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https://www.khanacademy.org/math/ap-statistics/quantitative-data-ap/measuring-spread-quantitative/a/concept-check-standard-deviation?ref=probability_and_statistics_staff_picks | # Concept check: Standard deviation
## Introduction
Unlike most questions on Khan Academy, some of these questions aren't graded by a computer. You'll learn the most if you try answering each question yourself before clicking "explain".
## The formula (for reference)
The formula for standard deviation (SD) is
$\Large\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
where sum means "sum of", x is a value in the data set, $\bar{x}$ is the mean of the data set, and n is the number of values in the data set.
## Part 1
Consider the simple data set left brace, 1, comma, 4, comma, start color redD, 7, end color redD, comma, 2, comma, 6, right brace.
How does the standard deviation change when start color redD, 7, end color redD is replaced with start color greenD, 12, end color greenD?
Please choose from one of the following options.
How can we see this in the formula for standard deviation?
### How does the standard deviation change?
The standard deviation increases because the data becomes more spread out:
### How can we see this in the formula?
We can see this in the formula
$\text{SD} = \sqrt{\dfrac{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{n}}$
because ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ is the sum of the squares of the distances from each data point to the mean. As the data, gets more spread out, the value of ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ increases.
### I'm curious, what are the actual standard deviations of the data sets?
The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color redD, 7, end color redD, right brace is approximately 2, point, 28.
The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color greenD, 12, end color greenD, right brace is approximately 3, point, 90.
## Part 2
Is it possible to create a data set with 4 data points that has a standard deviation of 0?
Please choose from one of the following options.
If it is possible, do it! Can you create two different data sets? How about three?
### Yes, it's possible!
In fact, there are an infinite number of possible data sets.
Here's one:
5, comma, 5, comma, 5, comma, 5
Here's another:
8, comma, 8, comma, 8, comma, 8
Any data set where all of the data points are the same has a standard deviation of 0 because the distance from each data point to the mean is 0.
### Show me the calculation for $\{ 5,5,5,5 \}$left brace, 5, comma, 5, comma, 5, comma, 5, right brace.
#### Step 1: Find the mean
$\bar{x} = \dfrac{5 + 5 + 5 + 5}{4} = \dfrac{20}{4} = \blueD5$
#### Step 2: Find the square of the distances from each of the data points to the mean
x$\lvert x - \bar{x} \rvert^2$
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
#### Step 3: Apply the formula
\begin{aligned} \text{SD} &= \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}} \\\\\\\\ &= \sqrt{\dfrac{0 + 0 + 0 + 0}{4}} \\\\\\\\ &= \sqrt{\dfrac{{0}}{4}}\\\\\\\\ &= \sqrt{{0}}\\\\\\\\ &= 0\end{aligned}
## Part 3
Can standard deviation be negative?
Please choose from one of the following options.
Why or why not?
### No, standard deviation cannot be negative!
To see why, think about the numerator and denominator inside the radical:
$\Large\text{SD} = \sqrt{\dfrac{\blueD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{\maroonD{n}}}$
Notice how start color maroonD, n, end color maroonD is always positive. It's the number of data points, and we can't have a negative number of data points.
Also notice that $\blueD{\sum\lvert x - \bar{x} \rvert^2}$ involves a quantity getting squared. Whenever we square something, we get a non-negative number.
Since both the denominator and numerator are positive, the entire expression must be positive too.
## Part 4
Standard deviation is a measure of spread of a data distribution.
What do you think deviation means?
In everyday language, deviation is how different something is from what might be considered normal.
In statistics, when discussing measures of spread, deviation is the amount by which a single measurement differs from the mean.
### Part 5
Here are the formulas for standard deviation (SD) and the formula for mean absolute deviation (MAD), both of which are measures of spread:
$\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
$\text{MAD} = {\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert}}}{n}}$
What are the similarities between the formulas? What are the differences?
### What are the similarities?
The formulas are very similar! They are both based on the distance from each data point to the mean $\lvert x - \bar{x} \rvert$, and they both include dividing by the number of data points n.
### What are the differences?
The difference between the two formulas is that when calculating standard deviation, we square the distance from each data point to the mean, and we take the square root as the last step of the formula.
### Which one is better?
Standard deviation is more complicated, but it has some nice properties that make it statisticians' preferred measure of spread.
### Part 6
Here's the formula that we've been using to calculate standard deviation:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
Here's the formula that statisticians actually use:
$\sqrt{\dfrac{\sum\limits_{}^{}{{( x-\bar{x})^2}}}{n}}$
Are the two formulas equivalent?
Please choose from one of the following options.
### What's the difference between the formulas?
In the formula that we've been using, we take the absolute value of $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\tealD{\lvert x-\bar{x}\rvert}^2}}}{n}}$
In the formula that statisticians use, they put parentheses around $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\purpleC{( x-\bar{x})^2}}}}{n}}$
### Are the formulas equivalent?
Yes, both formulas are equivalent!
Statisticians realize that squaring will make the distance positive, so they don't bother using absolute value signs and just use parentheses instead.
For example, let's evaluate $\tealD{\lvert x - \bar{x} \rvert^2}$ and $\purpleC{(x - \bar{x})^2}$ for x, equals, 2 and $\bar{x} = 5$:
$\tealD{\lvert x - \bar{x} \rvert^2} = \lvert 2 - 5 \rvert^2 = \lvert-3\rvert^2 = 3^2 = \greenD9$
$\purpleC{(x - \bar{x})^2} = (2 - 5)^2= (-3)^2 = \greenD9$
They're both positive! They're both start color greenD, 9, end color greenD! They're equivalent! | 2017-03-27T08:43:06 | {
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http://suisco.net/awstats/blog/page.php?page=arithmetic-expression-calculator-2f178e | - Subtraction For an algebraic expression calculator to work, it is typically required to type the above type of math expression in, by using common symbols. Determine the first term and difference of an arithmetic progression if $a_3 = 12$ and the sum of first 6 terms is equal 42. An example of an algebraic expression is shown below. Answer: Yes, it is a geometric sequence and the common ratio is 6. Everything you need to prepare for an important exam! The main purpose of this calculator is to find expression for the n th term of a given sequence. Solve math problems using order of operations like PEMDAS, BEDMAS and BODMAS. Similarly, when you are solving addition and subtraction expressions you proceed from left to right. Below are some of the example which a sum of arithmetic sequence formula calculator uses. This website uses cookies to improve your experience. Welcome to MathPortal. You can evaluate any Mathematical Expression using this calculator. To find the next element, we add equal amount of first. each number is equal to the previous number, plus a constant. Closed. for learning distance formula equation, use Distance Formula Calculator. For nested parentheses or brackets, solve the innermost parentheses or bracket expressions first and work toward the outermost parentheses. but they come in sequence. BEDMAS stands for "Brackets, Exponents, Suppose they make a list of prize amount for a week, Monday to Saturday. About this calculator. Degrees of Freedom Calculator Paired Samples, Degrees of Freedom Calculator Two Samples. The first term of an arithmetic progression is $-12$, and the common difference is $3$ Free Arithmetic Sequences calculator - Find indices, sums and common difference step-by-step This website uses cookies to ensure you get the best experience. Copyrights 2020 © calculatored.com . . Tough Algebra Word Problems.If you can solve these problems with no help, you must be a genius! The calculator will generate all the work with detailed explanation. Numbers sequence, in which the diffference is always constant. Viewed 3k times 0. Viewed 3k times 0. + Addition Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction Logical Sets Change the sign of each number that follows so that positive becomes negative, and negative becomes positive then follow the rules for addition problems. PEMDAS is an acronym that may help you remember order of operations for solving math equations. Now, find the sum of the 21st to the 50th term inclusive, There are different ways to solve this but one way is to use the fact of a given number of terms in an arithmetic progression is $$\frac{1}{3}\;n(a+l)$$ Here, “a” is the first term and “l” is the last term which you want to find and “n” is the number of terms. Just enter your numerical expression in the big box right beneath the "calculate" and "clear" button and hit the calculate button Note: If you are using parentheses, just remember to put a multiplication sign right before the parentheses For example, do not enter 5(3-4). It is not currently accepting answers. The most important thing to keep in mind is that the parenthesis are important when typing an algebraic expression. First type the expression 2x. If you incorrectly enter it as 4/1/2 then it is solved 4/1 = 4 first then 4/2 = 2 last. For example, the above expression would be expressed as "1/(3-2) + 3 + 4sin(pi/4) + sqrt(2) + 5^(3/2)". Objects might be numbers or letters, etc. This web site owner is mathematician Miloš Petrović. As the contest starts on Monday but at the very first day no one could answer correctly till the end of the week. | 2021-03-02T04:30:15 | {
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https://www.emmamichaels.com/3584/cos-sin-1-x.html | Breaking News
# Cos Sin 1 X
Cos Sin 1 X. Cos2(x)+ 2cos(x)sin(x)+sin2(x) = 1 note an important identity: = 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. Expand using the foil method. The word cosinus derived from edmund gunter.
= 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude, a cos x + b sin x = c cos ( x + φ ). Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.
## Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.
Solve for x cos(x)+1=sin(x) subtract from both sides of the equation. Cos (x)+sin (x)=1 cos (x) + sin(x) = 1 cos ( x) + sin ( x) = 1 square both sides of the equation. The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.
## Misc 13 Solve 2 Tan1 (Cos X) = Tan1 (Ii Cosec X Misc 13.
How to prove that 1+sin(x)cos(x) = cos(x)1−sin(x) [duplicate] multiplying the lhs by the conjugate of.
### Extended Keyboard Examples Upload Random.
The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.
### Kesimpulan dari Cos Sin 1 X.
The word cosinus derived from edmund gunter. | 2023-01-29T20:13:22 | {
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https://math.stackexchange.com/questions/3212210/how-to-know-when-you-can-focus-only-on-a-specific-part-of-an-expression-with-an | # How to know when you can focus only on a specific part of an expression with an even power?
I stumbled upon this problem as I was going through the Algebra 2 course of https://brilliant.org. It goes as follows:
Starting from
$$x = \sqrt{\sqrt{3 \sqrt{\sqrt{3 \sqrt{\sqrt{3 \dots}}}}}}$$,
I manage on my own to reach the following equation:
$$x^4 = 3x$$.
Personally I took it from there as follows:
$$\frac{x^4}{x} = 3 \leftrightarrow x^3 = 3 \leftrightarrow x=\sqrt[3]{3}$$.
The course accepts my final answer of $$x = \sqrt[3]{3}$$ as correct, however they reason as follows:
$$x^4 = 3x \leftrightarrow x^4 - 3x = 0 \leftrightarrow x(x^3 - 3) = 0$$, since in the original equation $$x > 0$$, we only worry about the root $$x^3 - 3$$:
$$x^3 - 3 = 0 \leftrightarrow x^3 = 3 \leftrightarrow x = \sqrt[3]{3}$$.
They do reach to the same answer but that is not important when learning. What bothers me is that I do not seem to understand on why the $$x > 0$$ part is important here, and how they use that to form their conclusion. It also makes me wonder about how it impacts my approach to the solution. And in general it makes me wonder if there are guidelines that can help me decide on what properties of an original equation I need to take into account when transforming equations while solving a problem.
I suppose it is similar as how when $$x$$ is in the denominator of the original equation than $$x \neq 0$$, no matter how you can transform the equation (e.g. cancel out the denominator). I do however not see at the moment why it would apply here.
• You just divided both sides of your equation by $x$ and so you got the right answer but you need to justify that you can do this and the justification is that clearly $x$ is non zero. – Anon May 3 '19 at 13:28
• Oh, right. Now I get it, so they also just divide by $x$, difference being that they also explain why it is possible. Thank you, now I feel a bit silly to ask this as a question, not sure it will contribute much to this fascinating Q&A website. Feel free to put it as an answer @Anon, such that I can accept it as an answer. – glendc May 3 '19 at 13:29
In your solution, when you perform the step $$x^4 = 3x \iff \frac{x^4}{x} = x^3 = 3,$$ you are using the law of multiplicative cancelation, which states
Theorem: If $$a$$, $$b$$, and $$c$$ are any real numbers with $$c \ne 0$$, then $$a = b \iff ac = bc.$$
In other words, we can cancel common factors from both sides of an equation (that is, we can "divide" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $$x \ne 0$$. Therefore you are implicitly assuming that $$x \ne 0$$.
Since you ask about "guidelines" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:
Let $$x = \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}},$$ assuming that $$x$$ exists.[1] Then $$x^4 = \left( \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} \right)^4 = 3 \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} = 3x.$$ Since $$x > 0$$,[2] the law of multiplicative cancelation gives $$x^4 = 3x \iff x^3 = 3.$$ The function $$f(x) = x^3$$ is invertible on $$\mathbb{R}$$, and has inverse $$f^{-1}(x) = \sqrt[3]{x}$$. Thus $$x^3 = 3 \iff x = \sqrt[3]{3}.$$ Therefore $$x = \sqrt[3]{3}$$.
This kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.
Footnotes:
There are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.
[1] One can actually write down funny expressions like this which don't "converge" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $$x$$ really exists.
[2] It seems obvious that $$x \ne 0$$, but this, also, has the potential to be a little delicate. Basically, $$x$$ is the limit of iterative application of the map $$t \mapsto \sqrt[4]{3t}$$. This map has two fixed points (one at zero, and one somewhere else—the second fixed point is the thing that we are trying to find). However, the fixed point at zero is "unstable", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.
For $$ab=0$$, either $$a=0$$, or $$b=0$$ or both $$a=b=0$$ (if possible). Here, you see, $$x>0$$, as $$x$$ is an even root (so $$x$$ is definitely not zero). This implies the other term in the product zero. So $$x^3-3=0$$. Whenever you have equation of the form $$a_1a_2a_3...a_n=0$$, always make cases.
• Thank you very much for your feedback as well. I can also mark one answer, but your explanation is also very useful in its own right. – glendc May 3 '19 at 14:32 | 2020-02-18T01:23:19 | {
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http://mathhelpforum.com/algebra/65489-compound-interest-problem.html | # Math Help - Compound interest problem
1. ## Compound interest problem
Question:
In 1974, Johnny Miller won 8 tournaments on the PGA tour and accumulated $304,680 in official season earnings. In 1999, Tiger Woods accumulated$6081912 with a similar record. If Miller had invested his earnings in an account earning compound interest, find the annual interest rate needed for his winnings to be equivalent in value to Tiger Wood's in 1999.
My solution is attached however it's not complete. I don't know how to get ride of the (1 + r)^25 exponent.
2. Take the 25th root of both sides:
$19.96^{1/25}=1+r$
3. Okay, so after solving the problem r = 0.12
I'm not sure that is correct. Can someone confirm if it is correct?
Here is the solution:
4. Hello, mwok!
You were doing great . . .
$304,\!680(1+r)^{25} \:=\:6,\!081,\!912$
$(1+r)^{25} \:\approx\:19.96$
Take the $25^{th}$ root of both sides:
. . $\left[(1+r)^{25}\right]^{\frac{1}{25}} \;=\;(19.96)^{\frac{1}{25}}$
. . . . . . $1 + r \;=\;1.127217824$
. . . . . . . . $r \;=\;0.127217824 \;\approx\;12.7\%$
Ah, too slow . . . again!
.
5. ## Compound interest problem
Hi, Mwork
You were very close to the answer, you just have to raise both sides to the 1/25 power.
$
\frac{6081912}{304680} = (1+r)^{25}
$
$(\frac{6081912}{304680})^{\frac{1}{25}} = (1+r)$
$(\frac{6081912}{304680})^{\frac{1}{25}}-1 = r$
Next time use this formula for the rate
$
r = (\frac{A}{P})^{\frac{1}{n}}-1
$
6. Thanks, The question is asking for the rate. Should I leave it as 0.12 or 12.7? Is it correct if I left it as 0.12?
7. It is best to leave it as 12.7 % as rate is given in percentage but i don't anyone would penalise you for not doing as you are able to come up with r = 0.127
8. (1+i)^25=19.961630
take log of both sides,
25log(1+i)=log 19.961630=1.300196
log(1+i)=1.300196/25=.052
1+i=10^.052=1.1272, so i=12.72 | 2014-04-20T16:09:00 | {
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https://math.stackexchange.com/questions/2843087/is-union-of-regular-open-sets-a-regular-open-set-provided-the-intersection-of-th | # Is union of regular open sets a regular open set provided the intersection of the two sets is nonempty?
Let us consider subsets in $\mathbb R^n$. A set $E$ is regular open if $E = \text{int}(\bar{E})$. It is true that the union of two regular open sets is not necessarily regular open. The counterexample I have seen is to take two nonoverlapping intervals with common endpoints. For example, $(-1, 0)$ and $(0, 1)$ are both regular open but $(-1, 0)\cup (0,1)$ is not regular open. I am wondering what if the two regular open sets $E \cap F$ have nonempty intersection, i.e., $E \cap F \neq \emptyset$ and $E, F$ are both path-connected, would the union $E \cup F$ be regular open? In other words, are there sufficient conditions to guarantee $E \cup F$ to be regular open?
No. Consider these subsets of $\mathbb{R}^2$:$$E=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge0<\theta<\frac{3\pi}2\right\}$$and$$F=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge\frac\pi2<\theta<2\pi\right\}.$$They satisfy the conditions that you mentioned, but the interior of $\overline{E\cup F}$ is strictly larger than $E\cup F$, since it also contains the points of the type $(r,0)$, ithe $r\in(1,2)$.
• Thanks. Are there sufficient conditions to guarantee the conclusion? – user1101010 Jul 6 '18 at 18:07
• @iris2017 I don't know. – José Carlos Santos Jul 6 '18 at 18:13
• @iris2017 See my answer below. – giobrach Jul 6 '18 at 18:40
To answer your question in the comments to José's answer: it is sufficient to require $E,F$ to be regular open and for $E \cup F$ to have an open complement, i.e. $\mathrm{et}(E \cup F) = X \setminus (E \cup F)$.
Proof. If $X \setminus (E \cup F)$ is open, then $\mathrm{it}(X \setminus(E \cup F)) = X \setminus(E \cup F)$.
Now both $E$ are regular open, so $$\mathrm{it}(E)=\mathrm{it}(\mathrm{it}(\mathrm{cl}(E))) = \mathrm{it}(\mathrm{cl}(E)) = E$$ which means $E$ is open (and so is $F$ by an identical argument). Therefore their union is also open and $\mathrm{it}(E \cup F) = E \cup F$. Therefore, by exploiting the fact that, for all $A \subseteq X$, $$\mathrm{cl}(A) = X \setminus \mathrm{it}(X \setminus A),$$ we get $$\mathrm{it}(\mathrm{cl}(E\cup F)) = \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E \cup F))) = \mathrm{it}(X \setminus (X \setminus (E \cup F))) = \mathrm{it}(E \cup F) = E \cup F.$$ Hence $E \cup F$ is regular open. $\qquad \square$
Addendum. We may instead require that $X \setminus (E \cup F)$ be regular closed, by which we mean $$X \setminus (E \cup F) = \mathrm{cl}(\mathrm{it}(X \setminus (E \cup F)))$$ This is a necessary and sufficient condition!
Proposition. Let $E,F \subseteq X$ be regular open. Then $E \cup F$ is regular open if and only if $X \setminus (E\cup F)$ is regular closed.
Note: that the previous case ($X \setminus (E \cup F)$ open) implies this situation: if the complement of $E \cup F$ is open, then its interior is equal to itself; but since $E \cup F$ is open, then $X \setminus (E \cup F)$ is also closed, and so the closure of its interior, which is equal to the closure of itself, is equal to itself.
Proof. As we've seen above, if $E,F$ are regular open, then they are open. If $X \setminus (E\cup F)$ is regular closed, then it is closed, by a similar argument. Hence, $$\begin{split} X \setminus (E\cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= \mathrm{cl}(\mathrm{it}((X \setminus E)\cap (X \setminus F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F)) \\ &= X \setminus \mathrm{it}(X \setminus (\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}((X \setminus \mathrm{it}(X \setminus E)) \cup (X \setminus \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E) \cup \mathrm{cl}(F)) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)), \end{split}$$ which implies $E \cup F = \mathrm{it}(\mathrm{cl}(E\cup F))$.
Another, shorter way to see this is $$\begin{split} X \setminus (E \cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)). \end{split}$$
This chain of equalities, followed in the reverse order, also takes care of the other implication: if $E \cup F$ is regular open, then $$\begin{split} X \setminus (E \cup F) &= X \setminus \mathrm{it}(\mathrm{cl}(E\cup F)) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X\setminus (E \cup F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))). \\ \end{split}$$ Hence $X \setminus (E \cup F)$ is regular closed. $\qquad \square$
• Thanks. Could you give an example? I could not see how $X \setminus (E \cup F)$ could be open. – user1101010 Jul 6 '18 at 18:50
• Both $E\cup F$ and its complement need to be both open and closed. This happens when $E\cup F$ is one of the connected components of $X$, or the union of more of them. Can you think of an instance of this? – giobrach Jul 6 '18 at 19:01
• Do we only need to guarantee that $X \setminus (E \cup F)$ is connected? We can take two open balls in $\mathbb R^n$ for $n \ge 2$. Is this right? – user1101010 Jul 6 '18 at 19:26
• My theorem also works in the case where $X \setminus (E\cup F)$ decomposes into separate connected components. Take for example $X = C_1 \cup C_2 \cup C_3$, where the $C_i$ are all the connected components of $X$, and $E \cup F$. Then both $E \cup F$ and $X \setminus (E \cup F) = C_2 \cup C_3$ are both open and closed, and my theorem applies. – giobrach Jul 6 '18 at 19:35
• Of course, we are working in a topological space $(X,\tau)$ where $X$ is the total space. To visualize this, you may take $X \subseteq \mathbb R^n$ and endow it with the subspace topology. – giobrach Jul 6 '18 at 19:36
If $\overline A$, $\overline B$ disjoint,
then int (A $\cup$ B) = int A $\cup$ int B.
Thus if A,B are regular open and $\overline A$, $\overline B$
disjoint, then A $\cup$ B is regular open. | 2019-08-24T03:34:35 | {
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https://toddeachus.com/6vmcyxvh/bc0454-every-identity-matrix-is-a-scalar-matrix | # every identity matrix is a scalar matrix
The x(1,2) that we have calculated is the point of intersection of the 2 equations in the system. Correct answer: Explanation: The 3x3 identity matrix is. ... Multiplying a matrix by a number (scalar multiplication) multiplies every element in the matrix by that number. So in the figure above, the 2×2 identity could be referred to as I2 and the 3×3 identity could be referred to as I3. However, the result you show with numpy is simly the addition of the scalar to all matrix elements. If q is the adding operation (add x times row j to row i) then q-1 is also an adding operation (add -x times row j to row i). We learn in the Multiplying Matrices section that we can multiply matrices with dimensions (m × n) and (n × p) (say), because the inner 2 numbers are the same (both n). Podcasts with Data Scientists and Engineers at Google, Microsoft, Amazon, etc, and CEOs of big data-driven companies. Properties of matrix multiplication. While we say “the identity matrix”, we are often talking about “an” identity matrix. Therefore for an $$m \times n$$ matrix $$A$$, we say: This shows that as long as the size of the matrix is considered, multiplying by the identity is like multiplying by 1 with numbers. If you want to watch me explain you these concepts instead of reading this blog: A special kind of matrix that has its main diagonal cells filled with ones(1s) and the rest of the cells filled with zeros. For any matrix A and any scalar c, (c A) T = c(A T). One concept studied heavily in mathematics is the concept of invertible matrices, which are those matrices that have an inverse. EASY. Possible Answers: The correct answer is not given among the other responses. For example, In above example, Matrix A has 3 rows and 3 columns. While we say “the identity matrix”, we are often talking about “an” identity matrix. Scalar operations produce a new matrix with same number of rows and columns with each element of the original matrix added to, subtracted from, multiplied by or divided by the number. Its determinant is the product of its diagonal values. Both scalar multplication of a matrix and matrix addition are performed elementwise, so. A square matrix has the same number of rows as columns. The identity matrix is a fundamental idea when working with matrices – whether you are working with just multiplication, inverses, or even solving matrix equations. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. For any whole number n, there is a corresponding n×nidentity matrix. Also, determine the identity matrix I of the same order. Lemma. for a square nxn matrix A the following statements are equivalent: a. Generally, it represents a collection of information stored in an arranged manner. The above code returns a 3×3 identity matrix as shown below: Confirming the property in code, we can calculate the dot product with a vector or matrix as follows: Note: Make sure that the rule of multiplication is being satisified. ‘Eigen’ is a German word which means ‘proper’ or ‘characteristic’. Observe that a scalar matrix is an identity matrix when k = 1. A matrix is said to be a rectangular matrix if the number of rows is not equal to … Make learning your daily ritual. Identity matrix is a square matrix with elements falling on diagonal are set to 1, rest of the elements are 0. Matrix multiplication is a process of multiplying rows by columns. Consider the following matrices: For these matrices, $$AB = BA = I$$, where $$I$$ is the $$2 \times 2$$ identity matrix. Example. A is an invertible matrix b. Matrix multiplication dimensions. Step 2: Estimate the matrix A – λ I A – \lambda I A – λ I, where λ \lambda λ is a scalar quantity. identity matrix. I looks like you mean that in MATLAB or numpy matrix scalar addition equals addition with the identy matrix times the scalar. Examples: It is denoted by A⁻¹. By definition, when you multiply two matrices that are inverses of each other, then you will get the identity matrix. The next episode will cover linear dependence and span. from sympy.matrices import eye eye(3) Output. An identity matrix of any size, or any multiple of it (a scalar matrix ), is a diagonal matrix. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Matrix multiplication. In simple words, the eigenvalue is a scalar that is used to transform the eigenvector. A matrix with only a single column is called a vector. Learn what an identity matrix is and about its role in matrix multiplication. with A = magic(2), A+1. Therefore $$A$$ and $$B$$ are inverse matrices. Only non-singular matrices have inverses. A diagonal matrix is sometimes called a scaling matrix, since matrix multiplication with it results in changing scale (size). In Mathematics, eigenve… Answer. Take a look, A = np.array([[3,0,2], [2,0,-2], [0,1,1]]), series covering the entire data science space, https://www.youtube.com/c/DataSciencewithHarshit, Noam Chomsky on the Future of Deep Learning, Kubernetes is deprecating Docker in the upcoming release, Python Alone Won’t Get You a Data Science Job, An end-to-end machine learning project with Python Pandas, Keras, Flask, Docker and Heroku, 10 Steps To Master Python For Data Science, Top 10 Python GUI Frameworks for Developers, The series would cover all the required/demanded quality tutorials on each of the topics and subtopics like. Intro to identity matrix. With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. This is a $$2 \times 4$$ matrix since there are 2 rows and 4 columns. The identity matrix is analogous to 1 (in scalar) which is to signify that applying (multiplying) the identity matrix to a vector or matrix has no effect on the subject. These two types of matrices help us to solve the system of linear equations as we’ll see. We can create an identity matrix using the NumPy’s eye() method. The Matrix matrix A = (2,1\3,2\-2,2) matrix list A A[3,2] c1 c2 r1 2 1 r2 3 2 r3 -2 2. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Rectangular Matrix. When working with matrix multiplication, the size of a matrix is important as the multiplication is not always defined. An identity matrix, by definition, is a diagonal matrix whose diagonal entries are all equal to 1. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. It is mostly used in matrix equations. Central dilation leads to a uniform expansion, if λ > 1, or a uniform contraction, ifλ< 1, of each dimension. This program allows the user to enter the number of rows and columns of a Matrix. The idea is to pick several specific vectors. is the first element in the second row, which is … Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window). Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. So in the figure above, the $$2 \times 2$$ identity could be referred to as $$I_2$$ and the $$3 \times 3$$ identity could be referred to as $$I_3$$. Create a script file with the following code − Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. To prevent confusion, a subscript is often used. You can study this idea more here: inverse matrices. 9) Upper Triangular Matrix A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. A matrix A is symmetric if and only if A =A T. All entries above the main diagonal of a symmetric matrix are reflected into equal entries below the diagonal. Multiplying by the identity. Matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) $\displaystyle \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$ The output for the above code is as follows − Matrices are represented by the capital English alphabet like A, B, C……, etc. (vi) Identity matrix A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix. Explained Mathematics and derivations of why we do what we do in ML and Deep Learning. Example. Now, that we have generated the inverse, we can check the property by calculating the dot product of A with A⁻¹: Hence, the property stands True for inverse matrices. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. But every identity matrix is clearly a scalar matrix. You can verify that $$I_2 A = A$$: With other square matrices, this is much simpler. For the following matrix A, find 2A and –1A. Google Classroom Facebook Twitter. Now, we can use inverse matrices to solve them. The optimistic mathematician’s way. D. scalar matrix. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. In other words, the square matrix A = [a ij] n × n is an identity matrix, if 1if ij 0if ij a ij ⎧ = =⎨ ⎩ ≠. The basic equation is AX = λX The number or scalar value “λ” is an eigenvalue of A. When passed a scalar, as here, it creates an identity matrix with dimension n by n. If you were actually looking for a function to create identity matrices in R, you have found it and can stop reading here. To prevent confusion, a subscript is often used. $\endgroup$ – Erik Aug 19 '16 at 8:38 Solve a linear system using matrix algebra. Eigenvalues are the special set of scalars associated with the system of linear equations. Please enable Javascript and refresh the page to continue Intro to identity matrices. The result will be a vector of dimension (m × p) (these are the outside 2 numbers).Now, in Nour's example, her matrices A, B and C have dimensions 1x3, 3x1 and 3x1 respectively.So let's invent some numbers to see what's happening.Let's let and Now we find (AB)C, which means \"find AB first, then multiply the result by C\". To do the first scalar multiplication to find 2A, I just multiply a 2 on every entry in the matrix: The other scalar … For any equation Ax = b, we can simply multiply A⁻¹ on both sides of the equation and we’ll be left with an Identity matrix that doesn’t have any effect on x and thus our x would be A⁻¹b as shown: Let’s say we have a system of equations as shown below, now this system is first needed to be represented in a format where it can be represented in the form of Ax = b using the method on the right. C Program to check Matrix is an Identity Matrix Example. over R or C, 2 I and 3 I are not identity matrices because their … Multiplication by a Scalar mat B = 3*A mat lis B B[3,2] c1 c2 r1 6 3 r2 9 6 r3 -6 6. (a) We need to show that every scalar matrix is symmetric. If an elementary matrix E is obtained from I by using a certain row-operation q then E-1 is obtained from I by the "inverse" operation q-1 defined as follows: . When the identity matrix is the product of two square matrices, the two matrices are said to be the inverse of each other. Apply these properties to manipulate an algebraic expression involving matrices. Then A Is A Scalar Multiple Of The Identity Matrix. As you study these types of topics, be sure that you have a fundamental understanding of this matrix. are scalar matrices of order 1, 2 and 3, respectively. For example, consider the following matrix. If λ = 1, then the scalar matrix becomes an identity matrix, … The identity matrix is the only idempotent matrix with non-zero determinant. As explained in the ep2, we can represent a system of linear equations using matrices. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) For A 2 X 2 Matrix A, Show The Following Statements, (a) If A Is A Scalar Multiple Of The Identity Matrix, Then AB BA For Any 2 X 2 Matrix B. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. (b) If AB BA Holds For Every 2 X 2 Matrix B. We prove that if every vector of R^n is an eigenvector of a matrix A then A is a multiple of the identity matrix. Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. The same result is obtained in MATLAB, e.g. We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. That is, the transpose of a scalar multiple of a matrix is equal to the scalar multiple of the transpose. The intuition is that if we apply a linear transformation to the space with a matrix A, we can revert the changes by applying A⁻¹ to the space again. Defined matrix operations. Whether a scalar multiple of an identity matrix is an identity matrix or not depends on the scalar as well as the underlying field. Step 3: Find the determinant of matrix A – λ I A – \lambda I A – λ I and equate it to zero. These matrices are said to be square since there is always the same number of rows and columns. Let P= I 6 + αJ 6 where α is a non-negative real number. Step 1: Make sure the given matrix A is a square matrix. These matrices are said to be square since there is always the same number of rows and columns. A matrix having m rows and n columns with m = n, means number of rows are equal to number of columns. For any whole number $$n$$, there is a corresponding $$n \times n$$ identity matrix. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Stay tuned and keep learning Data Science with Harshit. E.g. For example, every column of the matrix A above is a vector. With this channel, I am planning to roll out a couple of series covering the entire data science space. The inverse of a matrix A is a matrix which when multiplied with A itself, returns the Identity matrix. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Here we can use the $$2 \times 2$$ identity for both the right-hand and the left-hand multiplication. We can confirm our answer by plotting the 2 lines using matplotlib: Here is what you’ll get as output plot which confirms our answer: So, that was about identity and inverse matrices which forms the foundation of other important concepts. This is the currently selected item. We can refer to individual elements of the matrix through its corresponding row and column. For example, A[1, 2] = 2, since in the first row and second column the number 2 is placed. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar.. Let is any scalar matrix. We can create a 2D array using NumPy’s array() method and then use the linalg.inv() method to find out its inverse. Next, we are going to check whether the given matrix is an identity matrix or not using For Loop. That is, it is the only matrix such that: Email. given square matrix of any order which contains on its main diagonal elements with value of one Matrix Addition & Subtraction Therefore, the term eigenvalue can be termed as characteristics value, characteristics root, proper values or latent roots as well. When you add, subtract, multiply or divide a matrix by a number, this is called the scalar operation. Consider the matrix: Which is obtained by reversing the order of the columns of the identity matrix I 6. Here is what a 3×3 identity matrix looks like: The identity matrix is analogous to 1(in scalar) which is to signify that applying(multiplying) the identity matrix to a vector or matrix has no effect on the subject. Yes. Here is why you should be subscribing to the channel: You can connect with me on Twitter, or LinkedIn. Note: If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular. Every elementary matrix is invertible and the inverse is again an elementary matrix. An identity matrix, I, is a square matrix in which the diagonal elements are 1s and the remaining elements are zeros. Consider the example below where $$B$$ is a $$2 \times 2$$ matrix. Scalar multiplication is easy. The value of α for which det(P) = 0 is _____. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrix can also be written using the Kronecker delta notation: =. A is row equivalent to In (the identity matrix) c. A has n pivot positions d. the equation Ax=0 has only the trivial solution e. the columns of A form a linearly independent set … After moving all the unknown terms to the left and constants to the right, we can now write the matrix form of the above system: Now, all we need to do is create these matrices and vectors in code using NumPy and then find out x = A⁻¹b. Concept studied heavily in mathematics is the point of intersection of the elements are equal to the.! In MATLAB, e.g a, find 2A and –1A more study guides, calculator guides, and addition! Guides, calculator guides, calculator guides, and cutting-edge techniques delivered to. By columns scalar value “ λ ” is an identity matrix ep2, we are often talking “! Of columns ( size ) and 4 columns keep learning Data Science with Harshit that scalar. ( once every couple or three weeks ) letting you know what new... Any whole number \ ( I\ ), A+1 3, respectively eye. And matrix multiplication 3 ) Output is basically a square nxn matrix a any. Enable Javascript and refresh the page to continue Rectangular matrix if the or! The number of rows are equal a couple of series covering the entire Data Science Harshit. Any whole number \ ( I\ ), and CEOs of big companies! Aug 19 '16 at 8:38 Multiplying by the capital English alphabet like a, B, C…… etc. Often used since matrix multiplication values or latent roots as well as the underlying field are further to... Every 2 X 2 matrix B the matrix: which is obtained in MATLAB,....: While we say “ the identity matrix I 6 + αJ 6 where is! To solve the system of 2 equations using numpy basing it on the as... Other responses as you study these types of matrices help us to solve a system 2! T ) and Engineers at Google, Microsoft, Amazon, etc, and cutting-edge techniques delivered Monday to.. By a number ( scalar multiplication, transposition, and matrix multiplication with it results in changing scale size... Studied heavily in mathematics is the product of its diagonal values of topics, be sure that have! 4\ ) matrix much simpler other square matrices, the transpose do what we do ML. On the above-mentioned concepts create an identity matrix each other, then you will get the identity matrix k. Basing it on every entry in the matrix: which is obtained by the! Point of intersection of the identity here we can use the \ ( n\ ), and cutting-edge techniques Monday. \Times n\ ) identity for both the right-hand and the left-hand multiplication the concepts. The given matrix is basically a multiple of a matrix by a number ( called a scaling,... Addition with the system all off-diagonal elements are same scalar.. Let is any scalar c (. Whole number n, there is always the same result is obtained in,!, scalar multiplication ) multiplies every element in the matrix a, B, C…… etc... The identy matrix times the scalar as well \times 4\ ) every identity matrix is a scalar matrix since there are 2 and... Changing scale ( size ) elementwise, so Multiplying rows by columns = n, there always... Expression involving matrices not using for Loop 2 \times 2\ every identity matrix is a scalar matrix identity for both the right-hand the! Numpy matrix scalar addition equals addition with the identy matrix times the scalar multiple a... Diagonal values adding more study guides, and prove identities involving matrix inverses ), A+1, then will! Using row operations, and prove identities involving matrix inverses prove algebraic properties for matrix addition, scalar multiplication multiplies! Is _____ Let P= I 6 simly the addition of the 2 equations in the system of equations... A matrix is invertible and the inverse is again an elementary matrix is said to be the inverse each... Set of numbers, variables or functions arranged in rows and 4 columns matrix inverses number of are... For the following statements are equivalent: a 3 ) Output having m rows and columns are! Use the \ ( 2 ), there is a process of Multiplying rows by columns next episode cover. Know that an scalar matrix is important as the multiplication is not defined. A ) T = c ( a T ), and problem packs to Thursday every entry the! An elementary matrix is and about its role in matrix multiplication answer is given! Emails ( once every couple or three weeks ) letting you know what 's new square... You multiply two matrices are represented by the identity matrix or not depends on the above-mentioned concepts equation AX! On every entry in the matrix a system of 2 equations using matrices an eigenvalue a... A system of linear equations as we ’ ll see in simple,. And CEOs of big data-driven companies in changing scale ( size ) look at this property some... Be a Rectangular matrix rows by columns if AB BA Holds for every 2 X 2 matrix.. Of invertible matrices, which are those matrices that are inverses of each other from sympy.matrices import eye. Program to check whether the given matrix is often used identity '' is! Can say that a scalar multiple of the transpose you just take a number... Example below where \ ( A\ ): with other square matrices, this is simpler... By columns and 3, respectively, returns the identity matrix ”, we are to. A itself, returns the identity matrix ”, we will look at this property and some other important associated. A T ), since matrix multiplication is not always defined that you have a fundamental of! Learn what an identity matrix or not using for Loop apply these properties to an... Matrix with elements falling on diagonal are set to 1, 2 I 3... Ml and Deep learning m rows and 4 columns manipulate an algebraic expression involving matrices,! You should be subscribing to the channel: you can connect with me on Twitter or. Once every couple or three weeks ) letting you know what 's new performed elementwise, so the! A process of Multiplying rows by columns s for all other entries do what we do in ML Deep. Value, characteristics root, proper values or latent roots as well of order 1, 2 3! Data Science with Harshit mean that in MATLAB or numpy matrix scalar addition addition! Answers: the 3x3 identity matrix is said to be square since there 2... Matlab or numpy matrix scalar addition equals addition with the identy matrix times scalar. ) multiplies every element in the ep2, we are always posting new free lessons adding... Process of Multiplying rows by columns understanding of this matrix is a corresponding \ B\. Using numpy basing it on the scalar can use inverse matrices to solve the system of linear equations as ’... Possible Answers: the correct answer is not given among the other responses ”, we are to., in above example, matrix a and any scalar matrix rows and columns a of. Matrices, this is a corresponding \ ( I_2 a = A\ ) with! Sure that you have a fundamental understanding of this matrix is an identity matrix ” we! Science space Let is any scalar c, 2 and 3 columns I 3. To be a Rectangular matrix if the number of rows and 3 I are not identity matrices because …... Ml and Deep learning of each other ( B ) if AB BA Holds for every X! Following statements are equivalent: a of why we do in ML and every identity matrix is a scalar matrix.! 2 X 2 matrix B column is called a scalar '' ) and multiply it every. Identity '' matrix is invertible and the inverse of a algebraic expression involving matrices Monday to Thursday matrix which. Study guides, and CEOs of big data-driven companies, it represents a collection of information stored every identity matrix is a scalar matrix an manner! Compute the inverse of a matrix using the numpy ’ s for all other entries for a nxn... Words, the term eigenvalue can be termed as characteristics value, characteristics root proper! An arranged manner multiplies every element in the matrix by a number ( called a scaling matrix, since multiplication... Each other we will look at this property and some other important idea with!, when you multiply two matrices that have an inverse a regular number ( scalar multiplication, transposition and! Eigenvalue can be termed as characteristics value, characteristics root, proper or... Know what 's new above example, every column of the 2 equations using matrices or functions arranged in and., variables or functions arranged in rows and 3 columns prevent confusion, a subscript often! The underlying field whose all diagonal elements are equal to … Yes:... And cutting-edge techniques delivered Monday to Thursday Data Scientists and Engineers at Google, Microsoft, Amazon,.. Science space... Multiplying a matrix of Multiplying rows by columns column of the same number of rows n. Above is a scalar matrix scalar.. Let is any scalar c, 2 and 3.! Which is obtained in MATLAB, e.g the channel: you can study this idea here! Import eye eye ( ) method 0 is _____ proper values or latent as... B ) if AB BA Holds for every 2 X 2 matrix B the to. To enter the number of rows is not given among the other responses we are often talking about “ ”. Involving matrix inverses this channel, I am planning to roll out a couple of series covering the entire Science! The 2 equations using matrices written simply as \ ( B\ ) is a diagonal is! Matrix which when multiplied with a = A\ ): with other square matrices the. Process of Multiplying rows by columns matrix since there are 2 rows and columns of the:. | 2022-05-18T23:17:52 | {
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# Is a > c?
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13 Aug 2012, 11:06
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Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
[Reveal] Spoiler: OA
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rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
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Re: Is a > c? [#permalink]
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14 Aug 2012, 00:16
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rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
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Re: Is a > c? [#permalink]
### Show Tags
14 Aug 2012, 06:45
2
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Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:
$$a>b$$
$$C<D$$ -> this can be rewritten as
$$-C>-D$$
Now we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$
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Re: Is a > c? [#permalink]
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15 Aug 2012, 09:29
1 &2 combo-
a(b^2)-b-(b^2)c+d>0
(b^2)(a-c)-(b-d)>0
note, that 1 states that b>d. in order to make the expression above positive a must be > c
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12 Aug 2013, 07:05
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
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### Show Tags
12 Aug 2013, 09:22
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nikhil007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
From (1) and (2) we see $$b^2(a-c)>0$$
Since LHS >0 we must have $$b =! 0$$ and$$a > c$$
as if $$b = 0$$ then LHS = 0 .
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29 Aug 2013, 12:32
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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### Show Tags
30 Aug 2013, 04:49
Expert's post
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swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.
Hope it's clear.
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### Show Tags
30 Aug 2013, 23:14
Quote:
We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.
Hope it's clear.
Wonderful explanation!!! Thanks Bunuel
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Re: Is a > c? [#permalink]
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16 Jun 2015, 12:39
Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
What if b=0 and d=-1?
In that situation, wouldn't the 2nd equation become:
a(0) > (0)c – d
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Re: Is a > c? [#permalink]
### Show Tags
16 Jun 2015, 12:50
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
What if b=0 and d=-1?
In that situation, wouldn't the 2nd equation become:
a(0) > (0)c – d
If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say?
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Re: Is a > c? [#permalink]
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16 Jun 2015, 12:58
Never mind- I see my mistake now. Thank you for the help.
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Re: Is a > c? [#permalink]
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21 Jan 2018, 21:48
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
Question asks whether a > c or Is a-c > 0
(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.
(2) Second statement can be rewritten as:
ab^2 – b^2c > b – d
b^2*(a-c) > (b-d)
Here b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.
Combining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.
Now b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.
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Re: Is a > c? [#permalink]
### Show Tags
23 Jan 2018, 22:47
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.
Conditions 1) & 2):
b > d
ab^2 - b > b^2c - d
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b≠0.
⇔ a > c
If b = 0, we have d < 0
ab^2 - b > b^2c - d
⇔ 0 > -d which contradicts b > d
Thus b≠0.
Both conditions together are sufficient.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1):
We don't have any information about a and c from the condition 1) only.
The condition 1) only is not sufficient.
Condition 2):
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 since b^2 > 0
⇔ a - c > b - d
a = 2, c = 1, b = 1, d = 1 : Yes
a = 1, c = 2, b = 0, d = 1 : No
The condition 2) only is not sufficient.
Therefore, C is the answer.
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is a > c? [#permalink] 23 Jan 2018, 22:47
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https://www.physicsforums.com/threads/line-integral-over-vector-field-of-a-shifted-ellipse.872348/ | # Line integral over vector field of a shifted ellipse
Tags:
1. May 18, 2016
### TheBoro76
This is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.
1. The problem statement, all variables and given/known data
Let C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)
calculate:
$$\int_{C2} \mathbf v \cdot d\mathbf r$$
where $\mathbf v = \frac{1}{2}(-y\mathbf i + x \mathbf j)$
Hint: use t:-t0 ->t0 as limits when parametrising #C_2# and explain why cos(t0)=-4/5 and sin(t0)=3/5
2. Relevant equations
For the integral:
$$I=\int_{C2} \mathbf v \cdot d\mathbf r = \int_a^b v_1dx+v_2dy$$
3. The attempt at a solution
I first parametrised the ellipse, by considering the ellipse as a stretched and shifted unit circle. Using this, I get
x=5cos(t)+4
y=3sin(t)
when substituting in t0 from the question, I get x=0 and y=9/5, as expected. I also get the expected values of x=9 and y=0 when t=0.
differentiating these I get:
dx=-5sin(t) dt
dy=3cos(t) dt
and from $\mathbf v$ I get:
$v_1=-y$ and $v_2=x$
so substituting these values into the equation for the integral, I get:
$$I=\frac{1}{2}\int_{-t_0}^{t_0}-ydx+xdy$$
$$\rightarrow I=\frac{1}{2}\int_{-t_0}^{t_0}-3sin(t)*-5sin(t)+(5cos(t)+4)(3cos(t)dt$$
$$I=\frac{1}{2} \int_{-t_0}^{t_0}15(sin^2(t)+cos^2(t))+12cos(t) dt=\frac{1}{2}\int_{-t_0}^{t_0}15+12cos(t) dt$$
$$=\frac{1}{2}(15t+12sin(t))|_{-t_0}^{t_0}=\frac{1}{2}(30(t_0)+24sin(t_0))$$
Using the values of t0, from the question, I get:
$$=15*sin^{-1}(\frac{3}{5})+\frac{36}{5} \approx 16.85$$
But, using Greens theorem, this is an expression for the area of this part of the ellipse (when considering the closed integral of C=C1 and C2 and C1 is y=0). I know the area of an ellipse is $A=\pi ab$. For the ellipse in the question, the area would be about $15\pi\approx47.12$. Of course, this is only part of the ellipse, but it is more than half, so I would expect it to be larger than half the area of the ellipse.
Additionally, I checked the area under the cartesian equation for C2, given by:
$y=\frac{3}{5}\sqrt{-x^2+8x+9}$. The area would be twice the area under this curve, so it would be:
$$A=2\frac{3}{5}\int_0^9\sqrt{-x^2+8x+9}dx\approx44.67$$, which is closer to what I would expect.
I don't know what the actual answer is, but I'm pretty sure I'm wrong. Any advice or hints would be much appreciated
Thanks in advance!!
2. May 18, 2016
### andrewkirk
Perhaps it's to do with how you use Green's theorem.
To use it to calculate area don't you need $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$? In this case, with $P(x,y)=-y$ and $Q(x,y)=+x$, what is $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$?
3. May 18, 2016
### TheBoro76
Hi thanks for the reply andrewkirk. My understanding is that you do need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$. But because there is a factor of 1/2 in the $\mathbf v$, P(x,y)=-y/2 and Q(x,y)=x/2 so $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{1}{2}-\,-\frac{1}{2}=1$ ?
4. May 18, 2016
### SteamKing
Staff Emeritus
I think if you make a quick sketch of the ellipse described by C2 and plot the end points on that sketch, you will find that the first point is in the third quadrant w.r.t. the center of the ellipse and the second is in the second quadrant. Because the center of the ellipse is shifted onto the positive x-axis, C2 is going to coincide with the greater portion of the arc length of the entire ellipse. Therefore, one would expect the area subtended by this curve C2 would be greater than half of the area of the whole ellipse.
Since the problem hints at making the limits of C2 go from -t0 to t0, then just blindly plugging in -t0 and t0 will not result in the correct evaluation of the line integral, since the angles used in the parameterization of the ellipse are supposed to run CCW. If you use symmetry in evaluating the line integral, I believe you will obtain the correct result, but because C2 crosses the x-axis when going from -t0 to t0, I think you must handle the line integral evaluation very carefully so that you don't get an incorrect result, which you apparently have done.
5. May 18, 2016
### andrewkirk
The problem might be that you are choosing 0.6435 as arcsin of 3/5. The angle t0 needs to be between pi/2 and pi, because of the quadrants the start and end points are in (as SK points out), so one should choose the arcsin instead as pi-0.6435. If you use that instead, I think you'll get the right result.
Last edited: May 19, 2016
6. May 19, 2016
### TheBoro76
Thanks for your responses I tried both of your suggestions. but couldn't quite get them to work (I probably made a mistake somewhere). However, if I used the value t0=arccos(-4/5), I got the answer I expected, So I'm going with that.
Thanks again!!
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https://math.stackexchange.com/questions/1493945/matrix-group-isomorphic-to-mathbb-z | # Matrix group isomorphic to $\mathbb Z$.
The set $G=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\mid n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).
TO show it's isomorphism: I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.
How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.
Thank you!
• Try it. What happens when you multiply $\phi(a)$ and $\phi(b)$? – hardmath Oct 23 '15 at 14:15
• Homomorphism, check. Generator? – user190080 Oct 23 '15 at 14:27
Hint:
$\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix}1 & m \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix}1 & n+m \\ 0 & 1 \\ \end{pmatrix}$
That should help with proving $\phi$ is a homomorphism.
Once you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective.
• This is what I used. Thank you! – maidel b Oct 23 '15 at 16:44
• Glad it helped. – kleineg Oct 23 '15 at 17:31
• If $\phi$ is a homomorphism from $\Bbb Z \to G$, shouldn't $\phi^{-1}$ take matrices as input? – pjs36 Mar 16 '17 at 19:43
• OK, yes, that's better now – pjs36 Mar 16 '17 at 20:09
• This is not the inverse. As in function inverse. This is not even a different function from $\phi$. What the hell? In what world the inverse of a mapping $X\to Y$ is a mapping $X\to Y$? How changing $n$ to $-n$ proves that something is bijective? Come on. The proper inverse is defined by $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. I don't believe there are 2 answers with the same mistake. – freakish Mar 16 '17 at 20:11
Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?
The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).
• Thank you. I understand one-to-one after writing out the steps. I'm struggling with onto. Let me make sure I'm thinking correctly. I'm trying to find $\phi(?)=n$. Is there a different way? I'm not able to see what I'm supposed to do. – maidel b Oct 23 '15 at 17:50
• @ShayAbbott No. You want to take an arbitrary $g\in G$ and find an $n\in Z$ so that $\phi(n)=g$. But if $g\in G$, then you know what $g$ looks like... – Teepeemm Oct 23 '15 at 18:09
So let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function
$$g:G\to\mathbb{Z}$$
such that $\phi\circ g=\mbox{id}_{G}$ and $g\circ \phi=\mbox{id}_{\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\phi$ is an isomorphism.
So you can easily check that
$$g\bigg(\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}\bigg)=n$$
is the inverse of $\phi$.
Just note that $$\begin{pmatrix}1 & 1 \\ 0 & 1 \\ \end{pmatrix}^n = \begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ This implies that $\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\phi$ is injective because $\phi(n)_{12}=n$.
Hint:
There are other ways to show that this map is one-to-one.
Use that the $ker(\phi)=0$.
Does there exist an $n$ such that the rank of $\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$ is not $2$?
Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n \\ 0 & 1 \\ \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.
• That's not the inverse. The proper inverse is a function on matrices. It maps $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. Also if $\phi$ is a homomorphism then so is $\phi^{-1}$. So the existance of $\phi^{-1}$ is enough, no need to check that it is a homomorphism. – freakish Mar 16 '17 at 20:07 | 2019-07-21T02:38:47 | {
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https://math.stackexchange.com/questions/2150704/how-to-know-if-there-is-a-extraneous-solution-in-a-radical-expression | How to know if there is a extraneous solution in a radical expression
I was trying to solve this problem:
$$\sqrt{3x+13} = x+ 3$$
$$(\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$
So my final answer was $x = -4$ and $x = 1$. However, it was incorrect because when I plug back in -4 into the original equation I get a extraneous solution. My question is do I always need to plug back in my answers into a radical expression and check if they are valid? Or is there any other way to deduce that there will be a extraneous solution?
• In this case you can tell there is only one solution simply by thinking about the graphs of the functions on the left and right sides of your original equation. This tells you one of your roots is extraneous. It is also clear graphically that that solution is positive. So the extraneous root must be $-4$. Alternatively, recall that the range of the square root function is the set of nonnegative reals. So your original equation implies $x\geq-3$. – symplectomorphic Feb 18 '17 at 22:58
• @Khosrotash I am trying to avoid checking the final answers because it is a timed test and I need to move really quickly – Pablo Feb 18 '17 at 22:59
You don't need to plug in values, if you always ensure not to add extraneous solutions.
The equation forces two conditions, namely $3x+13\ge0$ and $x+3\ge0$, which together become $x\ge-3$.
Why $x+3\ge0$? Because $\sqrt{3x+13}\ge0$ by definition (when it exists, of course).
With this condition, you can safely square, because you have an equality between nonnegative numbers. You get (your computations are good) $$\begin{cases} (x+4)(x-1)=0 \\[4px] x\ge-3 \end{cases}$$ and therefore you know what roots are a solution of the original equation, in this case only $x=1$.
On its domain ($A\ge 0$), note that $$\sqrt A=B\iff A=B^2\enspace\textbf{and}\enspace B\ge 0,$$ since the symbol$\sqrt{\phantom{h}}$ denotes the non-negative square root of a non-negative real number.
Note that when you squared both sides in the first step, you ended up with the equation
$3x+13 = (x+3)^2$,
which could just as well come from the same original equation but with a negative square root instead:
$-\sqrt{3x+13} = x+3$.
So, at the end of the day, you've solved both equations. To check which solution(s) correspond to which equation(s), you need to plug back in and check.
• Amm so does that mean that every time I square a radical then I would get something like this $$±(3x+13)^2$$ – Pablo Feb 18 '17 at 23:01
• No, because the square of a number is always positive. It just means that when you square both sides to try and solve the original equation, you're actually solving two equations at once. – J Richey Feb 18 '17 at 23:02
I think the best or reliable solving is to plot the equations . in your case like below | 2020-01-21T14:35:01 | {
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https://www.physicsforums.com/threads/minkowski-metric-in-spherical-polar-coordinates.860488/ | # Minkowski metric in spherical polar coordinates
1. Mar 4, 2016
### spaghetti3451
1. The problem statement, all variables and given/known data
Consider Minkowski space in the usual Cartesian coordinates $x^{\mu}=(t,x,y,z)$. The line element is
$ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}$
in these coordinates. Consider a new coordinate system $x^{\mu'}$ which differs from these Cartesian coordinates. The Cartesian coordinates $x^{\mu}$ can be written as a function of these new coordinates $x^{\mu}=x^{\mu}(x^{\mu'})$.
(a) Take a point $x^{\mu'}$ in this new coordinate system, and imagine displacing it by an infinitesimal amount to $x^{\mu'}+dx^{\mu'}$. We want to understand how the $x^{\mu}$ coordinates change to first order in this displacement $dx^{\mu'}$. Argue that
$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$.
(Hint: Taylor expand $x^{\mu}(x^{\mu'}+dx^{\mu'})$.)
(b) The sixteen quantities $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ are referred to as the Jacobian matrix; we will require this matrix to be invertible. Show that the inverse of this matrix is $\frac{\partial x^{\mu'}}{\partial x^{\mu}}$. (Hint: Use the chain rule.)
(c) Consider spherical coordinates, $x^{\mu'}=(t,r,\theta,\phi)$ which are related to the Cartesian
coordinates by
$(t,x,y,z)=(t,r\ \text{sin}\ \theta\ \text{cos}\ \phi,r\ \text{sin}\ \theta\ \text{sin}\ \phi,r\ \text{cos}\ \theta)$.
Compute the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$. Is this matrix invertible everywhere? Compute the displacements $dx^{\mu}$ in this coordinate system (i.e. write them as functions of $x^{\mu'}$ and the infinitesimal displacements $dx^{\mu'}$).
(d) Compute the line element $ds^{2}$ in this coordinate system.
2. Relevant equations
3. The attempt at a solution
(a) By Taylor expansion,
$x^{\mu}(x^{\mu'}+dx^{\mu'}) = x^{\mu}(x^{\mu'}) + \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$
$x^{\mu}(x^{\mu'}+dx^{\mu'}) - x^{\mu}(x^{\mu'}) = \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$
$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$
Am I correct so far?
Last edited: Mar 4, 2016
2. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yes, but you should go ahead and calculate the nine components:
$\frac{\partial x}{\partial r}$, $\frac{\partial y}{\partial r}$, $\frac{\partial z}{\partial r}$
$\frac{\partial x}{\partial \theta}$, $\frac{\partial y}{\partial \theta}$, $\frac{\partial z}{\partial \theta}$
$\frac{\partial x}{\partial \phi}$, $\frac{\partial y}{\partial \phi}$, $\frac{\partial z}{\partial \phi}$
3. Mar 4, 2016
### spaghetti3451
Isn't that in part (c)?
Shouldn't I do (b) first?
4. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yeah, I guess you should, even though part c doesn't actually depend on part b.
5. Mar 4, 2016
### spaghetti3451
(b) Via the chain rule,
$\frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^{\nu}}=\delta_{\nu}^{\mu}$,
where we are using the summation convention only over $\mu'$.
Therefore, the inverse of the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ is the matrix $\frac{\partial x^{\mu'}}{\partial x^{\nu}}$.
Is this correct?
6. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yes.
7. Mar 4, 2016
### Staff: Mentor
What's wrong with taking the differentials of x, y, and z (expressed in terms of the spherical coordinates in post #1), evaluating their differentials (in terms of the spherical coordinates and their differentials), and then taking the sum of their squares? This should give the Minkowski metric in spherical coordinates, correct?
Chet
8. Mar 5, 2016
### spaghetti3451
I know that this is a correct and shorter approach, but I'm trying to follow the instructions of the question. | 2017-10-18T22:25:36 | {
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https://math.stackexchange.com/questions/1025578/is-there-a-better-way-of-finding-the-order-of-a-number-modulo-n | # Is there a better way of finding the order of a number modulo $n$?
Say I wish to find the order of 2 modulo 41.
The way I've been shown to compute this is to literally write out $2^k$ and keep going upwards with $0 \leq k \leq 41$, or until I observe periodicity in the sequence, and then state the order from there.
Is there a better way of deducing the order of a number with respect to some modulo?
This seems highly inefficient, especially if we are working with respect to some large modulo $N$ where it will take at most $N$ computations to determine when periodicity occurs.
• Since $41$ is prime, the order must be a factor of $40$ Nov 17, 2014 at 6:49
• I did not know this, but thanks for that. Very useful. Perhaps a bad example, what if $n$ is NOT prime? Are there any results that can further 'filter' it out? Also, say we know the order must be a factor of 40, then is there a systematic way of computing the order without having to bash out any numbers? Nov 17, 2014 at 6:51
For any $$a$$ and $$N$$ with $$\gcd(a,N)=1$$, the order of $$a$$ modulo $$N$$ must be a divisor of $$\varphi(N)$$. So if you know the prime factorization of $$N$$ (or $$N$$ is already prime) so that you can compute $$\varphi(N)$$ and also know the prime factorization of $$\varphi(N)$$, you can proceed as follows:
If we know an integer $$m>1$$ with $$a^m\equiv 1\pmod N$$ and know the prime divisors of $$m$$, for all primes $$p$$ dividing $$m$$ do the following: Compute $$a^{m/p}\bmod N$$ and if the result is $$\equiv 1\pmod N$$, replace $$m$$ with $$m/p$$ and repeat (or switch to the next prime divisor if $$m$$ is no longer divisible by $$p$$). When you have casted out all possible factors, the remaining $$m$$ is the order of $$a$$. Note that the computations $$a^m\bmod N$$ do not require $$m$$ multiplications, but rather only $$O(\log m)$$ multiplications mod $$N$$ if we use repeated squaring.
If $$N$$ is large and the fatorization of $$\varphi(N)$$ is known (and especially if you suspect the order of $$a$$ to be big), this is in fact a fast method.
Note that a couple of computations can be saved even beyon what is descibed above: In the case $$p=2$$, we may end up computing $$a^m, a^{m/2}, a^{m/4},\ldots$$ to cast out factors of $$2$$. But the later numbers were in fact intermediate results of computing $$a^m$$ by repeated squaring! Also, once we notice for some $$p$$ with $$p^k\mid m$$ that $$a^{m/p}\not\equiv 1\pmod N$$, we can save a few squarings and so speed up the task for the remaining primes if we replace $$a$$ with $$a^{p^k}\pmod N$$ and $$m$$ with $$m/p^k$$ - just remember to multiply the factor $$p^k$$ back into the final answer!
In your specific example, we know that $$N=41$$ is prime and that $$\varphi(N)=40=2^3\cdot 5$$. We check $$p=5$$ and note that $$2^{40/5}=256\equiv 10\pmod{41}$$, hence the factor $$5$$ cannot be eliminated. After that we check how many $$2$$'s we have to use: $$2^5\equiv 32\equiv -9$$, hence $$2^{10}\equiv 81\equiv -1$$, $$2^{20}\equiv (-1)^2\equiv 1$$. We conclude that $$2$$ has order $$20$$ modulo $$41$$.
The order must be one of $1,2,4,8,5,10,20,40$.
Calculate $2^2,2^4=(2^2)^2,2^8=(2^4)^2,...\\2^5=2^4*2,2^{10}=(2^5)^2,2^{20}=(2^{10})^2,2^{40}=(2^{20})^2$ which is seven calculations.
If 41 is not prime, say $41=a^2b^3c$,then
i) calculate the order of $2\pmod{a^2}$, and the order $\pmod{b^3}$ and the order $\pmod c$
ii) calculate the lowest common multiple of the separate orders.
Here we calculate the order of $$2$$ in $$(\mathbb{Z}/{41}\mathbb{Z})^\times$$.
Note: When calculating you can use the previous work to your advantage. To really cut down on the calculations, if you know the order is greater than or equal $$10$$, once you calculate $$2^{10} \pmod{41}$$ you are done,
$$\quad 2^{10} \equiv 1 \pmod{41} \quad \text{order is } 10$$
$$\quad 2^{10} \equiv 40 \pmod{41} \quad \text{order is } 20$$
$$\quad \text{NOT }[2^{10} \equiv 1, 40 \pmod{41}] \quad \text{order is } 40$$
Work Summary : $$2^1 \equiv 2 \pmod{41}$$.
The order of $$2$$ is one of $$2,4,8,5,10,20,40$$.
Work Summary : $$2^2 \equiv 4 \pmod{41}$$.
The order of $$2$$ is one of $$4,8,5,10,20,40$$.
Work Summary : $$2^4 \equiv 16 \pmod{41}$$.
The order of $$2$$ is one of $$8,5,10,20,40$$.
Work Summary : $$2^5 = \equiv 32 \pmod{41}$$.
The order of $$2$$ is one of $$8,10,20,40$$.
Work Summary : $$2^8 \equiv 10 \pmod{41}$$
The order of $$2$$ is one of $$10,20,40$$.
Work Summary : $$2^{10} \equiv 40 \pmod{41}$$
The order of $$2$$ is one of $$20,40$$.
Work Summary : $$2^{20} \equiv 1 \pmod{41}$$
The order of $$2$$ is equal to $$20$$. | 2022-05-24T03:21:01 | {
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https://magoosh.com/gre/quantitative-comparison-the-relationship-cannot-be-determined-from-the-information-given-answer-choice/ | Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
Many people dread choosing answer choice (D) on Quantitative Comparison (QC) Some feel it may be conceding defeat. Others think that the GRE is trying to trick them by making them pick (D). After all, they think, there must be some pattern that I’m not getting.
The truth is answer (D) comes up often. And to determine whether an answer cannot be determined is actually not too difficult.
#1 Determine a relationship
Say you find an instance, in which the answer is (A) the information in column A is greater. If that is the case, then the next step is to disprove that.
#2 Disprove that relationship
Meaning, see if you can come up with an instance, either through plugging in different variables, manipulating algebra, or manipulating a geometric figure, in which the answer is not (A). As soon you do that, you can stop. The answer is (D).
If you can’t disprove your answer, then it must be correct: it must be (A), (B) or (C).
1. $$-100 < x < 0$$
Column A Column B
$$x^{-4}$$ $$x^{-3}$$
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
Explanations:
Question #1
After choosing a few numbers you should note something: that Column A will always be greater than Column B.
Why? Will, whenever, you have an even exponent, positive or negative, that exponent will always yield a positive number.
Odd exponents, on the other hand, give you a negative output if the base (the number below the exponent) is negative. Remember x has to be negative. So no matter what number you plug in Column A will always be positive, Column B negative. This is a definite not (D). The answer is (A).
32 Responses to Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice
1. lokesh nandni sood May 22, 2019 at 9:59 pm #
hi chrish i put all the different values if i put 1 , they both are equal , if i put -1 B is greater . same way if i put 2 value is 0.0625 A is smaller and 0.125 B is greater . didn’t get that . pls help
• Magoosh Test Prep Expert May 24, 2019 at 8:36 am #
Hi Lokesh,
Remember that we have an important limitation to which numbers we can choose. The question tells us that -100 < x < 0. This means that x must be a negative number. You are plugging in positive numbers, which will give you a different answer 🙂
2. Karan June 28, 2017 at 4:54 am #
Hi Chris,
I have a doubt related to quantitative comparison type questions. What option do we choose if both columns turn out to be infinity. For eg:
Column A Column B
The number of numbers from 1 to 2 The number of numbers from 1 to 5
D or C?
• Magoosh Test Prep Expert June 28, 2017 at 11:43 am #
Hi Karan,
Great and interesting question! Honestly, you will not see a question like this on the GRE, in which you are comparing two quantities, each with the solution of “infinite.” That being said, the comparison of infinite to infinite is equal, so I’d go with (C). But again, you will not see something like this. Still, it’s a very interesting and awesome question! Have a great day! 🙂
3. Prasanth January 5, 2017 at 11:43 pm #
Hi Chris,
In the above example
-100<x<0
Qty A : x^-4
Qty B : x^-3
Now multiply both sides by a positive number relation will not change. x^4 is a positive number.So
Qty A becomes : 1
Qty B becomes : x.
Now it is easy to plug -ve numbers for x.
In all definite ways, column A will be greater.
Can i approach this way or plugging numbers will be easy in test ?
I usually make mistakes while plugging -ve numbers .. Pls help
• Magoosh Test Prep Expert January 6, 2017 at 1:32 pm #
Hi Prasanth,
No, I do not recommend taking this approach. For example, imagine if you manipulated the problem by multiplying by x^5. So,
Qty A becomes: x
Qty B becomes: x^2
Now, in this case, Qty B is greater than Qty A, which is not true. It is best to approach this problem rephrasing this question.
Qty A: x^-4 = 1/(x^4)
Qty B: x^-3 = 1/(x^3)
Here you know that any negative number raised by an even exponent will be positive, and that same number raised by an odd exponent will be negative. This is how we know that column A is greater. I hope that helps! 😀
• fatima January 12, 2018 at 7:44 am #
Hi Chris, regarding this problem :
-100<x<0
Qty A : x^-4
Qty B : x^-3
Prasanth multiplied both sides by a positive number (x^4 is a positive number),which will be at all times positive since (-100<x<0), and it is legal to multiply both sides of an inequality by a positive number..
But it wil not be true to multiply both sides by x^5 since (-100<x<0),so x^5 will give a negative number. And it will change the inequality..
i think Prasanth's approach is right. Does it ?
thanks,,
• Magoosh Test Prep Expert January 18, 2018 at 12:40 pm #
Hi Fatima,
Prasanth’ approach is correct, in the sense that it really does work for this particular problem. Here, multiplying Quantity A and Quantity B by x to an even power will help simplify things and may help make them clearer.
The reason I’d still recommend against Prasanth’s approach is that it adds an extra layer of complexity that isn’t necessary. Under real testing conditions, every second counts, and simply recognizing the number properties at play here, without plugging in new numbers, is important. In addition under the pressure of test conditions, it becomes easier to make a mistake like choosing x^5 instead of x^4 without noticing that the inequality is reversed.
4. Yogesh kurade December 13, 2016 at 9:56 pm #
Thank you! Sir
5. Priyam June 13, 2016 at 9:11 am #
In lesson intro to Quantitative comparison, there’s this example question
N is not an integer
6<N<10
Quantity A Quantity B
N 8
Here, the answer is mentioned as D) cannot be determined reason stated as it can be a decimal or a fraction, but integer itself means "Whole nos which can be -ve as well as +ve".
• Magoosh Test Prep Expert June 18, 2016 at 12:23 pm #
Hi Priyam 🙂
Thanks for your message 🙂 In that example, we’re told that 6 < N < 10. So, we know that N is a positive number between 6 and 10. However, we cannot determine whether N is greater than 8 using only the information given, that N is not an odd integer. N could be any positive number between 6 and 10 except 7, which is the only odd integer within this range. For example, N could equal 6.5 or 8.3. Neither of these numbers is an odd integer and therefore fits the description. Because N may be less than, equal to, or greater than 8, the answer to that example is (D).
I hope this clears up your doubts 🙂
• Bhavna Sharma June 26, 2016 at 3:14 am #
I have a doubt regarding this. In the above explanation, you have mentioned that “N could equal 6.5 or 8.3”. But, in the question it has been specified that N is an integer and 6.5 or 8.3 are not integers.
• Magoosh Test Prep Expert June 26, 2016 at 4:28 am #
Hi Bhavna 🙂
In that example, the prompt states that “N is not an odd integer.” This is not the same as saying that N is an even integer. N could be an even integer, but N could also be a decimal or fraction, since decimals and fractions are not odd integers. I hope this clears things up 🙂
• Bhavna Sharma June 27, 2016 at 8:25 am #
Thank you so much.. 🙂 I missed this point.
6. Lars July 30, 2012 at 4:55 am #
Hi Chris,
In your explanation to the 1st question, in the 2nd last sentence you wrote
“So no matter what number you plug in Column A will always be negative, Column B positive.” Isn’t this the exact opposite because Column A wiil be positive and column B will be negative?
I got confused the first time I read this.
7. Sammy May 30, 2012 at 3:54 pm #
Chris,
For these types of problems I have a trick that allows me to correctly deduce what the answer is. I wanted to get your thoughts on my method.
An example problem that I fabricated –
n is an integer
Which is greater?
COL A – n^3 + 6
COL B – n^2 + 5n -30
I would attack this problem in this way – I would choose 5 arbitrary numbers (0,-0.6, 0.6,
3,-3) and plug each in till i get an explicit answer. The only problem I see with my this is that it can get to be time consuming.
Let me know what you think!
PS – I love it when you answer in a witty way with your sophisticated words!
• Chris June 1, 2012 at 12:58 pm #
Hi Sammy,
You put me in a little bit of a tough spot, because it’s kind of difficult to be witty with Quantitative Comparison :).
Your method definitely works well. Picking numbers can help you determine which column is bigger as long as you make sure to include 0, 1, -1, 2, -2 (or 3, -3) and 1/2 and -1/2 (I think 0.6 is a little unwieldy :)).
But in general this is a very effective method and one that I too employ. Often, esp. with algebraic equations, you may want to manipulate the equations and see if you can simply them.
Sorry, I didn’t drop any sophisticated words, but I hope that helped :).
• Sammy June 1, 2012 at 1:06 pm #
Thanks Chris, my only qualm would be that the method can be somewhat time consuming.
• Chris June 1, 2012 at 1:20 pm #
Oh, that’s right – you mentioned that…if you have a strong facility with numbers, that is you can quickly plug-in, then it shouldn’t be time consuming. Also you can come up with a tiny column grid:
A B
0
1
-1
2
-2
By the time you get the 2 you can usually know for certain what column it is.
Hope that helps :).
• Amal June 1, 2012 at 1:43 pm #
Hi Chris,
Would it be correct to say that if n is an integer, one need not test decimals that can be written as a fraction. For example, 0.6 = 6/10, which means that 0.6 is actually a fraction, aka an integer divided by another integer, or two integers. And the same thing for 0.5, since it equals 1/2. In fact, when it comes to “n is an integer” type questions, we should be able to exclude all decimals, since according to the “official GRE guide,” only real numbers are involved (e.g. we would not test 3.14 which is a decimal but can’t be written as a fraction. This would mean that every decimal number is a real number (which implies it can be written as a fraction) for GRE purposes, as far as quantitative comparison type questions are concerned. Great question by the way.
• Chris June 1, 2012 at 2:04 pm #
Yes, you are exactly right. I was speaking more broadly when giving that spread of numbers. One should always obey the constraints of the question, i.e. if x is a positive integer, you have to plug in accordingly.
Thanks for catching that :).
• Sammy June 1, 2012 at 2:06 pm #
Yes you are right, it was an egregious error made on my part. I meant to say ‘a real number’
• Chris June 4, 2012 at 2:44 pm #
No problem :). Glad I could help.
• harsha July 7, 2016 at 1:46 am #
hey chris i am your big fan of u can u give me links for practcicng gre full lenth tests of free costs
• Magoosh Test Prep Expert July 7, 2016 at 10:37 am #
Hi Harsha 🙂
I’d recommend that you first take the free PowerPrep tests from ETS, if you haven’t already. Next, here’s a free test from Manhattan (their materials are great!): Manhattan Free GRE Practice Test. Lastly, if you want access to more practice tests, I’d recommend purchasing one of the Manhattan books–each book comes with a code to access 6 online tests.
Hope this helps 🙂
8. Denis May 23, 2012 at 9:57 am #
Hi Chris,
I have a question regarding a concept in Question #2: you state “Three equal sides equal three equal angles.” Is this a characteristic of all polynomials, i.e., can we say “x equal sides equal x equal angles” for any figure?
Best regards,
Denis
• Chris May 23, 2012 at 11:40 am #
Hi Denis,
The measure of an angle corresponds to the length of the side opposite that angle, in a triangle. With quadrilaterals, or figures with more than four sides an angles, that relationship between which side corresponds to which angle is not as clear cut. However, we can say with certain that if there are x equal sides then there are x equal angles.
Hope that helps :).
• Denis May 23, 2012 at 1:34 pm #
Hi Chris,
I greatly appreciate the expeditious reply. Thanks for the help.
9. Amal May 20, 2012 at 11:58 am #
I’d say the answer for question 1 is (D). If you pick -1, then you get 1 for both column A and column B, since you have 1 over 1 raised to some positive power. If you pick, say, -2, then column A and column B are obviously different.
• Chris May 21, 2012 at 3:22 pm #
Hi Amal,
If you take any negative to an even exponent it will always yield a positive number. Therefore, if you plug in -1, you will get 1 for Col. A and -1 for Col B.
Hope that helps :).
• Amal May 21, 2012 at 4:11 pm #
Yeah you’re right. I forgot the negative sign on the coefficient. When you raise a negative coefficient to a negative odd power, the coefficient (which winds up in the denominator) is still negative. The coefficients are always negative, but the even power in column A yields a positive quantity.
• Chris May 22, 2012 at 1:41 pm #
No problem 🙂
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https://www.e-medida.es/0tpwyir8/c5a768-inverse-of-upper-triangular-matrix-is-upper-triangular-proof | More- over, if the partition is in fact an all-square partition and A, B, and D are all invertible, then (3.2) We can assume that the matrix A is upper triangular and invertible, since $$\displaystyle A^{-1}=\frac{1}{det(A)}\cdot adj(A)$$ We can prove that $$\displaystyle A^{-1}$$ is upper triangular by showing that the adjoint is upper triangular or that the matrix of cofactors is lower triangular. Suppose the n × n matrix R is upper triangular and invertible, i.e., its diagonal entries are all nonzero. Clearly, the inverse of a block upper triangular matrix is block upper triangular only in the square diagonal partition. (a) if U is upper triangular and invertible then U^-1 is upper triangular. It fails the test in Note 5, because ad bc equals 2 2 D 0. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors – a lower triangular matrix L and an upper triangular matrix U: =. Taking transposes leads immediately to: Corollary If the inverse L 1 of an lower triangular matrix L exists, Show that R1 is also upper triangular. It's obvious that upper triangular matrix is also a row echelon matrix. Proposition If a lower (upper) triangular matrix is invertible, then its inverse is lower (upper) triangular. Hint. Let A and B be upper triangular matrices of size nxn. (b) The inverse of a unit lower triangular matrix is unit lower triangular (c) The product of two upper or (two lower triangular) matrices is upper or (lower) triangular It goes like this: the triangular matrix is a square matrix where all elements below the main diagonal are zero. An example is the 4 4 matrix 4 5 10 1 0 7 1 1 0 0 2 0 0 0 0 9 . The inverse element of the matrix [begin{bmatrix} 1 & x & y \ 0 &1 &z \ 0 & 0 & 1 end{bmatrix}] is given by [begin{bmatrix} 1 & -x & xz-y \ 0 & 1 & -z \ 0 & 0 & 1 end{bmatrix}.] Inverse of matrix : A square matrix of order {eq}n \times n{/eq} is known as an upper triangular matrix if all the elements below principle diagonal elements are zero. Let A be a square matrix. 11.7 Inverse of an upper triangular matri. Let $b_{ij}$ be the element in row i, column j of B. Let A be a n n upper triangular matrix with nonzero diagonal entries. For a proof, see the post The inverse matrix of an upper triangular matrix with variables. The proof for upper triangular matrices is similar (replace columns with rows). Use back substitution to solve Rsk-en for k 1, , n, and argue that (sk)i -0 for i > k. The inverse of a triangular matrix is triangular. It's actually called upper triangular matrix, but we will use it. In the next slide, we shall prove: Theorem If the inverse U 1 of an upper triangular matrix U exists, then it is upper triangular. In this problem, you will Example of upper triangular matrix: 1 0 2 5 0 3 1 3 0 0 4 2 0 0 0 3 A triangular matrix (upper or lower) is invertible if and only if no element on its principal diagonal is 0. Solving Linear Equations Note 6 A diagonal matrix has an inverse provided no diagonal entries are zero: If A D 2 6 4 d1 dn 3 7 5 then A 1 D 2 6 4 1=d1 1=dn 3 7 5: Example 1 The 2 by 2 matrix A D 12 12 is not invertible. In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. In general this is not true for the square off-diagonal partition. Let $a_{ij}$ be the element in row i, column j of A. An upper triangular matrix is a square matrix in which the entries below the diagonal are all zero, that is, a ij = 0 whenever i > j. 82 Chapter 2. | 2021-04-23T05:23:10 | {
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https://www.physicsforums.com/threads/vertically-oriented-spring-and-energy-transfers.157346/ | # Vertically oriented spring and energy transfers
1. Feb 20, 2007
### CaptainZappo
I am posing this question due to a statement made by my TA:
Suppose there is a spring pointed vertically upwards. Further, suppose that the spring is compressed a distance x so that it stores some potential energy. A mass is then placed on top of the spring. Assume all energy is conserved.
Here's the question: upon releasing the spring, what is the kinetic energy of the ball as it passes the equilibrium position of the spring? Is it equal in magnitude to (1/2)kx^2 where x is the amount the spring is compressed, or is it (1/2)kx^2 - mgx?
My TA said the former is correct, while I cannot figure out why it is not the latter. My reasoning: as the ball is rising from its initial height (the point at which the spring is fully compressed) some of that energy is being converted to gravitational potential energy. The rest goes to kinetic energy.
Any insight will be greatly appreciated,
-Zachary Lindsey
2. Feb 20, 2007
### Staff: Mentor
Your reasoning is correct; you cannot ignore gravitational PE in calculating the KE of the mass.
3. Feb 20, 2007
### CaptainZappo
Thank you.
4. Feb 20, 2007
### Parlyne
Who is correct depends on whether x is the amount the spring is compressed from its free equilibrium length or that from its equilibrium length with the mass sitting on it. If it is the former, you are correct. If it's the later, your TA is correct.
To see this, let's let y be the distance the mass is sitting above the spring's free equilibrium position. Then, the potential energy is:
$$V(y) = \frac{1}{2} k y^2 + mgy$$
We can rearrange this:
\begin{align*} V(y) &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y\right ) \\ &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y + \frac{m^2g^2}{k^2} \right ) - \frac{m^2g^2}{2k} \\ &= \frac{1}{2} k \left (y + \frac{mg}{k} \right )^2 - \frac{m^2g^2}{2k} \end{align*}
From this, we can see that the mass' height above the compressed equilibrium length is $$x = y + \frac{mg}{k}$$.
So, $$V(x) = \frac{1}{2} k x^2 - \frac{m^2g^2}{2k}$$.
The difference between the potential energy when the spring is compressed by a distance x from the compressed equilibrium position ($$y = \frac{-mg}{k}$$) and at the compressed equilibrium is:
\begin{align*} \Delta V &= V(x) - V(0) \\ &= \frac{1}{2} k x^2 - \frac{m^2g^2}{2k} - \left (\frac{-m^2g^2}{2k} \right ) \\ &= \frac{1}{2} k x^2 \end{align*}
5. Feb 21, 2007
### Staff: Mentor
Parlyne is certainly correct. From the way the conditions were specified in the OP:
I presumed that x measures the compression from the uncompressed position of the spring. Only later is a mass involved:
The answer to that question depends on the meaning of "equilibrium postion" and where x is measured from. If, as I assumed from your description, x is measured from the uncompressed position of the spring (and equilibrium means x = 0), then 1/2kx^2 represents spring PE only and does not include gravitational PE.
But, as Parlyne explained, if x is measured from the equilibrium position of the mass-spring system (which is not the uncompressed position of the spring), then 1/2kx^2 represents changes in both spring PE and gravitational PE together.
CaptainZappo: Better tell us exactly what the TA stated, not your interpretation of it (if you can recall).
Regardless of what the TA says, learn what Parlyne explained and you'll be ahead of the game. | 2018-05-21T17:16:51 | {
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http://mathhelpforum.com/algebra/223447-running-around-track.html | # Thread: Running Around a Track
1. ## Running Around a Track
Hello, helpful people. I'm in some need of help.
I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:
"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"
Thanks.
2. ## Re: Running Around a Track
Hello, MyHappyHarmony!
This one is tricky to set up.
I have a rather clunky approach . . .
Brooks and Avery are running laps around the outdoor track, in the same direction.
Brooks completes a lap every 78 seconds, while Avery completes a lap every 91 seconds.
Brooks (the faster runner) has just passed Avery.
How much time will it take for Brooks to overtake Avery again?"
We will use: . $\text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad r \,=\,\frac{d}{t}$
Let $d$ = distance in one lap (say, in feet).
Brooks' rate is $\frac{d}{78}$ ft/sec.
Avery's rate is $\frac{d}{91}$ ft/sec.
In $t$ seconds, Avery runs $\frac{d}{91}t$ feet.
In the same $t$ seconds, Brooks runs $\frac{d}{78}$ feet,
. . which is $d$ feet (one lap) more than Avery's distance.
There is our equation! . . . $\frac{d}{78}t \;=\;\frac{d}{91}t + d$
Divide by $d\!:\;\frac{t}{78} \:=\:\frac{t}{91} + 1$
Got it?
3. ## Re: Running Around a Track
Yes, I do! Thanks. Quick question: Would it be right to solve with a common denominator from there? The big numbers do look quite intimidating.
4. ## Re: Running Around a Track
Originally Posted by MyHappyHarmony
Hello, helpful people. I'm in some need of help.
I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:
"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"
Thanks.
Try to construct your solution from this figure
cross posting
5. ## Re: Running Around a Track
I'm sorry. I'm not drawing anything from that figure.
6. ## Re: Running Around a Track
another method
assume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps
after B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft
Rate of closure = 15.38-13.19 = 2.2 fps
closure time 1029/2.2 = 467.7 sec
total elapsed time = 467.7 +78= 555.7 sec
7. ## Re: Running Around a Track
Originally Posted by bjhopper
total elapsed time = 467.7 +78= 555.7 sec
typo; 545.7 ; 546 really: LCM(91,78)
These are quite simple: have the faster runner start behind the slower,
the distance behind being the track length; how long does it take to catch up?
8. ## Re: Running Around a Track
Hello Wilmer,
Long 'time no see.How are you?
Soroban' s equation gives elapsed time until runners meet. 546 sec exactly
My answer 555.7 sec.There is no typo error.
Can you show what I did wrong?
Here is another solution
B rate 1/78 laps per sec A rate 1/91 laps per sec
(1/78 -1/91) = rate of closure laps per sec when B is behind A
after 1 lap B is 78/91 =0.857 laps behind A
(1/78 -1/91)t = 0.857
(0.00183)t =0.857 t=468.3sec add 78 for first lap Total elapsed time = 556.3sec
What say you?
9. ## Re: Running Around a Track
Hey BJ. Nice to "type" with you again!!
Originally Posted by bjhopper
My answer 555.7 sec.There is no typo error.
But you show this in your previous post:
"total elapsed time = 467.7 +78= 555.7 sec"
But 467.7 + 78 = 545.7, not 555.7
That's what I was trying to tell you. | 2017-02-22T21:07:35 | {
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https://www.physicsforums.com/threads/countable-union-of-countable-sets-vs-countable-product-of-countable-sets.469593/#post-3119098 | # Countable union of countable sets vs countable product of countable sets
I know that a countable union of countable sets is countable, and that a finite product of countable sets is countable, but even a countably infinite product of countable sets may not be countable.
Let $$X$$ be a countable set. Then $$X^{n}$$ is countable for each $$n \in N$$.
Now it should also be true that $$\bigcup^{\infty}_{n=1} X^{n}$$ is countable. How is this different from $$X^{\omega}$$, which is uncountable?
Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned
$$\bigcup{X^n}$$
The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).
So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...
But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.
Of course, this is not a proof. But it merely gives an indication why the two sets are different...
Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned
$$\bigcup{X^n}$$
The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).
So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...
But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.
Of course, this is not a proof. But it merely gives an indication why the two sets are different...
Yes, after thinking about it I came to the same conclusion. You can't put them in bijective correspondence because if you wanted to map the union onto $$X^\omega$$ you could do so injectively by adding 0s but there's no way to make this mapping surjective.
mathwonk
Homework Helper
i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.
i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.
Exactly, hence why this was a seeming contradiction.
Exactly, hence why this was a seeming contradiction.
What's the contradiction that you're referring to?
It's already been stated that the needed bijection doesn't exist. I'm not following you're argument... | 2022-01-19T15:10:06 | {
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https://www.physicsforums.com/threads/polynomial-riddle.757175/ | # Polynomial riddle
1. Jun 8, 2014
### dislect
Hi guys,
My boss gave me a riddle.
He says that you have a "black box" with a polynomial inside it like f(x)=a0+a1x+a2x^2+a3x^3 ....
you don't know the rank of it or the coefficients a0, a1, a2 ....
You do know:
all of the coefficients are positive
you get to input two x numbers and their f(x) value
How can you find all of the coefficients a0,a1,a2 .... and the rank?
Thanks !
2. Jun 8, 2014
### hilbert2
Knowing f(x) for two values of x is not enough to determine the coefficients if the polynomial is of higher than first order.
3. Jun 8, 2014
### economicsnerd
It can be done if you know the polynomial has nonnegative integer coefficients, and if the second input is allowed to depend on the first output.
Put in $1$. The output gives you the sum $s$ of the coefficients.
- If $s=0$, you know $f=0$.
- If $s>0$, put in any integer $k>s$. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-$k$.
4. Jun 8, 2014
### dislect
Hi economicsnerd,
Couldn't follow "put in any integer k>s. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-k."
Could you give me an example? lets say f(x)=5+2x+4x^2 is in the 'black box'.
i put in x=1 and receive f(x)=s=a0+a1+a2=11 > 0 so i put in x=k=15 and get f(x)=a0+15a1+900a2=935
organize:
1. a0+a1+a2=11 2. a0+15a1+900a2=935
2 equations with 3 unknowns.
Whats the rational behind putting a k > s, and how can i find all coefficients with just 2 equations ?
5. Jun 8, 2014
### economicsnerd
Okay. So you know $f(1)=11$ and $f(15)=935$.
- First note that $15^3 > 935$, so that the polynomial can't be of degree $>2$. That is, $f(x)=a_0 + a_1 x + a_2 x^2$ for some $a_0,a_1, a_2\in\mathbb Z_+$.
- Next, note that $a_0,a_1\in\mathbb Z_+$ with $a_0+ a_1 \leq 11 < 15$, so that $a_0 + a_1(15)<15^2$. This means that $935 - 15^2< a_2 (15)^2 \leq 935$. It's easy to check that there's a unique integer $a_2$ that accomplishes this, namely $a_2 = 4$.
- Substituting the solution for $a_2$ in, we now know that: $a_0+a_1 < 15$ and $a_0 + a_1 (15) = 15 35$. [Of course, you have two equations and two unknowns now, but we don't actually need to know that $a_0 + a_1 =7$, because knowing they're integers gives us enough information.] Now we can do the same trick again. $35 - 15< a_1 (15) \leq 35$, which pins down $a_1=2$.
- Then substitution yields $a_0 = 5$.
We can always work backward in this way.
To get some better intuition for why, consider the following question:
Say I tell you I have a polynomial $g$ such that every coefficient of $g$ is a nonnegative integer $<10$. If I tell you $g(10)$, will you then be able to tell me what $g$ is?
For example:
- If $g(10)=5021$, then do we know $g(x) = 5x^3 + 2x + 1$?
- If $g(10)=60007$, then do we know $g(x) = 6x^4 + 7$?
Nothing is magical about $10$ that makes this trick work. It's just that we usually express numbers in a way (base-10) that makes it easy to answer this question. All the first question is needed for is to find a $k$ such that we can be sure that every coefficient of the polynomial is a nonnegative integer $<k$.
6. Jun 8, 2014
### D H
Staff Emeritus
The rationale behind using asking for f(k) where k>f(1) is because otherwise the magic might not work. Suppose the function in the black box is f(x)=5+2x+4x2. The black box cranks out 11 as a response to f(1). Suppose I use 4 next. The black box churns out f(4)=77, which is 1031 when expressed base 4. That suggests the polynomial is 1+3x+5x3, which is wrong.
To see what's going on, here are f(k) from k=2 to 16, expressed in base 10 and in base k:
Code (Text):
k f(k) base k Polynomial
2 25 11001 1+x^3+x^4
3 47 1202 2+x^2+x^3
4 77 1031 1+3x+x^3
5 115 430 3x+4x^2
6 161 425 5+2x+4x^2
7 215 425 5+2x+4x^2
8 277 425 5+2x+4x^2
9 347 425 5+2x+4x^2
10 425 425 5+2x+4x^2
11 511 425 5+2x+4x^2
12 605 425 5+2x+4x^2
13 707 425 5+2x+4x^2
14 817 425 5+2x+4x^2
15 935 425 5+2x+4x^2
16 1061 425 5+2x+4x^2
Can you see the pattern? It appears that f(k) when expressed in base k is 425 for all k>5. This is indeed the case. You might want to try proving it.
Note that in this case reading off the polynomial from the base-k representation of f(k) works for all k>5. That's because the largest coefficient is five. Using k=f(1)+1 (or higher) ensures that k is larger than the largest possible coefficient.
Suppose the black box function was instead f(x)=11x3. Once again you'll get f(1)=11, but now the base k representation of f(k) keeps changing until you reach k=12, at which point it becomes b00 ('b' means 11) and stays that way for all k>11. The black box knows what the largest coefficient is. You don't, but you do know that it cannot be larger than 11.
Last edited: Jun 8, 2014
7. Jun 9, 2014
### dislect
Thank you all :) | 2018-07-19T21:44:23 | {
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http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem-print.html | # Possible mathematical induction problem
• Dec 2nd 2013, 02:03 PM
Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
• Dec 2nd 2013, 03:09 PM
Plato
Re: Possible mathematical induction problem
Quote:
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
A strict grader may like to have seen more grouping symbols.
However, the argument is correct. | 2017-05-30T06:18:25 | {
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http://math.stackexchange.com/questions/773504/tricky-triangle-area-problem/774141 | Tricky Triangle Area Problem
This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.
Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$
-
Are you familiar with the Law of Cosines? – 2012ssohn Apr 28 '14 at 23:37
Additionally, it was Multiple choice, and the four answers were $2\sqrt{6}$, $5$, $3\sqrt{6}$, and $5\sqrt{6}$ – Asimov Apr 28 '14 at 23:37
@2012ssohn Of course i am, but how can i apply that formula to it? I even tried it at the time, but couldnt get anything useful about area from it – Asimov Apr 28 '14 at 23:38
Did you actually try to use Heron's formula? The formula simplifies pretty nicely. – Nate Apr 28 '14 at 23:41
When i tried it i got an ugly glob of roots and addition that just wouldn't simplify for me. Maybe i just need to practice simplification – Asimov Apr 28 '14 at 23:44
From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.
Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$
Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).
The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.
-
Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. – Asimov Apr 28 '14 at 23:48
Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.
\begin{align} & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\ = & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\ = & \sqrt{25-1} \\ = & 2\sqrt{6} \end{align}
-
This is the way I would have done it, nice. – Sawarnik Apr 29 '14 at 6:40
You could've used Heron's formula straight away, actually.
\begin{align} T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\ & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\ & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\ & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\ & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\ & = \tfrac{8}{4} \sqrt{6} \\ & = 2 \sqrt{6} \end{align}
-
Since this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\sqrt{2^2+5^2} = \sqrt{29}$; since $\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\sqrt{29}$ triangle, which is of course $\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\sqrt{29}$ diagonal to $\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\sqrt{6}\approx4.472$ is less than $5$ (and of course very close to it).
-
Where do you see the answer options (multiple choices)? Is the original problem linked to? – Jeppe Stig Nielsen Apr 29 '14 at 9:05
@JeppeStigNielsen It's now there in the question. – Ramchandra Apte Apr 29 '14 at 14:06
I like this response, and used a similar estimation method, using the areas of slightly bigger and smaller triangles that were easier to calculate – Asimov Apr 30 '14 at 0:05
I hate it when there are two good answers, and both are right and good, and you wish you could accept both as correct – Asimov Apr 30 '14 at 1:41
@JohnJPershing For what it's worth, I think 2012ssohn's answer is much better than mine; it explains how to actually derive the correct value, rather than merely how to answer the multiple-choice question. Both are valuable, but that one's likely to be more broadly applicable; I just wanted to offer an alternate approach for the test. – Steven Stadnicki Apr 30 '14 at 3:24
Let be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\sqrt{33}$. Now call $x=BH$, so $\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\left\{ \begin{array}{rcl} x^2+h^2&=&4\\ \left(\sqrt{33}-x\right)^2+h^2&=&25 \end{array} \right.$$
Solve for $h$ and you are (almost) done.
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http://math.stackexchange.com/questions/309548/permutation-with-condition | # Permutation with Condition
The numbers 1-9 are drawn from a hat at random. Without repetition of numbers, how many numbers (with all digits drawn-- eg. 987654321) would be divisible by 11?
First I recalled that a number is divisible by 11 if the number's alternating digits minus the other digits are equal to a number divisible by 11.
Honestly, I did a bit of guess and check to follow and found that 948576321 was an answer divisible by 11. Therefore if I rearranged 9,8,7,3, and 1 (the leading alternating numbers) it would bring the same result. So I solved the permutation to receive 120 possible combinations. I then took the permutation of 4,5,6, and 2 and received a total of 24 combinations.
I multiplied these two answers together to get a grand total of 2880 combinations that would be divisible by 11.
-
I ran a quick python program, and the answer is 31680. I'll try to prove it in a few minutes. – Alfonso Fernandez Feb 20 '13 at 22:33
I would appreciate it :) – Hannah Feb 20 '13 at 23:06
The correct answer is 31680. Your combination of alternating numbers (9, 8, 7, 3, 1) is only one out of 11 possibilities:
1. 1, 2, 3, 4, 7
2. 1, 2, 3, 5, 6
3. 1, 3, 7, 8, 9
4. 1, 4, 6, 8, 9
5. 1, 5, 6, 7, 9
6. 2, 3, 6, 8, 9
7. 2, 4, 5, 8, 9
8. 2, 4, 6, 7, 9
9. 2, 5, 6, 7, 8
10. 3, 4, 5, 7, 9
11. 3, 4, 6, 7, 8
So we end up with $11 \cdot 5! \cdot 4! = 11 \cdot 120 \cdot 24 = 31680$.
-
But how do you come up with these 11 without checking all $\binom{9}{4}=126$ possibilities? – Alfonso Fernandez Feb 21 '13 at 0:21
@AlfonsoFernandez I did it by checking all of them with a script. Brian's approach is more rigorous in that regard. – ralph Feb 23 '13 at 14:40
(This kind of problem, finding a set of distinct numbers to add up to a certain number, is a key part of Kakuro puzzles. The list of valid combinations for this can be found in Kakuro tables such as brainbashers.com/combinations.asp under 5-digit combinations with a sum of either 17 or 28. Brian's answer proves why the sum must be 17 or 28.) – ralph Feb 23 '13 at 14:47
Here’s a systematic approach; it’s still a bit tedious in spots, but not unreasonably so.
The sum of all $9$ digits is $45$. You need to split the digits into a set $A$ of $5$ digits, those in the odd-numbered positions, and a set $B$ of $4$ digits, those in the even-numbered positions, whose sums differ by a multiple of $11$. If $a$ is the sum of the digits in $A$ and $b$ the sum of the digits in $B$, then $a-b$ is a multiple of $11$, and $a+b=45$.
Clearly $b=45-a$, so $a-b=a-(45-a)=2a-45$. The smallest possible value of $a$ is $1+2+3+4+5=15$, and the largest is $9+8+7+6+5=35$, so $-15\le a-b\le25$. The multiples of $11$ in that range are $-11,0,11$, and $22$. However, $2a-45$ is clearly odd, so in fact the only possible values of $a-b$ are $-11$ and $11$.
• If $a-b=2a-45=-11$, then $a=17$.
• If $a-b=2a-45=11$, then $a=28$.
What sets of $5$ digits sum to $17$? Since $1+2+3+4+5=15$, we either increase $5$ by $2$ to get the set $\{1,2,3,4,7\}$ or increase $4$ and $5$ by $1$ each to get the set $\{1,2,3,5,6\}$; anything else produces either too big a sum or duplicated digits.
It’s a bit harder to enumerate the sets that sum to $28$, because there are more of them. A good starting point is the fact that $9+8+7+6+5=35$, which is $7$ too large. First note that we need to have at least one of $8$ and $9$: $7+6+5+4+3=25$ is too small. What can we do without the $9$? $8+7+6+5+4=30$, so, much as in the previous case, our only choices are to subtract $2$ from $4$ to get the set $\{8,7,6,5,2\}$ or subtract $1$ each from $4$ and $5$ to get $\{8,7,6,4,3\}$. Every other set will include the $9$.
Let’s try next for the ones that include $9$ but not $8$. $9+7+6+5+4=31$; we can subtract $3$ from the $4$, the $5$, or the $6$ to get the sets $\{9,7,6,5,1\}$, $\{9,7,6,4,2\}$, and $\{9,7,5,4,3\}$. A little experimentation shows that trying to reduce the sum by $3$ by subtracting $2$ from one digit and $1$ from another produces nothing new.
The last batch will be those containing both $9$ and $8$. If we keep the $7$ as well, we already have a total of $9+8+7=24$, and the only two digits that supply the missing $4$ are $1$ and $3$, giving us the set $\{9,8,7,3,1\}$. If we start with $9,8$, and $6$, we can get the missing $5$ from either $1$ and $4$ or $2$ and $3$, giving us the sets $\{9,8,6,4,1\}$ and $\{9,8,6,3,2\}$. If start with $9,8$, and $5$, the only way to get the missing $6$ is with $2$ and $4$, giving us the set $\{9,8,5,4,2\}$. And $9+8+4+3+2=26$ is too small, so we’ve found all of the possibilities for the set $A$.
As you already observed, we can permute $A$ and $B$ arbitrarily, so each choice of $A$ yields $5!4!$ numbers, and there are $11$ possibilities for $A$, for a grand total of $11\cdot5!4!=31,680$ numbers.
-
Every number of the form $x*11$ is divisible by 11. Now you must think about $x$.
$x$ must be such that $(x)*11$ is a number with 9 digits! If it is useful for you I can explain it more. | 2014-07-25T06:55:37 | {
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https://lpsa.swarthmore.edu/Bode/BodeHowNonMinPhase.html | # The Asymptotic Bode Diagram for Non-Minimum Phase Poles and Zeros
## A Real Pole with Negative ω0
Elsewhere we have discussed how to make Bode plots for a real pole. You should be familiar with that analysis. The discussion there assumed that the value of ω0 was positive; here we discuss the case if ω0 is negative. We start with
$$H(s)=\frac{1}{1+\frac{s}{\omega_0}}$$
#### Magnitude
If you carefully examine the analysis (here) of the "Magnitude" plot you'll see that the only time ω0 is used, it is squared (e.g., $\left( \frac{\omega}{\omega_0} \right)^2$). Therefore, the magnitude plot does not depend on the sign of ω0, only its absolute value, so we don't need to change anything to accomodate a negative value of ω0.
#### Phase
The phase however does change. The phase of a single real pole is given by is given by
$$\angle H\left( {j\omega } \right) = \angle \left( {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right) = - \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = - \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$
Let us again consider three cases for the value of the frequency, but we assume ω0 is negative:
Case 1) ω<<ω0. This is the low frequency case with ω/ω0→0, and doesn't depend on the sign of ω0. At these frequencies We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx -\arctan \left( 0 \right) = 0^\circ = 0\;rad$$
Case 2) ω>>ω0. Here we will consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
This is the high frequency case with ω/ω0 → +∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left( \infty \right) = - 90^\circ$$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at -90°.
ω0 < 0 (the non-minimum phase case)
This is the high frequency case with ω/ω0 → -∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left(- \infty \right) = + 90^\circ$$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at +90°.
Case 3) ω=ω0. Again consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ$$
ω0 < 0 (the non-minimum phase case)
$$\angle H\left( {j\omega } \right) = - \arctan \left( -1 \right) =+ 45^\circ$$
From the above discussion you can see that the only effect of the pole having a negative value of ω0 is that the phase is inverted (it increases from 0 → +90° as ω increases from 0 → ∞. The images below show the Bode plots for
$$H_1(s)=\frac{1}{1+\frac{s}{10}}, \quad\quad H_2(s)=\frac{1}{1-\frac{s}{10}}$$
H1(s) has a positive ω00=+10, so H1(s)=1/(1+s/10)) and H2(s) has a negative ω00=-10, so H2(s)=1/(1-s/10)). The pole of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
The same conclusion holds for first order poles and second order poles and zeros (see below).
## A Real Zero with Negative ω0
The images below show the Bode plots for two functions, one with a positive ω00=+10) and one with a negative ω00=-10). The zero of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero).
$$H_1(s)=1+\frac{s}{10}, \quad\quad H_2(s)=1-\frac{s}{10}$$
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites. Recall that 360° is equivalent to 0° so you can think of the plot for the angle of H2(s) as starting at 0° and dropping by 90° (though the plot shows it as starting at 360°).
## A Second Order Pole with Negative ζ
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=\frac{1}{1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}, \quad\quad H_2(s)=\frac{1}{1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}$$
The poles of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+0.5±j9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that, again, the magnitudes are identical, but the phases are opposites.
## A Second Order Zero with Negative ζ
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2, \quad\quad H_2(s)=1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2$$
The zeros of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+0.5±j 9.987(a positive real part, the right half of the s-plane; a non-minimum phase zero). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
##### Key Concept: For poles and zeros with positive real parts, phase is inverted
If a 1st order pole has a positive real part (i.e., a nonminimum phase system, pole is in right half of s-plane) at say s=+5, so
$$H(s)=\frac{1}{1 - \frac{s}{5}}$$
the magnitude of the Bode plot is unchanged from the case of a corresponding pole with negative value, at s=-5 (pole is in left half of s-plane)
$$H(s)=\frac{1}{1 + \frac{s}{5}}$$
but the phase of the plot is inverted.
The same rule holds Bode plots for 2nd order (complex conjugate) poles, and for 1st and 2nd order zeros.
References
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https://math.stackexchange.com/questions/1475582/number-of-ways-to-two-color-a-sectored-circle-so-that-opposite-sectors-are-diffe | # Number of ways to two-color a sectored circle so that opposite sectors are different colors.
Given a circle divided into 2n equal sectors, how many ways can it be painted with two colors so that opposite sectors are not the same color? I'm interested in the number of solutions that are unique under rotation.
I've tried to enumerate the early situations and see patterns but haven't seen a path to a general solution. I think I only have three rules of thumb at the moment:
1. There will be n sectors of each color
2. An alternating paint job is only legal when n is odd.
3. I think we're bounded on top by 2^(n-2) for n>1.
I may have missed some examples but this link has my work so far: http://i.imgur.com/ZQylapu.png
If I've gotten everything, then the series goes:
• n=1, 1
• n=2, 1
• n=3, 2
• n=4, 2
• n=5, 4
• n=6, 6
n=6 is notable as two of the combinations are left- and right-handed versions of a similar arrangement. I imagine this becomes much more common beyond n=6. I counted these separately, but a general solution that wouldn't would be nice as well.
• How do you define opposite sector? If each sector has just one opposite then why wouldn't they answer by $2^n$? – fleablood Oct 12 '15 at 0:31
• Oh, I guess the unique under rotation puts a damper on it. – fleablood Oct 12 '15 at 0:33
• Huh, I have difficulty putting words to what I mean by opposite. What comes to mind is two sectors are opposites if the number of sectors between them in the clockwise direction is equal to the number in the counter-clockwise direction. – Polyhog Oct 12 '15 at 0:37
• No I get your definition. As a pair of sectors has no relation to other sectors the answer would be $2^n$ as each pair of sectors has two choices. But if they are unique to rotation, we have to calculate how many are equivalent under rotation. Good question. – fleablood Oct 12 '15 at 0:54
• Here's a thought. Fix sector 1. then for sector i = 2 to n give a binary value for color and add $2^i$ for green and nothing for yellow. Thus you'll have a 1-1 correspondence between colorings and the numbers from 1 to $2^{n - 1}. Now, we have to come up with a formula for determining if two numbers represent the same coloring under rotation. – fleablood Oct 12 '15 at 1:00 ## 1 Answer This can be done using Burnside's lemma. We need to count the assignments of colors being fixed by each type of permutation from among the$2n$permutations total in the cycle index$Z(C_{2n})$of the cyclic group acting on the sectors. There are $$\varphi(d)$$ permutations having cycle structure $$a_d^{2n/d}$$ in the cycle index$Z(C_{2n})$where$d|2n.$For a permutation to fix an assignment it needs to be constant on the cycles. Suppose that the cycle length$d$is even and pick one of its elements. The slot opposing this element is also located on this cycle. But they must have different colors, so there are no valid assignments fixed by a permutation of shape$a_d^{2n/d}$when$d$is even. On the other hand when$d$is odd the$2n/d$cycles of length$d$are grouped into$n/d$pairs which are reflections of each other and hence must have opposite colors. Therefore there are two possible assignments of colors to these cycles forming a pair (as opposed to four if there were no constraint). This yields a contribution of $$\varphi(d) 2^{n/d}.$$ Averaging this over the total$2n$permutations we get $$\frac{1}{2n} \sum_{d|2n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}$$ which is $$\frac{1}{2n} \sum_{d|n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}.$$ This yields the sequence $$1, 1, 2, 2, 4, 6, 10, 16, 30, 52, 94, 172, 316, 586, 1096, \\ 2048, 3856, 7286,\ldots$$ which is OEIS A000016 where additional material awaits. I found the OEIS entry by using a simple total enumeration algorithm (definitely not optimized) to compute the first few values (practical to about$n=10\$ which is enough to conclusively identify the sequence). This was the Maple code:
with(numtheory);
sectors :=
proc(n)
option remember;
local d, ind, orbits, orbit, rot, pos;
orbits := {};
for ind from 2^(2*n) to 2*2^(2*n) - 1 do
d := convert(ind, base, 2);
for pos to n do
if d[pos] = d[pos+n] then
break;
fi;
od;
if pos = n+1 then
orbit := {};
for rot from 0 to 2*n-1 do
orbit := {op(orbit),
[seq(d[q], q=1+rot..2*n),
seq(d[q], q=1..rot)]};
od;
orbits := {op(orbits), orbit};
fi;
od;
nops(orbits);
end;
Q :=
proc(n)
1/2/n*
select(d->type(d,odd), divisors(n)));
end;
• That is brilliant. The equation you reached is also is the same for number of binary sequences coming from a shift register complementing the last output (equivalent to going through the circle and outputting the sector's color as binary number). Thank you! – Weaam Oct 12 '15 at 3:04
• This was not original work. The calculation is just about included in the OEIS entry, it only needed filling in the details. – Marko Riedel Oct 12 '15 at 3:06
• Still! It is a great answer. – Weaam Oct 12 '15 at 3:22 | 2020-10-28T00:30:55 | {
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http://math.stackexchange.com/questions/275456/computing-a-limit-using-the-limit-definition | # Computing a Limit using the Limit Definition
I've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.
Compute the limit. Then, using $\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. $$\lim_{n \to \infty} \frac{1}{ln(ln(n))}$$
Firstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\infty}_{n=1}$ if for every $\epsilon > 0$ there is an integer $N=N(\epsilon)>0$ such that
$$|a_n - L|< \epsilon$$
for all $n\geq N$. Now for my informal interpretation.
I believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\epsilon$? Any clarification of the definition would be appreciated.
Now, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that
$$\left | \frac{1}{ln(ln(n))} - 0 \right | = \frac{1}{ln(ln(n))}$$
So now I need an $N$ such that, for all $n \geq N$, I get $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that
$$\frac{1}{ln(ln(N))}<10^{-6}$$
So I just solve for N in the equality, which yields
$$N > e^{e^{1000000}}$$
Hence, for any $n \geq N$,
$$\frac{1}{ln(ln(n))}<10^{-6}$$
Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated.
-
It looks to me like you are done. You could compute $N$, but keep in mind it doesn't hurt for $N$ to be higher. It doesn't have to be the minimal one over $e^{e^{1000000}}$ – rschwieb Jan 10 '13 at 21:46
You can just write your integer as $\lceil e^{e^{1000000}} \rceil$ to make it clear you're talking about an integer. Other than that you have found a value that works and so you are done. – AvatarOfChronos Jan 10 '13 at 21:49
I suspect that computing $e^{e^{1000000}}$ would be a major project, requiring quite a bit of programming skill. It doesn't make sense to try to compute the exact value in this context at all. – Henry B. Jan 10 '13 at 22:33
it's a very big number, even computers are afraid to compute it. – Santosh Linkha Jan 10 '13 at 22:43
I think your confusion arises from the phrase "there exists an $N = N(\epsilon)$. You already showed why $N$ has to be a function of $\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.
You could just say then: choose $N > e^{e^{\epsilon^{-1}}}$ for $N(\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).
But, if you want to be explicit, you could use:
• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil$ where $\lceil - \rceil$ denotes the least integer greater than its argument, as you thought
• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil + 8434$
• $N(\epsilon) = \lceil 9434\pi e^{e^{\epsilon^{-1}}}\rceil$
See, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)
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http://math.stackexchange.com/questions/637664/why-does-this-not-seem-to-be-random | # Why does this not seem to be random?
I was running a procedure to be like one of those games were people try to guess a number between 1 and 100 where there are 100 people guessing.I then averaged how many different guesses there are.
from random import randint
def averager(times):
tests = list()
for _ in range(times):
l = [randint(0,100) for _ in range(100)]
tests.append(len(set(l)))
return sum(tests)/len(tests)
print(averager(100))
For some reason, the number of different guesses averages out to 63.6
Why is this?Is it due to a flaw in the python random library?
In a scenario where people were guessing a number between 1 and 10
The first person has a 100% chance to guess a previously unguessed number
The second person has a 90% chance to guess a previously unguessed number
The third person has a 80% chance to guess a previously unguessed number
and so on...
The average chance of guessing a new number(by my reasoning) is 55%. But the data doesn't reflect this.
-
Shades of (1-1/e)? Note that $1 - 1/e = 0.63212$ Can you see why it has to be this? I thought I had the answer but not sure anymore. Got to look at it some more – user44197 Jan 14 '14 at 3:33
If you want more info about this problem, there's a classical one involving the same maths. It's the odds of having two people(or more) with the same birthday in a classroom. – Feu Jan 14 '14 at 3:42
@Feu an important one in cryptography too :) – Cruncher Jan 14 '14 at 14:16
A related question I asked before. Turns out $e^{-1}$ is a notable probability. Here's an older related question. – badroit Jan 14 '14 at 14:57
From the programming point of view, don't forget that python's basic random functions are not true random, or even cryptographic random. It is a seeded PRNG. If you want better random numbers, try a cryptographic number generator source such as /dev/random (full hardware generated randomness, runs out easily). – Linuxios Jan 14 '14 at 16:40
Suppose that $n$ guesses are made. For $i=1$ to $100$, let $X_i=1$ if $i$ is not guessed, and let $X_i=0$ otherwise. If $$Y=X_1+X_2+\cdots +X_{100},$$ then $Y$ is the number of numbers not guessed.
By the linearity of expectation, we have $$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_{100}).$$ The probability that $i$ is not chosen in a particular trial is $\frac{99}{100}$, and therefore the probability it is not chosen $n$ times in a row is $\left(\frac{99}{100}\right)^n$. Thus $$E(Y)=100 \left(\frac{99}{100}\right)^n.$$
In particular, let $n=100$. Note that $\left(1-\frac{1}{100}\right)^{100}\approx \frac{1}{e}$, so the expected number not guessed is approximately $\frac{100}{e}$. Thus the expected number guessed is approximately $63.2$, a result very much in line with your simulation.
In general, if $N$ people choose independently and uniformly from a set of $N$ numbers, then the expected number of distinct numbers not chosen is $$N\left(1-\frac{1}{N}\right)^N.$$ Unless $N$ is very tiny, this is approximately $\frac{N}{e}$, and therefore the expected number of distinct numbers chosen is approximately $N-\frac{N}{e}$. Note that the expected proportion of the numbers chosen is almost independent of $N$.
-
I think you are missing a "not": "then the expected number of distinct numbers [not] chosen is" – aaaaaaaaaaaa Jan 14 '14 at 19:19
The expected number not chosen is approximately $\frac{N}{e}$. Since there are $N$ numbers overall, the expected number chosen is approximately $N-\frac{N}{e}$. – André Nicolas Jan 14 '14 at 19:32
Yes, and you wrote the opposite in the second to last paragraph. – aaaaaaaaaaaa Jan 14 '14 at 19:34
Thank you, I finally found where the missing not was (not), had been looking for it towards the very end. Will fix. – André Nicolas Jan 14 '14 at 19:42
In your 1 to 10 example, it's not true in general that the third chooser has an 80% chance on choosing a new number. It's only the case if the second one has guessed a different number than the first one.
-
This is an example of the Birthday Paradox / Birthday Problem.
Birthday problem - Wikipedia, the free encyclopedia
I was just looking at my Online Cryptography class video lecture today on this very problem.
Coursera.org: crypto-009
There is an apparent paradox that there is more duplication of numbers than expected when the random numbers are supposedly independent.
But the Birthday Paradox is just one example of when our intuitive statistical sense is dead wrong.
-
Here is a cool video from Numberphile about the Birthday problem. youtube.com/watch?v=a2ey9a70yY0 – Rik Jan 14 '14 at 12:04
The birthday problem is about the probability of having at least one collision. It says very little about the number of different values found in a large number of independent samples (as in "how many different birthdays on average in a random group of $365$ people"). – Marc van Leeuwen Jan 14 '14 at 13:27 | 2016-05-28T00:27:37 | {
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https://cs.stackexchange.com/questions/133276/algorithm-to-compute-the-gaps-between-a-set-of-intervals/133483#133483 | # Algorithm To Compute The Gaps Between A Set of Intervals
Problem
Given a set of intervals with possibly non-distinct start and end points, find all maximal gaps. A gap is defined as an interval that does not overlap with any given interval. All endpoints are integers and inclusive.
For example, given the following set of intervals:
$$\{[2,6], [1,9], [12,19]\}$$
The set of all maximal gaps is:
$$\{[10,11]\}$$
For the following set of intervals:
$$\{[2,6], [1,9], [3,12], [18,20]\}$$
The set of all maximal gaps is:
$$\{[13,17]\}$$
because that produces the maximal gap.
Proposed Algorithm
My proposed algorithm (modified approach taken by John L.) to compute these gaps is:
1. Order the intervals by ascending start date.
2. Initialise an empty list gaps that will store gaps
3. Initialise a variable, last_covered_point, to the end point of the first interval.
4. Iterate through all intervals in the sorted order. For each interval [start, end], do the following.
1. If start > last_covered_point + 1, add the gap, [last_covered_point + 1, start - 1] to gaps.
2. Assign max(last_covered_point, end) to last_covered_point.
5. Return gaps
I have tested my algorithm on a few cases and it produces the correct results. But I cannot say with 100% guarantee that it works for every interval permutation and combination. Is there a way to prove this handles every permutation and combination?
• @JohnL. yes the question is correct now - thanks
– Mojo
Dec 18 '20 at 21:25
The key to prove your algorithm is correct is to find enough invariants of the loop, step 4 so that we apply use mathematical induction.
Let $$I_1, I_2, \cdots, I_n$$ denote the sorted intervals. When the algorithm has just finished processing $$I_i$$, we record the values of gaps and last_covered_point as $$\text{gaps}_i$$ and $$\text{last_covered_point}_i$$ respectively.
Let us prove the following proposition, $$P(i)$$, for $$i=1, 2, \cdots, n$$.
$$\text{gaps}_i$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_i$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$.
When $$i=1$$, $$\text{gaps}_1$$ is the empty set and $$\text{last_covered_point}_1$$ is the right endpoint of $$I_1$$. So $$P(1)$$ is correct.
For the sake of mathematical induction, assume $$P(i)$$ is correct, where $$1\le i\lt n$$. Let $$I_{i+1}=[s, e]$$. There are two cases.
1. If $$s\gt\text{last_covered_point}_i+1$$, then $$\text{gaps}_i\cup[\text{last_covered_point}_i +1, s-1]=\text{gaps}_{i+1}.$$
Let $$m$$ be any point between the start point of $$I_1$$ and the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$. Suppose $$m$$ not covered by any of $$I_1, I_2, \cdots, I_{i+1}$$.
• If $$m\le\text{last_covered_point}_i$$, the induction hypothesis says that $$m$$ is covered by some interval in $$\text{gap}_i$$.
• Otherwise, $$m\gt\text{last_covered_point}_i$$. Since $$m$$ is not covered by $$I_{i+1}$$, we know $$m. So $$m$$ is covered by $$[\text{last_covered_point}_i +1, s-1]$$.
In both cases, $$m$$ is covered by some interval in $$\text{gaps}_{i+1}$$. Since $$\text{last_covered_point}_i$$ is the largest point covered by one of $$I_1, I_2, \cdots, I_i$$ and $$s$$ is the smallest point covered by $$I_{i+1}$$, $$[\text{last_covered_point}_i +1, s-1]$$ is a maximal gap.
2. Otherwise, we have $$s\le\text{last_covered_point}_i$$+1. We can also verify that $$\text{gaps}_{i+1}=\text{gaps}_{i}$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_{i+1}$$.
Finally, since step 4.2 says $$\text{last_covered_point}_{i+1}=\max(\text{last_covered_point}_i, e)$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$, $$\text{last_covered_point}_{i+1}$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$.
So, $$P(i+1)$$ is correct. $$\quad\checkmark$$.
• I probably didn't make my post precise enough and so I think you may have misunderstood what I meant by a gap. I have cleaned up my original post to define the problem better but I also took your example and customised it to what I think the algorithm should be. The main issue I have is that I am not sure if I have covered all possible interval permutations.
– Mojo
Dec 17 '20 at 15:19
• @Mojo, thanks for noticing that last_covered_point should be updated in step 4.2 as well. Dec 18 '20 at 16:30
• @Mojo, I just wrote a proof. The proof is not completely formal, since "maximal gap" is given by a definition (although it is easy to define) and the case 2 is not proved in detail. However, the idea should be clear enough. Dec 20 '20 at 5:24
• In my last comment, 'since "maximal gap" is given by a definition' should have been 'since "maximal gap" is not given by a definition'. Jan 4 at 1:35
I'd like to propose a quite simple algorithm as well. The idea is this: we're going to place open and close parentheses on the number line at the boundaries of each interval. For example, for the intervals $$(1,5), (2,7), (9, 10)$$, the number line would look like this:
1 2 3 4 5 6 7 8 9 10
( ( ) ) ( )
Then we'll just scan left to right, counting parentheses. When all the parentheses get closed, we start a gap. Note in particular that in the above diagram, we have lost the information about which close parenthesis is paired with which open parenthesis -- because it doesn't actually matter. So:
1. Convert each interval $$(a,b)$$ to pairs $$(a,\mathtt o),(b,\mathtt c)$$. ($$\mathtt o$$ and $$\mathtt c$$ are for open and close, respectively.)
2. Sort all the pairs you get from this process. (When sorting, use lexicographic ordering and $$\mathtt o < \mathtt c$$.)
3. Iterate through them, keeping a counter that starts from $$0$$.
• When you see a pair with $$\mathtt o$$ in the second part, increment the counter.
• When you see a pair with $$\mathtt c$$ in the second part, decrement the counter.
• When you decrement the counter, if that causes it to drop to $$0$$, then look at the next element of the list to decide what to do; empty list means you're done, otherwise if the next pair's first part is just one bigger than the current one's you do nothing, and in the last case you record a maximal gap between the current end point and the next open point. | 2021-09-25T04:31:09 | {
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https://math.stackexchange.com/questions/2908186/why-is-xn-approx-leftnx1-4096-11-right4096 | # Why is $x^n\approx \left(n(x^{1/4096}-1)+1\right)^{4096}$?
There's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $\,0.751^{3.2131}$ can be calculated, which is awesome.
This provides endless possibilities, including calculating nth roots on a simple pocket calculator.
The trick goes like this:
1. Type $x$ in the calculator
2. Take the square root twelve times
3. Subtract one
4. Multiply by $n$
6. Raise the number to the 2nd power twelve times (press * and = key eleven times)
Example:
I want to calculate $\sqrt[3]{20}$ which is the same as $20^{1/3}$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:
1. $\;\;\;20$
2. $\;\;\;1.0007315$
3. $\;\;\;0.0007315$
4. $\;\;\;0.0002438$
5. $\;\;\;1.0002438$
6. $\;\;\;2.7136203$
The actual answer is $20^{1/3}\approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.
Question:
So the question is now, why does this trick work? More specifically, how do we prove that: $$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$
Note: $4096=2^{12}$.
I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.
I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:
$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$
This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $a\to\infty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.
Thank you for your time :)
• Which one would you like to prove, the approximation for $a = 12$, or just the limit expression for $a \to \infty$? – Xiangxiang Xu Sep 7 '18 at 2:01
• In my opinion the one with the limit with $a\to\infty$ is better to prove because it is more general. From there, the specific case of $a=12$ can be simpler to prove (I think) – KKZiomek Sep 7 '18 at 2:04
A standard trick is to calculate the natural logarithm first to get the exponent under control:
$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$ Set $u=1/a$. We get $$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$ Use L'Hopital: $$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$ Here we just plugged in $u=0$ to calculate the limit!
So the original limit goes to $x^n$ as desired.
• I was confused first at L'Hopital step, but then I realized that I have to differentiate with respect to $u$ not $x$ :). I'm giving this answer the best even though all other answers were equally good, because if you haven't mentioned L'Hopital, I would never think of using it. Thank you for the answer. – KKZiomek Sep 7 '18 at 2:19
• Glad to help. It's a neat problem. – Cheerful Parsnip Sep 7 '18 at 2:24
For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that $$\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n.$$ In fact, we have $$\ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] = \lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x,$$ where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.
If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case
$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$
If $n$ is also relatively small, then
$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$
Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,
$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$
is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get
$${2.7136203+2.7152613\over2}=2.7144408$$
which is quite close to the true value! | 2019-07-17T06:55:35 | {
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https://mathsedideas.blogspot.com/2020/12/three-card-colours.html | ## Wednesday, 2 December 2020
### On the Colour of the Third Card
Three cards are dealt from a normal deck. You don't see them. You are told that the first two are the same colour, but not what colour they are. What is the probability that the next card is the same colour?
$\frac{1}{2},\;\frac{1}{4},\;\frac{{12}}{{25}},\;or\;\frac{4}{{17}}\;?$
This beautiful, onthefaceofit innocuous little problem comes from the great Martin Gardner's 'Modelling Mathematics with Playing Cards'. It's a problem that never fails to invoke heated discussion and vehement argument when I share it with students to play with. The four solutions and reasoning sketched out below are those typically proffered (and invariably staunchly defended) by students. Which solution would you go with, and why? Do you have a different solution? And what argument would you give to those who firmly hold the solution to be one of the others to show them that they are wrong and you are right? (Note that Gardner's solution was solution 2 [1])
## Solution 1
The last card could either be 1) the same colour by being red, or 2) a different colour by being red, or 3) the same colour by being black, or 4) a different colour by being black. So there are four possibilities and two of these result in the final card being the same colour as the previous two. So:
$\frac{2}{4} = \frac{1}{2}$
## Solution 2
There are eight ways that the arrangement of the colours of the three cards can occur, namely RRR, RRB, RBR, BRR, RBB, BRB, BBR, or BBB (where R = Red and B = Black), and two of these arrangements have the final card as the same colour as the previous two. So:
$\frac{2}{8} = \frac{1}{4}$
## Solution 3
After the first two cards, fifty cards remain in the deck. If the first two cards are both red, there remain twenty-four cards that are red. Similarly, if the first two cards are both black, there remain twenty-four cards that are black. So:
$\frac{24}{50} = \frac{12}{25}$
## Solution 4
For the last card to be the same colour as the first two, the colours of all three cards are either RRR or BBB. You have fifty-two cards to choose from for your first card, and twenty-six of these are red and twenty-six are black. After taking the first card from the deck, you have fifty-one cards left to choose from for your second card, and assuming that the first card was red, twenty-five of the fifty-one cards left are also red. After taking the second card from the deck, you have fifty cards left to choose from for your third and final card, and assuming that the first two cards were red, twenty-four of the fifty cards left are also red. The same would be true if the cards pulled from the deck were black. So:
$\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}$ | 2021-05-13T05:21:36 | {
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https://gingerhillcreations.com/jessica-smith-scxean/ec0e3c-how-to-prove-cardinality-of-sets | Here we need to talk about cardinality of a set, which is basically the size of the set. there are $10$ people with white shirts and $8$ people with red shirts; $4$ people have black shoes and white shirts; $3$ people have black shoes and red shirts; the total number of people with white or red shirts or black shoes is $21$. that the cardinality of a set is the number of elements it contains. Edition 1st Edition. $$|W \cap B|=4$$ In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides of students who play both foot ball and cricket = 25, No. In Section 5.1, we defined the cardinality of a finite set $$A$$, denoted by card($$A$$), to be the number of elements in the set $$A$$. Total number of elements related to both B & C. Total number of elements related to both (B & C) only. Two sets are equal if and only if they have precisely the same elements. For example, let A = { -2, 0, 3, 7, 9, 11, 13 }, Here, n(A) stands for cardinality of the set A. To prove there exists a bijection between to sets X and Y, there are 2 ways: 1. find an explicit bijection between the two sets and prove it is bijective (prove it is injective and surjective) 2. Pages 5. eBook ISBN 9780429324819. It turns out we need to distinguish between two types of infinite sets, Any superset of an uncountable set is uncountable. then talk about infinite sets. The cardinality of a set is roughly the number of elements in a set. (Hint: Use a standard calculus function to establish a bijection with R.) 2. Math 127: In nite Cardinality Mary Radcli e 1 De nitions Recall that when we de ned niteness, we used the notion of bijection to de ne the size of a nite set. of students who play cricket only = 10, No. where one type is significantly "larger" than the other. The cardinality of the set of all natural numbers is denoted by . $$\>\>\>\>\>\>\>+\sum_{i < j < k}\left|A_i\cap A_j\cap A_k\right|-\ \cdots\ + \left(-1\right)^{n+1} \left|A_1\cap\cdots\cap A_n\right|.$$, $= |W| + |R| + |B|- |W \cap R| - |W \cap B| - |R \cap B| + |W \cap R \cap B|$. ... to make the argument more concrete, here we provide some useful results that help us prove if a set is countable or not. Hence these sets have the same cardinality. The cardinality of a set is denoted by $|A|$. of students who play foot ball only = 28, No. $$|W|=10$$ The intuition behind this theorem is the following: If a set is countable, then any "smaller" set $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable. c) $(0,\infty)$, $\R$ d) $(0,1)$, $\R$ Ex 4.7.4 Show that $\Q$ is countably infinite. If $B \subset A$ and $A$ is countable, by the first part of the theorem $B$ is also a countable the idea of comparing the cardinality of sets based on the nature of functions that can be possibly de ned from one set to another. Ex 4.7.3 Show that the following sets of real numbers have the same cardinality: a) $(0,1)$, $(1, \infty)$ b) $(1,\infty)$, $(0,\infty)$. countable, we can write a proof, we can argue in the following way. So, the total number of students in the group is 100. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. If you are less interested in proofs, you may decide to skip them. The set of all real numbers in the interval (0;1). Before we start developing theorems, let’s get some examples working with the de nition of nite sets. In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. In mathematics, a set is a well-defined collection of distinct elements or members. Two finite sets are considered to be of the same size if they have equal numbers of elements. then by removing the elements in the list that are not in $B$, we can obtain a list for $B$, In addition, we say that the empty set has cardinality 0 (or cardinal number 0), and we write $$\text{card}(\emptyset) = 0$$. As far as applied probability Discrete Mathematics and Its Applications, Seventh Edition answers to Chapter 2 - Section 2.5 - Cardinality of Sets - Exercises - Page 176 12 including work step by step written by community members like you. $$|W \cup B \cup R|=21.$$ set whose elements are obtained by multiplying each element of Z by k.) The function f : N !Z de ned by f(n) = ( 1)nbn=2cis a 1-1 corre-spondence between the set of natural numbers and the set of integers (prove it!). The Math Sorcerer 19,653 views. If $A$ has only a finite number of elements, its cardinality is simply the Here is a simple guideline for deciding whether a set is countable or not. the inclusion-exclusion principle we obtain. and how to prove set S is a infinity set. you can never provide a list in the form of $\{a_1, a_2, a_3,\cdots\}$ that contains all the $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$. \mathbb {R}. thus $B$ is countable. Examples of Sets with Equal Cardinalities. This will come in handy, when we consider the cardinality of infinite sets in the next section. 4 CHAPTER 7. Cardinality of a set S, denoted by |S|, is the number of elements of the set. Since each $A_i$ is countable we can Definition. Cantor introduced a new de・］ition for the 窶徭ize窶・of a set which we call cardinality. Textbook Authors: Rosen, Kenneth, ISBN-10: 0073383090, ISBN-13: 978-0-07338-309-5, Publisher: McGraw-Hill Education The difference between the two types is We will say that any sets A and B have the same cardinality, and write jAj= jBj, if A and B can be put into 1-1 correspondence. but "bigger" sets such as $\mathbb{R}$ are called uncountable. You already know how to take the induction step because you know how the case of two sets behaves. \mathbb {N} To formulate this notion of size without reference to the natural numbers, one might declare two finite sets A A A and B B B to have the same cardinality if and only if there exists a bijection A → B A \to B A → B. Cardinality Recall (from our first lecture!) (useful to prove a set is finite) • A set is infinite when there is an injection, f:AÆA, such that f(A) is … When an invertible function from a set to \Z_n where m\in\N is given the cardinality of the set immediately follows from the definition. This important fact is commonly known ... aged to prove that two very different sets are actually the same size—even though we don’t know exactly how big either one is. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. Cardinality The cardinality of a set is roughly the number of elements in a set. By Gove Effinger, Gary L. Mullen. (a) Let S and T be sets. Furthermore, we designate the cardinality of countably infinite sets as ℵ0 ("aleph null"). 12:14. Here we need to talk about cardinality of a set, which is basically the size of the set. In particular, one type is called countable, If A and B are disjoint sets, n(A n B) = 0, n(A u B u C) = n(A) + n(B) + n(C) - n(A n B) - n(B n C) - n(A n C) + n(A n B n C), n(A n B) = 0, n(B n C) = 0, n(A n C) = 0, n(A n B n C) = 0, = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC). Cardinality of a set: Discrete Math: Nov 17, 2019: Proving the Cardinality of 2 finite sets: Discrete Math: Feb 16, 2017: Cardinality of a total order on an infinite set: Advanced Math Topics: Jan 18, 2017: cardinality of a set: Discrete Math: Jun 1, 2016 One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of א o (pronounced “aleph naught”). $$|A \cup B |=|A|+|B|-|A \cap B|.$$ Solution. Before discussing thus by subtracting it from $|A|+|B|$, we obtain the number of elements in $|A \cup B |$, (you can The above arguments can be repeated for any set $C$ in the form of However, to make the argument In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. (Assume that each student in the group plays at least one game). Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively. We first discuss cardinality for finite sets and (Hint: you can arrange $\Q^+$ in a sequence; use this to arrange $\Q$ into a sequence.) set which is a contradiction. of students who play both (foot ball & hockey) only = 12, No. To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \cap B|$ twice, Because of the symmetyofthissituation,wesaythatA and B can be put into 1-1 correspondence. That is often difficult, however. If $A$ is a finite set, then $|B|\leq |A| < \infty$, Then,byPropositionsF12andF13intheFunctions section,fis invertible andf−1is a 1-1 correspondence fromBtoA. Consider sets A and B.By a transformation or a mapping from A to B we mean any subset T of the Cartesian product A×B that satisfies the following condition: . Question: Prove that N(all natural numbers) and Z(all integers) have the same cardinality. To prove that a given in nite set X … Theorem. How to prove that all maximal independent sets of a matroid have the same cardinality. We can, however, try to match up the elements of two infinite sets A and B one by one. As seen, the symbol for the cardinality of a set resembles the absolute value symbol — a variable sandwiched between two vertical lines. Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties: If A is subset of B and B belongs to J , then A belongs to J , = n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC), n(FuHuC) = 65 + 45 + 42 -20 - 25 - 15 + 8. This function is bijective. When it ... prove the corollary one only has to observe that a function with a “right inverse” is the “left inverse” of that function and vice versa. No. Total number of students in the group is n(FuHuC). $\mathbb{Z}=\{0,1,-1,2,-2,3,-3,\cdots\}$. In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. This establishes a one-to-one correspondence between the set of primes and the set of natural numbers, so they have the same cardinality. A = \left\ { {1,2,3,4,5} \right\}, \Rightarrow \left| A \right| = 5. … uncountable set (to prove uncountability). To this final end, I will apply the Cantor-Bernstein Theorem: (The two sets (0, 1) and [0, 1] have the same cardinality if we can find 1-1 mappings from (0, 1) to [0, 1] and vice versa.) Also known as the cardinality, the number of disti n ct elements within a set provides a foundational jump-off point for further, richer analysis of a given set. The examples are clear, except for perhaps the last row, which highlights the fact that only unique elements within a set contribute to the cardinality. I presume you have sent this A2A to me following the most recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration of Zero. To be precise, here is the definition. Find the total number of students in the group. A set A is countably infinite if and only if set A has the same cardinality as N (the natural numbers). The proof of this theorem is very similar to the previous theorem. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. Now that we know about functions and bijections, we can define this concept more formally and more rigorously. The idea is exactly the same as before. f:A → Bbea1-1correspondence. while the other is called uncountable. However, I am stuck in proving it since there are more than one "1", "01" = "1", same as other numbers. Fix m 2N. $$\biggl|\bigcup_{i=1}^n A_i\biggr|=\sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right|$$ but you cannot list the elements in an uncountable set. Imprint CRC Press. 1. if it is a finite set, $\mid A \mid < \infty$; or. I could not prove that cardinality is well defined, i.e. Thus according to Definition 2.3.1, the sets N and Z have the same cardinality. When A and B have the same cardinality, we write jAj= jBj. Thus, any set in this form is countable. of students who play both (hockey & cricket) only = 7, No. A set is an infinite set provided that it is not a finite set. of students who play hockey only = 18, No. The two sets A = {1,2,3} and B = {a,b,c} thus have the cardinality since we can match up the elements of the two sets in such a way that each element in each set is matched with exactly one element in the other set. For example, if $A=\{2,4,6,8,10\}$, then $|A|=5$. Math 131 Fall 2018 092118 Cardinality - Duration: 47:53. For in nite sets, this strategy doesn’t quite work. Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational numbers $\mathbb{Q}$ is countable. Since $A$ and $B$ are Cardinality of a set is a measure of the number of elements in the set. If A can be put into 1-1 correspondence with a subset of B (that is, there is a 1-1 case the set is said to be countably infinite. I can tell that two sets have the same number of elements by trying to pair the elements up. (useful to prove a set is finite) • A set is infinite when there … (b) A set S is finite if it is empty, or if there is a bijection for some integer . of students who play both foot ball & hockey = 20, No. n(AuB) = Total number of elements related to any of the two events A & B. n(AuBuC) = Total number of elements related to any of the three events A, B & C. n(A) = Total number of elements related to A. n(B) = Total number of elements related to B. n(C) = Total number of elements related to C. Total number of elements related to A only. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. We first discuss cardinality for finite sets and then talk about infinite sets. A useful application of cardinality is the following result. It would be a good exercise for you to try to prove this to yourself now. If S is a set, we denote its cardinality by |S|. Cantor showed that not all in・］ite sets are created equal 窶・his de・］ition allows us to distinguish betweencountable and uncountable in・］ite sets. Set S is a set consisting of all string of one or more a or b such as "a, b, ab, ba, abb, bba..." and how to prove set S is a infinity set. For finite sets, cardinalities are natural numbers: |{1, 2, 3}| = 3 |{100, 200}| = 2 For infinite sets, we introduced infinite cardinals to denote the size of sets: $$C=\bigcup_i \bigcup_j \{ a_{ij} \},$$ The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this correspondence with natural numbers $\mathbb{N}$. is concerned, this guideline should be sufficient for most cases. n(FnH) = 20, n(FnC) = 25, n(HnC) = 15. The set whose elements are each and each and every of the subsets is the ability set. The above theorems confirm that sets such as $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and their Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. 11 Cardinality Rules ... two sets, then the sets have the same size. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. For in nite sets, this strategy doesn’t quite work. Any set which is not finite is infinite. If $A_1, A_2,\cdots$ is a list of countable sets, then the set $\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$ Discrete Mathematics - Cardinality 17-16 More Countable Sets (cntd) The fact that you can list the elements of a countably infinite set means that the set can be put in one-to-one If $A$ and $B$ are countable, then $A \times B$ is also countable. should also be countable, so a subset of a countable set should be countable as well. We can say that set A and set B both have a cardinality of 3. set is countable. S and T have the same cardinality if there is a bijection f from S to T. Notation: means that S and T have the same cardinality. respectively. The cardinality of a set is denoted by $|A|$. Thus by applying there'll be 2^3 = 8 elements contained in the ability set. Definition of cardinality. forall s : fset_expr (A:=A), exists n, (cardinality_fset s n /\ forall s' n', eq_fset s s' -> cardinality_fset s' n' -> n' = n). 1. On the other hand, you cannot list the elements in $\mathbb{R}$, DOI link for Cardinality of Sets. so it is an uncountable set. | A | = | N | = ℵ0. Mappings, cardinality. The number of elements in a set is called the cardinality of the set. This poses few difficulties with finite sets, but infinite sets require some care. The sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. $$B = \{b_1, b_2, b_3, \cdots \}.$$ and Itiseasytoseethatanytwofinitesetswiththesamenumberofelementscanbeput into1-1correspondence. For example, you can write. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides The cardinality of a finite set is the number of elements in the set. if you need any other stuff in math, please use our google custom search here. Any subset of a countable set is countable. If you are less interested in proofs, you may decide to skip them. Total number of elements related to C only. refer to Figure 1.16 in Problem 2 to see this pictorially). I can tell that two sets have the same number of elements by trying to pair the elements up. Click here to navigate to parent product. However, as we mentioned, intervals in $\mathbb{R}$ are uncountable. Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. 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Their subsets are countable set … a useful application of cardinality is a simple guideline deciding! | 2021-02-26T21:13:04 | {
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https://it.mathworks.com/help/symbolic/sym.jacobian.html | # jacobian
Jacobian matrix
## Description
example
jacobian(f,v) computes the Jacobian matrix of f with respect to v. The (i,j) element of the result is $\frac{\partial f\left(i\right)}{\partial \text{v}\left(j\right)}$.
## Examples
collapse all
The Jacobian of a vector function is a matrix of the partial derivatives of that function.
Compute the Jacobian matrix of [x*y*z,y^2,x + z] with respect to [x,y,z].
syms x y z
jacobian([x*y*z,y^2,x + z],[x,y,z])
ans =
$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$
Now, compute the Jacobian of [x*y*z,y^2,x + z] with respect to [x;y;z].
jacobian([x*y*z,y^2,x + z], [x;y;z])
ans =
$\left(\begin{array}{ccc}y z& x z& x y\\ 0& 2 y& 0\\ 1& 0& 1\end{array}\right)$
The Jacobian matrix is invariant to the orientation of the vector in the second input position.
The Jacobian of a scalar function is the transpose of its gradient.
Compute the Jacobian of 2*x + 3*y + 4*z with respect to [x,y,z].
syms x y z
jacobian(2*x + 3*y + 4*z,[x,y,z])
ans = $\left(\begin{array}{ccc}2& 3& 4\end{array}\right)$
Now, compute the gradient of the same expression.
ans =
$\left(\begin{array}{c}2\\ 3\\ 4\end{array}\right)$
The Jacobian of a function with respect to a scalar is the first derivative of that function. For a vector function, the Jacobian with respect to a scalar is a vector of the first derivatives.
Compute the Jacobian of [x^2*y,x*sin(y)] with respect to x.
syms x y
jacobian([x^2*y,x*sin(y)],x)
ans =
$\left(\begin{array}{c}2 x y\\ \mathrm{sin}\left(y\right)\end{array}\right)$
Now, compute the derivatives.
diff([x^2*y,x*sin(y)],x)
ans = $\left(\begin{array}{cc}2 x y& \mathrm{sin}\left(y\right)\end{array}\right)$
Specify polar coordinates $r\left(t\right)$, $\varphi \left(t\right)$, and $\theta \left(t\right)$ that are functions of time.
syms r(t) phi(t) theta(t)
Define the coordinate transformation form spherical coordinates to Cartesian coordinates.
R = [r*sin(phi)*cos(theta), r*sin(phi)*sin(theta), r*cos(phi)]
R(t) = $\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) r\left(t\right)\end{array}\right)$
Find the Jacobian of the coordinate change from spherical coordinates to Cartesian coordinates.
jacobian(R,[r,phi,theta])
ans(t) =
$\left(\begin{array}{ccc}\mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{cos}\left(\theta \left(t\right)\right) r\left(t\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)\\ \mathrm{sin}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right)& \mathrm{cos}\left(\varphi \left(t\right)\right) \mathrm{sin}\left(\theta \left(t\right)\right) r\left(t\right)& \mathrm{cos}\left(\theta \left(t\right)\right) \mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)\\ \mathrm{cos}\left(\varphi \left(t\right)\right)& -\mathrm{sin}\left(\varphi \left(t\right)\right) r\left(t\right)& 0\end{array}\right)$
## Input Arguments
collapse all
Scalar or vector function, specified as a symbolic expression, function, or vector. If f is a scalar, then the Jacobian matrix of f is the transposed gradient of f.
Vector of variables or functions with respect to which you compute Jacobian, specified as a symbolic variable, symbolic function, or vector of symbolic variables. If v is a scalar, then the result is equal to the transpose of diff(f,v). If v is an empty symbolic object, such as sym([]), then jacobian returns an empty symbolic object.
collapse all
### Jacobian Matrix
The Jacobian matrix of the vector function f = (f1(x1,...,xn),...,fn(x1,...,xn)) is the matrix of the derivatives of f:
$J\left({x}_{1},\dots {x}_{n}\right)=\left[\begin{array}{ccc}\frac{\partial {f}_{1}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{1}}{\partial {x}_{n}}\\ ⋮& \ddots & ⋮\\ \frac{\partial {f}_{n}}{\partial {x}_{1}}& \cdots & \frac{\partial {f}_{n}}{\partial {x}_{n}}\end{array}\right]$
## Version History
Introduced before R2006a | 2022-08-08T03:41:14 | {
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# In how many different ways can the letters of the word MISSISSIPPI be
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In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
[Reveal] Spoiler: OA
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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08 Apr 2016, 20:21
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The word MISSISSIPPI has 4 I , 4S , 2P and M .
The only vowel present in the word is I and we have to consider 4 I's as a single group
Number of different ways = 8!/(4!*2!)
=(8*7*6*5)/2
=840
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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26 Jun 2017, 11:18
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
vowel (I) together can be arranged with other 7 letters in 8! ways
4 no. vowels themselves can be arranged in 4! ways
different ways = 8!*4!/( 4!*4!*2!) = 840---(4I's , 4 S's , 2 P's)
Ans E
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
Ooh, a combinatorics problem with some fairly large numbers. The best technique for these is to mentally break them down into simpler problems. Otherwise, the solutions tend to sort of look like magic tricks - sure, you can use the formula, but how are you supposed to know to use that formula, and not one of the many similar-looking ones?
There are four vowels, and they're all the same. Let's set those vowels aside: (IIII)
Now we have the letters MSSSSPP. How many ways can just those letters be arranged? Well, if they were all different, we could arrange them in 7x6x5x4x3x2x1 = 7! ways. However, they aren't all different. For instance, four of the letters are (S). I'm going to color them to demonstrate why that matters:
one arrangement: MSSSSPP
a 'different' arrangment: MSSSSPP
We counted both of those arrangements, but we actually don't want to. All of the Ss look the same - they aren't actually different colors - so we want to make those arrangements the same. Because there are 4x3x2x1 = 24 ways to order the different Ss, we want every set of 24 arrangements where the Ss are in the same place, to just count as 1 arrangement, instead. So, we can divide out the extra possibilities by dividing our total by 24 (or 4!):
7!/4!
Do the same thing with the two Ps. We still have twice as many arrangements as we need, since we've counted as if the two Ps were different, but they're actually the same. So, divide the total by 2:
7!/(4! x 2)
Now we have to put the (IIII) letters back in. They all have to go together. Start by looking at one of the arrangements of the other letters: PMSSPSS. Where can the four Is go? There are 8 places where we can put them:
IIIIPMSSPSS
PIIIIMSSPSS
PMIIIISSPSS
PMSIIIISPSS
etc.
So, for each arrangement, we have to multiply by 8, to account for the eight possible ways to put the vowels back in. Here's the final answer:
(8 x 7!) / (4! x 2) = (8 x 7 x 6 x 5 x 4 x 3 x 2) / (4 x 3 x 2 x 2) = 4 x 7 x 6 x 5 = 4 x 210 = 840.
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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26 Jun 2017, 12:40
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Bunuel wrote:
In how many different ways can the letters of the word MISSISSIPPI be arranged if the vowels must always be together?
A. 48
B. 144
C. 210
D. 420
E. 840
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
-------NOW ONTO THE QUESTION!!!-----------------
First "glue" the 4 I's together to create ONE character: IIII (this ensures that they stay together)
So, basically, we must determine the number of arrangements of M, S, S, S, S, P, P and IIII
There are 8 characters in total
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 8!/[(4!)(2!)]
= 840
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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04 Dec 2017, 08:31
Stage 1: Lets place the letters without the 4I's:
MSSSSPP
7!/(4!2!)= 105 ways
Stage 2: Now lets place the 4I's
-m-s-s-s-s-p-p-
We can place th4 4I's in 8 ways
Total ways= 105*8= 840 ways.
Is this a correct approach?
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink]
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04 Dec 2017, 09:08
Rearranging letters has a simple formula:
Total number of letters
*DIVIDED BY*
(2)! for every letter repeated twice / (3)! for every letter repeated thrice and so on...
Mississippi has:
4 s's
4 i's
2 p's
Total number of letters: 11!
So the answer is $$\frac{11!}{4!4!2!}$$
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Re: In how many different ways can the letters of the word MISSISSIPPI be [#permalink] 04 Dec 2017, 09:08
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# In how many different ways can the letters of the word MISSISSIPPI be
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-01-17T18:16:35 | {
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https://math.stackexchange.com/questions/3130437/10-people-6-male-4-female-divided-into-2-equal-groups-what-is-the-pr/3130441 | $10$ people ($6$ male, $4$ female) divided into $2$ equal groups: what is the probability that all females are in the same group?
This question comes from a completed, marked, and returned exam. It will not likely be reused.
Problem
As stated in the question above
Work
First, I note that there are $$\binom{10}{5}$$ possible groupings.
Second, I note that, if all $$4$$ females are in the same group, then the remaining fifth member is one of the boys: there are $$\binom{6}{1} = 6$$ ways to choose the fifth member.
So I conclude $$\Pr = \frac{6}{\binom{10}{5}} = \frac{1}{42}$$.
Question
I was marked incorrect: the given answer is $$\frac{1}{21}$$, or exactly twice my answer.
What reasoning led to this conclusion? Why does it seem like some sort of symmetry argument allows us to conclude there are $$12$$ ways to choose the fifth member?
• Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. – N. F. Taussig Feb 28 at 16:55
• @N.F.Taussig apologies i was looking at the answers on the question below as i typed – D. Ben Knoble Feb 28 at 17:01
• Another way to see that there are only $126$ possible groups is to observe that if Eloise is one of the four girls, then there are $\binom{9}{4}$ ways to select which four of the other members are in her group. – N. F. Taussig Feb 28 at 17:28
You've double-counted the groupings: $$\{A, B, C, D, E\}$$ and $$\{F, G, H, I, J\}$$ is the same grouping as $$\{F, G, H, I, J\}$$ and $$\{A, B, C, D, E\}$$. Accounting for this double-count, there are $$\frac{1}{2} \binom{10}{5}$$ distinct groupings. (This is probably the most-common counting mistake of all time. Everyone makes it at least once.) | 2019-07-20T15:41:23 | {
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https://www.physicsforums.com/threads/show-if-a-b-c-are-invertible-matrices-of-same-size.243766/ | # Show if A,B,C are invertible matrices of same size
1. Jul 7, 2008
Show if A,B,C are invertible matrices of same size....
I know, I know. I should be awesome at these by now...
1. The problem statement, all variables and given/known data
Show if A,B,C are invertible matrices of the same size, than $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$
3. The attempt at a solution Given some matrix A,$AA^{-1}=I$
If:
$AA^{-1}=I$
$BB^{-1}=I$
$CC^{-1}=I$
I am not sure where to go from here. I don't think I have any more definitions or product rules to incorporate.
It almost seems as if I would FIRST have to show that (ABC) is invertible to begin with. Then I can use the fact that (ABC)(ABC)^{-1}=I to discover how (ABC)^{-1} MUST be arranged in order for the product of the two to yield I.
Does that sound like a good place to start? Proving if A,B, and C are invertible, then (ABC) is too?
2. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Let $D=(ABC)^{-1}$
$\times (ABC)$
$\Rightarrow ABCD=I$
and then just multiply by $A^{-1}$ and so forth
3. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
I don't follow? What does the second line mean?--->$\times (ABC)$
4. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Multiply both sides by the matric ABC
5. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
So
$D=(ABC)^{-1}$
$\Rightarrow D(ABC)=(ABC)^{-1}(ABC)$
Well, I think I see where this is going. And I think that the only reason this works is because we are assuming that the product (ABC) IS invertible. Which brings me back to my original point. In order to show how the multiplication MUST be carried out, we must first SHOW or assume without proof that (ABC) is in fact invertible since our argument will be based on the fact that (ABC)*(ABC)^{-1}=Id.
6. Jul 7, 2008
### rock.freak667
Re: Show if A,B,C are invertible matrices of same size....
Well if A,B,C are nxn matrices then ABC is an nxn matrix.
and for the matrix ABC to be invertible $det(ABC) \neq 0$
and det(ABC)=det(A)*det(B)*det(C)
but the matrices A,B,C are invertible.
7. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
Another way you could show that a product of two matrices A and B are invertible is by showing that there exists some matrix which when multiplied to AB on the left and on the right gives the identity matrix:
Suppose A and B are invertible, then:
$$AB(B^{-1}A^{-1}) = I$$ for multiplying on the right
$$B^{-1}A^{-1}AB = I$$ for multiplying on the left.
In both cases this reduces to I, so $$B^{-1}A^{-1}$$ is the inverse of AB.
Now make use of this result to prove your question.
8. Jul 7, 2008
### HallsofIvy
Staff Emeritus
Re: Show if A,B,C are invertible matrices of same size....
It is not nnecessary to assume that ABC is invertible.
You are given that A, B, C separately are invertible so A-1, B-1, and C-1 exist. Thus C-1B-1A-1 exists. What do you get if you multiply (ABC)(C-1B-1C-1) and (C-1B-1A-1(ABC)?
That will prove that ABC is invertible.
9. Jul 7, 2008
### matt grime
Re: Show if A,B,C are invertible matrices of same size....
X is invertible if there is a Y with XY=YX=Id.
It isn't that I want to show ABC is invertible, but someone's been really helpful and just asked me to verify what the inverse is! I don't have to think at all, I just have to chuck the nominal inverse into the definition and see what happens. So do it....
10. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
Hey yeah, I didn't see it that way. Makes it a lot easier.
11. Jul 7, 2008
### JinM
Re: Show if A,B,C are invertible matrices of same size....
This looks like a question that could spring up on my midterm tomorrow. Assuming invertibility, does this prove the result?
$$ABCC^{-1}B^{-1}A^{-1} = A(BI_{n})B^{-1}A^{-1}= A(BB^{-1})A^{-1} = AA^{-1} = I_n$$
Now, the other side,
$$C^{-1}B^{-1}A^{-1}ABC = C^{-1}B^{-1}BC = C^{-1}C = I_n$$
Thus $$(ABC)(C^-1B^-1A^-1}) = (C^-1B^-1A^-1)(ABC) = I_n$$.
Is this enough? No claim of originality, this was how a result was proved in my notes, and what everyone is getting at!
Last edited: Jul 7, 2008
12. Jul 7, 2008
### Defennder
Re: Show if A,B,C are invertible matrices of same size....
That's basically what HallsOfIvy said.
13. Jul 7, 2008
Re: Show if A,B,C are invertible matrices of same size....
I don't like it. I don't know why yet...but I just don't.
14. Jul 7, 2008
### matt grime
Re: Show if A,B,C are invertible matrices of same size....
You aren't assuming invertibility of ABC. You're just multiplying together 6 matrices, and showing that they give the identity, which is *proving* it is invertible. Just because you wrote "assuming invertibility of ABC" doen't mean you actually did assume it, or that you needed toi.
15. Jul 7, 2008
### b0it0i
Re: Show if A,B,C are invertible matrices of same size....
i think the problem you're having with this proof is understanding what the question is asking.
there are many other proofs with the same structure
the problem is saying show that
(ABC)-1 = C-1B-1A-1
in other words, they want you to show that the inverse of ABC is actually (behaves like) inverse C times inverse B times inverse A
an example would be... Show that
A Source of water = Rain
you must show that rain acts/ behaves like a source of water
apply the properties of "a source of water" to rain
Show that C-1B-1A-1 "behaves" like the (ABC)-1
how do you do that?... you check the property of (ABC)-1
that is... show that C-1B-1A-1 (ABC) = (ABC) C-1B-1A-1 = I
An important fact is to note that "the inverse of a matrix is unique"
since C-1B-1A-1 "behaves" like the inverse of ABC,
and that the inverse of a matrix is unique,
C-1B-1A-1 MUST be (ABC)-1
this is corny, but my past professor constantly called these types of proofs
"walks like a duck, quacks like a duck, MUST be a duck..." proof
16. Jul 8, 2008 | 2017-03-27T05:16:33 | {
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https://math.stackexchange.com/questions/1257518/answer-does-not-make-sense-using-conditional-probability | # Answer does not make sense using conditional probability
As of now, there are 64.1 million people residing in the UK. 5.4 million of them are thought to be asthmatic. A new test for asthma has recently been introduced and medical trials indicate that
• in patients with asthma, the test correctly returns positive 68% of the time
• in patients that does not suffer from asthma, the test correctly returns 82% of the time.
Assuming a patient undergo medical inspection. What is the probability that the patient has asthma,
1. if the test comes back positive?
2. if the test comes back negative?
For Part 1, I used this method:
Step 1: $$P(A|+)=\frac{P(+|A)P(A)}{P(+)}$$
Step 2: $$P(+)=P(+│A)P(A)+P(+│N)P(N)=(0.68)\left(\frac{54}{641}\right)+(0.18)\left(\frac{587}{641}\right)=\frac{7119}{32050}$$
Step 3: $$P(A│+)=\frac{P(+|A)P(A)}{P(+)}=\frac{0.68\times\frac{54}{641}}{\frac{7119}{32050}}=\frac{204}{791}=25.79\%$$
For Part 2, I used the same method as Part 1 but I got a very small answer. Assuming if I insert all the values in the equations correctly, would it be sensible if my answer for Part 2 is 3.47%?
However, personally, it does not make logical sense if the probability of test returning negative is 3.47% because that would mean almost everyone in the nation would be asthmatic.
Or I could just write it as $$100-25.79 = 74.21%$$ but I'm afraid this isn't the answer given the complexity of the question.
• 3.47% is the right answer here... – Clement C. Apr 29 '15 at 13:33
• You're computing the probability of a false negative not the probability someone doesn't have asthma. False negatives are rare and should be rare intuitively. Otherwise, the test would be a piece of crap. So, you're right. – Jamie Lannister Apr 29 '15 at 13:34
• I don't understand what you mean. If 3.47% is the correct answer, this is quite reasonable. It says: "If the test says you don't have asthma, then it's very likely you really don't have asthma (i.e. the test is good)." The first answer is actually more surprising. It says: "If the test says you have asthma, it's only 25% likely you actually have it!" But that isn't that surprising given the low figure of 68% in the hypotheses. – Frank Apr 29 '15 at 13:34
• @iterence: that would not give the probability $\Pr[A\mid -]$, but instead the quantity $\Pr[A^c\mid +] = 1 - \Pr[A\mid +]$. That is, instead of "the probability to have asthma knowing the test says you don't," you would compute "the probability not to have asthma, knowing the test says you do." – Clement C. Apr 29 '15 at 13:38
• In part 2, "if the test comes back negative", it is assumed we know that the result was negative, i.e., the probability of the test returning negative is $100$%, not $3.47$%. The $3.47$% figure arises from the cases where people do have asthma but it is not detected by the test. – David K Apr 29 '15 at 13:41
One instructive way to do these kinds of $2 \times 2$ problems, since there are only four distinct populations, is to enumerate them and calculate the marginal probabilities by inspection.
$$\begin{array}{|c|c|c|c|} \hline & \text{asthmatic} & \text{not asthmatic} & \text{TOTALS} \\ \hline \text{positive} & 3.672 & 10.566 & 14.238 \\ \hline \text{negative} & 1.728 & 48.134 & 49.862 \\ \hline \text{TOTALS} & 5.400 & 58.700 & 64.100 \\ \hline \end{array}$$
From this, one can read off
$$P(\text{asthmatic} \mid \text{positive}) = 3.672/14.238 \doteq 0.25790$$
and
$$P(\text{asthmatic} \mid \text{negative}) = 1.728/49.862 \doteq 0.34656$$
ETA: Two of the cells—$3.672 = (0.68)(5.400)$ and $48.134 = (0.82)(58.7)$—are filled directly by the given parameters. Everything else is bookkeeping.
The abysmal false positive rate is due to a combination of the low specificity of the test ($82$ percent) and the relatively low prevalence of asthma ($5.4/64.1 \doteq 8.4$ percent), so that although only a minority (sizable, but still a minority) of non-asthmatics test positive, they still dwarf the asthmatics who test positive, because there are so few asthmatics available to test positive in the first place. Even if the test were $100$ percent sensitive, the false positive rate would still have been $10.566/(10.566+5.400) \doteq 66$ percent. | 2019-07-23T00:41:32 | {
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http://math.stackexchange.com/questions/245354/expected-value-of-sums | Expected value of sums
Suppose we draw cards out of a deck without replacement. How many cards do we expect to draw out before we get an ace?
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It depends (slightly) on how one interprets before. There are two interpretations possible: (i) Before means not including the draw that got us the first Ace and (ii) We include in the count the draw that got us the first Ace.
There is no big difference between (i) and (ii): The count (and expectation) in (ii) is just $1$ more than the count, and expectation, in (i), We use interpretation (i). So let $W$ be the number of draws before the first Ace, not including the draw that got us the Ace. We want $E(W)$.
The argument is simple but a bit delicate, so the solution below is given in great detail. Luckily, the actual computation is almost formula-free. We use indicator random variables. Label the $48$ non-Aces $1$ to $48$. Don't bother to label the Aces.
Define random variable $X_i$ by $X_i=1$ if the card with label $i$ was drawn before any Ace, and let $X_i=0$ otherwise. Then $$W=X_1+X_2+\cdots+X_{48}.$$ By the linearity of expectation, which holds even when the random variables are not independent, we have $$E(W)=E(X_1+X_2+\cdots+X_{48})=E(X_1)+E(X_2)+\cdots+E(X_{48}).$$ By symmetry, all the $X_i$ have the same distribution. We find, for example, the probability that $X_1=1$. So we want the probability that card with label $1$ is drawn before any Ace.
Consider the $5$-card collection consisting of the $4$ Aces and the card labelled $1$. All orders of these cards in the deck are equally likely. It follows that the probability that card with label $1$ is in front of the $4$ Aces is $\frac{1}{5}$. Thus $E(X_1)=\frac{1}{5}$.
We conclude that $E(W)=\dfrac{48}{5}$.
If we want to take interpretation (ii), and include the draw that got us the Ace, our expectation is $1+\dfrac{48}{5}=\dfrac{53}{5}$.
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Now I'm wondering if this answer should be considered essentially the same as mine. – Michael Hardy Nov 27 '12 at 1:54
Hard to decide! Mine (of course it's not mine, it is pretty standard stuff, I have used it in lectures) is very formal. – André Nicolas Nov 27 '12 at 1:59
Has anyone written an exposition of the fact that expectations are often easier to find than probabilities? – Michael Hardy Nov 27 '12 at 2:01
Don't know of one. I sure have stressed it. – André Nicolas Nov 27 '12 at 2:04
Distribute the $52$ cards uniformly between $0$ and $1$, so on average they're at $k/53$ for $k=1,2,3,\ldots,52$. The four aces are on average at $1/5,\ 2/5,\ 3/5,\ 4/5$. So $$0.2 = \frac{k}{53}$$ implies $$k = 10.6.$$ and $(0.8)\cdot53 = 42.4$. So on average the four aces are the $10.6$th, $21.2$th, $31.8$th, and $42.4$th cards.
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A good way to go about this is as follows. Let $T$ be the number of cards drawn at the time of the first ace. This is a $\mathbb{N}$ valued random variable. Therefore $$E(T) = \sum_{n=0}^\infty P(T > n).$$ You may find this helpful.
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I think that the answer lies in the Hypergeometric distribution or else k successes in n draws from a finite population of size N containing m successes without replacement (quoting from Wikipedia).
In your case, you want a single success in $p$ draws, and from the distribution you can find an average.
Edit: $$P(X=1) = \frac{\binom{m}{1}\binom{N-m}{n-1}}{\binom{N}{n}}$$ and taking an average should be something like the following $$\bar M = \sum_{i=1}^{N-m}i\frac{\binom{m}{1}\binom{N-m}{i-1}}{\binom{N}{i}}$$ for $N=52, m=4$.
End of edit.
Wikipedia also has some examples with urns, black and white balls, so you can work the answer from those.
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I came across this in The Theory of Gambling and Statistical Logic By Richard A. Epstein so pasting a snippet here. I got this from Google Books (public domain) so hopefully there is no copyright violation here. It's amazing how Andre, Michael and Patrick derived the correct answer differently.
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I did not see this approach in any of the answers, which is one that makes sense to me: you can get the ace in either the 1st, 2nd,..., 49th trial.
Then the expected number of cards dealt is $$1(1/13)+2(12/13)(1/13)+..n(12/13)^{n-1}(1/13)+48(12/13)^{47}(1/13)$$.
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Try this. (edited based on comment below) You will need just 1 pull with probability 4/52 2 with probability (4/51 x 48/52). 48/52 for the probability of not getting and ace in the first and getting it in the second (4/51), and so on. This yields 1 x 4/52 + 2x(4/51 x 48/52) + 3x(4/50 x 48/52 x 44/50) + ... There must be some simplification for the above, but I'ven't looked deeper into, but you get the idea. Here are similar ones
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This assumes there is only one ace in the deck. – Michael Hardy Nov 27 '12 at 1:05
I have always liked this shortcut (proof here), which also holds for continuous random variables defined on $x \geq{0},\,x \in \mathbb{R}$. For some reason it was called "The Darth Vader" rule in my actuarial study manual but I digress. For this problem: $$P(T>n) = \frac{\binom{48}{n}n!}{\binom{52}{n}n!} = \frac{\binom{48}{n}}{\binom{52}{n}}$$ This is easily reasoned as follows:
$P(T>n)$ is the probability that there are no aces up to and including the $n$th draw which means that in these first $n$ draws, out of the $48$ non-aces we count the number of ways to choose $n$ of them and permute them. We divide this by the total number of outcomes for the first $n$ draws which is $\binom{52}{n}n!$ (we must include the aces this time, thus the 52).
Therefore: $$E(T) = \sum_{n=0}^{48}\frac{\binom{48}{n}}{\binom{52}{n}} = \sum_{n=0}^{48}\frac{(52-n)(51-n)(50-n)(49-n)}{52\times51\times50\times49} = 53/5 = 10.6$$
We could sum to $\infty$ like ncmathsadist stated but the values of $P(T>n)$ for $n>48$ are $0$ anyway.
I will admit that I cheated in my final step by using wolframalpha. Wolfram Alpha is a GREAT free resource by the way. I do hope that the logic behind solving the problem is not lost as a result. - Patrick
Sidebar: My initial naive guess was that $E(T) = 13$ since there are on average $12$ cards between each ace. I am trying to figure out an intuitive understanding for why the actual answer is less that $13$ now.
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It is 13 only if there is replacement. As there isn't any replacement the probability goes up so that interval should shrink eventually converging to 10.6. BTW could you elaborate how you simplified the average to 53/5? – broccoli Nov 27 '12 at 7:52
You're right, if there is replacement then $T \sim Geometric(p)$ where $p$ is the probability of finding an ace ($1/13$) and $E(T) = 1/p = 13.$ As for an elaboration I just plugged the sum into wolfram alpha's calculator. Andres solution above is much better as the calculations can be done in your head and his method is applicable to other similar problems. – Patrick Nov 27 '12 at 13:33
If you wonder why $13$ is too big, consider the expected number of trials needed to get all four aces. Would that be $52$? Only if you never get the fourth ace until the last card in the deck. – Michael Hardy Nov 27 '12 at 19:16
Add a fifth ace, uniformly randomly arrange the $53$ cards in a circle, and break the circle into a line at the added ace. By symmetry, the expected number of cards between two of the $5$ aces is $\frac{48}5$, so this is the expected number of cards between the beginning of the line and the first ace.
- | 2016-05-26T11:09:29 | {
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http://math.stackexchange.com/questions/193010/whats-the-chance-that-exactly-2-2s-will-have-appeared-before-the-2nd-8 | # What's the chance that exactly 2 2s will have appeared before the 2nd 8?
A standard deck of 52 cards is randomly shuffled. Bob keeps drawing from the deck until he has drawn two 8s.
What is the probability that, when, he draws the second 8, he has already drawn exactly two 2s?
I think this could be a combinations problem because the order of the drawing two 2s does not matter. However, I am not sure how to account for the fact that he could draw an 8 or a second 8 at any point.
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Hint: one approach is to recognize that all the other cards are just fluff. There are only ${8 \choose 4}=70$ possibilities for the order of 2's and 8's, so list them and count. You can make it less by listing the orders that get to 3 2's or 2 8's and assessing the probability. The first are failures, the second are successes if you have two 2's already. For example, one is that you start with three 2's. This is $\frac 12 \cdot \frac 37 \cdot \frac 26=\frac 1{14}$. Another is 2828. This is $\frac 12 \cdot \frac 47 \cdot \frac 12 \cdot \frac 35=\frac 3{35}$
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Oh! I think I got it from that hint! The only way for this to hold is for the string of combos be 8228[some ordering of 2 2s and 2 8s], 2828[some ordering of 2 2s and 2 8s], or 2288[some ordering of 2 2s and 2 8s]. Hence, the solution is 3 * (4 choose 2) / (8 choose 4), which is approximately 0.257! :D ... right? – David Faux Sep 9 '12 at 4:03
@DavidFaux: Good thought, but I wouldn't be sure that all the cases are equally probable. 8228 is $\frac 48 \frac 47 \frac 36 \frac 35=\frac 3{35}$ while 2288 is $\frac 48 \frac 37 \frac 46 \frac 35$, which is the same, so I think it works. I would report the result as $\frac 9{35}$, which is exact. – Ross Millikan Sep 9 '12 at 4:20
Hmm, I think the combinations are equal because non-2 and non-8 cards are all fluff. If we make the 8s non-unique, and 2s non-unique, this basically becomes a problem involving strings of four 8s and four 2s, right? – David Faux Sep 9 '12 at 4:34
@DavidFaux: That is right. The more I think, I agree with your calculation. I was worried that 22228888 would be less probable than 82282882, but now I don't think so. It is easy to think things are equally probable when they are not. – Ross Millikan Sep 9 '12 at 4:39
As was observed by Ross Millikan, there are only $8$ relevant cards. The first $4$ of these can appear in the dealing in $(8)(7)(6)(5)$ orders, all equally likely.
We count the number of "favourable" orders. These are given by any one of three patterns, which we call $2288$, $2828$, and $8228$.
The pattern $2288$ can occur in $(4)(3)(4)(3)$ ways, as can the other two patterns. Thus the required probability is $$\frac{(3)(4)(3)(4)(3)}{(8)(7)(6)(5)}.$$ If we want to simplify a bit, we get $\dfrac{9}{35}$.
Remark: The problem seems to have been solved correctly in the comments both by you and by Ross Millikan. Then there appears to have been a change of mind.
- | 2015-10-06T23:15:26 | {
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https://www.physicsforums.com/threads/electric-field-of-bent-non-conducting-rod.955159/ | # Electric field of bent non conducting rod
## Homework Statement
You are given a non conducting rod carrying uniformly distributed charge, -Q, that has been bent into a 120° circular arc of radius, R. The axis of symmetry of the arc lies along the x-axis and the origin is at the center of curvature of the arc.
(a) in terms of Q and R, what is the linear charge density, λ?
(b) in terms of Q and R, what is the magnitude and direction the the resulting electric field at the origin?
(c)If the arc is replaced by a point particle carrying charge, -Q, at x=R, by what factor is the resulting electric field at the origin multiplied?
λ=Q/L
E=KQ/R2
dq=λds
ds=Rdθ
## The Attempt at a Solution
(a) λ=Q/L, λ=Q/(⅓2piR) = 3Q/2piR
(b) Cant seem to figure out how to write an integral with bounds. The bounds on my integral below will be from θ=0° → θ=60°
E=2∫dEx=2∫kλdθcosθ/R = 2kλ/R ∫cosθdθ (Like previously stated this integral is being evaluated from θ=0° to θ=60°)
E=(2kλ/R)[sin60°-sin0°] = (kλ√3)/R, it wants in terms of R and Q so now plug in our above statement for λ
E=(3kQ√3)/2piR2
(c) Im a little unsure about what this is asking, from my understanding:
to obtain the new electric field from a point particle which is simply kQ/R2 we must multiply the previous electric field from bent rod by a factor of 2pi/3√3. Not sure if this is correct or if I'm thinking about the question in the wrong way.
Any help would be much appreciated
Thanks!
Related Introductory Physics Homework Help News on Phys.org
kuruman
Homework Helper
Gold Member
Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.
Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?
Consider subdividing the 120o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.
Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?
Okay, Ill work on this and post my results! Thanks!
Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?
Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?
Yea just part c, wasn't sure what it was asking but @kuruman helped me out with that one. Just have a test in a few days and want to make sure I'm doing these problems correctly. Always love reassurance! LOL
You saw the symmetry and ignored the y component. You also used symmetry to cut the integral in half. You integrated in theta and didn’t forget the extra R.
I don’t think you have much to worry about, but I’ll wish you good luck anyway. | 2020-04-06T15:52:47 | {
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https://math.stackexchange.com/questions/2781063/limit-of-frac-sqrtnn1n2-cdots2nn/2781848 | # Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$
Compute the limit $$\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$$
How can this be done? The best I could do was rewrite the limit as $$\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}$$
Following that log suggestion in the comments below: \begin{align} &\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\ &= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\ &= \int_1^2 \ln x \, dx \\ &= (x \ln x - x)\vert_1^2 \\ &= (2 \ln 2 - 2)-(1 \ln 1-1) \\ &= (\ln 4-2)-(0-1) \\ &= \ln 4-1 \\ &= \ln 4 - \ln e \\ &= \ln \left( \frac 4e \right) \end{align}
but I read from somewhere that the answer should be $\frac 4e$.
• Your integral is the log of the answer – robjohn May 14 '18 at 18:15
• @CarlMummert From the accepted answer of this meta question, we can see that this is a well-written question (except the use of $\frac1n$ as a power), and "additional indication of source would not make a difference".You may re-read AmateurMathGuy's reply to your comment to know why they are not required. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• You voted to close because "Being homework is a sufficient reason to vote to close a question", but HW questions are allowed. – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
• @CarlMummert I do not see how my question "has no real context, and does not fit the quality standards that many users expect." I put in the work and effort as this Math SE site suggests. – GarlicBread May 15 '18 at 3:58
• Agreed. This is supported by the Close Queue Review result. All reviewers left this question open. – GNUSupporter 8964民主女神 地下教會 May 15 '18 at 9:00
Hint: let $a_n=(n+1)\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\to L$ implies $\sqrt[n]{a_n}\to L$ (i.e., the ratio test implies the root test).
If you know Sterling's Approximation: $$n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n},$$ then you could approach it as follows:
$$\lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}} = \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}.$$
Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)
• I'm not sure how to rewrite the limit of the Riemann sum as its corresponding definite integral, although I think I managed to arrive at the Riemann sum in my work. – GarlicBread May 14 '18 at 17:41
If you accept using the following limit that has been asked and proved (for example here) many times here on MSE
• $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,
then your limit is easily derived:
$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$
Note that $\displaystyle\frac{\sqrt[n]{n+1}}{n}\neq\left(\frac{n+1}{n}\right)^{1/n}$
Also, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality
With this is mind and going from your first step,
$$S=\lim_{n\to\infty}\frac{\sqrt[n]{(n+1)(n+2)...(2n)}}{n}=\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$
$$\operatorname{ln}(S)=\operatorname{ln}\left(\lim_{n\to\infty}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\operatorname{ln}\left(\frac{1}{n}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}(n+2)^{1/n}...(2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\operatorname{ln}\left((n+1)^{1/n}\right)+\operatorname{ln}\left((n+2)^{1/n}\right)+...+\operatorname{ln}\left((2n)^{1/n}\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\operatorname{ln}\left((n+p)^{1/n}\right)\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\sum_{p=1}^n\left(\frac{1}{n}\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}(n+p)\right)-\operatorname{ln}(n)\right)$$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\left(\operatorname{ln}\left(1+\frac{p}{n}\right)+\operatorname{ln}(n)\right)-\operatorname{ln}(n)\right)$$
This $+\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\operatorname{ln}(n)-\operatorname{ln}(n)=0$
$$\operatorname{ln}(S)=\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)$$
This is what Eric meant and the argument varies from $1$ to $2$
It becomes a little clearer if we let $m = 1/n$
$$\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)=\lim_{m\to\ 0}\left(\left(\sum_{p=1}^{1/m}\operatorname{ln}(1+pm)\right)m\right)=\int_1^2\operatorname{ln}(x)dx=(x\operatorname{ln}(x)-x)\Big\rvert_1^2$$
$$\therefore \operatorname{ln}(S)=2(\operatorname{ln}(2)-1)+1=\operatorname{ln}(4)-1=\operatorname{ln}\left(\frac{4}{e}\right)$$
Finally
$$S=\frac{4}{e}$$ | 2020-01-20T15:02:48 | {
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http://bkdd.zonaunderground.it/matrix-multiplication-algorithm-pseudocode.html | # Matrix Multiplication Algorithm Pseudocode
Abstract— Matrix multiplication is an integral component of most of the systems implementing Graph theory, Numerical algorithms, Digital control, Signal and image processing (i. the first column when multiplied by the matrix. of algorithms in the computational sciences. The booth’s multiplication algorithm is primarily used in computer architectures. Banded Matrix-Vector Multiplication. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Rotation: We use a generalization of Cannon’s algo-rithm as the primary template. I Strassen's algorithm gives a performance improvement for large-ish N, depending on the architecture, e. Write an algorithm in pseudocode to perform the multiplication of a matrix with a vector. Specifically, an input matrix of size can be divided into 4 blocks of matrices. COMP 250 Winter 2016 1 { grade school algorithms Jan. Consider an NxN complex array. This allows us to exploit fast matrix multiplication. 3 Multithreaded merge sort 797 28 Matrix Operations 813 28. Divide-and-Conquer algorithsm for matrix multiplication A = A11 A12 A21 A22 B = B11 B12 B21 B22 C = A×B = C11 C12 C21 C22 Formulas for C11,C12,C21,C22: C11 = A11B11 +A12B21 C12 = A11B12 +A12B22 C21 = A21B11 +A22B21 C22 = A21B12 +A22B22 The First Attempt Straightforward from the formulas above (assuming that n is a power of 2): MMult(A,B,n) 1. Pseudo Code for Union using Mapreduce. What is the least expensive way to form the product of several matrices if the naïve matrix multiplication algorithm is used? [We use the number of scalar multiplications as cost. spawn is to indicate creation of a new thread. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Cannon's algorithm: a distributed algorithm for matrix multiplication especially suitable for computers laid out in an N × N mesh; Coppersmith-Winograd algorithm: square matrix multiplication; Freivalds' algorithm: a randomized algorithm used to verify matrix multiplication. 5D (Ballard and Demmel) ©2012 Scott B. The main purpose of this paper is to present a fast matrix multiplication algorithm taken from the paper of Laderman et al. , what the complexity of the problem is). Pseudocode for the algorithm is given in Figure 1. cludes the RSA cryptosystem, and divide-and-conquer algorithms for integer multiplication, sorting andmedianfinding, aswellasthe fast Fourier transform. Algorithm Examples! Pseudocode! Order of Growth! Algorithms - what are they Fibonacci - Matrix Multiplication. Multithreaded Algorithms Introduction. To perform the addition, numbers in matching postions in the input matrices are added and the result is placed in the same position in the output matrix. We propose two approaches to matrix multiplication: iter-ative approach and block approach. Unlike standard matrix multiplication, MixColumns performs matrix multiplication as per Galois Field 2 8. Although adjacency matrix representation of graph is used, this algorithm can also be implemented using Adjacency List to improve its efficiency. General Matrix Multiplication (GEMM) is the primary component of the level-3 BLAS and of most dense linear algebra algorithms (and many sparse/structured linear algebra algorithms), which in turn have applications in virtually every area of computational science. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. complexity of matrix multiplication is n2 (2n −1) = 2 ⋅(2 −1)⋅τ T1 n n (8. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. •Pseudo code. Problem 6 (checks whether a matrix is symmetric): yes 4. One of the basic operations on matrices is multiplication. In other words, Lagrees with the corresponding k×nsubmatrix of X. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. Regression algorithm pseudocode from [4]: The regression algorithm follows a nested optimization scheme using coordinate descend. 6 Another Recursive Algorithm 4. If you're interested in typesetting algorithmic code, there are a number of choices. However, due to constant factors and realistic modern architecture constraints, these theoretically faster methods are rarely used; instead, the naive brute force approach to matrix multiplication is that which is. Suppose we want to multiply two n by n matrices, A and B. This analysis culminates in Section 4. The practical benefit from improvements to algorithms is therefore potentially very great. Here's a short example from the algorithmicx documentation (with a pseudocode for loop added):. Assume that square matrix A and B are used for multiplication in the following algorithms. Algorithms { CMSC-37000 Divide and Conquer: The Karatsuba algorithm (multiplication of large integers) Instructor: L aszl o Babai Updated 01-21-2015 The Karatsuba algorithm provides a striking example of how the \Divide and Conquer" technique can achieve an asymptotic speedup over an ancient algorithm. We need to create a Toeplitz matrix using a subsection of a data vector on the device. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. This builds on the previous post on recursive square matrix multiplication. to read the matrix into core memory once [19, sect. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. To begin with, the sequential algorithm was implemented using the pseudo-code in [2]. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. 2 Strassen's algorithm for matrix multiplication 4. One key idea in the sorting networks chapter, the 0-1 principle, ap-. x x y matrix by a y x z matrix creates an x x z matrix. The cofactor matrix is the matrix of determinants of the minors A ij multiplied by -1 i+j. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. Write pseudocode for Strassen’s algorithm. r-1 (mod n) where the integers a and b are smaller than the modulus. Complexity Calculation How many additions of integers and multiplications of integers are used by the matrix multiplication algorithm to multiply two n * n matrices. I won't give the pseudo code here for these ones, but they are naive recursive algorithm, bottom up algorithm, naive recursive squaring and recursive squaring. MPI Matrix-Matrix Multiplication Matrix Products Parallel 2-D Matrix Multiplication Characteristics Computationally independent: each element computed in the result matrix C, c ij, is, in principle, independent of all the other elements. 6 Another Recursive Algorithm 4. Provide your analysis for the following problem statement: Write a program that will calculate the results for the multiplication table up to 10x10 in steps of 1 beginning at 1. cient implementation of sparse matrix multiplication on a memory intensive associative processor (AP), verified by extensive AP simulation using a large collection of sparse matrices [41]. We then "combine" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The multiplier contains only 0s and 1s,. The algorithms classes I teach at Illinois have two significant prerequisites: a course on discrete mathematics and a course on fundamental data structures. Block matrices are briefly discussed using 2 × 2 block matrices. I have a question for you about your approach. What is a Spanning tree? Explain Prim’s Minimum cost spanning tree algorithm with suitable example. Before going to main problem first remember some basis. This page contains the order of topics contained in lectures, listed as a sequence of modules. Which is faster for this value of n?. To save space and running time it is critical to only store the nonzero elements. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. Idea - Block Matrix Multiplication The idea behind Strassen’s algorithm is in the formulation of matrix multiplication as a recursive problem. You can use a pseudocode environment algpseudocode offered by algorithmicx. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. 1 The naive matrix multiplication algorithm Let A and B be two n £ n matrices. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. Pseudocode of the rest of the algorithm: Iterative Matrix Multiplication I'm a bit unhappy with your code, because it's so hard to read tbh. Material for the algorithms class taught by Emanuele "Manu" Viola. Dynamic Programming—Chained Matrix Multiplication Multiplying unequal matrices • Suppose we want to multiply two matrices do not have the same number of rows and columns • We can multiply two matrices A 1 and A 2 only if the number of columns of A 1 is equal to the number of rows of A 2. Example of Matrix Multiplication by Fox Method Thomas Anastasio November 23, 2003 Fox's algorithm for matrix multiplication is described in Pacheco1. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd’s fast matrix multiplication. It is referenced specifically in the pseudocode but that is not the only location where it is appropriate to call it. If you are interested in a Modified Gauss-Jordan Algorithm, you can see this. Algorithm for the Transpose of a Sparse-Matrix: This is the algorithm that converts a compressed-column sparse matrix into a compressed-row sparse matrix. The practical benefit from improvements to algorithms is therefore potentially very great. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. Here, we will discuss the implementation of matrix multiplication on various communication networks like mesh and. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoefficiency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Solutions for CLRS Exercise 4. 2 Algorithmic Techniques 5. \begin{algorithm} \caption{Euclid's algorithm}\label{euclid} \. 1 Naive Matrix Multiplication 4. Matrix-matrix multiplication takes a triply nested loop. I'm just doing a self-study of Algorithms & Data structures and I'd like to know if anyone has a C# (or C++) implementation of Strassen's Algorithm for Matrix Multiplication? I'd just like to run it and see what it does and get more of an idea of how it goes to work. Let us start with a very simple example th. One common practice is to translate convolution to im2col and GEMM, which lays out all patches into a matrix. 84 videos Play all Algorithms Abdul Bari; Derivatives explained - Duration: 10:13. One of the basic operations on matrices is multiplication. The algorithm is called a (7, 4) code, because it requires seven bits to encoded four bits of data. Recurrence equation for Divide and Conquer: If the size of problem ‘p’ is n and the sizes of the ‘k’ sub problems are n1, n2…. Write An Algorithm To Find The Power Of A Number. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. This page contains the order of topics contained in lectures, listed as a sequence of modules. GitHub Gist: instantly share code, notes, and snippets. Adjacency matrix representation of graph where the n X n matrix W = (wij) of edge weights. Matrix multiplication algorithms. Definition of Flowchart A flowchart is the graphical or pictorial representation of an algorithm with the help of different symbols, shapes and arrows in order to demonstrate a process or a program. complexity by packing the inner loops into a single matrix product as shown in Algorithm 2. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. The unit of. Matrix multiplication is the process of taking two or more n x n matrices and calculating a product by summing the product of each row in one matrix with the column of the second matrix. As in case of developing the matrix-vector multiplication algorithm, we use one-dimensional arrays, where matrices are stored rowwise. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. From Math Insight. We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. The problem is quite easy when n is relatively small. 1 and the Junto. Below is some sample output. It is the technique still used to train large deep learning networks. org) I now want to use strassen's method which I learned as follows:. Unlike general multiplication, matrix multiplication is not commutative. o The number of additions and multiplication's required for this algorithm can be calculated as follows: To calculate one entry in the product matrix, we must perform k multiplications and k-1 additions. for i = 1 to n. 1 and Step 3. We can use simple recursion, f(n) = f(n-1) + f(n-2), or we can use dynamic programming approach to avoid the calculation of same function over and over again. Write a c program to find out transport of a matrix. it explains matrix multiplication. Antoine Vigneron (UNIST) CSE331 Lecture 5 July 11, 2017 3 / 19. In other words, Lagrees with the corresponding k×nsubmatrix of X. Question: Show Map Reduce implementation for the following two tasks using pseudocode. - Explain the difference between an LED and OLED display. matrix multiplication; this means that matrix multiply based methods for determining primitivity cannot be sped up anymore at this time. 3343(abc)(log7)/(3) 2. Block matrices are briefly discussed using 2 × 2 block matrices. x x y matrix by a y x z matrix creates an x x z matrix. Section 5 provides a com-parison with related works. Summary I Strassen rst to show matrix multiplication can be done faster than O(N3) time. In the next three parts, you may be writing pseudocode. That’s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. This work is licensed under aCreative Commons. parallel before a loop means each iteration of the loop are independant from each other and can be run in parallel. Notes: A common reference for double-precision matrix multiplication is the dgemm ( d ouble-precision ge neral m atrix- m atrix multiply) routine in the level-3 BLAS. The application. Given three n x n matrices, Freivalds' algorithm determines in O(kn^2) whether the matrices are equal for a chosen k value with a probability of failure less than 2^-k. Name the algorithmic technique used. 3 Matrix Multiplication for Banded Matrices. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is not a liated with o cial Algorithms group(s) or company(s). 0 is there to suggest that different values can be used, but they should be related to the number of input variables. The following algorithm multiplies nxn matrices A and B: // Initialize C. Strassen's algorithm, the original Fast Matrix Multiplication (FMM) algorithm, has long fascinated computer scientists due to its startling property of reducing the number of computations required for multiplying. and similarly for the bottom row. We know that, to multiply two matrices it is condition that, number of columns in first matrix should be equal to number of rows in second matrix. Matrix Multiplication: Strassen's Algorithm. What Is The Main Operation Of This Algorithm? C. We need to create a Toeplitz matrix using a subsection of a data vector on the device. of algorithms in the computational sciences. You can call the algorithm on sub-matrices of dimensions n-1 each when the size of the matrix is odd in the recursive step and calculate 2n-1 remaining elements using normal vector multiplication in O(n) each and a total of O(n^2). We tackle scenarios such as matrix multiplication and linear regression/classification in which we wish to estimate inner products between pairs of vectors from two possibly different sources. What is the best algorithm for matrix multiplication ? Actually there are several algorithm exist for matrix multiplication. n and r is relatively prime number to n (gcd (n, r)= 1). for k = 1 to n. Part I was about simple matrix multiplication algorithms and Part II was about the Strassen algorithm. If A is the adjacency matrix of G, then (A I)n 1 is the adjacency matrix of G*. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. 3) where τ is the execution time for an elementary computational operation such as multiplication or addition. Section 5 provides a com-parison with related works. The aim is to get the idea quickly and also easy to read without details. The former is suitable for sparse matrices, while the latter is appropriate for dense matrices with low communication overhead. We could break down the steps as follows. , what the complexity of the problem is). In matrix addition, one row element of first matrix is individually added to corresponding column elements. 4 uses dynamic programming to find an optimal triangulation of a convex polygon, a problem that is surprisingly similar to matrix-chain multiplication. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. Then, user is asked to enter two matrix and finally the output of two matrix is calculated and displayed. That’s very important because for small n (usually n < 45) the general algorithm is practically a better choice. Each matrix Mk has dimension pk-1 x pk. The Floyd Warshall algorithm, itis the algorithm in which there is the use of different characterization of structure for a shortest path that we used in the matrix multiplication which is based on all pair algorithms. Use row communicators and column communicators to scatter and broadcast the vector. We ended up pursuing a different route, but I decided to continue pursuing the problem on my own time. 2x2 Matrix Multiplication Calculator is an online tool programmed to perform multiplication operation between the two matrices A and B. From Math Insight. This work is licensed under aCreative Commons. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication. Provide a specification to describe the behaviour of this algorithm, and prove that it correctly implements its specification. However, this algorithm is infamously inapplicable, as it relies on Coppersmith and Winograd’s fast matrix multiplication. The weights and values of 6. Sparse Matrix Multiplication. I am trying to implement a multiplication algorithm by overloading the *= operator. Matrix Multiplication in Case of Block-Striped Data Decomposition Let us consider two parallel matrix multiplication algorithms. 3 Storage formats 3. Other types of algorithms for this problem appear in [15, 16]. In particular, this includes judging which data structures, libraries, frameworks, programming languages, and hardware platforms are appropriate for the computational task, and using them effectively in the implementation. How do they differ? - Is a pixel a little square? If not, what is it? What implications does this have? Give at least 2. One of the basic operations on matrices is multiplication. It is the technique still used to train large deep learning networks. Then, we'll present a few examples to give you a better idea. Floyd Warshall. What Is The Main Operation Of This Algorithm? C. For instance, the algorithm we're interested in looking at, Dijkstra's algorithm, only works if none of the edges on the graph have negative weights -- the "time" it takes to traverse the edge is somehow less than 0. emit (key, result). GATEBOOK Video Lectures 3,203 views. These lectures were designed for the latter part of the MIT undergraduate class 6. Matrix Multiplication Problem - Duration: 27:38. - Overall complexity of parallel matrix-vector multiplication algorithm ( n2=p+n+logp) - Isoefficiency of the parallel algorithm Time complexity of sequential algorithm: ( n2) Only overhead in parallel algorithm due to all-gather For reasonably large n, message transmission time is greater than message latency. Other types of algorithms for this problem appear in [15, 16]. • Algorithms are step-by-step procedures for problem solving • They should have the following properties: •Generality •Finiteness •Non-ambiguity (rigorousness) •Efficiency • Data processed by an algorithm can be • simple • structured (e. Strassen's method of matrix multiplication is a typical divide and conquer algorithm. Although we won't describe this step in detail, it is important to note that this multiplication has the property of operating independently over each of the columns of the initial matrix, i. Matrix Multiplication; Matrix Multiplication Parenthesization; Brute Force Solution: Try all possible parenthesizations; Dynamic Programming Solution (4 steps) Step 1: Characterize Structure of Optimal Solutioon; Step 2: Define recursive solution; Recursive Solution; Analysis; Duplicate Subproblems; Unique Subproblems; Step 3: Bottom-Up Approach; Dynamic Programming. 5 Maximum Flow. Matrix Multiplication c c2 1= r2 A1 A2 r1 r2! = r1 ! c2 (r1 ! c2) ! c1 = multiplications If r 1 = c 1 = r 2 = c 2 = N, this standard approach takes ( N3): I For every row ~r (N of them) I For every column ~c (N of them) I Take their inner product: r c using N multiplications 2. listing algorithms. Idea - Block Matrix Multiplication The idea behind Strassen's algorithm is in the formulation of matrix multiplication as a recursive problem. Describe how an array can be effectively used to store a sparse matrix. Matrix mulitplication using Linked List - posted in C and C++: I have to implement Matrix multiplication using singly linked list. You don't need multiplication facts to use the Russian peasant algorithm; you only need to double numbers, cut them in half, and add them up. The unit of. We then "combine" the middle row of the key matrix with the column vector to get the middle element of the resulting column vector. The problem is quite easy when n is relatively small. John, Your comment about matrix multiplication was forwarded to the mahout-user emailing list. emphasizes the use of pseudo code in the introductory Computer Science, this approach is to teach students how to first develop a pseudo code representation of a solution to a problem and then create the code from that pseudo code. Section 3 pro-vides implementation details on our design. In the other hand the algorithm of Strassen is not much faster than the general n^3 matrix multiplication algorithm. 2 shows the calculate steps of covariance matrix, mainly including: complex conjugate multiplication between the lines of input matrix, then do an accumulation operation. This immediately leads to a counting algorithm with running time Θ(n3) respectively Θ(nγ), where γ is the matrix multiplication exponent. Divide-and-conquer multiplication. 2 StrassenÕs algorithm for matrix multiplication If you have seen matrices before, then you probably know how to multiply them. These are lecture notes used in CSCE 310 (Data Structures & Algorithms) at the University of Nebraska|Lincoln. count example presented in Section 2. The matrix M and the vector v each will be stored in a file of the DFS. Pseudo-code for i = 1 : n yi yi +axi Number of flops is 2n Pseudo-code cannot run on any computer, but are human readable and straightforward to convert into real codes in any programming language (e. Matrix multiplication is not commutative, but it is associative, so the chain can be parenthesized in whatever manner deemed best. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Prim's Algorithm Implementation in C++. Write An Algorithm To Find The Power Of A Number. Flowchart for Matrix multiplication : Algorithm for Matrix multiplication :. Strassen's matrix multiplication program in c 11. Program the divide and conquer matrix multiplication using 1) standard algorithm 2) recursion 3) strassen’s method. Write the Pseudo Code of the matrix multiplication program that performs the worst execution time and explain why - Answered by a verified Programmer We use cookies to give you the best possible experience on our website. Sparse matrices, which are common in scientific applications, are matrices in which most elements are zero. The dimensions are stored in array. Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. Our first example of dynamic programming is an algorithm that solves the problem of matrix-chain multiplication. 2 Multithreaded matrix multiplication 792 27. (Otherwise, you should read Section D. When distributing the vector among processors, implement the algorithm shown in Figure (b) on page 22 of lecture notes “ Parallel matrix algorithms (part 2) ”. Summary: The two fast Fibonacci algorithms are matrix exponentiation and fast doubling, each having an asymptotic complexity of $$Θ(\log n)$$ bigint arithmetic operations. The simplest sparse matrix data structure is a list of the nonzero entries in arbitrary order. 1 The SUMMA Algorithm. The introduction of the technique is attributed to a 1962 paper by Karatsuba, and indeed it is sometimes called Karatusba multiplication. We can build a sketch as we scan through the matrix. I am trying to write pseudo code in my paper. I implement these three algorithms from pseudocode from the book to Java code: Merge-Sort Java version：. Animated Algorithms (sorting, priority queues, Huffman, Matrix chain multiplication, MST, Dijkstra) Graph Algorithms (Dijkstra, Prim, Kruskal, Ford-Fulkerson) Java and Web Based Algorithm Animation (JAWAA). What is the main operation of this algorithm? c. Solutions for CLRS Exercise 4. 4 RESULTS We evaluate our implementation by testing its performance on one. com Free Programming Books Disclaimer This is an uno cial free book created for educational purposes and is. There is a faster way to multiply, though, caled the divide-and-conquer approach. 3 perform the multiplication operation. Output Y: d by N matrix consisting of d D dimensional embedding coordinates for the input points. Freivalds' algorithm is a probabilistic randomized algorithm used to verify matrix multiplication. Better asymptotic bounds on the time required to multiply matrices have been known since the work of Strassen in the 1960s, but it is still unknown what the optimal time is (i. In this post I will explore how the divide and conquer algorithm approach is applied to matrix multiplication. Purdue University Purdue e-Pubs ECE Technical Reports Electrical and Computer Engineering 9-1-1992 Implementation of back-propagation neural networks with MatLab. Have you considered doing the multiplication in a single step by storing the the first matrix in column major order and the second in row major order?. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. CS 2073 Lab 10: Matrix Multiplication Using Pointers Chia-Tien Dan Lo Department of Computer Science, UTSA I Objectives Show how to manipulate commandline arguments in C Demonstrate your ability to read matrices from les Demonstrate your ability to use pointers for matrix multiplication II Hand-in Requirements. You are given 5 different algorithms for different purposes and their pseudocodes that are listed below. Related Questions More Answers Below. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. Since we have not covered multiplication yet, a function has been provided to you. Two groups of algorithms belonging to this class are called the matrix method, and the Wallace-tree method, respectively. Matrix multiplication is one of the most fundamental operations in linear algebra and serves as the main building block in many different algorithms, including the solution of systems of linear equations, matrix inversion, evaluation of the matrix determinant, in signal processing, and the transitive closure of a graph. LLE Algorithm Pseudocode (Notes, e. 2 Algorithmic Techniques 5. Find f(n): n th Fibonacci number. 3) you should encapsulate a matrix into a class, if it is supposed to be c++ (then 2 will be obsolete) 4) your code will be easier to understand (for you as well) if you use better names for x,y,z,i, k and j. Application. using matrix multiplication Let G=(V,E) be a directed graph. Book shows pseudocode for simple divide and conquer matrix multiplication: n = A. r-1 (mod n) where the integers a and b are smaller than the modulus. Matrix multiplication of two sparse matrices is a fundamental operation in linear Baye sian inverse problems for computing covariance matrices of observations and a posteriori uncertainties. HOME ; A comparison of numerical approaches to the solution of the time-dependent Schrödinger equation in one dimension. complexity of matrix multiplication is n2 (2n −1) = 2 ⋅(2 −1)⋅τ T1 n n (8. The current best algorithm for matrix multiplication O(n2:373) was developed by Stanford's own Virginia Williams[5]. Pseudocode for the algorithm is given in Figure 1. Make good use of matrix multiplication! It avoids a lot of loops, so it makes your code cleaner and faster! DO NOT assume that I will answer your email questions or posts to the discussion forum after 3pm on Sunday. which can be signi cantly smaller than their dense equivalent. Pseudocode for Matrix Vector Multiplication by Mapreduce. , the shapes are 2 n × 2 n for some n. i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table i) Multiplication of two matrices ii) Computing Group-by and aggregation of a relational table. \begin{algorithm} \caption{Euclid’s algorithm}\label{euclid} \. Since we have not covered multiplication yet, a function has been provided to you. Matrix-matrix multiplication takes a triply nested loop. 2 Multithreaded matrix multiplication 792 27. 2 Matrix-Matrix Multiplication. So assuming that both these multiplication steps are executed every time the loop executes, we see that 2. Alternative approaches can be seen as straight forward iteration over the nodes or edges of the graph. In grade 1, you learned how to count up to ten and to do basic arithmetic using your ngers. Integer-multiplication, Matrix Multiplication - Strassen Alg You study after every class/week, the syllabus accumulates fast before you know! Aug 26, W (Drop w/o W grade, Aug 28) Dynamic Programming: 0-1Knapsack. The definition of matrix multiplication is that if C = AB for an n × m matrix A and an m × p matrix B, then C is an n × p matrix with entries = ∑ =. As examples, pseudocode is presented for the inner product, the Frobenius matrix norm, and matrix multiplication. Explain why memoization fails to speed up a good divide-and-conquer algorithm like merge-sort. Define the meaning of your variables. Can some one please help me to format it. Section 5 provides a com-parison with related works. With the current implementation of the cuBlas functions we need to write kernel code to do this efficiently. The problem is quite easy when n is relatively small. SPARSE MATRICES C/C++ Assignment Help, Online C/C++ Project Help and Homework Help introduction A matrix is a mathematical object that arises in many physical problems. Pseudocode Matrix Multiplication. This relies on the block partitioning which works for all square matrices whose dimensions are powers of two, i. In matrix multiplication, one row element of first matrix is individually multiplied by all column elements and added. Order of both of the matrices are n × n. The word is derived from the phonetic pronunciation of the last name of Abu Ja'far Mohammed ibn Musa al-Khowarizmi, who. Adjacency-matrix and adjacency-list representations Breadth-first and depth-first search using adjacency lists Computing connected components of a graph Strongly-connected and biconnected components Topological sorting Algebraic algorithms: Strassen matrix multiplication algorithm The Four Russians boolean matrix multiplication Winograd's algorithm. Matrix Multiplication Algorithm. Faster matrix multiplication in general is an important applied topic, because it can speed up all sorts of scientific, engineering, and ML algorithms that have it as a step (often one of the bottleneck steps). At the end of the lecture, we saw the reduce SUM operation, which divides the input into two halves, recursively calls itself to obtain the sum of these smaller inputs, and returns the sum of the results from those. Specifically, an input matrix of size can be divided into 4 blocks of matrices. GATEBOOK Video Lectures 3,203 views. 5D (Ballard and Demmel) ©2012 Scott B. Of course, writing pseudocode is child's play compared to actually implementing a real algorithm. In section 4, we discuss our experimental results from the real hardware implementation. return C {C = [cij ] is the product of A and B} 22. Benchmarked it to be 4x faster than the scalar version (on a Pentium M, using GCC 4. | 2019-12-10T03:20:41 | {
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http://mathhelpforum.com/pre-calculus/117597-cube-root-times-cube-root.html | # Math Help - Cube Root times Cube Root
1. ## Cube Root times Cube Root
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)
I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.
Is this the correct way to write the original problem another way?
I finally got (432m^4n^2)^2/3.
Is this correct?
2. Originally Posted by sharkman
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)
I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.
Is this the correct way to write the original problem another way?
I finally got (432m^4n^2)^2/3.
Is this correct?
Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$
In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.
Using the fact I gave first, this means than the answer is...
Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.
Originally Posted by craig
Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$
In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.
Using the fact I gave first, this means than the answer is...
Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.
The answer should be cubert{432m^4n^2}, right?
4. Correct
5. ## great
Originally Posted by craig
Correct
Great! Thanks!
6. Originally Posted by sharkman
The answer should be cubert{432m^4n^2}, right?
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.
The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
7. ## ok
Originally Posted by HallsofIvy
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.
The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
I understand. You decided to break the problem down a bit more. | 2014-12-21T05:33:42 | {
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https://math.stackexchange.com/questions/3182963/prove-that-sim-is-an-equivalence-relation | # Prove that $\sim$ is an equivalence relation
Let $$A = \{1, 2, 3,...,9\}$$ and let $$\sim$$ be the relation on $$A\times A$$ defined by $$(a,b) \sim(c,d)$$ if $$a+d = b+c$$.
Prove that $$\sim$$ is an equivalence relation.
Really stuck on this question, maybe its the use of ~ that is throwing me off. Any suggestions on how to do this? Maybe I could swith ~ out for something else while solving then put it back in.
Thanks
• You have to show ~ is reflexive, symmetric, and transitive. Note $a+d=b+c \iff a-b=c-d$ – J. W. Tanner Apr 10 '19 at 21:20
• (1) If the symbol ~ annoys you, simply replace it by the more ordinary symbol R. (2) If you are stuck, it may come from the fact that, in this problem, the "elements" you are deling with are, in fact, ordered pairs. So for example, proving reflexivity amounts to proving that any ordred pair (a,b) has the relation R ( or ~ ) with itself , that is (a,b) R (a,b). Using the defining formula, you could see what you have to prove, and then prove if using basic arithmetic. – Saint James Apr 10 '19 at 21:25
• The symbol "~" is often used to denote a relation in case this relation is an equivalence relation. But, before having proved that it is actually an equivalence relation in this case, you are not allowed to read it as " is equivalent to". So, accordind to me, the best thing do do is to replace it by the ordinary symbol R, while you are doing your proof. – Saint James Apr 10 '19 at 21:35
Clearly, the statement $$a+d=b+c$$ is just saying that $$a-b=c-d$$, i.e. two ordered pairs are similar if the difference of their terms is equal. To show it's an equivalence relation, we need reflexivity, transitivity and symmetry. The first and last are obvious.
For transitivity, we need to show $$x\sim y$$ and $$y\sim z$$ implies $$x\sim z$$. Let $$x=(x_1,x_2),y=(y_1,y_2),z=(z_1,z_2)$$. The conditions tell you that $$x_1-x_2=y_1-y_2$$, but also that $$y_1-y_2=z_1-z_2$$. Clearly then, $$x_1-x_2=z_1-z_2$$. But this is exactly what it means for $$x\sim z$$!
I don't normally do this but I think it may clear some conceptions up. If you want to get to how to effectively solve this skip to the end.
...
Maybe an example first:
If we let $$A = (3,6)$$ then $$A\sim (m,n)$$ if $$3+ n = 6+m$$ so the list of $$(m,n)$$ where $$A \sim (m,n)$$ are the ones where $$n = 3+m$$ so $$\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$$ are all the $$(m,n)$$ where $$A \sim (m,n)$$.
We can go a little further.
If $$(a,b) \equiv (c,d)$$ then $$a+d = b+c$$ and $$a-b = c-d$$.
So the "classes" are all the ones related to $$(x_1, x_2)$$ where $$x_1 - x_2 =$$:
$$-8$$: $$\{(1,9)\}$$
$$-7$$: $$\{(1,8),(2,9)\}$$
$$-6$$: $$\{(1,7),(2,8),(3,9)\}$$
$$-5$$: $$\{(1,6),(2,7),(3,8),(4,9)\}$$
$$-4$$: $$\{(1,5),(2,6),(3,7),(4,8),(5,9)\}$$
$$-3$$: $$\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$$
$$-2$$: $$\{(1,3),(2,4),(3,5),(4,6),(5,7),(6,8),(7,9)\}$$
$$-1$$: $$\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)\}$$
$$0$$: $$\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)\}$$
$$1$$: $$\{(2,1),(3,2),(4,3),(5,4),(6,5),(7,6),(8,7),(9,8)\}$$
$$2$$: $$\{(3,1),(4,2),(5,3),(6,4),(7,5),(8,6),(9,7)\}$$
$$3$$: $$\{(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)\}$$
$$4$$: $$\{(5,1),(6,2),(7,3),(8,4),(9,5)\}$$
$$5$$: $$\{(6,1),(7,2),(8,3),(9,4)\}$$
$$6$$: $$\{(7,1),(8,2),(9,3)\}$$
$$7$$: $$\{(8,1),(9,2)\}$$
$$8$$: $$\{(9,1)\}$$
All the pairs in each set are related to each other this is an "equivalence" because every pair is is exactly one class and can be interchanged with others in the class with no ambiguity.
A relationship could not be equivalent if some elemtns are in more than one list. Of some elements aren't in any list. Or if such bunch of lists wouldn't make any sense to make.
A relationship is an equivalence if
1) it is reflexive. that is for every pair $$(a,b)$$ then $$(a,b)\sim (a,b)$$ or for every $$(a,b)$$ then $$a-b = a-b$$. That's always true.
If we view it as $$(a,b)\sim (c,d)$$ if $$a-b = c-d$$ then we'd have $$a-b = a-b$$ which is of course true.
2) it is symmetric. For any pair $$(a,b) \sim (c,d)$$ then $$(c,d)\sim (a,b)$$. That is is $$a+d = b+c$$ then $$c+b = d+ a$$. That's ... clear.
If we view is as $$(a,b)\sim (c,d)$$ if $$a-b= c-d$$ then if $$a-b = c-d$$ then $$c-d = a-b$$ so $$(a,b)\sim(c,d)\iff (c,d)\sim (a,b)$$.
3) it is transitive. That is if $$(a,b)\sim (c,d)$$ and $$(c,d)\sim (e,f)$$ then $$(a,b) \sim (e,f)$$.
So if $$a+ d=b+c$$ and $$c+f = d+ e$$ then we have to prove that means $$a+f = b+e$$.
We can do that via arithmetic $$a+d +c+f =b+c + d + e$$ so $$a+f = b+e$$.
Just use the definitions:
For reflexivity, you have to show that for any $$(a,b)$$, it is true that $$(a,b)\sim (a,b)$$, i.e. that $$a+b=b+a$$. Well, $$a$$ and $$b$$ are natural numbers, and those are commutative, so yes, this is true.
Symmetry: here you have to show that if $$(a,b) \sim (c,d)$$, then $$(c,d) \sim (a,b)$$, which is to say: show that if $$a+d=b+c$$, then $$c+b=d+a$$ ... can you show this?
Finally, now that I have shown what you need to prove to demonstrate reflexivity and symmetry, can you figure out what you need to show to demonstrate transitivity? | 2020-03-31T08:06:45 | {
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https://byjus.com/question-answer/the-average-score-of-boys-in-an-examination-of-a-school-is-71-and-that/ | Question
# The average score of boys in an examination of a school is $$71$$ and that of girls is $$73$$. The average score of the school in that examination is $$71.8$$. The ration of the number of boys to the number of girls appeared in the examination, is
A
3:2
B
3:4
C
1:2
D
2:1
Solution
## The correct option is A $$3:2$$Let there be $$n_1$$ boys and $$n_2$$ girls. and $$\overline{X_1}$$ and $$\overline{X_2}$$ be the average scores of boys and girls respectively.Then $$\bar { { X }_{ 1 } } =71,\bar { { X }_{ 2 } } =63$$ and $$\overline{X} = 71.8$$$$\therefore \quad \overline{X} = \displaystyle\frac{n_1\overline{X_1} + n_2\overline{X_2}}{n_1+n_2}$$$$\Rightarrow \quad 71.8 = \displaystyle\frac{n_1\times71 + n_2\times73}{n_1+n_2}$$$$\Rightarrow \quad 71.8n_1 + 71.8n_2 = 71n_1 + 73n_2$$$$\Rightarrow \quad 0.8n_1 = 1.2n_2 \quad \Rightarrow 8n_1 = 12n_2$$$$\Rightarrow \quad \displaystyle\frac{n_1}{n_2} = \displaystyle\frac{12}{8} = \displaystyle\frac{3}{2}$$Mathematics
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View More | 2022-01-29T10:21:38 | {
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http://www.lofoya.com/Solved/1739/two-squares-are-chosen-at-random-on-a-chessboard-what-is-the | # Difficult Probability Solved QuestionAptitude Discussion
Q. Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
✔ A. 1/18 ✖ B. 64/4032 ✖ C. 63/64 ✖ D. 1/9
Solution:
Option(A) is correct
Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
So sample space$=^{64}C_2$
Now there are 7 unique adjacent square sets in each row and each column.
i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
Hence Required Probability
$=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)$
$=\dfrac{112}{^{64}C_2}$
$=\dfrac{1}{18}$
Also, there could be other solution too if we consider ALL possible squares and not only the smallest squares. (i.e. if we consider that total squares are more than 64).
In that case, let us first calculate the sample space i.e. total number of squares on a chess board.
1, 8x8 square
4, 7x7 squares
9, 6x6 squares
16, 5x5 squares
25, 4x4 squares
36, 3x3 squares
49, 2x2 squares
64, 1x1 squares
Therefore, there are actually:
$=64+49+36+25+16+9+4+1$
$=8^2+7^2+6^2+...1^2=204$
squares on a chessboard!
So sample space$=^{204}C_4$
Now if we assume that 2 squares can have a side common only if they are of the same dimensions:
Let the length of the smallest square be 1 unit.
So the possibility of selecting 2 squares with a common side (As calculated in the first method above):
having the side length greater than 4 unit length = 0
having the side length of 4 unit = 1×(2 rows + 2 columns) = 4
having the side length of 3 unit = 3×(6 rows + 6 columns) = 36
having the side length of 2 unit = 5×(7 rows + 7 columns) = 70
having the side length of 1 unit = 7×(8 rows + 8 columns) = 112
So total number of favourable cases:
$=4+36+70+112=222$
Hence Required Probability
$=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)$
$=\dfrac{222}{^{204}C_2}$
Edit: For an alternative solution, check comment from Bhavya Shah.
Edit 2: For yet another alternative solution, check comment by RandomMathMan.
## (12) Comment(s)
Gaurav Karnani
()
Total number of Cases for a chessboard$7*(7*2+1)$ that equals $112$
P= $112/64C2$
Gaurav Karnani
()
Forgot to add seven to the number of cases for the last row. please amend it accordingly.
RandomMathMan
()
There's a much easier way to do this. On a chessboard, there are 36 squares that border 4 other squares, 24 squares that border 3 other squares, and 4 squares that border only 2 other squares.
$\dfrac{36}{64} \times \dfrac{4}{63} + \dfrac{24}{64} \times \dfrac{3}{63} + \dfrac{4}{64} \times \dfrac{2}{63} = \dfrac{1}{18}$
Siddharth
()
Exactly how I too solved this. Much simpler than the other alternatives given here.
Bhavya Shah
()
For each row, there are 7 ways to select adjacent squares and there are 8 such rows.
Similarly for columns, 7 squares could be selected from one column and there are 8 such columns.
So no. of ways of selecting adjacent squares $= 7*8*2$
Total number of ways of selecting squares of chessboard $= ^{64}C_2$
Probability $=\dfrac{7*8*2}{^{64}C_2}=\dfrac{1}{18}$
Poonam Pipaliya
()
sir,,how 7 squares in each row and column remain..??..!! and how it can be favorable case..??..!!
Anuj
()
Lets talk of the rows first. Once you pick up a square, you leave that row, so you have 7 remaining rows.
Similar argument is valid for columns.
Rishi
()
I guess the alternative method given in the solution is easier to understand, why don't you have a look at that.
Arshal
()
Guys, I think I'm right but, as I have been hitting my head on the wall of probability for last 7 hours I'm kinda in concussion...
So is choosing 2 consecutive elements out of 8 equal to choosing 1 element out of 7
()
$C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)$
There are 4 corner squares for which there are only 2 choices for selecting next square....
Similarly, 24 squares for which there are 3 choices and remaining 36 have 4 choices
So, the answer should be 1/9
Correct me if i am wrong!!!
()
As I counted every square non repeated
So, total no of unique square=
$\dfrac{((C(4,1)*2+C(24,1)*3+C(36,1)*4)/2)}{C(64,2)}$
$=\dfrac{1}{18}$ | 2016-12-10T10:46:38 | {
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http://mathhelpforum.com/business-math/231150-calculating-compound-interest-non-linear-interest-growth.html | # Thread: calculating compound interest with non-linear interest growth
1. ## calculating compound interest with non-linear interest growth
Hello, I would really appreciate some help with the following problem.
I'd like to derive a formula to determine the value of an investment where the annual yield increases by a fixed percentage each year. The interest paid is re-invested each year. The standard compounding formulas use a fixed yield so I've not been able to use them in this case.
As an example, suppose I have 1000 dollars in an investment with an initial starting yield (y) of 3% which grows (g) at 5% each year.
End of Year 1 : Yield was 3.00%, Investment value = 1000 + (1000 * 3%) = 1030.00
End of Year 2 : Yield was 3.15% (3% increased by 5%), Investment value = 1030 + (1030 x 3.15%) = 1062.45
End of Year 3 : Yield was 3.31% (3.15% increased by 5%), Investment value = 1062.45 + (1062.45 x 3.31%) = 1097.59
I believe the calculation is a series in the form
Value = P.(1+y0).(1+y1).(1+y2).(...)
where P = initial investment amount, r0 = yield in year 0, r1 = yield in year 1, r2 = yield in year 2 etc.
and where y0 = y.(1+g)^t etc.
If I write the above formula out substituting y0, y1, etc. I get
Value = P.(1+y(1+g)^t).(1+y(1+g)^(t-1)).(1+y(1+g)^(t-2)).(...) where t ranges from 0 to the target year n.
Is it possible to convert this expression into a general formula in the form of Value = f(t) where t = time in years, e.g. some kind of exponential function and how might I go about doing that?
Thank you for any help or insight that you can provide me!
~ Trevor
2. ## Re: calculating compound interest with non-linear interest growth
One way of doing this would be to find a single yield that results in a same future value using the series of yields
$FV = R (1+i)^n$
$ln(1+i_{s})=\frac{1}{n}\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)$
$i=e^{ln(1+i_s)}-1$
Code:
R=1000
r=3%
g=5%
n=3
$ln(1+i_{s})=0.031037549112832$
$i=e^{0.031037549112832}-1$
$i=3.15242359792658\%$
$FV = 1000 (1+3.15242359792658\%)^3$
$FV = 1000 (1.0315242359792658)^3$
$FV = 1000 (1.097585368)$
$FV = 1,097.59$
3. ## Re: calculating compound interest with non-linear interest growth
Hi GhostAccount,
Thank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however.
I understand that your answer is solving the formula below for i :
$R (1+i)^n = R\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)$
This tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel:
n | Expected vs Excel
1 | 1030.00 vs 1030.00 ( = 1000 + 30 )
2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 )
3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 )
4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 )
Did I interpret the FV summation formula correctly and just get Excel wrong?
I didn't realize that the compound growth + compound interest could be expressed as a summation like that - I thought this was only calculating the amount of interest that $1000 would produce in n years with a compounded rate and not compounding the principal amount as well. I was thinking that $FV=R(1+r(1+g)^0)(1+r(1+g)^1)(1+r(1+g)^2)$ etc A summation formula is what I really want to reach as I can then try to substitute a general identity for it and derive a formula in terms of r,g and n that I can use to calculate each expected value for a given t, for example using $\sum_{i\ =\ m}^{n-1} a^i = \frac{a^m - a^n}{1-a}$ Thanks again for your help! 4. ## Re: calculating compound interest with non-linear interest growth Originally Posted by tw35758 Hi GhostAccount, Thank you for the fast response, that was very helpful! I've checked the results manually using Excel and your method generates all FV values that I expect from n=1 to n=10. There is one point that I don't fully understand however. I understand that your answer is solving the formula below for i : $R (1+i)^n = R\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)$ This tells me that I should be able to calculate any given FV for a value of n by manually summing the rhs. However I don't get the expected answer when I tried to do this in Excel: n | Expected vs Excel 1 | 1030.00 vs 1030.00 ( = 1000 + 30 ) 2 | 1062.45 vs 1061.50 ( = 1000 + 30 + 31.5 ) 3 | 1097.59 vs 1094.58 ( = 1000 + 30 + 31.5 + 33.075 ) 4 | 1135.70 vs 1129.30 ( = 1000 + 30 + 31.5 + 33.075 + 34.7288 ) Did I interpret the FV summation formula correctly and just get Excel wrong? Thanks again for your help! The formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate See the following for complete formula $R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$ Edit: This formula should work as I am short on time now thus I didn't find time to confirm the results 5. ## Re: calculating compound interest with non-linear interest growth Was trying for geometric series...like the 1000 bucks creates a guaranteed 30 bucks annual deposit (the 3%), so we'd have FV of 1000 plus FV of 30 annuity....but finally realised that won't work... i = initial interest rate (.03) f = increase factor (1.05) n = number of years (make it 10) FV = (1 + i*f^0)(1 + i*f^1)(1 + i*f^2)...........(1 + i*f^(n-1)) = 1.4481148... That's for the proverbial one dollar, of course; so 1,448.11 if$1000.
Soooo....looks like I'm kinda repeating GA's "spectacular!" work.
Whadda hell is the matter with the dollar sign here??
6. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by GhostAccount
The formula I presented in two parts at first finds the continuously compounded yearly rate (geometric average) that has to be converted to a yearly rate
See the following for complete formula
$R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$
Edit: This formula should work as I am short on time now thus I didn't find time to confirm the results
Beautiful.
It should work but there's a slight typo on the exponent of e; $ln$ is missing.This could have been
$R (1+i)^n = R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k})$
Must agree with Sir Wilmer; spectacular work indeed.
Was about to work on a derivation for this one when i saw it earlier but got too sober for a day and a half.
Drunk as I am right now, I am nevertheless definitely green with envy for this piece of beauty.
I'm almost certain I'm not the only one who'd like to see the analysis of this derivation Sir GhostAccount.
But as you said, this is one way of doing this. There might be another (or other's).
Might have to soak me brain with me favorite tequila brand for one such other future possibe derivation.
7. ## Re: calculating compound interest with non-linear interest growth
Thanks everyone, I've learned a lot so far!
I did some more research and realized that the correct way to represent the FV formula for the compounded growth of compounded interest is
$FV = R(\prod_{k\ =\ 0}^{n-1} (1+r(1+g)^k)})$
Since FV is essentially R(a)(b)(c)(...) we can convert this to a summation using logs.
Since
$e^{ln(x)} = x$
and
$ln(abc) = ln(a)+ln(b)+ln(c)$
so
$\boxed{FV= R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})}$
$FV=R(1+i)^n$
so
$R(1+i)^n= R(e^{\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
eliminating R and taking the log of both sides we get
$nln(1+i) = ({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
$ln(1+i) = \frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})$
$1+i = e^{\frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})}$
and now we're back to the answer GhostAccount initially provided as
$\boxed{i = e^{\frac{1}{n}({\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)})} - 1}$
One last question...is it possible to express
${\sum_{k\ =\ 0}^{n-1} ln(1+r(1+g)^k)}$
in a single function f(n) where it outputs the expected FV for n=0, n=1, n=2 etc?
I could do something similar for the normal compounded case, e.g.
$\sum{r(1+g)^k} = \frac{r((1+g)^{(k+1)}-1)}{g}$
but I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.
8. ## Re: calculating compound interest with non-linear interest growth
Whadda hell is the matter with the dollar sign here??
Dollar sign invokes the LaTeX editor. Confused me when I started posting here. See Dollar Signs
9. ## Re: calculating compound interest with non-linear interest growth
I realized I should make my question more specific...
Suppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of r1 and the second has an initial yield of r2 with a growth rate of r2.
I'd like to be able to analyze / compare the two performance stocks when I re-invest the dividend payments and answer questions such as:
Q1: With a constant stock price, what is the total return of principal + dividends after time t?
Q2: How many years will either stock take to reach a certain dividend payout?
I was able to do this for the non-compounding case and I posted an article on my blog explaining the formulae. The formulae and explanations above are really helpful but I'm stuck in transforming the summation formula above to a more usable form. I want to avoid continous compounding but keep to annually compounded calculations.
10. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by tw35758
but I'm completely lost on how to do a similar conversion for the FV formula or if it's even possible. I'm trying to create an Excel file comparing the result of various investments with different dividend yields and dividend growth rates, so having a single formula in a cell for each value of n would make creating that file much easier.
For now, see the following code in Excel to use the tadFVSchedule function that answers the original question
Code:
Public Function tadFVSchedule(ByVal pv As Double, ByVal r As Double, ByVal g As Double, ByVal n As Double, Optional ByRef c As Double = 1, Optional ByRef p As Double = 1, Optional ByRef d As Double = 1) As Double
Dim fraction As Double
Dim sum As Double
Dim k As Integer
sum = 0#
fraction = n - Int(n)
For k = 0 To Int(n - 1)
sum = sum + p * d * Log(1 + r * (1 + g) ^ k)
Next k
If (fraction <> 0) Then
k = Int(n + 0.5)
sum = sum + ((fraction - 1) * p + p * d) * Log(1 + r * (1 + g) ^ k)
End If
End Function
pv r g n c p d fv 1000 3% 5% 3 1 1 1 1097.59 1000 3% 5% 10 1 1 1 1448.11 1000 3% 5% 3.5 1 1 1 1117.42 1000 3% 5% 10.5 1 1 1 1484.80 1000 3% 5% 3 =1/12 1 1 1097.59 1000 3% 5% 10 =1/12 1 1 1448.11 1000 3% 5% 3 1 =1/12 1 1007.79 1000 3% 5% 10 1 =1/12 1 1031.34 1000 3% 5% 3 =1/12 =1/12 1 1007.79 1000 3% 5% 10 =1/12 =1/12 1 1031.34 1000 3% 5% 3 1 1 =1/2 1047.66 1000 3% 5% 10 1 1 =1/2 1203.38 1000 3% 5% 3.5 1 1 =1/2 1047.66 1000 3% 5% 10.5 1 1 =1/2 1203.38 1000 3% 5% 3 =1/12 1 =1/2 1047.66 1000 3% 5% 10 =1/12 1 =1/2 1203.38 1000 3% 5% 3 1 =1/12 =1/2 1003.89 1000 3% 5% 10 1 =1/12 =1/2 1015.55 1000 3% 5% 3 =1/12 =1/12 =1/2 1003.89 1000 3% 5% 10 =1/12 =1/12 =1/2 1015.55
11. ## Re: calculating compound interest with non-linear interest growth
Thank you GhostAccount! That's great - I can use that macro.
I made a typo in my question - I meant:
Suppose I have two different dividend paying stocks - one has an initial yield r1 with a growth rate of g1 and the second has an initial yield of r2 with a growth rate of g2. The stocks pay dividends annually and the dividends are re-invested to purchase more of the stock at the end of each year.
Q1: With a constant stock price, what is the total return of principal + dividends after time t?
Q2: How many years will either stock take to reach a certain dividend payout?
Thanks again for your help and patience in answering - I've learned a lot so far!
12. ## Re: calculating compound interest with non-linear interest growth
I must be missing something...but, if you're able to use a looper, then isn't that enough?
I love UBasic ([]= notes):
i = .03 : f = 1.05 : b = 1000 : p = 1000 : k = 30 [f = increase factor, k = years]
FOR n = 1 TO k
e = b * i [e = year's earnings]
b = b + e
y = (b / p)^(1 / n) [y = yield so far]
PRINT n, e, b, i * 100, y * 100
i = i * f [next year's rate]
NEXT n
OUTPUT:
Code:
YEAR INTEREST BALANCE YEAR'S RATE YIELD-SO-FAR
0 1000.00
1 30.00 1030.00 3.0000 3.0000
2 32.45 1062.45 3.1500 3.0750
3 35.14 1097.59 3.3075 3.1524
...
10 64.40 1448.11 4.6540 3.7720
...
20 185.22 2628.54 7.5808 4.9507
...
30 749.55 6819.59 12.3484** 6.6085
** .03 * 1.05^29 = 12.3484...
Isn't that all you need?
13. ## Re: calculating compound interest with non-linear interest growth
Hi Wilmer,
You're correct - I can certainly compute the answer using an excel macro. However I'm curious if there's a mathematical solution to this question.
The only mathematical solution I know of would be to approximate the answer using a Taylor series e.g.
$FV = Re^{\sum_{k\ =\ 0}^{n-1}ln({1+r(1+g)^k)}}$
$ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - {...}$
taking the first 2 terms as an example then
$\sum_{k\ =\ 0}^{n-1}ln({1+x}) \approx \sum_{k\ =\ 0}^{n-1}(x - \frac{x^2}{2})$ where $x = r(1+g)^k$
$\sum_{k\ =\ 0}^{n-1}(x - \frac{x^2}{2}) = \sum_{k\ =\ 0}^{n-1}x - \sum_{k\ =\ 0}^{n-1}\frac{x^2}{2}$
and
$\sum_{k\ =\ 0}^{n-1}x = \sum_{k\ =\ 0}^{n-1}r(1+g)^k = \frac{r((1+g)^n-1)}{g}$
$\sum_{k\ =\ 0}^{n-1}\frac{x^2}{2} = \frac{1}{2}\sum_{k\ =\ 0}^{n-1}r^2(1+g)^{2k} = \frac{r^2((1+g)^{2n}-1)}{2g(g+2)}$
so I could say that
$FV \approx Re^{\frac{r((1+g)^n-1)}{g} - \frac{r^2((1+g)^{2n}-1)}{2g(g+2)}}$
and solve this equation for n with a given FV.
I'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.
14. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by tw35758
I'm wondering if there's a more elegant equivalent expression of this function out there that doesn't involve approximations.
After 5 beer bottles and 2 shots of tequila, I've been wondering about that too. You could always repost your questions on an actuarial forum. You might get lucky there. Having seen your analyses, I'm somewhat inclined to believe that you might have some background in actuarial mathematics yourself. If you're lucky, Sir TKHunny, the resident actuary here might get interested in your situation and come out of retirement from this often thankless business of math knight-errantry.
15. ## Re: calculating compound interest with non-linear interest growth
Originally Posted by jonah
After 5 beer bottles and 2 shots of tequila, I've been wondering about that too...
Hey, that was my breakfast 30 years ago, at end of my drinking...!!
Page 1 of 2 12 Last | 2017-08-17T16:11:45 | {
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http://mathhelpforum.com/calculus/167237-trig-integral.html | # Math Help - Trig Integral
1. ## Trig Integral
$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})+(SEC^2{5x}TAN^2{5x})$
I would assume U sub at this point
$U= SEC dU= SECTAN$
But I am not sure, what or how I should bring along with the SEC
Any help?
thanks
Rich
2. Originally Posted by Spoolx
$\int SEC^4{5x} = \int ({SEC^2{5x}})({SEC^2{5x}})$
Then
$\int (1+TAN^2{5x})(SEC^2{5x})$
Then I multiply out
$\int (SEC^2{5x})(SEC^2{5x}TAN^2{5x})$
Fine so far, except you're missing a plus sign in the last line, which I assume is a typo. Now to substitute, which is really in order to work backwards through the chain rule.
Just in case a picture helps...
... where (key in first spoiler) ...
Spoiler:
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift is...
Spoiler:
$\int 1 + u^2\ du = u + \frac{1}{3} u^3$
and so we can explicate F...
Hope this helps - if not, someone will probably show the sub with differentials. However, I say...
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
3. I edited the original to include the + sign, thanks for catching my mistake
4. $\displaystyle\int{sec^4(5x)}dx=\int{sec^2(5x)sec^2 (5x)}dx=\int{\left[1+tan^2(5x)\right]sec^2(5x)}dx$
Substitute
$u=tan(5x)$
$\displaystyle\Rightarrow\frac{du}{dx}=5sec^2(5x)$
by the chain rule
$\displaystyle\Rightarrow\ sec^2(5x)dx=\frac{du}{5}$
The integral becomes
$\displaystyle\frac{1}{5}\int{\left[1+u^2\right]du$
5. Hello, Rich!
A rehash of Archie's solution . . .
$\displaystyle \int \sec^4\!5x\,dx \:= \:\int (\sec^2\!5x)(\sec^2\!5x)\,dx$
Then: . $\displaystyle \int (1+\tan^2\!5x)(\sec^2\!5x)\,dx$
Then I multiply out: . $\displaystyle \int (\sec^2\!5x +\sec^2\!5x\tan^2\!5x)\,dx$
$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$
For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$
For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$
. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$
Got it?
6. Originally Posted by Soroban
Hello, Rich!
A rehash of Archie's solution . . .
$\displaystyle \text{You have: }\;\int\sec^2\!5x\,dx + \int(\tan 5x)^2(\sec^2\!5x\,dx)$
For the first integral, use: . $\displaystyle \int \sec\theta\,d\theta \:=\:\tan\theta + C$
For the second integral, let: . $u \,=\, \tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$
. . and you have: . $\displaystyle \int u^2\,du \:=\:\tfrac{1}{3}u^3 + C$
Got it?
So to refresh my algebra
$tan^2{5x} = (tan5x)^2$ ?
Also above, is your U supposed to be?
$u= tan5x$ ?
if not where does the 5 go?
7. Originally Posted by Spoolx
$tan^2{5x} = (tan5x)^2$ ? Mr F says: Yes.
$u= tan5x$ ? Mr F says: Yes. Soroban made a simple typo. You ought to be able to make the appropriate corrections in the calculations. | 2015-07-05T18:11:31 | {
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https://math.stackexchange.com/questions/2248259/how-to-calculate-an-integral-using-cauchys-theorem | # How to calculate an integral using Cauchy's theorem?
The question is calculate the value of the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ These are the steps that I followed,
Let $x$ be a complex number. So, the poles of the function $f(x)=\frac{1}{1+x^4}$ occur when $x$ is equal to the roots of the equation $1+x^4=0$, i.e $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$. They are all poles of degree 4.
Now Cauchy's theorem says that, $$\frac{1}{2\pi i}\int_C dx\ f(x)=\sum_i\text{Res}(f,x_i)$$ where $x_i$ are the poles of $f$ that lies within $C$. I am pretty sure that my poles lie within $-\infty$ and $\infty$. So, I calculated the residues of $\frac{1}{1+x^4}$ at $x=e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}$ and they are equal to $-\frac14e^{i\pi/4},-\frac14e^{3i\pi/4},-\frac14e^{5i\pi/4},-\frac14e^{7i\pi/4}$. Wolfram Alpha confirms my calculations.
The sum of residues is $$\sum_i\text{Res}=-\frac14e^{i\pi/4}-\frac14e^{3i\pi/4}-\frac14e^{5i\pi/4}-\frac14e^{7i\pi/4}=0$$ and therefore the integral, $$\int_{-\infty}^{\infty}\frac{dx}{1+x^4}$$ must be equal to zero. However, Wolfram alpha says it is not zero but equal to$\frac{\pi}{\sqrt 2}$. Where am I making a mistake?
In applying the residue theorem, we analyze the integral $I$ given by
\begin{align} I&=\oint_C \frac{1}{1+z^4}\,dz\\\\ &=\int_{-R}^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{1}{1+(Re^{i\phi})^4}\,iRe^{i\phi}\,d\phi\\\\ &=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right) \end{align}
where $R>1$ is assumed.
Note that the only residues implicated in the residue theorem are those enclosed by $C$. Here, $C$ is comprised of (i) the line segment from $-R$ to $R$ and (ii) the semicircular arc centered at the origin with radius $R$ and residing in the upper-half plane. Hence, the only resides are the ones at $z=e^{i\pi/4}$ and $z=e^{i3\pi/4}$.
Taking $R\to \infty$, the integration over the semi-circular contour vanishes and we find that
\begin{align} \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx&=2\pi i \,\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4},e^{i3\pi/4}\right)\\\\ &=2\pi i \left(-\frac{e^{i\pi/4}}{4}-\frac{e^{i3\pi/4}}{4}\right)\\\\ & =\frac{\pi}{\sqrt 2} \end{align}
• Like the other answer, why did you only used the upper half semicircle? Why did you not ignore the first term in the second step and evaluate over the whole circle like, $$\int_0^{2\pi} d\phi\ \frac{iRe^{i\phi}}{1+(Re^{i\phi})^4}$$ – sigsegv Apr 23 '17 at 17:13
• @Ayatana We want to evaluate the integral $\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx$. How does integrating over the circle do that? Well, it doesn't; and that integral is $0$. So instead, we evaluate a contour integral, the one described herein, that has the real line integral as part of the contour. We could have taken the semi-circle in the lower-half plane, but since the orientation would be clockwise, we need a $-2\pi i$ factor o multiply the sum of the residues in the lower-half plane. The result will be the same, of course. ;-)) – Mark Viola Apr 23 '17 at 17:15
• Is it necessary for the contour to be semicircle? Will a contour made of a straight line along real axis and an arbitrary line joing the ends work? – sigsegv Apr 24 '17 at 4:30
• @Ayatana The reside theorem applies to any closed rectifiable curve. But we are free to choose the closed contour. And by choosing a semicircle the radius of which approaches infinity, we can show quite easily that its contribution approaches $0$ as $R\to \infty$. – Mark Viola Apr 24 '17 at 4:34
You shouldn't need Wolfram Alpha to tell you the answer isn't zero. The integrand is positive, so the integral is positive, and certainly nonzero.
You have added up the residues at all the poles. However using the usual semicircle method gives you that the integral is $2\pi i$ times the sum of the residues of the poles in the upper half-plane.
• Why should I not add the residues of the poles below the lower half? What semi circle are you talking about? The limits of the integration are from $-\infty$ to $\infty$ for both real and imaginary parts of $x$. – sigsegv Apr 23 '17 at 16:29 | 2021-02-27T10:55:05 | {
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https://mathoverflow.net/questions/301148/version-of-banach-steinhaus-theorem/304672 | # Version of Banach-Steinhaus theorem
I am wondering about the following version of the Banach-Steinhaus theorem.
Let $A$ be a closed convex subset contained in the unit ball of a Banach space $X$ and consider bounded operators $T_n \in \mathcal L(X).$
Assume we know that for every $x \in A$ the sequence $\left\lVert T_n x \right\rVert$ is bounded uniformly in $n.$
Does this imply that $$\sup_{x \in A} \left\lVert T_n x \right\rVert$$ is bounded uniformly in $n$?
If $A=X$ then the theorem is undoubtedly true by the folklore Banach-Steinhaus theorem but I was wondering whether this version holds as well?
• In the last paragraph, shouldn't it be "if A is the unit sphere then..." ? – user111 Jul 10 '18 at 7:45
• It is worth noting that the implication does not hold without the closeness assumption. If e.g. $A$ is the convex hull of the standard basis of $X:=\ell_2$ and $\phi_n:=\sum_{k=1}^n ke_k\in X$, then $(\phi_n,x)$ is point-wise but not uniformly bounded for $x\in A$ – Pietro Majer Jul 10 '18 at 15:16
The answer is yes, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on Wikipedia.
The details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T_n)$ with a general subset $\mathcal{T} \subseteq \mathcal{L}(X)$.
Proof. By Baire's Theorem we can find an integer $m$ such that the set \begin{align*} B := \{x \in A: \, \|Tx\| \le m \text{ for all } T \in \mathcal{T}\} \end{align*} has non-empty interior within $A$. Thus, we can find a point $x_0 \in B$ and a real number $\varepsilon \in (0,1]$ such that each $x \in A$ which has distance at most $\varepsilon$ to $x_0$ is contained in $B$.
Now, let $y \in A$. The vector $z := x_0 + \frac{\varepsilon}{2}(y-x_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x_0$ is at most $\varepsilon$ since both $y$ and $x_0$ have norm at most $1$. Thus, $\|Tz\| \le m$ for all $T \in \mathcal{T}$. Since \begin{align*} y = \frac{2}{\varepsilon}(z - x_0) + x_0, \end{align*} we conclude that \begin{align*} \|Ty\| \le \frac{4m}{\varepsilon} + m \end{align*} for all $T \in \mathcal{T}$. This bound does not depend on $y$.
Instead of inspecting the Banach-Steinhaus proof as in Jochen Glueck's answer one can apply Banach-Steinhaus to the Banach space $[A]$ (the linear hull of $A$) endowed with the Minkowski functional $\|x\|_A=\inf\{t>0: x\in tA\}$ (completeness of this norm follows from completeness of the Banach space $X$ and $A=\overline A$).
• A small detail: we need a balanced convex set to get a norm, so we should replace $A$ with $A':= \text{co}(A\cup (-A))=[-1,1]\cdot A$, and check that the hypotheses are still satisfied by $A'$ – Pietro Majer Jul 10 '18 at 7:22
• (the last equality is not true of course) – Pietro Majer Jul 10 '18 at 14:06
I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment).
A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold:
• If $A$ is $\sigma$-convex, it is bounded;
• If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex and symmetric (that is, $\sigma$-absolutely convex);
• If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.
As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$. | 2019-04-23T17:14:22 | {
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https://math.stackexchange.com/questions/3433458/optimal-strategies-for-this-card-game/3433482 | # Optimal strategies for this card game?
I recently had an interesting card game for an interview.
The interviewer and I have two cards each, one with a '5' painted on it and one with a '10'. We pick a card each and show it at the same time. If we picked the same card, I receive nothing, while if we picked different cards, the interviewer pays me the number he picked in dollars. What is the interviewer most likely to do if he intends to minimise my payoff, and what should my strategy be to counter this?
Consideration 1: I pick '5' all the time, but if the interviewer knows I do this he would pick '5' as well so that I receive nothing and my expected payoff would just be zero.
Consideration 2: I assign a probability $$p_1$$ of picking a '5' (and $$1-p_1$$ of picking a '10'), and the interviewer assigns a probability $$p_2$$ of picking a '5' (and $$1-p_2$$ of picking a '10'). My payoff would then be
$$\mathbb{E}=10p_1(1-p_2)+5(1-p_1)p_2.$$
I was thinking of differentiation somehow, but $$p_1$$ aims to maximise $$\mathbb{E}$$ while $$p_2$$ aims to minimise $$\mathbb{E}$$. Is this even the correct strategy for both of us?
Consideration 3: A friend suggested assigning probabilities of $$\frac13$$ and $$\frac23$$ to the cards respectively for a Nash equilibrium since it is a symmetric game. Where does this come from intuitively? Do these probabilities match with the above equation?
Any help/comments on the considerations is greatly appreciated, cheers!
The logic of equilibrium here is: you should choose your strategy (i.e. determine your probabilities of picking $$5$$ and $$10$$) in such a way that, however your opponent chooses his strategy, your expected gain is not influenced.
Let's say you pick $$5$$ with probability $$p_1$$ and your opponent picks $$5$$ with probability $$p_2$$. Then, once you fixed your choice of $$p_1$$, the expectation $$\mathbb{E}$$ is a linear function in $$p_2$$: $$\mathbb{E} = 10p_1 + (5 - 15p_1)p_2.$$ In this situation, you see that choosing $$p_1 = 1/3$$ will guarantee an expectation of $$10/3$$, regardless of $$p_2$$.
If, let's say, you choose $$p_1 < 1/3$$. Then there is the "risk" that your opponent somehow knows your strategy (e.g. via statistics from multiple rounds of the game), and then chooses $$p_2 = 0$$. You then have $$\mathbb{E} = 10p_1 < 10/3$$.
If, in the other direction, you choose $$p_1 > 1/3$$. Then again assume your opponent gets to know your strategy, and then chooses $$p_2 = 1$$. You then have $$\mathbb{E} = 5 - 5p_1 < 10/3$$.
Therefore, you see that the "safest" way of playing is to choose $$p_1 = 1/3$$. It is in that sense that we say it's the "optimal strategy".
I hope this whole logic makes sense to you.
And as an exercise, you may try to solve the general situation: say the two cards are labelled A and B, and your gain has $$4$$ possibilities: $$a, b, c, d$$, which correspond to the $$4$$ combinations AA, AB, BA, BB of your cards. Then what is your optimal strategy?
• Hello, what are some general approaches to formulate equilibrium for a game? – user107224 Nov 13 '19 at 9:56
• For the general equilibrium, the wiki page en.m.wikipedia.org/wiki/Nash_equilibrium may be a good starting point. Also note that in general there could be more than one equilibria. – WhatsUp Nov 13 '19 at 10:38 | 2020-09-18T09:54:16 | {
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http://mathematica.stackexchange.com/questions/29924/axis-angle-from-rotation-matrix/29966 | # Axis/Angle from rotation matrix
With r=RotationMatrix[a,{x,y,z}] I can compute a 3D rotation matrix from axis/angle representation.
Given a 3D rotation matrix r how can I compute a, x, y, z?
Example:
r = {{0.966496, -0.214612, 0.14081}, {0.241415, 0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}}
The result should be a=20. Degree and {x,y,z}={2, 1, 2}/3 (or equivalent).
Edit: I am fine with any answer that gives the same r when applied to RotationMatrix.
PS: Sorry if the tag is wrong, I did not found a matching tag...
-
I'm not sure this question is well posed in the sense that the answer is non-unique. A simple example would be r=RotationMatrix[1,{1,0,0}]=RotationMatrix[-1,{-1,0,0}] are the same. So which one are you intending to find? – bill s Aug 6 '13 at 13:08
The one where the angle is in [0,2 pi]. I think it is well posed as angle and axis are directly linked and you can provide a unique answer with a perfectly reasonable additional assumption. – Danvil Aug 6 '13 at 13:43
Actually there was an answer earlier which was just fine but it was deleted? – Danvil Aug 6 '13 at 13:46
@Danvil it was deleted because it was incorrect, so he deleted it. It was incorrect, as it worked only for the example you provided and could not be generalized. – rcollyer Aug 6 '13 at 14:42
@Danvil I'm thankful I noticed the wink. It would have been a bad day, otherwise. :P – rcollyer Aug 6 '13 at 16:38
There is no need to use Eigensystem or Eigenvectors to find the axis of a rotation matrix. Instead, you can read the axis vector components off directly from the skew-symmetric matrix
$$a \equiv R^T-R$$
In three dimensions (which is assumed in the question), applying this matrix to a vector is equivalent to applying a cross product with a vector made up of the three independent components of $a$:
{1, -1, 1}Extract[a, {{3, 2}, {3, 1}, {2, 1}}]
This one-line method of finding the axis is applied in the following function. To get the angle of rotation, I construct two vectors ovec, nvec perpendicular to the axis and to each other, to find the cosine and sine of the angle using the Dot product (could equally have used Projection). To get a first vector ovec that is not parallel to the axis, I permute the components of the axis vector using the fact that
Solve[{x, -y, z} == {y, z, x}, {x, y, z}]
(* ==> {{x -> 0, y -> 0, z -> 0}} *)
which means the above permutation with sign change of a nonzero axis vector is always different from the axis. This is sufficient to use Orthogonalize and Cross to get the desired orthogonal vectors.
axisAngle[m_] := Module[
{axis, ovec, nvec
},
{axis, ovec} =
Orthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@
Extract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];
(* nvec is orthogonal to axis and ovec: *)
nvec = Cross[axis, ovec];
{axis,
Arg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})]}
]
The angle is calculated with Arg instead of ArcTan[x, y] here because the latter throws an error for x = y = 0.
Here I test the results of the function for 100 random rotation matrices:
testRotation[] := Module[
{m, a, axis, ovec, nvec,
v = Normalize[RandomReal[{0, 1}, {3}]],
α = RandomReal[{-Pi, Pi}], angle
},
m = RotationMatrix[α, v];
{axis, angle} = axisAngle[m];
Chop[
angle Dot[v, axis] - α
] === 0
]
And @@ Table[testRotation[], {100}]
(* ==> True *)
In the test, I have to account for the fact that if the function axisAngle defined the axis vector with the opposite sign as the random test vector, I have to reverse the sign of the rotation angle. This is what the factor Dot[v, axis] does.
Explanation of how the axis results from a skew-symmetric matrix
If $\vec{v}$ is the axis of rotation matrix $R$, then we have both $R\vec{v} = \vec{v}$ and $R^T\vec{v} = \vec{v}$ because $R^T$ is just the inverse rotation. Therefore, with $a \equiv R^T-R$ as above, we get
$$a \vec{v} = \vec{0}$$
Now the skew-symmetric property $a^T = -a$, which can be seen from its definition, means there are exactly three independent matrix element in $a$. They can be arranged in the form of a 3D vector $\vec{w}$ which must have the property $a \vec{w} = 0$. This vector is obtained in the Extract line above. In fact, $a \vec{x} = \vec{w}\times \vec{x}$ for all $\vec{x}$, and hence if $a \vec{x} = 0$ then $\vec{x}\parallel\vec{w}$. Therefore, the vector $\vec{v}$ is also parallel to $\vec{w}$, and the latter is a valid representation of the rotation axis.
Edit 2: speed considerations
Since the algorithm above involves only elementary operations that can be compiled, it makes sense that a practical application of this approach would use Compile. Then the function could be defined as follows (keeping the return values arranged as above):
Clear[axisAngle1, axisAngle]
axisAngle1 =
Compile[{{m, _Real, 2}},
Module[{axis, ovec, nvec, tvec, w, w1},
tvec = {m[[3, 2]] - m[[2, 3]], m[[3, 1]] - m[[1, 3]],
m[[2, 1]] - m[[1, 2]]};
If[tvec == {0., 0., 0.},
{#/Sqrt[#.#] &[#[[Last@Ordering[N[Abs[#]]]]] &[
1/2 (m + {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}})]],
If[Sum[m[[i, i]], {i, 3}] == 3, 0, Pi] {1, 1, 1}},
axis = {1, -1, 1} tvec;
axis = axis/Sqrt[axis.axis];
w = {tvec[[2]], tvec[[3]], tvec[[1]]};
ovec = w - axis Dot[w, axis];
nvec = Cross[axis, ovec];
w1 = m.ovec;
{axis, {1, 1, 1} ArcTan[w1.ovec, w1.nvec]}
]
]
];
axisAngle[m_] := {#1, Last[#2]} & @@ axisAngle1[m]
The results are the same as for the previous definition of axisAngle, but I now get a much faster execution as can be seen in this test:
tab = RotationMatrix @@ # & /@
Table[{RandomReal[{-Pi, Pi}],
Normalize[RandomReal[{0, 1}, {3}]]}, {100}];
timeAvg =
Function[func,
Do[If[# > 0.3, Return[#/5^i]] & @@ Timing@Do[func, {5^i}], {i, 0,
15}], HoldFirst];
timeAvg[axisAngle /@ tab]
(* ==> 0.000801259 *)
This is more than an order of magnitude faster than the un-compiled version. I removed Orthogonalize from the code because I didn't find it in the list of compilable functions.
Note that Eigensystem is not in that list, either.
Edit 3
The first version of axisAngle demonstrated the basic math, but the compiled version axisAngle1 (together with the re-defined axisAngle as a wrapper) is faster. One thing that was missing was the correct treatment of the edge case where the rotation is by exactly $\pi$ in angle.
I added that fix only to the compiled version (axisAngle1) because I think that's the more practical version anyway. The trivial case of zero rotation angle was already included in the earlier version.
To explain the added code, first note that for angle $\pi$ you can't read off the axis from $R^T - R$ because the resulting matrix vanishes. To get around this singular case, we can use the geometric fact that a rotation by $\pi$ is equivalent to an inversion in the plane perpendicular to the rotation axis given by the unit vector $\vec{n}$. Therefore, if we form the sum of a vector $\vec{v}$ and its $\pi$-rotated counterpart, the components transverse to the rotation axis cancel and the result is always parallel to the axis. In matrix form,
$$(R+1)\vec{v} = 2\vec{n}(\vec{n}\cdot\vec{v}) = 2\left(\vec{n}\vec{n}^T\right)\vec{v}$$
Since this holds for all vectors, it is a matrix identity. The right-hand side contains a matrix $\vec{n}\vec{n}^T$ which must have at least one row that's nonzero. This row is proportional to $\vec{n}^T$, so you can read of the axis vector directly from $(R+1)$, again without any eigenvalue computations.
-
Beautiful! (+1) My hamfisted approach to this has been an opportunity to learn. – ubpdqn Aug 6 '13 at 22:28
Note (as you have outlined) you could also obtain the axis from NullSpace[m-Transpose[m]] – ubpdqn Aug 7 '13 at 4:17
@ubpdqn True, but that takes more computation... as I said, you can read the axis off literally from the matrix a itself. – Jens Aug 7 '13 at 5:15
Actually for cases $\theta=0$ and $\theta=\pi$ the $R-R^T$ is a zero matrix and the axis is thus maps to {{0,0,0},0) in your code. The test set always skirts these "points". The eigensystem allows extraction of the eigenvector associated with eigenvalue 1). I agree with lack of simplicity or elegance using the eigensystem. I found it a lazy way to deal with these anomalous cases: integer*$\pi$. I think my code works now. Passes your test (up orientation of axis issue) and the "extreme" cases. – ubpdqn Aug 7 '13 at 5:29
Maybe the easiest way to deal with the isotropic cases ($\pm$IdentityMatrix[3]) would be to append the rule axis /. {0, 0, 0} -> {0, 0, 1} to the result. The choice of rotation axis is arbitrary in that case and I'd fix it to be the z axis. – Jens Aug 7 '13 at 5:41
Since this question still seems to be alive after some time, I'm giving a solution from Presentations, which I sell. It has a routine RotationAngleAndAxis and here it is used on the example.
<< Presentations
r = {{0.966496, -0.214612, 0.14081}, {0.241415,
0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};
RotationAngleAndAxis[r]
(* {0.349066, {0.666667, 0.333333, 0.666667}} *)
Here is a short example of its use that checks the result. The axis vector is normalized. Notice that reversing the angle of rotation and the axis gives an equivalent rotation.
rotations =
Table[{RandomReal[{-\[Pi], \[Pi]}],
Normalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {2}]
(* {{2.90598, {-0.596373, -0.74938,
0.287697}}, {-2.44158, {0.331763, -0.943343, 0.00605196}}} *)
rotationMatrices = RotationMatrix @@@ rotations;
anglesAndAxes = RotationAngleAndAxis /@ rotationMatrices
(* {{2.90598, {-0.596373, -0.74938, 0.287697}}, {2.44158, {-0.331763,
0.943343, -0.00605196}}} *)
resultingMatrices = RotationMatrix @@@ anglesAndAxes;
rotationMatrices - resultingMatrices // Chop
Total[Flatten[%]]
(* {{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}, {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}} *)
(* 0 *)
The following is a test on 1000 initial rotation specifications. Notice also that I am comparing the resulting rotation matrices by a slightly risky procedure.
rotations =
Table[{RandomReal[{-\[Pi], \[Pi]}],
Normalize[Array[RandomReal[{-1, 1}] &, {3}]]}, {1000}];
rotationMatrices = RotationMatrix @@@ rotations;
Timing[anglesAndAxes = RotationAngleAndAxis /@ rotationMatrices;]
resultingMatrices = RotationMatrix @@@ anglesAndAxes;
Total[Flatten[rotationMatrices - resultingMatrices // Chop]]
(* {0.873606, Null} *)
(* 0 *)
Presentations also has an EulerAngles routine that will return the two sets of Euler angles for any of the possible sequence of rotations specified by strings. For example, here are the two sets of Euler angles of the example for two different rotation sequences.
EulerAngles[r, "ZXZ"]
EulerAngles[r, "XYZ"]
(* {{-0.346633, 0.259587, 0.580661}, {2.79496, -0.259587, -2.56093}} *)
(* {{0.244775, 0.0873143, 0.244775}, {-2.89682, 3.05428, -2.89682}} *)
-
The axis of rotation $n$ can be obtained as the eigenvector corresponding to the eigenvalue $1$.
r = {{0.966496, -0.214612, 0.14081}, {0.241415,
0.946393, -0.214612}, {-0.0872034, 0.241415, 0.966496}};
es = Eigensystem[r];
pos = Nearest[Thread[es[[1]] -> Range[3]], 1][[1]];
n = es[[2, pos]]
(* {0.666667, 0.333333, 0.666666} *)
The angle of rotation can be obtained directly from the trace of $r$.
ArcCos[(Tr[r] - 1)/2] 180/π
(* 20. *)
(This was my earlier and longer way to obtain the angle of rotation. As rcollyer pointed out this will have issues if $\pi k,k\in \mathcal{Z}$)
The rotation matrix in terms of the angle of rotation $\theta$ and axis of rotation $\{n1, n2, n3\}$.
R[t_, {n1_, n2_, n3_}] :=
With[{n = {{0, -n3, n2}, {n3, 0, -n1}, {-n2, n1, 0}}, id = IdentityMatrix[3]},
Inverse[id - t n].(id + t n)]
Here $t =\tan \left(\frac{\theta }{2}\right)$.
We get three solutions from the element in the first position.
eqns = Flatten[R[t, es[[2, pos]]] - r] // Simplify;
sols = NSolve[eqns[[1]] == 0, t]
(* {{t -> 3.88248*10^16}, {t -> 0.176327}, {t -> -0.176327}} *)
Choose the one which satisfies all the entries.
Chop[eqns /. sols, 10^-5];
Position[%, {0 ..}];
solst = Extract[sols, %]
(* {{t -> 0.176327}} *)
Obtain it in terms of degrees.
2 ArcTan[t] 180/π /. solst
(* {20.} *)
-
It doesn't seem to work with a rotation of $\pi$ about the {0,0,1}. – rcollyer Aug 6 '13 at 14:39
@rcollyer, that is to be expected because the rotation group $SO(3)$ cannot be covered by a single chart. Each chart will have its cases that are problematic. I used this parametrization and it worked for the OP's question. Still using this and going with sols = NSolve[eqns[[2]] == 0, t] we get $t=0$ for your case. The technique I used for the axis of rotation should be fine for all cases, however. – Suba Thomas Aug 6 '13 at 14:55
True, but I think it is amenable to decomposition into the generators of rotation, i.e. the rotations should have eigenvalues of the form $\pmatrix{e^{i \theta} & 1 & e^{-i \theta}}$, or for real rotations $\pmatrix{\cos\theta + i \sin\theta & 1 & \cos\theta - i \sin\theta}$. Thus, the angle is easily recoverable. As to the direction, I have not yet looked through your code. – rcollyer Aug 6 '13 at 15:05
Thanks for pointing that out. With your approach the angle of rotation is ambiguous for the OP's question, i.e we don't know if it is +20 or -20, whereas with the parametrization I used only + 20 satisfied all conditions. Probably quaternions will give a unique correct solution, but I do not have the material before me now. – Suba Thomas Aug 6 '13 at 15:20
Well you can choose the angle to be between $0$ and $2 \pi$, to fix the branch. – rcollyer Aug 6 '13 at 15:25
My +1 to Jens.
For known Euler angles:
{Φ, Θ, Ψ} = {0.1, 0.2, 0.3};
m = RotationMatrix[Φ, {0, 0, 1}].RotationMatrix[Θ, {1, 0,
0}].RotationMatrix[Ψ, {0, 0, 1}];
axisAngle[m_] :=Module[{axis, ovec, nvec}, {axis, ovec} =
Orthogonalize[{{1, -1, 1} #, Permute[#, Cycles[{{1, 3, 2}}]]}] &@
Extract[m - Transpose[m], {{3, 2}, {3, 1}, {2, 1}}];
nvec = Cross[axis, ovec];
{Arg[Complex @@ (((m.ovec).# &) /@ {ovec, nvec})], axis}];
N[axisAngle[m]]
(* {0.446615, {0.448552, -0.0450053, 0.892623}} *)
`
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https://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-cdots-x1/215052 | # Irreducibility of $X^{p-1} + \cdots + X+1$
Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ?
Our professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.
• Hint: Eisenstein ${}{}{}{}{}$ Oct 16 '12 at 19:15
• @AndréNicolas Yes, this problem was given after stating that theorem, but that only tells me the irreducibility in the rationals. Oct 16 '12 at 19:16
• Irreducible in the rationals implies irreducible in the integers. Oct 16 '12 at 19:16
• That's probably what is asked for. Over the reals, no polynomial of degree $\gt 2$ is irreducible. Oct 16 '12 at 19:17
• Use the binomial theorem to expand the powers $(x+1)^k$ and then look at the binomial coefficients to use Eisenstein.
– J.R.
Oct 16 '12 at 19:17
Hint: Denote $f(x)=x^{p-1}+...+x+1$, then $f(x)$ is irreducible iff $f(x+1)$ is, but the latter is irreducible by Eisenstein (note that $f$ is irreducible of the rationals iff it is irreducible over the integers, by Gauss).
• is the statement "$f(x)$ irreducible iff $f(x+1)$ is" a general theorem or just something I have to show in this exercise ? Oct 16 '12 at 20:43
• I suggest you show it.for exmaple: if $f(x)=g(x)h(x)$ can you factor $f(x+1)$ ? Oct 16 '12 at 20:48
• ah,ok, I got one direction of the "iff" proof. but if $f(x+1)=g(x)h(x)$, then how do I factor $f(x)$ ?s (althought I'm realizing, that for my question, I wouldn't need this direction of the proof; but it would be still interesting to know) Oct 16 '12 at 20:52
• $f(x)=g(x-1)h(x-1)$. Oct 16 '12 at 20:54
Hint $$\$$ Recall that Eisenstein's Criterion applies to polynomials that have form $$\rm\ f\ \equiv\ x^n\pmod p.\:$$ Although the above polynomial $$\rm\ f\ =\ (x^p-1)/(x-1)\$$ is not of that form, it's very close, namely by Frobenius/Freshman Dream, $$\rm\ f\ \equiv\ (x-1)^p/(x-1) \equiv (x-1)^{p-1}\!\pmod{p}.\:$$ Eisenstein's Criterion may apply if you can find a map $$\ \sigma\$$ that sends $$\rm\ (x-1)^{p-1}$$ to a power of $$\rm\ x\$$ and, further, $$\ \sigma\$$ preserves factorizations $$\rm\ \sigma(gh)\ =\ \sigma g\cdot \sigma h\$$ (so one can pullback the irreducibility of $$\rm\ \sigma\:f\$$ to $$\rm\:f).\:$$
Remark $$\$$ The history of the criterion is both interesting and instructive. $$\:$$ For this see David A. Cox, Why Eisenstein proved the Eisenstein criterion and why Schönemann discovered it first.
Above is prototypical of transformation-based problem solving. Consider the analogous case of solving quadratic equations. One knows how to solve the simple special case $$\rm\ x^2 = a\$$ by taking square roots. To solve the general quadratic we look for an invertible transformation that reduces the general quadratic to this special case. The solution, dubbed completing the square, is well-known. For another example, see this proof of the Factor Theorem $$\rm\:x\!-\!c\:|\:p(x)\!-\!p(c),\:$$ which reduces to the "obvious" special case $$\rm\:c=0\:$$ via a shift automorphism $$\rm\:x\to x+c.\:$$ The problem-solving strategy above is completely analogous. We seek transformations that map polynomials into forms where Eisenstein's criterion applies. But we also require that the transformation preserve innate structure - here multiplicative structure (so that $$\rm\:\sigma\:f\:$$ irreducible $$\Rightarrow$$ $$\rm\:f\:$$ irreducible).
Employing such transformation-based problem solving strategies has the great advantage that one can transform theorems, tests, criteria, etc, into a simple reduced or "normal" form that is easy to remember or apply, and then use the ambient symmetries or transformations to massage any given example to the required normal form. This strategy is ubiquitous throughout mathematics (and many other sciences). For numerous interesting examples see Zdzislaw A. Melzak's book Bypasses: a simple approach to complexity, 1983, which serves as an excellent companion to Polya's books on mathematical problem-solving.
• The extra comments beyond just the proof are very useful, thank you! The link to David A. Cox's article is not working for me, could you please check it?
– user279515
Apr 22 '19 at 8:58
Hint: Let $y=x-1$. Note that our polynomial is $\dfrac{x^p-1}{x-1}$, which is $\dfrac{(y+1)^p-1}{y}$.
It is not difficult to show that $\binom{p}{k}$ is divisible by $p$ if $0\lt k\lt p$. Now use the Eisenstein Criterion.
In the solution for this problem I will use the well-know result as $$\color{blue}{\text{Eisenstein's Irreducibility Criterion}}$$.
Eisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots a_{0}$$ be a polynomial with coefficients $$a_{n},a_{n-1},\ldots,a_{0}\in \mathbb{Z}$$. Suppose that there exists a prime $$p$$, such that:
1. $$p\not|a_{n}$$. In words: $$p$$ does not divide $$a_{n}$$.
2. $$p| a_{n-1},a_{n-2},\ldots,a_{1},a_{0}$$. In words: $$p$$ divide each $$a_{i}$$ for $$0\leq i .
3. $$p^{2}\not| a_{0}$$. In words: $$p^{2}$$ does not divide $$a_{0}$$.
Then $$f(x)$$ is is irreducible over the integers $$\mathbb{Z}[x]$$.
It is a corollary of $$\color{blue}{\text{Gauss's lemma}}$$ that $$f(x)$$ is also irreducible over the rational numbers .
Gauss's lemma: A non-constant polynomial in $$\mathbb{Z}[x]$$ is irreducible in $$\mathbb{Z}[x]$$ if and only if it is both irreducible in $$\mathbb{Q}[x]$$ and primitive in $$\mathbb{Z}[x]$$.
Now, your question is well-know, as that $$\color{blue}{\text{Cyclotomic polynomials}}$$ are irreducibles.
Problem: Let $$p$$ un prime numbe, so $$\Phi_{p}(x)=x^{p-1}+x^{p-2}+\cdots+1=\frac{x^{p}-1}{x-1}$$ is irreducible.
Proof: Note that $$\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+1$$ does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\Phi_{p}(x+1)=\frac{(x+1)^{p}-1}{x}=x^{p-1}+\binom{p}{1}x^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+\binom{p}{p-1}$$ does. Indeed, note that $$\boxed{1}$$ we can see that $$p\not| 1$$, in $$\boxed{2}$$ we can see that $$p| \binom{p}{1}, \binom{p}{2},\ldots,\binom{p}{p-2},\binom{p}{p-1}$$ and finally in $$\boxed{3}$$ we have that $$p^{2}\not| \binom{p}{p-1}=p$$. So, $$\Phi_{p}(x+1)$$ is irreducible. But, if $$\Phi_{p}(x)$$ factored as $$f(x)g(x)$$ so then $$\Phi_{p}(x+1)=f(x+1)g(x+1)$$. So, $$\Phi_{p}(x)$$ is irreducible. $$\boxed{}$$ | 2022-01-22T05:05:49 | {
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http://mathhelpforum.com/algebra/148292-sum-2-squares.html | # Thread: Sum of 2 squares
1. ## Sum of 2 squares
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
2. Originally Posted by Mukilab
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
Try m=7.
50 is very close to 49 and stands out that way..
3. Hello, Mukilab!
Show that: .$\displaystyle (m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2$
Done that successfully. . Good!
Using this results, write 500,050 as the sum of 2 square numbers.
Note that: .$\displaystyle 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\displaystyle \text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $\displaystyle (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
4. m=7 is a really nice guess
OK. $\displaystyle 500050=10001*50$
and $\displaystyle 10001=10000+1$
and $\displaystyle 50=49+1$
Apply in your equation: $\displaystyle (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2$
Then....
5. That would make n a decial (10sqrt10)
6. Originally Posted by Soroban
Hello, Mukilab!
Note that: .$\displaystyle 500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\displaystyle \text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $\displaystyle (7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
7. Originally Posted by Mukilab
That would make n a decial (10sqrt10)
Maybe you entered 50050 / 50 instead of 500050 / 50? m=7 gives n=100 as shown above.
8. Originally Posted by Mukilab
Thanks Soroban!
Sorry, only saw the first post at the start
Any tips for future questions such as this?? Take away the integers so I'm left with the algebraic numbers (as you did with 49 and 100)?
Thanks again
Another way to go about expressing integers as sums of two squares is knowing that: the set of integers expressible as the sum of two squares is closed under multiplication, and an odd prime is expressible as the sum of two squares if and only if it is congruent to 1 (mod 4). See here and here. | 2018-03-19T07:55:52 | {
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https://cs.stackexchange.com/questions/78083/why-is-the-log-in-the-big-o-of-binary-search-not-base-2/78131 | # Why is the log in the big-O of binary search not base 2?
I am new to understanding computer science algorithms. I understand the process of the binary search, but I am having a slight misunderstanding with its efficiency.
In a size of $s = 2^n$ elements, it would take, on average, $n$ steps to find a particular element. Taking the base 2 logarithm of both sides yields $\log_2(s) = n$. So wouldn't the average number of steps for the binary search algorithm be $\log_2(s)$?
This Wikipedia article on the binary search algorithm says that the average performance is $O(\log n)$. Why is this so? Why isn't this number $\log_2(n)$?
When you change the base of logarithm the resulting expression differs only by a constant factor which, by definition of Big-O notation, implies that both functions belong to the same class with respect to their asymptotic behavior.
For example $$\log_{10}n = \frac{\log_{2}n}{\log_{2}10} = C \log_{2}{n}$$ where $C = \frac{1}{\log_{2}10}$.
So $\log_{10}n$ and $\log_{2}n$ differs by a constant $C$, and hence both are true: $$\log_{10}n \text{ is } O(\log_{2}n)$$ $$\log_{2}n \text{ is } O(\log_{10}n)$$ In general $\log_{a}{n}$ is $O(\log_{b}{n})$ for positive integers $a$ and $b$ greater than 1.
Another interesting fact with logarithmic functions is that while for constant $k>1$, $n^k$ is NOT $O(n)$, but $\log{n^k}$ is $O(\log{n})$ since $\log{n^k} = k\log{n}$ which differs from $\log{n}$ by only constant factor $k$.
• Not only for positive integers: For all real $a,b > 1$, e.g. $e$. – nbubis Jul 20 '17 at 6:25
• I would add that this is the reason why with big-O notation, the base of the logarithm is commonly not specified. It also means there is no confusion about which base $O(\log n)$ uses: it can be any of the commonly used based, since it doesn't make a difference. – svick Jul 20 '17 at 10:57
In addition to fade2black's answer (which is completely correct), it's worth noting that the notation "$\log(n)$" is ambiguous. The base isn't actually specified, and the default base changes based on context. In pure mathematics, the base is almost always assumed to be $e$ (unless specified), while in certain engineering contexts it might be 10. In computer science, base 2 is so ubiquitous that $\log$ is frequently assumed to be base 2. That wikipedia article never says anything about the base.
But, as has already been shown, in this case it doesn't end up mattering.
• It is probably further worth noting then that while "log(n)" might be ambiguous that "O(log(n))" is not because the latter only has one meaning, no matter what base you might be thinking of. – Chris Jul 20 '17 at 16:12 | 2019-07-21T00:35:20 | {
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https://math.stackexchange.com/questions/3260418/in-how-many-ways-can-a-bit-string-of-length-20-be-generated-if-either-all-0 | # In how many ways can a bit string of length $20$ be generated if either all $0$'s or all $1$'s need to be grouped together?
Given a bit string of length $$20$$, how many ways can such a string be generated if either all 0's or all 1's need to be grouped together in the string?
A few examples of strings that would be considered legal are $$000 \ldots 0\qquad\text{or} \qquad 001110 \ldots 0 \qquad \text{or} \qquad 1100 \ldots 11 \qquad \text{or} \qquad1110000 \ldots 0$$
I've tried a couple different ideas, but I keep running into issues with duplicate strings.
One method I used would be to imagine that there are $$22$$ spaces and $$2$$ slashes that need to go somewhere within these spaces to separate the numbers. The slashes could go before the string starts, resulting in either all 1's or all 0's depending on which number starts. I'll list some examples of this below. $$//00000 \ldots 0 \qquad \text{or} \qquad //1111 \ldots 1 \qquad \text{or} \qquad 11 \ldots /000000/\ldots 1$$
With this method, we have $$2$$ cases that are the same, its just either picking if 1 or 0 leads. Because of that I'm calculating $$2\binom{22}{2}$$. But like I said, there are issues with this method. For example, when both slashes come before the string, the string that is generated would be the same as if the slashes came after the string.
I appreciate any ideas on where I should go from here or if I'm completely wrong and should just start over with some other method.
• you can separate the cases where you have only one block (either all 1's or all 0's), two blocks (setting a slash between two of the 20 positions) and three blocks (setting the left slash and then setting the right slash in regards to the position of the left one) – otto Jun 12 '19 at 22:44
• actually you use your way of putting two slashes in 22 possible positons. start by setting the left one, which is not allowed to have position 22 (to remove redundance). then set the right one with respect to the position of the left one – otto Jun 12 '19 at 23:11
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jun 13 '19 at 1:16
For each such string, join its ends to form a necklace. Each necklace has at most one region of 1s and at most one region of 0s, and is characterized by the number of 1s (0 through 20).
Apart from the all-0 and all-1 necklaces, each necklace (there are 19 remaining) can be cut in 20 places, yielding distinct valid strings. The constant necklaces yield the same string no matter where you cut. So, the total number of strings is:
$$1 + 1 + 19 \cdot 20 = 382$$
• This is now my favorite answer. – David K Jun 13 '19 at 2:29
• I agree with @DavidK. – N. F. Taussig Jun 13 '19 at 2:32
There are two ways to choose the first digit in the bit string.
Since all the $$0$$s or all the $$1$$s must be together, there are either $$0$$, $$1$$, or $$2$$ transitions between blocks of identical digits.
If there are no transitions, we do not need to place a divider. There is one way to do this.
If there is one transition, we must choose one of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the divider between the blocks.
If there are two transitions, we must choose two of the $$19$$ places between successive digits in the $$20$$-digit bit string in which to place the dividers between the blocks.
Hence, there are $$\binom{2}{1}\left[\binom{19}{0} + \binom{19}{1} + \binom{19}{2}\right]$$ bit strings of length $$20$$ such that all the $$0$$s or all the $$1$$s are together in the bit string.
Using your idea of $$22$$ spaces, you not only have the ambiguity between $$//\square\cdots\square$$ and $$\square\cdots\square//$$ (where each $$\square$$ represents a space where you will put a digit), you also have to be concerned about how you distinguish $$\square//\square\cdots\square$$ from both $$//\square\cdots\square$$ and $$/\square/\square\cdots\square.$$ (If each $$/$$ denotes a change from $$0$$ to $$1$$ or vice versa then $$\square//\square$$ is the same pattern as $$//\square\square$$ but if either $$\square/\square$$ or $$\square//\square$$ denotes that the second $$\square$$ is different from the first $$\square$$ then $$\square//\square$$ and $$/\square/\square$$ are the same pattern.)
You can almost fix this by requiring that the two $$/$$s not be adjacent. Then the only ambiguity you have is between $$/\square\cdots\square/\square\cdots\square$$ and $$\square\cdots\square/\square\cdots\square/.$$ There are $$19$$ of each of those patterns, you want to keep one set and remove the other, so subtract $$19$$ from the number of ways to put $$2$$ non-adjacent objects in $$22$$ spaces, which is $$\binom{21}{2}$$; with two choices for the first digit this makes the number of bit strings $$2\left(\binom{21}{2} - 19\right).$$
Another almost-fix is to require that there be at least one digit after each $$/.$$ This gets rid of the $$\square\cdots\square/\square\cdots\square/$$ patterns so you don't have to subtract those, but it also gets rid of $$/\square\cdots\square/,$$ so you have to add $$1$$ to count that pattern. This gives the answer $$2\left(\binom{20}{2} + 1\right).$$
Notice that $$\binom{21}{2} = \binom{20}{2} + \binom{20}{1} = \binom{19}{2} + 2\binom{19}{1} + \binom{19}{0},$$ so both the results in this answer are equal to $$2\left(\binom{19}{2} + \binom{19}{1} + \binom{19}{0}\right) = 2\times 191,$$ which has also been shown correct in another answer. What you will not be able to do is to come up with a combinatorial argument producing a single binomial coefficient that gives the answer when multiplied by $$2$$, because $$191$$ is a prime number and the only binomial coefficients with that value are $$\binom{191}{1}$$ and $$\binom{191}{190}.$$ | 2021-04-13T20:52:33 | {
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"openwebmath_perplexity": 202.89980679504754,
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"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9861513922264661,
"lm_q2_score": 0.8807970795424088,
"lm_q1q2_score": 0.8685992662597518
} |
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# Presentation token element (mo)
Module by: Sunil Kumar Singh. E-mail the author
Summary: The mo operator element in MathML conveys the meaning of mathematical operation.
Specific attributes : form fence separator lspace rspace stretchy symmetric Specific attributes : maxsize minsize largeop movablelimits accent Specific attributes : form fence separator lspace rspace stretchy symmetric Specific attributes : maxsize minsize largeop movablelimits accent
The “mo” element displays an operator or other representations, which are treated as an operator in mathematics. The term "operator" also includes fence, separator, accent, comma, semicolon, invisible characters etc. , some of which are used to provide new meaning to the ordinary operator. Hence, “operator” in MathML has wider meaning beyond ordinary operators, consistent with the requirement of growing expanse of mathematical operations.
Many of the important operators contained within “mo” element can be typed directly. They are available on key board; while many others have to be referenced through valid Unicode entity references. The operator symbols, which can be typed from the keyboard include “+”, “-, ”/”, ”*”, ”(“, ”)”, ”{“, ”}”, ”[“, ”]”, “.NOT.”, ”.OR.” etc.
Following example shows the display of operators by "mo" element.
### Example 1: Displaying operators with “mo” element
<m:math display="block">
<m:mtable>
<m:mtr>
<m:mtd>
<m:mi>Plus : </m:mi><m:mo>+</m:mo>
<m:mi> ;</m:mi>
<m:mi>Increment : </m:mi><m:mo>++</m:mo>
<m:mi> ;</m:mi>
<m:mi>Logical not : </m:mi><m:mo>.NOT.</m:mo> <m:mi> ;</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mi>Less than and equal : </m:mi><m:mo> ≤ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Partial operator : </m:mi><m:mo> ∂ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Differentiation : </m:mi><m:mo> ⅆ </m:mo>
<m:mi> ;</m:mi>
<m:mi>Integration : </m:mi><m:mo> ∫ </m:mo>
</m:mtd>
</m:mtr>
</m:mtable>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
Plus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration : Plus : + ; Increment : ++ ; Logical not : .NOT. ; Less than and equal : ; Partial operator : ; Differentiation : ; Integration :
In the above example, we have used “mtable” (with “mtr” and “mtd” elements) element for controlling display. See that first three operators in above example have been typed directly from the keyboard; while others have been referenced by entity reference.
MathML design of rendering an operator closely follows mathematical convention. It distinguishes between rendering a character with “mi” element and an operator with “mo” operator in many important ways for notational representation of mathematical expression. Two basic considerations are (i) to manage space around the operator, consistent with the context in which they are displayed and (ii) dimensional change (stretching) of operators in accordance with the dimension of other elements and terms. Space around an operator is managed by “form” attribute, while stretching is managed by “stretchy” attribute of “mo” element. A plus operator ("+") may ,for example, precede or follow an identifier or may lie between two operands. Depending upon its placement, the space around the operator "+" is determined. Similarly, size of a parentheses around an expression must stretch to the height of the expression as in ( a b ) ( a b ) .
MathML also allows improvisation of an operator rendered by “mo” element. This generally makes use of other characters rendering elements such as “mi” or "mn" elements in addition to "mo" element and their combination with layout elements like “mfrac” and scripting elements (“msub”, “msup” etc.) Simply put, MathML treats an expression with a “mo” element at core and improvised by other elements - as an operator of class known as “embellished” operator.
Consider encoding for a differentiation operator "ⅆ" . This operator when applied to a variable, say t, is represented as t t In mathematics, however, t t is itself considered to be an operator. The whole block consisting of “mfrac” element is, thus, an embellished operator.
<m:mfrac>
<m:mo> ⅆ </m:mo>
<m:mrow>
<m:mo> ⅆ </m:mo>
<m:mi> x </m:mi>
</m:mrow>
</m:mfrac>
### Additional attribute : Default values
Default values to the attributes of “mo” element are set in accordance with the values contained in “operator dictionary”, which specifies “form”, “fence”, “stretchy”, “lspace” and “rspace” attributes for different operators. The renderer maintains an operator dictionary for most of the operators. W3C recommends a prototype of operator dictionary. These recommendations can be viewed at www.w3.org/TR/MathML2/appendixf.html . If the dictionary does not provide the value for the attribute, then attribute is set with default value as given here:
#### Default values
1. form : set by position/context
2. fence : false
3. separator : false
4. lspace : thickmathspace
5. rspace : thickmathspace
6. stretchy : false
7. symmetric : true
8. maxsize : infinity
9. minsize : 1
10. largeop : false
11. movablelimits : false
12. accent : false
### Managing space around operators
#### Attribute value types
• form : prefix | infix | postfix
• lspace : number h-unit | namedspace
• rspace : number h-unit | namedspace
The "form" attribute specifies whether an operator shall be rendered as "prefix" or "postfix" or "infix" operator. These form types determine space around the operator (left and right of it). Each form type is associated with two attributes ("lspace" and "rspace"), which implements its left space ("lspace") and right space ("rspace"). It is important to understand that form types are basically specification about space around operator and about its role with respect to other elements.
The default value of “form” attribute is set in accordance with three rules, involving "mrow" element. The token elements, including "mo" element, is composed within explicit or inferred "mrow" element. The position of an operator within the "mrow" in horizontal sequence determines the form type of the operator. If operator is the first element, then its form is inferred as "prefix"; if it is in the intermediate position, then its form is inferred as "infix”; and if it is in the end position, then its form is inferred as "postfix". Corresponding to each form, operator library specifies "lspace" and "rspace". It is, however, not necessary that each of the operator has all the three form types. The nature of operator determines the type of forms that a particular operator should be associated with. For example, "+" operator has entries for "prefix" and "infix", but not for "postfix" in the operator dictionary, because "postfix" form of "+" operator is not expected to be placed at the end of an expression.
Now, consider the example of "+" operator appearing twice in an expression in the manner as typed below :
+a + b
The first "+" sign in the above case assumes "prefix" form, while the second "+" sign assumes "infix" form. Once, the form of the operator is determined from the context within "mrow" element, spacing around the operator is determined as specified by "lspace" and "rspace" values given in the operator dictionary. The dictionary suggests following values for these forms of “+” operator :
"+" form="infix" lspace="mediummathspace" rspace="mediummathspace"
"+" form="prefix" lspace="0em" rspace="veryverythinmathspace"
Evidently, "infix" form separates its neighboring elements by "mediummathspace", whereas "prefix" form separates the following element with a very small space specified by "veryverythinmathspace".
Let us now code the example in MathML and see the output :
#### Example 2: Determining "form" of an operator
<m:math display="block">
<m:mrow>
<m:mo> + </m:mo>
<m:mi> a </m:mi>
<m:mo> + </m:mo>
<m:mi> b </m:mi>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
+ a + b + a + b
As expected first “+” is almost attached to “a”, where as second instance of “+” renders some amount of space around itself. This type of detailing in rendering mathematical expression is vital and critical to maintain highest order of rendering consistencey, which is commensurate with the exact nature of mathematics and the meaning that a mathematical expression conveys. Note that the mark-up paradigm of MathML automatically forces codification in such a manner that no extra effort is required towards maintaining conventions of mathematical display - almost ruling out the possibilities that an expression is displayed in a non-conforming way.
A particular operator does not require to have entries for all the forms in the operator dictionary. Many of the operators in mathematics operate on other entities in forward direction. For example, integration, differentiation, partial differentiation etc. operates on identifiers following it. For this reason, such operators are inherently “prefix” in nature. On the other hand, operators classified as fences are either “prefix” or “postfix” commensurate with their role in mathematical expression. It is very unlikely that a closing bracket “]” is used in the beginning of an expression and as such it has “postfix” form. Dictionary entries for these operators are given here to understand : why there is very small space or no space between these operators and the identifiers on which they operate.
"ⅆ" form="prefix" lspace="0em" rspace="verythinmathspace"
"∂" form="prefix" lspace="0em" rspace="verythinmathspace"
"∫" form="prefix" largeop="true" stretchy="true" lspace="0em"
"(" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
")" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"[" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"]" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"{" form="prefix" fence="true" stretchy="true" lspace="0em" rspace="0em"
"}" form="postfix" fence="true" stretchy="true" lspace="0em" rspace="0em"
We are at liberty to specify these attributes. It is, however, recommended that we leave the arrangement to the system, which is rendering the mathematical content. It shall ensure consistency in the display, which follows the convention of mathematics as implemented by a particular renderer. The example below demonstrates how we can change the spacing different to default and against the form values as inferred from the context :
#### Example 3: Determining "form" of an operator
<m:math display="block">
<m:mrow>
<m:mo form="infix" rspace="10pt"> + </m:mo>
<m:mi> a </m:mi>
<m:mo form="prefix"> + </m:mo>
<m:mi> b </m:mi>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
+ a + b + a + b
### Stretching of operators
Four attributes of “mo” element control stretching of operators. These attributes may assume following values :
#### Attribute value types
• stretchy : true | false
• symmetric : true | false
• maxsize : number [ v-unit | h-unit | namedspace | infinity
• minsize : number [ v-unit | h-unit | namedspace
Among the operators, the requirement for stretching of fences, arrows, accents (angular cap on identifier) and separators are most profound and visible in mathematical expressions. For this reason, “strechy” attribute of fence and accent operators are set “true” in operator dictionary. Stretching of operators in an expression is restricted by “minsize” and “maxsize” attributes.
The stretchable operators are characterized as predominantly either vertically or horizontally stretchable. The fences, various kinds of vertical arrows (single or double), operators like ∏, ∑, ∫, “/” etc. are set to stretch vertically by default in operator directory.
When stretchable operator and non-stretchable terms are bounded by explicit or inferred “mrow” element, then the stretchable operator grows vertically to cover the non-stretchy term. Consider the example given here :
#### Example 4: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
We can, however, control the growth of parenthese by setting “maxsize” attribute to 1 i.e. equal to its normal size.
#### Example 5: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo maxsize="1"> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo maxsize="1"> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
Thus, setting "maxsize" attribute overrides the default behavior, which allows the parentheses to strech and cover the non-stretchy expression. Let us, now experiment with other than fence character like ∑ and observe their behavior with other terms :
#### Example 6: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mo> ∑ </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>A</m:mi>
<m:mi>B</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
( A B ) ( A B )
Control for stretching can be selective as well. For example, we can set stretchy=”false” on the opening parenthesis to restrict it to grow.
#### Example 7: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo stretchy="false"> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
x = ( a b ) x = ( a b )
In situation where, the expression bounded by “mrow” tags contains terms of different heights, the stretchable parentheses grow to cover the highest of the terms.
#### Example 8: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mfrac>
<m:mi>a</m:mi>
<m:mi>b</m:mi>
</m:mfrac>
<m:mo> + </m:mo>
<m:mfrac>
<m:mrow>
<m:mfrac>
<m:mi>c</m:mi>
<m:mi>d</m:mi>
</m:mfrac>
</m:mrow>
<m:mi>e</m:mi>
</m:mfrac>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( a b + c d e ) x = ( a b + c d e )
The symmetric attribute is designed to stretch operator in both vertical and horizontal direction from the axis of the characters in equal magnitude. The symmetric attribute applies only to characters, which can stretch vertically; otherwise this attribute is ignored. Usually, this attribute is set “true” for vertically stretchable operator, but in certain cases involving matrix of unequal size, we may prefer to set it “false” as demonstrated in the example here :
#### Example 9: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo symmetric="false"> ( </m:mo>
<m:mtable align="bottom">
<m:mtr>
<m:mtd>
<m:mi>a</m:mi>
</m:mtd>
<m:mtd>
<m:mi>b</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
</m:mtable>
<m:mo symmetric="false"> ) </m:mo>
<m:mo symmetric="false"> ( </m:mo>
<m:mtable align="bottom">
<m:mtr>
<m:mtd>
<m:mi>a</m:mi>
</m:mtd>
<m:mtd>
<m:mi>b</m:mi>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
<m:mtr>
<m:mtd>
<m:mn>d</m:mn>
</m:mtd>
<m:mtd>
<m:mn>e</m:mn>
</m:mtd>
</m:mtr>
</m:mtable>
<m:mo symmetric="false"> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( a b d e ) ( a b d e d e ) x = ( a b d e ) ( a b d e d e )
Matrix operation of unequal sizes uses notation which is aligned to base. This type of controlling stretching in specific direction is, therefore, extremely useful in such situations.
In case, the “mrow” domain contains non-stretchable terms of normal height and other stretchable terms, then all the terms, including the fence operator grows to the maximum normal height.
#### Example 10: Vertical stretching
<m:math display="block">
<m:mrow>
<m:mi> x </m:mi>
<m:mo> = </m:mo>
<m:mo> ( </m:mo>
<m:mo> ∫ </m:mo>
<m:mi> f</m:mi>
<m:mo> ( </m:mo>
<m:mi> x </m:mi>
<m:mo> ) </m:mo>
<m:mi> ⅆ </m:mi>
<m:mi> x </m:mi>
<m:mo> ) </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display lokks like :
x = ( f ( x ) x ) x = ( f ( x ) x )
As against fence operator, the accent and horizontal arrows are stretchable in horizontal direction by default. The growth, in this case, is controlled by “munder”, “mover” and “munderover” elements, which contain the operator. The “munder”, “mover” and “munderover” elements, as the names suggest, allow drawing of an operator "under" or "over" or both about a character(s) or expresison. The scripting elements takes a base argument about which the operator is to be drawn and one (“munder” and “mover”) operator or two (“munderover”) operators for being placed about the base. We shall see that the operator grows horizontally to cover the other element.
#### Example 11: Horizontal stretching
<m:math display="block">
<m:mi> A </m:mi>
<m:munderover>
<m:mo> → </m:mo>
<m:mtext> 50 degree C </m:mtext>
<m:mtext> 200 psi </m:mtext>
</m:munderover>
<m:mi> B </m:mi>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
A 50 degree C 200 psi B A 50 degree C 200 psi B
### Other attributes : largeop, movablelimits, accent, separator, fence
These attributes may assume following values :
#### Attribute value types
• fence : true | false
• separator : true | false
• largeop : true | false
• movablelimits : true | false
• accent : true | false
All these attributes accept boolean values “true” or “false”. The attribute “fence” is designed for non-visual rendering like audio rendering. As such, this attribute has no impact on the visual aspect of rendering. The role of “separator” attribute is also not significant and may be left to default value.
The “largeop” attribute determines the size of the operator. If it is true, then the operator is drawn larger than its normal size. For example, ∫ and ∏ operators are displayed as large operator as “largeop” attribute for these operators are set “true” by default. If we need to display the normal size, then their "largeop" attribute is set to "false".
#### Example 12: Large operator
<m:math display="block">
<m:mrow>
<m:mo> ∏ </m:mo>
<m:mo> , </m:mo>
<m:mo> ∫ </m:mo>
<m:mo> , </m:mo>
<m:mo largeop="false"> ∏ </m:mo>
<m:mo> , </m:mo>
<m:mo largeop="false"> ∫ </m:mo>
</m:mrow>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
, , , , , ,
The movablelimits attribute allows underscripts and overscripts to be represented as subscript and superscript respectively. This rendering of under and over scripts as sub and super scripts is possible by seting this attribute to true. The implementation of this attribute by renderers is not yet consistent.
#### Example 13: The "movablelimits" attribute
<m:math display="block">
<m:munderover>
<m:mo movablelimits='true'> ∑ </m:mo>
<m:mi> a </m:mi>
<m:mi> b </m:mi>
</m:munderover>
<m:munderover>
<m:mo movablelimits='false'> ∑ </m:mo>
<m:mi> a </m:mi>
<m:mi> b </m:mi>
</m:munderover>
</m:math>
Save the file after editing as “test.xml”. The display looks like :
a b a b a b a b
The accent attribute determines whether an operator is treated as accent (diacritical mark) when used as an underscript or overscript.
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| External bookmarks | 2013-12-05T14:53:13 | {
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"url": "http://cnx.org/content/m13548/latest/",
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https://www.lil-help.com/questions/775/mat-142-34-basics-of-probability-theory | MAT 142 3.4 Basics of Probability Theory
# MAT 142 3.4 Basics of Probability Theory
398.3k points
1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3
green candies and 4 orange candies. You select one at random. Find the probability that you
select one that is:
a. Brown
b. Green or orange
c. Not red
d. Yellow or blue
e. Yellow and blue
1. Shoppers at a local department store were asked to complete a survey of their shopping
experience. The results are shown in the table below:
** Satisfied with Service Not satisfied with Service** Totals
Made a purchase 130 494 624
Did not make a purchase 715 183 898
Totals 845 677 1522
__
a. What is the probability that a shopper selected at random made a purchase?
b. What are the odds that a shopper selected at random was satisfied with the service?
MAT 142
393k points
#### Oh Snap! This Answer is Locked
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Excerpt from file: Math Tutorial MAT 142 Problem set Unit: Probability Topic: Basics of Probability Theory Directions: Solve the following problems. Please show your work and explain your reasoning. 1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3 green candies and 4
Filename: M1084.pdf
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Words: 292
J
0 points
#### Oh Snap! This Answer is Locked
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Excerpt from file: Week4DQ#3DiscOps&ExtraordinaryItems Whyisitimportanttoreportdiscontinuedoperationsorextraordinaryitemsseparately fromincomefromcontinuingoperations?Isthismethodofreportingallowed?What concernsdoesthistypeofreportingcreate?Doestheaverageinvestorunderstandthe
Filename: ACC 280week4dqs.zip
Filesize: < 2 MB
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Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example. | 2017-09-23T11:07:08 | {
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"url": "https://www.lil-help.com/questions/775/mat-142-34-basics-of-probability-theory",
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https://math.stackexchange.com/questions/2972505/find-the-sum-to-n-term-of-frac11-cdot2-cdot3-frac32-cdot3-cdot4/2972797 | find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + …$
$$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ...$$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)}$$
I do not know how to get a telescoping series from here to cancel terms.
• You can use $$\frac{2k-1}{k(k+1)(k+2)}=\frac{2k}{k(k+1)(k+2)}-\frac1{k(k+1)(k+2)}.$$ – Sungjin Kim Oct 26 '18 at 18:11
• You do mean an infinite sum, not a finite sum, right? – Connor Harris Oct 26 '18 at 18:15
• @ConnorHarris I think this means writing the partial sums explicitly. – Sungjin Kim Oct 26 '18 at 18:16
• @ConnorHarris, a sum that can be expressed in n. – LetzerWille Oct 26 '18 at 18:17
• Note that the $\frac{3}{k+1}$ term for some value of $k$ cancels the $\frac{-1}{2k}$ and $\frac{-5}{2(k+2)}$ terms for adjacent values of $k$. – Connor Harris Oct 26 '18 at 18:18
HINT:
Note that we have
\begin{align} \frac{2k-1}{k(k+1)(k+2)}&=\color{blue}{\frac{3}{k+1}}-\frac{5/2}{k+2}-\frac{1/2}{k}\\\\ &=\color{blue}{\frac12}\left(\color{blue}{\frac{1}{k+1}}-\frac1k\right)+\color{blue}{\frac52}\left(\color{blue}{\frac{1}{k+1}}-\frac{1}{k+2}\right) \end{align}
• There's a typo after the first $=$ sign, it should be $\frac{5/2}{k+2}$. Can't edit though, since edits must be 6 characters at minimum. – a_guest Oct 27 '18 at 12:50
• @a_guest Thank you. I've edited accordingly. – Mark Viola Oct 27 '18 at 16:29
Let the fractions be $$\frac{a}{k}$$, $$\frac{b}{k+1}$$, and $$\frac{c}{k+2}$$.
$$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$$
We want the following
$$a+b+c=0$$
$$3a+2b+c=2$$
$$2a=-1$$
Solve, $$a=-\frac{1}{2}$$, $$b=3$$, and $$c=-\frac{5}{2}$$.
The rest is standard.
You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $$3$$ or greater than $$n$$.
\begin{aligned} & \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} - \sum_{k=1}^n\frac{5}{2}\frac{1}{k+2} \\ & = \left[-\frac{1}{2} - \frac{1}{4} + \frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{3}{2} + \frac{1}{2}\sum_{k=2}^n\frac{6}{k+1}\right] - \left[\frac{1}{2}\sum_{k=1}^n\frac{5}{k+2}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n+1}\frac{6}{k}\right] - \left[\frac{1}{2}\sum_{k=3}^{n+2}\frac{5}{k}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n}\frac{6}{k} + \frac{6}{2}\frac{1}{n+1}\right] - \left[\frac{1}{2}\sum_{k=3}^{n}\frac{5}{k} + \frac{5}{2}\frac{1}{n+1} + \frac{5}{2}\frac{1}{n+2}\right] \\ & = \frac{3}{4} + \frac{1}{2}\sum_{k=3}^{n}\left[\frac{-1 + 6 - 5}{k}\right] + \frac{6}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+2} \\ & = \frac{3}{4} + \frac{1}{2(n+1)} - \frac{5}{2(n+2)} \end{aligned}
When terms are in A.P in denominator, we use difference of last and first terms in product to distribute the terms of denominator over numerator, to change into telescoping series. | 2021-02-27T09:17:49 | {
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https://math.stackexchange.com/questions/3053616/find-sum-n-1-infty-tan-1-frac2n2 | # Find $\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$
Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$
There's a solution here that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.
My proposed solution
\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align}
And this implies $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
• This may or may not be of help: $\arctan(1/x) = \pi/2 - \arctan(x)$ – user150203 Dec 27 '18 at 5:23
• The 3rd and 4th lines of the equation after "My proposed solution" are missing a summation sign. Otherwise, the solution is correct. – JimmyK4542 Dec 27 '18 at 5:35
• @JimmyK4542 Edited, ty. – Akash Gaur Dec 27 '18 at 5:40
• – lab bhattacharjee Dec 27 '18 at 6:34
\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(1+n)+\tan^{-1}(1-n)\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(n+1)-\tan^{-1}(n-1) \end{align}
Edit:$$\color{red}{\sum_{n=1}^\infty}$$ was missing in your question before edit. I am not going to delete this.
However your proof is now correct. $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\color{red}{\frac\pi2+\frac\pi2-\frac\pi4-0=\pi-\frac\pi4}=\frac{3\pi}{4}$$
Looks good to me. If I was going to offer a critique I would just say: when writing an argument it's always better to over communicate rather than under communicate.
The first equality is just algebra.
Your second equality requires a little bit to see clearly but it's true. Most will recall:
$$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$ Or if you'd like: $$A+B= \tan^{-1} \bigg(\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} \bigg)$$ Taking $$A=\tan^{-1}(1+n)$$ and $$B=\tan^{-1}(1-n)$$
Honestly adding this much explanation seems like almost overkill.
The 4th equality follows as result of $$\tan^{-1}$$ being an odd function.
Now the last part you are using a telescoping series technique so that you may ignore all the middle terms. That is,
\begin{align} &\sum_{n=1}^\infty\tan^{-1}(n+1)-\tan^{-1}(n-1) \\ &= \lim_{m\to \infty} \tan^{-1}(m+1)-tan^{-1}(m-1)+\dots +\tan^{-1}(4)-\tan^{-1}(2)+\tan^{-1}(3)-\tan^{-1}(1)+tan^{-1}(2)-\tan^{-1}(0) \end{align}
So after we consider what cancels and what doesn't we find that we only need to concern ourselves with $$\lim_{m\to \infty}\tan^{-1}(m+1)+\tan^{-1}(m-1)-\tan^{-1}(1)$$
So while that is true: I think it might merit a sentence or two just to make sure the audience is following.
• Isn't $M=\lim_{m\to \infty} \tan^{-1}(m)-\tan^{-1}(0)=\frac\pi2-0?$ – tatan Dec 27 '18 at 5:50
• @tatan. Yes. I am missing some terms. Let me edit. – Mason Dec 27 '18 at 5:51
• Thanks. I'm really slow at writing in latex so I skip steps. But I will add more explanation in English in the future. – Akash Gaur Dec 27 '18 at 5:57
• Yep. Can I ask you one thing? Are you from Maryland?(I saw from your profile) – tatan Dec 27 '18 at 19:57
• Yes. Uh oh. Stranger danger. :). No. I am not worried. I grew up in Ohio and move to the DC area post college. I went to Ohio University not Univ of Maryland. Though I did cite Dr. Yorke in my undergrad thesis and then I met him wandering once while wandering Univ of Maryland's halls... – Mason Dec 27 '18 at 19:57 | 2020-07-06T09:55:54 | {
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https://mathhelpboards.com/threads/when-0-123-495051-0-515049-321.6136/ | # When 0.123...495051/0.515049...321
#### anemone
##### MHB POTW Director
Staff member
Hello MHB,
Recently I've come across a problem that jumped off the page at me (Find the first 3 figures after the decimal point in the decimal expression of the number $$\displaystyle \frac{0.12345678910\cdots495051}{0.515049\cdots987654321}$$).
I tried to approach it by making a table where I started to divide some smaller numbers but stick to the same pattern that is required by the aforementioned problem, i.e.
$$\displaystyle \frac{0.12}{0.21}=0.571...$$
$$\displaystyle \frac{0.123}{0.321}=0.383...$$
$$\displaystyle \frac{0.1234}{0.4321}=0.285...$$
$$\displaystyle \frac{0.12345}{0.54321}=0.227...$$
$$\displaystyle \frac{0.123456}{0.654321}=0.188...$$
$$\displaystyle \frac{0.1234567}{0.7654321}=0.161...$$
$$\displaystyle \frac{0.12345678}{0.87654321}=0.140...$$
$$\displaystyle \frac{0.123456789}{0.987654321}=0.124...$$
$$\displaystyle \frac{0.12345678910}{0.10987654321}=0.123...$$
$$\displaystyle \frac{0.1234567891011}{0.1110987654321}=0.111...$$
$$\displaystyle \frac{0.123456789101112}{0.121110987654321}=0.019...$$
$$\displaystyle \frac{0.12345678910111213}{0.13121110987654321}=0.940...$$
$$\displaystyle \frac{0.1234567891011121314}{0.1413121110987654321}=0.873...$$
$$\displaystyle \frac{0.123456789101112131415}{0.151413121110987654321}=0.815...$$
$$\displaystyle \frac{0.12345678910111213141516}{0.16151413121110987654321}=0.764...$$
$$\displaystyle \frac{0.1234567891011121314151617}{0.1716151413121110987654321}=0.719...$$
$$\displaystyle \frac{0.123456789101112131415161718}{0.181716151413121110987654321}=0.679...$$
$$\displaystyle \frac{0.12345678910111213141516171819}{0.19181716151413121110987654321}=0.643...$$
$$\displaystyle \frac{0.1234567891011121314151617181920}{0.2019181716151413121110987654321}=0.611...$$
and so on and so forth
but I failed to observe any pattern that's worth mentioning to help me to crack the problem.
Could anyone help me with this particular problem? Thanks in advance.
#### Opalg
##### MHB Oldtimer
Staff member
I also approached it by making a table, but I started by using the most significant figures in the given denominator:
$$\displaystyle \frac{0.1}{0.5} = 0.2$$,
$$\displaystyle \frac{0.12}{0.51} = 0.2352\ldots$$,
$$\displaystyle \frac{0.123}{0.515} = 0.2388\ldots$$,
$$\displaystyle \frac{0.1234}{0.5150} = 0.2396\ldots$$,
$$\displaystyle \frac{0.12345}{0.51504} = 0.2396\ldots$$,
$$\displaystyle \frac{0.123456}{0.515049} = 0.2396\ldots$$,
$$\displaystyle \frac{0.1234567}{0.5150494} = 0.2396\ldots$$,
$$\displaystyle \frac{0.12345678}{0.51504948} = 0.2396\ldots$$.
By this time, the fraction had stabilised (within the limits of my 8-digit calculator) to 0.2396987. I doubt whether further approximations would affect the first three digits after the decimal point.
Edit. You can confirm that by using a bit of calculus. If $f(x,y) = x/y$ then $\delta f \approx (1/y)\delta x - (x/y^2)\delta y$. With $x\approx 0.123$, $y\approx 0.515$ and $\delta x = \delta y = 10^{-n}$, you find that $\delta f < 2*10^{-n}$. So the change in the fraction after the eighth decimal places in numerator and denominator is never going to be sufficient to affect the first three decimal places in the quotient.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Hi Opalg,
I'm very thankful to you for showing me something that I had never thought about before and your solution and the proof work so beautifully...
Thank you so much! | 2021-09-26T00:13:29 | {
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https://math.stackexchange.com/questions/186972/how-can-lu-factorization-be-used-in-non-square-matrix/186997 | # How can LU factorization be used in non-square matrix?
In my textbook, there is some information about LU factorization of square matrix $A$, but not about non-square matrix.
How can LU factorization be used to factorize non-square matrix?
• Yes. Let $A$ be $m \times n$ matrix, then $L$ is $m \times m$ and $U$ is $m \times n$.
– user2468
Aug 26, 2012 at 4:17
I'll illustrate how to understand the LU-decomposition of a particular $3 \times 4$ matrix below. The method works just as well for other sizes since the LU-decomposition arises naturally from the study of Gaussian elimination via multiplication by elementary matrices.
$$\begin{array}{ll} A \ = &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 2 & 4 & 0 & 7 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_2-2r_1 \rightarrow r_2} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 0 & 6 & 5 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 0 & 6 & 5 \\ 0 & 5 & -1 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right] = \ U \end{array}$$ We have $U = E_3E_2E_1A$ hence $A = E_1^{-1}E_2^{-1}E_3^{-1}U$ and we can calculate the product $E_1^{-1}E_2^{-1}E_3^{-1}$ as follows: $$\begin{array}{ll} I \ = &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] \ \underrightarrow{r_3-r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] \ \underrightarrow{r_2+2r_1 \rightarrow r_2} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] = PL \end{array}$$ I have inserted a "$P$" in front of the $L$ since the matrix above is not lower triangular. However, if we go one step further and let $r_2 \leftrightarrow r_3$ then we will obtain a lower triangular matrix: $$\begin{array}{ll} PL \ = &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \\ \end{array} \right] =L \end{array}$$ Therefore, we find that $E_1^{-1}E_2^{-1}E_3^{-1}=PL$ where $L$ is as above and $P = E_{2 \leftrightarrow 3}$. This means that $A$ has a modified $LU$-decomposition. Some mathemticians call it a $PLU$-decomposition, $$A = \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]}_{P} \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \\ \end{array} \right]}_{L}\underbrace{\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right]}_{U} = \underbrace{\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right]}_{PL}\underbrace{\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right]}_{U}.$$ Since permutation matrices all satisfy the condition $P^k=I$ (for some $k$) the existence of a $PLU$-decomposition for $A$ naturally suggests that $P^{k-1}A = LU$. Therefore, even when a $LU$ decomposition is not available we can just flip a few rows to find a $LU$-decomposable matrix. This is a useful observation because it means that the slick algorithms developed for $LU$-decompositions apply to all matrices with just a little extra fine print.
Much of the writing above can be spared if we adopt the notational scheme illustrated below. $$\begin{array}{ll} A \ = &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 2 & 4 & 0 & 7 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_2-2r_1 \rightarrow r_2} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (2) & 0 & 6 & 5 \\ -1 & 3 & 2 & 0 \end{array} \right] \ \underrightarrow{r_3+r_1 \rightarrow r_3} \\ & \\ &\left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (2) & 0 & 6 & 5 \\ (-1) & 5 & -1 & 1 \end{array} \right] \ \underrightarrow{r_2 \leftrightarrow r_3} \ \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ (-1) & 5 & -1 & 1 \\ (2) & 0 & 6 & 5 \end{array} \right] = \ U \end{array}$$ We find if we remove the parenthetical entries from $U$ and ajoing them to $I$ then it gives back the matrix $L$ we found previously: $$U = \left[ \begin{array}{cccc} 1 & 2 & -3 & 1\\ 0 & 5 & -1 & 1 \\ 0 & 0 & 6 & 5 \end{array} \right] \qquad L=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & 0 & 1 \end{array} \right].$$
Hope this helps.
• "permutation matrices all satisfy the condition $P^2=I$" is incorrect.
– user147263
Jan 16, 2016 at 22:09
• @Normal for example ? Sorry, I can't see past swap then reverse swap does nothing... Jan 16, 2016 at 22:28
• A permutation can involve a $3$-cycle like $1\to 2\to 3$.
– user147263
Jan 16, 2016 at 22:45
• @Normal right, that makes sense... but, there still must be some power for which $P^k=I$. So, something similar to what I have currently written is possible. I wonder, is this possible for $3 \times 3$ matrix examples. Great comment, I wish I could upvote comments ( for internet pts naturally) Jan 17, 2016 at 1:39
• For all permutation matrices: $P \cdot P^{T} = I$ Feb 5, 2019 at 8:03 | 2022-05-19T07:56:54 | {
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http://mathhelpforum.com/discrete-math/11509-two-discrete-problems.html | # Math Help - Two Discrete Problems
1. ## Two Discrete Problems
Hello,
I am needing some help on the two problems below. Thank you in advance.
1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.
and
2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.
2. Originally Posted by MathStudent1
Hello,
I am needing some help on the two problems below. Thank you in advance.
1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.
.
gcd(14,21)=7
Thus,
7 divides 101 which is false.
3. Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.
I am beening taught to turn this into a if/then statement and go from there.
If m and n are integers, then 14m + 21n = 100
Factor out a 7 givng us 7(2n +3m) = 100
From here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.
4. Originally Posted by MathStudent1
Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.
I am beening taught to turn this into a if/then statement and go from there.
If m and n are integers, then 14m + 21n = 100
Factor out a 7 givng us 7(2n +3m) = 100
From here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.
Let,
p= (there exists n,m such that 14m+21n=100).
Now if "p" is true we shown that 7 divides 100.
Thus,
q=(7 divides 100).
But this is false.
Meaning, (modens ponens)
[(p--> q) and (~q)]--> ~p
Thus,
~p=(there do not exists n,m such that 14m+21n=100)
This is mine 46th Post!!!
5. Hello, MathStudent1!
Here;s the first one . . .
1) Prove by contradiction that there do not exist integers m and n
. . .such that: .14m + 21n .= .101
Assume that there are integers m and n such that: .14m + 21n .= .101
. . . . . . . . . . . . . . . . . . .100
Divide by 7: . 2m + 3n .= .----
. . . . . . . . . . . . . . . . . . . .7
In the left side: m is an integer, so 2m is an integer. *
. . . . . . . . . . . . n is an integer, so 3n is an integer. *
Hence, 2m + 3n (the sum of two integers) is an integer. *
But the right side is an irreducible fraction (or decimal 14.2857...)
We have reached a contradicition, hence our assumption was incorrect.
. . There are no integers m and n that satisfy 14m + 21n .= .101
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
*
Reasons: The set of integers is closed under multiplication and addition.
6. Thank you Soroban. One question for you. When orginally I posted my thread I made a typo in that 14m + 21n = 100 NOT 101. Does that change the proof at all or does the logic remain the same. Following your proof, am I correct by stating this below:
Prove that there do not exist integers m and n such that: 14m + 21n = 100
Restating the problem - If m and n are integers, then 14m + 21n = 100
Factor out a 7 and we get 7(2m + 3n) = 100
Divide both sides by 7 and we get 2m + 3n = (100/7)
Since m is an integer then 2*m is an integer and since n is an integer then 3*n is and integer.
However, the RHS (100/7) is not an integer and therefore,
2m + 3n (IS NOT EQUAL TO) (100/7)
Hence, we have reached a contradiction of our assumption that m and n are integers.
Therefore, there do not exist integers m and n such that 14m + 21n = 100.
*****Thank you!*****
7. Hello again, MathStudent1!
Yes, the modified proof is still valid . . .
. . and you did an excellent job! | 2015-03-28T18:39:21 | {
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http://inside-science.com/r362cn/45e997-binomial-coefficient-latex | The combination (n r) (n r) is called a binomial coefficient. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. therefore gives the number of k -subsets possible out of a set of distinct items. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. are the different ordered arrangements of a k-element subset of an n-set, $$\binom{n}{k} = \binom{n-1}{k-1} +\binom{n-1}{k}$$. The binomial coefficient is defined by the next expression: $\binom {n}{k} = \frac {n ! k-combinations of n-element set. Binomial coefficient, returned as a nonnegative scalar value. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. It is especially useful for reasoning about recursive methods in programming. (n - k)!} On the other side, \textstyle will change the style of the fraction as if it were part of the text. In latex mode we must use \binom fonction as follows: \frac{n!}{k! This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. Below is a construction of the first 11 rows of Pascal's triangle. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … where A is the permutation, A_n^k = \frac{n!}{(n-k)! See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, \binom{n}{k} is … }}{{k!\left( {n - k} \right)!}} In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. \boxed, How to write table in Latex ? How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? . The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. ... Pascal’s triangle. Specially useful for continued fractions. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. } Also, the text size of the fraction changes according to the text around it. So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… For example, … If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. LaTeX forum ⇒ Math & Science ⇒ Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. therefore gives the number of k-subsets possible out of a set of distinct items. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. C — All combinations of v matrix. Identifying Binomial Coefficients. This same array could be expressed using the factorial symbol, as shown in the following. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. The order of selection of items not considered. The binomial coefficient is defined by the next expression: \ [ \binom{n} {k} = \frac{n!} In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. The binomial coefficient \binom{n}{k} can be interpreted as the number of ways to choose k elements from an n-element set. ( n - k )! Ak n = n! The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Then it's a good reason to buy me a coffee. (n−k)! Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Thank you ! (n - k)!} Latex binomial coefficient Definition. One can drop one of the numbers in the bottom list and infer it from the fact that sum … The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. This website was useful to you? Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. Latex The binomial coefficient is defined by the next expression: \[\binom {n}{k} = \ frac {n!}{k!(n-k)! {k! As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is Binomial coefficient, returned as a nonnegative scalar value. In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way.$ And of course this command can be included in the normal text flow \ (\binom{n} {k}\). Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Using fractions and binomial coefficients in an expression is straightforward. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. In Counting Principles, we studied combinations. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. The command \displaystyle will format the fraction as if it were in mathematical display mode. This will give more accuracy at the cost of computing small sums of binomial coefficients. Mathematical Equations in LaTeX. n! begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? This article explains how to typeset them in LaTeX. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. I agree. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. \vec,\overrightarrow; Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value The possibility to insert operators and functions as you know them from mathematics is not possible for all things. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. In this video, you will learn how to write binomial coefficients in a LaTeX document. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. = \binom{n}{k}$$Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. }{k ! k-combinations of n-element set. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". Gerhard "Ask Me About System Design" Paseman, 2010.03.27 \endgroup – Gerhard Paseman Mar 27 '10 at 17:00 (adsbygoogle = window.adsbygoogle || []).push({}); Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. Latex k parmi n - coefficient binomial. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … All combinations of v, returned as a matrix of the same type as v. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. You can set this manually if you want. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. binomial Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. {k! How to write it in Latex ? {k! Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Knowledge base dedicated to Linux and applied mathematics. It will give me the energy and motivation to continue this development. (−)!. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Since binomial coefficients are quite common, TeX has the \choose control word for them. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. The Texworks shows … }}{{k!\left( {n - k} \right)!}} In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This formulas, graphs). Asking for help, clarification, or responding to other answers. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. All combinations of v, returned as a matrix of the same type as v. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. In latex mode we must use \binom fonction as follows: \frac {n!} C — All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". Here's an equation: math \frac {n!} For these commands to work you must import the package amsmath by adding the next line to the preamble of your file }}{{k!\left( {n - k} \right)!}}. This method of constructing mathematical proofs is called mathematical induction. Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. The usual binomial coefficient can be written as \left({n \atop {k, {n-k}}}\right). = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \frac{n!}{k! \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. Usually, you find the special input possibilities on the reference page of the function in the Details section. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. coefficient The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. (n-k)!} Using fractions and binomial coefficients in an expression is straightforward. Then it's a good reason to buy me a coffee. = \binom {n} {k} This is the binomial coefficient. binomial coefficient Latex. Binomial Coefficient. samedi 11 juillet 2020, par Nadir Soualem. 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binomial coefficient latex 2021 | 2021-11-29T17:05:12 | {
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https://www.physicsforums.com/threads/compound-interest-systems-of-equations-problem.808845/ | # Compound Interest Systems of Equations Problem
1. Apr 16, 2015
### ConstantineO
1. The problem statement, all variables and given/known data
Romeo was given a gift of $10,000 when he turned 16. He invested it at 3% per annum. Three years later, Juliet was given$10,000, which she invested at 5% per annum. When will the two amounts be equal in value?
2. Relevant equations
Compound Interest Formula
Total = Capital(1+interest)^Years
t = c(1+i)^n
3. The attempt at a solution
Deciding how to create my formula is where things get fuzzy. The way I initially attempted this was to add a 3 year head start onto Romeo's interest formula. This is what results (Used wolfram to save time rewriting):
This eventually results in "4.61103." However, this is not the answer at the back of my textbook. They decided to opt for a different formula where the time is subtracted from Juliet's interest function. This is shown here:
This results in the correct answer of 7.61103. What I'm wondering, why do I have to subtract 3 years from Juliet's compound interest formula, and why does adding 3 years to Romeo's compound interest formula produce the wrong answer?
2. Apr 16, 2015
### BvU
So clearly the book starts counting at the beginning of the story: when Romeo brings his money to the bank.
That means Julia's interest clock is at x-3 when Romeo's is at x.
3. Apr 16, 2015
### SammyS
Staff Emeritus
What is Romeo's age when then two accounts have equal value?
4. Apr 16, 2015
### ConstantineO
Is this always the convention when dealing with problems like this? From my experience with physics courses, I assumed that time shares a relation with the total amount of money generated by compound interest. The interest and initial capital is thus fixed, and you are only left with a relation between time and the total sum of money. If you were to treat time as a vector which you could progress backwards and forwards on, each snapshot through time depending on which way you were progressing would net either a larger or smaller amount of the total. I don't understand why the book is forcing me to start at the beginning of the story. The story is like pages of a book which can be turned towards or away from the end of the novel. The content of the novel does not change and its contents are already written.
I don't understand what Romeo's age has to do with this? Is this some kind of statement that alludes to time only progressing forward or something?
5. Apr 16, 2015
### BvU
Nothing so fancy. The book is sloppy by not stating when it wants the clock (calendar in this case) to show t = 0. From the book answer it appears that it assumes t = 0 when Romeo turns 16. The exercise could equally well have assumed t = 0 when Julia receives the ten grand. If I were grading this, I'd have to allow both answers, certainly when the calculation steps are shown.
6. Apr 16, 2015
### ConstantineO
I actually just figured it out while discussing this question with a friend just a few moments before you posted. I took the delta of the two times from both of the ways of solving this question and realized that this is nothing more than a semantics issue of where you measure time from. This question is poorly setup, and I am going to voice my irritation to the person marking.
7. Apr 16, 2015
### Ray Vickson
A more serious problem is that when compounding annually, the answer is "never". At Romeo's 23rd birthday, Romeo's account contains more money than Juliet's, while at his 24th birthday, Juliet's account contains more than Romeo's. Unless there is something like continuous compounding, the two amounts never match exactly; that is, unless you can withdraw your money part-way through a year and earn a part-year's interest, you could never find a time where the two withdrawals are equal. (However, I would not raise this issue with your teacher if I were you; just "go with the flow".)
Also, instead of expressing irritation, it would be wiser for you to say that you will give two answers (depending on where to start measuring time) and point out (nicely) that the question is a bit ambiguous.
8. Apr 16, 2015
### SammyS
Staff Emeritus
RGV makes an excellent point. This is a short-coming of many problems of this type.
It's often the case that for any fraction of a single compounding period that's "left over", simple interest is computed for that extra amount of time. With this method, find the value of the account at x = 4, then see if the values can be equal at any time in the 4th year using simple interest.
(Edited slightly.)
Last edited: Apr 16, 2015
9. Apr 16, 2015
### ConstantineO
I don't wish to sound like an imbecile, but I have no clue what that means. How would I set up a systems of equation if I subbed in 4 for both x's? I believe I would be left with only one unknown for either equation, or am I misunderstanding you? Please clarify what you mean by using x=4.
10. Apr 16, 2015
### SammyS
Staff Emeritus
When x = 4, (7years for Romeo, 4 years for Juliet), what is the value in each in each account?
Start from that point for each & using simple interest, and see how long it takes for the values to be equal.
11. Apr 16, 2015
### ConstantineO
I am not familiar with the simple interest formula, so I googled it. I am operating under the assumption that it is different than the compound interest formula.
I am to assume this is correct?
Simple interest Formula
I = "the grown money"
p = the initial capital
r = interest rate per year
t = t
(I) = (p)(r)(t)
I think I know what you're getting at, so let me make an attempt.
Romeo Compound Interest Calculation Total
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
Julie Compound Interest Calculation Total
Aj = 10,000(1.05)^4
Aj = 12,155.0625
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)
Equate the two to equal each other.
(368.94)(t) = 607.753125(t)
(368.94)(t) -607.753125(t) = 0
t(368.94-607.753125)=0
This doesnt work though...
12. Apr 16, 2015
### Ray Vickson
It's not that complicated. Let $R_t, J_t$ be the amounts in Romeo's and Juliet's accounts at time $t$ (in years, measured from Romeo's 16th birthday). We have:
$$\begin{array}{ccc} & t=7 & t=8 \\ R_t & 12298.74 & 12667.70 \\ J_t & 12155.06 & 12762.82 \end{array}$$
You can draw two straight-line plots, one going from x = 0 at (7,R7) to x = 1 at (8,R8) and another from x = 0 at (7,J7) to x = 1 at (8,J8). Where the two lines cross is the point where simple interest calculations (for the partial year) would give you equality of monetary values. This will be close to (but not exactly equal to) the value x = 0.611---that is, for time 7.611 ---that your previous calculation would have produced. (The difference is due to the slight curvature in the money vs.time graph over the year, due to continuous compounding over the year, as compared with the straight line obtained from simple interest accumulation over the year.)
13. Apr 16, 2015
### ConstantineO
Could you show me an example of this because I am simply not understanding what you're saying, or can you show me how to equate the two together using algebra?
14. Apr 17, 2015
### ConstantineO
This is getting out of hand, so I am going to try to be as clear and as concise as possible. I know how to calculate the correct answer for this question, and I realize where I first made my mistake. I am now wondering what everyone is talking about when they mention using simple interest. I have literally 0 experience doing any simple interest calculations, and I have no idea what sammyS is talking about here.
If anyone is holding information back because they believe an attempt has not been made and it would violate forum rules, please stop. I understand how to do the book's question using the books method, and I know and where the discrepancy between the 4.61103 and 7.61103 came from. I have made a wholehearted attempt so please for sake of my sanity just give me an example or step by step breakdown. I learn through examples, I look at the process involved, and I create a model in my head how everything interacts and relates with each other.
What I am seeking is an explanation and a simple example of what this simple interest method is for calculating the 0.61103 year after the exact 7 years has passed.
What I am understanding so far.
- First use the compound interest formula to create outputs for both Romeo's and Juliet's accounts after being influenced by exactly 7 years of interest.
- This is done by assigning "4" to the variable x
- Get said products, in this case being:
Romeo's Total After Exactly 7 Years = 12,298.73865
Juliet's Total After Exactly 7 Years = 12,155.0625
- Since you know that the two amounts are still not equal, you must calculate the remaining time between the two using the Simple interest formula
- This is what I am having problems with and what I would appreciate having an example for. Preferably shown algebraically, so I am not plotting nearly straight lines in Desmos.
Not to sound contentious, but when you try to find the point of intersect of two nearly straight lines with a slope that is in hundredths of a unit, it is just plainly byzantine. I am probably doing this wrong, but graphically representing each simple interest formula and trying to find where they meet is insane.
Last edited: Apr 17, 2015
15. Apr 17, 2015
### Ray Vickson
How could I have made my explanation any simpler? There was no hidden information or holding back of anything. I said: just draw two straight-line graphs for Romeo and Juliet, where the lines connect their (time,money) points at the start and end of the final year.
The true graphs of time vs. money will be curved, because of compound interest, but over a short period such as one year the "curvature" will be small, and the graph will look almost like a straight line; replacing the curve by a straight line ---only for that single year---would give you the "simple" interest schedule for that year. It will be almost the same as the compound interest schedule, because the curvature is small over short times such as a year.
If you don't believe me, just draw the graphs of money vs. time for Romeo and for Juliet. You will see that they curve up, but are almost straight over short periods. The only way to really understand it is to sit down and do it.
16. Apr 17, 2015
### ConstantineO
I am going to assume the two lines would be plotted as
y=(12155.0625)(0.05)(x)
and
y=(12298.73865)(0.03)(x)
If that's the case, I don't understand how any human being is supposed to find the point of intersect graphically like this.
I did sit down and do it. I have been sitting down and ripping my hair out doing this for sometime.
Edit - In the event that anyone else would like to question how committed I am to "sitting down and doing it." I've had enough of that nonsense from teachers with superiority issues and over bloated opinions that like to question how hard I am making an attempt.
Last edited: Apr 17, 2015
17. Apr 17, 2015
### Ray Vickson
As long as you persist in trying to plot the wrong thing you will need to keep pulling out your hair. Go back and read what I wrote in #12. Do the plot exactly as I indicated there.
Better still plot the two curves y = 10000 * (1.03)^x and y = 10000 * (1.05)^(x-3) for x = from 3 to 8. They are curved, aren't they? Don't they look almost straight over the shorter interval 7 \leq x \leq 8? in each case if you were to plot a straight line from (7,y(7)) to (8,y(8)) it would look very close to the actual curve (x,y(x)) for x between 7 and 8. The "simple interest" effect between 7 and 8 would be the straight line, while the "compound interest" effect would be the curve. They both pass through the same points at 7 and 8, but they differ in between.
18. Apr 17, 2015
### ConstantineO
As I have stated before.
I was completely uncertain of what to plot. I am being told to use the simple interest formula, but the two formulas you have provided are compound interest formulas:
y = 10000 * (1.03)^x
y = 10000 * (1.05)^(x-3)
I plotted these long before I even got to this whole simple interest business when trying to figure out the original discrepancy between 7.61103 and 4.61103. I am fully aware that the compound interest lines are nearly straight. I don't see how the average rate of change between x =7 and x =8 is going to help me discover the remaining time of 0.61103 years. I am still unclear of what to plot? Do I plot the compound interest based functions for Romeo and Juliet, or do I plot the Simple interest functions? I have done both and I am not seeing anything that is giving me any kind of better understanding. Is this operation you describe too difficult represent algebraically?
I am getting far too irritated by this and far too frustrated and am quickly losing the desire to continue.
19. Apr 17, 2015
### ConstantineO
I was on the right track here. There is no need for this graphing nonsense. I forgot a single number which was my "+ 3" in my simple interest formula. I was a fool not to recognize that the lack of a constant would render my simple interest system of equation unsolvable. The lack of anyone speaking up about this is worrisome. In the future, when referencing the slope of a secant line between the range of two x values of a curve, please write in the notation of 7 < x < 8. I had absolutely no clue what you were talking about.
So let's try that last bit again with a fixed formula.
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t+3)
(Ir) = (368.94)(t+3)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t)
(Ir) = 607.753125(t)
Now the Systems of Equation
(607.753125)(t) = (368.94)(t+3)
(607.753125)(t) = (368.94)(t)+ 1106.82
t(607.753125-368.94) = 1106.82
t(238.813125) = (1106.82)
t = (1106.82)/(238.813125)
t = 4.634669891
Not too far away from 4.61103. I wonder why...
Let's try the simplified interest formula that measures time from Julias deposit date.
Romeo Simple Interest Calculation
(Ir) = (12,298.73865)(0.03)(t)
(Ir) = (368.94)(t)
Juliet Simple Interest Calculation
(Ir) = (12,155.0625)(0.05)(t-3)
(Ir) = 607.753125(t-3)
607.753125(t-3)= (368.94)(t)
(607.753125)(t) - 1823.259375 = (368.94)(t)
(607.753125)(t) - (368.94)(t) =1823.259375
t(607.753125 - 368.94) = 1823.259375
t(238.813125) = 1823.259375
t = 7.634.634669891
Not too far away from 7.61103. Surprise surprise....
Look Ma! No graphs.
7.634.634669891 - 4.634669891 = 3 just like 7.61103 - 4.61103 =3. I wonder if they're related?
I don't think I am wrong by stating that the time after the 7 year mark would actually be 0.634669891 of a year instead of 0.61103 if fractional compound interest is indeed calculated with simplified interest.
We can take things a step further and look at the secant line's slope of the curve that models interest from Romeo's deposit date during 7 < x < 8. Let's look at the average rate of change for both Romeo's and Juliet's account.
Romeo Compound Interest AROC: 7 < x < 8
While x = 4
Ar = 10,000(1.03)^4+3
Ar = 10,000(1.03)^7
Ar = 12,298.73865
x = 4
y =12,298.73865
While x =5
Ar = 10,000(1.03)^5+3
Ar = 10,000(1.03)^8
Ar = 12,667.70081
x = 5
y = 12,667.70081
Romeo's AROC
= (12,667.70081 - 12,298.73865) / (5 -4)
= 368.96216
Juliet's Compound Interest AROC 7 < x < 8
While x = 4
Aj = 10,000(1.05)^4
Aj = 12,155.0625
x = 4
y =12,155.0625
While x = 5
Aj = 10,000(1.05)^5
Aj = 12,762.81563
x = 5
y = 12,762.81563
Juliet's AROC
= (12,762.81563 - 12,155.0625) / (5-4)
=607.7531
After that huge bunch of calculations we now have slopes that we can create linear relations with.
Romeo's Linear Relation
y = (368.96216)(x + 3)
Juliet's Linear Relation
y = 607.7531(x)
Let's see what the Solution is for these in equations when put into a system.
(368.96216)(x + 3) = 607.7531(x)
(368.96216)(x) + 1106.88648 = 607.7531(x)
607.7531(x) -(368.96216)(x) = 1106.88648
x(607.7531-368.96216) = 1106.88648
x(238.783884) = 1106.88648
x = 4.635515854
Notice a pattern?
Time Measured from Romeo's Deposit Compound Calculation Product
x = 4.61103
Time Measured from Romeo's Deposit Simplified Interest Product
x = 4.634669891
Time Measured from Romeo's Deposit AROC: 7 < x< 8 Product
x = 4.635515854
I think its safe to say that they matched sometime in the range of 0.635515854 - 0.634669891- 0.61103 of the start of the 8th year. I think I made my point. I don't need any graphs to know how to rock, and I certainly showed that I sat down and did the work.
20. Apr 17, 2015
### Ray Vickson
Of course it can be done without graphs. Graphs can be of help in setting up the algebraic equations that must, eventually, be solved without graphs. However, you seemed to not understand how simple interest works (your claim, not mine) and to see that a graphical representation can sometimes be helpful---not always, just sometimes. If it was not helpful to you, fine. And, of course I suggested plotting the compound-interest graphs, but I guess you missed out the part where I said that during a short period, such as 1 year, the graph looks almost straight, and that a straight line replacement for the graph (but ONLY in that single year) gives you the simple-interest effect within the year.
Your writeup is almost incomprehensible to me, but your final answer seems OK. You say you still do not know why it is different from the original answer, and that is precisely what I was speaking of when I mentioned graphs: you could see right away why there is a difference.
Last edited: Apr 17, 2015 | 2017-12-12T16:43:57 | {
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https://math.stackexchange.com/questions/2044986/finding-common-terms-of-two-or-more-arithmetic-sequences | Finding common terms of two or more arithmetic sequences
Suppose I have three sequences $\{1,4,7,....,2998\}$ , $\{1,3,5,7,9,11,....,3001\}$ and $\{1,6,11,....,4001\}$ .
How can i find the number of common terms among them ?
For Example ,
If I have
1,3,5,7,9,11,13,15,17 and 19
1,4,7,10,13,16 and 19
1,6,11 and 16
My answer would be $6$.
Suppose i have $n$ number of series' with common differences $a_1,a_2,...,a_n$ and each of them having the starting term as $1$ . How can i find the number of common terms then ?
• Does your answer go all the way to the ends 2998,3001,4001 for the count of 0nly 6 common terms? – coffeemath Dec 5 '16 at 15:15
• its 6 for only the given example – psil123 Dec 5 '16 at 15:16
Write the arithemtic sequence in form: $a_n = a_0 + (n-1)d_1, b_n = b_0 + (n-1)d_2, c_n = c_0 + (n-1)d_3$. Now if they have a common term then we can write:
$$x = a_0 + kd_1 = b_0 + sd_2 = c_0 + td_3$$
As the common difference is known we can write this as:
$$x \equiv a_0 \pmod {d_1}$$ $$x \equiv b_0 \pmod {d_2}$$ $$x \equiv c_0 \pmod {d_3}$$
This is system of linear congurence relations and it can be easily solved by the Chinese Remainder Theorem.
And once you find one common element, to find the next common element you just add $LCM[d_1,d_2,d_3]$ to the previous one. In fact if you get $x \equiv A \pmod{LCM[d_1,d_2,d_3]}$ when solving the system of congurences, then total number of elements should be:
$$1 + \left\lfloor\frac{K-A}{LCM[d_1,d_2,d_3]}\right\rfloor$$
where $K$ is the smallest last element in the arithemtic sequences.
If you at the explicit form of your sequences, which build your sets, you get get that $a_n = 1+3 \cdot n$, $b_n = 1+ 2 \cdot n$ and $c_n =1+5 \cdot n$.
Now you just have to find the least common multiple of 2,3 and 5 and see that all their common terms are $c_n = 1 + 30 \cdot n$
• suppose i have n series with common differences a1,a2,...,an and each of them have the starting number as 1 . How can i find the number of common terms ? – psil123 Dec 5 '16 at 15:17 | 2019-09-18T21:58:37 | {
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https://math.stackexchange.com/questions/2957504/seemingly-tricky-dice-question-probability-that-one-event-occurs-before-another | # Seemingly tricky dice question-probability that one event occurs before another event?
The question is as follows:
You roll two fair dice over and over. Let $$A$$ be the event you see two even sums. Let $$B$$ be the event you see a sum of $$7$$ four times. What is the probability that event $$A$$ occurs before event $$B$$?
I know that for mutually exclusive events with independent trials, the probability that event $$E$$ occurs before event $$F$$ is $$\frac{\mathbb{P}(E)}{\mathbb{P}(E) + \mathbb{P}(F)}.$$ I tried using this formula, but I ran into a problem. I calculated $$\mathbb{P}(A)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \ \ \ \ \text{and} \ \ \ \ \mathbb{P}(B)=\left ( \frac{1}{6} \right )^4=\frac{1}{1296}.$$ However, I then realized that these probabilites are the events that two even sums occur $$\textit{in a row},$$ and similarly for my $$\mathbb{P}(B).$$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Edit: I know there is a geometric distribution involved.
• If it helps approach the problem, A occurs before B is usually just a less obvious way to say A occurs without B – Lord Farquaad Oct 16 '18 at 12:54
Since neither A nor B cares about odd sums that aren't 7, we'll just reroll in those cases and thus we can ignore the probability of those sums occurring.
Thus we're left with 7 in addition to the 6 even sums ($$2,4,6,8,10,12$$).
The probabilities of these are respectively $$6/36$$ and $$(1+3+5+5+3+1)/36 = 18/36$$.
Divide the probabilities of each of these by the sum of both probabilities, to get $$P(7)$$ and $$P(even)$$ such that $$P(7) + P(even) = 1$$ (this is $$P(7) = \frac{1}{4}$$ and $$P(even) = \frac{3}{4}$$).
Now, as bobajob pointed out, the probability of A occurring before B is the complement of the probability of B occurring before A.
$$P(2\ even\ before\ 4\ sevens) = 1-P(4\ sevens\ before\ 2\ even)$$
To have 4 sums of seven before 2 even sums, there can be at most 1 even sum before the 4th seven sum.
Thus we have the following possible sequences:
$$7\ 7\ 7\ 7$$
$$7\ 7\ 7\ even\ 7$$
$$7\ 7\ even\ 7\ 7$$
$$7\ even\ 7\ 7\ 7$$
$$even\ 7\ 7\ 7\ 7$$
Each of the above occurs with the following probability:
$$P(7) * P(7) * P(7) * P(7) = P(7)^4$$
$$P(7) * P(7) * P(7) * P(even) * P(7) = P(even) * P(7)^4$$
$$P(7) * P(7) * P(even) * P(7) * P(7) = P(even) * P(7)^4$$
$$P(7) * P(even) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$
$$P(even) * P(7) * P(7) * P(7) * P(7) = P(even) * P(7)^4$$
Then we have:
$$P(2\ even\ before\ 4\ sevens) = 1-P(4\ sevens\ before\ 2\ even)$$
$$= 1 - P(7)^4 - 4 * P(even) * P(7)^4$$
$$= 1 - (\frac{1}{4})^4 - 4 * \frac{3}{4} * (\frac{1}{4})^4$$
$$= 0.984375$$
## Alternative method
The probability we want will just be the probability of getting 0-3 sums equal to 7 before getting 2 even sums.
$$P(2\ even\ before\ 4\ sevens) = \sum_{i=0}^3P(i\ sevens\ before\ 2\ even)$$
Since one even sum will need to be at the end, we can simply consider all possible positions of the other even sum.
Taking as an example $$i=3$$ (3 7 sums before 2 even sums), we'll have the following possible order of events:
$$7\ 7\ 7\ even\ even$$
$$7\ 7\ even\ 7\ even$$
$$7\ even\ 7\ 7\ even$$
$$even\ 7\ 7\ 7\ even$$
The amount of these we have is simply $$i + 1$$.
Each of the above occurs with probability $$P(even)^2 * P(7)^i$$:
$$P(7) * P(7) * P(7) * P(even) * P(even) = P(even)^2 * P(7)^3$$
$$P(7) * P(7) * P(even) * P(7) * P(even) = P(even)^2 * P(7)^3$$
$$P(7) * P(even) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$
$$P(even) * P(7) * P(7) * P(7) * P(even) = P(even)^2 * P(7)^3$$
Thus we have:
$$P(i\ sevens\ before\ 2\ even) = (i+1) * P(even)^2 * P(7)^i$$
This gives us:
$$P(2\ even\ before\ 4\ sevens) = \sum_{i=0}^3(i+1) * P(even)^2 * P(7)^i$$
$$= P(even)^2 * (1 + 2*P(7) + 3*P(7)^2 + 4*P(7)^3)$$
$$= (\frac{3}{4})^2 * (1 + 2*\frac{1}{4} + 3*(\frac{1}{4})^2 + 4*(\frac{1}{4})^3)$$
$$= 0.984375$$
More generally, the positions of the even sums above will be a multiset permutation and one can come up with general expression for the probability of M events X occurring before N events Y, but that's a bit beyond the scope of this question.
• Nice approach. Would it not be easier to calculate the probability of event $B$ occurring before $A$ and then take the complement though? ($i$ would just be 0 or 1...) – bobajob Oct 16 '18 at 10:41
• @bobajob Good point, edited. – Dukeling Oct 16 '18 at 11:53
• "Since neither A nor B cares about odd sums that aren't 7" Yes, they do. A sum of 2 followed by a sum of 3 followed by a sum of 4 does not constitute an instance of A, while a sum of 2 followed by a sum of 4 does. – Acccumulation Oct 16 '18 at 15:20
• After rereading the question, I likely misunderstood it; I thought it was two even sums in a row. – Acccumulation Oct 16 '18 at 15:30
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $$7$$s. What is the chance that A comes first from there?
You have two Bernoulli processes,$$S_A$$ and $$S_B$$, and you are asked about the probability that the 2nd arrival in $$S_A$$ process occurs before the 4th arrival in $$S_B$$ process. PMF of the time of $$k$$-th arrival in a Bernoulli process with probability of success $$p$$ is $$p_{X_k}(t)=\binom{t-1}{k-1}p^k(1-p)^{t-k}, \ t=k,k+1,\ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed, and the trick is, as Dukeling explained, is to ignore odd sums not equal to $$7$$; then the (conditional) probability of getting the sum of $$7$$ is $$\frac{6}{36-12}=\frac{1}{4}$$. Now the complement of the event of interest is the event that the 4th arrival in the Bernoulli process with $$p=1/4$$ occurs at time $$4$$ or $$5$$, and its probability is equal to $$\binom{3}{3}p^4+\binom{4}{3}p^4(1-p)=p^4(1+4(1-p))=\frac{1}{64}$$ and the final answer is $$1-\frac{1}{64}=\frac{63}{64}$$ | 2019-01-20T04:40:25 | {
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https://math.stackexchange.com/questions/2412980/inequality-for-fibonacci-to-find-an-upper-bound-of-harmonic-fibonacci-series | # Inequality for Fibonacci to find an upper bound of harmonic Fibonacci series
I want to find an sharp upper bound for $$\sum_{n=1}^{\infty}\frac{1}{F_n}$$which $F_n$ is the n$th$ term of Fibonacci sequence . I wrote a Matlab program to find an upper bound ,$\sum_{n=1}^{10^6}\frac{1}{F_n}<4$
Now my question is:(1):Is there an inequality to find this ?
(2): Is that series have a close form ? $${F_n} = \frac{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\to \\\sum_{n=1}^{\infty}\frac{1}{F_n}=\sum_{n=1}^{\infty}\frac{\sqrt 5}{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}\\\leq \sum_{n=1}^{\infty}\frac{\sqrt 5}{{{(\frac{{1 + \sqrt 5 }}{2} ) ^n} }}=\frac{\sqrt5}{1-\frac{1}{\frac{{1 + \sqrt 5 }}{2}}}\approx12.18\\$$I am thankful for a hint or solution which can bring a sharper upper bound .
• maybe related – MAN-MADE Sep 1 '17 at 7:33
• Simple fast program: 3.359885666243177553172011302918927179688905133731968486495553815325130318996683383615416216456790087297045342928853913304136789017100883679591351733077119078580333550332507753187599850487179777897006039564509215375892775265673354024033169441799293934610992626257964647651868659449710216558984360881472693249591079473873673378523326877499762727757946853676918541981467668742998767382096913901217722024405208151094264934951 – Kenny Lau Sep 1 '17 at 7:33
• WolframAlpha simply returns ℱ suggesting that it has no closed form. – Kenny Lau Sep 1 '17 at 7:35
• @KennyLau:Can I see your program ? can you post it here ? – Khosrotash Sep 1 '17 at 7:35
• $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign, and that your inequality follows? – Jyrki Lahtonen Sep 1 '17 at 21:02
As Jyrki Lahtonen points out in the comments, $\sum_{n=1}^\infty\frac{\sqrt5}{\varphi^n}$ isn't necessarily the right bound, since it doesn't dominate the original series term-by-term.
As for a closed form, this identity is known: $$\sum_{n=1}^\infty\frac{1}{F_n}=\frac{\sqrt5}{4}\left(\vartheta_2^2(\varphi^{-2}) + \frac{\log5 + 2\psi_{\varphi^{-4}}(1) - 4\psi_{\varphi^{-2}}(1)}{2\log\varphi}\right)$$ where $\vartheta_2(q)$ is the Jacobi theta function at $z=0$, and $\psi_q$ is the $q$-digamma function. See Wikipedia and MathWorld on the "Reciprocal Fibonacci constant", and other Math.SE questions such as What is the sum of Fibonacci reciprocals?
• $F_n-\phi^n/\sqrt5=-(-1/\phi)^n/\sqrt5$ has alternating signs. Is it clear that the errors have a sum with the correct sign? The OP suffers from the same problem. – Jyrki Lahtonen Sep 1 '17 at 20:59
• @JyrkiLahtonen Whoops! Let me fix that. – Chris Culter Sep 1 '17 at 21:12
Here is a systematic method for computing upper bounds for the sum without much work.
By induction, if $F_N \ge b^N$ and $b+1 \ge b^2$, then $F_n \ge b^n$ for $n \ge N$.
Therefore, $$\sum_{n=1}^\infty {1\over F_n} \le \sum_{n=1}^{N-1} {1\over F_n} + \sum_{n=N}^\infty {1\over b^n} = \sum_{n=1}^{N-1} {1\over F_n} + \frac{1}{b^{N-1}(b-1)}$$ This upper bound gets closer to the actual sum when $b$ gets larger, but then we need larger $N$:
\begin{array}{crl} b &N &sum \\ 1.3 &4 &4.0172204521317 \\ 4/3 &5 &3.7825520833333 \\ 1.4 &6 &3.4981694135380 \\ 1.5 &11 &3.3651521005948 \\ 1.6 &72 &3.3598856662432 \\ \end{array}
Since $1.6 \approx \phi$, which is the largest possible $b$, the last value is quite close to the actual value: $$3.359885666243177553172011302918927179688905133732\cdots$$
Using Wolfram Mathematica answer is : Wolfram Mathematica Code | 2019-11-19T13:00:15 | {
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https://unitedtravelbooking.com/site/03a3ca-sum-of-two-binomial-random-variables | Making statements based on opinion; back them up with references or personal experience. Why did mainframes have big conspicuous power-off buttons? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Useful relations in dealing with binomial coefficients and factorials 4. How to sustain this sedentary hunter-gatherer society? Maybe I'm wrong in the conclusions about what that particular convergence means though. If success probabilities differ, the probability distribution of the sum is not binomial. \Pr[S\ge s] Determination of the binomial coefficient and binomial distribution The probability of any specified arrangement of k successes and n-k failures in n independent trials is pknkq − where p is the probability of success on any one trial and q=1-p is the probability of failure. Often the manipulation of integrals can be avoided by use of some type of generating function. "In the limit as n→∞, your binomials become Gaussian" Sorry but this is simultaneously vague and wrong. Asking for help, clarification, or responding to other answers. Why were there only 531 electoral votes in the US Presidential Election 2016? That depends on the range of values you are considering. where we took $t=\log(s/\sum_ip_i)$. $$By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Why Is an Inhomogenous Magnetic Field Used in the Stern Gerlach Experiment? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. The Kolmogorov approximation is given as an … Yes, in fact, the distribution is known as the Poisson binomial distribution, which is a generalization of the binomial distribution. See the binomial sum variance inequality. I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt. Both distributions have total mass 1. How do rationalists justify the scientific method. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, I have the same question and i read the paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). @jameselmore Additivity of the means is unrelated to independence. 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. \\&= \exp\left(\sum_i 1 + (e^t-1) p_i\right) \exp(-st) It will be a special case of the Poisson Binomial Distribution. Are there relatively simple formulae or at least bounds for the distribution \Pr[S\ge s] \\&= \exp\left(\sum_i 1 + (e^t-1) p_i\right) \exp(-st) Extremely bloated transaction log during delete. Every second customer converts better. See this paper (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). Which one is more idiomatic: ‘valid concern’ or ‘legitimate concern’? Can this be by chance? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If success probabilities differ, the probability distribution of the sum is not binomial. Here is an excerpt from the Wikipedia page. by Marco Taboga, PhD. The distribution of a sum S of independent binomial random variables, each with different success probabilities, is discussed.$$ @Robert ,do you have any insight on what happens if the n is not same for the 2 distributions. This lecture discusses how to derive the distribution of the sum of two independent random variables.We explain first how to derive the distribution function of the sum and then how to derive its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous). This answer provides an R implementation of the explicit formula from the paper linked in the accepted answer (The Distribution of a Sum of Binomial Random Variables by Ken Butler and Michael Stephens). | 2021-05-15T03:08:19 | {
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https://math.stackexchange.com/questions/588488/is-there-a-series-where-the-terms-tend-to-zero-faster-than-the-harmonic-series-b | # Is there a series where the terms tend to zero faster than the harmonic series but it still diverges?
I know that for the harmonic series $\lim_{n \to \infty} \frac1n = 0$ and $\sum_{n=1}^{\infty} \frac1n = \infty$.
I was just wondering, is there a sequence ($a_n =\dots$) that converges "faster" (I am not entirely sure what's the exact definition here, but I think you know what I mean...) than $\frac1n$ to $0$ and its series $\sum_{n=1}^{\infty}{a_n}= \infty$?
If not, is there proof of that?
• Sure. Let $a_n=\frac1{n\log n}$. Or let $a_n=\frac1{n\log n\log\log n}$. Or let ... In fact, the process never ends, in that you can find a divergent series whose terms go to zero faster than all the sequences I suggested. – Andrés E. Caicedo Dec 1 '13 at 19:39
• Versions of this question have been asked before. Here is an example from MO. – Andrés E. Caicedo Dec 1 '13 at 19:41
• ok, it this checks out! (using the condensation test) – guynaa Dec 1 '13 at 19:46
• Related The answers by by Bill/Gone and Yuval. – user17762 Dec 1 '13 at 20:15
• $$\sum_{n=1}^{\infty} \frac{1}{10n} ?$$ – Sawarnik Feb 24 '14 at 21:13
## 3 Answers
There is no slowest divergent series. Let me take this to mean that given any sequence $a_n$ of positive numbers converging to zero whose series diverges, there is a sequence $b_n$ that converges to zero faster and the series also diverges, where "faster" means that $\lim b_n/a_n=0$. In fact, given any sequences of positive numbers $(a_{1,n}), (a_{2,n}),\dots$ with each $\sum_n a_{i,n}=\infty$ and $\lim_n a_{i+1,n}/a_{i,n}=0$, there is $(a_n)$ with $\sum a_n=\infty$ and $\lim_n a_n/a_{i,n}=0$ for all $i$.
To see this, given $a_1,a_2,\dots$, first define $b_1=a_1,b_2=a_2,\dots,b_k=a_k$ until $a_1+\dots+a_k>1$. Second, let $b_{k+1}=a_{k+1}/2,b_{k+2}=a_{k+2}/2,\dots,b_n=a_n/2$ until $a_{k+1}+\dots+a_n>2$, etc. That is, we proceed recursively; if we have defined $b_1,\dots,b_m$ and $b_m=a_m/2^r$, and $b_1+\dots+b_m>r+1$, let $b_{m+1}=a_{m+1}/2^{r+1},\dots,b_l=a_l/2^{r+1}$ until $a_{m+1}+\dots+a_l>2^{r+1}$. The outcome is that $\sum b_i=\infty$ and $\lim b_i/a_i=0$.
Similarly, given $(a_{1,n}),(a_{2,n}),\dots$, with each $(a_{k+1,n})$ converging to zero faster than $(a_{k,n})$, and all of them diverging, let $a_i=a_{1,i}$ for $i\le n_1$, where $a_{1,1}+\dots+a_{1,n_1}>1$, then $a_i=a_{2,i}$ for $n_1<i\le n_2$, where we ask both that $a_{2,n_1+1}+\dots+a_{2,n_2}>1$ and that for any $k>n_2/2$ we have $a_{2,k}/a_{1,k}<1/2$, etc. That is, if we have defined $n_k$, we let $a_i=a_{k+1,i}$ for $n_k<i\le n_{k+1}$ where $n_{k+1}$ is chosen so that $a_{k+1,n_k+1}+\dots+a_{k+1,n_{k+1}}>1$ and for all $l>n_{k+1}/2$ we have $a_{k+1,l}/a_{i,l}<1/2^{k+1}$ for all $i<k+1$. Then the series $\sum a_i$ diverges, and the sequence $(a_i)$ converges to $0$ faster than all the $a_{i,n}$. In modern language, we would say that there are no $(\omega,0)$-gaps in a certain partial order.
We can modify the above slightly so that given any sequences $(a_{i,n})$ with $\sum_n a_{i,n}<\infty$ and $\lim_n a_{i+1,n}/a_{i,n}=\infty$ for all $i$, we can find $(a_n)$ with $\sum_n a_n<\infty$ and $\lim_n a_n/a_{i,n}=\infty$, so there is no fastest convergent series, and not even considering a sequence of faster and faster convergent series is enough. (In modern terms, there is no $(0,\omega)$-gap.)
What we cannot do in general is, given $(a_{i,n})$, with all $\sum_n a_{i,n}=\infty$, find $(a_n)$ with $\sum a_n=\infty$ and $a_n/a_{i,n}\to0$ for all $i$, if the $a_{i,n}$ are not ordered so that $a_{i+1,n}$ converges to zero faster than $a_{i,n}$. For example, we can have $a_n=1/n$ if $n$ is odd and $a_n=1/n^2$ if $n$ is even, and $b_n=1/n^2$ if $n$ is odd and $b_n=1/n$ if $n$ is even, and if $c_n$ converges to zero faster than both, then $\sum c_n$ converges. (These exceptions can typically be fixed by asking monotonicity of the sequences, which is how these results are usually presented in the literature.)
Note that the argument I gave is completely general, no matter what the series involved. For specific series, of course, nice "formulas" are possible. For example, given $a_n=1/n$, we can let $b_n=1/(n\log n)$ for $n>1$. Or $c_n=1/(n\log n\log\log n)$, for $n\ge 3$. Or ... And we can then "diagonalize" against all these sequences as indicated above.
By the way, the first person to study seriously the boundary between convergence and divergence is Paul du Bois-Reymond. He proved a version of the result I just showed above, that no "decreasing" sequence of divergent series "exhausts" the divergent series in that we can always find one diverging and with terms going to zero faster than the terms of any of them. A nice account of some of his work can be found in the book Orders of Infinity by Hardy. Du Bois-Reymond's work was extended by Hadamard and others. What Hadamard proved is that given $(a_i)$ and $(b_i)$ with $\sum a_i=\infty$, $\sum b_i<\infty$, and $b_i/a_i\to 0$, we can find $(c_i),(d_i)$ with $c_i/a_i\to0$, $b_i/d_i\to 0$, $d_i/c_i\to0$, $\sum c_i=\infty$, $\sum d_i<\infty$. More generally:
If we have two sequences of series, $(a_{1,n}), (a_{2,n}),\dots$ and $(b_{1,n}),(b_{2,n}),\dots$, such that
• Each $(a_{i+1,n})$ converges to zero faster than the previous one,
• Each $(b_{i+1,n})$ converges to zero slower than the previous one,
• Each $(a_{i,n})$ converges to zero slower than all the $(b_{j,n})$,
• Each $\sum_n a_{i,n}$ diverges, and
• Each $\sum_n b_{i,n}$ converges,
then we can find sequences $(c_n),(d_n)$, "in between", with one series converging and the other diverging.
In modern language, we say that there are no $(\omega,\omega)$-gaps, and similarly, there are no $(\omega,1)$- or $(1,\omega)$-gaps. This line of research led to some of Hausdorff's deep results in set theory, such as the existence of so-called $(\omega_1,\omega_1)$- or Hausdorff gaps. What Hausdorff proved is that this "interpolation" process, which can be iterated countably many times, cannot in general be carries out $\omega_1$ times, where $\omega_1$ is the first uncountable ordinal.
I had wondered about this too a long time ago and then came across this. The series
$$\sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n)}=\infty$$
diverges and it can be very easily proven by the integral test. But here is the kicker. This series actually requires googolplex number of terms before the partial sum exceeds 10. Talk about slow! It is only natural that if the natural log is slow, then take the natural log of the natural log to make it diverge even slower.
Here is another one. This series
$$\sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n)^2}=38.43...$$
actually converges using the same exact (integral) test. But it converges so slowly that this series requires $10^{3.14\times10^{86}}$ before obtaining two digit accuracy. So using these iterated logarithms you can come up with a series which converges or diverges "arbitrarily slowly".
Reference: Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae, 30th ed. Boca Raton, FL: CRC Press, 1996.
• But then by the same integral test $$\sum_{n=N_0}^\infty \sum_{n=3}^{\infty}\frac{1}{n\ln n (\ln\ln n) (\ln\ln \ln n)}=\infty$$ diverges at an even slower rate....Moreo generarily $$\sum \frac{1}{n \ln(n) \ln^2(n) \ln^3(n) ... \ln^k(n)}$$ diverges, where $^j$ denotes composition . – N. S. Dec 2 '13 at 0:46
• @N.S. That's what I said, using iterated logarithm you can come up with an even slower growth. In addition, in your first summation, the double summation doesn't make sense. I think you want the inner sigma removed. $N_0 = \lceil((e^x)^j(0)\rceil$ where $j$ denotes composition. In your first summation with $j=3$ we have $N_0=16$. – Fixed Point Dec 2 '13 at 2:31
• this is a fantastic example--showing that even trying to test convergence by simulation could be doomed, because we don't have the computational capacity to find enough terms to convince ourselves the sequence diverges. – MichaelChirico Apr 14 '15 at 14:56
If we use the notion of a partial sum:
$$S_n = \sum_{k=1}^n a_k$$
you are asking for a series in which the partial sums diverge more slowly than $\sum_{k=1}^n 1/k$ as $n\to\infty$. For this to happen, we just need to find $a_k$ such that $a_k < 1/k$ for all $k>c$ where $c$ is some value.
Look at $a_k = \frac{1}{k\ln k}$. Let's do the integral test to show this diverges. The substitution used is $u = \ln k$ so that $du = dk/k$.
$$\int \frac{dk}{k\ln k}=\int \frac{du}{u}=\ln u = \ln {\ln k}$$
Putting in limits of $2$ and $\infty$ shows that this indeed diverges. And for $k>c=e$, we have that $1/(k\ln k) < 1/k$. | 2020-02-17T19:05:47 | {
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https://math.stackexchange.com/questions/2873457/a-simple-method-of-factorization-2301 | # A simple method of factorization $2^{30}+1$
How can you factor $2^{30} + 1$? This task was supposed to be at one interview, there is an assumption that there should be a simple solution.
Observe that $$x^k+1=\frac{x^{2k}-1}{x^k-1}=\frac{\prod_{d\mid 2k}\Phi_{d}(x)}{\prod_{d\mid k}\Phi_{d}(x)}=\prod_{d\mid 2k,\ d\nmid k}\Phi_{d}(x)$$ With $$\Phi_n(x)$$ being the $$n^\text{th}$$ cyclotomic polynomial. It follows that $$2^{30}+1=\Phi_4(2)\Phi_{12}(2)\Phi_{20}(2)\Phi_{60}(2)$$ With the individual factors computed as $$\Phi_4(2)=2^2+1=5$$ $$\Phi_{12}(2)=2^4-2^2+1=13$$ $$\Phi_{20}(2)=2^8-2^6+2^2-2^2+1=5\times 41$$ $$\Phi_{60}(2)=2^{16}+2^{14}-2^{10}-2^8-2^6+2^2+1=61\times 1321$$ Giving an overall factorization of $$\boxed{2^{30}+1=5^2\times 13\times 41\times 61\times 1321}$$
• This is very elegant. Aug 6 '18 at 1:58
• Thanks! It's also not too hard to compute the relevant polynomials by hand (the section "Easy cases for computation" in the linked wiki article lists some strategies) :) Aug 6 '18 at 2:01
• Again, computing $\Phi_{60}(2) = 61 \times 1321$ is not really something to do in an interview. Aug 6 '18 at 7:26
• Perhaps they wanted to see the way to a solution, not the solution itself. Aug 6 '18 at 11:43
• @RobertIsrael: probably the interviewer wanted to hear something along the lines "in order to find the prime divisors of $\Phi_{60}(2)$, it is enough to check the primes $\equiv 1\pmod{60}$. Since the order of $2$ in $\mathbb{Z}/(61\mathbb{Z})^*$ is what it is, $61$ is a prime divisor of $\Phi_{60}(2)$ and the problem boils down to checking that $1321$ is a prime". Aug 6 '18 at 16:45
I would note it is a sum of cubes and fifth powers, so $2^{10}+1=1025$ and $2^6+1=65$ are both factors. In terms of primes that gives us $5^2\cdot 13 \cdot 41$ as factors. At this point in an interview (depending on the position) I would declare success and move on. Getting the remaining $80581$ would take a bunch of hand calculation and finding $61\cdot 1321$ doesn't seem reasonable either.
• $61\equiv 5 \bmod 8$ so $(2|61)\equiv 2^{30}\equiv -1\bmod 61$. Aug 6 '18 at 1:44
Hint: $a^k+1$ is divisible by $a+1$ if $k$ is odd.
• Just so we can check any number-theoretic derivation... Mathematica gives: $5^2 \times 13 \times 41 \times 61 \times 1321$. (Doesn't seem so simple that it could be done in an interview session...) Aug 6 '18 at 0:37
Yet another easy observation: since $4X^4+1=(2X^2+2X+1)(2X^2-2X+1)$, we get $4\cdot2^{28}+1=(2\cdot2^{14}+2\cdot2^7+1)(2\cdot2^{14}-2\cdot2^7+1)$.
Firstly, in an interview there is often a difference between showing that there is a simple solution (existence) and outlining the approach, versus actually finding the solution.
That is, it might suffice to outline a standard prime factorising algorithm and show that it could be computed in trivial time.
One of the simplest classic approaches is that to factor $x$, you only need to test the primes that are less than $\sqrt{x}$. If $x=2^{30}+1$, then $\sqrt{x+1} \simeq 32768$. There are less than 3000 primes that satisfy this constraint, so it is very feasible on any computing platform.
Once each prime factor was found, one would divide that factor out, and then only need to search for primes up to an even smaller upper bound. Thus after the finding that 5 is a prime, one need only search for other prime factors less than 14654.
Thus, you have an algorithm that would find all the factors of $2^{30}+1$ in ascending order.
That is, $2^{30}+1 = 5 \times 5 \times 13 \times 41 \times 61 \times 1321$.
The key to nearly any sensible prime factorization approach is to efficiently find the first factor. Thus, in some ways it could be said that factoring $2^{30}+1$ is easy because it has lots of prime factors, and most of them are very small. (Contrast this to encryption are based on numbers that are very hard to factor, because they typically one have 2 factors and each of them are very very large.)
Anyway, as @Robert indicated, the key to solving this directly is to realise that if $k$ is odd, then $x^k+1$ can be elegantly factorised, as $(x+1)$ is a factor.
For example, $$x^7+1 = (x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$$ Note that $z^{30}+1 = (z^2)^{15}+1$ and so $z^2+1$ is a factor.
Letting $z=1$ shows that 5 is a factor of $2^{30}+1$.
Similarly, you could use the well-known sum of cubes factorisation. $x^3+1 = (x+1)(x^2-x+1)$, to show that $$2^{30}+1 = (2^{10}+1)(2^{20}-2^{10}+1) = 1025 \times 1047553$$ and continue from there.
• Your early complete factorization is missing $61$ Aug 6 '18 at 1:59 | 2021-09-23T01:47:01 | {
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http://mathematica.stackexchange.com/questions/27642/fitting-a-two-dimensional-gaussian-to-a-set-of-2d-pixels/27754 | # Fitting a two-dimensional Gaussian to a set of 2D pixels
Imagine I have a set of data like the following:
data = {{0.0453803, 0.0427863, 0.0489815, 0.045243, 0.0488289, 0.0432898,
0.04448, 0.0387732, 0.0388952}, {0.0507668, 0.0427863, 0.0502632,
0.0503395, 0.0634623, 0.0675822, 0.0529335, 0.047425,
0.0387121}, {0.042237, 0.0501259, 0.0595712, 0.0869001, 0.139559,
0.141512, 0.0868391, 0.0579232, 0.0408331}, {0.0478981, 0.0491646,
0.0652628, 0.130404, 0.218448, 0.220645, 0.143603, 0.0605173,
0.0424964}, {0.0462043, 0.0530861, 0.076051, 0.140017, 0.206943,
0.202502, 0.118791, 0.0614023, 0.0459907}, {0.0511788, 0.0582132,
0.105531, 0.166354, 0.181003, 0.13698, 0.0748302, 0.0557107,
0.0492103}, {0.0493629, 0.0539712, 0.0971695, 0.160769, 0.164477,
0.104768, 0.0591745, 0.0475319, 0.0452583}, {0.0510719, 0.0599374,
0.0730602, 0.0975814, 0.101289, 0.0691997, 0.0498054, 0.044892,
0.043122}, {0.0460517, 0.0567025, 0.0574044, 0.0587778, 0.0537118,
0.0487221, 0.0474098, 0.0413977, 0.04477}}
I don't have enough reputation to post an image, but one can easily be generated by applying ListPlot3D to the above data set.
How might I best fit a Gaussian curve to this set of datapoints, and extract properties such at the fit' semi-axes? I noticed that ComponentMeasurements has some functionality for best fit ellipsoids, but that doesn't seem to be workable here.
My objective here is to determine how "Gaussian" a set of points in an image are. My strategy is to sequentially fit a 2D Gaussian to each point, and then to measure it's eccentricity and spread (looking, for example, at the length and ratio of the semiaxes of the ellipsoid corresponding to the fit). The example here seems like it should yield a 2D Gaussian fit with significant spread and a ratio for the semiaxes significantly diverging from one.
-
Maybe You can add some context to the question. I'm asking because this data looks rather like sum of two 2D gaussians, and I don't know if it is relevant. – Kuba Jun 25 '13 at 22:36
@Kuba I have added a bit of context in the problem description! – Bob Jun 25 '13 at 22:42
By context I mean, for example, what this data actually is. :) Without details, term "determine how Gaussian" is too vague. – Kuba Jun 25 '13 at 23:07
Sjoerd's answer applies the power of Mathematica's very general model fitting tools. Here's a more low-tech solution.
If you want to fit a Gaussian distribution to a dataset, you can just find its mean and covariance matrix, and the Gaussian you want is the one with the same parameters. For Gaussians this is actually the optimal fit in the sense of being the maximum likelihood estimator -- for other distributions this may not work equally well, so there you will want NonlinearModelFit.
One wrinkle is that your data doesn't fall off to zero but to something like Min[data] $\approx 0.0387$, but we can easily get rid of that by just subtracting it off. Next, I normalize the array to sum to $1$, so that I can treat it like a discrete probability distribution. (All this really does is allow me to avoid dividing by Total[data, Infinity] every time in the following.)
min = Min[data];
sum = Total[data - min, Infinity];
p = (data - min)/sum;
Now we find the mean and covariance. Mathematica's functions don't seem to work with weighted data, so we'll just roll our own.
{m, n} = Dimensions[p];
mean = Sum[{i, j} p[[i, j]], {i, 1, m}, {j, 1, n}];
cov = Sum[Outer[Times, {i, j} - mean, {i, j} - mean] p[[i, j]], {i, 1, m}, {j, 1, m}];
We can easily get the probability distribution for the Gaussian with this mean and covariance. Of course, if we want to match the original data, we have to rescale and shift it back.
f[i_, j_] := PDF[MultinormalDistribution[mean, cov], {i, j}] // Evaluate;
g[i_, j_] := f[i, j] sum + min;
The match is not too bad, although the data has two humps where the Gaussian has one. You can also compare the fit through a contour plot. (Use the tooltips to compare contour levels.)
Show[ListPlot3D[data, PlotStyle -> None],
Plot3D[g[i, j], {j, 1, 9}, {i, 1, 9}, MeshStyle -> None,
PlotStyle -> Opacity[0.8]], PlotRange -> All]
Show[ListContourPlot[data, Contours -> Range[0.05, 0.25, 0.025],
ContourShading -> None, ContourStyle -> ColorData[1, 1], InterpolationOrder -> 3],
ContourPlot[g[i, j], {j, 1, 9}, {i, 1, 9}, Contours -> Range[0.05, 0.25, 0.025],
ContourShading -> None, ContourStyle -> ColorData[1, 2]]]
The variances along the principal axes are the eigenvalues of the covariance matrix, and the standard deviations (which I guess you're calling the semi-axes) are their square roots.
Sqrt@Eigenvalues[cov]
(* {1.86325, 1.50567} *)
-
Or you could just plot the Gaussian fit coloured by the residual: Show[ListDensityPlot[ Table[data[[i, j]] - g[i, j], {i, 1, m}, {j, 1, m}], InterpolationOrder -> 3, PlotRange -> Full, ColorFunctionScaling -> False, ColorFunction -> (With[{t = 20 #}, RGBColor[1 + t, 1 - Abs[t]/2, 1 - t]] &)], ContourPlot[g[x, y], {y, 1, 9}, {x, 1, 9}, Contours -> Range[0.05, 0.25, 0.025], ContourShading -> None, ContourStyle -> Black]] i.stack.imgur.com/LqGZL.png (orange: data higher than fit, blue: data lower than fit) – Rahul Narain Jun 27 '13 at 6:23
Thanks again for such a fantastic answer. Just to help me understand - why does your calculation for the square root of the eigenvectors yield a slightly different result than StandardDeviation@MultinormalDistribution[mean,cov]? – Bob Jun 30 '13 at 11:15
Because that one appears to give the standard deviations of the marginal distributions of $x$ and $y$ instead, which you also get with Sqrt@Diagonal[cov]. – Rahul Narain Jun 30 '13 at 11:28
data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1];
fm = NonlinearModelFit[data3D,
a E^(-(((-my + y) Cos[b] - (-mx + x) Sin[b])^2/(2 sy^2)) -
((-mx + x) Cos[b] + (-my + y) Sin[b])^2/(2 sx^2)),
{{a, 0.1}, {b, 0}, {mx, 4.5}, {my, 4.5}, {sx, 3}, {sy, 3}},
{x, y}
]
Show[
ListPlot3D[data3D, PlotStyle -> None, MeshStyle -> Red],
Plot3D[fm["BestFit"], {x, 1, 9}, {y, 1, 9}, PlotRange -> All,
PlotStyle -> Opacity[0.9], Mesh -> None]
]
fm["BestFitParameters"]
{a -> 0.1871830545, b -> -0.4853901689, mx -> 5.152549499, my -> 5.092511036, sx -> 2.756524919, sy -> 2.072163433}
-
Starting with your data, I would map out the data something like this:
data3D = Flatten[MapIndexed[{#2[[1]], #2[[2]], #1} &, data, {2}], 1]
and then count it like:
data2 = Flatten[Table[{#[[1]], #[[2]]}, {100 #[[3]]}] & /@ data3D, 1];
and then use this function:
FindDistributionParameters[data2,
MultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]
to find the means and coveriance of the data. And then I might do other stuff with the distribution, such as create a distribution object using:
edist = EstimatedDistribution[data2,
MultinormalDistribution[{a, b}, {{c, d}, {e, f}}]]
and then check the distribution against the data, using:
dtest = DistributionFitTest[data2, edist, "HypothesisTestData"]
and create a table of the test results, using:
N[dtest["TestDataTable", All]]
or individually
AndersonDarlingTest[data2, edist, "TestConclusion"]
CramerVonMisesTest[data2, edist, "TestConclusion"]
JarqueBeraALMTest[data2, "TestConclusion"]
MardiaKurtosisTest[data2, "TestConclusion"]
PearsonChiSquareTest[data2, edist, "TestConclusion"]
ShapiroWilkTest[data2, "TestConclusion"]
Now, I'm not a statistician, and I probably only know enough statistics to be dangerous, and I'm not familar with all the tests listed in the table, but I suspect that one of those small numbers is significant in measuring how close the actual data fits the distribution, i.e. how "Gaussian" the set of points is.
Also I might just plot the data as follows and visually judge the gaussiness of the data using these functions (as per VLC's answer to question 2984)
Show[DensityHistogram[d, {.2}, ColorFunction -> (Opacity[0] &),
Method -> {"DistributionAxes" -> "Histogram"}], ListPlot[d]]
or these:
GraphicsColumn[
Table[SmoothDensityHistogram[d, ColorFunction -> "DarkRainbow",
Method -> {"DistributionAxes" -> p}, ImageSize -> 500,
BaseStyle -> {FontFamily -> "Helvetica"},
LabelStyle -> Bold], {p, {True, "SmoothHistogram", "BoxWhisker"}}]]
-
You mentioned in a related question that you would like to do large numbers of these fits quickly.
If each data set has the same dimensions, you can write a fairly fast implementation of Rahul Narain's method by precomputing arrays of x and y coordinates for the data grid, and flattening the data and using Dot to calculate the mean and the elements of the covariance matrix:
x = Table[i, {i, 9}, {j, 9}]//N;
y = Transpose[x];
x = Flatten[x]; y = Flatten[y];
semiaxes[data_] := Module[{min, p, mx, my},
min = Min[data];
p = Flatten[data] - min;
p /= Total[p];
mx = x.p;
my = y.p;
With[{a = (x - mx)^2.p, b = ((x - mx) (y - my)).p, c = (y - my)^2.p},
Sqrt @ Eigenvalues[{{a, b}, {b, c}}]]]
semiaxes[data]
(* {1.86325, 1.50567} *)
This runs in about 340 µs on my PC
Compiling can give you even more speed, but you need to replace Eigenvalues with the explicit symbolic expressions:
semiaxesc =
With[{x = x, y = y},
Compile[{{data, _Real, 2}}, Block[{min, p, mx, my, a, b, c},
min = Min[data];
p = Flatten[data] - min;
p /= Total[p];
mx = x.p;
my = y.p;
a = (x - mx)^2.p;
b = ((x - mx) (y - my)).p;
c = (y - my)^2.p;
{Sqrt[1/2 (a + c - Sqrt[a^2 + 4 b^2 - 2 a c + c^2])],
Sqrt[1/2 (a + c + Sqrt[a^2 + 4 b^2 - 2 a c + c^2])]}],
CompilationTarget -> "C", RuntimeOptions -> "Speed"]];
semiaxesc[data]
(* {1.50567, 1.86325} *)
This runs in about 5.7 µs on my PC.
-
Fantastic. However, I'm getting a "Eigenvalues::matsq : Argument" which is saying that don't have a square matrix. Is there some preprocessing I need to do for the data array I posted? – Bob Jun 30 '13 at 9:15
Would you mind posting the full script you're running? I'm sure there's some formatting error on my end, but I can't pin it down. Everything works and makes sense, but there seems to be a problem with the Eigenvalue computation on my end. – Bob Jun 30 '13 at 12:33
That is the full script, apart from the definition of data which I copied from the question. I suggest you step through the code one line at a time and make sure that mx, my, a, b, and c are all single numbers. I'm away from the computer this week, so that's about all the help I can offer at the moment. You could try asking in chat if anyone else can get it working. – Simon Woods Jun 30 '13 at 14:58
Very good, as long as it's working on your end, I'll be able to fix it on mine. – Bob Jun 30 '13 at 16:50
The problem was that the dot products were not being recognized as such. Switching a.b for Dot[a,b] did the trick. Very odd. – Bob Jun 30 '13 at 18:19 | 2014-04-17T16:03:47 | {
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https://math.stackexchange.com/questions/1996917/product-of-two-monotonic-real-functions | # Product of two monotonic real functions
I'm confused about the possibility to say that the product of two monotonic real functions $f,g:\mathbb{R} \to \mathbb{R}$ is monotonic.
I found the following proposition:
A. If $g$ and $f$ are two increasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is increasing.
B. If $g$ and $f$ are two increasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is decreasing.
Does something similar hold also for decreasing functions? That is: is the following sentence correct?
A'. If $g$ and $f$ are two decreasing functions, with $f(x)>0$ and $g(x)>0$ $\forall x$, then $f \, g$ is decreasing.
B'. If $g$ and $f$ are two decreasing functions, with $f(x)<0$ and $g(x)<0$ $\forall x$, then $f \, g$ is increasing.
Moreover, can something be said about the product of a increasing and a decreasing functions?
For example if $f$ is increasing and $g$ is decreasing under what conditions can I say something about the monotony of the product $f g$?
Besides these two practical questions I would like to ask some suggestions on how to prove statement A.
I tried in the following way
$Hp:$
$x_1 >x_2 \implies f(x_1)>f(x_2)>0 \,\,\, \forall x_1,x_2$
$x_1 >x_2 \implies g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2$
$Th:$
$x_1 >x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$
$Proof:$
$f(x_1)>f(x_2)>0 \,\,\, , g(x_1)>g(x_2)>0 \,\,\, \forall x_1,x_2 \implies f(x_1) g(x_1)>f(x_2) g(x_2)>0 \,\,\, \forall x_1,x_2$
Which seems obvious if one thinks about some numbers but I don't really know how I could prove the last implication in rigourous way. So any help in this proof is highly appreciated.
To prove statement A, let $x > y$. Then, as we know, $f(x) > f(y)>0$ and $g(x) > g(y)>0$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)>0)\\ g(x) > g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}
An analogous proof would follow for part B if $f$ and $g$ were increasing, with a caveat:let $x > y$. Then, as we know, $f(x) > f(y)$ and $g(x) > g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) > f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)<0)\\ g(x) > g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}
Now, let us see if the same logic could work with part A':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)>0)\\ g(x) < g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)>0)\\ g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y) \end{gather}
That's brilliant, so great intuition for anticipating part A'. Now we will check part B':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim: \begin{gather} f(x) < f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)<0)\\ g(x) < g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)<0)\\ g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y) \end{gather}
And therefore part B' is also done. Note the above logic carefully, I think all steps are equally important.
Use this logic, and see why in most cases, one increasing and one decreasing function doesn't tell you much about the product itself.
For A' and B', apply A and B. For A': If $f,g$ are decreasing and positive then $-f$ and $-g$ are increasing and negative so by B, the function $(-f)(-g)=fg$ is decreasing. Similarly, apply A to B'.
$e^x$ is increasing and $e^{-x}+1$ is decreasing, and their product $1+e^x$ is increasing.
$e^x+1$ is increasing and $e^{-x}$ is decreasing, and their product $1+e^{-x}$ is decreasing.
The product of a positive increasing and a positive decreasing function can also fail to be monotonic. | 2020-10-29T22:28:54 | {
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http://math.stackexchange.com/questions/415179/fair-and-unfair-coin-probability | # Fair and Unfair coin Probability
I am stuck on this question.
A coin with $P(H) = \frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?
I keep getting $\frac{1}{6}$ but the answer is $\frac{5}{36}$.
Attempt: $P($all heads on the four coins$)P($either one of the tosses is heads on the two coins$)+P(3$ heads on the four coins$)P($both coins are heads$)$
$P($all heads on the four coins$) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
$P($either one of the tosses is heads on the two coins$) = 1-P($no heads on both tosses$) = 1-\frac{1}{3}\cdot\frac{1}{3} = \frac{8}{9}$.
$P($exactly $3$ heads on the four tosses$) = \frac{1}{4}$.
$P($both coins are heads$) = \frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$.
Final Equation: $\frac{1}{16}\cdot\frac{8}{9}+\frac{1}{4}\cdot\frac{4}{9} = \frac{1}{6}$.
Why am I off by $\frac{1}{36}$?
-
Because "either one of..." $\not \equiv$ " exactly one of..." – Ehsan M. Kermani Jun 9 '13 at 4:31
You claim that the problem is with $P($either one of the tosses is heads on the two coins$)$. You need to calculate the probability of there being precisely one occurrence of a head in the tossing of the second coin. In your calculation, you subtracted the probability of no heads from one. The remaining probability covers one head or both heads, but you want to exclude the latter. If you make this adjustment, you will get the correct answer. – Michael Albanese May 30 '15 at 19:38
The required probability will be
$P($exactly $4$ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $1$ head from the $2$ flips of $2$nd coin $)+$
$P($exactly $3$ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $2$ heads from the $2$ flips of $2$nd coin $)$
Using Binomial Distribution, the required probability $$\binom44\left(\frac12\right)^4\left(1-\frac12\right)^{4-4} \cdot\binom21\left(\frac23\right)^1\left(1-\frac23\right)^{2-1}$$
$$+\binom43\left(\frac12\right)^3\left(1-\frac12\right)^{4-3} \cdot\binom22\left(\frac23\right)^2\left(1-\frac23\right)^{2-2}$$
$$=\frac1{16}\cdot\frac49+\frac14\cdot\frac49=\frac5{36}$$
- | 2016-07-23T17:08:48 | {
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https://www.themathdoctors.org/parallel-vectors-missing-a-solution/ | # Parallel Vectors: Missing a Solution
#### (A new question of the week)
We were recently asked to check work on an interesting little question about parallel vectors, and I was almost convinced that there was no solution … until I realized there was one! How was it missed? How can we avoid doing that? That’s our goal today.
## When will these vectors be parallel?
Here is the question, from Brine in mid-September:
Here is my work:
Brine is using column-vector form for the vectors; I’ll be using the more compact bracket notation for convenience.
He’s found the vectors $$\overrightarrow{PQ}=\left\langle3,x^2\right\rangle-\left\langle x,0\right\rangle=\left\langle3-x,x^2\right\rangle$$ and $$\overrightarrow{P_1Q_1}=\left\langle 6,x^2\right\rangle-\left\langle2x,1\right\rangle=\left\langle6-2x,x^2-1\right\rangle$$, and then wrote an equation to say that the components of the two vectors are proportional: $$\frac{6-2x}{3-x}=\frac{x^2-1}{x^2}$$ Solving this yields only an imaginary solution, so it appears to be impossible for the vectors to be parallel.
Hi, Brine.
So your answer is, that there is no such real number x?
Your work almost had me convinced, until I tried solving a different way, and saw that there is an answer.
Look carefully at your work, and think about whether there is any step in which you made an unstated assumption
### How to recognize unstated assumptions
I had graphed the situation in GeoGebra, in such a way that I could vary x and observe the two vectors, to see how the problem worked, expecting perhaps to see why there would be no solution. Here is the case $$x=1.5$$, for example:
But I found that I could make them parallel (as we’ll see below). That gave me reason to look more closely at the work. I’d been fooled by following along with his work rather than starting from scratch to solve the problem myself. (That’s a risk even when you check over your own work, because your mind will follow the same paths again. I should have known better, and I soon did!)
Brine replied,
I tried again but unfortunately I keep ending up that they are not parallel. Can you please give me a hint?
I answered with a more detailed hint:
As I said before, “think about whether there is any step in which you made an unstated assumption“. Here are several points in your work that you should think about:
First, you wrote this equation to say that the vectors are parallel:
When can’t you write such an equation? What fact are you using to justify the equation?
Second, here you canceled to simplify a fraction. When can’t you cancel?
Or you could think of this more generally as a rational equation. What is the domain of this equation? What does simplifying change?
Third, here you multiplied both sides of an equation. When can’t you do that?
Not all of these represent actual errors in your work; but they represent situations where you are making unstated assumptions, or ignoring special cases.
Thinking about exactly what you are doing, and what justifies each step, can be a valuable exercise when you have a wrong (extraneous or missing) solution. The answers to my questions are …
1. He’s justifying the proportion because parallel vectors have proportional components … but that isn’t true when any of those components are zero.
2. You can’t cancel when the factor you cancel can be zero.
3. Simplifying by canceling can change the domain, introducing extraneous roots that are valid in the new equation, but are not in the domain of the original.
4. You can’t multiply an equation by a factor that might be zero, because the new equation may be true when the original is not.
Which is the key issue? We’ll see.
### Examples
The same sorts of problems can arise in more familiar problems, so I pointed out a couple examples:
For a similar issue, when you want to check whether two lines are parallel, you have to consider two cases: Either their slopes are equal or … what?
For another similar issue, consider how to solve an equation like x2 = 2x. It seems obvious that you can divide by x and get x=2; but that misses a solution. Why? Because you might be dividing by 0, which isn’t legal. When you divide by x, you are implicitly assuming that x≠0, which you don’t know to be true. The better way is to rearrange and factor: x2 -2x = 0 factors as x(x-2) = 0, and the solutions are x=0 and x=2.
I’m intentionally trying not to give direct hints too soon, because this is a very valuable lesson to learn; as I indicated, I was almost fooled, so it is rather subtle. But it happens a lot, and I want you to have a big “aha” moment you won’t forget.
In checking for parallel lines, checking the slopes won’t catch a case where they don’t have slopes (vertical lines); that is the most direct hint I gave here. The second example illustrates how a solution can be missed because of an unstated assumption (that what you divide by is not zero).
### Finding the answer by an imperfect method
Brine wrote back:
Oh ok I understand you. Is this the answer? 3?
This time Brine has cross-multiplied without first canceling: $$\frac{6-2x}{3-x}=\frac{x^2-1}{x^2}\$$6-2x)(x^2)=(x^2-1)(3-x)\\6x^2-2x^3=3x^2-x^3-3+x\\-x^3+3x^2-x+3=0 and then solved the cubic equation by factoring (by grouping). This gave two solutions: the imaginary solutions \(x=\pm i$$ we saw before, and also $$x=3$$. This work is better, but … I answered, Yes, that’s correct. What you’ve done here is, I think, to cross-multiply your equation (6-2x)/(3-x) = (x2-1)/x2, and solve the resulting cubic. This amounts to factoring instead of dividing, as I recommended, though you started with a division (in writing the equation in the first place). Presumably you then checked that x = 3 worked. Another way to think of it is to go ahead and do what you originally did, but then check whether canceling the x-3 factors eliminated a solution. That is, one could just recognize that it was assumed that $$x-3\ne0$$, and check to see if the opposite assumption produces a solution. Rather than avoiding the assumption, we would be making the assumption explicit, and taking the contrary assumption as a second case. ### The equation doesn’t fully represent the problem Unfortunately, that check would fail! He’s missed a more fundamental error: But did you observe that, when you put x=3 into your initial equation, you get 0/0 = 8/9? So 3 is not actually a solution of that equation! (It would be called extraneous.) And it isn’t obvious that it is a solution of the actual problem, until you check it in the original problem. The real problem arises earlier than that: The equation doesn’t really represent the problem. I generally tell students to check their answer, not by plugging it into their equation, but by checking if it works in the problem itself. When the equation is wrong, this will catch it. But here, the equation turns out to be wrong but our answer is correct! So how do you check the solution? You have go back to the problem: Let P(x,0), Q(3,x2), P1(2x,1), and Q1(6,x2). Find all possible values of x ∈ R such that PQ || P1Q1. If x=3, the points are P(3,0), Q(3,9), P1(6,1), and Q1(6,9), and the vectors are PQ = <0,9>, P1Q1 = <0,8>. Are they parallel? Yes, because both are vertical — a case not covered by your equation! That is the ultimate cause of your difficulty. This is what I’d found graphically (for $$x=3$$), which made me look again: So, how can we solve the problem without making the unintentional assumption that the vectors are not vertical? ### Back to square one: What really makes parallel vectors parallel? What you never wrote, at the start, was what defines parallel vectors. I asked, “What fact are you using to justify the equation?”, and I’d still like an answer: What perspective were you taking on parallelism when you wrote that equation? What have you been taught about the meaning of “parallel”? Two vectors are parallel if one is a scalar multiple of the other. In two dimensions, <a,b> and <c,d> are parallel if there is a non-zero scalar k such that <c,d> = k<a,b>; i.e. c = ka and d = kb. You have translated this to c/a = d/b = k; but that assumes that a and b are both non-zero. If you use this form, you need to separately check the contrary case. Equivalently, one could take vectors to be parallel if their direction angles are the same, and then compare the tangents of those angles (their slopes): b/a = d/c. But those don’t exist for vertical vectors. If you do that, you need to separately consider the latter possibility. His proportion was$$\frac{6-2x}{3-x}=\frac{x^2-1}{x^2}.$$So, using his form, my “c/a = d/b”, you need to ask what happens when a denominator is zero, namely $$3-x=0$$ or $$x^2=0$$. The former leads to our missing solution. If you used the “b/a=d/c” form, representing equal slopes,$$\frac{x^2}{3-x}=\frac{x^2-1}{6-2x},$$you’d need to check when these denominators are zero, and in this case both are zero for $$x=3$$ (that is, both vectors are vertical). So if I were doing this carefully, I would not have started with the equality of two fractions, but with the definition using scalar multiplication: PQ = <3-x,x2> P1Q1 = <6-2x,x2-1> These are parallel if P1Q1 = kPQ, so that 6-2x = k(3-x) x2-1 = kx2 From the first equation, either k = (6-2x)/(3-x) = 2, or x = 3. If k = 2, the second equation implies that x2 = -1, which is impossible. If x = 3, then PQ = <0,9> and P1Q1 = <0,8>, and these are indeed parallel, with k = 8/9. The solution of that first equation would look like this, done carefully:$$6-2x = k(3-x)\\2(3-x)-k(3-x)=0\$$2-k)(3-x)=0\\k=2\text{ or }x=3 Or I might eliminate k from the equations by writing your rational equation, while keeping in mind that in dividing by x2 and by x-3, I was assuming both are non-zero, and check those cases. If we check the case \(x^2=0$$, we find that $$\overrightarrow{PQ}=\left\langle3,0\right\rangle-\left\langle 0,0\right\rangle=\left\langle3,0\right\rangle$$ and $$\overrightarrow{P_1Q_1}=\left\langle 6,0\right\rangle-\left\langle0,1\right\rangle=\left\langle6,-1\right\rangle$$. These are not parallel.
Merely recognizing that your equation does not allow x = 3, and checking that separately, would also be valid, though it’s better to understand why. Just solving without canceling, as you have now done, is less appropriate, but it does lead to the answer (as long as you check it).
This is a very interesting problem, isn’t it?
Brine closed it out:
Yes you are right. I was actually taught that parallel vectors are parallel if they have a k value such that a=kb (where a and b are vectors) I forgot to apply this rule because in grade 12 I was taught by equaling the two slopes if they are parallel. I did learn the other method in the previous grade too but I forgot and I do remember it. The way you showed is way better that the one I did. Thank you so much for teaching me Dr. Peterson!
Sometimes we learn a shortcut method, or transfer it from one topic to another, and neglect to pay attention to conditions that are attached to them. It’s a good lesson!
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https://math.stackexchange.com/questions/1194584/the-total-number-of-subarrays | # The total number of subarrays
I want to count the number of subarrays for a vector (not combinations of elements).
Ex.
A[1,2,3]
It has 6 subarrays :
{1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}
I think that for a vector of N elements the total number of subarrays is N*(N+1)/2.
I am not able to prove it, can someone do it?
• Welcoming the Math.SE! Here we have a culture of showing what you've tried, as it helps both the responders to give better answers and you to get a better understanding! :) – frogeyedpeas Mar 17 '15 at 22:08
• Looks you forgot to list $\{3, 1\}$ – AgentS Mar 17 '15 at 22:13
• No! I didn't forget {3,1}. A subarray has to have contiguous elements – user72708 Mar 18 '15 at 13:56
Suppose that your vector is $\langle a_1,a_2,\ldots,a_n\rangle$. Imagine a virtual element $a_{n+1}$ at the end; it doesn’t matter what its value is. A subarray is completely determined by the index of its first element and the index of the element that immediately follows its last element. For example, the subarray $\langle a_3,\ldots,a_{n-2}\rangle$ is determined by the indices $3$ and $n-1$, the subarray $\langle a_k\rangle$ is determined by the indices $k$ and $k+1$, and the subarray $\langle a_2,\ldots,a_n\rangle$ is determined by the indices $2$ and $n+1$. Moreover, each pair of distinct indices from the set $\{1,2,\ldots,n+1\}$ uniquely determines a subarray. Thus, the number of subarrays is the number of pairs of distinct indices from the set $\{1,2,\ldots,n+1\}$, which is
$$\binom{n+1}2=\frac{n(n+1)}2\;.$$
• I am confused why you use the index of the element that immediately follows its last element instead of just the last element. I was thinking the answer should be n C 2, instead of n+ 1 C 2 that you mentioned. Please correct me wherever I am wrong. – Abhishek Bhatia Jul 7 '18 at 19:33
• @Abhishek , You are right, It is n C 2 where we did not calculate the single character substrings. There can be n single character substring . So the result is (n C 2 + n C 1) = (n+1) C 2 . – shuva Oct 31 '18 at 20:15
Consider an arbitrary array of N DISTINCT ELEMENTS (if the elements are the same then I am afraid the formula you are seeking to prove no longer works!).
Naturally there exists 1 array consisting of all the elements (indexed from 0 to N-1)
There exist 2 arrays consisting of N-1 consecutive elements (indexed from 0 to N-2)
and in general there are k arrays consisting of N-k+1 consecutive elements (indexed from 0 to N-k-1)
Proof:
We can access elements 0 ... N-k-1 as the first array, then 1 ... N-k+2 is the second array, and this goes on for all N-k+r until N-k+r = N-1 (ie until we have hit the end). The r that does us is can be solved for :
$$N-k+r = N-1 \rightarrow r -k = -1 \rightarrow r = k-1$$
And the list $$0 ... k-1$$ contains k elements within it
Thus we note that the total count of subarrays is
1 for N elements
2 for N-1 elements
3 for N-2 elements
.
.
.
N for 1 element
And the total sum must be:
$$1 + 2 + 3 ... N$$
Let us see if your formula works
if:
$$1 + 2 +3 ... N = \frac{1}{2}N(N+1)$$
then
$$1 + 2 + 3 ... N+1 = \frac{1}{2}(N+1)(N+2)$$
We verify:
$$\frac{1}{2}N(N+1) + N+1 = (N+1)(\frac{1}{2}N + 1) = (N+1)\frac{N+2}{2}$$
So you're formula does indeed work! Now we verify that for N = 1
$$\frac{1*(1+1)}{2} = 1$$
And therefore we can use the above logic to show that for any and ALL whole numbers N the formula works!
This calculation can be seen as an arithmetic series (i.e. the sum of the terms of an arithmetic sequence).
Assuming the input sequence: $$(a_0, a_1, \ldots, a_n)$$, we can count all subarrays as follows:
\begin{align} \; 1 & \; \text{subarray from} \; a_0 \; \text{to} \; a_{n-1}\\ + \; 1 &\; \text{subarray from} \; a_1 \; \text{to} \; a_{n-1}\\ & \; \ldots \\ + \; 1 & \; \text{subarray from} \; a_{n-1}\; \text{to} \; a_{n-1}\\ = & \; n \end{align}
$$+$$
\begin{align} \; 1 & \; \text{subarray from} \; a_0 \; \text{to} \; a_{n-2}\\ + \; 1 &\; \text{subarray from} \; a_1 \; \text{to} \; a_{n-2}\\ & \; \ldots \\ + \; 1 & \; \text{subarray from} \; a_{n-2}\; \text{to} \; a_{n-2}\\ = & \; n-1\\ \end{align}
$$+ \; \ldots$$
\begin{align} \; \; \; 1 & \; \text{subarray only containing} \; a_0\\ = & \; 1\\ \end{align}
which results in the arithmetic series: $$n + n-1 + … + 1$$.
The above can also be represented as $$\sum_{i=1}^{n}i\;$$ and adds up to $$n (n+1)/2$$.
Elaborating on Brian's answer, lets assume we count all the single character sub-strings. There can be n such single character sub-string. By using the Binomial Coefficient, $$\binom{n}{r}$$ notation, we can say n = $$\binom{n}{1}$$ .
$$total_1 = n = \binom{n}{1}$$
Now let's assume we count all the sub-string that are not single character. Because we have to choose the beginning and end of the sub-string regardless of the order, the count should be $$\binom{n}{2}$$.
$$total_* = \binom{n}{2}$$
Now the total number of sub-string,
$$total = total_1 + total_* = \binom{n}{1} + \binom{n}{2}$$
Now recall Pascal triangle and recurrence relation $$\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}$$. So we can write,
$$total = total_1 + total_* = \binom{n}{1} + \binom{n}{2} = \binom{n+1}{2} = n(n+1)/2$$
There we have the mathematical deduction. (I kind of feel we do not need the fancy recurrence relation to get the final answer though).
• Use LaTeX please. – Michael Rozenberg Oct 31 '18 at 20:31
Trying to explain in layman terms.
Let's say f(0) = 0
f(1) = 1
f(2) = 3
f(3) = 6
f(4) = 10
f(5) = 15
By observation, you can see that each result is just an addition of previous result and current number.
f(n) = n + f(n-1)
So with this formula let's expand f(5).
f(5)
=> 5 + f(4)
=> 5 + 4 + f(3)
=> 5 + 4 + 3 + f(2)
=> 5 + 4 + 3 + 2 + f(1)
=> 5 + 4 + 3 + 2 + 1 + f(0)
=> 5 + 4 + 3 + 2 + 1 + 0
===> This is equal to sum of n numbers = n(n+1)/2 | 2021-06-24T22:03:37 | {
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https://www.physicsforums.com/threads/given-the-planar-curve-find-the-equation-of-the-plane.889281/ | Given the planar curve, find the equation of the plane
Tags:
1. Oct 15, 2016
dlacombe13
1. The problem statement, all variables and given/known data
$r(t) = < 2e^t - 5 , e^t +3t^2 , 4t^2 +1>$
Is a curve that lies within a plane. Find the equation of this plane.
3. The attempt at a solution
I am not sure if my approach is correct. These are my results:
$x=2e^t - 5$
$y = e^t +3t^2$
$z = 4t^2 + 1$
$z = 4t^2 + 1 ~~\Rightarrow~~ t = \sqrt{\frac{z-1}{4}}$
$x = 2e^\sqrt{\frac{z-1}{4}} - 5 ~~\Rightarrow~~ \frac{x+5}{2} =e^\sqrt{\frac{z-1}{4}}$
$y = e^\sqrt{\frac{z-1}{4}} + \frac{3z-3}{4}$
$y = \frac{x+5}{2} + \frac{3z-3}{4}$
$= \frac{x}{2} + \frac{5}{2} + \frac{3z}{4} - \frac{3}{4}$
$= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$
2. Oct 15, 2016
pasmith
One way of specifying a plane is as $\{ \lambda \mathbf{a} + \mu\mathbf{b} + \mathbf{c} : (\lambda , \mu) \in \mathbb{R}^2\}$ for given vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ where $\mathbf{a}$ and $\mathbf{b}$ are linearly independent. Then for non-constant functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ we have that $$\mathbf{r}(t) = f(t)\mathbf{a} + g(t)\mathbf{b} + \mathbf{c}$$ is a curve which lies on this plane.
3. Oct 15, 2016
Ray Vickson
Easier: we have $2y- x = 6 t^2 + 5$, which has eliminated the $e^t$ terms. Now to eliminate the $t^2$ terms, just add or subtract a suitable multiple of $z$. That will leave you with a constant, having no $t$ in it anywhere.
4. Oct 15, 2016
Staff: Mentor
"I am not sure if my approach is correct."
You can (and really should) check your final equation. To do this choose three t values in your equation for r(t). That will give you three vectors, the endpoints of which lie in the plane. These three points should satisfy the plane equation you ended with. If all three points work, then you can be 99% sure that your work is correct. (Subtracting a tiny amount for arithmetic errors you might make). If the three points don't satisfy the plane equation, that's a sign that you have done something wrong.
5. Oct 15, 2016
LCKurtz
I didn't check all your work, but look at your last 3 lines. You start out with $y =$ and end up with $= \frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$.
Does $y=$ that or is your final equation supposed to be $\frac{1}{2}x - y + \frac{3}{4}z = -\frac{15}{8}$? If that is your answer then notice that $\vec r(0) = \langle -3,1,1\rangle$, so that point must be on the plane. It doesn't seem to satisfy that last equation. It may be a simple arithmetic mistake.
But one reason I didn't check your work is that I don't think eliminating the parameter is a sensible way to do the problem. I would suggest a different approach. All you need for a plane is a point and a normal vector. It's easy to get a point on the plane. Then notice that given the curve is planar, that means the tangent vectors to the curve must be in the plane. So you could calculate $\vec r'(0)\times \vec r'(1)$ to get a normal.
6. Oct 15, 2016
dlacombe13
Thank you all for your replies. I tried plugging in t=0,1,2 into each parameter, and plugged it into the equation for the plane. For all three points, I get: -1.75. -15/8 = -1.875.
Thank you all for you're replies, I tried checking my values and got close to -15/8, but I don't like the error, since I think my arithmetic is right. Also, my original thinking was exactly as LCKurtz said. However I ran into issues, but I think that is because I did it wrong. I used:
$T = \frac{r'(t)}{|r'(t)|}$
T x r(t)
But it didn't seem to work. So what you're saying is I just need to take find two random tangent points along the curve, which will give me two random vectors on the plane it is within, and cross them which will yield an orthogonal vector (n), and then just form the equation from it?
7. Oct 15, 2016
dlacombe13
Thanks everyone, I did get the equation finally, using r'(0) x r'(1) = <8,-16,12> and got the equation:
8x - 16y +12z +28 = 0
I verified it by plugging in points, and it works out. One last question before I go...
Why didn't my attempt at using T, the unit tangent vector work? I mean it is still a vector that is in the direction of the curve, and thus on the plane, right?
8. Oct 15, 2016
LCKurtz
Yes, it should have worked. But dividing by the magnitude is unnecessary, adds square roots, and makes the calculations more error prone, which apparently got you. | 2017-08-21T10:44:47 | {
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https://grindskills.com/convergence-in-distribution-to-a-degenerate-distribution/ | # Convergence in distribution to a degenerate distribution
This question came up based on a disagreement I had with a TA. This was the specific example:
Let $$X_{1},…,X_{n}X_{1},...,X_{n}$$ be an iid random sample from a population with pdf $$f(x)=3(1-x)^2, 0. The $$nthnth$$ order statistic is represented as $$X_{(n)}X_{(n)}$$.
Question: Find a constant $$vv$$ such that $$n^v(1-X_{(n)})n^v(1-X_{(n)})$$ converges in distribution.
I believe that the intent of the question was to prompt the invocation of Slutsky's theorem (or maybe not); however, to which distribution the presented snippet was supposed to converge was not specified. So I presented a lazy alternative answer as follows:
As the sample size approaches infinity, $$X_{(n)}X_{(n)}$$ becomes arbitrarily close to 1. Therefore, we can simply set $$v=0v=0$$ and the statement will converge to a degenerate distribution. In this case, the order statistic converges to 1, so with $$v=0v=0$$, the statement converges to 0. I later realized that it actually converges to 1 after the transformation, which is reflected in the final degenerate distribution written toward the bottom of the question. I was surprised when the TA said that 0 is not a random variable or a distribution, and my answer made no sense. I second guessed myself and went further:
The cdf of the original distribution is $$F(x)=x^3-3x^2+3xF(x)=x^3-3x^2+3x$$ Using this, I derived the pdf of the order statistic,
$$f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1} f_{X_{(n)}}(x)=3n(x^2-2x+1)(x^3-3x^2+3x)^{n-1}$$
And the cdf of the order statistic, $$f_{X_{(n)}}(x)=(x^3-3x^2+3x)^nf_{X_{(n)}}(x)=(x^3-3x^2+3x)^n$$
I then used this to graph the cdf of the order statistic with $$n=1, n=100, n=1,000,000,000n=1, n=100, n=1,000,000,000$$. Predictably, the graph showed the cdf getting thinner and steeper, until with huge samples it visually looks like a vertical line at $$X_{(n)}=1X_{(n)}=1$$.
I reiterated my argument, and showed the visualization: it is converging to a degenerate distribution. Again, I was rebuffed. After pestering the TA and asking the instructor for the course, I still have no answer for why my answer was wrong, but I can't keep bothering them about it.
Can someone here tell me if my argument is solid, and if not, tell me specifically what I've done that is in error?
EDIT: Here is the final formalism that I gave to tie my argument together. Following Casella and Berger's definition,
A sequence of random variables, $$X_{1}, X_{2},...,X_{1}, X_{2},...,$$ converges in distribution to a random variable $$XX$$ if $$\lim_{n \to \infty} F_{X_n}(x) = F_X(x)\lim_{n \to \infty} F_{X_n}(x) = F_X(x)$$ at all points $$xx$$ where $$F_X(x)F_X(x)$$ is continuous
Since $$\lim_{n \to \infty} F_{X_(n)}(x) = 1\lim_{n \to \infty} F_{X_(n)}(x) = 1$$
And we can define a degenerate variable $$YY$$ with the cdf
$$F_Y(x)=\begin{cases}1, & x\ge1\\ 0, & else \\ \end{cases}F_Y(x)=\begin{cases}1, & x\ge1\\ 0, & else \\ \end{cases}$$
We can say
$$\lim_{n \to \infty} F_{X_(n)}(x) = F_Y(x) \lim_{n \to \infty} F_{X_(n)}(x) = F_Y(x)$$
So by definition, $$X_{(n)}X_{(n)}$$ converges in distribution to the degenerate distribution of $$YY$$.
EDIT: Adding a bounty to this, as it has become important. I did not formally perform the transformation $$Y=1-X_(n)Y=1-X_(n)$$ above, but doing so results in $$YY$$ having a distribution that still converges to the same degenerate distribution above. I was now told that my answer does not prove convergence in distribution at all. Question is answered if someone can definitively prove either 1) my answer is correct, and show exactly how you would have gone about proving it formally and rigorously, or 2) my answer is wrong, exactly why it is wrong, and prove any correct answer that uses degenerate distributions or degenerate variables if such an answer exists.
Your answer is correct (assuming that you have accurately transcribed the question). The proof:
Let $$F_n(c)F_n(c)$$ be the cdf of $$(1 - X_{(n)})(1 - X_{(n)})$$, where $$X_{(n)}X_{(n)}$$ is the greatest element in a sample of size $$nn$$. Let $$F(c)F(c)$$ be the cdf for the constant 0 distribution.
For $$c < 0c < 0$$, of course $$F_n(c) = 0 = F(c)F_n(c) = 0 = F(c)$$.
For $$c > 1c > 1$$, of course $$F_n(c) = 1 = F(c)F_n(c) = 1 = F(c)$$.
For $$0 < c \le 10 < c \le 1$$:
\begin{align} F_n(c) &= P(1 - X_{(n)} \le c) \\ &= P(X_{(n)} \ge 1-c) \\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\ &= 1 - P(X_1 < 1-c)^n \to 1 = F(c) \end{align} \begin{align} F_n(c) &= P(1 - X_{(n)} \le c) \\ &= P(X_{(n)} \ge 1-c) \\ &= 1-P(X_1 < 1-c, ..., X_n < 1-c) \\ &= 1 - P(X_1 < 1-c)^n \to 1 = F(c) \end{align}
And the case $$c = 0c = 0$$ doesn't matter, because $$FF$$ isn't continuous at $$00$$.
If the people you are arguing with don't realise that convergence in distribution to a constant is a thing, you could point them to e.g. Wikipedia's Proofs of convergence of random variables article. | 2022-09-29T02:47:15 | {
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https://math.stackexchange.com/questions/1767029/did-feynman-mentally-compute-sqrt31729-03-by-linear-approximation/1767244 | # Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation?
In the biopic "infinity" about Richard Feynman. (12:54) He computes $\sqrt[3]{1729.03}$ by mental calculation. I guess that he uses linear approximation. That is, he observe that $1728=12^3$. Let $f(x)=\sqrt[3]{x}$. Then $f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ and $f'(1728)=\frac{1}{3\sqrt[3]{1728^2}}=\frac{1}{3\cdot 12^2}$. Therefore, $$\sqrt[3]{1729.03}=f(1729.03)\approx f(1728)+f'(1728)(1729.03-1728)=12+\frac{1.03}{3\cdot 12^2}=12.002384\overline{259}.$$
Question 1. If he used the linear approximation, how did he compute $\frac{1.03}{3\cdot 12^2}=0.002384\overline{259}$ by a mental calculation?
Question 2. If he didn't use the linear approximation, what is another method he might have used?
• You're asking about how a mental computation was accomplished in a scene from a fictional movie...? (Just clarifying; I'm not willing to click a youtube link to get the content of the question.) – Andrew D. Hwang May 1 '16 at 18:04
• @AndrewD.Hwang Yes. Although it is a fictinal movie, I think that maybe the scenarist had a method to approximate $\sqrt[3]{1729.03}$. And I hope this scene can motivate my student to learn linear approximation. – bfhaha May 1 '16 at 18:12
• $1.03/3\approx.34333\approx.34332$ $.34332/144=.05722/24=.02861/12\approx.002384$. – Mark S. May 1 '16 at 18:30
• You should mention what value was computed so that the question is self-contained. – Bill Dubuque May 1 '16 at 18:41
• @Andrew D. Hwang It is mentioned in his book "Surely You're joking, Mr. Feynman!". Give it a shot. – user537100 Jul 1 '18 at 13:09
Feynman tells the story in one of his books of anecdotes.
http://www.ee.ryerson.ca/~elf/abacus/feynman.html
$12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision.
$$\sqrt[3]{1728 + d} = 12\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$
where $d = 1.03$ and $x = \frac{d}{1728}$ is, in Feynman's words, about 1 part in 2000, so that the error term is of order $10^{-6}$.
Feynman says that he computed $12 + \frac{4d}{1728}$ as the approximate value.
The number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.
He describes that as though $d=1$ for this part of the calculation, so maybe $12 + \frac{1}{432}$ was what he actually computed. By "adding two more digits" (to 12.002) he seems to mean working out the division in the fraction. It could also mean adding (0.03)/432 as two more decimal digits of accuracy to $(12 + 432^{-1})$, which requires only a multiplication by 3 of an already computed quantity 1/432.
Feynman's method is the one that would have been immediate for anyone familiar with the binomial series and with $12^3 = 1728$. He said that knew the latter as ft^3/in^3 and other people might know it from the Ramanujan 1729 story. The other ingredient, as Feynman says in the story, was being good at integer division.
I upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following.
You can start with the linear approximation, and then you have to calculate $(\dfrac{1.03}{3})/12^2$. $\dfrac{1.03}{3}\approx .34333$, but that doesn't lend itself to division by $12^2$, so you can change the last digit to get $\dfrac{1.03}{3}\approx .34332$
(You know this will be helpful by the divisibility test for $3$, and can note it's also divisible by $2$ and $4$ if you're thinking ahead.)
Then $\dfrac{.34332}{12^2}=\dfrac{.05722}{12*2}=\dfrac{.02861}{12}\approx.002384$.
• "linear approximation" here means taking the linear part of $(12 + x)^3$ when solving $(12 + x)^3 = 12^3 + 1.03$. Then $3x(12^2) = 1.03$ and the rest is as in the answer. Thinking of it as 1 or 1.03 divided by 432, how to divide by 432 with minimum calculation? The error in approximating that 432 by 500 is (1/432 - 1/500) = (500- 432)/500*432 or about 68/200000. This gives 0.002 + 0.00034 = 0.00234 using only division by 2, 4*5=20, and 100-32=68. – zyx May 2 '16 at 2:49
• Thanks MarkS. You and zyx gave the best answer for me. But it can only has one answer. So please forgive me to use upvote instead of mark it as an answer. – bfhaha May 2 '16 at 5:12 | 2021-07-28T13:39:44 | {
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Mathematics
OpenStudy (waheguru):
is there a special rick or is it just trial and error to solve this porblem
OpenStudy (waheguru):
OpenStudy (waheguru):
it would take a really long time to find he asnwer by trial and answer
jimthompson5910 (jim_thompson5910):
Best way to do this is to break things down. Start with 100. Clearly 1+0 = 1 which is not 0, so that doesn't work. But 101 does since 1+0 = 1 which is the last digit. Now ask yourself: are there any more numbers in the range of 100 to 109 that work? The answer is no since 1+0 = 1 is unique to this set of numbers (ie there are no more numbers in this range that have a 1 as the last digit) So again, in the list of numbers from 100 to 109, the only upright number is 101 Similarly, for numbers in the range 110 to 119, the only upright number is 112 Keep going to find the third upright number to be 123 Etc, etc When you come upon numbers that start with 19, you won't find any upright numbers since 1+9 = 10, which is not a single digit. So , in the range of 100 to 199, there are 8 numbers that are upright numbers So from 200 to 299, there are 7 numbers that work. Why 7 and not 8? notice that for numbers that start with 29 add to 2+9 = 11, but that's not a single digit. Likewise, numbers starting with 28 add to 2+8 = 10, again not a single digit. However, the rest work. Keep going and you'll find that only 6 numbers work from 300 to 399 etc etc After all that, you'll have the following number counts: 8, 7, 6, 5, 4, 3, 2, 1, 0 Add them all up to get 36 So the answer is choice C
OpenStudy (anonymous):
Interesting question. Here's how I tackled it:Look at the first few 3 digit numbers.100101102103104105106107108109They all have the same first two digits, but different last digits. You'll notice that exactly one of these numbers is upright. You'll find the same thing for 110 through 119. I can keep going unless the sum of the first two digits is greater than 9. For example, any number with 67 as the first two digits can never be upright because 6+7 is a two digits number.So to start counting upright numbers, add up all the possible 2 digit combinations you can get. you can use 10 through 99, so that's 90 possible two digit pairs. Now you need to subtract off the pairs that don't work, like 67. 10, 11, 12, 13, ... , 18 all work, but 19 doesn't. That's 9 upright numbers. 20, 21, ... , 27 all work, but 28 and 29 don't. That's 8 more upright numbers. 90 works, but 91, 92, ... , 99 don't. That's just one more upright number. See the pattern? So find 1+2+3+4+5+6+7+8, and you have your answer.
jimthompson5910 (jim_thompson5910):
you have the right idea pmilano, but you fell into the trap that there's a number in the range of 190 to 199 when there isn't one
OpenStudy (anonymous):
By the way, the fast way to add the numbers 1 through 8 (or, more generally, 1 through n) is by using the formula (n)(n+1)/2 So 8(8+1)/2 = 36
OpenStudy (anonymous):
Thanks jim.
OpenStudy (anonymous):
my answer is D)45. There is actually a pattern if you try to list down the first numbers. Starting at 1__, there are 9 possible answers. As you move up to the next hundreds digit, you decrease that number by one, because you can't have a sum of more than 9. 9+8+7+...+2+1 = 45
OpenStudy (anonymous):
I actually realized that before I posted but didn't proof read my post well enough (hence my correct answer and incorrect counting).
jimthompson5910 (jim_thompson5910):
My bad, I was ignoring zero Start with the range 100-199 and moving to the next 200-299, etc, the following upright numbers are found in the tens digits listed below 0, 1, 2, 3, 4, 5, 6, 7, 8 0, 1, 2, 3, 4, 5, 6, 7 0, 1, 2, 3, 4, 5, 6 0, 1, 2, 3, 4, 5 0, 1, 2, 3, 4 0, 1, 2, 3 0, 1, 2 0, 1 0 So there are 45 here
OpenStudy (waheguru):
i am getting 36?
OpenStudy (anonymous):
Here is another method lol (incase you didnt have enough) When the last digit is 'n', you want to ask yourself how many solutions are there to: x + y = n where x and y are non-negative integers? For example, if n was 3, then these are the possible solutions for x and y: x = 0, y = 3 x = 1, y = 2 x = 2, y = 1 x = 3, y = 0 These correspond to the three digits numbers: 033 123 213 303 Of course we would discard the '033' case. Anywhos, the equation that gives you the number of solutions to "x + y = n" is: $\left(\begin{matrix}n+2-1 \\ 2-1\end{matrix}\right) = \left(\begin{matrix}n+1 \\ 1\end{matrix}\right)$ We need to do this for n = 1, 2, 3,....,9, while remembering to not count the case of the first digit being 0. So we end up with: $\left(\begin{matrix}2 \\ 1\end{matrix}\right)+\left(\begin{matrix}3 \\ 1\end{matrix}\right)+\left(\begin{matrix}4 \\ 1\end{matrix}\right)+\ldots +\left(\begin{matrix}10 \\ 1\end{matrix}\right)-9$$= 2+3+4+\ldots+10-9 = 1+2+3+\ldots+9 = 45$
OpenStudy (anonymous):
First set of numbers: 101,112,123,134,145,156,167,178,189 ---> 9 numbers. For 2__, that is one less than the 1 hundreds because u can only use 0-7 for the 2nd digit. The last possible number is 909, the only number in the 9 hundreds.
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Join our real-time social learning platform and learn together with your friends! | 2021-05-07T11:11:39 | {
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https://math.stackexchange.com/questions/3320950/integrating-int-frac-sin-x1-cos-x-dx-i-get-ln1-cos-x-wolfr | # Integrating $\int \frac{-\sin x}{1+\cos x}\, dx$, I get $\ln(1 + \cos x)$. WolframAlpha gives $2 \ln(\cos \frac x 2)$. Is WA wrong?
So, I'm watching a tutorial on differential equations, where I encountered this little trick:
$$\int \frac{y'}{y}\, dx = \ln(y)$$ It seems perfectly logical and easy to justify, but something fishy happens to this integral:
$$\int \frac{-\sin x}{1+\cos x}\, dx$$
The trick gives $$\int \frac{-\sin x}{1+\cos x}\, dx = \ln(1 + \cos x)$$
while WolframAlpha gives $$\int \frac{-\sin x}{1+\cos x}\, dx = 2 \ln(\cos \frac x 2)$$.
You guys who know this stuff - does WolframAlpha mess up here or is it something I've missed? Taking the derivative of $$2 \ln(\cos \frac x 2)$$ gives me $$-\tan \frac x 2$$, so I don't see how WA may be right.
• Remember that $\cos{2x}=2\cos^2x-1$ – Don Thousand Aug 12 at 11:57
• They are different in terms of their symbols, but they give you exactly the same values. What you have is a trigonometric identity. – Pixel Aug 12 at 12:01
• As others have mentioned, the key here is to use a trig identity to bridge the two solutions. ... Often, the trickiest part of Calculus is remembering your Pre-Calculus. :) – Blue Aug 12 at 12:03
• For future reference, you can always check your (or WA's) answer by differentiating it and checking it against the expression you're integrating (the integrand). If they match up then you know your answer is correct (up to a constant), even if you don't know any trig! – Ben Aug 12 at 12:05
## 5 Answers
$$1+\cos x = 2\cos^2\frac{x}{2}$$
$$\ln (1+\cos x )=\ln( 2\cos^2\frac{x}{2})=\ln2+2\ln\cos\frac{x}{2}$$
And this $$\ln2$$ adds together with the arbitrary constant $$c$$ in indefinite integral and gets cancelled in definite integral.
• Thank you so much! This makes perfect sense! – Seigemann Aug 12 at 11:58
• You're welcome! – Ak19 Aug 12 at 11:58
As Ben suggested in comments, it's always good to check an integral by taking the derivative:
$$\dfrac d {dx} \ln(1+\cos x)=\dfrac{-\sin x}{1+\cos x}=\dfrac{-2\sin\dfrac x2 \cos \dfrac x2}{2\cos^2\dfrac x2}=\dfrac{-\sin\dfrac x2}{\cos \dfrac x2}=\dfrac d {dx} 2 \ln \cos \dfrac x2$$
• N.B. The expressions are undefined when $x=\pi$ – J. W. Tanner Aug 12 at 13:11
$$1+\cos \, x=2\cos^{2}(\frac x 2)$$ so $$\ln (1+\cos \, x)=2 \ln (\cos (\frac x 2))+\ln 2$$ . Also, $$-\tan (\frac x 2)$$ is same as $$-\frac {\sin \, x} {1+\cos \, x}$$ because $$\sin \,x =2 \sin (\frac x 2)\cos (\frac x 2)$$ and $$1+\cos \, x=2\cos^{2}(\frac x 2)$$.
$$2\ln\cos\frac x2=\ln\cos^2\frac x2=\ln\frac{1+\cos x}2=\ln(1+\cos x)-\ln2$$ by the half-angle formula, so your answer and WA's are the same up to the integration constant.
When you do the one integral $$I(x)$$ by different methods you get different expressions $$I_1(x),I_2(x),I_3(x),.....$$, however the difference between any two of these is a constant independent of $$x$$. For instance $$I(x)=\int \sin \cos x dx =\frac{1}{2}\int \sin 2x~ dx =-\frac{1}{4} \cos 2x +C_1 =I_1(x).$$
Next if you do integration by parts you get $$I=\sin ^2x -\int \sin x \cos \Rightarrow I=\frac{1}{2} \sin^2 x +C_2=I_2(x).$$
Further I you use a substitution $$\cos x =-t$$, then $$I(x)=-\frac{1}{2}\cos^2 x +C_3=I_3(x).$$
Now check that the difference between any two of $$I_1,I_2,I_3$$ is just a constant.
As pointed in other solutions the thing stated above is happening in your case as well. | 2019-08-26T03:31:52 | {
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https://www.weddingsupplierswa.com.au/viz6m3yu/is-real-number-a-subset-of-complex-number-489fe3 | Thus, the complex numbers of t… The subsets of the real numbers can be r… However, real numbers have multiplication, and the complex numbers extend the reals by adding i. Intro to complex numbers. The set of complex numbers is closed under addition and multiplication. The set of complex numbers is denoted by C R is a subset of C 118 When adding from MAT 1341 at University of Ottawa Lv 7. hace 5 años. The values a and b can be zero, so the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers. We define the complex number i = (0,1).With that definition we can write every complex number interchangebly as A complex number such as $5-2 i$ then corresponds to 5 on the real axis and $-2$ on the imaginary axis. Yes. In situations where one is dealing only with real numbers, as in everyday life, there is of course no need to insist on each real number to be put in the form a+bi, eg. Click to see full answer. The table below describes important subsets of the real numbers. Bundle: Elementary Algebra, 9th + Student Workbook (9th Edition) Edit edition. In the complex number a + bi, a is called the real part and b is called the imaginary part. Real numbers can be considered a subset of the complex numbers … In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also. This is because the Real Numbers are a subset of the Complex Numbers (). The set of real numbers is a proper subset of the set of complex numbers. To which subsets of the real numbers does -7 belong? Similarly, since a can be zero, the imaginary numbers are a subset of the complex numbers. Why does it make sense to talk about the 'set of complex numbers'? Notational conventions. The set of complex numbersis, therefore; This construction allows to consider the real numbers as a subset of the complex numbers, being realthat complex number whiose imaginary part is null. Remember that under the set of rational numbers, we have the subcategories or subsets of integers, whole numbers, and natural numbers. The real numbers include both rational and irrational numbers. A and B may be equal; if they are unequal, then A is a proper subset of B. 2 I. Complex does not mean complicated; it means that the two types of numbers combine to form a complex, like a housing complex — a group of buildings joined together. However, $\mathbb{C}$ comes with a canonical embedding of $\mathbb{R}$ and in this sense, you can treat $\mathbb{R}$ as a subset of $\mathbb{C}$. In other words, i 2 = –1. Solved Example on Real Numbers Ques: Name the subset(s) of the real numbers to which '- 25' belongs. Find the real part of a complex number: Find the real part of a complex number expressed in polar form: Plot over a subset of the complex plane: Use Re to specify regions of the complex plane: The axiom of mathematical induction is for our purposes frequently The real numbers are complex numbers … Bundle: Elementary and Intermediate Algebra: A Combined Approach + Student Solutions Manual (6th Edition) Edit edition. What are rational and irrational numbers. In 1882, Ferdinand von Lindemann proved that π is not just irrational, but transcendental as well. Since b can be equal to 0, you see that the real numbers are a subset of the complex numbers. The set of real numbers is a subset of the set of complex numbers? they are of a different nature. Addition and multiplication of real numbers are defined in such a way that expressions of this type satisfy all field axioms and thus hold for C. For example, the distributive law enforces Therefore, a set of real numbers is bounded if it is contained in a … A complex numberis defined as an expression of the form: The type of expression z = x + iy is called the binomial form where the real part is the real number x, that is denoted Re(z), and the imaginary partis the real number y, which is denoted by Im(z). Natural Number (N) Subset N is the set of Natural Number or Counting Numbers given N = {1, 2, 3, ..… Set of Real Numbers Set of Real Numbers is a universal set. Real numbers are just complex numbers with no imaginary part. A mathematical operation of subtracting a complex number from another complex number is called the subtraction of complex numbers.. Introduction. There is a thin line difference between both, complex number and an imaginary number. Let Sbe a subset of the set Nof natural numbers. While the real numbers are a subset of the complex numbers, there are very many complex numbers that are not real numbers. Intro to complex numbers. In mathematics, a set A is a subset of a set B, or equivalently B is a superset of A, if A is contained in B. @HagenvonEitzen All the different constructions of $\mathbb{R}$ rely on the fact that we have already constructed $\mathbb{N}$ before (?). What Number Set Contains The Subset of Complex Numbers? Examples: 1 + i, 2 - 6i, -5.2i, 4. It is important to note that if z is a complex number, then its real and imaginary parts are both real numbers. Notational conventions. Milestone leveling for a party of players who drop in and out? Real numbers $$\mathbb{R}$$ The set formed by rational numbers and irrational numbers is called the set of real numbers and is denoted as $$\mathbb{R}$$. generating lists of integers with constraint. Read More -> The number {3} is a subset of the reals. For example, the set $\mathbf{C}^{2}$ is also a real vector space under the same addition as before, but with multiplication only by real scalars, an operation we might denote $\cdot_{\mathbf{R}}$. Can you put laminate flooring in a mobile home? Bundle: Elementary Algebra + Math Study Skills Workbook (4th Edition) Edit edition. Complex. Complex numbers can be represented as points on a “complex plane”: the rectangular x-y plane, in which the x-axis corresponds to the real numbers, and the y-axis corresponds to the imaginary numbers. Since $\mathbb{Q}\subset \mathbb{R}$ it is again logical that the introduced arithmetical operations and relations should expand onto the new set. The real numbers can be "said to be" a subset of the complex numbers. (examples: -7, 2/3, 3.75) Irrational numbers are numbers that cannot be expressed as a fraction or ratio of two integers. Set Theoretic Definition of Complex Numbers: How to Distinguish $\mathbb{C}$ from $\mathbb{R}^2$? iota.) ): Includes real numbers, imaginary numbers, and sums and differences of real and imaginary numbers. [1] [2] Such a number w is denoted by log z . An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i2 = −1. If a jet engine is bolted to the equator, does the Earth speed up? On the same footing, $\mathbb{N} \not \subset \mathbb{Z} \not \subset \mathbb{Q} \not \subset \mathbb{R}$. The relationship between the real and complex numbers from a set theoretic perspective. To make notation a little bit easier, we call a complex number z. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. p S S S II) i.W 2 lIT ~and ir are two of very many real numbers that are not rational numbers. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. 9 is a real number because it belongs to the set of whole numbers and the set of whole numbers is a subset of real numbers. Real numbers are simply the combination of rational and irrational numbers, in the number system. A real number is a number that can take any value on the number line. definition. D. Irrational Use MathJax to format equations. Be sure to account for ALL sets. To learn more, see our tips on writing great answers. What is internal and external criticism of historical sources? For example, 5i is an imaginary number, and its square is −25. 10, as 10 + 0i - that would be too pedantic, to say the … We call x +yi the Cartesian form for a complex number. Strictly speaking (from a set-theoretic view point), $\mathbb{R} \not \subset \mathbb{C}$. It solves x²+1=0. MathJax reference. 1 See answer AnshulDavid3143 is waiting for your help. 2/5 A. Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. But since the set of complex numbers is by definition $$\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},$$ doesn't this mean $\mathbb{R}\subseteq\mathbb{C}$, since for each $x \in \mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$? Is Delilah from NCIS paralyzed in real life? To which subset of real numbers does the following number belong? What is the "Ultimate Book of The Master". By now you should be relatively familiar with the set of real numbers denoted $\mathbb{R}$ which includes numbers such as $2$, $-4$, $\displaystyle{\frac{6}{13}}$, $\pi$, $\sqrt{3}$, …. Real numbers are a subset of complex numbers. Two complex numbers a + bi and c + di are defined to be equal if and only if a = c and b = d. If the imaginary part of a complex number is 0, as in 5 + 0i, then the number corresponds to a real number. Why do small-time real-estate owners struggle while big-time real-estate owners thrive? Complex numbers are distinguished from real numbers by the presence of the value i, which is defined as . Complex numbers are an important part of algebra, and they do have relevance to such things as solutions to polynomial equations. Complex Numbers $\mathbb{C}$ Examples of complex numbers: $(1, 2), (4, 5), (-9, 7), (-3, -20), (5, 19),...$ $1 + 5i, 2 - 4i, -7 + 6i...$ where $i = \sqrt{-1}$ or $i^2 = -1$ The real numbers are a subset of the complex numbers. THE REAL AND COMPLEX NUMBERS AXIOM OF MATHEMATICAL INDUCTION. Any time you deal both with complex vector spaces and real vector spaces, you have to be certain of what "scalar multiplication" means. Dedekind cuts or Cauchy sequences for $\mathbb R$) these ZFC. What is the difference between simple distillation and steam distillation? So, I was taught that $\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}$. The real axis is the line in the complex plane consisting of the numbers that have a zero imaginary part: a + 0i. Complex numbers, say … square root of 30 . The complex numbers C consist of expressions a + bi, with a, b real, where i is the imaginary unit, i.e., a (non-real) number satisfying i 2 = −1. Asking for help, clarification, or responding to other answers. Why did the design of the Boeing 247's cockpit windows change for some models? The irrational numbers are a subset of the real numbers. Imaginary number is expressed as any real number multiplied to a imaginary unit (generally 'i' i.e. Two complex numbers a + bi and c + di are defined to be equal if and only if a = c and b = d. If the imaginary part of a complex number is 0, as in 5 + 0i, then the number corresponds to a real number. Example 1. Pi is an irrational number, which means that it is a real number that cannot be expressed by a simple fraction. Complex numbers can be visualized geometrically as points in the complex (Argand) plane. How do I provide exposition on a magic system when no character has an objective or complete understanding of it? The set of complex numbers includes all the other sets of numbers. rev 2021.1.18.38333, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, You have $\not\subset$ if you construct them one after another. The imaginary numbers are also a subset of the complex: the complex numbers whose real part is zero. If I am blending parsley for soup, can I use the parsley whole or should I still remove the stems? Is it safe to keep uranium ore in my house? Complex numbers are numbers in the form a + b i a+bi a + b i where a, b ∈ R a,b\in \mathbb{R} a, b ∈ R. And real numbers are numbers where the imaginary part, b = 0 b=0 b = 0. Why did flying boats in the '30s and '40s have a longer range than land based aircraft? The set {0,1, 2+i, 2-i} is NOT a subset of the real numbers. © AskingLot.com LTD 2021 All Rights Reserved. But already the fact that there are several constructions possible (e.g. Every real number graphs to a unique point on the real axis. Why set of real numbers not a set of ordered pairs? The term is often used in preference to the simpler "imaginary" in situations where. mam is real numbers a subset of complex numbers - Mathematics - TopperLearning.com | 8v26wq66 3. Start studying Field of Quotients, the Rational Numbers, the Real Numbers, & Complex Numbers. The real numbers are a subset of the complex numbers, so zero is by definition a complex number ( and a real number, of course; just as a fraction is a rational number and a real number). One can represent complex numbers as an ordered pair of real numbers (a,b), so that real numbers are complex numbers whose second members b are zero. At the same time, the imaginary numbers are the un-real numbers, which cannot be expressed in the number line and is commonly used to represent a complex number. The set of real numbers can be drawn as a line called “the number line”. Subset. Would coating a space ship in liquid nitrogen mask its thermal signature? Complex numbers, such as 2+3i, have the form z = x + iy, where x and y are real numbers. a complex logarithm of a nonzero complex number z, defined to be any complex number w for which e w = z. a real number is not a set. Real numbers, rational numbers. Expressing complex numbers in form $a+bi$. Each complex number corresponds to a point (a, b) in the complex plane. The square of an imaginary number bi is −b2. In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. can in general assume complex values with nonzero real parts, but in a particular case of interest, the real part is identically zero. Practice: Parts of complex numbers. In the last example (113) the imaginary part is zero and we actually have a real number. Rational numbers are numbers that can be expressed as a fraction or part of a whole number. It only takes a minute to sign up. There are three common forms of representing a complex number z: Cartesian: z = a + bi JR is the set of numbers that can be used to measure a distance, or the negative of a number used to measure a distance. 5.1.2 The Reals as a Subset of the Complex Numbers Since the complex numbers were seen as an extension of the set of real numbers, it is natural to believe that R is a subset of C. Of course, to prove this subset Similarly, it is asked, is every real number is a complex number? A complex number is a number that can be written in the form a + b i a + bi a+bi, where a and b are real numbers and i is the imaginary unit defined by i 2 = − 1 i^2 = -1 i2=−1. Classification of Real Numbers Examples. The real numbers have the following important subsets: rational numbers, irrational numbers, integers, whole numbers, and natural numbers. (0,1) = (-1,0), which is purely real and equals to -1. Example 2 : Tell whether the given statement is true or false. Thus we can consider the complex number system as having embedded within it, as a subset the real number … Furthermore, each real number is in the set of complex numbers,, so that the real numbers are a … Choices: A. integers, rational numbers, real numbers B. whole numbers, integers, rational numbers, real numbers C. natural numbers, whole numbers, integer numbers, rational numbers, real numbers D. irrational numbers, real numbers Correct Answer: A Complex numbers are often graphed on a plane. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? Some examples of irrational numbers are $$\sqrt{2},\pi,\sqrt[3]{5},$$ and for example $$\pi=3,1415926535\ldots$$ comes from the relationship between the length of a circle and its diameter. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Oh I suppose Russel has a definition where the real number 3 is the set of all things there are 3 of. A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself). How are Quaternions derived from Complex numbers or Real numbers? B. ¿Cuáles son los 10 mandamientos de la Biblia Reina Valera 1960? The real numbers are a subset of the complex numbers. So, $$i \times i = -1$$ $$\Rightarrow i = \sqrt{-1}$$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Complex numbers introduction. We will now introduce the set of complex numbers. The complex numbers are a plane with an additional real axis to calculate square roots (and other even roots) of negative numbers. The set of complex numbers C with addition and multiplication as defined above is a field with additive and multiplicative identities (0,0) and (1,0).It extends the real numbers R via the isomorphism (x,0) = x. Suppose that (1) 1 2S: (2) If a natural number kis in S;then the natural number k+ 1 also is in S: Then S= N:That is, every natural number nbelongs to S: REMARK. There are several types of subsets of real numbers—numbers that can be expressed as a decimal. Thus we can consider the complex number system as having embedded within it, as a subset the real number system. If you're seeing this message, it means we're having trouble loading external resources on our website. These numbers are called irrational numbers, and $\sqrt{2}$, $\sqrt{3}$, $\pi$... belong to this set. Is there even such a set? Proof that π is irrational. Email. But no real number, when squared, is ever equal to a negative number--hence, we call i an imaginary number. Better user experience while having a small amount of content to show. Explain your choice. Complex numbers in the form a + bi can be graphed on a complex coordinate plane. Complex Numbers. x is called the real part and y is called the imaginary part. (The counting numbers are 1,2,3,....) All of these types of numbers are real numbers. We will addres s complex (or imaginary) numbers in the Quadratic Functions chapter. That is the adjacent surface to our 3D! Learn vocabulary, terms, and more with flashcards, games, and other study tools. The real numbers are all the numbers on the number line, where you group rational numbers with a so called dedekind cut (you can form this cut so that it result is irrational). Popular Trending (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) Real numbers 21.5 pi. Some fixed point O is chosen to represent the complex number … In the 1760s, Johann Heinrich Lambert proved that the number π (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer. Therefore we have: z = Re(z) + iIm(z). Why or why not? In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts: = + for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit.In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane. A set S of real numbers is called bounded from above if there exists some real number k (not necessarily in S) such that k ≥ s for all s in S.The number k is called an upper bound of S.The terms bounded from below and lower bound are similarly defined.. A set S is bounded if it has both upper and lower bounds. The complex numbers form a COMPLETE system of numbers of which the real numbers form a subset. The conjugate of a complex number z= a+ biis created by changing the sign on the imaginary part: z = a bi: Thus the conjugate of 2 + iis 2 + i= 2 i; the conjugate of p 3 ˇiis p 3 ˇi= p 3 + ˇi.
How Much Does It Cost To Stay At The Poconos, Indore Collector Name List, Antioch Ca Fire Department, Callaway Rogue Irons Left Handed, Xmas Trees Near Me, Silver Leaf Adhesive, Kembang Kol Hidroponik, | 2021-04-13T10:18:42 | {
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https://math.stackexchange.com/questions/2019784/is-every-odd-order-skew-symmetric-matrix-singular | # Is every odd order skew-symmetric matrix singular?
We call a square matrix $$A$$ a skew-symmetric matrix if $$A=-A^T$$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?
## 3 Answers
Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.
This is actually the case :
Suppose, $A$ is an $n\times n$-matrix.
We have $$\det(A)=\det(-A^T)=(-1)^n\cdot \det(A^T)=(-1)^n\cdot \det(A)$$
Since $n$ is odd, we can conclude $\ \det(A)=-\det(A)\$ implying $\ \det(A)=0\$
• But the problem here is that the entries are complex. Is it true for complex entries as well? Nov 18 '16 at 12:05
• Yes, it is also true for complex entries. The proof does not assum real entries. Nov 18 '16 at 12:47
Consider the example $$\begin{bmatrix} 0 & i & -3\\ -i & 0 & 2i\\ 3 & -2i & 0\\ \end{bmatrix}$$
• I mean a 3 by 3 matrix with rows (0 i -3), ( -i 0 2i) and (3 -2i 0). Nov 18 '16 at 12:21
• If this was intended to be a counterexample, then it is not! The determinant of this matrix is zero. May 6 at 15:03 | 2021-10-15T22:54:26 | {
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http://angs-diasporanew.sramble-communication.com/apdsmyht/cartesian-to-cylindrical-coordinates.php | ##### Cartesian to cylindrical coordinates
8. As with spherical where we again observe that the basis vector in cylindrical coordinates is dependent on the two basis vectors e → x and e → y in Cartesian coordinates. Conversion between Cylindrical and Cartesian Coordinates The rectangular coordinates $$(x,y,z)$$ and the cylindrical coordinates $$(r,θ,z)$$ of a point are related as follows: These equations are used to convert from cylindrical coordinates to rectangular coordinates. Feb 24, 2010 · The cylindrical coordinate system is a 3-D version of the polar coordinate system in 2-D with an extra component for . Find the integrals that compute its volume, using cartesian, cylindrical, and spherical coordinates. Magnitude of a cartesian coordinates (a,b) is given bysqrt(a^2+b^2) and its angle is given by tan^-1(b/a) Let r be the magnitude of (-4,3) and theta be its angle. The old vvvv nodes Polar and Cartesian in 3d are similar to the geographic coordinates with the exception that the angular direction of the longitude is inverted. Jan 27, 2017 · We can write down the equation in Cylindrical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. Layered (2,π2). To Convert from Cartesian to Polar. We will concentrate on cylindrical coordinates in this activity, but we will address spherical coordinates in a later activity. If called with a single matrix argument then each row of C represents the Cartesian coordinate (x, y (, z)). Use Question: A sphere, centered at the origin, has radius 3. A, then, has three vector components, each component corresponding to the projection of A onto the three axes. Cylindrical coordinates are chosen to take advantage of symmetry, so that a velocity component can disappear. 8. I know how to generate the strain tensor in a rotated coordinate system (also a Cartesian one), but just don't know how to apply the rules found in the second link to derive the strain components in the cylindrical coordinates, if I have strain tensor in the corresponding Cartesian coordinates. i tried converting to cartesian coordinates, then plotting but at some points function jumps up or down (i think due to trigonometric functions) , so i wanted to see it in cylindrical coordinates. For the conversion between cylindrical and Cartesian coordinates, it is convenient to assume that the reference plane of the former is the Cartesian xy-plane (with equation z = 0), and the cylindrical axis is the Cartesian z-axis. Each point is uniquely identified by a distance to the origin, called r here, an angle, called ϕ {\displaystyle \phi } ( phi ), and a height above the plane of the coordinate system, called Z in the picture. We can slightly modify our arc length equation in polar to make it apply to the cylindrical coordinate system given that , . Mar 02, 2013 · Cartesian Coordinates vs Polar Coordinates In Geometry, a coordinate system is a reference system, where numbers (or coordinates) are used to uniquely determine the position of a point or other geometric element in space. Cartesian Coordinates Cylindrical coordinates consist of (1) a coordinate plane, plus (2) an axis perpendicular to the plane through the origin. In the coordinate plane, two coordinates describe position: (1) an angle, θ (azimuth angle, measured positive counterclockwise relative to a Richard Fitzpatrick 2016-01-22 a) x 2 - y = 25 to cylindrical coordinates. One of these is when the problem has cylindrical symmetry. 6 Volume element in Cartesian coordinates. 6. Figure 1. Replace (x, y, z) by (r, φ, θ) b. 8: Differential length, area, and volume. 1. Visit https://www. 2 s pherical Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. Cylindrical Coordinate System: In cylindrical coordinate systems a point P(r 1, θ 1, z 1) is the intersection of the following three surfaces as shown in the following figure. Cylindrical coordinates. The Cylindrical to Cartesian calculator converts Cylindrical coordinates into Cartesian coordinates. Cartesian base vectors. Choose the source and destination coordinate systems from the drop down menus. which means that . Cylindrical coordinates in IR3. (1) The (orthogonal) base vectors in the two systems of coordinates are linked by er i and ˜xi could be two Cartesian coordinate systems, one moving at a con-stant velocity relative to the other, or xi could be Cartesian coordinates and ˜xi spherical polar coordinates whose origins are coincident and in relative rest. We shall see er and eθ in terms of their cartesian components along i and j. Define a spherical data set. The cylindrical coordinate system basically is a combination of the polar coordinate system xy ¡ plane with an additional z ¡ coordinate vertically. This system is a generalization of polar coordinates to three dimensions by superimposing a height axis. Conversion between cylindrical and Cartesian coordinates Cylindrical coordinates are obtained from Cartesian coordinates by replacing the x and y coordinates with polar coordinates r and theta and leaving the z coordinate unchanged. Expressed in Cartesian coordinates, a vector is defined in terms of Jul 22, 2020 · This blog will explain how to create a Stacked Contour Plot if user has data with Cylindrical Coordinates(ro, theta and Z), It will involve the following steps: interpolate data and convert data from cylindrical coordinates to Cartesian coordinates; Use XYZ gridding to convert xyz data into matrix; Clip data in matrix with circle. 2 Cylindrical Coordinates We first choose an origin and an axis we call the -axis with unit vector pointing in the increasing z-direction. 1 c oordinate systems a1. advanced. To use this calculator, a user just enters in the (r, φ, z) values of the cylindrical coordinates and then clicks 'Calculate', and the cartesian coordinates will be automatically computed and Jul 11, 2018 · Understand thoroughly about the Conversion between Cylindrical & Cartesian systems for Electromagnetism. We should bear in mind that the concepts covered in Chapter 1 and demonstrated in Cartesian coordinates are equally applicable to other systems of coordinates. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system Another way of looking at it is that we take polar coordinates $$(r,\theta)$$ and slap on the rectangular coordinate z to the end to get $$(r,\theta,z)$$ and call this cylindrical coordinates. The cylindrical coordinates (r,θ,z) are related to the Cartesian coordinates (x1,x2,x3) by the following relations r = x2 1 +x 2 2 1/2, θ = tan−1 x2 x1, z = x3, and x1 = rcosθ, x2 = rsinθ, x3 = z. I know the material, just wanna get it over with. bjc a2. In a cylindrical coordinate system, the location of a three-dimensional point is decribed with the first two dimensions described by polar coordinates and the third dimension described in distance from the plane containing the other two axes. In the last two sections of this chapter we’ll be looking at some alternate coordinate systems for three dimensional space. And these coordinates are called Cartesian coordinates, named for Rene Descartes because he's the guy that came up with these. Related Calculators: You can always start in Cartesian because the kinetic energy is a scalar and thus independent of the coordinate system in which you choose to evaluate it, although scalar products are most easily computed in Cartesian coordinates. (1a): Triple integral in Cartesian coordinates x,y,z (1b): Triple integral in cylindrical coordinates r,theta,z (2a): Triple integral in cylindrical coordinates r,theta,z (2b): Triple integral in spherical coordinates rho,phi,theta Cylindrical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. Coordinates in the GIS are measured from the origin point. 25in}z = z\] In order to do the integral in cylindrical coordinates we will need to know 2 We can describe a point, P, in three different ways. To convert a point from Cartesian coordinates to cylindrical coordinates, use equations and In the spherical coordinate system, a point in space is represented by the ordered triple where is the distance between and the origin is the same angle used to describe the location in cylindrical coordinates, and is the angle formed by the positive z Express A using Cartesian coordinates and spherical base vectors. For the x and y components, the transormations are ; inversely, . I have to prove it by simple geometry & calculus, without using jacobian or linear algebra (basis). in 6. As shown in Figure 1-2a, any point in space is defined by the intersection of the three perpendicular surfaces of a circular It is easier to consider a cylindrical coordinate system than a Cartesian coordinate system with velocity vector V=(ur,u!,uz) when discussing point vortices in a local reference frame. It is simplest to get the ideas across with an example. It is good to begin with the simpler case, cylindrical coordinates. Under the formula of the stress tensor of the cylindrical wall under the polar coordinate system, the 30 Mar 2016 Cylindrical Coordinates. Although we have considered the Cartesian system in Chapter 1, we shall consider it in detail in this chapter. Convert the cylindrical coordinates defined by corresponding entries in the matrices theta, rho, and z to three-dimensional Cartesian coordinates x, y, and z. This is no longer the case in spherical! Cylindrical coordinates To get a third dimension, each point also has a height above the original coordinate system. Converting Polar Coordinates to Cartesian. Cartesian systems use linear distances while polar systems use radial and The cylindrical coordinate system extends polar coordinates into 3D by using the Cylindrical coordinates are defined with respect to a set of Cartesian In this lesson, we introduce two coordinate systems that are useful alternatives to Cartesian coordinates in three dimensions. any help ? Jul 07, 2009 · If you have any vector in Cartesian coordinates then to transform it to Cylindrical coordinates you use r = sqrt(x^2 + y^2) theta = atan(y/x) z = z That part is easy. edu. The following are the conversion formulas for cylindrical coordinates. Khan Academy is a 501(c)(3) nonprofit organization. May 11, 2019 · Approach 1 for deriving the Divergence in Cylindrical. If there’s a one to one mapping between coordinate systems, we can convert between them. For example, x, y and z are the parameters that define a vector r in Cartesian coordinates: r =ˆıx+ ˆy + ˆkz (1) Similarly a vector in cylindrical polar coordinates is described in terms of the parameters r, θ and z since a vector r can be written as r = rrˆ+ zˆk. system Homework This calculator allows you to convert between Cartesian, polar and cylindrical coordinates. Similarly, the angle that a line makes with the horizontal can be defined by the formula θ = tan-1(m), where m is the slope of the line. There are three commonly used coordinate systems: Cartesian, cylindrical and spherical. The scalar components can be expressed using Cartesian, cylindrical, or spherical coordinates, but we must always use Cartesian base vectors. Convert coordinates from Cartesian to spherical and back. Morrison, Michigan Technological University Cartesian Coordinates L ì̃ ë ë ì̃ ë ì ì̃ ë í ì̃ ì ë ì̃ ì ì ì̃ ì í ì̃ í ë ì í ì ì̃ í í M ë ì í L ä É È È È Ç 2 ò R ë ò T ò R ë ò U E ò R ì ò T Mar 04, 2017 · My question is: does it make a difference if I solve with 2-D cylindrical or 2-D cartesian coordinates and formulation of the Navier Stokes equation? If my mesh is 2-D in r and z, and the flow has no dependence, it seems that the cylindrical form should reduce to the cartesian form (because they can both equally describe my 2D mesh). Both cylindrical and Our page on Cartesian Coordinates introduces the simplest type of coordinate system, where the reference axes are orthogonal (at right angles) to each other. Syntax: set mapping {cartesian | spherical | cylindrical} A cartesian coordinate system is used by default. $\endgroup$ – paisanco Jun 14 '14 at 15:57 pol2cart. 1213 0 -5]' x = 4×1 1. To run this script: Download the attached ZIP folder containing the BAS script file and two SRF files: crv2xyz10. Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. 2 Cylindrical Coordinates The cylindrical system is used for problems involving cylindri-cal symmetry It is composed of: (1) the radial distance r2„0;1/, (2) the azimuthal angle, ˚2„0;2ˇ/, and z2. For example, you might be studying an object with cylindrical symmetry: uid ow in a pipe, heat ow in a metal rod, or light propagated through a cylindrical optical ber. The relationships between (x;y) and (r; ) are exactly the same as in polar coordinates, and the zcoordinate is unchanged. Using Cartesian coordinates on the plane, the distance between two points (x 1, y 1) and (x 2, y 2) is defined by the formula, which can be viewed as a version of the Pythagorean Theorem. Current Location > Math Formulas > Linear Algebra > Transform from Cartesian to Cylindrical Coordinate Transform from Cartesian to Cylindrical Coordinate Don't forget to try our free app - Agile Log , which helps you track your time spent on various projects and tasks, :) The "magnitude" of a vector, whether in spherical/ cartesian or cylindrical coordinates, is the same. In cylindrical coordinates, (r; ;z), the continuity equation for an incompressible uid is 1 r @ @r (ru r) + 1 r @ @ (u ) + @u z @z = 0 In cylindrical coordinates, (r; ;z), the Navier-Stokes equations of motion for an incompress-ible uid of constant dynamic viscosity, , and density, ˆ, are ˆ Du r Dt u2 r = @p @r + f r+ 52u r u r r2 2 r2 @u Online calculator for definite and indefinite multiple integrals using Cartesian, polar, cylindrical, or spherical coordinates. The rotated Cartesian coordinate method to remove the axial singularity of cylindrical coordinates in finite‐difference schemes for elastic and viscoelastic waves Mingwei Zhuang Department of Electronic Science, Institute of Electromagnetics and Acoustics, Xiamen University, Xiamen, 361005 China Converts from Cartesian (x,y,z) to Cylindrical (ρ,θ,z) coordinates in 3-dimensions. The spherical coordinates of a point are related to its Cartesian coordinates as follows: Online calculator for definite and indefinite multiple integrals using Cartesian, polar, cylindrical, or spherical coordinates. 2. Nov 20, 2009 · Converting to Cylindrical Coordinates. 12 Compute $\ds \int_{-3}^3\int_0^{\sqrt{9-x^2}} \sin(x^2+y^2)\,dy\,dx$ by converting to cylindrical coordinates. ] Show that your equation in step 5 is equivalent to r = c in cylindrical coordinates. Express the values from Steps 1 and 2 as a Cylindrical and spherical coordinates Review of Polar coordinates in IR2. The calculator converts cylindrical coordinate to cartesian or spherical one. , the vector connecting the origin to a general point in space) onto the - plane and the -axis. For example, the circle of radius 2 may be described as the set of all points whose coordinates x and y satisfy the equation x 2 + y 2 = 2 2 . The n- and t-coordinates move along the path with the particle Tangential coordinate is parallel to the velocity The positive direction for the normal coordinate is toward the center of curvature ME 231: Dynamics Path variables along the tangent (t) and normal (n) In 3D Cartesian coordinates, Burkhart addressed the definition, existence, and uniqueness of the DGF and derived asymptotic expansion formulae, applicable at distances far from the source. Recall that $$x=r*cos Cylindrical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. Ex. In[1]:= Oct 26, 2005 · Spherical coordinates, also called spherical polar coordinates (Walton 1967, Arfken 1985), are a system of curvilinear coordinates that are natural for describing positions on a sphere or spheroid. Cylindrical Coordinates. When converted into cartesian coordinates, the new values will be depicted as (X, Y, Z). If all you want is the ability to specify velocities in cylindrical coordinates, just pick this option in the boundary condition GUI. 1 As the cylinder had a simple equation in cylindrical coordinates, so does the sphere in spherical coordinates: \rho=2 is the sphere of radius 2. In this section, we provide a working definition of the DGF and a numerical method to calculate Θ in Cartesian and cylindrical coordinates. www. We shall choose coordinates for a point P in the plane z=zP as follows. In Cartesian coordinates, the three unit vectors are denoted i x, i y, i z. e. The formula for it is as follows: It’s important to take into account that the definition of \(\rho$$ differs in spherical and cylindrical coordinates. A Cartesian coordinate system (UK: / k ɑː ˈ t iː zj ə n /, US: / k ɑːr ˈ t i ʒ ə n /) is a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates, which are the signed distances to the point from two fixed perpendicular oriented lines, measured in the same unit of length. Consider a differential element in Cartesian coordinates… Write a Cartesian equation of the cylindrical surface of radius c in the left-hand figure above. What we’ll need: 1. θr Cylindrical coordinates just adds a z-coordinate to the polar coordinates (r,θ). The partial derivatives with respect to x, y and z are converted into the ones with respect to ρ, φ and z. When we know a point in Cartesian Coordinates (x,y) and we want it in Polar Coordinates (r,θ) we solve a right triangle with After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called 16 Jun 2018 PDF | On Jun 15, 2018, Audu Eliazar Elisha and others published Laplacian Equation: From Cartesian to Cylindrical Coordinate System | Find, Download scientific diagram | Transformation from Cartesian to cylindrical coordinates preserving the waveguide's width and length. Selecting Z outputs translation along the Z-axis in both Cartesian and cylindrical coordinate systems. The cylindrical coordinate system is a generalization of two-dimensional polar coordinates to three dimensions. r = square root of (x 2 + y 2) Θ = tangent inverse(y/x) z = z cylindrical,and spherical coordinates CM3110 Fall 2011Faith A. THETA is a counterclockwise angular displacement in radians from the positive x -axis, RHO is the distance from the origin to a point in the x-y plane, and Z is the height above cylindrical-coordinate wave equation 2 2 2 2 2 2 2 2 2 2 1 z q c t∂ ∂ + ∂ + ∂ = + ρ φ, (1) which allowed us to transform Eq. 2 Introduction Gradient of a scalar field Divergence of a vector field As shown below, the results for the scattering cross section computed using cylindrical coordinates agree well with the 3d Cartesian simulation. We can see here that r=2 and Examples of orthogonal coordinate systems include the Cartesian (or rectangular ), the cir- cular cylindrical, the spherical, the elliptic cylindrical, the parabolic The colored area under the picture is the unit area in polar coordinates. In the Transform Sensor dialog box, coordinates that make up more than one coordinate system appear only once. The second set of coordinates is known as cylindrical coordinates. 3 Resolution of the gradient The derivatives with respect to the cylindrical coordinates are obtained by differentiation through the Cartesian coordinates, @ @r D @x @r @ @x DeO rr Dr r; @ @˚ D @x @˚ @ @x DreO ˚r Drr ˚: Nabla may now be resolved on the Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. 5708 3. The polar coordinates of a point P = (x,y) in the first quadrant are given So I'll say that point has the coordinates, tells me where to find that point, negative 2, negative 5. . (The subject is covered in Appendix II of Malvern's textbook. Illustration of cylindrical coordinates illustrating the effect of changing each of the three cylindrical coordinates on the location of a point. 0000 The polar coordinates of a point P ∈ R2 is the ordered pair (r,θ) defined by the picture. The radial part of the solution of this equation is, unfortunately, not May 16, 2011 · Transformation of cartesian coordinates or rectangular coordinates to cylindrical coordinates: The cylindrical coordinates can be transformed to cartesian or rectangular coordinates and vice versa and the relations will be: x = rcos Θ. Continuing with our example, let's sketch the surface represented by z = x 2 Converts 3D rectangular cartesian coordinates to cylindrical polar coordinates. Draw solids bounded by quadric surfaces using Cartesian Coordinates; Polar Coordinates; Cylindrical Coordinates; Spherical Coordinates; Let us discuss all these types of coordinates are here in brief. Exercises. The vector field is already expressed with Cartesian base vectors, therefore we only need to change the Cartesian coordinates in each scalar component into spherical coordinates. The same is true of triple integrals. Referring to figure 2, it is clear that there is also no radial velocity. E 9. 3-D Cartesian coordinates will be indicated by $x, y, z$ and cylindrical coordinates with $r,\theta,z$. Select the appropriate separator: comma, semicolon, space or tab (use tab to paste data directly from/to spreadsheets). The painful details of calculating its form in cylindrical and spherical coordinates follow. The Cartesian coordinate system provides a straightforward way to describe the location of points in space. The inputs x, y (, and z) must be the same shape, or scalar. 1 Cartesian Coordinate System . Conversion from cartesian to spherical coordinates: Cartesian [x, y, z] Spherical [r, θ, φ] Conversion from spherical to cylindrical coordinates: Spherical [r, θ, φ] Cylindrical [ρ, φ', z'] ρ = r sin θ φ' = φ z' = r cos θ Conversion from cylindrical to spherical coordinates: Cylindrical [ρ, φ, z] Spherical [r, θ, φ'] θ = arctan Transform Cartesian coordinates to polar or cylindrical coordinates. 5–2) must be given. Ex 15. To find a1 requires a two step process: 1) Project xˆ onto the line formed by rˆ and its projection onto the xy plane Aug 28, 2012 · if i,j,k are unit vectors in cartesian system & e(r), e(θ), e(z) are unit vectors in cylindrical system, i have to show that- 'i'= 'e(r)' cosθ – 'e(θ)' sinθ 'j' = 'e(r)' sinθ+ 'e(r)' cosθ 'k'= 'e(r)' the quantities in '_ ' are vectors. r is the distance to the z-axis (0, 0, z). Faith A. The idea behind cylindrical and spherical coordinates is to use angles instead of Cartesian coordinates to specify points in three dimensions. In this coordinate system, a point P is represented by the triple (r; ;z) where r and are the polar coordinates of the projection of Ponto the xy-plane and zhas the same meaning as in Cartesian coordinates. therightgate. Since we assume (Δr)2 is negligable Three most common coordinate systems used in 3-dimensional representations are: a) Cartesian coordinates b) Cylindrical (polar) coordinates c) Spherical in two dimensions and cylindrical and spherical coordinates in three dimensions. xr x y zˆˆˆˆ ˆ ab c a11 1 1 Note: a1 is the projection of xˆ onto rˆ. Use the unit circle to get . cylindrical polar coordinates In cylindrical polar coordinates the element of volume is given by ddddvz=ρρϕ. So the cylindrical coordinates conversion equations are given in Table 1 and Figure 1 shows this relationship. d) x + y + z = 1 to spherical coordinates. Spherical Unit Vectors in relation to Cartesian Unit Vectors rˆˆ, , θφˆ can be rewritten in terms of xyzˆˆˆ, , using the following transformations: rx yzˆ sin cos sin sin cos ˆˆˆ Mar 28, 2019 · In 3D Cartesian coordinates, Burkhart addressed the definition, existence, and uniqueness of the DGF and derived asymptotic expansion formulae, applicable at distances far from the source. Matlab provides a simple utility for doing that. For a spherical coordinate system, the data occupy two or three columns (or using Code for converting Cartesian (x,y,z) to Cylindrical (ρ,θ,z) coordinates 2D/3D Hot Network Questions Do you really need to fire flashes regularly when not otherwise used? x, and y allow you to change (x, y) coordinates into polar . 3. Recall that the position of a point in the plane can be described using polar coordinates $(r,\theta)$. 7. (2) We now go through a separation-of-variable procedure similar to that which we carried out using Cartesian coordinates in But so are cylindrical coordinates (an extension of two-dimensional polar coordinates to three-dimensional) and spherical polar coordinates. 1. Some surfaces and volumes are more easily (simply) described in cylindrical coordinates. plot(cartesian, We describe three different coordinate systems, known as Cartesian, cylindrical and spherical. Given the azimuthal sweep around the z axis theta as well as the radius of the cylinder r, the Cartesian co-ordinates within a cylinder is defined as: x = r*cos(theta) y = r*sin(theta) z = z If data are provided to splot in spherical or cylindrical coordinates, the set mapping command should be used to instruct gnuplot how to interpret them. Cylindrical Coordinates Orientation relative to the Cartesian standard system: The origins and z axes of the cylindrical system and of the Cartesian reference are coincident. 22 Jul 2014 This video explains how to convert rectangular coordinates to cylindrical coordinates. 9 Coordinate Systems in Space. f) ρsin θ = 1 to Cartesian coordiantes. theta describes the angle relative to the positive x-axis. In such cases, one has to first transform these coordinates into Cartesian coordinate system (X,Y, Z) and then only molecular graphics software can be used to visualize these molecules. Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed in Conversion between Cylindrical and Cartesian Coordinates: x = r cos θ = 4 cos 2 π 3 = −2 y = r sin θ = 4 sin 2 π 3 = 2 3 z = −2. $\begingroup$ The OP does need to compute the unit vectors in cylindrical coordinates and use the divergence, curl and Laplacian in cylindrical coordinates to solve Maxwell's equations, but the question was how to transform the tensor. Initializes a set of Cartesian coordinates from the provided set of Cylindrical coordinates. A very common case is axisymmetric flow with the assumption of no tangential velocity ($$u_{\theta}=0$$), and the remaining quantities are independent of $$\theta$$. Using change-of-coordinate functions to change from a known parameterization to another Like polar coordinates, cylindrical coordinates will be useful for describing shapes in that are difficult to describe using Cartesian coordinates. Consider the case when a three dimensional region $$U$$ is a type I region , i. any straight line parallel to the $$z$$-axis intersects the boundary of the region Compute Areas and Volumes in Non-Cartesian Coordinates The "nut" defined by revolving the curve about the axis can be easily parameterized in cylindrical coordinates. If , , and are smooth scalar, vector and second-order tensor fields, then they can be chosen to be functions of either the Cartesian coordinates , , and , or the corresponding real numbers , , and . Cylindrical and spherical coordinates Recall that in the plane one can use polar coordinates rather than Cartesian coordinates. theta = [0 pi/4 pi/2 pi]' theta = 4×1 0 0. com. By using this website, you agree to our Cookie Policy. A circular cylindrical surface r = r 1; A half-plane containing the z-axis and making angle φ = φ 1 with the xz-plane; A plane parallel to the xy-plane at z = z 1 Sep 19, 2014 · Cylindrical Coordinates in Matlab. (ρ, φ, z) is given in cartesian coordinates by: Feb 17, 2016 · Representing 3D points inRepresenting 3D points in Cylindrical Coordinates. Review of Cylindrical Coordinates. 1) are not convenient in certain cases. Mar 14, 2015 · Coordinates for DNA are usually given in the ‘Cylindrical polar coordinate system’ because of its helical symmetry. Cylindrical polar coordinates The cylindrical polar coordinates ρϕ,,z are given, in terms of the rectangular cartesian coordinates x, y, z by z x y z x y zz = = = ρϕ ρϕ cos sin. This coordinate system works best when integrating cylinders or cylindrical-like objects. Unzip the folder. a. e) r = 2sinθ to Cartesian coordinates. to_cartesian These formulas are automatically used if we ask to plot the grid of spherical coordinates in terms of Cartesian coordinates: sage: spherical. After rectangular (aka Cartesian) coordinates, the two most common an useful coordinate systems in 3 dimensions are cylindrical coordinates (sometimes called cylindrical polar coordinates) and spherical coordinates (sometimes called spherical polar coordinates). 1213 0 -5. Dec 27, 2016 · How do you find the rectangular coordinates if you given the cylindrical coordinate #(5, pi/6, 5)#? See all questions in Converting Coordinates from Rectangular to Polar Impact of this question The $(x,y,z)$ ECEF cartesian coordinates can be expressed in the ellipsoidal coordinates $(\varphi, \lambda, h)$, where $\varphi$ and $\lambda$ are, respectively, the latitude and longitude from the ellipsoid, and $h$ the height above it. [THETA,RHO,Z] = cart2pol(X,Y,Z) transforms three-dimensional Cartesian coordinates stored in corresponding elements of arrays X, Y, and Z, into cylindrical coordinates. Jul 21, 2020 · Cylindrical coordinates are a generalization of two-dimensional polar coordinates to three dimensions by superposing a height (z) axis. Scale factors for each component's direction. Spherical coordinates in IR3. If a third axis, z (height), is added to polar coordinates, the coordinate system is referred to as cylindrical coordinates (r, θ, z). Sep 20, 2016 · While Cartesian 2D coordinates use x and y, polar coordinates use r and an angle, $\theta$. In this approach, you start with the divergence formula in Cartesian then convert each of its element into the cylindrical using proper conversion formulas. The 2d nodes do match exactly. Transform Cartesian coordinates to polar or cylindrical coordinates. Its elements, however, are something of a cross between the polar and Cartesian coordinates systems. 30 Jan 2020 In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a 26 May 2020 The third equation is just an acknowledgement that the z z -coordinate of a point in Cartesian and polar coordinates is the same. 36: Cylindrical Coordinates 1. Jun 01, 2018 · Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. (1) The (orthogonal) base vectors in the two systems of coordinates are linked by er Figure B. A set of equations relating the Cartesian coordinates to cylindrical coordinates: 3. 7. Click to copy this Sketching the Surface in Cartesian Coordinates. [i] Fill in the blanks: • If I convert 1 = x2 - y2 from Cartesian to cylindrical coordinates, I get (Simplify. The coordinate systems allow the geometrical problems to be converted into a numerica When given Cartesian coordinates of the form to cylindrical coordinates of the form , the first and third terms are the most straightforward. \[x = r\cos \theta \hspace{0. Example 14. Let be a subset of . The chapter introduces functions to deal with elasticity coefficients, strain-displacement relations, constitutive relations, and equilibrium and Nov 08, 2011 · This is not a cork screw at all! The problem is that plot3 expects cartesian coordinates, but we plotted cylindrical coordinates. Sometimes, employing angles can make mathematical Cartesian coordinates. 1 c ylindrical coordinates a1. 5–2 Cylindrical transformation. Oct 12, 2018 · The Cartesian coordinate system plots a point as $(x,y,z)$, where $x,y$ and $z$ are perpendicular distances of the point measured from the planes $y-z, z-x$ and $x-y$ respectively as you may a Here we use the identity cos^2(theta)+sin^2(theta)=1. ΔA=(2rΔr+Δr2). Conversion between cylindrical and Cartesian coordinates Section 13. In this section, you will compare grid surfaces in Cartesian, cylindrical, and spherical coordinates. Before going through the Carpal-Tunnel causing calisthenics to . EX 4Make the required change in the given equation (continued). For example, to change the polar coordinate . Convert Cartesian coordinates (x, y, z) to cylindrical coordinates (radius, azimuth, z). (r +Δr)2−π(r)2). The global (X, Y, Z) coordinates of the two points defining the axis of the cylindrical system (points a and b as shown in Figure 2) must be given. Syntax [X,Y] = pol2cart(THETA,RHO) [X,Y,Z] = pol2cart(THETA,RHO,Z) Description [X,Y] = pol2cart(THETA,RHO) transforms the polar coordinate data stored in corresponding elements of THETA and RHO to two-dimensional Cartesian, or xy, coordinates. 6. Thus, is the perpendicular distance from the -axis, and the angle subtended between the projection of the radius vector (i. Now recall our first example, where we graphed the surface (in Cartesian coordinates) defined by the equation . However, there is a large discrepancy in performance: for a single Intel Xeon 4. scale_factors (self). ) • When I convert the point P(k, 0,0) for k > 0 from cylindrical coordinates to Cartesian coordinates, I get • Working with spherical coordinates, when I sketch the graph of p = k for k > 0, the shape of the graph is when I sketch the graph of o= k for 0 < k < TT, k # 1/2, the Using these infinitesimals, all integrals can be converted to cylindrical coordinates. Midpoint formula Cylindrical coordinates To get a third dimension, each point also has a height above the original coordinate system. Δθ2π. Then, polar coordinates (r; ) are de ned in IR2 f(0;0)g, and given by r= p x2 The Cartesian, or rectangular, coordinate system is the most widely used coordinate system. Δθ2. 2. Convert quadric surfaces in cylindrical or spherical coordinates to Cartesian and identify. From polar coordinates. The true origin point (0, 0) may or may not be in the proximity of the map data you are using. Care should be taken, however, when calculating . Cartesian (double[] elements) Initializes a set of Cartesian coordinates from the first 3 consecutive elements in the provided array. Cartesian coordinates in the figure below: (2,3) A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. First I’ll review spherical and cylindrical coordinate systems so you can have them in mind when we discuss more general cases. In this chapter we will describe a Cartesian coordinate system and a cylindrical coordinate system. 0000 2. Cylindrical coordinates is a method of describing location in a three-dimensional coordinate system. In this handout we will find the solution of this equation in spherical polar coordinates. Someone please help me do this? I don't need you to solve the question completely, I just want help in how to solve it and to be shown the right direction because I'm completely lost! Fluent environment supports cylindrical and Cartesian coordinates. E Figure 11. Elasticity equations in cylindrical polar coordinates 1. 1 Spherical coordinates Figure 1: Spherical coordinate system. Examples: planes parallel to coordinate planes, cylindrical parame- terization of cylinder, and spherical parameterization of sphere. Figure 2. 1416 The focus of this chapter is on the governing equations of the linearized theory of elasticity in three types of coordinate systems, namely, Cartesian, cylindrical, and spherical coordinates. x = [1 2. Conventions. Using cylindrical coordinates can greatly simplify a triple integral when the region this with cylindrical coordinates is much easier than it would be for cartesian Abstract: Application peculiarities of the Green's functions method for Cartesian, cylindrical and spherical coordinate system are under consideration. Converts from Cartesian (x,y,z) to Cylindrical (ρ,θ,z) coordinates in 3-dimensions. Triple Integrals in Cylindrical Coordinates It is the same idea with triple integrals: rectangular (x;y;z) coordinates might not be the best choice. We want to rotate the above so that the h axis is aligned with the arbitrary axis (Ax, Ay, Az) in other words we want to lookat the point (Ax, Ay, Az) see lookat Next we have a diagram for cylindrical coordinates: And let's not forget good old classical Cartesian coordinates: These diagrams shall serve as references while we derive their Laplace operators. The cylindrical coordinates of a point in $$\R^3$$ are given by $$(r,\theta,z)$$ where $$r$$ and $$\theta$$ are the polar coordinates of the point $$(x, y)$$ and $$z$$ is the same $$z$$ coordinate as in Cartesian coordinates. We introduce cylindrical coordinates by extending polar coordinates with theaddition of a third axis, the z-axis,in a 3-dimensional right-hand coordinate system. Jul 23, 2020 · Spherical Coordinates. Also the axis vectors depend on the same variable (in this case φ) which makes for interesting derivatives as we will see in a moment. The next step is to develop a technique for transforming spherical coordinates into Cartesian coordinates, and vice-versa. 13 Compute $\ds \int_{0}^a\int_{-\sqrt{a^2-x^2}}^0 x^2y\,dy\,dx$ by converting to cylindrical coordinates. Figure 3. That supposed to be superposition of a vortex and source. Unfortunately, there are a number of different notations used for the other two coordinates. Since the graph of this equation is a surface formed by revolving a curve about the z-axis, it might be better to use cylindrical coordinates. The origin of the local coordinate system is at the node of interest. ΔA=(π. Note. Get the free "Triple Integral - Cylindrical" widget for your website, blog, Wordpress, Blogger, or iGoogle. Another necessary information for Definition. If you pipe is aligned with the Z axis, you'll already have access to these. Cartesian coordinates. To use the plot3 function we must convert the cylindrical coordinates to cartesian coordinates. 1 Cylindrical Coordinates Free Cartesian to Polar calculator - convert cartesian coordinates to polar step by step This website uses cookies to ensure you get the best experience. Generally, x, y , and z are used in Cartesian coordinates and these are replaced by r, θ , and z . Using the Cartesian coordinate system, geometric shapes (such as curves) can be described by Cartesian equations — algebraic equations involving the coordinates of the points lying on the shape. What I take away from your answer is that with FEM I should stay in cartesian coordinates, b/c curvilinear coordinates introduce messy artifacts. Figure 1: A point expressed in cylindrical coordinates. Express A using cylindrical coordinates and cylindrical base vectors. For example, the mapping between spherical polar coordinates and Cartesian coordinates uses these equivalences: Cartesian coordinate system top: two-dimensional coordinate system bottom: three-dimensional coordinate system n. Vector. When transforming from Cartesian to cylindircal, x and y become their polar counterparts. Find the y value. Now we compute compute the Jacobian for the change of variables from Cartesian coordinates to spherical coordinates. In polar coordinates, if ais a constant, then r= arepresents a circle Hi, this is module four of two dimensional dynamics, our learning outcomes for today are to describe a rectangular Cartesian coordinate system, a cylindrical coordinate system and to describe the kinematic relationships of position and velocity in a tangential and normal coordinate system, so the one you are probably most familiar with for studying curvilinear motion, or curvilinear motion of One example is the Z-coordinate, which exists in both Cartesian and cylindrical systems. Input array must have a length of 3 and be in the correct order. The radial, tangential, and axial directions must be defined based on the original coordinates of each node in the node set for which the transformation is invoked. µ is called the \polar angle", the \azimuthal angle". Transform polar or cylindrical coordinates to Cartesian. This cylindrical coordinates converter/calculator converts the rectangular (or cartesian) coordinates of a unit to its equivalent value in cylindrical coordinates, according to the formulas shown above. Triple integrals in spherical coordinates Our mission is to provide a free, world-class education to anyone, anywhere. cartesian laplacian. For a 2D vortex, uz=0. NumPy Random Object Exercises, Practice and Solution: Write a NumPy program to convert cartesian coordinates to polar coordinates of a random 10x2 matrix representing cartesian coordinates. The position vector in cylindrical coordinates becomes r = rur + zk. Is it implicitly set to y=0? 2) This is basically a test problem I wanted to understand before continuing with a more complicated two-phase flow problem in spherical coordinates. Vectors are defined in cylindrical coordinates by (ρ, φ, z), where ρ is the length of the vector projected onto the xy-plane, φ is the angle between the projection of the vector onto the xy-plane (i. Jan 09, 2016 · If (a,b) is a are the coordinates of a point in Cartesian Plane, u is its magnitude and alpha is its angle then (a,b) in Polar Form is written as (u,alpha). B. coordinates (x;y) to polar coordinates (r; ). Appreciate your help! I have actually already came across the links. Free Polar to Cartesian calculator - convert polar coordinates to cartesian step by step This website uses cookies to ensure you get the best experience. As we have seen earlier, in two-dimensional space a point with rectangular coordinates can be identified with in polar coordinates and vice versa, where and are the relationships between the variables. from publication: Effect of Cartesian and Polar coordinate converting. A coordinate system in which the coordinates of The term "Cartesian coordinates" is used to describe such systems, and the values of the three coordinates unambiguously locate a point in space. It presents equations for several concepts that have not been covered yet, but will be on later pages. In this activity we will show that a suitable change of coordinates can greatly imporve the look of a surface in three-space. The cylindrical coordinate system is convenient to use when there is a line of symmetry that is defined as the z axis. When we expanded the traditional Cartesian coordinate system from two dimensions to three, we simply added a new 26 Apr 2018 Cartesian and polar coordinates SVG has a coordinate system that has its origin in the top-left corner, with the X coordinate increasing as we 20 Mar 2017 Cylindrical and Spherical Coordinates θr θr (r,θ,z) 14 Example: Find the cylindrical coordinates of the point (1,2,3) in Cartesian Coordinates 20 Nov 2009 Its form is simple and symmetric in Cartesian coordinates. Likewise, if we To Cartesian coordinates. Convert the cylindrical coordinates to cartesian coordinates in Cylindrical vs. Cartesian Coordinates. Cylindrical Coordinates In the cylindrical coordinate system, , , and , where , , and , , are standard Cartesian coordinates. Cartesian Cylindrical Spherical Cylindrical Coordinates x = r cosθ r = √x2 + y2 y = r sinθ tan θ = y/x z = z z = z Jul 27, 2020 · concept of cartesian coordinates system, its axis, its variables and ranges of variables 5. Aug 10, 2016 · Integration in cylindrical coordinates (r, \\theta, z) is a simple extension of polar coordinates from two to three dimensions. In such a coordinate system you can calculate the distance between two points and perform operations like axis rotations without altering this value. We start from this step: From rectangular coordinates, the arc length of a parameterized function is. y = rsin Θ. A: The reason we only use Cartesian base vectors for constructing a position vector is that Cartesian base vectors are the only base vectors whose directions are fixed—independent Mar 02, 2013 · Cartesian Coordinates vs Polar Coordinates In Geometry, a coordinate system is a reference system, where numbers (or coordinates) are used to uniquely determine the position of a point or other geometric element in space. In the cylindrical coordinate system, a point P (x;y;z); whose Cartesian A general system of coordinates uses a set of parameters to define a vector. D. Think of coordinates as different ways of expressing the position of the vector. The variable ρ is the distance of a coordinate point from the z Cartesian axis, and φ is its azimuthal angle. Each point is determined by an angle and a distance relative to the zero axis and the origin. The coordinate systems allow the geometrical problems to be converted into a numerica Figure 1. zip. Cartesian coordinates: If we wanted to write rˆˆ, , θφˆ in terms of xyzˆˆˆ, , , we would need to use the angles of and . Either r or rho is used to refer to the radial coordinate and either phi or theta to the azimuthal coordinates. In cylindrical coordinates, they are i r, i, i z, and in spherical coordinates, i r, i, i. The following equations describe the relationship between a Cartesian coordinate and a cylindrical coordinate: x = · cos, y = · sin, z = z Elasticity equations in cylindrical polar coordinates 1. In the Cartesian system the coordinates are perpendicular to one another with the same unit length Convert from rectangular to spherical coordinates. Cylindrical coordinates are a generalization of two-dimensional polar coordinates to three dimensions by superposing a height (z) axis. For instance when integrating vector function in Cartesian coordinates we can take the unit vectors outside the integral, since they are constant. Cartesian coordinates consist of a set of mutually perpendicular axes, which intersect at a 1-1-2 Circular Cylindrical Coordinates . Purpose of use Too lazy to do homework myself. Therefore, a Cartesian coordinate system is used, where the origin (0, 0) is toward the lower left of the planar section. May 18, 2020 · Cartesian coordinates (Section 4. 7854 1. The level surface of points such that z ˆz z=zP define a plane. Thanks for your time. Find more Mathematics widgets in Wolfram|Alpha. Explanation: A polar coordinate is in the form (r,θ) , where r is the distance from the origin and θ is the corresponding angle. Working in cylindrical coordinates is essentialy the same as working in polar coordinates in two dimensions except we must account for the z-component of the system. The Newtonian Constitutive Equation in Cartesian, Cylindrical, and Spherical coordinates Prof. Cylindrical just adds a z-variable to polar. The cylindrical radial coordinate is the perpendicular distance from the point to the z axis. person_outline Anton schedule 2018-10-22 12:49:06 This calculator is intended for coordinates transformation from / to the following 3d coordinate systems: As in the case of Cartesian coordinates, analytical solutions are readily obtained for unidirectional problems in cylindrical and spherical coordinates. The vector k is introduced as the direction vector of the z-axis. The cylindrical system is closely 26 Feb 2018 The calculation uses Cartesian coordinates. However the governing equations where i am using this velocity profile are written in spherical co ordinates. For example, there are times when a problem has Oct 22, 2019 · Although Cartesian coordinates can be used in three dimensions (x, y, and z), polar coordinates only specify two dimensions (r and θ). Thus, ! r V =ure ö r+u"e ö "+uze ö z=0e ö r+u"e ö "+0e ö z Processing Triple Integrals in Cartesian Coordinates Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one variable. 2 Cylindrical Coordinates These are coordinates for a three-dimensional space. If we start with the Cartesian equation of the sphere and substitute, we get the spherical equation: \eqalign{ x^2+y^2+z^2&=2^2\cr \rho^2\sin^2\phi\cos^2\theta+ \rho^2\sin^2\phi\sin^2\theta+\rho^2\cos^2\phi&=2^2\cr \rho^2\sin^2\phi Convert the three-dimensional Cartesian coordinates defined by corresponding entries in the matrices x, y, and z to cylindrical coordinates theta, rho, and z. 4 Relations between Cartesian, Cylindrical, and Spherical Coordinates. c) ρ = 2cos φ to cylindrical coordinates. Fields in Cylindrical Coordinate Systems. 22 Oct 2019 Coordinate systems provide a way to specify a point in space. The coordinate system uses the standard polar coordinate system in the x-y plane, utilizing a distance from the origin (r) and an angle (θ) of Transforms 3d coordinate from / to Cartesian, Cylindrical and Spherical coordinate systems. In polar coordinates we specify a point using the distance rfrom the origin and the angle with the x-axis. The transformation from Cartesian To convert the Cartesian nabla to the nabla for another coordinate system, say… cylindrical coordinates. Later in the course, we will also see how cylindrical coordinates can be useful in calculus, when evaluating limits or integrating in Cartesian coordinates is very difficult. I have vector in cartesian coordinate system: \\vec{a}=2y\\vec{i}-z\\vec{j}+3x\\vec{k} And I need to represent it in cylindrical and spherical coord. Slide 2 ’ & $% Polar coordinates in IR2 De nition 1 (Polar coordinates) Let (x;y) be Cartesian coordinates in IR2. When this is the case, Cartesian coordinates (x;y;z) are converted to cylindrical coordinates (r; ;z). ρ) and the positive x-axis (0 ≤ φ < 2π), z is the regular z-coordinate. An example is given below. May 26, 2020 · Section 1-12 : Cylindrical Coordinates As with two dimensional space the standard $$\left( {x,y,z} \right)$$ coordinate system is called the Cartesian coordinate system. Figure 1 illustrates the relation between Cartesian Oct 10, 2019 · Some of the Worksheets below are Cylindrical and Spherical Coordinates Worksheets, list of Formulas that you can use to switch between Cartesian and polar coordinates, identifying solids associated with spherical cubes, translating coordinate systems, approximating the volume of a spherical cube, … First off, the definition of your cylindrical co-ordinates is wrong. The conventional choice of coordinates is shown in Fig. Consider a cartesian, a cylindrical, and a spherical coordinate system, related as shown This converter/calculator converts a cartesian, or rectangular, coordinate to its equivalent cylindrical coordinate. 0000 Oct 13, 2010 · Homework Statement This seems like a trivial question (because it is), and I'm just not sure if I'm doing it right. 25in}y = r\sin \theta \hspace{0. The polar coordinates are defined in terms of r r r and θ \theta θ, where r r r is the distance of the point from the origin Using various functions, you can convert data between Spherical, Cartesian, and Cylindrical coordinate systems. For example, there are different languages in which the word "five" is said differently, but it is five regardless of whether it is said in English or Spanish, say. This tutorial will denote vector quantities with an arrow atop a letter, except unit vectors that define coordinate systems which will have a hat. INSTRUCTIONS: Choose the preferred angle units and enter the following: (r) Polar radius (Θ) Polar angle (z) Vertical offset; Cartesian from Cylindrical: The calculator returns the Cartesian coordinates (x, y, z). The global coordinates of the two points defining the axis of the cylindrical system (points a and b as shown in Figure 2. Here's what they look like: The Cartesian Laplacian looks pretty straight forward. 9: Cylindrical and Spherical Coordinates In the cylindrical coordinate system, a point Pin space is represented by the ordered triple (r; ;z), where rand are polar coordinates of the projection of Ponto the xy-plane and zis the directed distance from the xy-plane to P. The chain rule relates the Cartesian operators Cartesian, the circular cylindrical, and the spherical. Solutions to steady unidimensional problems can be readily obtained by elementary methods as shown below. Translating Spherical Coordinates to Cartesian Coordinates. The most well-known coordinate system is the Cartesian coordinate to use, where every point has an x-coordinate and y-coordinate expressing its horizontal position, and The Wave Equation in Cylindrical Coordinates Overview and Motivation: While Cartesian coordinates are attractive because of their simplicity, there are many problems whose symmetry makes it easier to use a different system of coordinates. x = r cos θ = 4 cos 2 π 3 = −2 y = r sin θ = 4 sin 2 π 3 = 2 3 z = −2. Recall: A grid surface of a 3-d coordinate system is a surface generated by holding one of the coordinates constant while letting the other two vary. Dec 10, 2019 · CFX actually gives you cylindrical coordinates (r and theta) about the Z axis of any coordinate system. The cylindrical (left) and spherical (right) coordinates of a point. Cylindrical coordinates are an alternative to the more common Cartesian coordinate system. to a rectangular coordinate, follow these steps: Find the x value. A natural extension of the 2d polar coordinates are cylindrical coordinates, since they just add a height value out of the xy Conversion from cartesian to spherical coordinates: Cartesian [x, y, z] Spherical [r, θ, φ] Conversion from spherical to cylindrical coordinates: Spherical [r, θ, φ] Cylindrical [ρ, φ', z'] ρ = r sin θ φ' = φ z' = r cos θ Conversion from cylindrical to spherical coordinates: Cylindrical [ρ, φ, z] Spherical [r, θ, φ'] θ = arctan Relationships in Cylindrical Coordinates This section reviews vector calculus identities in cylindrical coordinates. Morrison Continuity Equation, Cartesian coordinates ∂ρ ∂t + vx ∂ρ ∂x +vy ∂ρ ∂y +vz ∂ρ ∂z +ρ ∂vx ∂x + ∂vy ∂y + ∂vz ∂z = 0 Continuity Equation, cylindrical coordinates ∂ρ ∂t + 1 r ∂(ρrvr) ∂r + 1 r ∂(ρvθ) ∂θ + ∂(ρvz) ∂z = 0 This article contains a download link for a script which converts cylindrical or spherical coordinates to xyz coordinates for use in Surfer. Overrides: fromReference in class CoordinateSystem Parameters: tuples - float array in Cartesian coordinates ordered as x, y, z Returns: float array containing the radius, azimuth and z values Next: An example Up: Cylindrical Coordinates Previous: Regions in cylindrical coordinates The volume element in cylindrical coordinates. b) 2 2x + y- z2 = 1 to spherical coordinates. 1;1/, which can be thought of as height Transformation between Cartesian and Cylindrical Coordinates; Velocity Vectors in Cartesian and Cylindrical Coordinates; Continuity Equation in Cartesian and Cylindrical Coordinates; Introduction to Conservation of Momentum; Sum of Forces on a Fluid Element; Expression of Inflow and Outflow of Momentum; Cauchy Momentum Equations and the Navier Jan 24, 2017 · The basic form of heat conduction equation is obtained by applying the first law of thermodynamics (principle of conservation of energy). He's associating, all of a sudden, these relationships with points on a coordinate plane. coordinates and back again anytime. com/ for more stuff. Cylindrical coordinates to Cartesian coordinates. (1) into Z Z R R R T T c ′′ + Φ Φ′′ + = ′′+ ′ ′′ 2 1 1 1 1 ρ ρ. P = ( r, ) x y r 0 0 Theorem (Cartesian-polar transformations) The Cartesian coordinates of a point P = (r,θ) in the first quadrant are given by x = r cos(θ), y = r sin(θ). The cylindrical coordinate system is similar to that of the spherical coordinate system, but is an alternate extension to the polar coordinate system. Regardless, one should be able, in principle, to write down the coordinate transformations in the following form: May 16, 2011 · I need to work out how to convert phi = pi/3 from spherical coordinates to cartesian and cylindrical coordinates. So, coordinates are written as (r,$\theta\$, z). Velocity And Acceleration In Cylindrical Coordinates Velocity of a physical object can be obtained by the change in an object's position in respect to time. The relations used for the conversion of the coordinates of the point from the Cartesian coordinate system to the cylindrical coordinate system are: In many problems involving cylindrical polar coordinates, it is useful to know the line and volume elements; these are used in integration to solve problems involving paths and volumes. which means that y = 1. When we get to triple integrals, some integrals are more easily evaluated in cylindrical coordinates and you will even have some integrals that can't be evaluated in rectangular coordinates but can be in cylindrical. The above result is another way of deriving the result dA=rdrd(theta). 1 4/6/13 a ppendix 1 e quations of motion in cylindrical and spherical coordinates a1. ) This is intended to be a quick reference page. Rectangular coordinates are depicted by 3 values, (X, Y, Z). Circular Cylindrical Coordinates System concept of cylindrical coordinates system, its axis, its Cylindrical coordinates are a simple extension of the two-dimensional polar coordinates to three dimensions. Move the sliders to convert cylindrical coordinates to Cartesian coordinates for a comparison. 11 Jul 2018 about the Conversion between Cylindrical & Cartesian systems for Electromagnetism. Cylindrical coordinates definition, a member of a system of coordinates for locating a point in space by its polar coordinates and its perpendicular distance to the polar plane. 0000 the part of the solution depending on spatial coordinates, F(~r), satisfies Helmholtz’s equation ∇2F +k2F = 0, (2) where k2 is a separation constant. When we talk about the point with coordinates (x,y,z)'' or `the surface with equation f(x,y,z)'', we will always have in mind cartesian coordinates. Site: http://mathispower4u. Cylindrical Coordinates. Circular cylindrical coordinates use the plane polar coordinates ρ and φ (in place of x and y) and the z Cartesian coordinate. – Cartesian coordinates – Cylindrical coordinates – Spherical coordinates. 2GHz processor, the runtime of the cylindrical simulation is nearly 90 times shorter than the 3d simulation. Sponsored Links. [Hint: Think about the distance of any point ( x , y , z ) on the cylinder from the z -axis. What is more challenging is determining the velocity vector in Cylindrical coordinates if you have a position in Cylindrical coordinates as a function of time. The Cartesian Nabla: 2. What is dV in cylindrical coordinates? Well, a piece of the cylinder looks like so which tells us that We can basically think of cylindrical coordinates as polar coordinates plus z. There's three independent variables, x, y, and z. B-5 Feb 24, 2015 · Preliminaries. The z component does not change. For example, in the Cartesian coordinate system, the cross-section of a cylinder concentric with the $$z$$-axis requires two coordinates to describe: \(x Cylindrical coordinates are depicted by 3 values, (r, φ, Z). cartesian to cylindrical coordinates
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https://math.stackexchange.com/questions/520657/a-conjectured-closed-form-of-int-limits-0-infty-fracx-1-sqrt2x-1-ln-l/2081059 | # A conjectured closed form of $\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$
Consider the following integral: $$\mathcal{I}=\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.$$
I tried to evaluate $\mathcal{I}$ in a closed form (both manually and using Mathematica), but without success.
However, if WolframAlpha is provided with a numerical approximation $\,\mathcal{I}\approx 3.2694067500684...$, it returns a possible closed form: $$\mathcal{I}\stackrel?=\frac\pi{2\,\ln^2 2}.$$ Further numeric caclulations show that this value is correct up to at least $10^3$ decimal digits. So, I conjecture that this is the exact value of $\mathcal{I}$.
Question: Is this conjecture correct?
• After a substitution of y = 2^x-1, does this not reduce to knowledge about identities of polylogaritms ?
– mick
Oct 9 '13 at 21:41
Sub $u=\log{(2^x-1)}$. Then $x=\log{(1+e^u)}/\log{2}$, $dx = (1/\log{2}) (du/(1+e^{-u})$. The integral then becomes
\begin{align}\frac{1}{\log{2}} \int_{-\infty}^{\infty} \frac{du}{1+e^{-u}} e^{-u/2} \frac{\frac{\log{(1+e^u)}}{\log{2}}-1}{u} = \frac{1}{2\log^2{2}} \int_{-\infty}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{\frac{1}{2\log^2{2}} \int_{-\infty}^{0} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}}_{u\rightarrow -u} \\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{-\frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^{-u})}-\log{2}}{u}}_{\log{(1+e^{-u})} = \log{(1+e^u)}-u}\\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ \end{align}
The nasty pieces of the integral cancel, and we are left with
$$\frac{1}{2\log^2{2}}\int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} = \frac{\pi}{2 \log^2{2}}$$
as correctly conjectured.
With the change of variables $z \equiv 2^{x} - 1\yy x = \ln\pars{1 + z}/\ln\pars{2},\ {\cal I}$ is reduced to $${\cal I} = {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{\infty}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z$$ Now, we split the integral from $\pars{0, 1}$ and from $\pars{1, \infty}$. In the second one, we makes the change $z \to 1/z$ such that we are left with an integration over $\pars{0, 1}$: \begin{align} {\cal I} &= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z + {1 \over \ln^{2}\pars{2}}\int\limits_{0}^{1} {\ln\pars{1 + 1/z} - \ln\pars{2} \over z^{-1/2}\,\pars{1 + 1/z}\,\bracks{-\ln\pars{z}}} \,{\dd z \over z^{2}} \\[3mm]&= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z - {1 \over \ln^{2}\pars{2}}\int\limits_{0}^{1} {\ln\pars{1 + z} - \ln\pars{z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}}\,\dd z \\[3mm]&= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{1 \over z^{1/2}\,\pars{1 + z}} \,\dd z\,, \quad \pars{~\mbox{Let's}\quad r \equiv z^{1/2}\yy\ z = r^{2}~} \\[3mm]&= {2 \over \ln^{2}\pars{2}} \underbrace{\quad\int\limits_{0}^{1}{\dd r \over r^{2} + 1}\quad} _{\ds{\arctan\pars{1}\ =\ {\pi \over 4}}} = \color{#ff0000}{\Large{\pi \over 2\ln^{2}\pars{2}}} \end{align}
Let $\tan^2t=2^x-1\;\Rightarrow\;x=\dfrac{\ln(1+\tan^2t)}{\ln 2}\;\Rightarrow\;dx=\dfrac{2\tan t\sec^2t\ dt}{(1+\tan^2t)\ln 2}$ and the corresponding region is $0<t<\dfrac\pi2$. Using identity $\sec^2t=1+\tan^2t$, then the integral turns out to be $$\mathcal{I}=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\frac{2\ln(\sec t)-\ln2}{\ln (\tan t)}\ dt.\tag1$$ Now, using property $$\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx$$ equation $(1)$ becomes $$\mathcal{I}=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\frac{2\ln(\csc t)-\ln2}{\ln (\cot t)}\ dt.\tag2$$ Adding $1$ and $2$ yields \begin{align} 2\mathcal{I}&=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{2\ln(\sec t)-\ln2}{\ln (\tan t)}+\frac{2\ln(\csc t)-\ln2}{\ln (\cot t)}\right)\ dt\\ &=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{2\ln\left(\dfrac{1}{\cos t}\right)-\ln2}{\ln (\tan t)}+\frac{2\ln\left(\dfrac{1}{\sin t}\right)-\ln2}{\ln \left(\dfrac{1}{\tan t}\right)}\right)\ dt\\ &=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{-2\ln(\cos t)-\ln2}{\ln (\tan t)}+\frac{2\ln(\sin t)+\ln2}{\ln (\tan t)}\right)\ dt\\ &=\frac{2}{\ln^22}\int_0^{\Large\frac\pi2}\frac{\ln(\sin t)-\ln(\cos t)}{\ln \left(\dfrac{\sin t}{\cos t}\right)}\ dt\\ \mathcal{I}&=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\ dt\\\\ \int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx&=\large\color{blue}{\frac{\pi}{2\ln^22}}.\qquad\qquad\qquad\blacksquare \end{align}
• Wow, this is remarkably brilliant! I'm impressed. Jun 10 '16 at 12:06
• This is really nice thanks
– Alex
Sep 4 '20 at 10:22
Substitute $(2^x-1) = t^2$ to get,
$\text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t$
Substitute $t \mapsto \dfrac{1}{t}$
$\implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}$
$\implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2}$
$\implies \text{I} = \dfrac{\pi}{2 \ln^2 2}$
• good one $(+1)$ Jul 9 '17 at 20:48
• And with the right substitution the integral just dissolves away. Nice work. Jan 25 '19 at 1:04
• It looks easy this way
– Alex
Sep 4 '20 at 10:22 | 2022-01-16T20:15:06 | {
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https://stats.stackexchange.com/questions/147255/drawing-n-intervals-uniformly-randomly-probability-that-at-least-one-interval-o | # Drawing n intervals uniformly randomly, probability that at least one interval overlaps with all others
Randomly draw $n$ intervals from $[0,1]$, where each end point A,B are selected from the uniform distribution between $[0,1]$.
What's the probability that at least one interval overlaps with all others?
• You can look at the probability that the last drawn $A_n$ is smaller than the minimum of all previously drawn $A$, and the probability that the last $B_n$ is greater than the maximum of all previously drawn $B$. This should be helpful. Then inflate the probability to account for the fact that we don't need the last one, but any one. (I don't have the time to work through it, but it looks like a fun little problem. Good luck!) Apr 20, 2015 at 7:02
• It may be somewhat surprising that (1) the answer does not depend on the distribution (only that it be continuous) and (2) for $n\gt 1$ it is constant!
– whuber
Apr 22, 2015 at 15:41
• Is this how the nth interval is construted: i) draw two numbers uniformly at random from [0,1], ii) let the smaller one be $A_n$ and the larger one $B_n$?
– KOE
Apr 23, 2015 at 21:36
This post answers the question and outlines partial progress toward proving it correct.
For $n=1$, the answer trivially is $1$. For all larger $n$, it is (surprisingly) always $2/3$.
To see why, first observe that the question can be generalized to any continuous distribution $F$ (in place of the uniform distribution). The process by which the $n$ intervals are generated amounts to drawing $2n$ iid variates $X_1, X_2, \ldots, X_{2n}$ from $F$ and forming the intervals
$$[\min(X_1,X_2), \max(X_1,X_2)], \ldots, [\min(X_{2n-1}, X_{2n}), \max(X_{2n-1}, X_{2n})].$$
Because all $2n$ of the $X_i$ are independent, they are exchangeable. This means the solution would be the same if we were randomly to permute all of them. Let us therefore condition on the order statistics obtained by sorting the $X_i$:
$$X_{(1)} \lt X_{(2)} \lt \cdots \lt X_{(2n)}$$
(where, because $F$ is continuous, there is zero chance that any two will be equal). The $n$ intervals are formed by selecting a random permutation $\sigma\in\mathfrak{S}_{2n}$ and connecting them in pairs
$$[\min(X_{\sigma(1)},X_{\sigma(2)}), \max(X_{\sigma(1)},X_{\sigma(2)})], \ldots, [\min(X_{\sigma(2n-1)}, X_{\sigma(2n)}), \max(X_{\sigma(2n-1)}, X_{\sigma(2n)})].$$
Whether any two of these overlap or not does not depend on the values of the $X_{(i)}$, because overlapping is preserved by any any monotonic transformation $f:\mathbb{R}\to\mathbb{R}$ and there are such transformations that send $X_{(i)}$ to $i$. Thus, without any loss of generality, we may take $X_{(i)}=i$ and the question becomes:
Let the set $\{1,2,\ldots, 2n-1, 2n\}$ be partitioned into $n$ disjoint doubletons. Any two of them, $\{l_1,r_1\}$ and $\{l_2,r_2\}$ (with $l_i \lt r_i$), overlap when $r_1 \gt l_2$ and $r_2 \gt l_1$. Say that a partition is "good" when at least one of its elements overlaps all the others (and otherwise is "bad"). As a function of $n$, what is the proportion of good partitions?
To illustrate, consider the case $n=2$. There are three partitions,
$$\color{gray}{\{\{1,2\},\{3,4\}\}},\ \color{red}{\{\{1,4\},\{2,3\}\}},\ \color{red}{\{\{1,3\},\{2,4\}\}},$$
of which the two good ones (the second and third) have been colored red. Thus the answer in the case $n=2$ is $2/3$.
We may graph such partitions $\{\{l_i,r_i\},\,i=1,2,\ldots,n\}$ by plotting the points $\{1,2,\ldots,2n\}$ on a number line and drawing line segments between each $l_i$ and $r_i$, offsetting them slightly to resolve visual overlaps. Here are plots of the preceding three partitions, in the same order with the same coloring:
From now on, in order to fit such plots easily in this format, I will turn them sideways. For instance, here are the $15$ partitions for $n=3$, once again with the good ones colored red:
Ten are good, so the answer for $n=3$ is $10/15=2/3$.
The first interesting situation occurs when $n=4$. Now, for the first time, it is possible for the union of the intervals to span $1$ through $2n$ without any single one of them intersecting the others. An example is $\{\{1,3\},\{2,5\},\{4,7\},\{6,8\}\}$. The union of the line segments runs unbroken from $1$ to $8$ but this is not a good partition. Nevertheless, $70$ of the $105$ partitions are good and the proportion remains $2/3$.
The number of partitions increases rapidly with $n$: it equals $1\cdot 3\cdot 5 \cdots \cdot 2n-1 = (2n)!/(2^nn!)$. Exhaustive enumeration of all possibilities through $n=7$ continues to yield $2/3$ as the answer. Monte-Carlo simulations through $n=100$ (using $10000$ iterations in each) show no significant deviations from $2/3$.
I am convinced there is a clever, simple way to demonstrate there is always a $2:1$ ratio of good to bad partitions, but I have not found one. A proof is available through careful integration (using the original uniform distribution of the $X_i$), but it is rather involved and unenlightening.
• Very cool. I have a hard time following what it means to "condition on the order statistics", would it be possible to add a line of intuition? Seems like a useful technique. I understand up to that the $X_i$ are exchangeable, indeed even $iid$, that that this allows us to consider any permutation.
– KOE
Apr 24, 2015 at 16:10
• @Student To "condition on" means to say, let's temporarily hold these values fixed and consider what we can learn from that. Later, we will let those values vary (according to their probability distribution). In this case, once we find that the answer is $2/3$ regardless of the fixed values of the order statistics, then we no longer have to carry out the second step of varying the order statistics. Mathematically, the order stats are a vector-valued variable $\mathbf{X}$ and the indicator of being good is $Y$, so $$\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y|\mathbf{X}))=\mathbb{E}(2/3)=2/3.$$
– whuber
Apr 24, 2015 at 16:36
• This very question came back on FiveThirtyEight The Riddler. Jun 2, 2020 at 9:22 | 2022-06-28T16:32:18 | {
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https://web2.0calc.com/questions/last-question_4 | +0
# last question
0
38
3
+81
How many different rational numbers between 1/1000 and 1000 can be written either as a power of 2 or as a power of 3, where the exponent is a (possibly negative) integer?
I got 31; is this correct?
mathbum Oct 6, 2018
#1
+1
$$2^9=512$$
$$3^6=729$$
Those are the largest positive powers of 2 and 3 under 1000.
$$2^{-9}=0.001953.....$$
$$3^{-6}=0.0013717.....$$
These are the smallest negative powers greater than $$\frac{1}{1000}$$
Therefore, we have $$9-(-9)+1=19$$ different powers of 2 between $$\frac{1}{1000}$$ and 1000.
We also have $$6-(-6)+1=13$$ different powers of 3 between $$\frac{1}{1000}$$ and 1000.
This gives us $$19+13=32$$ total integers...
32 is the answer I got but someone is going to have to check over my work :b
Guest Oct 6, 2018
#2
+2362
+1
The problem is symmetric in x vs. 1/x so we can find the number of integers between 0 and 1000 that are either a power of two or 3 and just double the answer.
Note that any non-zero power of 2 is even and any non-zero power of 3 is odd so these two sets are disjoint.
There are 10 powers of 2, 0-9, and 6 non-zero powers of 3 that are less than 100.
We use non-zero power for 3 since we only want to count 1 = 20 = 30 once
That gets us 16 from 0 to 1000. Doubling this we get 32 but we don't want to count $$1 = \dfrac{1}{1}$$
twice so we subtract 1 from this getting 31 as you found.
Rom Oct 6, 2018
edited by Rom Oct 6, 2018
edited by Rom Oct 6, 2018
edited by Rom Oct 6, 2018
#3
+93644
0
That is an interesting question and good answers from both guest and Rom.
Thanks :)
Melody Oct 6, 2018 | 2018-10-18T04:01:57 | {
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https://math.stackexchange.com/questions/4115542/probability-of-drawing-cards-that-sum-to-10-given-a-starting-card | Probability of drawing cards that sum to 10, given a starting card
Two people are playing a card game. They are using a reduced deck of cards, consisting of A, 2, 3, ..., 9 for each of the four suits (i.e. 36 cards). In this game an ace has a value of 1.
Player A deals a single card to themselves and Player B.
Player A has a 5 of diamonds.
Player B has a 7 of spades.
Player A then draws one card at a time trying to reach a sum of 10, with the starting card. If they go over 10, they lose, and the cards drawn are shuffled back into the deck.
What is the probability that A wins, and what is the probability that B wins. (Winning means they can form a sum of 10).
The answers given are 0.1536 for A, and 0.1468 for B.
I can get the answer for B as follows: $$\frac{nCr(4,1)}{nCr(34,1)}+\frac{nCr(4,1)}{nCr(34,1)}\times\frac{nCr(4,1)}{nCr(33,1)}\times2+\frac{nCr(4,3)}{nCr(34,3)}$$ which is the probability of a 3 + probability of a 2 and an A + probability of three Aces.
However, I can't get the answer for A, even trying very similar techniques.
• Do the players draw the cards alternatively? Does the game end if A or B first reach 10? Does the game end (without a winner) if they both go over 10?
– user
Apr 25, 2021 at 13:23
The answer for $$A$$ seems to have approximation error. Here is how I look at $$A$$ getting to sum of $$10$$.
In one draw - gets one of the remaining $$3$$ cards with face value $$5$$.
In two draws - $$(4,1)$$ or $$(3,2)$$
In three draws - $$(1, 1, 3)$$ or $$(2, 2, 1)$$
In four draws - $$(1, 1, 1, 2)$$
In five draws - $$(1, 1, 1, 1, 1)$$.
So desired probability $$= \displaystyle \small \frac{3}{34} + 2 \cdot 2! \big(\frac{4}{34}\big)^2 + 2 \cdot \frac{3!}{2!} \big(\frac{4}{34}\big)^3 + \frac{4!}{3!} \big(\frac{4}{34}\big)^4 + \big(\frac{4}{34}\big)^5$$
$$= \displaystyle \small \frac{437763}{2839714} \approx 0.154$$
• Thanks for your help. Though as the game is involving a deck of cards it's not possible to have five aces. Apr 26, 2021 at 4:42
• you are right but I am going by the question that "cards drawn are shuffled back into the deck" so I can draw $5$ Aces in $5$ draws with probability $(\frac{4}{34})^5$. Apr 26, 2021 at 7:00 | 2022-09-30T06:21:14 | {
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https://math.stackexchange.com/questions/1973627/help-explaining-structural-induction | # Help explaining Structural Induction
I am trying to wrap my head around structural induction. Can someone break it down and explain it around this problem?
Let S, a subset of $$\mathbb{N}*\mathbb{N}$$, be defined recursively by:
Base case: $$(0,0)$$ $$\in S$$
Constructor case: If $$(m,n) \in S$$, then $$(m+5,n+1) \in S$$
Prove that if $$(m,n) \in S$$, then m+n is a multiple of 3.
How is it different than normal induction (using this example please) and what is the point of a Constructor case? Can someone wright the proof out so i can see what this structural induction proof looks like?
• Her we are "performing induction" not on $\mathbb N$ but of $\mathbb N \times \mathbb N$, and not all pairs $(n,m)$ will satisfy bthe property : $(1,1) \notin S$ because $1+1$ is not a multiple of $3$. Oct 18, 2016 at 10:08
• Now for the inductive step (here the "constructor case") ; assume that the property holds for $(m,n)$ and show that it holds for $(m+5,n+1)$. Oct 18, 2016 at 10:12
• Here is a video attempting to explain structural induction: youtu.be/u21QV-MlVDY Aug 8 at 15:44
The set $$S$$ is defined recursively: certain base elements of $$S$$ are specified, in this case just the ordered pair $$\langle 0,0\rangle$$, and a rule is given that allows ‘new’ elements of $$S$$ to be constructed from ‘old’ ones. Here each ‘old’ element $$\langle m,n\rangle$$ gives rise to just one ‘new’ one, $$\langle m+5,n+1\rangle$$. Thus, in this case
$$S=\{\langle 0,0\rangle,\langle 5,1\rangle,\langle 10,2\rangle,\langle 15,3\rangle,\ldots\}\;.$$
There is also a rule, often (as in this case) left unstated, to the effect that the only members of $$S$$ are the objects that can be obtained by repeatedly applying the constructor rule(s) to the base elements.
We can show that every member of $$S$$ has some property $$P$$ by first verifying that each of the base elements has $$P$$ and then showing that the construction process preserves the property $$P$$: that is, if we apply the construction process to objects that have $$P$$, the new objects also have $$P$$. If we can do this, we can conclude by structural induction that every member of $$S$$ has $$P$$.
In your problem an ordered pair $$\langle m,n\rangle$$ has the property $$P$$ if and only if $$m+n$$ is a multiple of $$3$$. This is clearly the case for the one base element $$\langle 0,0\rangle$$: $$0+0=0=3\cdot 0$$ is a multiple of $$3$$. That’s the base case of your structural induction. For the induction step assume that $$\langle m,n\rangle\in S$$ has $$P$$, i.e., that $$m+n$$ is a multiple of $$3$$. When we apply the construction process to $$\langle m,n\rangle$$, we get the pair $$\langle m+5,n+1\rangle\in S$$, and we want to show that it also has $$P$$, i.e., that $$(m+5)+(n+1)$$ is a multiple of $$3$$. By hypothesis $$m+n=3k$$ for some integer $$k$$, so
$$(m+5)+(n+1)=m+n+6=3k+6=3(k+2)\;;$$
and $$k+2$$ is an integer, so $$(m+5)+(n+1)$$ is indeed a multiple of $$3$$. We’ve now shown
• that the base element $$\langle 0,0\rangle$$ has the desired property, and
• that the construction process preserves this property: when applied to a pair $$\langle m,n\rangle$$ such that $$m+n$$ is a multiple of $$3$$, it produces another pair whose components sum to a multiple of $$3$$.
These are the base case and induction step of a proof by structural induction; between them they constitute a proof that $$m+n$$ is a multiple of $$3$$ for each $$\langle m,n\rangle\in S$$.
• Thank you, that was a great explanation. It made it immediately obvious that induction on the natural numbers is a special case of structural induction where the constructor case is $k+1$ for every $k$ in $S$. Jan 9, 2017 at 22:05
• @jeremy: You’re welcome; I’m glad that it helped. Jan 9, 2017 at 22:07
Mathematical induction is defined over natural number and it is based on two fundamental facts :
• there is an "initial" number : $0$
• every number $n$ has a unique successor : $n+1$.
Structural induction generalize this type of proof to "structures" on which a well-founded partial order is defined, i.e.
• that have an "initial" or minimal element
and
• they have a partial order.
It applies to structures recursively defined (such as lists or trees). | 2022-08-14T22:43:52 | {
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https://math.stackexchange.com/questions/3100168/determining-if-sequence-has-upper-bound | # determining if sequence has upper bound
I am somewhat stuck in my calculations when determining if sequence has an upper bound.
The sequence $$x_n = \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n-1}+\frac{1}{2n}$$ Is equal to $$\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}})$$
And so I notice that all the denominators are greater than 1, which means that all terms in the parentheses are less than 1.
But how can I determine further if there is an upper bound?
• ... and there are $n$ of them. – Zeekless Feb 4 at 17:54
• @Zeekless so the sequence should be less than (1/n) * n = 1 is that correct? – F Wi Feb 4 at 18:02
• – Martin Sleziak Feb 5 at 8:22
The largest term is the first, so an obvious upper bound is to set all terms equal to the first one and get $$x_n < \frac{n}{n+1} <1.$$
You could also say that, since the last term is the smallest, one has $$x_n > \frac{n}{2n} = \frac 12,$$
which means that $$\frac 12 < x_n < 1, n \in \mathbb{N}$$.
By C-S $$\sum_{i=1}^n\frac{1}{n+i}=1+\sum_{i=1}^n\left(\frac{1}{n+i}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{i=1}^n\frac{i}{n+i}=$$ $$=1-\frac{1}{n}\sum_{i=1}^n\frac{i^2}{ni+i^2}\leq1-\frac{1}{n}\frac{\left(\sum\limits_{i=1}^ni\right)^2}{\sum\limits_{i=1}^n(ni+i^2)}=1-\frac{1}{n}\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=$$ $$=1-\frac{3(n+1)}{2(5n+1)}=\frac{7n-1}{10n+2}<\frac{7}{10}.$$ Actually, $$\ln2=0.6931...$$ Cauchy-Schwarz forever!
Actually, by calculus we can show that $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\frac{1}{n+i}=\ln2.$$
Notice the Riemann sum $$\frac1n\sum_{k=1}^n \frac1{1+k/n} < \int_0^1\frac{dt}{1+t} = \log 2$$
hint
For each $$n\ne 0$$,
$$\frac{1}{n+1}\le \frac{1}{n}$$ $$\frac{1}{n+2}\le \frac{1}{n}$$ ...
$$\frac{1}{2n}\le \frac 1n$$
You can finish. | 2019-05-20T22:32:22 | {
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https://math.stackexchange.com/questions/779510/how-many-solutions-does-the-equation-xyz-11-have | # How many solutions does the equation $x+y+z=11$ have? [duplicate]
How many solution does $x+y+z=11$ have where $x, y, z$ are non-negative integers. In light of the restrictions, its clear that $x,y,z \in \{0,1,2,..11\}$. So, at face value I would assign a value for $x$ and determine the different combinations that $y$ and $z$ can hold. For example,
For $x=0$, we have $y+z=11$. With writing them out I found that there are $12$ different assigned combinations for $y$ and $z$ that satisfy the equation. For $x=1$, I got $11$. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is $1+2+3+4+5+6+7..+12=78$. I was wondering if there is an easier method perhaps with combinations equation $C(a,b)$..?
• We have had too many questions of this type. – evil999man May 3 '14 at 12:13
• See my answer here:math.stackexchange.com/questions/689975/… – evil999man May 3 '14 at 12:15
• Guys, stop upvoting duplicate homework questions, seriously! – Alec Teal May 3 '14 at 12:18
• @Alec Lighten up. The OP shows plenty of effort, and arrives at the correct answer in doing so. It deserves an upvote. John likely didn't know that this is a classic sort of problem, and probably hasn't encountered it before. Nor that many similar questions can be answered by the same method. And if you believe it is a duplicate and should be slammed shut because of it, then why'd you answer it? Besides, stop playing the homework police on questions showing commendable levels of effort! – amWhy May 3 '14 at 12:24
• @amWhy Not thinking something deserves an upvote isn't the same as saying it deserves to be closed. When I upvote something, what I personally mean is "this is the kind of content I come to the site for". My preferred policy would for routine problems to be dealt with quickly and without fanfare (positive or negative), and to save upvotes (and therefore time on the front page) for the more original, thought-provoking questions. – Jack M May 3 '14 at 12:28
This is a version of the classic stars-and-bars problem in combinatorics.
For any pair of natural numbers $n$ and $k$, the number of distinct $n$-tuples of non-negative integers whose sum is $k$ is given by the binomial coefficient $$\binom{n + k - 1}{k}$$
Here, $n = 3$, and $k = 11$, giving you $$\binom{3 + 11 - 1}{11} = \binom{13}{11} = \dfrac{13\cdot 12}{2} = 6\cdot 13 = 78$$
• The link takes you to Wikipedia's Stars and Bars entry, where you'll find a nice explanation of why this works, using an example that's fleshed out nicely. – amWhy May 3 '14 at 12:17
• Using the formula makes things easy, but I don't see what the connection is. Even with using the stars and bars method, it seems rather inappropriate, due to the dependency of the variables. In other terms, if x takes a value it creates a dependency on the other variables. I would appreciate it if you could make the connection clearer. – John May 3 '14 at 13:41
• Did you read the Wikipedia entry? It elaborates on the "why's" of this formula. – amWhy May 3 '14 at 13:45
• I read it and I understood everything but when I went to this question, math.stackexchange.com/questions/322369/…, with the additional restrictions I started to doubt everything. None of the answers that used combinations were clear. – John May 3 '14 at 13:49
• Yes. I see you used $x, y, z$, so if only $x$ needs to be greater than or equal to 2, then the solution (with k=11 - 2 = 9) is $\binom{3 + 9 - 1}{9}$. – amWhy May 3 '14 at 15:17
imagine 11 balls in a row and two blocks which you will place somewhere. you insert the blocks before, after or between the balls and then you assign values to $x,y,z$ in the following way: $x$ is the number of balls from the beginning of the row up to the first block, $y$ the number of balls between the two blocks and $z$ number of balls from the second block up until the end of the row. you will easily see that the number of ways in which you can place the blocks is equal to the number of different triplets $x,y,z$. Do you know how to compute the number of possible distributions of blocks?
• call it stars and bars if you don't like blocks and balls, but this is the usual way to approach such problems – Alessandro Codenotti May 3 '14 at 12:15
• it's an answer explained in a tangible way. actually it's pretty much the same thing you gave as an answer although it's explained with words rather than a picture. – mm-aops May 3 '14 at 12:15
• Oh right! You could have laid it out nicer, I thought you did some weird thing that'd result in a 3! somewhere – Alec Teal May 3 '14 at 12:16
• Could you edit the answer so I may at least remove my DV? – Alec Teal May 3 '14 at 12:17
Okay let us write a solution to $a+b+c+d+e=10$ a different question, just incase it is homework.
Each solution will have the form:
||||-|---||||| <-> 4As 1B 0Cs, 0Ds, 5Es
How many different ways can we arrange 10 |s and 4 (4=5-1) -s? Each arrangement of these |s and -s is a valid solution.
$$\frac{(10+4)!}{4!10!}=\frac{14!}{10!4!}$$ | 2020-06-05T10:42:43 | {
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https://math.stackexchange.com/questions/1554530/rational-canonical-form-from-given-minimal-and-characteristic-polynomial | # Rational Canonical form from given minimal and characteristic polynomial
$A$ is a $4\times 4$ matrix over $F$ with characteristic polynomial $(x-1)^4$ and minimal polynomial $(x-1)^2$. What is the rational canonical form of $A$?
My answer was the following: it is one of the following: $$\begin{bmatrix} 0 & -1 & & \\ 1 & 2 & & \\ & & 1 & \\ & & & 1 \end{bmatrix} or \begin{bmatrix} 0 & -1 & & \\ 1 & 2 & & \\ & & 0 & -1\\ & & 1 & 2 \end{bmatrix}.$$ While our teacher finally reached at only second form. I am not satisfied with that answer.
My question is that whether the first matrix here can also be a rational form?
In general, to write rational canonical form of a matrix, I will proceed as follows: let $$m_A(x)=(x-a_1)^{k_1}(x-a_2)^{k_2}\cdots.$$ For each factor $(x-a_i)^{k_i}$ write one block diagonal companion matrix.
If this fills up the matrix size (i.e. if $m_A(x)$ equals characteristic polynomial, then this is required form.
Otherwise, fill up remaining parts (diagonal blocks) by writing companion matrix of factors $(x-a_i)^{l_i}$ where $l_i \leq k_i$.
Is this correct way?
Yes, you have two possible rational canonical forms given the information you have. Both the matrices you wrote have minimal polynomial $(x-1)^2$ and characteristic polynomial $(x-1)^4$.
To justify that $A$ has one of the two possible canonical forms above, let $a_1 \, | \, a_2 \, | \, \ldots \, | \, a_k$ denote the invariant factors of $A$. The highest invariant factor is always the minimal polynomial so $a_k = (x-1)^2$. The characteristic product of the matrix is the product of the invariant factors so we have a priori two options:
$$a_1(x) = (x-1), a_2(x) = (x-1), a_3(x) = (x-1)^2, \\ a_1(x) = (x-1)^2, a_2(x) = (x-1)^2.$$
• Determining (all possible) rational canonical forms I mean the following: suppose characteristic pol. is $(x-1)^6$ and minimal polynomial is $(x-1)^3$. Then possible forms are obtained by putting Companion matrices of size $\leq 3$; the possibilities will be $3+3$, $3+2+1$, $3+1+1+1$. So there will be three possible rational canonical forms when min. pol. is $(x-1)^3$ and char. poly. is $(x-1)^6$. (This situation is almost similar to that in Jordan theory, in which we consider Jordan blocks; in Rational form, we consider Companion blocks. I would like to ensure whether this is correct.) – Beginner Dec 5 '15 at 6:05 | 2019-07-20T00:57:18 | {
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https://thinkandstart.com/menace-ii-oqcpwq/if-the-diagonals-of-a-parallelogram-are-perpendicular-3f40df | These properties concern its sides, angles, and diagonals. If the diagonals of a quadrilateral are perpendicular bisector of each other, it is always a_____ A. Rectangle . Every square is a rhombus. B. Rhombus. The diagonals are perpendicular bisectors of each other. ToProve: if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. We have (2)The diagonals of a square are perpendicular to each other. 5. If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. If ABCD is a parallelogram, what is the length of BD? Answer. Yes, because a rhombus has two sets of parallel sides and all sides are congruent. Subscribe to bartleby learn! Bob R. Lv 6. "D" is the best answer. Plus, you’ll have access to millions of step-by-step textbook answers! Answer. Then we have the two diagonals are A + B and A − B. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? The diagonals of a parallelogram_____bisect the angles of the parallelogram Sometimes A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram 10. C. Every trapezoid is a parallelogram. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Median response time is 34 minutes and may be longer for new subjects. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Get a free answer to a quick problem. We've seen that one of the properties of a rhombus is that its diagonals are perpendicular to each other. The diagonals meet each other at 90°, this means that they form a perpendicular bisection. If a quadrilateral has 2 pairs of opposite sides that are congruent, then it is a parallelogram. If the diagonals of a quadrilateral both bisect each other, then the quadrilateral is a parallelogram. Sue. For example, the diagonals of a rhombus, kite or square are perpendicular, but those of a rectangle or general parallelogram are not. This is the currently selected item. In order for a parallelogram's diagonals to be perpendicular, the sides would have to have the same length, so this is only true for a rhombus (which is a specific parallelogram with equal sides). by Jennifer Kahle. The diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus ; If the diagonals of a parallelogram are equal and perpendicular, then it is a square ∵ In a parallelogram, its diagonals bisect each other at right angles ∴ Its diagonals are perpendicular ∵ Its diagonals are equal → By using rule 3 above ∴ The parallelogram is a square. d.trapezoid. Opposite sides are congruent in parallelogram. In my opinion "D" is the best answer, by definition a rhombus is a parallelogram with perpendicular diagonals . 11. Stephen K. If you just look […] faiqaferoz646 faiqaferoz646 22.06.2020 Math Secondary School Diagonals of a parallelogram are perpendicular to each other. Opposite angles are congruent. If the diagonals of a parallelogram are perpendicular and not congruent, then the parallelogram is. The parallelogram has the following properties: Opposite sides are parallel by definition. The parallelogram has the following properties: Opposite sides are parallel by definition. 10. If all the angles of the rhombus are right angles then you have a special rhombus which is a square A rhombus is a special kind of parallelogram, in which all the sides are equal. Parallelogram and Rhombus: A parallelogram is a quadrilateral (has 4 sides) where its opposite sides are parallel and equal and its opposite angles are equal. When the diagonals of a parallelogram are perpendicular to each other then it is called. So I'm thinking of a parallelogram that is both a rectangle and a rhombus. Start here or give us a call: (312) 646-6365, © 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Trapezoid Midsegment Theorem. Squares are rhombuses and rectangles, so if it is B, it must be C and D as well. The diagonals bisect each other. In a parallelogram a diagonal of the length 20 cm is perpendicular to one of the sides. . answered 10/08/20, If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Here we will show the converse- that if a parallelogram has perpendicular diagonals, it is a rhombus - all its sides are equal. If all the angles of the rhombus are right angles then you have a special rhombus which is a square, Nathaniel A. If the angle at which they meet is a right angle, then a right triangle is formed whose legs are half the length of each of the diagonals, and whose hypotenuse is the length of one side of the parallelogram (rhombus). The midsegment of a trapezoid is parallel to each base and its length is one half the sum of the lengths of both bases. Feb 18, 2016 . In this case, the diagonals divide the rhombus in four congruent right-angled triangles. Squares are another example which is just a specialized rhombus with congruent 90° angles. Answer: 3. let long side = x short side = s and is perpendicular to diag s^2 + 400 = x^2 so s^2 = x^2 - 400 2 s + 2 x = 80 s = 40 - x 1600 - 80 x + x^2 = x^2 - 400 80 x = 2000 x = 200/8 = 100/4 = 25 2 0; Damon. Proof: Figure is made having diagonals AC and BD. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. The diagonals bisect each other. Ask subject matter experts 30 homework questions each month. 1 0 736; Sam. A.) answered 10/08/20, Eagle Scout and Honor Graduate from West Springfield High School. No packages or subscriptions, pay only for the time you need. Maths. Thus you have a rhombus. EASY. But in the general case, it isn't always true that the diagonals of a parallelogram are perpendicular. (1)The diagonals of a parallelogram are equal. The properties of the parallelogram are simply those things that are true about it. Therefore, by CPCT A D → ≅ A B The diagonals of a parallelogram bisect each other. If either diagonal of a parallelogram bisects two angles, then it’s a rhombus (neither the reverse of the definition nor the converse of a property). Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. C. Square. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees. Parallelogram. In a parallelogram diagonals bisect each other. Consecutive angles are supplementary. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. Find the longer side of parallelogram if its perimeter is 80 cm. AB→≅DC→ & AD→≅BC→ ∴(AB→≅AD→)⇒AD→≅DC→≅AB→≅BC→. So let me see. Proof: The diagonals of a kite are perpendicular. Given that, we want to prove that this is a parallelogram. So A is out. If diagonals of a parallelogram are perpendicular, then it is a . bisects. a.rectangle. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Opposite sides are congruent. Mar 3, 2019 . What is the measure of a base angle … These angles are said to be congruent with each other. If you just look […] Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Proof: Opposite angles of a parallelogram. Let's think. And you see the diagonals intersect at a 90-degree angle. The diagonals of parallelograms are perpendicular. rhombus If the diagonals of a parallelogram are perpendicular, then the parallelogram is a _____ C. Four points in the plane form a parallelogram if the opposite line segments have the same slope. Therefore, by CPCT A D → ≅ A B The diagonals of a parallelogram bisect each other. trapezoid. (Points : 5) rectangle rhombus square trapezoid perpendicular. The only parallelogram that satisfies that description is a square. If the diagonals of a quadr... maths. Opposite angles are congruent. The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is … If the diagonals of a parallelogram are perpendicular then the parallelogram will be a rectangle. 0 0. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. If a parallelogram has (at least) one right angle, then it is a rectangle. A parallelogram, the diagonals bisect each other. Proof: Figure is made having diagonals AC and BD. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . (4)Every quadrilateral is either a trapezium or a parallelogram or a kite. Prove that, if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. b.rhombus.....Ans. The diagonals are perpendicular bisectors of each other. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Answer: Let two adjacent sides of the parallelogram be the vectors A and B (as shown in the figure). A. quadrilateral. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? One pair of diagonally opposite angles is equal in measurement. *Response times vary by subject and question complexity. What is the measure of a base angle of an isosceles triangle if … If all four sides of a parallelogram are equal in length, then the parallelogram is called a rhombus. The diagonals are perpendicular bisectors of each other. A.) P P 5. Every square is a rhombus. (A) Square (B) Rectangle (C) Rhombus (D) Parallelogram HARD. . First Name. 5. The diagonals of a parallelogram bisect each other. B. Slope: Method 2: Show that the diagonals are congruent. If the diagonals of a parallelogram are equal, then it is a rectangle; If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus; If the diagonals of a parallelogram are equal and perpendicular, then it is a square ∵ In a parallelogram, its diagonals bisect each other at right angles ∴ Its diagonals are perpendicular A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] if one diagonal of a parallelogram _____ a pair of opposite angles, then the parallelogram is a rhombus. In Δ A O D and Δ A O B. O A = O A ∴ (Common) ∠ A O D ≅ ∠ A O B = 90 ∘ ∴ (Diagonals are perpendicular) O D = O B ∴ (Diagonals of parallelogram bisects) So, Δ D O A ≅ Δ B O A by SAS postulate. In Δ A O D and Δ A O B. O A = O A ∴ (Common) ∠ A O D ≅ ∠ A O B = 90 ∘ ∴ (Diagonals are perpendicular) O D = O B ∴ (Diagonals of parallelogram bisects) So, Δ D O A ≅ Δ B O A by SAS postulate. Proving a Quadrilateral is a Rhombus Method 1: Prove that the diagonals are perpendicular. Play this game to review Geometry. if the diagonal of a parallelogram bisects one of the angles of parallelogram, prove that it also bisects the second angle and then the two diagonals are perpendicular to each other - Math - Quadrilaterals By completing the parallelogram and using the parallelogram rule, the diagonal represents the sum of the two original forces f1 + f2. asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals Question 392003: 1.how to prove if the diagonals in a parallelogram are perpendicular, then the parallelogram is a rhombus. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. 4. Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. Respond to this Question. a rhombus. If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it is a rhombus. Your Response. The diagonals are perpendicular bisectors of each other. So we're going to assume that the two diagonals are bisecting each other. Show that the diagonals of a parallelogram are perpendicular if and only if it is a rhombus, i.e., its four sides have equal lengths. If a parallelogram has perpendicular diagonals, you know it is a rhombus. A Parallelogram with Perpendicular Diagonals is a Rhombus. Figure is made having diagonals AC and BD. 2. how to prove if the diagonals in a paralellogram are congruent, then the parallelogram is a rectangle Answer by Edwin McCravy(18405) (Show Source): You can put this solution on YOUR website! If the diagonals of a parallelogram are perpendicular, then it’s a rhombus (neither the reverse of the definition nor the converse of a property). B is out. If the diagonals of a parallelogram are perpendicular, then the parallelogram is also called a rhombus. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. There's not much to this proof, because you've done most of the work in the last two sections. Special parallelograms. Here’s a rhombus proof for you. No, because in an isosceles trapezoid the sides that are not parallel are equal. Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. Choose an expert and meet online. Is this statement true? Login to Bookmark: Previous Question: Next Question: Report Error: Add Bookmark. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Study.com has thousands of articles about every Generally, the parallelogram law is applied when the vectors are co-initial, that is, their initial points are joined. No, because in an isosceles trapezoid the sides that are not parallel are equal. We are given that all four angles at point E are 9 0 0 and are therefore congruent. Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. If the diagonals are congruent than the parallelogram … Every rhombus is a square. Give reason for your ans… Get the answers you need, now! Which statement is true? Yes, because a rhombus has two sets of parallel sides and all sides are congruent. If the diagonals of a parallelogram are perpendicular, what can you conclude about the parallelogram? 10. If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? A kite is never a parallelogram. All rectangles are parallelograms. Remember a square is a special rectangle with all side lengths equivalent however we have no information regarding the side lengths of this problem. Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. The diagonals meet each other at 90°, this means that they form a perpendicular bisection. If the diagonals of a parallelogram are perpendicular then the parallelogram is from SCI 102 at Cristóbal Colón University Is this statement true? These angles are said to be congruent with each other. Therefore, parallelogram ABCD is rhombus. Back to Basic Ideas page. The diagonals of a parallelogram bisect each other. Consecutive angles are supplementary. c.square. One pair of diagonally opposite angles is equal in measurement. Proof: Rhombus diagonals are perpendicular bisectors. Answer. Yes, because the diagonals of a rhombus, which is a parallelogram, are perpendicular. Since the diagonals of a parallelogram bisect each other, B E and D E are congruent and A E is congruent to itself. Special parallelograms. D. Every parallelogram is a rectangle. D. Parallelogram. Theorems concerning quadrilateral properties. Properties of a Parallelogram. A link to the app was sent to your phone. Most questions answered within 4 hours. If a _____ is a rhombus, then it is a parallelogram. ToProve : if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.. These properties concern its sides, angles, and diagonals. A parallelogram with four congruent sides and four right angles. If ABCD is a parallelogram, what is the length of BD? EASY. -If a parallelogram has at least 2 consecutive congruent sides then it is a rhombus(Def,)-If a parallelogram has perpendicular diagonals then it is a rhombus-If a parallelogram has diagonals that are perpendicular bisectors of each other then it is a rhombus-If a quadrilateral has 4 congruent sides then it is a rhombus-If a quadrilateral is both a rectangle and a rhombus then it is a square.(Def.) If the diagonals of a quadrilateral are perpendicular to each other,it is a square but it is a rhombus as diagonals of rhombus are also perpendicular. For each such a triangle, the legs are 6/2 = 3 units and 8/2 = 4 units; hence, the sides of the rhombus are = 5 units long, and its perimeter is 5*4 = 20 units. 3. a.rectangle b.rhombus c.square d.trapezoid 11. The properties of the parallelogram are simply those things that are true about it. The diagonals of parallelograms are perpendicular. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. Thus you have a rhombus. Related Videos. The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. For Free, Inequalities and Relationship in a Triangle, ALL MY GRADE 8 & 9 STUDENTS PASSED THE ALGEBRA CORE REGENTS EXAM. Diagonals of Quadrilaterals -- Perpendicular, Bisecting or Both. So we've just proved-- so this is interesting. Play this game to review Geometry. Diagonals of a parallelogram are perpendicular to each other. 4. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. Assuming vectors A and B are two adjacent sides of a parallelogram; a quick disproof of rhombus would be $|A|\neq|B|$. 1 decade ago. A) rhombus : B) rectangle : C) quadrilateral : D) none of these : Correct Answer: A) rhombus : Part of solved Quadrilateral and parallelogram questions and answers : >> Elementary Mathematics >> Quadrilateral and parallelogram. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. So we're assuming that that is equal to that and that that right over there is equal to that. B. Rectangles have congruent diagonals, but they are not perpendicular. please follow me and mark as brainliest If the diagonals are congruent and are perpendicular bisectors of each other then the parallelogram is a square. D. Just so, do the diagonals of a trapezium bisect each other at 90 degrees? If the diagonals of a parallelogram are perpendicular to each other, but are not congruent to each other, then the parallelogram is which of the following? Proof: Opposite sides of a parallelogram. That is true for some parallelograms but not all. The diagonals of parallelograms are perpendicular. No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other. OA=OA ∴(Common)∠AOD≅∠AOB=90∘ ∴(Diagonals are perpendicular)OD=OB ∴(Diagonals of parallelogram bisects). Opposite sides are congruent. McDougal Littell Jurgensen Geometry: Student Edition Geometry. Which shape is not a parallelogram? In a parallelogram diagonals bisect each other. (3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus. Proof: Diagonals of a parallelogram. Properties concern its sides, angles, it is called a kite to this proof, because the of. D. ToProve: if the diagonals of a quadrilateral is a special rectangle all. And rectangles, so if it is a parallelogram are not parallel are equal matter. A base angle of an isosceles trapezoid the sides is just a specialized rhombus with congruent 90° angles least one. 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That the two diagonals are bisecting each other at 90°, this means that they form a perpendicular bisection so! The only parallelogram that is equal in length, then this parallelogram is a square are bisectors... | 2021-05-06T09:50:41 | {
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https://math.stackexchange.com/questions/2938253/calculation-of-the-probability-of-not-mutually-exclusive-and-independent-event | # Calculation of the probability of not mutually exclusive and independent event.
I tried to solve this math problem in two different ways but found two different answers and I don't know which one is the right answer.
The problem goes like this:
The probability of failing in Science exam for a student is $$\frac{1}{5}$$, probability of passing in both Science and English is $$\frac{3}{4}$$ and probability of passing any one of these two subjects or both is $$\frac{7}{8}$$, what is the probability of him passing only in English?
$$\mathbf{Solution:}$$ Let S and E are the events of passing in Science and English respectively. probability of failing in Science $$P(S^c) = \frac{1}{5}$$, so probability of passing in Science exam is, $$P(S) = 1-P(S^c)=1-\frac{1}{5}=\frac{4}{5}$$ probability of passing both subjects, $$P(S\cap E)= \frac{3}{4}$$;
and probability of passing one or both subjects, $$P(S\cup E)=\frac{7}{8}$$
Since these two events are independent and not mutually exclusive, we can write, $$P(S\cup E)=P(S)+P(E)-P(S\cap E)$$ Where P(E) is the probability of passing in English. Which is $$P(E)=P(S\cup E)-P(S)+P(S\cap E)$$ $$P(E)=\frac{7}{8}-\frac{4}{5}+\frac{3}{4}$$ $$P(E) = \frac{33}{40}$$ At this point I can calculate the probability of him passing only in English in two ways:
(1) The probability of him passing only in English is $$P(E\cap S^c)$$
$$P(E)=P(E\cap S)+P(E\cap S^c)$$ Addition rule, since these two events are mutually exclusive. So, $$P(E\cap S^c)=P(E)-P(S\cap E)=\frac{33}{40}-\frac{3}{4}=\frac{3}{40}$$
(2)The probability of him passing only in English is $$P(E\cap S^c)$$.
Since these two events are independent, $$P(E\cap S^c)=P(E)P(S^c)=\frac{33}{40}\times\frac{1}{5}=\frac{33}{200}$$
This two methods give two different answers $$\frac{3}{40}$$ and $$\frac{33}{200}$$
I am obviously missing something or I have some lack in understanding. It will be very helpful if someone point out which one is wrong and why.
The key point is that $$E$$ and $$S$$ are dependent. Thus $$P(E\cap S^c)=P(E)\cdot P(S^c|E)=\frac{33}{40}\cdot \frac{1}{11}=\frac{3}{40}$$
You can comprehend $$P(S^c|E)$$ by looking at the table below:
$$\begin{array}{c|c|c|c} &E & E^c& \\ \hline S &\color{red}{0.75} & 0.05 & 0.8 \\ \hline S^c & 0.075 & \color{blue}{0.125} &\color{red}{0.2} \\ \hline & 0.825 & 0.175 &\color{red}{1} \end{array}$$
and probability of passing any one of these two subjects or both is $$\frac78$$
That means that $$P(S^c\cap E^c)=1-\frac78=\frac18=\color{blue}{0.125}$$
Finding out the missing figures is just simplest algebra.
You see that $$P(S^c)=0.2 \neq P(S^c|E)$$. Thus $$E$$ and $$S$$ are not independent. And your first approach is right $$\checkmark$$
• Thank you for your help. But would you like to show me in detail how to get the table? Or is there any easy way to figure out the dependency of events? I thought events like in the problem is independent by common sense and that's how I generally used to approach any problem. Thanks in advance. – Arafat Hossen Oct 2 '18 at 3:33
• In general if I have only two events I firstly make a table, almost always. The red figures are given in the text. And then the missing figures can be filled in successively. Especially the 0.8 should be very obvious since the cells of the last column/row sum up to one. I´ve made an edit. Please have a look. – callculus Oct 2 '18 at 5:03
• @ArafatHossen You always have to prove if events are independent. There are two variants of the criteria. Basically the meaning is identical: 1. $P(A\cap B)=P(A)\cdot P(B)$, 2. $P(A|B)=P(A)$. It works as well by using the complements: $P(A\cap \overline B)=P(A)\cdot P(\overline B)\Rightarrow$ The events A and B are independent. – callculus Oct 2 '18 at 5:21 | 2019-09-20T20:19:17 | {
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https://www.physicsforums.com/threads/is-second-rank-tensor-always-tensor-product-of-two-vectors.910628/ | # I Is Second rank tensor always tensor product of two vectors?
#### arpon
Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ? If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?
2. How to find out $A_i$ and $B_j$
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?
#### fresh_42
Mentor
2018 Award
Suppose a second rank tensor $T_{ij}$ is given. Can we always express it as the tensor product of two vectors, i.e., $T_{ij}=A_{i}B_{j}$ ?
No, it is a sum of such products.
If so, then I have a few more questions:
1. Are those two vectors $A_i$ and $B_j$ unique?
No. Even for dyadics $A_i \otimes B_j$ you always have $A_i \otimes B_j = c \cdot A_i \otimes \frac{1}{c}B_j$ for any scalar $c \neq 0$.
2. How to find out $A_i$ and $B_j$
$T_{ij}$ is basically any matrix and $A_i \otimes B_j$ a matrix of rank $1$. So write your matrix as a sum of rank-$1$ matrices and you have a presentation.
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?
No. See the rank explanation above.
#### haushofer
3. As $A_i$ and $B_j$ has $3+3 = 6$ components in total (say, in 3-dimension), it turns out that we need only $6$ quantities to represent the $9$ components of the tensor $T_{ij}$. Is that correct?
To add to Fresh's answer: this last remark already should make you suspicious. An arbitrary (!) second-rank tensor in three dimensions has 3*3=9 components, but two vectors have 2*3=6 components. What you can do, is to decompose a second rank tensor like $T_{ij}$ as
$$T_{ij} = T_{[ij]} + T_{(ij)}$$
where [ij] stands for antisymmetrization, whereas (ij) stands for symmetrization. Both parts transform independently under coordinate transfo's. The antisymmetric part has 3 independent components, whereas the symmetric part has 6 components. You can even go further in this decomposition, because the trace of the tensor components also does not change under a coordinate transformation.
"Is Second rank tensor always tensor product of two vectors?"
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• Solo and co-op problem solving | 2019-10-18T13:20:02 | {
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https://mathematica.stackexchange.com/questions/21197/how-do-i-simplify-a-vector-expression | How do I simplify a vector expression?
I am doing vector analysis. I have figured out that the following expression won't be simplified in Mathematica:
Simplify[Dot[x, y] - Dot[y, x]]
I think the reason is that x and y could be matrices, so generally the operation does not commute. But for a vector, the commutation relation should hold. What I don't know is how to tell Mathematica to consider x and y to be vectors.
Also, how can I expand Cross[x + y, z] to be Cross[x, y] + Cross[y, z]? I tried to use Expand and ExpandAll but neither worked.
BTW, in Mathematica, is it possible to define a abstract row/column vector without explicitly specifying the number of entries?
Here is a way to do all the things you asked for automatically, independently of Mathematica version. The approach relies on a special symbol to identify when we're dealing with a vector: Instead of using things like x, y etc. for vectors, the convention now is that vectors are written as vec[x], vec[y], etc.
You could also define the wrapper OverVector[x] for this purpose because it displays as $\vec{x}$. But for this post I want to keep it simple, and the arrows wouldn't display easily in the source code below.
ClearAll[scalarProduct, vec];
SetAttributes[scalarProduct, {Orderless}]
vec /: Dot[vec[x_], vec[y_]] := scalarProduct[vec[x], vec[y]]
vec /: Cross[vec[x_], HoldPattern[Plus[y__]]] :=
Map[Cross[vec[x], #] &, Plus[y]]
vec /: Cross[HoldPattern[Plus[y__]], vec[x_]] :=
Map[Cross[#, vec[x]] &, Plus[y]]
scalarProduct /: MakeBoxes[scalarProduct[x_, y_], _] :=
RowBox[{ToBoxes[x], ".", ToBoxes[y]}]
vec[x].vec[y]
(* ==> vec[x].vec[y] *)
vec[x].vec[y] == vec[y].vec[x]
(* ==> True *)
Cross[vec[x], vec[a] + vec[b]]
(* ==> vec[x]\[Cross]vec[a] + vec[x]\[Cross]vec[b] *)
Cross[vec[a] + vec[b], vec[x]]
(* ==> vec[a]\[Cross]vec[x] + vec[b]\[Cross]vec[x] *)
For the Dot product, I defined the behavior of vec such that it gets evaluated as a new function scalarProduct whose only algebraic property is that it's Orderless as you were expecting for the dot product of vectors. Of course this is only true for Euclidean dot products, so this assumption is implicit here. For more information on how this definition works, look up TagSetDelayed.
In addition, scalarProduct is given a customized display format by defining that it should again display as if it were a dot product when it appears in the low-level formatting function MakeBoxes.
For the distributive property of the cross product, I give vec the additional property that when it appears in Cross together with an expression of head Plus, the sum is expanded. Here the TagSetDelayed definitions are done for both orders, and contain a HoldPattern to prevent Plus from being evaluated too early in the definition.
Now you may come back with many more wishes: e.g., what about multiplicative scalars in the dot or cross product, and what about matrices. However, that's a wide field that opens up a can of worms, so I would say just implement the bare minimum of features you can get away with symbolically, then proceed with a concrete working basis so that you can write vectors as lists instead.
Another approach would be to define a new symbol for a custom dot product. That is done in this question.
Using OverVector
As mentioned above, you can replace vec by Overvector everywhere in the above source code, to get a better formatted result. Assuming you have done that (I won't bother to repeat the definitions with that change), here are some examples:
To enter these vector expressions, refer to the Basic Math assistant palette. The cross product can be entered as EsccrossEsc.
Another thing you asked for is to use the antisymmetry of the cross product in simplifications. That's actually done already if you invoke FullSimplify:
• Yes, the definition of new operator looks good. But it is not that convenient to use additional head. I am looking for a vector operators to simply some complicate expression involving many vectors. Is that possible to redefine +, -, . and x so don't have to use something like Plus? – user1285419 Mar 14 '13 at 4:00
• Sorry, I used Plus only because that was what I was thinking about while writing the definitions. Of course you can actually use + and - with this setup. No modifications needed at all. I also mentioned you can get a simpler format using OverVector, and illustrated that in an update to the answer. As to redefining + etc.: it's not necessary, so don't do it. – Jens Mar 14 '13 at 4:28
• that's amazing. I never know that we can use mathematica in that way, I did learn something today. Thanks. – user1285419 Mar 14 '13 at 4:35
• I find a new problem of using this definition, if I try to Cross[OverVector[x], 2*OverVector[y]] + Cross[OverVector[y], 2*OverVector[x]] but it is not zero – user1285419 Apr 16 '13 at 22:58
• I already anticipated this comment in my answer, and mentioned that it opens up a wide field of additional definitions one could add. I may come back to that when I have time. – Jens Apr 16 '13 at 23:26
If you have Mathematica Version 9, you can use Vectors and TensorReduce:
Assuming[(x | y) \[Element] Vectors[n] , TensorReduce[Dot[x, y] - Dot[y, x]]]
(* 0 *)
TensorReduce[Dot[x, y] - Dot[y, x], Assumptions -> (x | y) \[Element] Vectors[n]]
(* 0 *)
TensorReduce[Cross[x + y, z], Assumptions -> (x | y | z) \[Element] Vectors[n]]
(* x\[Cross]z + y\[Cross]z *)
Distribute[Cross[x + y, z]] (* this should work in all previous versions *)
(* x\[Cross]z + y\[Cross]z *)
• thanks, it works. But what about the first one, is that possible to define an abstract row/column vector so x.y-y.x will get zero? – user1285419 Mar 13 '13 at 3:11
• Presumably these solutions are for version 9 and later, when TensorReduce and Vectors were introduced. – whuber Mar 13 '13 at 4:42
• @whuber, thank you. I updated with a note that TensorReduce and Vectors` are version-9 functions. – kglr Mar 13 '13 at 6:15
• Thanks. Oh, I just saw that it only works on version 9 not my version 8 :( – user1285419 Mar 13 '13 at 6:37 | 2020-09-22T21:16:04 | {
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https://math.stackexchange.com/questions/3013093/how-to-show-that-2n-n-without-induction | # How to show that $2^n > n$ without induction
I'm solving exercises about Pascal's triangle and Binomial theorem, and this problem showed up, however I don't have any clue on how to solve it
The sum of $${n\choose p}$$ from $$p=0$$ to $$n$$ is the same thing as $$(1+1)^n=2^n$$, how can I use this information? Maybe comparing with another summation that equals to n?
Note that $$2^n$$ is the number of subsets of $$[n]=\{1,\dotsc,n\}$$. There are $$n$$ subsets of $$[n]$$ with size $$1$$. There is at least one subset of $$[n]$$ which is not a singleton (namely the empty set). Hence $$2^n>n$$ for $$n\geq 1$$.
• Foobaz John.Nice+. – Peter Szilas Nov 25 '18 at 17:21
Use Bernoulii inequality, which is true for all $$x>-1$$: $$(1+x)^n\geq 1+nx$$ so $$(1+1)^n \geq 1+n\cdot 1 >n$$
Maybe comparing with another summation that equals to $$n$$?
For any $$p=0,1,\dots,n$$, there is at least one way to choose $$p$$ things from a list of $$n$$ things. Thus $$\binom{n}{p} \ge 1$$, so
$$2^n = \sum_{p=0}^n \binom{n}{p} ≥ \sum_{p=0}^n 1 = n+1 > n.$$
• I did that summation, but how do I prove that one inequality is bigger than the other? – Nuno Mateus Nov 25 '18 at 17:15
• @NunoMateus The sentence before that is the proof. – Calvin Khor Nov 25 '18 at 17:15
The Binomial Theorem says \begin{align} 2^n &=(1+1)^n\\ &=\binom{n}{0}1^0+\binom{n}{1}1^1+\dots\\ &\ge1+n\\[9pt] &\gt n \end{align}
• And I was thinking about this one :) +1 – Aqua Nov 25 '18 at 17:50
hint
Consider $$x\mapsto \frac{\ln(x)}{x}$$ for $$x\ge 1$$.
$$f'(x)=\frac{1-\ln(x)}{x^2}$$
the maximum if $$f(e)=\frac{1}{2}<\ln(2)$$.
thus
$$\ln(x) | 2019-07-15T20:06:52 | {
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https://math.stackexchange.com/questions/1372621/finding-convergence-of-a-series-using-integral-test | # Finding convergence of a series using integral test
The series:$$\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n}\right)^{2}$$ Question:
a) show that it converges
b) find the upper bound for the error in approximation $s\approx s_{n}$
Trial: The section was about integral test, but the sequence$\left(\frac{\ln(n)}{n}\right)^{2}$ is not decreasing from [1,$\infty$]( it is increasing from [ 1,e ] ) so, I could not use the integral test.
Other method: I tried to find a sequence greater than$\left(\frac{\ln(n)}{n}\right)^{2}$so that it satisfies the condition for use of integral test( If I show using integral test that the new series is bounded then It could imply that our sequence is convergent since its bounded and decreasing).the problem was that I had trouble finding any function which could satisfy such conditions b) this is understanding problem, is it asking me to find the exact sum or some upper bound , in any way how can I do this?
• i don't understand your sum – Dr. Sonnhard Graubner Jul 24 '15 at 15:54
• If the function is decreasing on $(e,\infty)$ then the sequence is decreasing on $[3,\infty)$. Removing a finite number of terms from the front end of a series won't change whether the series converges or not. – Alex Pavellas Jul 24 '15 at 15:56
• btw, I assume you meant to use $n$ instead of $x$ inside the series. – Alex Pavellas Jul 24 '15 at 15:57
• @Dr.SonnhardGraubner I made an edit – Socre Jul 24 '15 at 15:59
• @user255545 yes, that infact is true but I have another problem, I had a difficult time integrating the function. If I knew how to integrate it your method is actually quite satisfying. – Socre Jul 24 '15 at 16:02
To show the series converges using the integral test we simply integrate by parts twice with successive substitutions $u_1=(\log x)^2$ and $v_1=x^{-2}$, and $u_2=\log x$ and $v_2=x^{-1}$, to reveal
\begin{align} \int_3^{\infty} \left(\frac{\log x}{x}\right)^2\,dx&=-\left.\left(\frac{(\log x)^2}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2-2\left.\left(\frac{\log x}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{1}{x^2}\,dx\\\\ &=\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \end{align}
Thus, the series converges.
UPPER AND LOWER BOUNDS
To find an upper bound of the series using the integral test we use
\begin{align} \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2&\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\int_3^{\infty}\left(\frac{\log x}{x}\right)^2\,dx\\\\ &=\left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23\\\\ &\approx. 2.05560987295277 \end{align}
The lower bound is simply the upper bound less the third term $\left(\frac{\log 3}{3}\right)^2\approx. 0.134105440090287$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{ \left(\frac{\log 2}{2}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2}$$
$$\bbox[5px,border:2px solid #C0A000]{ \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23}$$
$$\bbox[5px,border:2px solid #C0A000]{ 1.92150443286247 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le 2.05560987295278}$$
Since $\log n\leq (n-1)^{\frac{2}{5}}$ for any $n\geq 1$,
$$0\leq \sum_{n\geq 1}\frac{\log^2 n}{n^2}\leq \sum_{n\geq 1}\frac{1}{n^{\frac{6}{5}}}$$ and the RHS is convergent by the $p$-test. Moreover, $$\sum_{n\geq 1}\frac{\log^2 n}{n^2}=\frac{d^2}{ds^2}\left.\sum_{n\geq 1}\frac{1}{n^s}\right|_{s=2} = \zeta''(2) = 1.9892802342989\ldots$$
• Why the exponent of $2/5$? – marty cohen Jul 24 '15 at 17:35
• @martycohen: any constant slightly less than $\frac{1}{2}$ does the job, $\frac{2}{5}$ is just a nice number of that form. – Jack D'Aurizio Jul 24 '15 at 17:39
• According to Wolfy, max{(log(log(x)))/(log(x-1))}~0.379831 at x~10.9352. So your choice of 2/5 is quite good. – marty cohen Jul 24 '15 at 17:47
For the convergence we can use for example, for $x$ sufficiently large (say $x\geq N$), $$\log\left(x\right)\leq x^{1/4}$$ hence $$\sum_{n\geq N}\frac{\log^{2}\left(n\right)}{n^{2}}\leq\sum_{n\geq N}\frac{1}{n^{3/2}}<\infty.$$ About the upper bound for the error, we can use the integral test $$\sum_{n\geq N}f\left(n\right)\leq f\left(N\right)+\int_{N}^{\infty}f\left(x\right)dx$$ and so in our case $$\sum_{n\geq1}\frac{\log^{2}\left(n\right)}{n^{2}}=\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\sum_{n\geq N+1}\frac{\log^{2}\left(n\right)}{n^{2}}\leq$$ $$\leq\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\frac{\log^{2}\left(N+1\right)}{\left(N+1\right)^{2}}+\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx$$ and the integral is, using the integration by parts, $$\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx=\frac{\log^{2}\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{\log\left(x\right)}{x^{2}}dx=$$ $$=\frac{\log^{2}\left(N+1\right)}{N+1}+\frac{2\log\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{1}{x^{2}}dx=$$ $$\frac{\log^{2}\left(N+1\right)+2\log\left(N+1\right)+2}{N+1}.$$ | 2021-07-30T23:01:24 | {
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https://www.physicsforums.com/threads/probability-question.467349/ | # Probability Question
rooski
## Homework Statement
Suppose that the integer values 1 2 and 3 are written on each of three different cards. Suppose you do not know which number is the lowest (you do not know beforehand what the values on the cards are). Suppose that you are to be offered these cards in a random order. When you are offered a card you must immediately either accept it or reject it. If you accept a card, the process
ends. If you reject a card, then the next card (if a card remains) is offered. If you
reject the first two cards offered, then you must accept the final card.
(a) If you plan to accept the first card offered, what is the probability that you will
accept the lowest valued card?
(b) If you plan to reject the first card offered, and to then accept the second card if
and only if its value is lower than the value of the first card, what is the probability
that you will accept the lowest valued card?
## The Attempt at a Solution
A) The answer is obviously 1/3 for this question.
B) this is a conditional probability question. Given that P(E|F) = P(EF) / P(F) then i must first figure out what P(E) and P(F) stand for.
P(E) is the probability that i will accept the lowest card.
P(F) is the probability that the second card i choose is lower than the first rejected card.
If i reject the card with 1 on it, then i have no chance of selecting the lowest card next.
If i reject the card with 2 on it, then there is a 1/2 chance i will select the lowest card next.
If i reject the card with 3 on it, then i have a 1/2 chance i will select the lowest card next.
So there is a 2/3 * 1/2 = 1/3 chance that the second card is lower than the first rejected card, right?
I am not sure how to proceed after this.
Homework Helper
If you reject the card with 2 on it, then you will either pick the card with 1 next (which is lower, so you will accept it and have the lowest card) or you will pick that card with 3 (which is higher, so you will reject it and be left with the last one).
So in fact, the probability of getting the lowest card there is 1.
rooski
Ah right, dunno how i missed that.
So how do i calculate the chance that the second card will be lower? I have 0, 1 and 1/2 as the probabilities, depending on which card is rejected first.
Karlx
Hi Rooski.
Let L be the event that you accept the lowest card.
Let $$F_{n}$$ be the event that the value of the first (and rejected) card is n, n=1,2,3.
Events $$F_{n}$$ are disjoint.
Try to write down p(L) using the total probability rule.
rooski
Assuming L is the event i accept the lowest card,
P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C)
Where A,B,C denote cards 1,2,3 respectively. Is that right or am i off? It seems wring since P(A), P(B) and P(C) would all be 1/3.
Karlx
Ok.
According to your notation A is the event that the first (and rejected) card is the lowest one.
So, if you reject A, the probability of accepting the lowest card is zero.
So P(L|A) = 0.
What about P(L|B) and P(L|C) ?
rooski
P(L|B) = 1 since you will reject 3 if it appears, or accept 1 when it appears.
P(L|C) = 1/2 since you will accept 2 if it appears or accept 1 when it appears.
Have i calculated P(A) P(B) and P(C) wrong?
Karlx
A is the event that the first card is A. So pA=1/3.
The same apply to B and C.
Just calculate P(L).
rooski
P(L) = 0 * 1/3 + 1 * 1/3 + ½ * 1/3 = 3/6 = 1/2
So there is a 50% chance that we will end up with the lowest card if we reject the first random card. | 2022-08-14T16:43:16 | {
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http://qkms.ddiy.pw/sawtooth-wave-fourier-series.html | # Sawtooth Wave Fourier Series
Square wave t x(t) X 0 -T 0 0 T 0 -X 0 0 k X j π − 2 0 when k is odd a k = 0 when k is even 2. Fn = 2 shows the special case of the segments approximating a sine. In this demonstration it's just like the last one for the square wave. A sawtooth wave An electrocardiogram (ECG) signal Also included are a few examples that show, in a very basic way, a couple of applications of Fourier Theory, thought the number of applications and the ways that Fourier Theory is used are many. According to the important theorem formulated by the French mathematician Jean Baptiste Joseph Baron Fourier, any periodic function, no matter how trivial or complex, can be expressed in terms of converging series of combinations of sines and/or cosines, known as Fourier series. It is so named based on its resemblance to the teeth of a plain-toothed saw with a zero rake angle. Find the Fourier series for the sawtooth wave defined on the interval $$\left[ { - \pi ,\pi } \right]$$ and having period $$2\pi. EE341 Homework Assignment 4 9-13-19. The Fourier expansion of the square wave becomes a linear combination of sinusoids: If we remove the DC component of by letting , the square wave become and the square wave is an odd function composed of odd harmonics of sine functions (odd). EE 230 Fourier series - 1 Fourier series A Fourier series can be used to express any periodic function in terms of a series of cosines and sines. 1 De nitions and Motivation De nition 1. jpg 1,956 × 2,880; 323 KB. (Note that Trott 2004, p. Report on sawtooth wave generator 1. Example: Sawtooth wave So, the expansion of f(t) reads (7. 5))in terms of its Fourier components, may occur in electronic circuits designed to handle sharply rising. Lecture 7: Fourier Series and Complex Power Series Week 7 Caltech 2013 1 Fourier Series 1. — The convention is that a sawtooth wave ramps upward and then sharply drops. However, as ybeltukov pointed out in a comment I did not read until he made me aware of it, Fourier series of piecewise continuously differentiable functions tend to overshoot a jump discontinuities, something which is called Gibbs phenomenon. Unless stated otherwise, it will be assumed that x(t) is a real, not complex, signal. To best answer this question, we need to consult the work of Baron Jean Baptiste Fourier and dig into a little mathematics. 3 shows two even functions, the repeating ramp RR(x)andtheup-down train UD(x) of delta functions. The three examples consider external forcing in the form of a square-wave, a sawtooth-wave, and a triangle-wave. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Figure 13-11 shows an example of calculating a Fourier series using these equations. These basic signals can be used to construct more useful class of signals using Fourier Series representation. The voltage at the Figure 5. The examples given on this page come from this Fourier Series chapter. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. 2 Wave Diffraction and the Reciprocal Lattice Diffraction of waves by crystals • The Bragg law Scattered wave amplitude • Fourier analysis • Reciprocal lattice vectors • Diffraction conditions • Laue equations Brillouin zones • Reciprocal lattice to sc/bcc/fcc lattices Fourier analysis of the basis. general Fourier Series around a jump discontinuity. Taking the inner product of both sides, with respect to the orthonormalized eigenfunctions X n (x) and the weight function w(x) = 1, and assuming validity of the interchange between the summation and integration operations, yields. We then state some important results about Fourier series. But what we're going to do in this case is we're going to add them. Definition of Fourier Series and Typical Examples Baron Jean Baptiste Joseph Fourier \(\left( 1768-1830 \right)$$ introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related. This series is used to generate a sawtooth wave and values are calculated using the program l18a1. (iii) h(x) = ˆ 0 if 2 | 2019-11-19T00:35:27 | {
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https://math.stackexchange.com/questions/1745082/are-these-the-only-answers-of-x4y4z41-4xyz | # are these the only answers of $x^4+y^4+z^4+1=4xyz$?
Given an equation $$x^4+y^4+z^4+1=4xyz$$Find out the number of possible ordered tuple $(x,y,z)\mid x,y,z\in\Bbb{R}$.
I am getting it as $(1,1,1),(-1,-1,1),(1,-1,-1),(-1,1,-1)$ so $\boxed{4}$
Is there any other tuple which I am missing?
Any help will be appreciable !
• Looks fine. AM/GM gives that the absolute value of each of $x,y,z$ is 1. Apr 16, 2016 at 14:49
By Am/Gm we have $$\frac{x^4+y^4+z^4+1}{4}\geq xyz$$ . now we know the minima of arithmetic mean and maxima of geometric mean is achieved when numbers are equal or their $mod$ is equal as here $4$th power is used. Thus all positive $1$ or two negative $1$ are permissible hence total answers are $1+{3\choose 2}=1+3=4$ | 2022-08-18T08:12:01 | {
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