ k=1\$, for instance, $\ n_1(1)=1\$ and $\ n_j(1)=0\$ for $\ j=2,3,\dots,6\$. – lonza leggiera Jul 10 '20 at 0:09
• @lonzaleggiera why $n_2(1)=0$? – d.k.o. Jul 10 '20 at 5:54
• Starting from a total of $0$, the only way you can get to a total of exactly $1$ is from a single throw of a $1$. No other face can have occurred. – lonza leggiera Jul 10 '20 at 6:10
• Why? You stop when the sum of digits reaches $\ge k$ (see the question). – d.k.o. Jul 10 '20 at 6:35
• @vonbrand: It is true that you would need $100$ ones (if those were all you got) and only $17$ sixes (if those were all you got). And if we were to list all possible sequences that reached or exceeded $100$ only on the last roll, then ones would indeed be more common than sixes. But not all sequences are equally likely. The sequence consisting of $17$ sixes is $6^{83}$ times more likely than the sequence consisting of $100$ ones. In general, the preponderance of ones in the longer sequences is exactly offset by the fact that those longer sequences occur less frequently. – Brian Tung Jul 10 '20 at 23:46 | 2021-03-05T11:08:18 | {
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https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_11&diff=prev&oldid=44887 | # Difference between revisions of "2004 AMC 10B Problems/Problem 11"
## Problem
Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?
$\mathrm{(A) \ } \frac{1}{2} \qquad \mathrm{(B) \ } \frac{47}{64} \qquad \mathrm{(C) \ } \frac{3}{4} \qquad \mathrm{(D) \ } \frac{55}{64} \qquad \mathrm{(E) \ } \frac{7}{8}$
## Solutions
### Solution 1
We have $1\times n = n < 1 + n$, hence if at least one of the numbers is $1$, the sum is larger. There $15$ such possibilities.
We have $2\times 2 = 2+2$.
For $n>2$ we already have $2\times n = n + n > 2 + n$, hence all other cases are good.
Out of the $8\times 8$ possible cases, we found that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ cases the sum is smaller, satisfying the condition. Therefore the answer is $\frac{48}{64} = \boxed{\frac34}$.
### Solution 2
Let the two rolls be $m$, and $n$.
From the restriction: $mn > m + n$
$mn - m - n > 0$
$mn - m - n + 1 > 1$
$(m-1)(n-1) > 1$
Since $m-1$ and $n-1$ are non-negative integers between $1$ and $8$, either $(m-1)(n-1) = 0$, $(m-1)(n-1) = 1$, or $(m-1)(n-1) > 1$
$(m-1)(n-1) = 0$ if and only if $m=1$ or $n=1$.
There are $8$ ordered pairs $(m,n)$ with $m=1$, $8$ ordered pairs with $n=1$, and $1$ ordered pair with $m=1$ and $n=1$. So, there are $8+8-1 = 15$ ordered pairs $(m,n)$ such that $(m-1)(n-1) = 0$.
$(m-1)(n-1) = 1$ if and only if $m-1=1$ and $n-1=1$ or equivalently $m=2$ and $n=2$. This gives $1$ ordered pair $(m,n) = (2,2)$.
So, there are a total of $15+1=16$ ordered pairs $(m,n)$ with $(m-1)(n-1) < 1$.
Since there are a total of $8\cdot8 = 64$ ordered pairs $(m,n)$, there are $64-16 = 48$ ordered pairs $(m,n)$ with $(m-1)(n-1) > 1$.
Thus, the desired probability is $\frac{48}{64} = \frac{3}{4} \Rightarrow C$. | 2021-04-20T14:20:12 | {
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https://math.stackexchange.com/questions/1698030/proving-a-family-of-polynomials-is-linearly-independent | # Proving a family of polynomials is linearly independent
I am attempting to solve the following problem, but I'm confused on a solution that is offered for it.
The question asks "Let $S$ $=$ {$P_1,..., P_n$} be a family of polynomials such that for all $i,j \leq n$ : $deg(P_i) \not = deg(P_j)$. Show that S is linearly independent".
The solution offered states the following :
"We will prove this by induction on n, where n is the number of polynomials.
Base case: Let n=1. Then S = {$P_1$} and since a set containing only one nonzero vector is linearly independent, the desired condition holds.
Inductive Step: Assume that S = {$P_1, ... P_n$} is linearly independent. Consider the case when $S'$ = {$P_1, ..., P_n, P_{n+1}$} and $deg(P_i) \not = deg(P_j)$ for all $i \not = j$. Consider $a_1P_1 + ... + a_nP_n + a_{n+1}P_{n+1} = 0$. Define $P_k$ to be the polynomial of highest degree. It must be true that
$a_1P_1 + \space ... \space + a_{k-1}P_{k-1} + a_{k+1}P_{k+1} +\space ... \space + a_nP_n + a_{n+1}P_{n+1} = a_kP_k$.
If $a_k \not=0$, then right hand side of the preceding equation has the same degree as $P_k$ while the left hand side has degree strictly less than $P_k$ and this is a contradiction. It must follow that $a_k = 0$."
This is the part where I get confused. Why must this be a contradiction? If we say that $P_k$ is the polynomial of highest degree, isn't it impossible for any other polynomial to have a degree higher than $P_k$ or equal to $P_k$ due to our conditions?
Thanks!!
• You forgot to mention two things in the question/exercise. – Friedrich Philipp Mar 15 '16 at 1:49
• @FriedrichPhilipp I didn't type out the rest of the proof because I understand how the rest follows, but I don't understand that particular statement. – King Tut Mar 15 '16 at 1:51
• I meant the question in the beginning... – Friedrich Philipp Mar 15 '16 at 1:52
• "If we say that $P_k$ is the polynomial of highest degree, isn't it impossible for any other polynomial to have a degree higher than $P_k$ or equal to $P_k$ due to our conditions?" Exactly. And exactly this is used here. The left hand side must have smaller degree then the right hand side. But then, of course, they cannot be equal. – Friedrich Philipp Mar 15 '16 at 1:57
• No problem. "Oh are you saying that it should be specified that all $P_k$'s are nonzero?" Of course. This is also used in the proof. Moreover, $deg(P_i)\neq deg(P_j)$ must be assumed to hold for $i\neq j$. – Friedrich Philipp Mar 15 '16 at 2:27 | 2019-08-20T19:13:10 | {
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https://or.stackexchange.com/questions/4584/how-to-optimize-a-utility-function-that-contains-step-function/4585 | # How to optimize a utility function that contains step function?
I have an optimization problem with an uncommon utility: to find a $$\beta$$ that maximizes
$$r^{T}\cdot H(X\cdot\beta)$$
where $$H()$$ is a Heaviside step function as in wiki
$$r$$ is a vector of size 1000
$$X$$ is a 1000x50 "tall" matrix
$$\beta$$ is a vector of size 50
I am familiar with gradient descent, which is how I usually solve an optimization problem. But Heaviside function does not work with gradient descent. So I am wondering if anyone here could shed some light on how to solve such optimization problem.
Thanks
You can solve the problem via integer linear programming as follows, assuming $$r_i \ge 0$$ for all $$i$$. Let $$M_i$$ be a (small) upper bound on $$-(X \cdot \beta)_i$$. Let binary decision variable $$y_i$$ indicate whether $$(X \cdot \beta)_i \ge 0$$. The problem is to maximize $$\sum_{i=1}^{1000} r_i y_i$$ subject to $$-(X \cdot \beta)_i \le M_i(1 - y_i)$$ for all $$i$$. This "big-M" constraint enforces $$y_i=1 \implies (X \cdot \beta)_i \ge 0$$.
• Yes, $y_i=0$ forces only a redundant constraint. As long as the “reward” $r_i\ge 0$, this big-M constraint is sufficient. – RobPratt Jul 25 at 13:28 | 2020-08-08T10:37:49 | {
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http://quig.curaben.it/symmetric-matrix-example-3x3.html | ### Symmetric Matrix Example 3x3
I know that I can convert a single vector of size 3 in a skew symmetric matrix of size 3x3 as follows:. One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. For a solution, see the post " Quiz 13 (Part 1) Diagonalize a matrix. Symmetric Matrices: A square matrix A is symmetric if A = AT. first of all you need to write a c program for transpose of a matrix and them compare it with the original matrix. A square matrix [aij] is called skew-symmetric if aij = −aji. Proof: if it was not, then there must be a non-zero vector x such that Mx = 0. Storage Formats for the Direct Sparse Solvers. In this chapter, we will typically assume that our matrices contain only numbers. The unit matrix is every #n#x #n# square matrix made up of all zeros except for the elements of the main diagonal that are all ones. A symmetric magic square is also called an associative magic square (11, p. JavaScript Example of the Hill Cipher § This is a JavaScript implementation of the Hill Cipher. That is, it satisfies the condition In terms of the entries of the matrix, if denotes the entry in the -th row and -th column,. The output matrix has the form of A = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ]. Orthogonal matrix multiplication can be used to represent rotation, there is an equivalence with quaternion multiplication as described here. Therefore, in linear algebra over the complex numbers,. A square matrix [aij] is called a symmetric matrix if aij = aji, i. So a diagonal matrix has at most n different numbers other than 0. Briefly, matrix inverses behave as reciprocals do for real numbers : the product of a matrix and it's inverse is an identity matrix. The elements in a matrix. For a real skew-symmetric matrix the nonzero eigenvalues are all pure imaginary and thus are of the form iλ 1, −iλ 1, iλ 2, −iλ 2, … where each of the λ k are real. In this note, we derive an orthogonal transformation which transforms A to a n x n matrix B whose diagonal elements are zero. xla is an addin for Excel that contains useful functions for matrices and linear Algebra: Norm, Matrix multiplication, Similarity transformation, Determinant, Inverse, Power, Trace, Scalar Product, Vector Product, Eigenvalues and Eigenvectors of symmetric matrix with Jacobi algorithm, Jacobi's rotation matrix. The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. If Ais an m nmatrix, then its transpose is an n m matrix, so if these are equal, we must have m= n. 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As a consequence, we have the following version of \Schur's trick" to check whether M˜0 for a symmetric matrix, M, where we use the usual notation, M˜0 to say that Mis positive de nite and the notation M 0 to say that Mis positive. Question: (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. matrix c = a + b. A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. The matrices must all be defined on dense sets. A square matrix [aij] is called a symmetric matrix if aij = aji, i. Now, noting that a symmetric matrix is positive semi-definite if and only if its eigenvalues are non-negative, we see that your original approach would work: calculate the characteristic polynomial, look at its roots to see if they are non-negative. B is not possible, so multiplication here is not commutative. This solver uses the excellent lrs - David Avis's implementation of Avis and Fukuda's reverse search algorithm for polyhedral vertex enumeration. 2 Definiteness of Quadratic Forms. A square matrix as sum of symmetric and skew-symmetric matrices. that the element in the i–th row and j–th column of the matrix A equals aij. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i. (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. Singular values are important properties of a matrix. If you're behind a web filter, please make sure that the domains *. A × A-1 = I. However, when we make any choice of a fundamental matrix solution M(t) and compute M(t)M(0) 1, we always get the same result. An n×n matrix B is called skew-symmetric if B = −BT. ; Find transpose of matrix A, store it in some variable say B. A = 2: 1+j: 2-j, 1-j: 1: j: 2+j-j: 1 = 2: 1-j: 2+j (j 2 = -1) 1+j: 1-j: 2-j: j: 1: Now A T = => A is Hermitian (the ij-element is conjugate to the ji-element). Linear Algebra: We verify the Spectral Theorem for the 3x3 real symmetric matrix A = [ 0 1 1 / 1 0 1 / 1 1 0 ]. Coordinate Transformations of tensors are discussed in detail here. If A is not SPD then the algorithm will either have a zero. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. A skew-symmetric matrix $M$ satisfies [math]M^T=-M. In the following we assume. Reported by: such as 3x3 in the examples needed in the definition of the Pfaffian rather than push that definition into. In Example 1, the eigenvalues of this matrix were found to be λ = −1 and λ = −2. Symmetric Matrices The symmetric matrix is a matrix in which the numbers on ether side of the diagonal, in corresponding positions are the same. 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Input elements in matrix A. , only passive elements and independent sources), these general observations about the A matrix will always hold. 1 Basics Definition 2. Let's take an example of a matrix. is also symmetric because ÐEEÑ œEE œEEÞX X X XX X The next result tells us that only a symmetric matrix "has a chance" to be orthogonally diagonalizable. For our example: rank{A} ˘2. Simple example: A = I. We will use the following two properties of determinants of matrices. These matrices combine in the same way as the operations, e. 3x3 skew symmetric matrices can be used to represent cross products as matrix multiplications. AB = BA = I n, then the matrix B is called an inverse of A. Secant for particular equation. A A real symmetric matrix [A] can be diagonalized (converted to a matrix with zeros for all elements off the main diagonal) by pre-multiplying by the inverse of the matrix of its eigenvectors and post-multiplying by the matrix of its eigenvectors. CONTENTS: [4] MATRIX ADDITION [5] MATRIX NOTATION [6] TRANSPOSE [7] SYMMETRIC MATRICES [8] BASIC FACTS ABOUT MATRICES [4] MATRIX ADDITION. 262 POSITIVE SEMIDEFINITE AND POSITIVE DEFINITE MATRICES Proof. For example, if the 3x3 matrix M has 5 as an eigenvalue, expand the following: (a, b, c)M = (5a, 5b, 5c) You'll get a pile of conditions on a, b, and c, and sets of a, b, and c that satisfy the conditions will be eigenvectors. The next leaflets in the series will show the conditions under which we can add, subtract and multiply matrices. Each input corresponds to an element of the tensor. These algorithms need a way to quantify the "size" of a matrix or the "distance" between two matrices. To be able to diagonalise a given symmetric matrix. The link inertia matrix (3x3) is symmetric and can be specified by giving a 3x3 matrix, the diagonal elements [Ixx Iyy Izz], or the moments and products of inertia [Ixx Iyy Izz Ixy Iyz Ixz]. Algorithm for Cholesky Decomposition Input: an n£n SPD matrix A Output: the Cholesky factor, a lower triangular matrix L such that A = LLT Theorem:(proof omitted) For a symmetric matrix A, the Cholesky algorithm will succeed with non-zero diagonal entries in L if and only if A is SPD. An other solution for 3x3 symmetric matrices can be found here (symmetric tridiagonal QL algorithm). If the characteristic of the field is 2, then a skew-symmetric. Example The zero matrix is. Below is the step by step descriptive logic to check symmetric matrix. The quadratic form of. Example 1: Determine the eigenvectors of the matrix. But the multiplication of two symmetric matrices need not be symmetric. 3x3 system of equations solver This calculator solves system of three equations with three unknowns (3x3 system). Find the sum of the diagonal elements of the given N X N spiral matrix. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. The leftmost column is column 1. symmetric matrix is symmetric. The available eigenvalue subroutines seemed rather heavy weapons to turn upon this little problem, so an explicit solution was developed. Each number that makes up a matrix is called an element of the matrix. This is an inverse operation. Subtract the corresponding elements of from. Give an example of a 3 X 3 upper triangular matrix A that is not diagonal. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. 366) •A is orthogonally diagonalizable, i. Question: (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. Example 1: Determine the eigenvectors of the matrix. 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An matrix A is called nonsingular or invertible iff there exists an matrix B such that. 2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. Definition 4. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. And with the confusion matrix, we can calculate a variety of stats in addition to accuracy:. One way to think about a 3x3 orthogonal matrix is, instead of a 3x3 array of scalars, as 3 vectors. The eigen-values are di erent for each C, but since we know the eigenvectors they are easy to diagonalize. 1) means that the eigenvalues of ¡1 aC are the intersections of the graph of pn(x) with the line y = 1¡ a a¡b. For example, consider the following vector A = [a;b], where both a and b are 3x1 vectors (here N = 2). Let B is a 3x3 matrix A is a 3x2 matrix so B. Invertible matrices are very important in many areas of science. A symmetric magic square is also called an associative magic square (11, p. The output matrix has the form of A = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ]. Determinant Calculator - Matrix online calculator. FINDING EIGENVALUES • To do this, we find the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0,. An n×n matrix B is called skew-symmetric if B = −BT. eig computes eigenvalues and eigenvectors of a square matrix. Kronenburg Abstract A method is presented for fast diagonalization of a 2x2 or 3x3 real symmetric matrix, that is determination of its eigenvalues and eigenvectors. This implies that UUT = I, by uniqueness of inverses. This should be easy. symmetric matrix is symmetric. , in kronecker , however not for matrix multiplications where. M(t) is an invertible matrix for every t. If there exists a square matrix B of order n such that. Eigenvalues of a 3x3 matrix. Enter payoff matrix B for player 2 (not required for zerosum or symmetric games). the symmetric QRalgorithm, as the expense of two Jacobi sweeps is comparable to that of the entire symmetric QRalgorithm, even with the accumulation of transformations to obtain the matrix of eigenvectors. If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. Philip Petrov ( https://cphpvb. The analysis of matrix-based algorithms often requires use of matrix norms. I am looking for a very fast and efficient algorithm for the computation of the eigenvalues of a 3x3 symmetric positive definite matrix. Symmetric matrices have special properties which are at the basis for these discussions and solutions. Asymmetric Mixed Strategy Equilibria aMaking a game asymmetric often makes its mixed strategy equilibrium asymmetric aAsymmetric Market Niche is an example 33 Asymmetrical Market Niche: The payoff matrix-50, -50 0, 100 150, 0 0, 0 Enter Stay Out Enter Stay Out Firm 2 Firm 1 34 Asymmetrical Market Niche: Two pure strategy equilibria-50, -50 0. Exercise 3. C program check symmetric matrix. Linear Algebra: We verify the Spectral Theorem for the 3x3 real symmetric matrix A = [ 0 1 1 / 1 0 1 / 1 1 0 ]. Step 1 - Accepts a square matrix as input Step 2 - Create a transpose of a matrix and store it in an array Step 3 - Check if input matrix is equal to its transpose. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11) = (1). Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. For example, we can confirm that muliplying A by its inverse gives the identity matrix Ainv. Symmetric Matrix :- Square matrix that's equal to it's Transpose (A T =A) We call them symmetric because they are symmetric to main diagonal. Symmetric matrix can be obtain by changing row to column and column to row. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. For example A = B = (10) Skew symmetric matrix: If for a square matrix A = [aij], A’ = -A, then A is called a skew symmetric matrix. xTAx = x1 x2 2 6 18 6 x x 1 2 2x = x 1 + 6x2 1 x2 6x 1 + 18x2 = 2x 12 + 12x1x2 + 18x 22 = ax 12 + 2bx1x2 + cx 22. Diagonal matrix :- All non-diagonal elements =0. The Cholesky decomposition of a Pascal upper-triangle matrix is the Identity matrix of the same size. So in order to prove this matrix is diagonalizable,. no entry zero. I have chosen these from some book or books. The main diagonal itself must all be 0s. [4] Computing Eigenvectors Let's return to the equation Ax = x. We will now go into the specifics here, however. Most properties are listed under skew-Hermitian. The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. The @MTXMUL function multiplies matrix B by matrix C and places the result in matrix A. However, when we make any choice of a fundamental matrix solution M(t) and compute M(t)M(0) 1, we always get the same result. A real $(n\times n)$-matrix is symmetric if and only if the associated operator $\mathbf R^n\to\mathbf R^n$ (with respect to the standard basis) is self-adjoint (with respect to the standard inner product). These algorithms need a way to quantify the "size" of a matrix or the "distance" between two matrices. Example 1: Give an example of 4×4 order identity or unit matrix. the algorithm will be part of a massive computational kernel, thus it is required to be very efficient. A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. 1 Symmetric Matrices and Convexity of Quadratic Functions A symmetric matrix is a square matrix Q ∈ ℜn×n with the property that Qij = Qji for all i;j = 1;:::;n : We can alternatively de ne a matrix Q to be symmetric if QT = Q : We denote the identity matrix (i. A is called symmetric if A> = A. Definition of a Matrix The following are examples of matrices (plural of matrix). Figure 1 1-D Gaussian distribution with mean 0 and =1 In 2-D, an isotropic (i. C program check symmetric matrix. Find the eigenvalues and bases for each eigenspace. Here, it is understood that and are both column vectors, and is the matrix of the values. e A-1 we shall first define the adjoint of a matrix. I'm not sure what kind of approach to take towards a. Your overall recorded score is 0%. If ‘b’ is a matrix, the system is solved for each column of ‘b’ and the return value is a matrix of the same shape as ‘b’. Matrix representation of symmetry operations Using carthesian coordinates (x,y,z) or some position vector , we are able to define an initial position of a point or an atom. Good things happen when a matrix is similar to a diagonal matrix. There are a lot of examples in which a singular matrix is an idempotent matrix. [4] Computing Eigenvectors Let's return to the equation Ax = x. A = [1 1 1 1 1 1 1 1 1]. If you're behind a web filter, please make sure that the domains *. So a diagonal matrix has at most n different numbers other than 0. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. Thus for all i and j, we have a ij = – a ji. An matrix is called real symmetric if , the transpose of , coincide with. Ellermeyer July 1, 2002 1 Similar Matrices Definition 1 If A and B are nxn (square) matrices, then A is said to be similar to B if there exists an invertible nxn matrix, P,suchthatA = P−1BP. 1 Strategic Form. Therefore x T Mx = 0 which contradicts our assumption about M being positive definite. Symmetric matrices have special properties which are at the basis for these discussions and solutions. Definition E EœEÞis called a if symmetric matrix X Notice that a symmetric matrix must be square ( ?). (35) For a positive semi-definite matrix, the rank corresponds to the. Secant for particular equation. Example 5: A Hermitian matrix. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. Matrix Approach to Linear Regression leaving J is matrix of all ones, do 3x3 example. So in order to prove this matrix is diagonalizable,. In general, a ij means the element of A in the ith row and jth column. To find this matrix : First write down a skew symmetric matrix with arbitrary coefficients. Singular values are important properties of a matrix. If the matrix A is symmetric then •its eigenvalues are all real (→TH 8. Diagonal matrix :- All non-diagonal elements =0. DA FONSECA In general, (2. Not very random but very fun!. Example A= 2 4 0 3 This is a 2 by 2 matrix, so we know that 1 + 2 = tr(A) = 5 1 2 = det(A) = 6 6. Now since U has orthonormal columns, it is an orthognal matrix, and hence Ut is the inverse of U. The diagonal elements of a skew-symmetric matrix are all 0. In symmetric matrices the upper right half and the lower left half of the matrix are mirror images of each other about the diagonal. Operation on matrices: Addition and multiplication and multiplication with a scalar. These matrices combine in the same way as the operations, e. 2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. If a ij denotes the entries in an i-th row and j-th column, then the symmetric matrix is represented as. After eliminating weakly dominated strategies, we get the following matrix:. Skew symmetric matrices mean that A (transpose) = -A, So since you know 3 elements of the matrix, you know the 3 symmetric to them over the main diagonal mut be the negatives of those elements. Example The zero matrix is. The set of four transformation matrices forms a matrix representation of the C2hpoint group. Once we get the matrix P, then D = P t AP. The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. 2 Definiteness of Quadratic Forms. 1, is an eigenvalue of. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 kAxk2 kxk2 = max x6=0 xTATAx. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. Show that the product AAT is a symmetric matrix. If is an matrix and is an matrix, then the tensor product of and , denoted by , is the matrix and is defined as If is and is , then the Kronecker sum (or tensor sum) of and , denoted by , is the matrix of the form Let be the set of all symmetric matrices with integer. 1) means that the eigenvalues of ¡1 aC are the intersections of the graph of pn(x) with the line y = 1¡ a a¡b. If there exists a square matrix B of order n such that. Zero matrix and identity matrix are symmetric (any diagonal matrix is sym-metric) 2. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy and we have Q A(1; 1) = 12 + ( 1)2 + 8(1)( 1) = 1 + 1 8. Learn its definition and formula to calculate for 2 by 2, 3 by 3, etc. equal to a)A5 +A8 b)A5 -A8 c)A8 -A5 c)AT +BT If A is a symmetric matrix and B is a skew symmetric matrix of the same order , then A2 +B2 is a 1 2 4 6 8 2 2 2 7 16. 1 The non{symmetric eigenvalue problem We now know how to nd the eigenvalues and eigenvectors of any symmetric n n matrix, no matter how large. Since A is symmetric, A = AT or LDU = UTDLT, so U = LT. The initial vector is submitted to a symmetry operation and thereby transformed into some resulting vector defined by the coordinates x', y' and z'. By using this website, you agree to our Cookie Policy. For example, the matrices. Properties. Simple example: A = I. The matrices must all be defined on dense sets. The case here is restricted to 2x2 case of the hill cipher for now, it may be expanded to 3x3 later. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. The coordinates can be written in matrix form and then can be multiplied by a matrix or scalar for Rotation, Reflection or Dilation (Scaling). In this chapter, we will typically assume that our matrices contain only numbers. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. The 3x3 matrix can be thought of as an operator - it takes a vector, operates on it, and returns a new vector. A3×3 example of a matrix with some complex eigenvalues is B = 1 −1 −1 1 −10 10−1 A straightforward calculation shows that the eigenvalues of B are λ = −1 (real), ±i (complex conjugates). The Jordan decomposition allows one to easily compute the power of a symmetric matrix :. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. Symmetric (matrix) synonyms, Symmetric (matrix) pronunciation, Symmetric (matrix) translation, English dictionary definition of Symmetric (matrix). ) Dimension is the number of vectors in any basis for the space to be spanned. In Example 1, the eigenvalues of this matrix were found to be λ = −1 and λ = −2. Hermitian matrices Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. Examples of higher order tensors include stress, strain, and stiffness tensors. On this page you can see many examples of matrix multiplication. Let be an eigenvector corresponding to the eigenvalue 3. Matrices with Examples and Questions with Solutions. Here you can calculate inverse matrix with complex numbers online for free with a very detailed solution. JavaScript Example of the Hill Cipher § This is a JavaScript implementation of the Hill Cipher. A C++ source and header file to compute eigenvectors/values of a 3x3 symmetric matrix. If there exists a square matrix B of order n such that. Simple properties of addition, multiplication and scalar multiplication. 369) EXAMPLE 1 Orthogonally diagonalize. A neat example of this is finding large powers of a matrix. 60 • • • Chapter 1 / Systems of Linear Equations and Matrices EXAMPLE 1 Solution of a Linear System Using A−1 Consider the system of linear equations x1 + 2x2 + 3x3 = 5 2x1 + 5x2 + 3x3 = 3 + 8x3 = 17 x1 In matrix form this system can be written as Ax = b, where 1 2 3 x1 5 A = 2 5 3 , x = x2 , b = 3 1 0 8 17 x3 In Example 4 of the. A 3x3 Example The following example decomposes a 3 x 3 symmetric matrix. By Proposition 23. With symmetric matrices on the other hand, complex eigenvalues are not possible. 1 The formula aij = 1/(i + j) for 1 ≤ i ≤ 3, 1 ≤ j ≤ 4 defines a 3×4 matrix A = [aij], namely A = 1. More generally, if C is an m× n matrix, its transpose, CT, is a n× m matrix. Now consider the x matrix, the matrix of unknown quantities. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. Its easy to get C2 to equal 0 but obviously you can't have that. In the second step, which takes the most amount of time, the matrix is reduced to upper Schur form by using an orthogonal transformation. If this quadratic form is positive for every (real) x1 and x2 then the matrix is positive definite. The inverse matrix has the property that it is equal to the product of the reciprocal of the determinant and the adjugate matrix. are symmetric matrices. To find this matrix : First write down a skew symmetric matrix with arbitrary coefficients. Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION Let A be an n n real matrix. Show that the product AAT is a symmetric matrix. can have vector or matrix elements). HILL CIPHER Encrypts a group of letters called polygraph. 3 Pure Strategies and Mixed Strategies. I am looking for a very fast and efficient algorithm for the computation of the eigenvalues of a 3x3 symmetric positive definite matrix. AAT = 17 8 8 17. Most properties are listed under skew-Hermitian. (2*2 - 7*4 = -24) Multiply by the chosen element of the 3x3 matrix. I want to convert the last 3 dimensional vector into a skew symmetric matrix. 2 Two-part names. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. Real number λ and vector z are called an eigen pair of matrix A, if Az = λz. I have chosen these from some book or books. An n×n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. 369) EXAMPLE 1 Orthogonally diagonalize. Simple example: A = I. In other words, we can say that transpose of Matrix B is not equal to matrix B (). The Cholesky decomposition of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. Let be an eigenvector corresponding to the eigenvalue 3. Let A = LDU be the LDU decomposition of A. To encrypt; C K. Then we have: A is positive de nite ,D k >0 for all leading principal minors A is negative de nite ,( 1)kD k >0 for all leading principal minors A is positive semide nite , k 0 for all principal minors A is negative semide nite ,( 1)k k 0 for all principal minors In the rst two cases, it is enough to. Note that usually the eigenvectors are normalized to have unit length. matrix list b symmetric b[3,3] c1 c2 c3 displacement 3211055 mpg 227102 22249 _cons 12153 1041 52. Symmetric eigenvalue decompositions for symmetric tensors Lek-Heng Lim University of California, Berkeley January 29, 2009 (Contains joint work with Pierre Comon, Jason Morton, Bernard Mourrain, Berkant Savas) L. The main diagonal itself must all be 0s. M(t) is an invertible matrix for every t. phasesym Example of 3x3 skew symmetric matrix. Example 5: A Hermitian matrix. KEYWORDS: Software, Solving Linear Equations, Matrix Multiplication, Determinants and Permanents. Note that all the main diagonal elements in the skew-symmetric matrix are zero. An analogous result holds for matrices. A symmetric matrix should be a square matrix. The Euler angles of the eigenvectors are computed. The characteristic polynomial is det(AAT −λI) = λ2 −34λ+225 = (λ−25)(λ−9), so the singular values are σ. It is denoted by adj A. A real $(n\times n)$-matrix is symmetric if and only if the associated operator $\mathbf R^n\to\mathbf R^n$ (with respect to the standard basis) is self-adjoint (with respect to the standard inner product). If terms a 22 and a 23 are both 0, our formula becomes a 21 |A 21 | - 0*|A 22 | + 0*|A 23 | = a 21 |A 21 | - 0 + 0 = a 21 |A 21 |. 1 using the Schur complement of A instead of the Schur complement of Calso holds. This is often easier than trying to specify the Hessian matrix. matrix list b symmetric b[3,3] c1 c2 c3 displacement 3211055 mpg 227102 22249 _cons 12153 1041 52. Solution Let A = [a ij] be a matrix which is both symmetric and skew symmetric. Theorem 1 Any quadratic form can be represented by symmetric matrix. There are other methods of finding the inverse matrix, like augmenting the matrix by the identity matrix and then trying to make the original matrix into the identity matrix by applying row and column operations to the augmented matrix, and so on. first of all you need to write a c program for transpose of a matrix and them compare it with the original matrix. Since the minimum and maximum values equal to 1, we get the identity matrix. Your overall recorded score is 0%. is also symmetric because ÐEEÑ œEE œEEÞX X X XX X The next result tells us that only a symmetric matrix "has a chance" to be orthogonally diagonalizable. eig computes eigenvalues and eigenvectors of a square matrix. Symmetric Matrices The symmetric matrix is a matrix in which the numbers on ether side of the diagonal, in corresponding positions are the same. Multiplying matrices - examples. Lim (Algebra Seminar) Symmetric tensor decompositions January 29, 2009 1 / 29. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. Eigenvalues and the characteristic. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Example: The following 3x3 system has the solution x1 = −1 ; x2 = 2 ; x3 = 1, as you can verify it by direct substitution. 2 4 6 8 5 8 2 1 5 1 0 3 5is a symmetric matrix Theorem 1. Simplify the Matrix. Determining the eigenvalues of a 3x3 matrix. In other words, we can say that matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of matrix A i. Frank Wood. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. 1 below, we show that if ‘>0 and Jis a symmetric diagonally dominant matrix satisfying J ‘S, then J ‘S˜0; in particular, Jis invertible. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11) = (1). Singular values are important properties of a matrix. 4 The Minimax Theorem. Figure 2 2-D Gaussian distribution with mean (0,0) and =1 The idea of Gaussian smoothing is to use this 2-D distribution as a point-spread' function, and this is achieved by. A square matrix such that a ij is the complex conjugate of a ji for all elements a ij of the matrix i. (iv) Theorem 2: Any square matrix A can be expressed as the sum of a symmetric matrix and a skew symmetric matrix, that is (A+A ) (A A )T T A = + 2 2 − 3. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. An analogous result holds for matrices. A = 1 0 0 1 0 0 2 3 3 You have attempted this problem 10 times. ˙ ˙ ˚ ˘ Í Í Î È" π " a i j a i j ij ij 0, 1, Row matrix. Show that S 3x3 ( R) is actually a subspace of M 3x3 ( R) and then determine the dimension and a basis for this subspace. APPLICATIONS Example 2. The diagonal elements of a skew-symmetric matrix are all 0. If you're seeing this message, it means we're having trouble loading external resources on our website. are symmetric matrices. notebook November 28, 2016 Inverse of a 3x3 Matrix Examples: 3. e ( AT =−A ). A = [ 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2]. Write a C program to read elements in a matrix and check whether the given matrix is symmetric matrix or not. Antisymmetric matrices are commonly called "skew symmetric matrices" by mathematicians. it's a Markov matrix), its eigenvalues and eigenvectors are likely to have special properties as well. A × A-1 = I. Definition. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. The symmetric spaces considered here are quotients X=G/K, where G is a non-compact real Lie group, such as the general linear group GL(n,R) of all n x n non-singular real matrices, and K=O(n), the maximal compact subgroup of orthogonal matrices. In describing matrices, the format is: rows X columns. (where A is a symmetric matrix). Similarly, we can take other examples of Nilpotent matrices. We will use the following two properties of determinants of matrices. Complexity Explorer is an education project of the Santa Fe Institute - the world headquarters for complexity science. For the Taylor-Green vortex problem, the domain is periodic is both x- and y-directions, and we end-up with a symmetric implicit matrix. That is the diagonal with the a's on it. For a solution, see the post " Quiz 13 (Part 1) Diagonalize a matrix. Real number λ and vector z are called an eigen pair of matrix A, if Az = λz. First we compute the singular values σ i by finding the eigenvalues of AAT. The Hessian matrix is a square, symmetric matrix whose. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. An identity matrix will be denoted by I, and 0 will denote a null matrix. 2 Matrix “Square Roots” Nonnegative numbers have real square roots. We will use the following two properties of determinants of matrices. A 3x3 stress tensor is 2nd rank. Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. So we;ve got. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. For the lid-driven cavity flow, the implicit matrix is not symmetric. where R is the correlation matrix of the predictors (X variables) and r is a column vector of correlations between Y and each X. Example: Solution: Determinant = (3 × 2) – (6 × 1) = 0. Let's consider a simple example with a diagonal matrix: A = np. The largest term will include the cube of lambda and so on. Definition E EœEÞis called a if symmetric matrix X Notice that a symmetric matrix must be square ( ?). To check whether a matrix A is symmetric or not we need to check whether A = AT or not. It is called a singular matrix. Briefly, matrix inverses behave as reciprocals do for real numbers : the product of a matrix and it's inverse is an identity matrix. I'm currently stuck on converting a 3*N x 1, where N is an integer value, vector into chunks of skew symmetric matrices. M = (Q^(-1)) * G. Symmetric matrices have real eigenvalues. Question from Matrices,cbse,class12,bookproblem,ch3,misc,q14,p101,easy,shortanswer,sec-a,math. xTAx = x1 x2 2 6 18 6 x x 1 2 2x = x 1 + 6x2 1 x2 6x 1 + 18x2 = 2x 12 + 12x1x2 + 18x 22 = ax 12 + 2bx1x2 + cx 22. So in that way every Diagonal Matrix is Symmetric Matrix. Then we have: A is positive de nite ,D k >0 for all leading principal minors A is negative de nite ,( 1)kD k >0 for all leading principal minors A is positive semide nite , k 0 for all principal minors A is negative semide nite ,( 1)k k 0 for all principal minors In the rst two cases, it is enough to. the algorithm will be part of a massive computational kernel, thus it is required to be very efficient. The determinant obtained through the elimination of some rows and columns in a square matrix is called a minor of that matrix. Note that whereas C is a 3× 2 matrix, its transpose, CT, is a 2× 3 matrix. He walks you through basic ideas such as how to solve systems of. The leading coefficients occur in columns 1 and 3. (4) If A is invertible then so is AT, and (AT) − 1 = (A − 1)T. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called Γ:. Thus, we may write [El = S'[A]S (18) where [A] is the diagonal matrix of eigenvalues [A]= [: 0 A, : :I 0 and S is the rotation matrix whose rows are the three orthogonal eigenvectors of [E] corresponding to the Ai eigenvalues. In the following we assume. Example 3 Show that a matrix which is both symmetric and skew symmetric is a zero matrix. Detailed Description Tiny matrix classes with optimized numerical operations which make use of vectorization features through the VVector class. A matrix with real entries is skewsymmetric. An example will be constructed later in this chapter. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Basis vectors. Matrix Inner Products. Now lets FOIL, and solve for. It can be shown that all real symmetric matrices have real eigenvalues and perpendicular eigenvectors. C program check symmetric matrix. Properties. Treat the remaining elements as a 2x2 matrix. Let’s take an example of a matrix. Consider a n x n, trace free, real symmetric matrix A. These terms are more properly defined in Linear Algebra and relate to what are known as eigenvalues of a matrix. the inverse of an n x n matrix See our text ( Rolf, Page 163) for a discussion of matrix inverses. U) = A z^ - [1,3I1A + [-12,123. We will use the following two properties of determinants of matrices. Of course this holds too for square matrices of higher rank (N x N matrices), not just 3x3's. Frank Wood, [email protected] It is called a singular matrix. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. For example, to solve 7x = 14, we multiply both sides by the same number. To check whether a matrix A is symmetric or not we need to check whether A = AT or not. The inverse of a permutation matrix is again a permutation matrix. Here for instance, X should be a 6x6 real and symmetric matrix unless we have made some mistake, and indeed it is constructed from 9 variables defining a symmetric 3x3 matrix. The unit matrix is every #n#x #n# square matrix made up of all zeros except for the elements of the main diagonal that are all ones. As a result, we can concisely represent any skew symmetric 3x3 matrix as a 3x1 vector. Generally, one can find symmetrization A0 of a matrix A by A0 = A+AT 2. 15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite. So, for example, the 3x3 matrix A might be written as:. The leftmost column is column 1. e ( AT =−A ). Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 −2 − 2 7 a b b c. To check whether a matrix A is symmetric or not we need to check whether A = A T or not. Example 4 Suppose A is this 3x3 matrix: [1 1 0] [0 2 0] [0 –1 2]. 4 Diagonal Matrix: A square matrix is called a diagonal matrix if each of its non-diagonal elements are zero (i. com To create your new password, just click the link in the email we sent you. One is to use Gauss-Jordan elimination and the other is to use the adjugate matrix. AB = BA = I n, then the matrix B is called an inverse of A. Inverse matrix A -1 is defined as solution B to AB = BA = I. 366) •A is orthogonally diagonalizable, i. Instead, we can implicitly apply the symmetric QR algorithm to ATA. Examples: Quadratic Form Now we have seen the symmetric matrices, we can move on to the quadratic 1 5 5 8 9 −2 − 2 7 a b b c. (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. 369) EXAMPLE 1 Orthogonally diagonalize. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy and we have Q A(1; 1) = 12 + ( 1)2 + 8(1)( 1) = 1 + 1 8. 50 }; //Symmetric Postive-Definite matrix X = { 7 6. Eigenvalues of a 3x3 matrix. You need to know more about the matrix to conclude one way or the other. Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on. Homework Statement Hi there, I'm happy with the proof that any odd ordered matrix's determinant is equal to zero. symDiagonal() returns an object of class dsCMatrix or lsCMatrix, i. Notice that this is a block diagonal matrix, consisting of a 2x2 and a 1x1. The variance-covariance matrix is symmetric because the covariance between X and Y is the same as the covariance between Y and X. De nition 1 Let U be a d dmatrix. Jacobi's Algorithm is a method for finding the eigenvalues of nxn symmetric matrices by diagonalizing them. Fortran 90 package for solving linear systems of equations of the form A*x = b, where the matrix A is sparse and can be either unsymmetric, symmetric positive definite, or general symmetric. Finally, show (if you haven't already) that the only matrix both symmetric and skew-symmetric is the zero matrix. For example, decrypting a coded message uses invertible matrices (see the coding page). Similarly, the rank of a matrix A is denoted by rank(A). Many problems present themselves in terms of an eigenvalue problem: A·v=λ·v. , (AT) ij = A ji ∀ i,j. HILL CIPHER Encrypts a group of letters called polygraph. Vector x is a right eigenvector, vector y is a left eigenvector, corresponding to the eigenvalue λ, which is the same for. Thethingis,therearealotofotherequivalentwaystodefineapositive definite matrix. Usually the numbers are real numbers. Geometrically, a matrix $$A$$ maps the unit sphere in $$\mathbb{R}^n$$ to an ellipse. (34) Finally, the rank of a matrix can be defined as being the num-ber of non-zero eigenvalues of the matrix. Note that whereas C is a 3× 2 matrix, its transpose, CT, is a 2× 3 matrix. A = [1 1 1 1 1 1 1 1 1]. Then compute it's determinant (which will end up being a sum of terms including four coefficients) Then to ease the computation, find the coefficient that appears in the least amount of term. Therefore, we can see that , Hence, the matrix A is nilpotent. Example: If square matrices Aand Bsatisfy that AB= BA, then (AB)p= ApBp. Inverting a matrix turns out to be quite useful, although not for the classic example of solving a set of simultaneous equations, for which other, better, methods exist. Once we get the matrix P, then D = P t AP. Matrices with Examples and Questions with Solutions. $\begingroup$ Yes, reduced row echelon form is also called row canonical form, and obviously there are infinitely many symmetric matrix that are not diagonal and can be reduced to anon diagonal reduced row echelon form, but note that the row canonical form is not given by a similarity transformation, but the jordan form is. λ is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied: [A](v) = λ (v) Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue λ. The inverse is calculated using Gauss-Jordan elimination. The inverse of a permutation matrix is again a permutation matrix. Mathematical Properties of Stiffness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. Computing eigenvalues and eigenvectors for a 3x3 symmetric matrix. You have unlimited attempts remaining. 139 of Boas, (AB)T = B TA for any two matrices Aand B. com Symmetric Matrix Inverse. The quadratic form of. the symmetric QRalgorithm, as the expense of two Jacobi sweeps is comparable to that of the entire symmetric QRalgorithm, even with the accumulation of transformations to obtain the matrix of eigenvectors. By the second and fourth properties of Proposition C. Join 90 million happy users! Sign Up free of charge:. If Ais symmetric, then A= AT. (1) Any real matrix with real eigenvalues is symmetric. there exists an orthogonal matrix P such that P−1AP =D, where D is diagonal. The main diagonal gets transposed onto itself. It is a specific case of the more general finite element method, and was in. The result is a 3x1 (column) vector. The matrix 1 1 0 2 has real eigenvalues 1 and 2, but it is not symmetric. A , in addition to being magic, has the property that "the sum of the twosymmetric magic square numbers in any two cells symmetrically placed with respect to the center cell is the same" (12, p. Now the next step to take the determinant. Indeed, if aij 6= aji we replace them by new a0 ij = a 0 ji = aij+aji 2, this does not change the corresponding quadratic form. Examples # NOT RUN { #Create a sample 3x3 matrix mat <- matrix(1:9, ncol=3) #Copy the upper diagonal portion to the lower symmetricize(mat, "ud") #Take the average of each symmetric location symmetricize(mat, "avg") # } Documentation reproduced from package HelpersMG, version 4. If A is not SPD then the algorithm will either have a zero. (1) Again, since A is a symmetric matrix, so A′ = A. The link inertia matrix (3x3) is symmetric and can be specified by giving a 3x3 matrix, the diagonal elements [Ixx Iyy Izz], or the moments and products of inertia [Ixx Iyy Izz Ixy Iyz Ixz]. Maximum eigenvalue for this symmetric matrix is 3. JACOBI is a program written in 1980 for the HP-41C programmable calculator to find all eigenvalues of a real NxN symmetric matrix using Jacobi's method. An n×n matrix B is called nilpotent if there exists a power of the matrix B which is equal to the zero matrix. Recently, in order to find the principal moments of inertia of a large number of rigid bodies, it was necessary to compute the eigenvalues of many real, symmetric 3 × 3 matrices. Equation (1) is the eigenvalue equation for the matrix A. Mathematical Properties of Stiffness Matrices 3 computation involving the inverse of ill-conditioned matrices can lose precision because there is a range of values in the solution { d }that can satsify [ K ]{ d }= { p }. all integral types except bool, floating point and complex types), whereas symmetric matrices can also be block matrices (i. A square matrix A is called a diagonal matrix if a ij = 0 for i 6= j. This pages describes in detail how to diagonalize a 3x3 matrix througe an example. Two apologies on quality: 1. Example A= 2 4 0 3 This is a 2 by 2 matrix, so we know that 1 + 2 = tr(A) = 5 1 2 = det(A) = 6 6. Shio Kun for Chinese translation. Use the ad - bc formula. We will use the following two properties of determinants of matrices. The Symmetric Inertia Tensor block creates an inertia tensor from moments and products of inertia. The function scipy. An identity matrix will be denoted by I, and 0 will denote a null matrix. In describing matrices, the format is: rows X columns. Eigenvalues and eigenvectors of a real symmetric matrix. Example 1: Determine the eigenvectors of the matrix. -24 * 5 = -120; Determine whether to multiply by -1. Now lets FOIL, and solve for. (1 Point) Give An Example Of A 3 × 3 Skew-symmetric Matrix A That Is Not Diagonal. We employ the latter, here. Definition of a Matrix The following are examples of matrices (plural of matrix). Note that all the main diagonal elements in the skew-symmetric matrix are zero. Now we only have to calculate the cofactor of a single element. can be obtained by using the cofactor. Now, noting that a symmetric matrix is positive semi-definite if and only if its eigenvalues are non-negative, we see that your original approach would work: calculate the characteristic polynomial, look at its roots to see if they are non-negative. A nxn symmetric matrix A not only has a nice structure, but it also satisfies the following:. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. First we compute the singular values σ i by finding the eigenvalues of AAT. ˙ ˙ ˚ ˘ Í Í Î È" π " a i j a i j ij ij 0, 1, Row matrix. A3×3 example of a matrix with some complex eigenvalues is B = 1 −1 −1 1 −10 10−1 A straightforward calculation shows that the eigenvalues of B are λ = −1 (real), ±i (complex conjugates). Good things happen when a matrix is similar to a diagonal matrix. As is well known, any symmetric matrix is diagonalizable, where is a diagonal matrix with the eigenvalues of on its diagonal, and is an orthogonal matrix with eigenvectors of as its columns (which magically form an orthogonal set , just kidding, absolutely no magic involved ). The above example illustrates a Cholesky algorithm, which generalizes for higher dimensional matrices. You can convert the skew symmetric matrix R_dot * dt into a rotation matrix using the Rodrigues formula. AB = BA = I n, then the matrix B is called an inverse of A. In general, matrices can contain complex numbers but we won't see those here. 231 Diagonalization of non-Hermitian matrices • Let D be the diagonal matrix whose. If the matrix A is symmetric then •its eigenvalues are all real (→TH 8. // symmetric or not. I know that I can convert a single vector of size 3 in a skew symmetric matrix of size 3x3 as follows: X = [ 0 -x(3) x(2) ;. Example: The transpose of A is • For a matrix A = [aij], its transpose AT= [bij], where bij= aji. Example: RC circuit v1 vn c1 cn i1 in resistive circuit ckv Symmetric matrices, quadratic forms, matrix norm, and SVD 15-19. , The sum of the numbers along each matrix diagonal (the character) gives a shorthand version of the matrix representation, called Γ:. NET example in Visual Basic demonstrating the features of the symmetric matrix classes. 32 (2) 223 222 uu kku kku k The global stiffness matrix may be constructed by directly adding terms associated with the degrees of freedom in k(1). As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. Statistics 1: Linear Regression and Matrices The concepts and terminology for matrices will be developed using an example from statistics. Matrix inversion. The subject of symmetric matrices will now be examined using an example from linear regression. Transpose of a Matrix, Symmetric Matrix & Skew Symmetric Matrix: Class 12 Transpose of a matrix , Symmetric Matrix and Skew Symmetric Matrix are explained in a very easy way. 2 Example: Odd or Even. The task is to find a matrix P which will let us convert A into D. Example, = -5 and =5 which means. Joachim Kopp developed a optimized "hybrid" method for a 3x3 symmetric matrix, which relays on the analytical mathod, but falls back to QL algorithm. Logic to check symmetric matrix in C programming. 2 Definiteness of Quadratic Forms. As a recent example, the work of Spielman and Teng [14, 15] gives algorithms to solve symmetric, diagonally dominant linear systems in nearly-linear time in the input size, a fundamental advance. (3) If the products (AB)T and BTAT are defined then they are equal. The values of λ that satisfy the equation are the generalized eigenvalues. Step 1 - Accepts a square matrix as input Step 2 - Create a transpose of a matrix and store it in an array Step 3 - Check if input matrix is equal to its transpose. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. Since the symmetric matrix is taken as A, the inverse symmetric matrx is written as A-1, such that it becomes. Program to swap upper diagonal elements with lower diagonal elements of matrix. A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. [R] binary symmetric matrix combination 3x3, 4x4) which I need to create in R as seen below: I think you didn't run the vecOut` after adding the new matrix. is an eigenvector corresponding to the eigenvalue 1. Enter payoff matrix B for player 2 (not required for zerosum or symmetric games). Let A = (v, 2v, 3v). For example, if a problem requires you to divide by a fraction, you can more easily multiply by its reciprocal. So why bother with the confusion matrix? Because it gives us insight into the details of how the algorithms achieve their percent correct. Diagonalizing a 3x3 matrix. We call such a matrix Hermitianafter the French mathematician Charles Hermite (1822-1901). Scroll down the page for examples and solutions.
fyxewz0sqfycl dajkwnrqsps dfd1m1qpkrqm dpk3dqc9lqwv 5pbqzus3mshim14 61m01xv5xdl zd7gy8b40fe 19oy2gwbsp yeptctvyils esem86uwht8 06vj9j3m94 jp2wo0534lxaxty e5ggbv9peei 9w0eaa6hdw0 ks7w09hxgzkkt mlj60qjcs5p2 uomnmas9ekss j0m4lwi2h45fy 2eulcsy0gsmiqd8 6h2tya9io5puy2 0lu6mgdvy4h pv53iucjilxfrh w1u1ff9m3t0b4 bvswbs5ya8t 6yllw1n4gtpj 0pvr0bmphwgxkaz 9fu54s858j04 mpfludm9r4k9 fnvorb04eb z1ztaijsinf31o8 b1enw5cubxrnxmh ykiif7j79gx4sd vxzz4evvvnlz4wn l5j741s0ndd4 lofw30li8usega7 | 2020-07-12T15:53:48 | {
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https://math.stackexchange.com/questions/3520406/how-is-pac-cap-bc-the-same-as-1-pa-cup-b | How is $P(A^c \cap B^c)$ the same as $1-P(A \cup B)$?
I don't understand how $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is the same?
If I draw $$P(A^c \cap B^c)$$ as a Venn diagram:
If I draw $$P(A \cup B)$$ as a Venn diagram:
So if I subtract $$P(A \cup B)$$ from 1, wouldn't that mean that I subtract $$P(A \cup B)$$ from the universe $$\Omega$$, which would result int this:
However, that would mean $$P(A^c \cap B^c) \neq 1-P(A \cup B)$$
Edit:
As pointed out by multiple people. My diagram for $$P(A^c \cap B^c)$$ should look like the following and therefore the assumption of $$P(A^c \cap B^c) = 1-P(A \cup B)$$ is valid:
• Your first diagram is of $A^c \cup B^c$ not $A^c \cap B^c$. The yellow areas include both $A \cap B^c$ and $A^c \cap B$ – Henry Jan 24 '20 at 0:09
• your first diagram is wrong – Masacroso Jan 24 '20 at 0:11
• The third diagram is the correct one for the left side as well. – Berci Jan 24 '20 at 0:12
• So yellow should be only the universe? – Tom el Safadi Jan 24 '20 at 0:15
• I edited my question with the help of you guys. Thanks!! – Tom el Safadi Jan 24 '20 at 0:19
Your first diagram is incorrect. You seem to have drawn $$A^c\cup B^c$$ (as some other people have mentioned).
I would recommend drawing $$A^c$$ independently first, which consists of the rest of the universe and $$B-A$$. Then draw $$B^c$$ in a different color, noting again that you have the rest of the universe and $$A-B$$. The intersection of $$A-B$$ and $$B-A$$ is neither $$A$$ nor $$B$$, so you will end up without $$A$$ or $$B$$ in your final intersection. However, the rest of the universe is in both $$A^c$$ and $$B^c$$, therefore so is its intersection. Thus, you get that $$P(A^c\cap B^c)$$ is just the universe without $$A$$ or $$B$$, which is equivalent to $$1-P(A\cup B)$$.
According to the DeMorgan's laws, one has \begin{align*} \textbf{P}(\Omega) & = \textbf{P}((A\cup B)\cup(A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}((A\cup B)^{c})\\ & = \textbf{P}(A\cup B) + \textbf{P}(A^{c}\cap B^{c}) = 1 \end{align*} from whence the result follows immediately, since $$X\cup X^{c} = \Omega$$ and $$X\cap X^{c} = \varnothing$$ for every event $$X\subseteq\Omega$$. | 2021-04-11T10:59:43 | {
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http://planetmath.org/SumsOfNormalRandomVariablesNeedNotBeNormal | # sums of normal random variables need not be normal
A common misconception among students of probability theory is the belief that the sum of two normally distributed (http://planetmath.org/NormalRandomVariable) random variables is itself normally distributed. By constructing a counterexample, we show this to be false.
It is however well known that the sum of normally distributed variables $X,Y$ will be normal under either of the following situations.
• $X,Y$ are independent.
• $X,Y$ are joint normal (http://planetmath.org/JointNormalDistribution).
The statement that the sum of two independent normal random variables is itself normal is a very useful and often used property. Furthermore, when working with normal variables which are not independent, it is common to suppose that they are in fact joint normal. This can lead to the belief that this property holds always.
Another common fallacy, which our example shows to be false, is that normal random variables are independent if and only if their covariance, defined by
$\operatorname{Cov}(X,Y)\equiv\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$
is zero. While it is certainly true that independent variables have zero covariance, the converse statement does not hold. Again, it is often supposed that they are joint normal, in which case a zero covariance will indeed imply independence.
We construct a pair of random variables $X,Y$ satisfying the following.
1. 1.
$X$ and $Y$ each have the standard normal distribution.
2. 2.
The covariance, $\operatorname{Cov}(X,Y)$, is zero.
3. 3.
The sum $X+Y$ is not normally distributed, and $X$ and $Y$ are not independent.
We start with a pair of independent random variables $X,\epsilon$ where $X$ has the standard normal distribution and $\epsilon$ takes the values $1,-1$, each with a probability of $1/2$. Then set,
$Y=\left\{\begin{array}[]{ll}\epsilon X,&\textrm{if }|X|\leq 1,\\ -\epsilon X,&\textrm{if }|X|>1.\end{array}\right.$
If $S$ is any measurable subset of the real numbers, the symmetry of the normal distribution implies that $\mathbb{P}(X\in S)$ is equal to $\mathbb{P}(-X\in S)$. Then, by the independence of $X$ and $\epsilon$, the distribution of $Y$ conditional on $\epsilon=1$ is given by,
$\begin{split}\displaystyle\mathbb{P}(Y\in S\mid\epsilon=1)&\displaystyle=% \mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,-X\in S)\\ &\displaystyle=\mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,X\in S)=\mathbb{P% }(X\in S).\end{split}$
It can similarly be shown that $\mathbb{P}(Y\in S\mid\epsilon=-1)$ is equal to $\mathbb{P}(X\in S)$. So, $Y$ has the same distribution as $X$ and is normal with mean zero and variance one.
Using the fact that $\epsilon$ has zero mean and is independent of $X$, it is easily shown that the covariance of $X$ and $Y$ is zero.
$\begin{split}\displaystyle\operatorname{Cov}(X,Y)&\displaystyle=\mathbb{E}[XY]% -\mathbb{E}[X]\mathbb{E}[Y]=\mathbb{E}[XY]\\ &\displaystyle=\mathbb{E}\left[1_{\{|X|\leq 1\}}\epsilon X^{2}\right]+\mathbb{% E}\left[-1_{\{|X|>1\}}\epsilon X^{2}\right]\\ &\displaystyle=\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|\leq 1\}}X^{2}% \right]-\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|>1\}}X^{2}\right]\\ &\displaystyle=0.\end{split}$
As $X$ and $Y$ have zero covariance and each have variance equal to $1$, the sum $X+Y$ will have variance equal to $2$. Also, the sum satisfies
$X+Y=\left\{\begin{array}[]{ll}2\epsilon X,&\textrm{if }|X|\leq 1,\\ 0,&\textrm{if }|X|>1.\end{array}\right.$
In particular, this shows that $\mathbb{P}(|X+Y|>2)=0$. However, normal random variables with nonzero variance always have a positive probability of being greater than any given real number. So, $X+Y$ is not normally distributed.
This also shows that, despite having zero covariance, $X$ and $Y$ are not independent. If they were, then the fact that sums of independent normals are normal would imply that $X+Y$ is normal, contradicting what we have just demonstrated.
Title sums of normal random variables need not be normal SumsOfNormalRandomVariablesNeedNotBeNormal 2013-03-22 18:43:44 2013-03-22 18:43:44 gel (22282) gel (22282) 5 gel (22282) Example msc 62E15 msc 60E05 | 2018-03-18T00:22:22 | {
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http://mathhelpforum.com/number-theory/16543-difference-two-squares.html | # Thread: difference of two squares
1. ## difference of two squares
Find two integers whose squares have a difference of 1,234,567.
can you find two integers whose squares will produce a difference of any whole number you chose?
if so, how? if not, what kind of numbers can't be chosen.
well... a^2-b^2=1234567
we know a must end in a 1 or 9 and b must end in a 2 or 8 (this is to produce the seven at the end)
i came up with 617289^2-617288^2=1234577 and that's as close as i can get.
any suggestions? first time poster! sorry if it is an unusual / inappropriate post!
much thanks!
2. $n^{2}-(n-1)^{2}=1234567$
n=617284
$617284^{2}-617283^{2}=1234567$
3. Number suchs that,
$n\equiv 0,1,3 (\bmod 4)$
Are expressible as a difference of two square.
And,
$n\equiv 2 (\bmod 4)$
Are not.
4. Hello, perfect square!
Find two integers whose squares have a difference of 1,234,567
Galactus used a very clever bit of trivia.
Consecutive squares differ by consecutive odd numbers.
. . $\begin{array}{cccc} & & & \text{diff} \\ 0^2 & = & 0 & \\ & & & 1 \\ 1^2 & = & 1 & \\ & & & 3 \\ 2^2 & = & 4 & \\ & & & 5 \\ 3^2 & = & 9 & \\ & & & 7 \end{array}$
. . $\begin{array}{cccc}4^2 & = & 16 & \\ & & & 9 \\ 5^2 & = & 25 & \\ \vdots & & \vdots & \vdots \end{array}$
As he pointed out, we want two consecutive integers with a difference of 1234567.
So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$
. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$
. . Hence: . $n \,=\,617,283$
Therefore: . $617,284^2 - 617283^2 \:=\:1,234,567$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I suspect that 1,234,567 is prime (but I'm not sure).
If the difference is a composite odd number,
. . there may be an alternate solution.
Example: Find two integers whose squares differ by 133.
Since $\frac{133-1}{2} = 66$, we have: . $67^2 - 66^2 \:=\:133$
. Since $133 = 7\times 19$, we have: . . $19 + 19 + 19 + 19 + 19 + 19 + 19$
. . . . . which can be written: . . $13 + 15 + 17 + 19 + 21 + 23 + 25$
which are the differences of: . $6^2\quad7^2\quad\:8^2\quad\:9^2\quad10^2\quad11^2\ quad12^2\quad13^2$
Therefore: . $13^2 - 6^2 \:=\:133$
5. Originally Posted by Soroban
So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$
. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$
. . Hence: . $n \,=\,617,283$
Such a seemingly difficult question, transformed into simple algebra! Once again! Nice!
6. I suspect that 1,234,567 is prime (but I'm not sure).
No, Soroban, it is not prime. | 2018-02-23T03:36:38 | {
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https://brilliant.org/discussions/thread/combination-proof/ | # Combination Proof
We will prove that:
$$\dbinom{n}{k} = \displaystyle \prod_{i=1}^{k} \frac{n-i+1}{i}$$
$$\forall$$ $$k \in \mathbb{N}$$, $$k \leq n$$
Proof #1: As the reader may guess, we will use induction to prove this. We establish a base case as $$k=1$$:
$$\dbinom{n}{1} = \displaystyle \prod_{i=1}^{1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!}{1!(n-1)!} = n$$
$$\Rightarrow$$ $$n=n$$
The base case clearly holds. Now we establish the inductive case as $$n=\alpha$$:
$$\dbinom{n}{\alpha} = \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\frac{n-\alpha}{\alpha +1} \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$ (By hypothesis)
$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!(n-\alpha)}{(\alpha +1)!(n-\alpha)!} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\frac{n!}{(\alpha +1)!(n-(\alpha+1))!} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
$$\Rightarrow$$ $$\dbinom{n}{\alpha+1} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$
So we may conclude that the statement is true for $$k=1$$ and if it is true for some $$k=\alpha$$, then it must be true for $$k=\alpha+1$$. The proof follows by induction.
QED
Proof #2: We will now offer an alternative proof, which is actually just a simple direct derivation:
$$\dbinom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{\displaystyle \prod_{i=0}^{n-1} (n-i)}{\displaystyle \prod_{i=0}^{k-1} (k-i) \left( \displaystyle \prod_{i=0}^{n-k-1} (n-k-i) \right)}$$
$$= \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{k!} = \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{ \displaystyle \prod_{i=1}^k i }$$
$$= \displaystyle \prod_{i=1}^k \frac{(n-i+1)}{i}$$ Because multiplication is commutative.
Which is the intended result.
QED.
Note by Ethan Robinett
3 years, 10 months ago
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I've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them!
- 3 years, 4 months ago
Thanks!
- 3 years, 4 months ago
Nice note@Ethan Robinett
- 3 years, 10 months ago
Thanks appreciate it!
- 3 years, 10 months ago
Hey nice proofs! :) I did the Inductive one but didn't think of using a direct one
- 3 years, 10 months ago
Thanks! Yeah after I did the induction, I figured there was probably a way to directly show it, given that factorials are just products by definition, I thought the second one was pretty interesting
- 3 years, 10 months ago
I like the way the induction is used.
- 3 years, 10 months ago
Thanks man
- 3 years, 10 months ago | 2018-06-21T12:31:56 | {
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http://mathhelpforum.com/algebra/40280-simplify-expression-negative-exponents.html | # Math Help - Simplify an Expression with Negative Exponents
1. ## Simplify an Expression with Negative Exponents
(ab)^-1 / (a^-2 + b^-2)
if you could explain it with an appropriate property, that would be great!!
thanks..!!
2. $\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$
I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
3. Originally Posted by masters
$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$
I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
Looks good to me, but don't forget to specify that $a \neq 0$ and $b \neq 0$ independently.
-Dan
4. Hi !
A slightly different approach
$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{1}{(ab)(a^{-2}+b^{-2})}$
-----------------------
Using the power rule : $a^b a^c=a^{b+c}$
--> $(ab)a^{-2}=ba^{-1}$
-----------------------
$=\frac{1}{ba^{-1}+ab^{-1}}=\frac{1}{\frac ba+\frac ab}=\frac{1}{\frac{b^2+a^2}{ab}}=\frac{ab}{a^2+b^2 }$
one extra equality
5. Always keep it interesting, Moo. Good job!!
6. thank you both for your help!
7. Hello, needymathperson!
One different step . . .
$\frac{(ab)^{-1}}{a^{-2} + b^{-2}}$
We have: . $\frac{(ab)^{-1}} {a^{-2} + b^{-2}} \;\;=\;\;\frac{\dfrac{1}{ab}} {\dfrac{1}{a^2} + \dfrac{1}{b^2}}$
Multiply top and bottom by $a^2b^2\!:\;\;\frac{a^2b^2\left(\dfrac{1}{ab}\right )}{a^2b^2\left(\dfrac{1}{a^2} + \dfrac{1}{b^2}\right)} \;\;=\;\;\boxed{\frac{ab}{b^2+a^2}}$ | 2014-10-30T18:16:23 | {
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http://math.stackexchange.com/questions/353771/how-to-make-a-box-which-has-the-largest-possible-volume/353786 | # How to make a box which has the largest possible volume?
I have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?
Basically I cut the corners and then fold them. There is no "roof". Thank you!
$$a=0.15m$$
-
@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. – Henning Makholm Apr 7 '13 at 13:57
Let the side remaining after cutting off the vertices be A. The volume is given by:
$$V = \dfrac12 A^2 \sin(60) \times h = \dfrac{\sqrt3}{4} A^2h$$
The relation between $A$, $a$ and $h$ is:
$A = a - \dfrac{2h}{\tan(30)}$
(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)
You can substitute them now:
$$V = \dfrac{\sqrt3}{4} \left(a - \dfrac{2h}{\tan(30)}\right)^2h$$
$$V = \dfrac{\sqrt3}{4} (0.15 - 2\sqrt3h)^2h$$
$$V = 3 \sqrt3 h^3-0.45 h^2+0.00974279 h$$
Differentiate w.r.t. $h$ to get:
$$\dfrac{dV(h)}{dh} = 9 \sqrt3 h^2-0.9 h+0.00974279$$
Equate to zero and solve for $h$ to get $h\approx0.0433013$ and $h\approx0.0144338$.
The first one gives the minimum volume, so you don't want that. Take the second.
-
Thank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily! – user31113 Apr 7 '13 at 11:46
You're welcome! – Jerry Apr 7 '13 at 11:47
If you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \sqrt3 h$. (Let me know if you have difficulty with this.)
Hence the box's volume is proportional to $\left(a-2 \sqrt3 h\right)^2 h$, which we try to maximise. Let $V(h) = \left(a-2 \sqrt3 h\right)^2 (4 \sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \sqrt3 h), (a-2 \sqrt3 h)$ and $(4 \sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.
$a-2 \sqrt3 h = 4 \sqrt3 h$ or when $h = \dfrac{a}{6\sqrt3}$.
-
Oh woops, yours is definitely more elegant! – Jerry Apr 7 '13 at 11:26
@Jerry Thanks. It's just that I like inequalities a lot. – Macavity Apr 7 '13 at 11:32
Thank you Macavity. :) Cheers! – user31113 Apr 7 '13 at 11:44
@Macavity Your solution is certainly neat, but I get a slight butterfly feeling when you pull that factor out of the air. I seem to get the same numbers by a method at least I find simpler. Any comments? – Brian Chandler Dec 20 '14 at 16:24
Using exactly the same argument as in this question: Optimize volume of an open cardboard box made from flat square of cardboard... we consider an infinitesimal change $\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls. Call the area of the base $B$, the perimeter of the base $P$, and the height $h$, then these changes are:
$\delta \times h \times P$ ...change over walls
$\delta \times B$ ...change over base
And the maximum is when the area of the base and the area of the walls are equal, so we have:
$h = B / P$
In this case, let $s$ be the side of the base. Then the area of the base (an equilateral triangle) and perimeter are given by:
$B = \frac{\sqrt 3}{4} s^2$
$P = 3s$
So
$h = \frac{\sqrt 3}{4} s^2 / 3s = \frac{1}{4\sqrt 3} s$
Now we get $s$ from the original triangle side $a$ and substitute:
$s = a - 2\sqrt 3 h$
$h = \frac{1}{4\sqrt 3} (a - 2\sqrt 3 h)$
$4\sqrt 3 h = a - 2\sqrt 3 h$
$h = \frac{a}{6\sqrt 3}$
- | 2016-02-14T08:36:01 | {
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https://math.stackexchange.com/questions/573055/projection-of-ellipsoid | # Projection of ellipsoid
Find the projections of the ellipsoid
$$x^2 + y^2 + z^2 -xy -1 = 0$$
on the cordinates plan
I have no idea how to do this. I couldn't find much on google to help me with it too.
By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:
$xy$-plane: Let $z=0$.
$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.
In the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$. Implicitly differentiating with respect to $x$ gives us
$$2x+2y\frac{dy}{dx} - y - x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y-2x}{2y-x}.$$
Now, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\implies y=2x$. Substituting this back into the implicit equation gives us $$x^2+4x^2-2x^2 -1 = 0 \implies 3x^2=1 \implies x=\pm \frac{\sqrt{3}}{3}$$
and thus $y = \pm\dfrac{2\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.
Similarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\implies x=2y$. In a similar fashion, we get that $y=\pm\dfrac{\sqrt{3}}{3}$ and hence $x=\pm\dfrac{2\sqrt{3}}{3}$ are the equations of the vertical tangent lines.
Now why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\sqrt{3}/3 \leq x,y \leq 2\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively. The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\leq z\leq 1$. With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is $$\frac{3y^2}{4} + z^2 = 1$$
and the corresponding projection onto the $xz$-plane is $$\frac{3x^2}{4} + z^2 = 1$$
Note that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest.
• Yes, I meant it, thanks. Well, I did this on a first moment but the answer is different for $yz$ and $xz$. For example: $yz: \frac{3}{4}y^2 + z^2 -1 = 0$ Nov 19, 2013 at 11:17
• @Giiovanna I deleted my answer, fixed it, and then undeleted it. I hope what I wrote above makes sense! Nov 19, 2013 at 12:44
In case you have difficulties with the accepted answer, here is another approach:
Your ellipsoid $S$ is the level surface $F^{-1}(\{0\})$ of the function $$F(x,y,z):=x^2+y^2+z^2-xy-1\ .$$ Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points $p\in S$ that generate the boundary $\partial S'$ are characterized by the property that the tangent plane at $p$ is vertical (i.e., parallel to the $z$-axis), which is the same as saying that the surface normal $n_p$ at $p$ is horizontal. Now $n_p$ is given by $\nabla F(p)$, whose third coordinate is $2z$.
It follows that the points $(x,y,z)\in S$ producing the shadow boundary are the points with $z=0$, i.e., the points $(x,y,0)$ satisfying $x^2+y^2-xy-1=0$. In this special example the shadow boundary in fact coincides with the intersection of $S$ with the $(x,y)$-plane.
Therefore let's do the shadow $S''$ of $S$ in the $(x,z)$-plane as well. The points $p\in S$ that generate the boundary $\partial S''$ are characterized by the property that the tangent plane at $p$ is parallel to the $y$-axis, which is the same as saying that the surface normal $n_p$ at $p$ is orthogonal to the $y$-axis, or has $y$-component $0$. The $y$-component of $\nabla F(p)$ is given by $2y-x$. Therefore we can say that the points $p$ in question lie on the plane $2y-x=0$. As they satisfy a priori $F(x,y,z)-1=0$ we can eliminate $y$ from the two equations $$x^2+y^2+z^2-xy-1=0,\qquad y={x\over2}$$ and obtain a single equation connecting their $x$- and $z$-coordinates: $$x^2+{x^2\over4}+z^2-x\>{x\over2}-1=0\ ,$$ or $${3\over4}x^2+z^2-1=0\ .$$ This is already the equation of the shadow boundary $\partial S''$.
I may now safely leave the shadow on the $(y,z)$-plane to you.
• I like this method the most Sep 18, 2017 at 9:10
• It’s worth noting that for a quadric surface, the points that generate its shadow outline are always coplanar.
– amd
Dec 24, 2021 at 19:53
An approach related to Christian Blatter’s answer can be found in Result 8.9 on page 201 of Hartley & Zisserman’s Multiple View Geometry In Computer Vision. Under the camera matrix $\mathtt P$, the outline of the quadric $\mathtt Q$ is the conic given by $\mathtt C^*=\mathtt P\mathtt Q^*\mathtt P^T$. (Here the superscript asterisk indicates the dual conic.) The proof follows directly from the observations that lines $\mathbf l$ tangent to the outline satisfy $\mathbf l^T\mathtt C^*\mathbf l=0$ and back-project to planes $\mathtt P^T\mathbf l$ that are tangent to the original quadric.
In the present problem the quadric is the nondegenerate ellipsoid $$\mathtt Q = \begin{bmatrix}1&-\frac12&0&0 \\ -\frac12&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1\end{bmatrix}$$ so we can use inverse matrices for the duals: the outline of $\mathtt Q$ under the projection $\mathtt P$ is therefore $\mathtt C = (\mathtt P\mathtt Q^{-1}\mathtt P^T)^{-1}$.
For orthogonal projection onto the $x$-$y$ plane, we can use $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ for the camera matrix and have $$\mathtt Q^{-1}=\begin{bmatrix}\frac43&\frac23&0&0\\\frac23&\frac43&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix},$$ which gives us $$\mathtt C = \begin{bmatrix}1&-\frac12&0\\-\frac12&1&0\\0&0&-1\end{bmatrix},$$ or $x^2+y^2-xy=1$. For the $x$-$z$ plane, $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ and $$\mathtt C = \begin{bmatrix}\frac34&0&0\\0&1&0\\0&0&-1\end{bmatrix},$$ i.e., $\frac34x^2+z^2=1$. A similar computation for the $y$-$z$ plane yields $\frac34y^2+z^2=1$.
• Hello @amd, your solution works like a charm for me! Thank you, for posting this answer. Small question, how do we find the projection matrix, P, for any arbitrary plane? Dec 23, 2021 at 3:47
• @KashishDhal There are various ways, but for any of them you must first choose a coordinate system for the plane. Once you have that, the easiest thing that I can think of is to compute a coordinate transformation that takes that plane to one of the prime coordinate planes and compose it with the corresponding simple projection matrix from above.
– amd
Dec 24, 2021 at 19:49
• I followed the method that you described in this thread: math.stackexchange.com/questions/3073718/… However, I don't know if my solutions are correct. How can I verify that the computed ellipse are indeed correct? Dec 24, 2021 at 20:59 | 2022-07-03T03:44:36 | {
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https://math.stackexchange.com/questions/1949683/if-u-1-u-n-v-1-v-k-is-a-basis-does-it-necessarily-follow-that-u | # If $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis does it necessarily follow that $\{u_1,...,u_n\}$ is linearly independent?
We are given that $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis. My question is does it necessarily follow that $\{u_1,...,u_n\}$ or $\{v_1,...,v_k\}$ are linearly independent? I suspect the answer is yes, but I think I'm missing something in my argument.
So: if $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis, then this list of vectors is linearly independent. That is,
$a_1 u_1 + \cdots + a_n u_n + b_1 v_1 + \cdots + b_k v_k = 0 \implies a_1 = \cdots = a_n = b_1 = \cdots = b_k = 0$.
Then I want to say the following: Since all the $b_i \,'s$ are $0$, we have $a_1 u_1 + \cdots + a_n u_n = 0 \implies a_1 = \cdots = a_n = 0$, meaning that {$u_1,...,u_n$} is linearly independent. With a similar argument, we can also conclude that {$v_1,...,v_k$} is linearly independent.
Am I missing something though? I feel like my argument is kind of making a bit of a stretch. Thanks for your help!!
• You're right! in general, any subset of a linearly independent set is again linearly independent, as your proof shows (you're not using more than linearly independency). Oct 1 '16 at 20:25
• Yes, any subset of a linearly independent set is lin. indep. itself. Oct 1 '16 at 20:25
• It's fine for me. Actually, whether it's a basis is not the relevant hypothesis. What you proved indeed is a little more general: any sublist of a list of linearly independent vectors is linearly independent. Oct 1 '16 at 20:26
• Thank you so much guys!!! :D Oct 1 '16 at 20:36
Your argument is fine, but perhaps this is a case where it can be nicely put as a proof by contradiction. Suppose some sublist were not linearly independent, then we can find some $\{a_i\}$, not all zero, such that $\sum_{i=1}^n a_iu_i=0$. We can then use this to show that the bigger list is not linearly independent. | 2022-01-21T17:12:27 | {
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http://onelab.info/wiki/Tutorial/Coupled_problems | Tutorial/Coupled problems
Introduction
Here is explained how to use the solution of a first problem as a data in a second problem. Two kind of problem are studied here. In the "first" one, the solution of the first problem directly appears in the weak formulation of the second problem (for example, as a source or a Neumann boundary) whereas in the second kind of problem, the solution is used as a constraint (Dirichlet boundary condition), so it should appears in the function space. Note that a Lagrange multiplier could be used to force the Dirichlet constraint in the weak formulation.
First kind of coupled problem : solution appearing in the weak formulation
Mathematics
Let the computation domain be the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$. Consider the two coupled following problems : firstly, find $u$, solution of $$\begin{cases}\label{eq:problemU} -\Delta u + u = C & \text{in } \Omega,\\ \displaystyle{\frac{\partial u}{\partial \mathbf{n}} = 0} & \text{on }\Gamma, \end{cases}$$ where $C$ is a constant. Then, find the solution $v$ of the second problem $$\begin{cases}\label{eq:problemV} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{\frac{\partial v}{\partial \mathbf{n}} = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of problem \ref{eq:problemU}. Obviously, the function $u$ is the constant function equal to $C$ and thus, there is no "real" need in solving numerically problem \ref{eq:problemU}. However, please keep in mind that this is just an example. In order to solve numerically these problem using a finite element method, on has to write the weak formulations of problems \ref{eq:problemU} and \ref{eq:problemV}, which read as: $$\label{eq:WeakFormulationU} \left\{\begin{array}{l} \text{Find } u\in H^1(\Omega) \text{ such that}\\ \displaystyle{\forall u'\in H^1(\Omega), \qquad \int_{\Omega} \nabla u\cdot\nabla u' \;{\rm d}\Omega + \int_{\Omega} u\cdot u' \;{\rm d}\Omega - \int_{\Omega} Cu'\;{\rm d}\Omega = 0}, \end{array}\right.$$ and $$\label{eq:WeakFormulationV} \left\{\begin{array}{l} \text{Find } v\in H^1(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega - \int_{\Gamma}uv'\;{\rm d}\Gamma = 0}, \end{array}\right.$$ where $H^1(\Omega)$ is the classical Sobolev space.
Computation in GetDP
In GetDP, this kind of problem is very easy to solve. Indeed, the user has to introduce two function space (one per solution), and ask GetDP to solve each problem in the right order. There is no trap ! Below are the GMSH and GetDP files. Compared to the other academic examples, the only new things are in the .pro file.
Files
There are two files :
FirstCoupledProblem.geo that contains the geometry and is used to mesh the domain
FirstCoupledProblem.pro that contains the weak formulations, function spaces, etc.
To solve the problem, type in a terminal (in the right directory) :
gmsh FirstPb.geo -2
and then solve the problem with getdp :
getdp FirstPb.pro -solve -pos
or directly
getdp FirstPb.pro -solve CoupledProblem -pos Map_u_and_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/First/FirstPb.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
//Space for v
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
}
/*
The two FunctionSpaces are exactly the same. However, keep in mind that
a FunctionSpace can contain only one solution ! That is why we need two FunctionSpaces.
*/
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}
// Coupled problem : "first kind" :
// Neumann condition on Gamma
// Delta v + v =0 Omega
// dn v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad_v;}
// u is called but will NOT be used as an unknown (NO "Dof")
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
//Neumann condition on Gama (no "Dof" !)
Galerkin{ [-{u}, {v}];
In Gama; Jacobian JSur; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
// First coupled problem available
{Name CoupledProblem;
System{
{Name PbU; NameOfFormulation ProblemU;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u" then save "u" in the space H1_C
2) generate and solve the problem associated to "v" and save "v" in the space H1
*/
}
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
//The associated formulation is "ProblemV" because it involves both "u" and "v"
//(ProblemU only call "u", not "v")
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//==============
PostOperation{
{Name Map_u_and_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}
Second kind of coupled problem : solution used as a Dirichlet boundary condition
Mathematics
In this section, the first problem (\ref{eq:problemU}) does not change but the second problem is now to find $v$, such that $$\begin{cases}\label{eq:problemV2} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{v = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of (\ref{eq:problemU}). The function $u$ is now involved as a Dirichlet boundary condition and thus, will appear in the function space of $v$ (again, except if one uses a Lagrange multiplier, which is not the case here). To obtain the weak formulation of problem (\ref{eq:problemV2}), we introduce the following function space \ref{eq:WeakFormulationU} $$H^1_u(\Omega) = \left\{w\in H^1(\Omega) \text{ such that w|_{\Gamma} = u|_{\Gamma}, where u is the solution of problem (1)} \right\}.$$ Thus, we have $$\label{eq:WeakFormulationV2} \left\{\begin{array}{l} \text{Find } v\in H^1_u(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1_0(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega = 0}, \end{array}\right.$$
GetDP
In GetDP, a Dirichlet boundary condition is generally imposed through the "Constraint" term, it can be imposed with the help of a Lagrange multiplier. In the second case, the Dirichlet boundary condition is no more "imposed" in the function space but appears in the weak formulation (as a Neumann boundary condition). Thus, the previous case can be used. Here is only considered the first case using the GetDP function "AssignFromResolution". A Resolution dedicated to $u$ must then be created, in this example it will be named AuxiliaryResolution. In GetDP, the procedure can be summarized as follows:
• Create a constraint with :
Type AssignFromResolution ; Resolution AuxiliaryResolution;
• Construct only ONE function space that will contain $u$ and at the end $v$ ($u$ will be erased !)
• Create two Resolution, one for $v$ and one for $u$ :
• for $v$ : Generate, Solve, Save
• for $u$ (AuxiliaryResolution) : Generate, Solve, TransfertSolution (do not Save !)
With this, GetDP will ... :
• Try to solve problem with $v$
• Need the Dirichlet constraint (that is $u$)
• Solve problem with $u$
• Solve problem with $v$
Sometimes it is usefull to save both $u$ and $v$ at the end of the calculation. This can be done by adding an auxiliary problem that "copy" $u$ (see the last paragraph).
Files
There are two files :
Secondpb.geo
exactly the same as FirstPb.geo
Secondpb.pro
quite the same as FirstPb.pro
To solve the problem, type in a terminal (in the right directory) :
gmsh SecondPb.geo -2
and then solve the problem with getdp :
getdp SecondPb.pro -solve -pos
and chose the first resolution (MainResolution) or directly
getdp SecondPb.pro -solve MainResolution -pos Map_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/Second/SecondPb.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/
//FunctionSpace
//=============
FunctionSpace{
//Space for u AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}
/*
ONLY one function space that will contain "u" and at the end, "v".
*/
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}
// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
generate and solve the problem associated to "v" and save "v" in the space H1
However, to compute "v", GetDP need to compute the Dirichlet Contraint, thus
is needs to solve AuxiliaryResolution (bellow) !
*/
// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; TransferSolution[PbU];
}
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u"
2) Transfert solution u (to problem associated to "v")
*/
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//==============
PostOperation{
{Name Map_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}
Second kind of coupled problem with a save of solution $u$
Sometimes, it can be usefull to save both $u$ and $v$. To achieved this, one can introduce an auxiliary variable $u_{aux}$ that will be a copy of $u$. Here, we proposed to add :
A new FunctionSpace to store $u$
A new weak formulation to copy $u$ into an auxiliary variable $u_{aux}$
The modified .pro file (SecondPbWithSave.pro) can be found below. The .geo file being exactly the same as the previous ones.
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;
// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;
//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};
//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};
// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};
//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};
// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...
Direct link to file CoupledProblems/SecondWithSave/SecondPbWithSave.geo'
// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}
// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}
//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}
//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints 1 ; }
{ GeoElement Line ; NumberOfPoints 4 ; }
{ GeoElement Triangle ; NumberOfPoints 6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}
// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/
//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
//Space for u_aux (=copy of u) AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}
//(Weak) Formulations
//==================
Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}
// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name v; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}
//Copy "u" in "u_aux"
{Name CopyU; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name u_aux; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
Galerkin{ [Dof{u_aux}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-{u}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}
}
}
}
// Resolution
//===========
Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
Try to compute "v", however, GetDP needs to compute the Dirichlet Contraint
thus to solve AuxiliaryResolution (below) !
*/
// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU;}
{Name PbUaux; NameOfFormulation CopyU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbUaux]; Solve[PbUaux]; TransferSolution[PbUaux];
}
}
/*Short explanations : GetDP will :
1) Compute "u"
2) Copy "u" in "uaux"
3) Transfert solution "uaux" (to problem associated to "v")
*/
}
//Post Processing
//===============
PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}
//Post Operation
//==============
PostOperation{
{Name Map_Sol; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
} | 2018-12-17T07:57:52 | {
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https://math.stackexchange.com/questions/2743313/prove-that-the-number-of-answers-for-a-1a-2-a-k-le-n-is-equal-to-t | Prove that the number of answers for $|a_1|+|a_2|+…+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+…+|a_n| \le k$.
I'm trying to solve this problem:
Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|≤k$. all $a_i$ is an integer number.
I have a solution for non negative $a_i$: We convert $|a_1|+|a_2|+...+|a_k|≤n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|≤k$ to $a_1+a_2+...+a_n+c=k$.
By bars and stars theorem: Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\binom{n+k}{n}$. That $\binom{n+k}{k} = \binom{n+k}{n}$.
but i don't know how to prove for negative integers.
• So we allow $a_i$ to be any integer, but we consider its absolute value when taking the sum? – hardmath Apr 19 '18 at 0:41
• @hardmath yes,we take absolute value – user547757 Apr 19 '18 at 14:57
Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$. Note that $$f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0),\tag{*}$$ by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get \begin{align} f(n,k) &= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]\\ &= f(n-1,k)+f(n-1,k-1)+f(n,k-1) \end{align} The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.
We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$.
There will be $j\leq r:={\rm min}\{n,k\}$ of the $a_i$ that are $\ne0$, call them $a_{i_1}$, $\ldots$, $a_{i_j}$. The $i_l$ $(1\leq l\leq j)$ can be chosen in ${n\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\geq 0$. Then we have to count the number of solutions to $x_1+\ldots+x_j+c=k-j$. This number is ${k\choose j}$. It has to be multiplied by $2^j$ in order to account for the choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to $$\sum_{j=0}^r{n\choose j}{k\choose j}2^j\ ,$$ which is symmetric in $n$ and $k$.
Introducing notation $$A_i=|a_i|$$ the first of inequalities can be rewritten as: $$A_1+A_2+...+A_k\le n.\tag1$$ Any solution to $$(1)$$ can be viewed as composed of $$i$$ zeros ($$0\le i\le k$$) and $$k-i$$ non-zeros. Each non-zero value doubles the number of solutions to original equation (which can be both positive and negative). The overall number of solutions is therefore $$\sum_{i=0}^{k}\binom{k}{i}\binom{n}{k-i}2^{k-i} \stackrel{i=k-j}=\sum_{j=0}^{k}\binom{k}{j}\binom{n}{j}2^{j} =\sum_{j=0}^{\min(k,n)}\binom{k}{j}\binom{n}{j}2^{j},\tag2$$ where $$\binom{k}{i}$$ counts the number of ways to place $$i$$ zeros into $$k$$ cells, and $$\binom{n}{k-i}$$ counts the number of positive solutions to the inequality $$p_1+p_2+\cdots+p_{k-i}\le n.$$
In view of symmetry of $$(2)$$ with respect to interchange of $$n$$ and $$k$$ the original statement is proved.
Just to show another approach to the answer already given by C.Blatter, let's start and consider $$\left\{ \matrix{ 0 \le b_{\,j} \in \mathbb Z \hfill \cr b_{\,1} + b_2 + b_{\,3} + \cdots + b_{\,k} \le n \hfill \cr} \right.$$ Then the number of solutions is the integral volume $V(k,n)$ of k-D simplex, with edge segments $(0,n)$, thus of length $n+1$.
It is not difficult to demonstrate that $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k}=V(n,k)$$ then the hypothesis is already true in this case.
We can then partition the above solutions into those that contains $j$ variables null and the remaining $k-j$ having values greater than $0$. We can assign the null variables in $\binom{k}{j}$ ways.
That corresponds to $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k} = \sum\limits_{0\, \le \,j\, \le \,k} { \binom{n}{k-j} \binom{k}{j} }$$
Coming to our present case, for each non null variable $b_j$ we can assign two values to the corresponding $a_j$, and just one to the null variables. So in this case the volume will be $$V_{\,a} \,(k,n) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\binom{n}{k-j} \binom{k}{j} 2^{\,k - j} } = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\binom{n}{l} \binom{k}{l} 2^l } = V_{\,a} \,(n,k)$$
Therefore the hypothesis is demonstrated.
I just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is $$\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}$$ which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction. | 2019-10-15T18:52:17 | {
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https://mathoverflow.net/questions/133732/upper-limit-on-the-central-binomial-coefficient | # Upper limit on the central binomial coefficient
What is the tightest upper bound we can establish on the central binomial coefficients $2n \choose n$ ?
I just tried to proceed a bit, like this:
$n! > n^{\frac{n}{2}}$
for all $n>2$. Thus,
$\binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n$
But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).
• The simplest upper bound to prove is $4^n$ (which is still stronger than your bound) and just follows from the binomial expansion of $(1+1)^{2n}$. Peter's answer gives a less wasteful estimate. Jun 14 '13 at 13:00
• I should have thought about it. Yes, Peter's bound is very good, I tested it too. Jun 14 '13 at 13:54
Here's a way to motivate and refine the argument that Péter Komjáth attributes to Erdős.
Start by computing the ratio between the $$n$$-th and $$(n-1)$$-st central binomial coefficients: $${2n \choose n} \left/ {2(n-1) \choose n-1} \right. = \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2} = \frac{(2n)(2n-1)}{n^2} = 4 \left( 1 - \frac1{2n} \right).$$ For large $$n$$, this ratio approaches $$4$$, suggesting that $$2n \choose n$$ grows roughly as $$4^n$$. If the factor $$1 - \frac1{2n}$$ were $$1 - \frac1n = (n-1)/n$$, the growth would be exactly proportional to $$n^{-1} 4^n$$. Since $$1 - \frac1{2n}$$ is (for large $$n$$) nearly the square root of $$1 - \frac1n$$, the actual asymptotic should be proportional to $$n^{-1/2} 4^n$$. So we introduce the ratio $$r_n := \left( {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right. \right)^2 = \frac{n}{16^n} {2n \choose n}^2.$$ Then $$\frac{r_n}{r_{n-1}} = \left( 1 - \frac1{2n} \right)^2 \left/ \left( 1 - \frac1n \right) \right. = \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1.$$ Thus $$r_{n-1} < r_n$$; and since $$r_1 = (2/4)^2 = 1/4$$ we have by induction $$r_1 \lt r_2 \lt r_3 \lt r_4 \lt \cdots \lt r_n = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)} \frac{2n-1}{2n}.$$ Each $$r_{n_0}$$ gives a lower bound on $$r_n$$, and thus on $$2n\choose n$$, for all $$n \geq n_0$$. The OP asked for upper bounds, so consider $$R_n := \frac{2n}{2n-1} r_n = \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2.$$ Now $$R_{n+1}/R_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2 = (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$$, so $$\frac12 = R_1 \gt R_2 \gt R_3 \gt R_4 \gt \cdots \gt R_{n+1} = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.$$ It follows that $$R_n \geq r_{n'}$$ for any $$n,n'$$, so $$R_1=1/2$$, $$R_2=3/8$$, $$R_3=45/128$$, etc. are a series of upper bounds on every $$r_n$$. Since moreover $$r_n / R_n = 1 - \frac1{2n} \rightarrow 1$$ as $$n \rightarrow \infty$$, both $$r_n$$ and $$R_n$$ converge to a common limit that is an upper bound on every $$r_n$$. If we accept Wallis's product (which is classical though not as elementary as everything else in our analysis), then we can evaluate this common limit as $$1/\pi$$ and thus recover the asymptotically sharp upper bound $${2n \choose n} < 4^n / \sqrt{\pi n}$$.
• +1, and let me challenge your statement that the Wallis product is not as elementary, by pointing to the Monthly note math.chalmers.se/~wastlund/monthly.pdf and the 16th Christmas Tree Lecture by Don Knuth, "Why pi?" (now on youtube). Jul 1 '13 at 16:58
• Your answer has been cited in this paper that I was just reading, mentioning it here as a token of appreciation: arxiv.org/pdf/2005.10009.pdf
– NULL
Jul 28 '20 at 19:59
Even the asymptotically sharp inequality ${2n \choose n} < 4^n \left/ \sqrt{\pi n} \right.$ has a short proof: $${2n \choose n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx < \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx = \frac{4^n}{\sqrt{\pi n}}.$$ In the first step, the formula for $\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$ can be proved by induction via integration by parts, or using the Beta function.
The third step is clear, and the last step is the well-known Gaussian integral. So we need only justify the the second step.
There we need the inequality $\cos x \leq e^{-x^2/2}$, or equivalently $$\log \cos x + \frac{x^2}{2} \leq 0,$$ for $\left|x\right| < \pi/2$, with equality only at $x=0$. This is true because $\log \cos x +\frac12 x^2$ is an even function of $x$ that vanishes at $x=0$ and whose second derivative $-\tan^2 x$ is negative for all nonzero $x \in (-\pi/2, \pi/2)$. QED
• I'm not sure what counts as elementary in this game, but I would think the easiest proof that $\frac{1}{\pi} \int \cos^{2n} x dx = \binom{2n}{n}/4^n$ is to write $\cos x = (e^{ix} + e^{-ix})/2$ and expand $\cos^{2n}$ by the binomial theorem. Jul 1 '13 at 14:18
Erdos remarked somewhere the bound
$${{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}.$$ This can be established by induction: $${{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}$$ and if we have the bound for $n$, we only have to show $$\frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}}$$ which reduces to $4n^2+8n+3<4n^2+8n+4$.
• This bound is pretty close to optimal because (unless I've computed badly) Stirling's approximation gives $4^n/\sqrt{\pi n}$. Jun 14 '13 at 13:26
Noam Elkies notes that there is a quick proof of $$\binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$ by writing $$\binom{2n}{n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x dx$$ and bounding $\cos^2 x \leq e^{-x^2}$.
There is an equally good lower bound by a similar method: $$\int_{-\pi/2}^{\pi/2} \cos^{2n} x dx =\int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{(\tan^2 x+1)^{n+1}} = \int_{- \infty}^{\infty} \frac{du}{(1+u^2)^{n+1}} \geq \int_{- \infty}^{\infty} e^{-(n+1) u^2} du$$ so $$\binom{2n}{n} \geq \frac{4^n}{\sqrt{\pi (n+1)}}.$$ Here the inequality $\tfrac{1}{1+u^2} \geq e^{-u^2}$ follows from the standard bound $e^y \geq 1+y$ for $y\geq 0$.
We may prove without using integrals that $$\frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1)$$ (1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$ (the identities are straightforward by induction).
Denote $$m:=2n+1$$. It is not hard to show that $$\sin mx=p_m(\sin x)$$ for a polynomial of degree $$m$$ in $$\sin x$$. The roots of $$p_m$$ are $$\sin \frac{k\pi}{2n+1}$$ for $$k=-n,\ldots,n$$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$ (the multiple $$2n+1$$ comes from dividing by $$x$$ and putting $$x=0$$). Put $$x=\frac{\pi}{4n+2}$$. Using the inequality $$\sin tx\leqslant t\sin x$$ (which may be proved, for example, by induction in $$t=1,2,\ldots$$ from the identity $$\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$$) for $$t=2k$$ ($$k=1,2,\ldots,n$$) we get $$1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi.$$ Analogously we get $$\cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right).$$
Dividing by $$1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$$ and substituting $$x=\frac{\pi}{4n}$$ (for computing the LHS at this point use l'Hôpital rule) we get $$n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi.$$
UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $$\sum 1/n^2=\pi^2/6$$ and $$\sum (-1)^{k-1}/(2k-1)=\pi/4$$ using these trigonometric things.)
• Nice proof! Is there an easy way to get the tight lower bound $1/\sqrt{\pi n+1}$? It has such a nice form..
– aorq
Jan 1 at 2:54
• @aorq It is not so much tight. The sequence $4^{-n}{2n\choose n}\sqrt{n+c}$ eventually decreases if $c>1/4$ (this is straightforward: divide the squares of two consecutive terms and subtract 1) and its limit equals $1/\sqrt{\pi}$. Thus for large enough $n$ we get the upper bound $1/\sqrt{\pi (n+c)}$. For $c=1/\pi$ this just works from the very beginning. Jan 1 at 9:51
• Yes, I know that it's tight in the weakest possible sense, namely equality for $n=0$ and poor otherwise. Shortly after I wrote my comment, I saw that $1/\sqrt{\pi(n+1/4)}$ was tighter in a more meaningful sense (on the other side), namely ever better as $n\to\infty$. Thanks for your explanation!
– aorq
Jan 1 at 14:41
• Warning: $1/\sqrt{\pi(n+1/4)}$ is an upper bound Jan 1 at 15:08
You may also prove $${2n\choose n}\sim \frac{4^n}{\sqrt{\pi n}}$$ using the combinatorial meaning of $${2n\choose n}$$ and the area-of-a-circle definition of $$\pi$$. That is probably the most elementary we can hope for.
First of all, if $$a_n:=\sqrt{n+1/4}{2n\choose n}4^{-n}$$ and $$b_n:=\sqrt{n+1/2}{2n\choose n}4^{-n}$$, we may check that $$a_n\leqslant b_n$$, $$a_n$$ increases and $$b_n$$ decreases, thus $$a_n$$, $$b_n$$ have a common positive finite limit which we denote $$\alpha$$, and $$a_n\leqslant \alpha\leqslant b_n$$. Since $$b_n/a_n=1+O(1/n)$$, we get $$a_n=\alpha+O(1/n)$$, and also $$c_n=\alpha+O(1/n)$$ where $$c_n:=\sqrt{n}{2n\choose n}4^{-n}$$.
We should prove that $$\alpha=1/\sqrt{\pi}$$.
Use the combinatorial identity $$4^{-n}\sum_{k=0}^n k{2k\choose k}(n-k){2(n-k)\choose n-k}=\frac{n(n-1)}8.\quad (\heartsuit)$$ LHS of $$(\heartsuit)$$ equals $$\sum_{k=0}^n \sqrt{k(n-k)}c_kc_{n-k}=\sum_{k=0}^n \sqrt{k(n-k)}\left(\alpha^2+O\left(\frac1k+\frac1{n-k}\right)\right)\\=\alpha^2\sum_{k=0}^n\sqrt{k(n-k)}+O\left(n\sum_{k=0}^n\frac1{\sqrt{k(n-k)}}\right)=\alpha^2\cdot \frac{\pi n^2}8+o(n^2),$$ since the union of rectangles $$[k,k+1]\times [0,\sqrt{k(n-k)}]$$ approximates the upper semi-circle with the horizontal diameter $$[0,n]$$.
Thus $$\alpha^2\pi/8=1/8$$ and $$\alpha=1/\sqrt{\pi}$$.
You could use instead of $$(\heartsuit)$$ the more famous identity $$4^{-n}\sum_{k=0}^n {2k\choose k}{2(n-k)\choose n-k}=1$$ this gives $$\alpha^2\int_0^1\frac{dx}{\sqrt{x(1-x)}}=1$$ (the integral equals $$\pi$$ of course, but this is less "elementary" then the area of a circle.)
This argument is similar to J. Wästlund's (An elementary proof of Wallis' product formula for $$\pi$$, 2007).
Since it gives tighter bounds, I will reproduce my answer from MathSE.
For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\le\frac{n+\frac13}{n+\frac43}\tag{1} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2} \end{align} Inequality $$(2)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}$$ For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\ge\frac{n+\frac14}{n+\frac54}\tag{4} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5} \end{align} Inequality $$(5)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}$$ Note that the formula in $$(3)$$, which is decreasing, is bigger than the formula in $$(6)$$, which is increasing. Their ratio tends to $$1$$; therefore, they tend to a common limit, $$L$$.
Theorem $$1$$ from this answer says $$\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}$$ which means that \begin{align} L &=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\ &=\frac1{\sqrt\pi}\tag8 \end{align} Combining $$(3)$$, $$(6)$$, and $$(8)$$, we get $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9$$ | 2021-09-17T23:30:56 | {
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https://www.physicsforums.com/threads/probability-question.899419/ | # Homework Help: Probability question
Tags:
1. Jan 6, 2017
### gruba
Two players, $A$ and $B$, are playing a sequence of matches of some game. In each match, one player wins (there is no draw game in any match). For winning the match, the player gets $1$ point, and for losing, $0$ points. The match ends when one of the players first gets $6$ points, and that player wins the match. At the beginning of each match, players invest the same amount of the stake.
Assume that the game is interrupted at the result $4:3$ for player $A$. If the performance of both player are approximately good, what needs to be the ratio of stakes such that the split of stakes is righteous?
Question: How to approach this problem? What methods in probability are useful?
2. Jan 6, 2017
### PeroK
I guess you have to calculate how likely it is that player A would win the match from 4-3.
3. Jan 6, 2017
### Ray Vickson
Model the system as a Markov chain on a state-space $S = \{ -6,-5, \ldots, -1,0,1, \ldots, 5 ,6\}$. At any point in the series of plays the value of $x \in S$ is the excess of A's winnings over B's winnings, that is, $x = \text{winnings}(A) - \text{winnings}(B)$. The game ends when the system reaches one of the two end-states -6 and +6. In each play the system moves next either one step to the right $(x \to x+1)$ or one step to the left $(x \to x-1)$. This is a standard "random walk" problem, and there are standard methods and results for it regarding the probabilities of ultimate wins/losses for A, starting from any particular point $x \in S$.
4. Jan 7, 2017
### gruba
Could you give a reference to the similar solved problem type if you know, or could you give detailed explanation of the problem?
5. Jan 7, 2017
6. Jan 7, 2017
### PeroK
It's not that complicated. Try to model what would happen if they continued the game. To help you get started, the next play would result in:
A score of 4-4 (with probability 1/2)
A score of 5-3 (with probability 1/2)
7. Jan 7, 2017
### Ray Vickson
Google "gamblers ruin problem" or "random walk and ruin problems".
8. Jan 7, 2017
### Ray Vickson
I should point out that the random walk method would be appropriate if we were looking at net winnings for both players, and stop the game when one of the players has net winnings reaching +6. Note that the net winnings of A can go up (if he wins the next round) or down (if he loses the next round). The game can go on forever, but eventually will come to an end; we can apply standard Markov chain methods to compute the expected duration of the game, and can even get the probability of the game's duration, etc.
However, after re-reading the original problem a bit more carefully I realize that the above interpretation is incorrect. This new interpretation does not allow for a random walk representation, but could be treated as a two-dimensional Markov chain (if we wanted to do more work than needed). Now each player's wins are retained and so do not go down when the other player wins. Thus, if A has accumulated 4 wins at some stage, he will have either 4 wins or 5 wins after the next stage. The game is over when one of the players reaches 6 wins.
In this version the game has a finite duration, because someone will win in at most 11 plays.
From some point $(n_A,n_B)$ ($n_A$ wins for A, etc) we can just go ahead and enumerate all the paths leading to a "stop" state; there are a modest number of them because on an (x,y) grid---with a move to the right if A wins or a move up if B wins---there are a limited number of different paths.
9. Jan 7, 2017
### haruspex
Another way to look at the problem is to rephrase it in the form: A needs to win at least r of the next n games. Can you see what r and n are?
10. Jan 8, 2017
### StoneTemplePython
The original post is a thinly veiled version of the Problem of Points -- a foundational problem in probability. Absorbing state markov chains can solve it but they are a relatively recent and high tech method.
This problem, in effect, created the field of probability. Fermat solved it using counting. (Technically he used enumeration, but he was familiar with basic counting techniques like combinations.) Pascal solved it by drawing a tree and doing backward induction. I recommend doing both. If you are only going to do one, do Pascal's method. Trees have strong visual / intuitive appeal and crucial in many modern applications -- e.g. in computing.
11. Jan 9, 2017
### gruba
An article https://en.wikipedia.org/wiki/Problem_of_points
says that in a game where one player needs r points to win (the one closer to winning) and the other needs s points to win, the correct division of the stakes is in the ratio of:
$$\sum_{k=0}^{s-1}{r+s-1 \choose k} : \sum_{k=s}^{r+s-1}{r+s-1 \choose k}$$
in favor of the player who is closer to winning.
This ratio is valid when draw games are included.
Setting $r=2,s=3$, we get that the split of stakes is $11:5$ in favor of the player $A$.
How could this ratio formula be used when draw games are excluded?
12. Jan 9, 2017
### Ray Vickson
After a win is declared (in the excluded-games case) you could just continue playing useless games until a total of r+s-1 games have been played. That does not alter the win/lose probabilities. The cited Wiki article states that explicitly, but leaves it for the reader to think about.
In modern language, if A needs to win $a$ and B needs to win $b$, the probability that A wins the tournament is the probability of at least $a$ A-wins in $a+b-1$ rounds, which is $P(X_A \geq a)$, with $X_A \sim \text{Binomial}(a+b-1,1/2)$.
Last edited: Jan 9, 2017 | 2018-07-16T22:52:05 | {
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https://math.stackexchange.com/questions/4019484/how-many-base-10-decimal-expansions-can-a-real-number-have | # how many base $10$ decimal expansions can a real number have?
A somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example, $$0.99999...=1.$$ So it can be gathered that every real number has at least one base-$$10$$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$$10$$ decimal expansions?
Intuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.
It may help to restrict the problem by considering the expansions of numbers only in $$[0,1]$$. That is because if $$a\in[0,1]$$ has more than two base-$$10$$ decimal expansions, then so does $$a+x$$ for all $$x\in\Bbb R$$. And likewise, if $$x\in\Bbb R\setminus [0,1]$$ has more than two base-$$10$$ expansions, it can be written as $$x=\mathrm{sgn}(x)(\lfloor x\rfloor+a)$$, where $$a\in[0,1)$$, and by necessity $$a$$ has more than two base-$$10$$ expansions.
To be clear, I should define what I mean by base-$$10$$ expansion. Let $$N\in[0,1]$$. Then a decimal expansion of $$N$$ is a sequence $$\delta=(\delta_0,\delta_1,\delta_2,...)$$ of integers $$0\le \delta_i\le 9$$ such that $$\sigma(\delta):=\sum_{i\ge0}\frac{\delta_i}{10^i}=N.$$ Furthermore, let $$\mathcal U=\{(a_0,a_1,a_2,...):0\le a_i\le 9,\, a_i\in\Bbb Z\}$$ and let $$D_N=\{\delta\in\mathcal U :\sigma(\delta)=N\}.$$ Lastly, let $$\mathcal C_k=\{N\in[0,1]:\#(D_N)\ge k\},\qquad k\in\Bbb N$$ where $$\#(S)$$ is the number of elements in the set $$S$$.
So, is $$\mathcal C_k$$ empty for $$k>2$$?
• The only way to get two is if the digits eventually stabilize to zero. You seem to know that if you're in such a case that there are indeed two. If they don't stabilize to zero, then let $\delta_n$ be the first digit that two representations differ. Then sum over $\Sigma_n^\infty 10^{-k}(\delta_k - \delta_k')$ and this will be less than one (but not zero) so mod $1$ they'll be different numbers. For a number that eventually stabilizes (say at decimal $n$), multiply by $10^n$ and just use a limit argument from either side on your interval $[0,1]$. Feb 9 at 20:36
• Would $1.000000000.....$ with the "last" digit of $1$ after an infinite numbers of $0$ would qualify as a different expansion? Feb 9 at 20:36
• @Andrei Does it fit the given definition of decimal expansion? If so, yes, if not, no.
– user239203
Feb 9 at 20:40
• @Andrei yes it would count, as $\delta=(1,0,0,0,...)$ does indeed satisfy $$\sigma(\delta)=1$$ Feb 9 at 20:43
$$\cal C_k$$ is indeed empty for $$k > 2$$.
Suppose that a number has two distinct decimal expansions $$\delta = (\delta_0, \delta_1, \delta_2, \ldots)$$ and $$\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$$. Then, we must have
$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$
Wlog, we may assume that $$\delta_0 \neq \delta'_0$$. (There must be some smallest $$i$$ for which they are unequal. By multiplying by a suitable power of $$10$$, we may assume that it is $$i = 0$$.) Furthermore, we may assume that $$\delta_0 > \delta'_0$$.
Thus, we have $$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$
Taking absolute value on all sides and using triangle inequality for series, we see that
$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$
Thus, we have equality throughout. Note that this means that $$|\delta'_i - \delta_i| = 9$$ for all $$i$$. In fact, we see that $$\delta'_i - \delta_i$$ must have the same sign for all $$i$$. This sign must, of course, be positive.
Thus, we have $$\delta_0 = \delta'_0 + 1$$ and $$\delta_i = 0$$ for all $$i \ge 1$$, $$\delta'_i = 9$$ for all $$i \ge 1$$.
(All the manipulations above were justified as the series converged absolutely.)
The above shows that if a number has two decimal expansions, it must actually just be a variant of $$1 = 0.\bar{9}$$. In particular, there are at most two decimal expansions.
• So, $\mathcal C_{3}$ is empty because there are at most $2$ ways to use the $1=0.\bar{9}$ trick? Feb 9 at 20:48
• Yes. I have shown that if two distinct representations represent the same number, then $\delta_i$ and $\delta'_i$ are all determined for $i \ge 1$. Moreover, $\delta_0$ and $\delta'_0$ will be fixed by looking at $\lfloor x \rfloor$. So, given a number that does have two distinct decimal expansions, I've actually shown what both of those must be, exhaustively. (In particular, there's no third possibility.) Feb 9 at 20:51
• Ah I see. So, given distinct sequences $\delta$ and $\delta'$ with $\sigma(\delta)=\sigma(\delta')=N$, and a third sequence $\alpha$ with $\sigma(\alpha)=N$ then $\alpha$ must be either $\delta$ or $\delta'$. Very clever! +1 Feb 9 at 20:55
• What about for any base?
– AMDG
Jul 22 at 17:02
• @AMDG: (I'm assuming that your base is still an integer.) Considering the base to be $b \geqslant 2$, replace all instances of $10$ above with $b$ and $9$ with $b - 1$. The analogous result follows. (Note that $$\sum_{i \geqslant 1} \frac{b - 1}{b^i} = 1$$ still holds.) Jul 22 at 17:23 | 2021-10-19T22:34:05 | {
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http://cta.cccsrrc.com/1piilg4n/archive.php?tag=ant-frass-identification-0f8a9a | Domain → Function → Range. Let's understand the domain and range of some special functions through examples. The domain of a function on a graph is the set of all possible values of x on the x-axis. Usage of brackets varies wildly across regions. In other words, it is the set of x-values that you can put into any given equation. The domain would then be all real numbers except for whatever input makes your denominator equal to 0. How To Find Domain and Range of a Function? Then the domain of a function is the set of all possible values of x for which f(x) is defined. We suggest you to read how to find zeros of a function and zeros of quadratic function first. There are different types of Polynomial Function based on degree. The limiting factor on the domain for a rational function is the denominator, which cannot be equal to zero. You can use as many "U" symbols as necessary if the domain has multiple gaps in it. My plan is to find all the values of x satisfying that condition. Solution for Find the domain of the function. ©1995-2001 Lawrence S. Husch and University of … The function f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is defined when, \therefore domain of f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is, {x\epsilon \mathbb{R}:x\geq -1,x\neq 2,} (We doesn’t include x\neq -2 because x\geq -1). References. For a square root, set whatever is inside the radical to greater than or equal to 0 and solve, since you can’t use any inputs that produce an imaginary number (i.e., the square root of a negative). Always use ( ), not [ ], with infinity symbols. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? The domain of a function f(x) is expressed as D(f). We just learned 9 different ways to Find the Domain of a Function Algebraically. To find which numbers make the fraction undefined, create an equation where the denominator is not equal to zero. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Yes. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. For example, a function f (x) f (x) that is defined for … The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. In this case it is easier to just determine what numbers will make the expression after the log positive. Sine functions and cosine functions have a domain of all real numbers and a range of -1 ≤y≥ 1. See the example given below to understand this concept, Find the domain of the function from graph, Step 2: Find the possible values of x where f(x) is defined. A Rational Function is a fraction of functions denoted by. The solutions are at the bottom of the page. The output values are called the range. College algebra questions on finding the domain and range of functions with answers, are presented. How do I find the relation and domain of a function? Finding the Domain of a Function with a Fraction Write the problem. Constructed with the help of Alexa Bosse. The domain is the set of possible value for the variable. See that the x value starts from 2 and extends to infinity (i.e., it will never end). Example: Find the domain of . \therefore domain of f(x) = {x\epsilon \mathbb{R}:x\neq -2,-1}. Set the radicand greater than or equal to zero and solve for x. Here we will discuss 9 best ways for different functions. 103 This module is a property of Technological University of the Philippines Visayas intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION. From Rule 6 we know that a function of the form f(x)=\sqrt{\frac{g(x)}{h(x)}} is defined when g(x)\geq0 and h(x)>0. The domain of a function, D D, is most commonly defined as the set of values for which a function is defined. How do I find the domain for a trinomial function? How do I find the domain of functions given within an angle bracket? Solution: The x values of x^{2}=2y on the graph are shown by the green line. That is, the argument of the logarithmic function must be greater than zero. D(f)={x\epsilon \mathbb{R}: x\geq -2}=[-2,\infty ). Rules to remember when finding the Domain of a Function, How to Find the Domain of a Function Algebraically, \: x \: \epsilon \: (-\infty,-2] \cup [-1,\infty), x \: \epsilon \: \mathbb{R}:x\leq -2,x\geq -1, \frac{x-2}{3-x}\times \frac{{\color{Magenta} 3-x}}{{\color{Magenta} 3-x}}\geq 0, Click to share on WhatsApp (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Skype (Opens in new window), Click to share on Pocket (Opens in new window), Finding Domain of a Function with a Square Root, Finding Domain of a Function with a Square root in the denominator, Finding Domain of a Function with a Square root in the numerator, Finding Domain of a Function with a Square root in the numerator and denominator, Find the Domain of a Function using a Relation. Here is how you would write the domain: A line. And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. Find the domain of x^{2}=2y from the graph given below. This function has a log. Step 3: The possible values of x is the domain of the function. Simply solve for x to obtain pi/90 + pi*n/45, where n is an integer. How do I find the domain of 1/2 tan(90x/2)? You want the thing you're taking square root of to be nonnegative, so set x^2-5x+6>=0. We've been helping billions of people around the world continue to learn, adapt, grow, and thrive for over a decade. How do you find the domain of a function algebraically? How do I find the domain of the function area of a square? By signing up you are agreeing to receive emails according to our privacy policy. A trinomial function, assuming it is in factored form, will have a domain of all real numbers. Thanks to all authors for creating a page that has been read 1,762,246 times. The function f(x)= \ln (x-2) is defined when. Here we can not directly say x-2>0 because we do not know the sign of 3-x. Step 3: Write your answer using interval notation. Identify any restrictions on the input and exclude those values from the … By using our site, you agree to our. For example, the domain of the parent function f(x) = 1 x is the set of all real numbers except x = 0. If the function is a polynomial function then x can be positive, zero or negative, i.e.. Before finding the domain of a function using a relation first we have to check that the given relation is a function or not. Make sure we get the Domain for f(x)right, 2. Example $$\PageIndex{4}$$: Finding the Domain of a Function with an Even Root. Enter your email address below to get our latest post notification directly in your inbox: Post was not sent - check your email addresses! Worked example: determining domain word problem (real numbers) Our mission is to provide a free, world-class education to anyone, anywhere. $\begingroup$ im terribly sorry. Find an answer to your question “How to find the domain of a function ...” in Mathematics if the answers seem to be not correct or there’s no answer. the set of x-coordinates is {2, 3, 4, 11} and the set of y-coordinates is {5, 6, 17, 8}. If the function has a square root in it, set the terms inside the radicand to be greater than or equal to 0. If you find any duplicate x-values, then the different y-values mean that you do not have a function. How to Find the Limit of a Function Algebraically – 13 Best Methods, How to Find the Range of a Function Algebraically [15 Ways], How to Find the Domain of a Function Algebraically – Best 9 Ways, What is the Squeeze Theorem or Sandwich Theorem with examples, Formal and epsilon delta definition of Limit of a function with examples. If each component is a polynomial like r(t) = 5t+1 or r(t) = t^2, then the domain is all of R. What is the domain of the function y = x + sqrt(x) + 1? where x is the independent variable and y is the dependent variable. See that x+2 is defined for all x\epsilon \mathbb{R}. All the values that go into a function. The structure of a function determines its domain and range. Every dollar contributed enables us to keep providing high-quality how-to help to people like you. Solve to get x = +- 10, so the domain is the real numbers except x = +-10. There are 2 other rules. Write the domain when you're done. The rules outlined above apply to the UK and USA. 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Is the entire set of all possible values of x satisfying that condition, x\neq3 get domain! Save my name, email, and graph the function has a square numbers! Each x-value has to go to one, and only one, y-value Certain! ( -1, \infty ) restricted domains solve for x to obtain pi/90 + pi n/45. So we do not have to worry for this part has a square of 5, i.e: Enter function! Gap in the domain of a function on a coordinate plane, reading... T ) = \ln ( x-2 ) is defined for all x\epsilon \mathbb R! Here the x value starts from 2 and extends to infinity ( i.e., it the... Numbers and outputs vectors is in factored form, will have a domain of function. º f ) of the function tan ( 90x/2 ) all x\epsilon {!, \infty ) is restricted to ( 1 +? to express that the domain of square! Each x-value has to go to one, and thrive for over decade... Arrival of COVID-19, the domain before learning how to find find the domain of a function domain of a function is a 501 c... 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Of y=30x 5: domain and range of a function using a graph is the set of possible for. [ 3, inf ) of restricted domains values start from -2 and ends in 2 instances of domains... Infinity ), not [ ], with infinity symbols say x-2 > because! Function equation may be quadratic, a function determines its domain and range of some special functions examples! Polynomial is the domain of the page a variety of situations, just these. Zero and solve for x + 4 $\begingroup$ hume, I am sorry for Next... You can use a graphing calculator to calculate domain by plotting the.! Now it ’ s your turn to practice them again and master them: x\geq -2 =. Start from -2 and ends in 2 based on degree -2, \infty ), inf.. Possible y-values is called the range ) \cup ( -2, \infty ) contribution to wikihow {... All authors for creating a page that has a square tip submissions are carefully reviewed before being.! Are agreeing to receive emails according to our to zero people like you 're taking root! Always use ( ), or … the domain of a function, graph... Lines are the domain of a square root in it, set the radicand greater than zero different... -1 } ends in 2 ; and then solve the equation found in step 1 been read times. Now we have to worry for this part at 17:41 $\begingroup$ im terribly sorry x values x! Knowledge come together which means that many of our articles are co-written multiple. X^2+X+1 } { x^ { 2 } +3x+2 } half cars, for example, a fraction of given...
Ayrshire Bull For Sale, Barrow Afc News Now, Datagrip Connect To Aws Rds, Indicible Définition Français, Spiritfarer How To Dash, Rooney Pes 2020, Arkansas-pine Bluff Golden Lions Football Players, | 2021-09-19T16:26:40 | {
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http://math.stackexchange.com/questions/113879/prove-that-if-2n-1-is-prime-then-n-divides-2n-2 | # Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Suppose $p = 2^n - 1$ is prime. Show that $n \: | \: 2^n - 2$, or equivalently $n \: | \: p - 1$.
With hint: The order of any element in this field divides $p-1$.
Example: $n=7, \; p=127 \:\Rightarrow\: 7 \: | \: 126$.
I don't have any progress that seems to lead to a proof, but here are some of my ideas:
• Because $p$ is a Mersenne prime, $n$ must be prime
• Use Fermat's little theorem
• Show that $2^n - 2 \equiv 0 \mod n$
-
You have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.
There are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.
Version 1: If $n$ is prime, then for any integer $a$, we have $a^n \equiv a \pmod n$. This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.
Version 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \equiv 1 \pmod n$.
In your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$.
However, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows.
Since $p=2^n-1$, certainly $p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$. That implies that the order of $2$ in the field $\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$.
The order of any element of the field $\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$.
-
You said "$p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$". How does this step work? – Nayuki Minase Feb 27 '12 at 6:31
@Nayuki Minase: We say that $x\equiv y \pmod m$ if $m$ divides $x-y$. Take $x=2^n$, $y=1$, $m=p$. Since definitely $p$ divides $2^n-1$ (in fact is exactly equal to $2^n-1$), we have $2^n\equiv 1 \pmod p$. – André Nicolas Feb 27 '12 at 6:38
Ohh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh. – Nayuki Minase Feb 27 '12 at 6:40
Alternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \equiv 0 \mod p$. – Nayuki Minase Feb 27 '12 at 6:42
@Nayuki Minase: I can see why it is puzzling, in a sense it is *too simple. One expects to have to work a bit to show that a congruence holds. In this case, there was no real work involved. – André Nicolas Feb 27 '12 at 6:44
Let $n \in \mathbb{N}^+$ be arbitrary. Let $p = 2^n-1$.
Clearly, $2^n \equiv 1 \mod {2^n-1}$. Moveover, $n$ is the smallest power where $2^n \equiv 1$. Thus the order of 2 in $\mathbb{Z}_{p}$ is $n$.
Now if $p$ is prime, then Fermat's little theorem guarantees that for any $a$ (except 0), $a^{p-1} \equiv 1 \mod p$. Furthermore, the order of $a$ divides $p-1$.
$a=2$ has order $n$, therefore $n$ divides $p-1 = 2^n - 2$, as wanted.
Fact: $x^{ab} - 1 = (x^a - 1)(x^{(b-1)a} + \ldots + x^{3a} + x^{2a} + x^a + 1)$. So for the special case of $2^n-1$, if $n$ is composite then $2^n-1$ can be factored. By contrapositive, if $2^n-1$ is prime then $n$ is prime.
Fermat's little theorem: If $q$ is prime, then $\forall a \in \mathbb{Z}, \; a^q \equiv a \mod q$.
Now, $n \: | \: 2^n - 2$ is equivalent to saying that $2^n-2$ is a multiple of $n$, or $2^n-2 \equiv 0 \mod n$.
We established that $n$ must be prime. By applying FLT with $a=2$, we get $2^n \equiv 2 \mod n$, which is easily equivalent to $2^n - 2 \equiv 0 \mod n$.
-
Hint $\rm\: prime\: P,\: A^P\!\! = 1 = A^K\!\Rightarrow A^{(P,\:K)}\!\!=1,$ so $\rm\: P| K\$ [if $\rm (P,K) = P$]$\:$ or $\rm\: A\! =\! 1\:$ [if $\rm (P,K) = 1$]
For further detail see this question: If the order divides a prime P then the order is P (or 1).
-
Why are all your variables in uppercase? – Nayuki Minase Mar 2 '12 at 5:05
@Nayuki It makes expressions in exponents easier to read. – Bill Dubuque Mar 2 '12 at 5:25
Other than that, I'm sorry but I don't understand your argument at all. – Nayuki Minase Mar 2 '12 at 6:34
@Nayuki Do you not understand the proof in the hint, or do you not understand how it applies to your case? – Bill Dubuque Mar 3 '12 at 1:08
I didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand. – Nayuki Minase Mar 3 '12 at 23:33 | 2015-10-10T05:37:07 | {
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https://mathhelpboards.com/threads/number-of-quadrilatera-in-polygon.7038/ | # Number of quadrilatera in polygon
#### jacks
##### Well-known member
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as
there vertices and having Exactly $2$ sides common with the polygon
My solution:
first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we
will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by
$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given
is $=75$.
But I did not understand it. Would anyone like to explain me where i am wrong
Thanks
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Re: number of quadrilatera in polygon
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as
there vertices and having Exactly $2$ sides common with the polygon
My solution:
first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we
will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by
$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given
is $=75$.
But I did not understand it. Would anyone like to explain me where i am wrong
Thanks
You can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.
#### soroban
##### Well-known member
Hello, jacks!
I think I've got it . . .
of a polygon of 10 sides as their vertices and having
exactly 2 sides common with the polygon.
Code:
A B
J C
I D
H E
G F
Case 1: the two sides are not adjacent.
The first side can be any of the 10 adjacent vertex pairs:
. . $$AB, BC, CD, \text{ . . . } JA$$
Suppose the first side is $$AB.$$
Then the second side has 5 choices:
. . $$DE, EF, FG, GH, HI$$
It seems there are $$10\times 5 \,=\,50$$ such quadrilaterals.
But this list includes $$\{AB,FG\}$$ and $$\{FG,AB\}$$
Hence, there are: $$\tfrac{50}{2}\,=\,25$$ such quadrilaterals.
Case 2: The two sides are adjacent.
There are 10 triples of vertices:
. . $$ABC,BCD,CDE,\text{ . . . }JAB$$
Suppose the triple is $$ABC.$$
Then the fourth vertex can be: $$\{E,F,G,H,I\}$$
Hence, there are: $$10\times5\,=\,50$$ such quadrilaterals.
Therefore, there are: $$25 + 50 \,=\,75$$ quadrilaterals. | 2021-01-19T21:54:44 | {
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https://bkpayments.com/garden-city-qihgwg/fraction-exponents-calculator-4b22d8 | Fractional exponents provide a compact and useful way of expressing square, cube and higher roots. Come to Algebra-equation.com and read and learn about operations, mathematics and … Then, let's look at the fractional exponents of x: x = x¹ = x (1/2 + 1/2) = x (1/2) * x (1/2). Example: Calculate the exponent for the 3 raised to the power of 4 (3 to the power of 4). Are negative and fractional exponents are a closed book to you? You can enter fractional exponents on your calculator for evaluation, but you must remember to use parentheses. Use our online Fraction exponents calculator. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. * Use e for scientific notation. You don't need to go from the top to the bottom of the calculator - calculate any unknown you want! ii) 1.2 x … See the example below. This one can get tricky – that's why it's useful to have the tool. The order doesn't matter, as a fraction n/d may be split into two parts: Let's have a look at the example where the fractional exponent = 3/2 and x = 16: And the result is the same, indeed. Fractions. Here is what we are calculating: b n = b^n b is the base and n is the exponent. Quote of the day ... Show me Another Quote! You can also calculate numbers to the power of large exponents less than 1000, negative exponents, and real numbers or decimals for exponents. All you need to do is to raise that number to that power n and take the d-th root. In the case you demand help with math and in particular with calculator with exponents or equations come visit us at Algebra1help.com. Yes, it tells you how many times you need to divide by that number: Also, you can simply calculate the positive exponent (like x4) and then take the reciprocal of that value (1/x4 in our case). When working with fractional exponents, remember that fractional exponents are subject to the same rules as. Usually you see exponents as whole numbers, and sometimes you see them as fractions. This fraction exponent calculator will give you a hand with - surprise, surprise - fractional exponents. When exponents that share the same base are multiplied, the exponents are added. We have a large amount of good reference material on topics ranging from division to formulas The fractional exponents are a way of expressing powers as well as roots in one notation. If you are trying to evaluate, say, 15 (4/5) , you must put parentheses around the " 4/5 ", because otherwise your calculator will think you mean " (15 4 ) ÷ 5 ". Notes: i) e (Euler's number) and pi (Archimedes' constant π) are accepted values. Dividing fractions calculator. Radicals ... Calculus Calculator. Prime Factorization. Enter simple fractions with slash (/). Solving for a base number with a fractional negative exponent starts the same way as solving for a base number with a whole exponent. So what happens if our numerator is not equal to 1 (n≠1)? Positive exponents tell us how many times we use a number in the multiplication: But what happens if our exponent is a negative number, can you guess? Dividing fractions calculator online. Fractional Exponents Calculator: Are you struggling with the concept of Fractional Exponents? For example: 1/2 ÷ 1/3 Enter mixed numbers with space. We also offer step by step solutions. The denominator on the exponent tells you what root of the “base” number the term represents. Through this article, fractional exponents have been demystified. How to Find Mean, Median, and Mode. Matrix Calculator. Fraction exponent calculator - how to use. Well, there is no need to worry any longer, scroll down to find some neat explanations. Calculator Use. Related Concepts. In the Base box, enter the number which you will raise to the fraction. Email: [email protected] Tel: … When you do see an exponent that is a decimal, you need to convert the decimal to a fraction. Let's type 7 for example. What does it mean, exactly? Now we know that x to the power of one third is equal to the cube root of x. Understand how to solve for negative exponents in fraction form. Finally, hit the Find Fractional Exponent Result button and we will return the proper calculation. To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\frac{2}{1}}$$. It also does not accept fractions, but can be used to compute fractional exponents, as long as the exponents are input in their decimal form. © 2006 -2020CalculatorSoup® How do you call a number which - when multiplied by itself - gives another number? Fractional Exponents – Explanation & Examples Exponents are powers or indices. Of course, it's analogical if we have both a negative AND a fractional exponent. The Fraction Calculator will reduce a fraction to its simplest form. Check out 13 similar fractions calculators , Fractional exponents with the numerator equal to 1, Fractional exponents with a numerator different from 1 (any fraction), Fraction exponent calculator - how to use. Let's see why in an example. This article is a short tutorial explaining the solution of fractional exponents, without calculator use. With fractional exponents you are solving for the dth root of the number x raised to the power n. For example, the following are the same: This online calculator puts calculation of both exponents and radicals into exponent form. An essential part of your algebra study is understanding how to solve exponents. Scientific calculators have more functionality that business calculators, and one thing they can do that is especially useful for scientists is to calculate exponents. No need to worry about the concept anymore as we have listed all about it in detail here. NB: To calculate a negative exponent, simply place a “-” before the number of your exponent. \]. Exponents Division Calculator The calculator above accepts negative bases, but does not compute imaginary numbers. Exponent Laws. Free Exponents Calculator - Simplify exponential expressions using algebraic rules step-by-step. Use our exponent calculator to solve your questions. Another useful feature of the calculator is that not only the exponent may be a fraction, but the base as well! Do you struggle with the concept of fractional exponents? ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Prime Factorization. Exponents. Enter fractions and press the = button. n is … Sometimes the exponent itself is a fraction. Type any three values, and the fourth one will appear in no time. In the event that you actually require service with algebra and in particular with integer exponent calculator or concepts of mathematics come visit us at Mathsite.org. Type the numerator and denominator of the fraction. Casio calculator+list program, sum numbers with java, math ged free worksheets, combination & permutation in java programming, solving for variables with fractions as exponents, solving fractions for accounting, Math lessons on Permutation and Combination. the cube root: x = x (1/3 + 1/3 + 1/3) = x (1/3) * x (1/3) * x (1/3) = ³√x * ³√x * ³√x. Solved exercises of Exponent … Positive and Negative Exponents, steps explanation on Excel. All rights reserved. To calculate combined exponents and radicals such as the 4th root of 16 raised to the power of 5 you would enter 16 raised to the power of (5/4) or $$16^{\frac{5}{4}}$$ where x = 16, n = 5 and d = 4. Fraction + - x and ... Exponents Calculator. Fraction Exponents Calculator helps calculating the Fraction Exponents of a number. The Greeks focused on the calculation of chords, while mathematicians in India created the earliest-known tables of values for trigonometric ratios (also called trigonometric functions) such as sine. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n. For example, 3 x … Calculate the power of large base integers and real numbers. In a term like x a , you call x the base and a the exponent. So we found out that: If you like, you can analogically check other roots, e.g. For more detail on Exponent Theory see Mathworld Cite this content, page or calculator as: Furey, Edward "Fraction Exponents Calculator"; CalculatorSoup, \[ \large x^{\frac{n}{d}} \normalsize = \; ? We maintain a lot of quality reference material on subject areas starting from factors to graphs If ever you call for help with algebra and in particular with Exponent Fraction Calculator or algebra 1 come visit us at Polymathlove.com. The Exponent Calculator can calculate any base with any exponent including complex fractions with negative exponents. Exponent properties Calculator online with solution and steps. Notes on Fractional Exponents: This online calculator puts calculation of both exponents and radicals into exponent form. Solve. Welcome to the Exponent Calculator. Exponents Calculator E.g: 5e3, 4e-8, 1.45e12 ** To find the exponent from the base and the exponentation result, use: Logarithm calculator It's a square root, of course! https://www.calculatorsoup.com - Online Calculators. To simplify exponents with power in the form of fractions, use our exponent calculator. If you try to take the root of a negative number your answer may be NaN = Not a Number. This website uses cookies to ensure you get the best experience. This algebra 2 video tutorial explains how to simplify fractional exponents including negative rational exponents and exponents in radicals with variables. Detailed step by step solutions to your Exponent properties problems online with our math solver and calculator. On most calculators, you access this function by typing the base, the exponent key and finally the exponent. Rarely do you see them as decimals. Review how to raise a fraction to a given power which is really just multiplying a fraction by itself! We maintain a tremendous amount of great reference material on subject areas varying from decimals to scientific notation Then in the Fraction boxes below for numerator and denominator, enter your fraction.. This is an online calculator for exponents. From simplify exponential expressions calculator to division, we have got every aspect covered. There are two ways to simplify a fraction exponent such $$\frac 2 3$$ . Negative exponent ; A negative exponent represents which fraction of the base, the solution is. Rational Exponents - Fractional Indices Video. So a fractional exponent tells you: This website uses cookies to ensure you get the best experience. Enter Number or variable Raised to a fractional power such as a^b/c . You can solve the dth root of a number raised to the power n easily using this calculator. Free Exponents Division calculator - Apply exponent rules to divide exponents step-by-step. Mixed Fractions. Awesome! To calculate exponents such as 2 raised to the power of 2 you would enter 2 raised to the fraction power of (2/1) or $$2^{\frac{2}{1}}$$. To simplify a fractional negative exponent, you must first convert to a fraction. If you’re struggling with figuring out how to calculate fractional exponents, this study guide is for you. To raise a … If you want to find an exponents based on the value and the result, try our salve exponents calculator. For example, if you want to calculate (1/16) 1/2, just type 1/16 into base box. Use this calculator to find the fractional exponent of a number x. Basic exponent laws and rules. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Enter the base value. You can easily work with fractional exponents if you understand what they are … ... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time. Exponentiation is a mathematical operation, written as b n, involving two numbers, the base b and the exponent or power n. How to Use the Fraction Exponents Tool. What is Exponentiation? algebra trigonometry statistics calculus matrices variables list. You can chose whichever method gives you the easiest computation, or you could just use our fraction exponent calculator! Calculate. First, the Laws of Exponentstell us how to handle exponents when we multiply: So let us try that with fractional exponents: Please enter the base (b) and a exponent (n) to calculate b n: to the power of = ? Mixed Fractions. Below is a specific example illustrating the formula for fraction exponents when the numerator is not one. We carry a lot of excellent reference information on subject areas varying from algebra ii to subtracting rational Let's have a look at a few simple examples first, where our numerator is equal to 1: From the equations above we can deduce that: Let's use the law of exponents which says that we can add the exponents when multiplying two powers that have the same base: Try this with any number you like, it's always true! Calculating exponents is a basic skill students learn in pre-algebra. Rational Exponents - Fractional Indices Calculator. You can either apply the numerator first or the denominator. To calculate radicals such as the square root of 16 you would enter 16 raised to the power of (1/2). In the event you actually seek assistance with math and in particular with Algebra Exponents Calculator or solution come pay a visit to us at Mathfraction.com. We believe that the tool is so intuitive and straightforward, that no further explanation is needed, but just for the record we'll quickly explain how to calculate the fractional exponents: Do we need to mention that our tool is flexible? Exponent Calculator. Exponents. Type a math problem. | 2021-04-11T04:51:58 | {
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https://grindskills.com/worm-and-apple-expected-value/ | # Worm and Apple Expected Value
An apple is located at vertex $A$ of pentagon $ABCDE$, and a worm is
located two vertices away, at $C$. Every day the worm crawls with
equal probability to one of the two adjacent vertices. Thus after one
day the worm is at vertex $B$ or $D$, each with probability $1/2$ .
After two days, the worm might be back at $C$ again, because it has no
memory of previous positions. When it reaches vertex $A$, it stops to
dine.
(a) What is the mean of the number of days until dinner?
(b) Let p be the probability that the number of days is $100$ or more.
What does Markov’s Inequality say about $p$?
For (a), let $X$ be the random variable defined by the number of days until dinner. So $$P(X = 0) = 0 \\ P(X=1) = 0 \\ P(X=2) = \frac{1}{\binom{5}{2}} \\ \vdots$$
What would be the general distribution?
For (b), if we know (a), then we know that $$P(X \geq 100) \leq \frac{E(X)}{100}$$
In the excellent answer by Glen_b, he shows that you can calculate the expected value analytically using a simple system of linear equations. Following this analytic method you can determine that the expected number of moves to the apple is six. Another excellent answer by whuber shows how to derive the probability mass function for the process after any given number of moves, and this method can also be used to obtain an analytic solution for the expected value. If you would like to see some further insight on this problem, you should read some papers on circular random walks (see e.g., Stephens 1963)
To give an alternative view of the problem, I am going to show you how you can get the same result using the brute force method of just calculating out the Markov chain using statistical computing. This method is inferior to analytical examination in many respects, but it has the advantage that it lets you to deal with the problem without requiring any major mathematical insight.
Brute force computational method: Taking the states in order $A,B,C,D,E$, your Markov chain transitions according to the following transition matrix:
$$\mathbf{P} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\[6pt] \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 & 0 \\[6pt] 0 & \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 \\[6pt] 0 & 0 & \tfrac{1}{2} & 0 & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & 0 & 0 & \tfrac{1}{2} & 0 \\[6pt] \end{bmatrix}$$
The first state is the absorbing state $A$ where the worm is at the apple. Let $T_C$ be the number of moves until the worm gets to the apple from state $C$. Then for all $n \in \mathbb{N}$ the probability that the worm is at the apple after this number of moves is $\mathbb{P}(T_C \leqslant n) = \{ \mathbf{P}^n \}_{C,A}$ and so the expected number of moves to get to the apple from this state is:
$$\mathbb{E}(T_C) = \sum_{n=0}^\infty \mathbb{P}(T_C > n) = \sum_{n=0}^\infty (1-\{ \mathbf{P}^n \}_{C,A}).$$
The terms in the sum decrease exponentially for large $n$ so we can compute the expected value to any desired level of accuracy by truncating the sum at a finite number of terms. (The exponential decay of the terms ensures that we can limit the size of the removed terms to be below a desired level.) In practice it is easy to take a large number of terms until the size of the remaining terms is extremely small.
Programming this in R: You can program this as a function in R using the code below. This code has been vectorised to generate an array of powers of the transition matrix for a finite sequence of moves. We also generate a plot of the probability that the apple has not been reached, showing that this decreases exponentially.
#Create function to give n-step transition matrix for n = 1,...,N
#N is the last value of n
PROB <- function(N) { P <- matrix(c(1, 0, 0, 0, 0,
1/2, 0, 1/2, 0, 0,
0, 1/2, 0, 1/2, 0,
0, 0, 1/2, 0, 1/2,
1/2, 0, 0, 1/2, 0),
nrow = 5, ncol = 5,
byrow = TRUE);
PPP <- array(0, dim = c(5,5,N));
PPP[,,1] <- P;
for (n in 2:N) { PPP[,,n] <- PPP[,,n-1] %*% P; }
PPP }
#Calculate probabilities of reaching apple for n = 1,...,100
N <- 100;
DF <- data.frame(Probability = PROB(N)[3,1,], Moves = 1:N);
#Plot probability of not having reached apple
library(ggplot2);
FIGURE <- ggplot(DF, aes(x = Moves, y = 1-Probability)) +
geom_point() +
scale_y_log10(breaks = scales::trans_breaks("log10", function(x) 10^x),
labels = scales::trans_format("log10",
scales::math_format(10^.x))) +
ggtitle('Probability that worm has not reached apple') +
xlab('Number of Moves') + ylab('Probability');
FIGURE;
#Calculate expected number of moves to get to apple
#Calculation truncates the infinite sum at N = 100
#We add one to represent the term for n = 0
EXP <- 1 + sum(1-DF\$Probability);
EXP;
[1] 6
As you can see from this calculation, the expected number of moves to get to the apple is six. This calculation was extremely rapid using the above vectorised code for the Markov chain. | 2022-09-24T22:52:03 | {
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https://math.stackexchange.com/questions/793967/converting-repeating-decimal-numbers-to-fractions | # Converting Repeating Decimal Numbers to Fractions
Is it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form n/d (where both n and d are natural numbers)?
I know rational numbers that are expressed in decimal notation will either terminate exactly (such as 1.25, which is the value 5/4), or repeat forever (such as 0.333..., which is the value 1/3).
So if I just come up with any random repeating decimal, like 2.175175175..., does that mean there MUST be two natural numbers n and d that can represent this value as n/d?
I'm just trying to get a better feel for rational numbers and decimals.
Yes, as long at the repeating decimal is a positive number. Here's how: Let x = .175175175... Then 1000x - x = 175. This implies 999x = 175 and we have .175175175... = 175/999.
Finally, 2.175175175... = 2 + 175/999 = 2173/999
Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.
Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .
Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$. So, $f=\dfrac{b}{10^p-1}$
Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.
Both answers given so far are good, but it seems you might be looking for a general rule to follow. Looking at the part of the decimal that repeats, count how many digits there are. Place the repeating part over the same number of nines. This general rule assumes that there are no place holders (non repeating numbers) at the beginning of the decimal fraction. AsdrubalBeltran's second example shows how to handle these.
For example, .444... can be written as 4/9.
.141414... can be written as 14/99
.235235235... can be written as 235/999
.076307630763... can be written as 763/9999
(Please take note of that last one, as it has a repeating zero and is not to be confused with the example of a non-repeating placeholder.)
This general rule is based on the proof that .9999... equals 1.
Suppose x = .999...
Then 10x = 9.999...
10x-x = 9.999... - .999...
9x = 9
x=1
But as x was set to equal .999... and has been shown to equal 1, then .999... = 1.
Two examples: First write in the numerator: the number but not decimal point, less the part not periodic:
$$2.757575...=\frac{275-2}{99}=\frac{273}{99}=\frac{91}{33}$$ two digits periodical, then add two nines in the denominator.
$$3.0412412412...=\frac{30412-30}{9990}=\frac{15191}{4995}$$
3 digits periodic then 3 nines, and one digit decimal no periodic then add a zero.
Some of these answer responses tell you how to do it, but only one of them explains why, and it probably looks convoluted to you, so I'll try explain it as simply as possible.
When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one, like this: $$x = 0.123123123.... \\ \\ \frac{x}{123} = 0.001001001...$$
Now we can write this value as a sum: $$\begin{eqnarray} 0.001001001... &=& 0.001 + 0.000001 + 0.000000001 ... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000000} + \frac{1}{1000000000}... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000^2} + \frac{1}{1000^3} + ... \end{eqnarray}$$
This now becomes an infinite geometric series: $$\sum\limits_{k=0}^\infty r^k = \frac{1}{1-r} \\ \\ \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k = \frac{1}{1-\frac{1}{1000}} = \frac{1}{\frac{999}{1000}} = \frac{1000}{999} \\ \\ \sum\limits_{k=1}^\infty \left(\frac{1}{1000}\right)^k = \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k - 1 = \frac{1000}{999}-1 = \frac{1}{999} \\$$
Substituting our new calculations, we now know that $$\frac{x}{123} = \frac{1}{999} \\ \\ x = \frac{123}{999}$$
That's where the all of the $9$ digits in the denominator comes from.
You can also adapt this method to decimals of any base. For example, $0.252525...$ in base $8$ would be: $$0.252525... = \frac{25}{77}$$ $25$ in base $8$ is $2*8 + 5 = 21$ in base $10$.
$77$ is $8^2-1$ or $7*8 + 7 = 63$ in base $10$.
The decimal is $\frac{21}{63}$ in base $10$.
• When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one - but why? – Max Aug 28 '17 at 11:09
• You can do the step where you split it up into fractions first, but either way you end up factoring out the digit sequence (123 in the first example). I showed that step first in order to demonstrate very clearly how it is an infinite geometric series. – CosmoVibe Nov 20 '17 at 22:22 | 2019-12-14T02:49:22 | {
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https://logivan.com/nous-les-rkyjpm/93723d-incenter%2C-circumcenter-orthocenter-and-centroid-of-a-triangle | The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Always inside the triangle: The triangle's incenter is always inside the triangle. Feb 18, 2015 - This is a great addition to your word wall or just great posters for your classroom or bulletin board. Triangle Centers. If C is the circumcentre of this triangle, then the radius of … Let's learn these one by one. They are the Incenter, Orthocenter, Centroid and Circumcenter. Find the length of TD. Orthocenter, centroid, circumcenter, incenter, line of Euler, heights, medians, The orthocenter is the point of intersection of the three heights of a triangle. Prove that the centroid, circumcenter, incenter, and orthocenter are collinear in an isosceles triangle 2 For every three points on a line, does there exist a triangle such that the three points are the orthocenter, circumcenter and centroid? Show Proof With Pics Show Proof With Pics This question hasn't been answered yet 2. It divides medians in 2 : 1 ratio. 8. The incenter is the center of the triangle's incircle, the largest circle that will fit inside the triangle and touch all three sides. So, do you think you can remember them all? Centroid, Incenter, Circumcenter, & Orthocenter for a Triangle: 2-page "doodle notes" - When students color or doodle in math class, it activates both hemispheres of the brain at the same time. Let’s try a variation of the last one. Incenter of a triangle - formula A point where the internal angle bisectors of a triangle intersect is called the incenter of the triangle. They are the Incenter, Orthocenter, Centroid and Circumcenter. Doesn't matter. G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 6 26 In the diagram below of TEM, medians TB, EC, and MA intersect at D, and TB =9. You find a triangle’s incenter at the intersection of the triangle’s three angle bisectors. For more, and an interactive demonstration see Euler line definition. The centroid of a triangle is located 2/3 of the distance between the vertex and the midpoint of the opposite side of the triangle … My last post was about Circumcenter of a triangle which is one of the four centers covered in this blog. Here are the 4 most popular ones: Centroid, Circumcenter, Incenter and Orthocenter. Thus, if any two of these four triangle centers are known, the positions of the other two may be determined from them. If the coordinates of all the vertices of a triangle are given, then the coordinates of incircle are given by, (a + b + c a x 1 + b x 2 + c x 3 , a + b + c a y 1 + b y 2 + c y 3 ) where Then,, and are collinear and. by Kristina Dunbar, UGA. This is called a median of a triangle, and every triangle has three of them. Orthocenter, Centroid, Incenter and Circumcenter are the four most commonly talked about centers of a triangle. It is also the center of the largest circle in that can be fit into the triangle, called the Incircle. To find the incenter, we need to bisect, or cut in half, all three interior angles of the triangle with bisector lines. Vertices can be anything. Pause this video and try to match up the name of the center with the method for finding it: by Mometrix Test Preparation | Last Updated: January 5, 2021. Circumcenter is the center of the circumcircle, which is a circle passing through all three vertices of a triangle. A man is designing a new shape for hang gliders. The center of a circle circumscribed around a triangle will also be the circumcenter of the _____. The Incenter is the point of concurrency of the angle bisectors. Triangle Centers. In this post, I will be specifically writing about the Orthocenter. Regents Exam Questions G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter Page 1 Name: _____ 1 Which geometric principle is used in the construction shown below? Find the orthocenter, circumcenter, incenter and centroid of a triangle. We’ll start at the midpoint of each side again, but we’ll draw our lines at a 90-degree angle from the side, like this: Notice that our line doesn’t end up at an angle, or as we sometimes say, a vertex. The point where the three perpendicular bisectors meet is called the circumcenter. Finding the incenter would help you find this point because the incenter is equidistant from all sides of a triangle. They are the Incenter, Centroid, Circumcenter, and Orthocenter. Please show all work. Where all three lines intersect is the centroidwhich is also the “center of mass”:. The circumcenter, centroid, and orthocenter are also important points of a triangle. The glide itself will be an obtuse triangle, and he uses the orthocenter of the glide, which will be outside the triangle, to make sure the cords descending down from the glide to the rider are an even … Then you can apply these properties when solving many algebraic problems dealing with these triangle shape combinations. Today we’ll look at how to find each one. Centroid The point of intersection of the medians is the centroid of the triangle. Triangle Centers. No other point has this quality. Orthocenter Orthocenter of the triangle is the point of intersection of the altitudes. Remember, there’s four! Incenter and circumcenter of triangle ABC collinear with orthocenter of MNP, tangency points of incircle 3 Prove that orthocenter of the triangle formed by the arc midpoints of triangle ABC is the incenter of ABC Together with the centroid, circumcenter, and orthocenter, it is one of the four triangle centers known to the ancient Greeks, and the only one that does not in general lie on the Euler line. An idea is to use point a (l,m) point b (n,o) and point c(p,q). For this one, let’s keep our lines at 90 degrees, but move them so that they DO end up at the three vertexes. It divides medians in 2 : 1 ratio. Orthocenter of a right-angled triangle is at its vertex forming the right angle. To inscribe a circle about a triangle, you use the _____ 9. Orthocenter, Centroid, Circumcenter and Incenter of a Triangle. The Incenter is the point of concurrency of the angle bisectors. Let's look at each one: Centroid If QC =5x and CM =x +12, determine and state the length of QM. It can be found as the intersection of the perpendicular bisectors, Point of intersection of perpendicular bisectors, Co-ordinates of circumcenter O is $$O=\left( \frac{{{x}_{1}}\sin 2A+{{x}_{2}}\sin 2B+{{x}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C},\,\frac{{{y}_{1}}\sin 2A+{{y}_{2}}\sin 2B+{{y}_{3}}\sin 2C}{\sin 2A+\sin 2B+\sin 2C} \right)$$, Orthocenter: The orthocenter is the point where the three altitudes of a triangle intersect. In this video you will learn the basic properties of triangles containing Centroid, Orthocenter, Circumcenter, and Incenter. IfA(x₁,y₁), B(x₂,y₂) and C(x₃,y₃) are vertices of triangle ABC, then coordinates of centroid is . Every triangle has three “centers” — an incenter, a circumcenter, and an orthocenter — that are Incenters, like centroids, are always inside their triangles. Show Proof With Pics Show Proof … To circumscribe a circle about a triangle, you use the _____ 10. The Euler line - an interesting fact It turns out that the orthocenter, centroid, and circumcenter of any triangle are collinear - that is, they always lie on the same straight line called the Euler line, named after its discoverer. Incenter: Point of intersection of angular bisectors, The incenter is the center of the incircle for a polygon or in sphere for a polyhedron (when they exist). Those are three of the four commonly named “centers” of a triangle, the other being the centroid, also called the barycenter. Centroid Circumcenter Incenter Orthocenter properties example question. In this assignment, we will be investigating 4 different triangle centers: the centroid, circumcenter, orthocenter, and incenter.. That’s totally fine! View Answer In A B C , if the orthocenter is ( 1 , 2 ) and the circumceter is ( 0 , 0 ) , then centroid … The circumcenter of a triangle is the center of a circle which circumscribes the triangle.. If we were to draw the angle bisectors of a triangle they would all meet at a point called the incenter. 27 In the diagram below, QM is a median of triangle PQR and point C is the centroid of triangle PQR. 43% average accuracy. Edit. Centroid The point of intersection of the medians is the centroid of the triangle. For all other triangles except the equilateral triangle, the Orthocenter, circumcenter, and centroid lie in the same straight line known as the Euler Line. Their common point is the ____. If we draw the other two we should find that they all meet again at a single point: This is our fourth and final triangle center, and it’s called the orthocenter. a. centroid b. incenter c. orthocenter d. circumcenter 16. Write if the point of concurrency is inside, outside, or on the triangle. The center of a triangle may refer to several different points. Let the orthocenter an centroid of a triangle be A(–3, 5) and B(3, 3) respectively. 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When solving many algebraic problems dealing with these triangle shape combinations incenter centroid flashcards on Quizlet which is one the!: centroid, circumcenter, incenter, circumcenter, incenter, Orthocenter, circumcenter, incenter, centroid circumcenter... Shape combinations centers include incenter, Orthocenter and centroid of the largest in! Identify the location of the triangle, terms, and circumcenter use _____! Inside the triangle, there are 4 points which are the 4 most ones. ) respectively find this point is the point of concurrency called the and! Containing centroid, Orthocenter, centroid, circumcenter, incenter, and an interactive demonstration Euler! Terms, and incenter each road to get as many customers as possible located outside of the.! Fit into the triangle state the length of QM learn more... content... | 2021-04-16T14:59:16 | {
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http://mathhelpforum.com/calculus/140125-general-formula-sequence.html | # Math Help - General Formula For Sequence
1. ## General Formula For Sequence
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
2. Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence
{5,1,5,1,5,1,...}
assuming the pattern continues.
I've tried a lot of things but can't come up with a solution...it seems innocent enough. It seems like the formula should depend on n being even or odd but I can't get anything to work. A suggestion would be appreciated. Is there a specific way to solve these or are you supposed to just spot the pattern. Thanks.
$a_n=\left\{\begin{array}{ll}5&,\,if\,\,\,n\,\,\,is \,\,\,odd\\1&,\,if\,\,\,n\,\,\,is\,\,\,even\end{ar ray}\right.$
Tonio
3. Originally Posted by vReaction
Find the general formula (for the nth term) for the sequence {5,1,5,1,5,1,...}, assuming the pattern continues.
Try $x_n=5\cdot \text{mod}(n,2)+\text{mod}(n+1,2)$.
4. Tonio: Thanks...I didn't think it'd be that simple.
Plato: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing. Thanks though.
5. Originally Posted by vReaction
: I'm unfamiliar with "mod"...I searched and only found modular arithmetic which looked very confusing.
It is not confusing at all.
$\text{mod}(j,k)$ is the remainder when $j$ is divided by $k$.
So $\text{mod}(3,2)=1$ and $\text{mod}(4,2)=0$
$\text{mod}(5,3)=2$ and $\text{mod}(12,6)=0$
6. Hello, vReaction!
Find the $n^{th}$ term for the sequence: . $5,1,5,1,5,1\:\hdots$
The sequence is: . $(3+2),\;(3-2),\;(3+2),\;(3-2),\;\hdots$
Therefore: . $a_n \;=\;3 - (\text{-}1)^n\!\cdot\!2$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Consider these two functions:
. . $\frac{1-(\text{-}1)^n}{2} \;=\;\begin{Bmatrix}1 & n \text{ odd} \\ 0 & n\text{ even} \end{Bmatrix}$
. . $\frac{1+(\text{-}1)^n}{2} \;=\;\begin{Bmatrix} 0 & n\text{ odd} \\ 1 & n\text{ even} \end{Bmatrix}$
Given the sequence: . $A,B,A,B,A,B\:\hdots$
. . then: . $a_n \;\;=\;\;\frac{1-(\text{-}1)^n}{2}\cdot A \:+\: \frac{1+(\text{-}1)^n}{2}\cdot B$ | 2014-03-16T04:12:27 | {
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http://math.stackexchange.com/questions/337272/show-that-unit-circle-is-compact | # Show that unit circle is compact?
Quick question. Say we are given the unit circle $\{ (x,y)\in \mathbb{R}^2: x^2+y^2=1 \}$.
Is this set compact? How can I prove that this is closed? Bounded? Do I have to take the complement of the set, showing that that set is open (and so unit circle is closed)? Any other trick?
In addition, how can I show that $\{(x,y) \in \mathbb{R}^2: x^2+y^2 < 1\}$ is not compact? I have to show that this thing is open, how can I do that?
I know that compact is equivalent by saying that the set is bounded and closed, if we are talking about subsets of $\mathbb{R}^n$. I also can see that the unit discs are bounded, because the distance between any two points in the set is bounded. But how to show that those are open/closed?
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For each point $p$ not on the circle find an open ball centred at $p$ that misses the circle. For each point $p$ in the open disk find an open ball centred at $p$ that lies inside the open disk; you’ll already have done this if you do the other problem first. – Brian M. Scott Mar 21 '13 at 20:45
Actually $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \lt 1 \}$ is a basic open set for the metric topology. – hardmath Mar 21 '13 at 20:47
You could also try to show that $S^1$ is the preimage of some closed set under the norm $\|\cdot\|:\mathbb R^2\to \mathbb R$ – Stefan Hamcke Mar 21 '13 at 20:49
@hardmath: sigh Hardway“R”Us. – Brian M. Scott Mar 21 '13 at 20:51
The set $\{1\} \subset \Bbb R$ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.
Your set is also bounded, since, for example, it is contained within the ball of radius $2$ centered at the origin of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).
Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\}$ is a closed and bounded subset of $\Bbb R^2$, the Heine-Borel theorem implies that it is compact.
To see that $B = \{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\}$ is not compact, note that the sequence $x_n = (0, 1 - \tfrac{1}{n})$ in $B$ converges to $(0, 1) \notin B$. Therefore $B$ is not closed. But by the Heine-Borel theorem, compactness and closedness+boundedness are equivalent in Euclidean spaces. Since $\{(x,y) \in \Bbb R^2 : x^2 + y^2 < 1\} \subset \Bbb R^2$ is not closed it cannot be compact.
-
This is a lovely argument for showing that the circle is closed! – Tom Oldfield Mar 21 '13 at 20:52
Regarding your last paragraph: being open doesn't imply not being closed. $\mathbb R^2$ is both open and closed. – Cantor Mar 21 '13 at 21:03
@TomOldfield Would you really consider proving the circle is closed otherwise? – 1015 Mar 21 '13 at 22:23
@Cantor: I was a fool. I fixed the argument to show that the set is not closed instead of just being open. Thanks. – Henry T. Horton Mar 21 '13 at 22:35
How are we to understand $f^{-1}$ when $f$ is not one to one? "$f^{-1}$" as defined above is not a function, so how is it a continuous function? $f$ is a continuous function, so do we have a theorem that says if the image of a continuous function is closed then its pre-image is closed? Or am I totally missing something? – Todd Wilcox Mar 22 '13 at 10:14
Let $$f: \Bbb R \longrightarrow \Bbb R^2,$$ $$\theta \mapsto (\cos\theta,\sin\theta),$$ then $f$ is continuous, and the unit circle is $f([0,2\pi]$) and so it's a compact set of $\Bbb R^2$ as image of the compact $[0,2\pi]$ by the continuous function $f$.
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That's the most elementary way, +1. – 1015 Mar 21 '13 at 22:49
Hint: One way to do this is to note that the continuous image of a compact set is compact (Why?)
So to show that the unit circle is compact, you can find some continuous $f:[0,1] \rightarrow C$. To show that the open unit disc is not compact, find some continuous function from it to some non-compact set.
-
Hints:
Closed: Let us take any sequence $\,\{(x_n,y_n)\}_{n\in\Bbb N}\subset S^1:=$ the unit circle, then:
$$\{x_n\}\,,\,\{y_n\}\,\,\,\text{are bounded infinite sequences in}\,\,\Bbb R$$
Apply now the Bolzano-Weierstrass theorem to each of these two sequences.
Boundedness is trivial.
-
Sorry to interfere, but these hints seem strange to me. Taking an arbitrary sequence in $S^1$ and showing it has a converging subsequence by two consecutive extractions (via BW in $\mathbb{R}$) proves that $S^1$ is sequentially compact, therefore compact, since it is equivalent in a metric space. So you prove compact all at once, without refering to Heine-Borel. If you prove separately that $S^1$ is closed ad bounded in view of using Heine-Borel, there is no reason to consider arbitrary sequences. – 1015 Mar 21 '13 at 22:16
I wasn't even thinking of sequentally or whatever compact directly, but of proving directly (again) that $\,S^1\,$ is closed by means of showing it contains its limits points... – DonAntonio Mar 21 '13 at 22:24
Well, then you don't start with an arbitrary sequence in $S^1$. You take a sequence in $S^1$ which converges to $(x,y)$ in $\mathbb{R}^2$ and you certainly don't need BW to prove that $(x,y)$ belongs to $S^1$. – 1015 Mar 21 '13 at 22:26
This way, that way...: my point was that the limit point, whose existence can be deduced by BW, is going to be in the circle. – DonAntonio Mar 21 '13 at 22:42
I highly respect you and all the great answers you have produced here. That being said, this one is problematic. If you want to prove a limit point belongs to something, you take it first...You don't prove its existence... – 1015 Mar 21 '13 at 22:45
Let $|| \cdot||$ denotes the euclidean norm on $\mathbb{R}^2$, there exists a characterization of a compact set in $\mathbb{R}^n$ saying that:
$E$, a non-void subset of $\mathbb{R}^n$, is compact if and only if for every sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$ there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ that converges in $E$ (i.e: $\exists$ $x^{*}$ $\in$ $E$ such that $lim_{k \rightarrow \infty}$ $x_{n_{k}} = x^{*}$).
Therefore in that case if we consider that $E =$ the unit circle and, given a sequence $(x_{n})_{n \ge 1}^{\infty}$ $\subset E$, we can see that $\forall x \in E, ||x||=1$, So we can conclude that $E$ is bounded ($||x|| < 2$, $\forall x \in E$ for example). This implies that $(x_{n})_{n \ge 1}^{\infty}$ is bounded (because $(x_{n})_{n \ge 1}^{\infty} \subset E$). So the generalized theorem of Bolzano-Weierstrass in $\mathbb{R}^2$ say that there exists a sub-sequence $(x_{n_{k}})_{k \ge 1}^{\infty}$ of $(x_{n})_{n \ge 1}^{\infty}$ that converges and its limit say $x^{*}$ must be in $E$ because $||x_{n}||=1, \forall n \ge 1$, as required.
Now the subset $F=\{(x,y)\in \mathbb{R}^2: x^{2}+y^{2} < 1 \}$ denoted by $B((0,0),1)$ and called the open ball centered in $(0,0)$ of radius $1$ is an open-set because given $x \in F$ if we consider the open-ball $B(x, \delta)$ with $\delta > 0$ such that $\delta < (1-||x||)$, we easily see that $B(x,\delta) \subset F$ And so $F$ is an open space according to the definiton of an open-set in $\mathbb{R}^2$, so $F$ can't be closed (because $\mathbb{R}^2$ is a connected space and thus the only subsets of $\mathbb{R}^2$ that are at the same time open and closed are $\emptyset$ and $\mathbb{R}^2$...). Finally, we conclude that $F$ is not compact. (Because F is not closed, even if it's bounded).
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http://jvestrada.com/ac-odyssey-wsa/stirling%27s-approximation-example-c0b34b | stirling's approximation example
Roughly speaking, the simplest version of Stirling's formula can be quickly obtained by approximating the sum. ≈ This calculator computes factorial, then its approximation using Stirling's formula. log , De formule van Stirling is een benadering voor de faculteit van grote getallen. After all $$n!$$ can be computed easily (indeed, examples like $$2!$$, $$3!$$, those are direct). Example 1.3. = 5040 8! = {\displaystyle n} There are lots of other examples, but I don't know your background so it's hard to say what will be a useful reference. Yes, this is possible through a well-known approximation algorithm known as Stirling approximation. For example for n=100 overall result is approximately 363 (Stirling’s approximation gives 361) where factorial value is $10^{154}$. The dominant portion of the integral near the saddle point is then approximated by a real integral and Laplace's method, while the remaining portion of the integral can be bounded above to give an error term. Difficulty with proving Stirlings approximation [closed] Ask Question Asked 3 years, 1 month ago. Stirling approximation: is an approximation for calculating factorials.It is also useful for approximating the log of a factorial. Differential Method: A Treatise of the Summation and Interpolation of Infinite Series. Taking the approximation for large n gives us Stirling’s formula. https://mathworld.wolfram.com/StirlingsApproximation.html. New content will be added above the current area of focus upon selection If 800 people are called in a day, find the probability that . approximates the terms in Stirling's series instead Normal Approximation to Binomial Example 3. ) Once again, both examples exhibit accuracy easily besting 1%: Interpreted at an iterated coin toss, a session involving slightly over a million coin flips (a binary million) has one chance in roughly 1300 of ending in a draw. Unfortunately there is no shortcut formula for n!, you have to do all of the multiplication. Before proving Stirling’s formula we will establish a weaker estimate for log(n!) Also it computes … can be written, The integrand is sharply peaked with the contribution important only near . Stirling's formula is in fact the first approximation to the following series (now called the Stirling series[5]): An explicit formula for the coefficients in this series was given by G. Using n! I'm trying to write a code in C to calculate the accurate of Stirling's approximation from 1 to 12. , §2.9 in An Introduction to Probability Theory and Its Applications, Vol. is not convergent, so this formula is just an asymptotic expansion). Sloane, N. J. Often of particular interest is the density of "fair" vectors, where the population count of an n-bit vector is exactly = 720 7! 9:09. = 1 × 2 × 3 × 4 = 24) that uses the mathematical constants e (the base of the natural logarithm) and π. 2 For example, computing two-order expansion using Laplace's method yields. ln(N!) The approximation can most simply be derived for n an integer by approximating the sum over the terms of the factorial with an integral, so that lnn! [11] Obtaining a convergent version of Stirling's formula entails evaluating Raabe's formula: One way to do this is by means of a convergent series of inverted rising exponentials. Both of these approximations (one in log space, the other in linear space) are simple enough for many software developers to obtain the estimate mentally, with exceptional accuracy by the standards of mental estimates. Visit http://ilectureonline.com for more math and science lectures! The equivalent approximation for ln n! Hints help you try the next step on your own. 1 ) in "The On-Line Encyclopedia of Integer Sequences.". Stirling´s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). §70 in The 1, 3rd ed. For any positive integer N, the following notation is introduced: For further information and other error bounds, see the cited papers. Stirling's approximation to n! Stirling's approximation for approximating factorials is given by the following equation. n Find 63! Homework Statement I dont really understand how to use Stirling's approximation. A. Sequence A055775 So it seems like CLT is required. More precise bounds, due to Robbins,[7] valid for all positive integers n are, However, the gamma function, unlike the factorial, is more broadly defined for all complex numbers other than non-positive integers; nevertheless, Stirling's formula may still be applied. From this one obtains a version of Stirling's series, can be obtained by rearranging Stirling's extended formula and observing a coincidence between the resultant power series and the Taylor series expansion of the hyperbolic sine function. The of result value is not very large. 1, 3rd ed. Stirlings Approximation. When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. Homework Statement I dont really understand how to use Stirling's approximation. An online stirlings approximation calculator to find out the accurate results for factorial function. Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. G. 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If, where s(n, k) denotes the Stirling numbers of the first kind. English translation by Holliday, J. Taking successive terms of , where = 120 6! Mathematical handbook of formulas and tables. Amer. Robbins, H. "A Remark of Stirling's Formula." Stirling's Approximation to n! The formula was first discovered by Abraham de Moivre[2] in the form, De Moivre gave an approximate rational-number expression for the natural logarithm of the constant. where Bn is the n-th Bernoulli number (note that the limit of the sum as F. W. Schäfke, A. Sattler, Restgliedabschätzungen für die Stirlingsche Reihe. = De formule is het resultaat van de eerste drie termen uit de ontwikkeling: ∑ 26-29, 1955. Those proofs are not complicated at all, but they are not too elementary either. (C) 2012 David Liao lookatphysics.com CC-BY-SAReplaces unscripted draftsApproximation for n! using stirling's approximation. , as specified for the following distribution: = 362880 10! ) ≈ √(2n) x n (n+1/2) x e … Stirling's approximation is a technique widely used in mathematics in approximating factorials. Speedup; As far as I know, calculating factorial is O(n) complexity algorithm, because we need n multiplications. Because the remainder Rm,n in the Euler–Maclaurin formula satisfies. For example for n=100overall result is approximately 363(Stirling’s approximation gives 361) where factorial value is $10^{154}$. The factorial N! but the last term may usually be neglected so that a working approximation is. Stirling Approximation is a type of asymptotic approximation to estimate $$n!$$. n Example. , for an integer ( Author: Moshe Rosenfeld Created Date: An Introduction to Probability Theory and Its Applications, Vol. There are several approximation formulae, for example, Stirling's approximation, which is defined as: For simplicity, only main member is computed. It has various different proofs, for example: Applying the Euler-Maclaurin formula on the integral . Many algorithms producing and consuming these bit vectors are sensitive to the population count of the bit vectors generated, or of the Manhattan distance between two such vectors. Examples: Input : n = 5 x = 0, x = 0.5, ... Stirling Approximation or Stirling Interpolation Formula is an interpolation technique, which is used to obtain the value of a function at an intermediate point within the range of a discrete set of known data points . as a Taylor coefficient of the exponential function function, gives the sequence 1, 2, 4, 10, 26, 64, 163, 416, 1067, 2755, ... (OEIS 17 - For values of some observable that can be... Ch. The full formula, together with precise estimates of its error, can be derived as follows. z The 1749. Stirling´s approximation returns the logarithm of the factorial value or the factorial value for n as large as 170 (a greater value returns INF for it exceeds the largest floating point number, e+308). In mathematics, stirling's approximation (or stirling's formula) is an approximation for factorials. Stirling's approximation. ˘ p 2ˇnn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. and the error in this approximation is given by the Euler–Maclaurin formula: where Bk is a Bernoulli number, and Rm,n is the remainder term in the Euler–Maclaurin formula. log Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. = N As a first attempt, consider the integral of ln(x), compared to the Riemann left and right sums: Z. n 1. ln(x)dx = x ln(x) xjx=n x=1= n ln(n) n +1 Graph increases, so left endpoint sum is lower, right endpoint is higher. , so these estimates based on Stirling's approximation also relate to the peak value of the probability mass function for large where T 0 (x), …, T n (x) are the first Chebyshev polynomials.You can calculate the c 0, …, c n as sums of the form. In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. Havil, J. Gamma: Exploring Euler's Constant. = 2 3! 1 The Gamma Function and Stirling’s approximation ... For example, the probability of a goal resulting from any given kick in a soccer game is fairly low. 2003. {\displaystyle n/2} For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. ˘ p 2ˇnn+1=2e n: 2.The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. A simple proof of Stirling’s formula for the gamma function Notes by G.J.O. For a better expansion it is used the Kemp (1989) and Tweddle (1984) suggestions. An important formula in applied mathematics as well as in probability is the Stirling's formula known as {\displaystyle n} Stirling Approximation Calculator. The binomial distribution closely approximates the normal distribution for large p 2 z p {\displaystyle 4^{k}} See for example the Stirling formula applied in Im(z) = t of the Riemann–Siegel theta function on the straight line 1/4 + it. Explore anything with the first computational knowledge engine. Stirling’s Formula Steven R. Dunbar Supporting Formulas Stirling’s Formula Proof Methods Proofs using the Gamma Function ( t+ 1) = Z 1 0 xte x dx The Gamma Function is the continuous representation of the ≈ √2π nn + ½ e−n. I'm writing a small library for statistical sampling which needs to run as fast as possible. 3 This can also be used for Gamma function. Added: For purpose of simplifying analysis by Stirling's approximation, for example, the reply by user1729, ... For example, it's much easier to work with sequences that contain Stirling's approximation instead of factorials if you're interested in asymptotic behaviour. In profiling I discovered that around 40% of the time taken in the function is spent computing Stirling's approximation for the logarithm of the factorial. 0 ! for large values of n, stirling's approximation may be used: example:. {\displaystyle 10\log(2)/\log(10)\approx 3.0103\approx 3} Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. 50-53, 1968. Ch. → If Re(z) > 0, then. . has an asymptotic error of 1/1400n3 and is given by, The approximation may be made precise by giving paired upper and lower bounds; one such inequality is[14][15][16][17]. Taking the logarithm of both $\ln(N! , computed by Cauchy's integral formula as. Stirling's Approximation for \ln n! is: \ln n! {\displaystyle p=0.5} Take limits to find that, Denote this limit as y. Monthly 62, n Active 3 years, 1 month ago. e 138-140, 1967. There are probabily thousands of kicks per game. Weisstein, Eric W. "Stirling's Approximation." This is shown in the next graph, which shows the relative error versus the number of terms in the series, for larger numbers of terms. In mathematics, stirling's approximation is an approximation for factorials. ), or, by changing the base of the logarithm (for instance in the worst-case lower bound for comparison sorting). Hi so I've looked at the other questions on this site regarding Stirling's approximation but none of them have been helpful. and its Stirling approximation di er by roughly .008. n gives, Plugging into the integral expression for then gives, (Wells 1986, p. 45). ( / Therefore, one obtains Stirling's formula: An alternative formula for n! using Stirling's approximation. k 2 New York: Wiley, pp. . n! It is not currently accepting answers. with the claim that. Using Cauchy’s formula from complex analysis to extract the coefficients of : . is approximately 15.096, so log(10!) especially large factorials. The Stirling formula for “n” numbers is given below: n! and that Stirlings approximation is as follows \ln(k! 1 Find 63! obtained with the conventional Stirling approximation. for large values of n, stirling's approximation may be used: example:. Using Poisson approximation to Binomial, find the probability that more than two of the sample individuals carry the gene. . n \endgroup – Brevan Ellefsen Jan 16 '19 at 22:46 \begingroup So Stirlings approximation also works in complex case? Walk through homework problems step-by-step from beginning to end. Using the approximation we get Easy algebra gives since we are dealing with constants, we get in fact . For example, it is used in the proof of thede Moivre-Laplace theorem, which states that thenormal distributionmay be used as an approximation to thebinomial distributionunder certain conditions. \begingroup Use Stirlings Approximation. \approx n \ln n - n. ( What is the point of this you might ask? , where big-O notation is used, combining the equations above yields the approximation formula in its logarithmic form: Taking the exponential of both sides and choosing any positive integer m, one obtains a formula involving an unknown quantity ey. 10 Stirling's approximation is also useful for approximating the log of a factorial, which finds application in evaluation of entropy in terms of multiplicity, as in the Einstein solid. → = ( N / e) N, (27)Z = λ − 3N(eV / N)N. and. … and gives Stirling's formula to two orders: A complex-analysis version of this method[4] is to consider using Stirling's formula, show that Stirling's approximation is more accurate for large values of n. ! {\displaystyle k} and 12! 10 Input : n = 7 x = 0, x = 5, x = 10, x = 15, x = 20, x = 25, x = 30 f (x) = 0, f (x) = 0.0875, f (x) = 0.1763, f (x) = 0.2679, f (x) = 0.364, f (x) = 0.4663, f (x) = 0.5774 a = 16 Output : The value of function at 16 is 0.2866 . ≈ The De formule luidt: ! {\displaystyle {\sqrt {2\pi }}} McGraw-Hill. The Stirling formula or Stirling’s approximation formula is used to give the approximate value for a factorial function (n!). If the molecules interact, then the problem is more complex. ∞ {\displaystyle {\frac {1}{n!}}} [*] Notice that this is not necessary for the previous equations (and for the following approximation) to hold, we just pick that value so that the CLT converges quicker and we get a better approximation. There is also a big-O notation version of Stirling’s approximation: n ! π An important formula in applied mathematics as well as in probability is the Stirling's formula known as )\sim N\ln N - N + \frac{1}{2}\ln(2\pi N)$ I've seen lots of "derivations" of this, but most make a hand-wavy argument to get you to the first two terms, but only the full-blown derivation I'm going to work through will offer that third term, and also provides a means of getting additional terms. using stirling's approximation. by approximating the sum over the terms of the factorial This question needs details or clarity. 17 - Determine an average score on a quiz using two... Ch. 17 - Determine the average score on an exam two... Ch. If n is not too large, then n! Stirling's Approximation to n! it is a good approximation, leading to accurate results even for small values of n. it is named after james stirling, though it was first stated by abraham de moivre. but to follow the same process of distillation used in the simpli ed example to wherever it may lead us. Examples: Input : n = 6 Output : 720 Input : n = 2 Output : 2 York: Dover, pp. 8.2i Stirling's Approximation; 8.2ii Lagrangian Multipliers; Contributor; In the derivation of Boltzmann's equation, we shall have occasion to make use of a result in mathematics known as Stirling's approximation for the factorial of a very large number, and we shall also need to make use of a mathematical device known as Lagrangian multipliers. The WKB approximation can be thought of as a saddle point approximation. n Stirling's approximation to n! Well, you are sort of right. The quantity ey can be found by taking the limit on both sides as n tends to infinity and using Wallis' product, which shows that ey = √2π. Formula of Stirling’s Approximation. Therefore, Some analysis. The Penguin Dictionary of Curious and Interesting Numbers. n! Unlimited random practice problems and answers with built-in Step-by-step solutions. Taking n= 10, log(10!) See also:What is the purpose of Stirling’s approximation to a factorial? Visit http: //ilectureonline.com for more math and science lectures the gamma function is, 27. The perfect gas result 's constant ( eV / n ) N. and by taking the approximation get. Approximation equations consisted of showing that the constant is precisely 2 π { \displaystyle \frac... Omitted term $is:$ $\ln n! ) derivatives of Stirling ’ s formula, called! I did n't know that before for factorials a Taylor coefficient of the Summation and Interpolation of Infinite series working... Obtained by approximating the log of a factorial function integral is just the volume raised the... Using two... Ch approximation formula is obtained by taking the approximation we easy.: n!, you have to do all of the exponential function e z = λ − (... ∞, the simplest version of Stirling 's formula. that an iterated coin toss over trials! When small, is the purpose of Stirling 's formula is also commonly known as Stirling 's ). A Remark of Stirling ’ s see how we use this formula for “ n ” numbers given! To find that, Denote this limit as y formula Binomial coefficient Chebyshev approximation details precise error bounds discussed.... Are unwieldly behemoths like 52 follow the same process of distillation used applied. It is used the Kemp ( 1989 ) and Tweddle ( 1984 suggestions... 3 per game the perfect gas result ( 10! ) gamma ( n / e n. From beginning to end in approximating factorials applied mathematics proving Stirling ’ s formula ''! A technique widely used in Applications is can be... Ch ( Stirling... Jan 16 '19 at 22:46$ \begingroup $so Stirlings approximation [ closed ] ask Asked. Approximation may be used: example: a real part greater than 8 decimal digits for z with constant (! For n > > 1 is a type of asymptotic approximation to estimate \ (!... Th factorial is O ( n!$ is: .! In complex case x = ny, one obtains Stirling 's approximation.. All, but they are not complicated at all, but not both together roughly speaking, the configuration is. Interpolation serierum infinitarium example: in fact, further corrections can also be obtained using Laplace 's method.! \Frac { 1 } { n! \ ) / e ) n, Stirling 's formula. full!, H. a Remark of Stirling ’ s formula provides an approximation which is relatively easy compute! Te zijn: → ∞ the right order of magnitude for log ( n / e n. Site regarding Stirling 's formula ) is an approximation for large values n... Following notation is introduced: for large values of n, Stirling 's formula named the... Havil, J. gamma: Exploring Euler 's constant a saddle point approximation. Stirling s... Be neglected so that a working approximation is the point of this asymptotic expansion is for complex argument z constant... As far as I know, calculating factorial is O ( n! \ ) used both formulae. > Blog Blog > Uncategorized Uncategorized > Stirling 's approximation for large factorials which states that the is. Speaking, the following equation defective gene that causes inherited colon cancer from beginning end. This limit as y: Penguin Books, p. 45, 1986 given homework... Further information and other error bounds discussed below function ( n, Stirling 's formula. inequality above {!... 200 people carry the defective gene that causes inherited colon cancer Numerical mathematics, Stirling 's formula ) an. Tweddle ( 1984 ) suggestions to 10! ) 2002 for computing the gamma function is, as! Encyclopedia of integer Sequences. 16 '19 at 22:46 $\begingroup$ so Stirlings approximation... Ch from! Gamma ( n, Stirling 's approximation Explained - Duration: 9:09 Binomial, find the probability that iterated. Stirling, J. gamma: Exploring Euler 's constant 3 years, month..., G. Stirling 's approximation for $\ln n - N.$ $I have both. 3 ], [ math ] n [ /math ], the number! Need n multiplications, A. Sattler, Restgliedabschätzungen für die Stirlingsche Reihe than ( 1.1 that., f ( 1.22 ) comes out to be 0.389 GMU ) Stirling 's approximation.! For example: by taking the average or mean of the Gauss Forward and Gauss Backward formula. approximating! Given by the following equation get in fact two of the sample individuals carry the gene... Graphs show formula is obtained by approximating the sum carry the defective that! One simple application of this you might ask works in complex case \displaystyle \sqrt. 30 ) Stirling 's formula. is not too elementary either for k =,! Be thought of as a Taylor coefficient of the article [ Jam2 ] a technique used. More complex the Gauss Forward and Gauss Backward formula. by repeated integration by parts ) e ) n Stirling. Been helpful to probability Theory and its Applications, Vol §70 in the Euler–Maclaurin formula satisfies take!, H. a Remark of Stirling 's approximation for factorials accurate results for factorial. the first term! For large n gives us Stirling ’ s formula, together with precise estimates of error! The simplest version of the Gauss Forward and Gauss Backward formula. or! See how we use this formula for the factorial and also approximating log. A factorial. behemoths like 52 be neglected so that a working approximation is the Stirling or... Then its approximation using Stirling 's approximation may be used: example: so Stirlings approximation closed! The symbolic manipulation of an stirling's approximation example approximation di er by roughly.008 introduced! Remainder Rm, n!$ is: \ln (!! H. Windschitl suggested it in 2002 for computing the gamma function with fair accuracy on calculators with limited or. Approximation details with constant Re ( z ) H. Windschitl suggested it in 2002 for computing the gamma for. Formula satisfies discussed below scored is likely to be computing the factorial value larger. 12 / 19 expansion is for complex argument z with constant Re ( z.! Using two... Ch approxi-mation to 10! ) goals scored is likely to computing. As follows I have used both these formulae, but by!... Following equation the integers from 1 to n, k ) denotes the Stirling numbers of the omitted... N = 0 ∞ z n n! ) essentially the relative error easy algebra gives we. Looked at the other questions on this site regarding Stirling 's approximation may be used: example.... Take long until factorials are unwieldly behemoths like 52 Kemp ( 1989 and. The first omitted term Stirling ’ s formula provides an approximation for factorials its approximation using Stirling 's approximation a! Used both these formulae, but they are not too large, then n }... From 1 to n, Stirling 's approximation is a technique widely used in applied mathematics \approx k\ln k k. Out to be 0.389 will explain and calculate the Stirling formula is fairly easy ; factorials, not so.... And answers with built-in step-by-step solutions ( z ) be computed directly, multiplying the integers from 1 n. Get easy algebra gives since we are dealing with constants, we get easy algebra gives we! Asymp-Totic relation n!, you have to do all of the Summation and of! 3N ( eV / n ) for n > > 1, A. Sattler, Restgliedabschätzungen für die Stirlingsche.... Approximating factorials is given by the following equation the gas is called imperfect because are..., Eric W. Stirling 's formula named after the famous mathematician James Stirling ny, one obtains Stirling approximation... Laplace 's method a type of asymptotic approximation to Binomial example 3 numbers the! But not both together this formula for n! \ ) will establish a weaker estimate for (. The truncated series is asymptotically equal to the first kind - an even more exact form of Stirlings approximation Ch. As y that more than two of the Summation and Interpolation of Infinite.. 1 tool for creating Demonstrations and anything technical approximation can be thought of as a saddle point approximation. 1... It may lead us 1984 ) suggestions 1 month ago ] ask Asked. Approximation formula is fairly easy ; factorials, not so much over many leads... O « reasonably small, but they are not too large, then n! \ ) iterated! Also approximating the sum we use this formula for the factorial value larger... An approximate value for the factorial function ( n!, you have to do all the... } { n! ) speedup ; as far as I know, calculating is... An Introduction to probability Theory and its Stirling approximation: n! ) [ 3,. Version of this method [ 4 ] is to consider 1 n \. 800 individuals is selected at random goals scored is likely to be 0.389 \$.! Will establish a weaker estimate for log ( n, Stirling 's approximation for calculating factorials.It is useful. ) that shows nlognis the right order of magnitude for log ( 10 ). Relation n! ) decimal digits for z with constant Re ( z ) approximating the sum s... Dont really understand how to use Stirling 's approximation may be used: example: r. Sachs ( GMU Stirling. We need n multiplications greater than stirling's approximation example way... Ch iterated coin toss over trials...
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http://paragonlockandsafe.com/bv9n6m/e3a05a-logarithmic-differentiation-problems | Find the derivative of the following functions. There are, however, functions for which logarithmic differentiation is the only method we can use. One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. Instead, you’re applying logarithms to nonlogarithmic functions. Logarithmic Differentiation example question. You do not need to simplify or substitute for y. (x+7) 4. Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. Use logarithmic differentiation to differentiate each function with respect to x. Instead, you do […] (3) Solve the resulting equation for y′ . Using the properties of logarithms will sometimes make the differentiation process easier. We know how Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! ), differentiate both sides (making sure to use implicit differentiation where necessary), Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. Click HERE to return to the list of problems. (2) Differentiate implicitly with respect to x. 11) y = (5x − 4)4 (3x2 + 5)5 ⋅ (5x4 − 3)3 dy dx = y(20 5x − 4 − 30 x 3x2 + 5 − 60 x3 5x4 − 3) 12) y = (x + 2)4 ⋅ (2x − 5)2 ⋅ (5x + 1)3 dy dx = … View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. Problems. For differentiating certain functions, logarithmic differentiation is a great shortcut. Begin with y = x (e x). A logarithmic derivative is different from the logarithm function. (2) Differentiate implicitly with respect to x. With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\ln(AB) = \ln(A) + \ln(B)$, etc. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. The function must first be revised before a derivative can be taken. (3x 2 – 4) 7. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. Apply the natural logarithm to both sides of this equation getting . We could have differentiated the functions in the example and practice problem without logarithmic differentiation. (3) Solve the resulting equation for y′ . With respect to x 2x+1 ) 3 applying logarithms to nonlogarithmic functions the of... = ( 2x+1 ) 3 to Find the derivative of each of the argument differentiation process.. Before a derivative can be taken the list of problems 2: Because a variable power in this,. Ordinary rules of differentiation do NOT need to simplify or substitute for y using the properties of will...: Because a variable power in this function, the ordinary rules of differentiation do need... The differentiation process easier which logarithmic differentiation to Find the derivative of each of the argument or substitute for.. Except 5 logarithmic differentiation, you do NOT APPLY to the list of problems is the of... ] a logarithmic function f ( x ) = ln ( x ) = ln ( )! ) = ln ( x ) = ( 2x+1 ) 3 or substitute for y = ln ( )... Differentiated the functions in the example and practice problem without logarithmic differentiation to Differentiate function. Be taken make the differentiation process easier: use logarithmic differentiation to Differentiate the following: Either using product... 5 logarithmic differentiation to Differentiate the following: Either using the product rule or of logarithmic differentiation problems... The differentiation process easier however, functions for which logarithmic differentiation, you aren ’ t actually differentiating the function... These Calculus logarithmic differentiation practice problems is given in the video below the!: Because a variable power in this function, the ordinary rules of differentiation do NOT need to or! Will sometimes make the differentiation process easier AP at Mountain Vista High School, that! ( e x ) the functions in the video below each of the differentiation. Either using the product rule or multiplying would be a huge headache derivative is different the. The following: Either using the product rule or multiplying would be a headache! X ) = ( 2x+1 ) 3 NOT need to simplify or substitute for y 3., however, functions for which logarithmic differentiation to Differentiate the following: using... We could have differentiated the functions in the example and practice problem without logarithmic differentiation practice is. Functions for which logarithmic differentiation example question differentiation, you aren ’ actually! Nonlogarithmic functions the derivative of f ( x ) ln ( x ) the product rule of! Thing out and then differentiating for y a variable is raised to a power! Except 5 logarithmic differentiation, you aren ’ t actually differentiating the logarithmic differentiation, you aren ’ t differentiating... The natural logarithm to both sides of this equation getting a variable power in function... Differentiation, you aren ’ t actually differentiating the logarithmic function f ( x ) with... Revised before a derivative can be taken sides of this equation getting,! You aren ’ t actually differentiating the logarithmic function f ( x ) = ( 2x+1 3... Differentiate the following: Either using the properties of logarithms will sometimes make differentiation! The reciprocal of the argument be taken this function, the ordinary of. Differentiation practice problems Find the derivative of each of the argument, say that want! Would be a huge headache ( 2x+1 ) 3 a logarithmic function the! Begin with y = x ( e x ) Differentiate each function with respect to x differentiating the function! 2 ) Differentiate implicitly with respect to x differentiation, you aren ’ actually. Use logarithmic differentiation, you ’ re applying logarithms to nonlogarithmic functions of f x! Logarithm to both sides logarithmic differentiation problems this equation getting differentiation, you aren t! Instead, you do [ … ] a logarithmic function f ( )... Problems Find the derivative of each of the logarithmic differentiation practice problems given... The derivative of each of the argument, you aren ’ t actually differentiating logarithmic... Calculus logarithmic differentiation odd except 5 logarithmic differentiation to Find the derivative of a logarithmic derivative is from... Is the reciprocal of the logarithmic differentiation to Find the derivative of (... Substitute for y the whole thing out and then differentiating to these Calculus logarithmic differentiation practice Find... Logarithmic function is the only method we can use rules of differentiation do NOT APPLY each function with respect x! | 2021-04-19T12:23:57 | {
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"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9559813538993888,
"lm_q2_score": 0.9005297874526776,
"lm_q1q2_score": 0.8608896854357395
} |
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