Search is not available for this dataset
url
string | text
string | date
timestamp[s] | meta
dict |
|---|---|---|---|
https://stats.stackexchange.com/questions/358367/calculating-multivariate-integrals-between-lower-and-upper-bounds | # Calculating multivariate integrals between lower and upper bounds
Suppose $$\vec{X}=(x_1,x_2,...,x_n)$$ follows some continuous multivariate distribution, such that $$x_i\in{\rm I\!R}, i=1,...,n$$.
• $$\phi(\vec{x})$$, which gives me the pdf at point $$\vec{x}=(x_1,x_2,...,x_n)$$
• $$\Phi^{upper}(\vec{x})$$, which calculates the CDF from $$-\infty$$ to $$\vec{x}$$. More specifically: $$\Phi^{upper}(\vec{x})=\int_{\vec{k}=(-\infty,...,-\infty)}^{(x_1,...,x_n)}\phi(\vec{k})d\vec{k}$$
My main question is: how can I use $$\Phi^{upper}(\vec{x})$$ to calculate the CDF between lower and upper bounds?
In other words, how can I calculate integral of $$\phi$$ from $$\vec{x}_{lower}$$ to $$\vec{x}_{upper}$$ using $$\Phi^{upper}$$?
In mathematical terms, how can I use $$\Phi^{upper}$$ to calculate the following integral?
$$\Phi^{upper}_{lower}(\vec{x}_{lower},\vec{x}_{upper})=\int_{\vec{k}=(x_1^{lower},...,x_n^{lower})}^{(x_1^{upper},...,x_n^{upper})}\phi(\vec{k})d\vec{k}$$
I know that if there are only 2 dimensions in $$\vec{X}$$, I can calculate $$\Phi^{upper}_{lower}(\vec{x}_{lower},\vec{x}_{upper})$$ like this:
$$\Phi^{upper}_{lower}(\vec{x}_{lower},\vec{x}_{upper}) = \Phi^{upper}(x_1^{upper},x_2^{upper}) - \Phi^{upper}(x_1^{upper},x_2^{lower}) - \Phi^{upper}(x_1^{lower},x_2^{upper}) + \Phi^{upper}(x_1^{lower},x_2^{lower})$$
However, I don't know how to generalize this result for an N-dimensional case.
Does anyone have any insight on how to do this?
Thank you very much!
• Not exactly, no. Consider the 2 dimensional example I gave. If you just do $\Phi^{upper}(\vec{x}_{upper}) - \Phi^{upper}(\vec{x}_{lower})$, you're leaving in a few areas that shouldn't be considered. Here is an image to illustrate my point. – Felipe D. Jul 22 '18 at 3:35
• Assuming you're talking about a rectangular area, you probably want to apply the inclusion-exclusion rule here. – Ben Jul 22 '18 at 3:38
• I can see how this touches on what I'm getting at, but I'm having a bit of a hard time understanding how I would apply this concept from set theory to this particular problem. Could you maybe elaborate a little bit further on how to do so? Thanks! – Felipe D. Jul 22 '18 at 3:45
This problem involves the application of a probability measure over a union of non-disjoint sets, and so it can be solved by application of the inclusion-exclusion rule.
To facilitate this analysis, we will let $\mathcal{A}(x) \equiv (-\infty, x]$ denote the real numbers up to $x$ so that $\mathcal{A}(x_1) -\mathcal{A}(x_0) = (x_0,x_1]$ is a bounded interval. (If you are dealing with a continuous random variable you do not have to worry about whether the ends of the intervals are open or closed.) We will also simplify the problem by using a slight abuse of notation, treating each $\mathcal{A}(x_i)$ as a subset of $\mathbb{R}^n$ with an upper bound in dimension $i$ and free in the other dimensions. This means that for an input vector $\mathbf{x} = (x_1,...,x_n)$ you have CDF values of the form:
$$\Phi(\mathbf{x}) = \mathbb{P} (\mathbf{X} \leqslant \mathbf{x}) = \mathbb{P} \Bigg( \bigcap_{k=1}^n \mathcal{A}(x_k) \Bigg).$$
You want to calculate the probability of a rectangular area, which can be written as an intersection of bounded intervals as $\mathcal{R}_n = \bigcap_{k=1}^n (\mathcal{A}(\bar{x}_{k}) -\mathcal{A}(\underline{x}_{k}))$, where you have lower and upper bounds $\underline{\mathbf{x}} < \bar{\mathbf{x}}$. You want to be able to write the probability of this rectangular event using your CDF $\Phi$. To apply the inclusion-exclusion rule we will let $\mathfrak{N}_k$ denote the class of all subsets of $\{ 1,...,n \}$ with exactly $k$ elements. Using this rule, and some other set algebra, we have:
\begin{aligned} \mathbb{P}(\mathcal{R}_n) &= \mathbb{P} \Bigg( \bigcap_{k=1}^n (\mathcal{A}(\bar{x}_{k}) -\mathcal{A}(\underline{x}_{k})) \Bigg) \\[6pt] &= \mathbb{P} \Bigg( \bigcap_{k=1}^n \mathcal{A}(\bar{x}_{k}) \Bigg) - \mathbb{P} \Bigg( \bigcap_{k=1}^n \mathcal{A}(\bar{x}_{k}) \cap \bigcup_{k=1}^n \mathcal{A}(\underline{x}_{k}) \Bigg) \\[6pt] &= \mathbb{P} \Bigg( \bigcap_{k=1}^n \mathcal{A}(\bar{x}_{k}) \Bigg) - \sum_{k=1}^n \Bigg[ (-1)^{k-1} \sum_{\mathcal{D} \in \mathfrak{N}_k} \mathbb{P} \Bigg( \bigcap_{i \notin \mathcal{D}} \mathcal{A}(\bar{x}_{i}) \cap \bigcap_{i \in \mathcal{D}} \mathcal{A}(\underline{x}_i) \Bigg) \Bigg] \\[6pt] &= \Phi (\bar{\mathbf{x}}) - \sum_{k=1}^n (-1)^{k-1} \sum_{\mathcal{D} \in \mathfrak{N}_k} \Phi(\mathbf{x_\mathcal{D}}), \\[6pt] \end{aligned}
where the data vector $\mathbf{x}_\mathcal{D}$ uses the lower bounds $\underline{x}_i$ for all $i \in \mathcal{D}$ and the upper bounds $\bar{x}_i$ for all $i \notin \mathcal{D}$. This gives you a general mathematical form for calculating the probability of a rectangle directly using a multivariate CDF.
Special cases: Application of the general rule yields and expression with $2^n$ terms. For small $n$ we can write this out explicitly without using summation notation and it is not too long. For $n=2$ we get the special case:
$$\mathbb{P}(\mathcal{R_2}) = \Phi(\bar{x}_1, \bar{x}_2) - \Phi(\underline{x}_1, \bar{x}_2) - \Phi(\bar{x}_1, \underline{x}_2) + \Phi(\underline{x}_1, \underline{x}_2).$$
For $n=3$ we get the special case:
\begin{aligned} \mathbb{P}(\mathcal{R_3}) &= \Phi (\bar{x}_1, \bar{x}_2, \bar{x}_3) - \Phi(\underline{x}_1, \bar{x}_2, \bar{x}_3) - \Phi(\bar{x}_1, \underline{x}_2, \bar{x}_3) - \Phi(\bar{x}_1, \bar{x}_2, \underline{x}_3) \\[4pt] &\quad + \Phi(\underline{x}_1, \underline{x}_2, \bar{x}_3) + \Phi(\underline{x}_1, \bar{x}_2, \underline{x}_3) + \Phi(\bar{x}_1, \underline{x}_2, \underline{x}_3) - \Phi(\underline{x}_1, \underline{x}_2, \underline{x}_3). \end{aligned}
For larger $n$ the number of terms increases exponentially and so it becomes cumbersome to write the expression out without using the summation notation in the general form.
• Thanks for the clear example. The last special case really helped me see what was going on in those nested summations. Just got a python code up and running and spitting out the correct results. Thanks again! – Felipe D. Jul 22 '18 at 20:38
• That's great that you got the code working. The easiest way to code something like this is to sum over all the $2^n$ binary values of length $n$, using the binary values as representations of whether you are using an upper or lower bound for an argument in the CDF. – Ben Jul 22 '18 at 22:17
As suggested by @Ben, this problem boils down to an application of the inclusion-exclusion principle. For notational simplicity, let's replace $x_i^{lower}$ with $0$, and $x_i^{upper}$ with $1$, so that for example the vector $\vec{x}=(x_1^{lower},x_2^{upper},x_3^{upper},x_4^{lower})$ becomes $\vec{x}=(0,1,1,0)$.
Then the general formula for $\Phi^{upper}_{lower}(\vec{x}_{lower},\vec{x}_{upper})$ with $n$-dimensional $\vec{x}$ is:
$$\Phi^{upper}_{lower}\big((0,\dots,0),\ (1,\dots,1)\big) \ = \ \Phi^{upper}(1,\dots,1) \ - \sum_{x_1+\cdots+x_n=n-1} \Phi^{upper}(x_1,\dots,x_n) \ + \cdots \\+ \ (-1)^{n-1} \cdot \sum_{x_1+\cdots+x_n=1} \Phi^{upper}(x_1,\dots,x_n) \ + \ (-1)^n \cdot\Phi^{upper}(0,\dots,0)$$
• Thanks for clearing things up! But I'm still a bit confused. First of all, if we only have two dimensions, $\vec{x}=(x_1,x_2)$. So I believe you meant to say that $\vec{x}_{lower}=(0,0)$ and $\vec{x}_{upper}=(1,1)$. Or more generally, for an n-dimensional case, $\vec{x}_{lower}=(0,...,0)$ and $\vec{x}_{upper}=(1,...,1)$. But besides that, I'm still getting confused with the limits of the summations. What do you mean when you use $x_1+...+x_n=n-1$ as the limits/bounds of your summation? Is it summing all possible combinations where $x_1+...+x_n=n-1$ is true? – Felipe D. Jul 22 '18 at 9:23
• Furthermore, what are the terms $x_1, ..., x_n$ in your example? Are they ones or zeros? This seems a bit ambiguous. Thanks again for the help! – Felipe D. Jul 22 '18 at 9:27
• @FelipeD. My reference to $\vec{x}=(x_1^{lower},x_2^{upper},x_3^{upper},x_4^{lower})$, which I assume is what your confusion is about in your first comment, was just meant to show an example of the notation that I'm using, where it just so happens that $n=4$. You're correct that using this notation $\vec{x}_{lower}=(0,\dots,0)$ and $\vec{x}_{upper}=(1,\dots,1)$. You are also correct in your understanding of the summation notation, that it's referring to all combinations of $x_i$ such that the equality holds. This is a commonly-used notational convention. – jon_simon Jul 22 '18 at 18:33
• Regarding your second comment, yes, all of the $x_i$ are either $0$ (denoting $x_i^{lower}$) or $1$ (denoting $x_i^{upper}$) – jon_simon Jul 22 '18 at 18:36 | 2019-09-17T15:27:52 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/358367/calculating-multivariate-integrals-between-lower-and-upper-bounds",
"openwebmath_score": 0.9776192903518677,
"openwebmath_perplexity": 280.550548157201,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9921841100551559,
"lm_q2_score": 0.8539127529517043,
"lm_q1q2_score": 0.847238664852135
} |
https://mathematica.stackexchange.com/questions/192628/defining-a-recursive-function-with-additional-parameters-that-can-be-used-in-a-m | # Defining a recursive function with additional parameters that can be used in a Manipulated ListPlot
I'm trying to code up some plots for Autocorrelation Functions in Time Series Analysis, which can often be defined recursively. The goal is then to have Manipulate sliders that allow you to dynamically change the controlling parameters. Here's an example of the sort of plots I'm trying to emulate:
(here, $$\phi_1$$ and $$\phi_2$$ are the parameters I want to manipulate.)
For this $$AR(2)$$ model, the ACF, denoted $$\rho_k$$, can be defined recursively as follows:
$$\rho_0 = 1$$
$$\rho_1 = \frac{\phi_1}{1-\phi_2}$$
$$\rho_k = \phi_1 \rho_{k-1} + \phi_2 \rho_{k-2}$$
How do I implement this recursion and ListPlot in Mathematica?
I've tried doing:
ρ[0,φ1_,φ2_]:= 1
ρ[1,φ1_,φ2_]:= φ1/(1-φ2)
ρ[k,φ1_,φ2_]:= φ1 * ρ[k-1,φ1,φ2] + φ2 * ρ[k-2,φ1,φ2]
But then the ListPlot, even when plotted for e.g. {k, 0, 10}, only shows $$ρ[0]$$ and $$ρ[1]$$. I'm still a bit new to coding larger projects in Mathematica (only basic ODE StreamPlots in the past), so I think I'm misunderstanding the proper way to handle defining functions.
• You can use \[Rho][k_, phi1_, phi2_] := If[k == 0, 1, If[k == 1, phi1/(1 - phi2), phi1*\[Rho][k - 1, phi1, phi2] + phi2*\[Rho][k - 2, phi1, phi2]]] – ulvi Mar 5 at 6:05
• You are missing an _ after your k in your third definition. Thus φ1=1/2; φ2=1/4; ρ[0,φ1_,φ2_]:= 1; ρ[1,φ1_,φ2_]:= φ1/(1-φ2); ρ[k_,φ1_,φ2_]:= φ1 * ρ[k-1,φ1,φ2] + φ2 * ρ[k-2,φ1,φ2]; ListPlot[Table[ρ[k,φ1,φ2],{k, 0, 10}]] plots just fine – Bill Mar 5 at 6:16
The underlying difference equation for $$\text{AR}(2)$$ is a three-term recurrence with constant coefficients. RSolve[] can directly solve this, after which you can use the solution along with DiscretePlot[] for visualization.
I will let someone else implement that approach. Instead, let me show how to use LinearRecurrence[] along with ListPlot[]:
Manipulate[ListPlot[Rest[LinearRecurrence[{φ1, φ2}, {1, φ1/(1 - φ2)}, n + 1]],
Filling -> Axis, Frame -> True,
FrameLabel -> {"Lag", "\!$$\*SubscriptBox[\(ρ$$, $$k$$]\)"},
PlotRange -> All, PlotStyle -> Black],
{{φ1, 1}, 0, 3}, {{φ2, -1/2}, -1, 1}, {{n, 12}, 2, 20, 1}]
• ρ[k_, φ1_, φ2_] = ρ[k] /. FullSimplify[First[RSolve[{ρ[0] == 1, ρ[1] == φ1/(1 - φ2), ρ[k] == φ1*ρ[k - 1] + φ2*ρ[k - 2]}, ρ[k], k]]] and then Manipulate[DiscretePlot[ρ[k, φ1, φ2], {k, 10}], {φ1, .5, 1.5}, {φ2, -.75, .25}] – Roman Mar 5 at 10:26
• @Roman, very nice. Why not edit your answer to include it? – J. M. is away Mar 5 at 12:07
• Superb. I'm certainly not wedded to implementing it using the approach I was initially using, so LinearRecurrence[] is a great solution I didn't know about. – Jack Gallagher Mar 5 at 17:06
With recursions I'd recommend to always use memoizing functions; have a look at the tutorial on the distinction between immediate = and delayed := assignments.
You're doing a recursion where k is the recursion index and φ1,φ2 are variables, so there is a big distinction between these two kinds of parameters of ρ. In this case, I would do the recursion in terms of functions of φ1,φ2, not in terms of concrete numerical values φ1,φ2:
ρ[0] = Function[{φ1, φ2}, 1];
ρ[1] = Function[{φ1, φ2}, φ1/(1 - φ2)];
ρ[k_Integer /; k >= 2] := ρ[k] = Function[{φ1, φ2},
Evaluate[φ1*ρ[k-1][φ1, φ2] + φ2*ρ[k-2][φ1, φ2] // FullSimplify]]
This way of defining the recursion has the advantage that the formula for each ρ[k] is only evaluated once, and the recursion is not re-traversed for every numerical value of φ1 and φ2 anew. FullSimplifying at each step in the recursion makes sure that the resulting formulas for the ρ[k] stay reasonably small, and they can be evaluated quickly in a Manipulated plot.
plot:
Manipulate[DiscretePlot[ρ[k][φ1, φ2], {k, 10}], {φ1, .5, 1.5}, {φ2, -.75, .25}]
As @JM points out, this specific recursion has a closed-form solution:
ρ[k_, φ1_, φ2_] = ρ[k] /. FullSimplify[First[
RSolve[{ρ[0] == 1, ρ[1] == φ1/(1-φ2), ρ[k] == φ1*ρ[k-1] + φ2*ρ[k-2]}, ρ[k], k]]]
$$2^{-k-1} \left(\left(\varphi_1-\sqrt{\varphi_1^2+4 \varphi_2}\right)^k+\left(\sqrt{\varphi_1^2+4 \varphi_2}+\varphi_1\right)^k+\frac{\varphi_1 (\varphi_2+1)\left(\left(\varphi_1-\sqrt{\varphi_1^2+4 \varphi_2}\right)^k-\left(\sqrt{\varphi_1^2+4 \varphi_2}+\varphi_1\right)^k\right)}{(\varphi_2-1) \sqrt{\varphi_1^2+4\varphi_2}}\right)$$
Manipulate[DiscretePlot[ρ[k, φ1, φ2], {k, 10}], {φ1, .5, 1.5}, {φ2, -.75, .25}]
• For \[CurlyPhi]2 == (-\[CurlyPhi]1)/4 the closed-form solution needs to be defined in the Limit – Bob Hanlon Mar 5 at 14:40
• Great! Nice to learn what memoizing functions are! That's what I like about this forum. If I were to need to solve this problem myself, I'd come up with my solution and that would be that. There is a lot to learn by seeing different approaches (such as your approach, and J. M.'s, for example)! – mjw Mar 5 at 17:03
• Also, I like the notation where you split the arguments based on function k an index and phi1 and phi2 parameters defining the recurrence. I noticed m_goldberg took a similar approach in another post. I'll have to try to remember to use this going forward! – mjw Mar 5 at 17:32
ρ[k_, φ1_, φ2_] :=
Which[k == 0, 1, k == 1, φ1/(1 - φ2),
k > 1, φ1*ρ[k - 1, φ1, φ2] + φ2*ρ[k - 2, φ1, φ2]]
Manipulate[
ListPlot[Table[{k, ρ[k, φ1, φ2]}, {k, 10}],
PlotRange -> {-.5, 1}, Filling -> Axis ],
{φ1, .5, 1.5}, {φ2, -.75, .25}] | 2019-08-25T10:41:21 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/192628/defining-a-recursive-function-with-additional-parameters-that-can-be-used-in-a-m",
"openwebmath_score": 0.4585334062576294,
"openwebmath_perplexity": 5109.867439166618,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9532750413739076,
"lm_q2_score": 0.8887587846530938,
"lm_q1q2_score": 0.8472315672116019
} |
https://www.physicsforums.com/threads/when-is-the-frobenius-norm-of-a-matrix-equal-to-the-2-norm-of-a-matrix.599237/ | # When is the Frobenius norm of a matrix equal to the 2-norm of a matrix?
What conditions most be true for these two norms to be equal? Or are they always equal?
What conditions most be true for these two norms to be equal? Or are they always equal?
I'm far from being a specialist in this, but it seems to me that "Frobenius norm of a matrix" is just the name given to the 2-norm...
Don
Well, in the applied linear algebra course I'm taking currently, the Frobenius norm of a matrix A is defined as the square root of the trace of A'A and the 2-norm is defined as the square root of the largest eigenvalue of A'A. I'm just not sure if they're always the same.
The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1 (i.e. if and only if the matrix can be represented as A=c r, where r is a row and c is a column).
You can see that from the fact that Frobenius norm is $\left( \sum_k s_k^2\right)^{1/2}$ and the 2-norm is $\max s_k$, where $s_k$ are singular values. So equality happens if and only if there is only one non-zero singular value, which is equivalent to the fact that the rank is 1.
The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1 (i.e. if and only if the matrix can be represented as A=c r, where r is a row and c is a column).
You can see that from the fact that Frobenius norm is $\left( \sum_k s_k^2\right)^{1/2}$ and the 2-norm is $\max s_k$, where $s_k$ are singular values. So equality happens if and only if there is only one non-zero singular value, which is equivalent to the fact that the rank is 1.
Excellent, thank you. The matrix in a proof I'm working on involves a rank 1 matrix, so this equality of the two norms applies perfectly.
AlephZero
Homework Helper
The Frobenius and 2-norm of a matrix coincide if and only if the matrix has rank 1
More generally, ##||A||_2 \le ||A||_F \le \sqrt{r}||A||_2## where r is the rank of A.
More generally, ##||A||_2 \le ||A||_F \le \sqrt{r}||A||_2## where r is the rank of A.
May you shed some light on this? Or quote any possible reference? Thanks
jbunniii
Homework Helper
Gold Member
May you shed some light on this? Or quote any possible reference? Thanks
Assuming you accept Hawkeye18's formulas, namely
$$\|A\|_F = \left( \sum_k s_k^2\right)^{1/2}$$
and
$$\|A\|_2 = \max{s_k}$$
then we have
$$\|A\|_2 = \max{s_k} = \left( (\max{s_k})^2\right)^{1/2} \leq \left( \sum_{k} s_k^2 \right)^{1/2} = \|A\|_F$$
For the second inequality, note that the rank of ##A## is precisely the number of nonzero singular values. Let's sort the singular values so that the nonzero ones all come first. Then for a rank ##r## matrix, we have
$$\|A\|_F = \left( \sum_{k=1}^{r} s_k^2\right) ^{1/2} \leq \left( \sum_{k=1}^{r} (\max s_k)^2 \right)^{1/2} = (r (\max s_k)^2)^{1/2} = \sqrt{r} \|A\|_2$$
Equality holds if and only if the ##r## nonzero singular values are all equal.
hairetikos and jim mcnamara
when matrix A is Singular which means det(A)=0. | 2021-07-26T05:40:51 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/when-is-the-frobenius-norm-of-a-matrix-equal-to-the-2-norm-of-a-matrix.599237/",
"openwebmath_score": 0.9488903880119324,
"openwebmath_perplexity": 235.0284951136759,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9566342012360932,
"lm_q2_score": 0.8856314738181875,
"lm_q1q2_score": 0.8472253575456058
} |
http://spot.pcc.edu/math/APEXCalculus/sec_dot_product.html | The previous section introduced vectors and described how to add them together and how to multiply them by scalars. This section introduces a multiplication on vectors called the dot product.
##### Definition10.3.1Dot Product
1. Let $\vec u = \la u_1,u_2\ra$ and $\vec v = \la v_1,v_2\ra$ in $\mathbb{R}^2\text{.}$ The dot product of $\vec u$ and $\vec v\text{,}$ denoted \dotp uv, is \begin{equation*} \dotp uv = u_1v_1+u_2v_2. \end{equation*}
2. Let $\vec u = \la u_1,u_2,u_3\ra$ and $\vec v = \la v_1,v_2,v_3\ra$ in $\mathbb{R}^3\text{.}$ The dot product of $\vec u$ and $\vec v\text{,}$ denoted \dotp uv, is \begin{equation*} \dotp uv = u_1v_1+u_2v_2+u_3v_3. \end{equation*}
Note how this product of vectors returns a scalar, not another vector. We practice evaluating a dot product in the following example, then we will discuss why this product is useful.
##### Example10.3.2Evaluating dot products
1. Let $\vec u=\la 1,2\ra\text{,}$ $\vec v=\la 3,-1\ra$ in $\mathbb{R}^2\text{.}$ Find \dotp uv.
2. Let $\vec x = \la 2,-2,5\ra$ and $\vec y = \la -1, 0, 3\ra$ in $\mathbb{R}^3\text{.}$ Find \dotp xy.
Solution
The dot product, as shown by the preceding example, is very simple to evaluate. It is only the sum of products. While the definition gives no hint as to why we would care about this operation, there is an amazing connection between the dot product and angles formed by the vectors. Before stating this connection, we give a theorem stating some of the properties of the dot product.
The last statement of the theorem makes a handy connection between the magnitude of a vector and the dot product with itself. Our definition and theorem give properties of the dot product, but we are still likely wondering “What does the dot product mean?” It is helpful to understand that the dot product of a vector with itself is connected to its magnitude.
The next theorem extends this understanding by connecting the dot product to magnitudes and angles. Given vectors $\vec u$ and $\vec v$ in the plane, an angle $\theta$ is clearly formed when $\vec u$ and $\vec v$ are drawn with the same initial point as illustrated in Figure 10.3.4(a). (We always take $\theta$ to be the angle in $[0,\pi]$ as two angles are actually created.)
The same is also true of 2 vectors in space: given $\vec u$ and $\vec v$ in $\mathbb{R}^3$ with the same initial point, there is a plane that contains both $\vec u$ and $\vec v\text{.}$ (When $\vec u$ and $\vec v$ are co-linear, there are infinite planes that contain both vectors.) In that plane, we can again find an angle $\theta$ between them (and again, $0\leq \theta\leq \pi$). This is illustrated in Figure 10.3.4(b).
The following theorem connects this angle $\theta$ to the dot product of $\vec u$ and $\vec v\text{.}$
When $\theta$ is an acute angle (i.e., $0\leq \theta \lt \pi/2$), $\cos(\theta)$ is positive; when $\theta = \pi/2\text{,}$ $\cos(\theta) = 0\text{;}$ when $\theta$ is an obtuse angle ($\pi/2\lt \theta \leq \pi$), $\cos(\theta)$ is negative. Thus the sign of the dot product gives a general indication of the angle between the vectors, illustrated in <<Unresolved xref, reference "fig_dotpsign"; check spelling or use "provisional" attribute>>.
\captionof{figure}{Illustrating the relationship between the angle between vectors and the sign of their dot product.} We can use Theorem 10.3.7 to compute the dot product, but generally this theorem is used to find the angle between known vectors (since the dot product is generally easy to compute). To this end, we rewrite the theorem's equation as \begin{equation*} \cos(\theta) = \frac{\dotp uv}{\norm{\vec u}\norm{\vec v}} \Leftrightarrow \theta = \cos^{-1}\left(\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right). \end{equation*}
We practice using this theorem in the following example.
##### Example10.3.9Using the dot product to find angles
Let $\vec u = \la 3,1\ra\text{,}$ $\vec v = \la -2,6\ra$ and $\vec w = \la -4,3\ra\text{,}$ as shown in Figure 10.3.8. Find the angles $\alpha\text{,}$ $\beta$ and $\theta\text{.}$
Solution
We see from our computation that $\alpha + \beta = \theta\text{,}$ as indicated by Figure 10.3.8. While we knew this should be the case, it is nice to see that this non-intuitive formula indeed returns the results we expected.
We do a similar example next in the context of vectors in space.
##### Example10.3.11Using the dot product to find angles
Let $\vec u = \la 1,1,1\ra\text{,}$ $\vec v = \la -1,3,-2\ra$ and $\vec w = \la -5,1,4\ra\text{,}$ as illustrated in Figure 10.3.10. Find the angle between each pair of vectors.
Solution
All three angles between these vectors was $\pi/2\text{,}$ or $90^\circ\text{.}$ We know from geometry and everyday life that $90^\circ$ angles are “nice” for a variety of reasons, so it should seem significant that these angles are all $\pi/2\text{.}$ Notice the common feature in each calculation (and also the calculation of $\alpha$ in Example 10.3.9): the dot products of each pair of angles was 0. We use this as a basis for a definition of the term orthogonal, which is essentially synonymous to perpendicular.
##### Definition10.3.12Orthogonal
Vectors $\vec u$ and $\vec v$ are orthogonal if their dot product is 0.
The term perpendicular originally referred to lines. As mathematics progressed, the concept of “being at right angles to” was applied to other objects, such as vectors and planes, and the term orthogonal was introduced. It is especially used when discussing objects that are hard, or impossible, to visualize: two vectors in 5-dimensional space are orthogonal if their dot product is 0. It is not wrong to say they are perpendicular, but common convention gives preference to the word orthogonal.
##### Example10.3.13Finding orthogonal vectors
Let $\vec u = \la 3,5\ra$ and $\vec v = \la 1,2,3\ra\text{.}$
1. Find two vectors in $\mathbb{R}^2$ that are orthogonal to $\vec u\text{.}$
2. Find two non–parallel vectors in $\mathbb{R}^3$ that are orthogonal to $\vec v\text{.}$
Solution
An important construction is illustrated in Figure 10.3.14, where vectors $\vec u$ and $\vec v$ are sketched. In part (a), a dotted line is drawn from the tip of $\vec u$ to the line containing $\vec v\text{,}$ where the dotted line is orthogonal to $\vec v\text{.}$ In part (b), the dotted line is replaced with the vector $\vec z$ and $\vec w$ is formed, parallel to $\vec v\text{.}$ It is clear by the diagram that $\vec u = \vec w+\vec z\text{.}$ What is important about this construction is this: $\vec u$ is decomposed as the sum of two vectors, one of which is parallel to $\vec v$ and one that is perpendicular to $\vec v\text{.}$ It is hard to overstate the importance of this construction (as we'll see in upcoming examples).
The vectors $\vec w\text{,}$ $\vec z$ and $\vec u$ as shown in Figure 10.3.14 (b) form a right triangle, where the angle between $\vec v$ and $\vec u$ is labeled $\theta\text{.}$ We can find $\vec w$ in terms of $\vec v$ and $\vec u\text{.}$
Using trigonometry, we can state that $$\norm{\vec w} = \norm{\vec u}\cos(\theta) . \label{eq_proj1}\tag{10.3.1}$$
We also know that $\vec w$ is parallel to to $\vec v$ ; that is, the direction of $\vec w$ is the direction of $\vec v\text{,}$ described by the unit vector $\frac{1}{\norm{\vec v}}\vec v\text{.}$ The vector $\vec w$ is the vector in the direction $\frac{1}{\norm{\vec v}}\vec v$ with magnitude $\norm{\vec u}\cos(\theta)\text{:}$ \begin{align*} \vec w \amp = \Big(\norm{\vec u}\cos(\theta) \Big)\frac{1}{\norm{\vec v}}\vec v.\\ \end{align*} Replace $\cos(\theta)$ using Theorem 10.3.7: \begin{align*} \amp = \left(\norm{\vec u}\frac{\dotp uv}{\norm{\vec u}\norm{\vec v}}\right)\frac{1}{\norm{\vec v}} \vec v\\ \amp = \frac{\dotp uv}{\norm{\vec v}^2}\vec v.\\ \end{align*} Now apply Theorem 10.3.3. \begin{align*} \amp = \frac{\dotp uv}{\dotp vv}\vec v. \end{align*}
Since this construction is so important, it is given a special name.
##### Definition10.3.17Orthogonal Projection
Let $\vec u$ and $\vec v$ be given. The orthogonal projection of $\vec u$ onto $\vec v$, denoted $\proj uv\text{,}$ is \begin{equation*} \proj uv = \frac{\dotp uv}{\dotp vv}\vec v. \end{equation*}
##### Example10.3.18Computing the orthogonal projection
1. Let $\vec u= \la -2,1\ra$ and $\vec v=\la 3,1\ra\text{.}$ Find $\proj uv\text{,}$ and sketch all three vectors with initial points at the origin.
2. Let $\vec w = \la 2,1,3\ra$ and $\vec x = \la 1,1,1\ra\text{.}$ Find $\proj wx\text{,}$ and sketch all three vectors with initial points at the origin.
Solution
Consider Figure 10.3.20 where the concept of the orthogonal projection is again illustrated. It is clear that $$\vec u = \proj uv + \vec z. \label{eq_orthogproj}\tag{10.3.2}$$
As we know what $\vec u$ and $\proj uv$ are, we can solve for $\vec z$ and state that \begin{equation*} \vec z = \vec u - \proj uv. \end{equation*}
This leads us to rewrite Equation (10.3.2) in a seemingly silly way: \begin{equation*} \vec u = \proj uv + (\vec u - \proj uv). \end{equation*}
This is not nonsense, as pointed out in the following Key Idea. (Notation note: the expression “$\parallel \vec y$” means “is parallel to $\vec y\text{.}$” We can use this notation to state “$\vec x\parallel\vec y$” which means “$\vec x$ is parallel to $\vec y\text{.}$” The expression “$\perp \vec y$” means “is orthogonal to $\vec y\text{,}$” and is used similarly.)
##### Key Idea10.3.21Orthogonal Decomposition of Vectors
Let $\vec u$ and $\vec v$ be given. Then $\vec u$ can be written as the sum of two vectors, one of which is parallel to $\vec v\text{,}$ and one of which is orthogonal to $\vec v\text{:}$ \begin{equation*} \vec u = \underbrace{\proj uv}_{\parallel\ \vec v}\ +\ (\underbrace{\vec u-\proj uv}_{\perp\ \vec v}). \end{equation*}
We illustrate the use of this equality in the following example.
##### Example10.3.22Orthogonal decomposition of vectors
1. Let $\vec u = \la -2,1\ra$ and $\vec v = \la 3,1\ra$ as in Example 10.3.18. Decompose $\vec u$ as the sum of a vector parallel to $\vec v$ and a vector orthogonal to $\vec v\text{.}$
2. Let $\vec w =\la 2,1,3\ra$ and $\vec x =\la 1,1,1\ra$ as in Example 10.3.18. Decompose $\vec w$ as the sum of a vector parallel to $\vec x$ and a vector orthogonal to $\vec x\text{.}$
Solution
We give an example of where this decomposition is useful.
##### Example10.3.23Orthogonally decomposing a force vector
Consider Figure 10.3.24(a), showing a box weighing 50lb on a ramp that rises 5ft over a span of 20ft. Find the components of force, and their magnitudes, acting on the box (as sketched in part (b) of the figure):
1. in the direction of the ramp, and
2. orthogonal to the ramp.
Solution
# Subsection10.3.1Application to Work
In physics, the application of a force $F$ to move an object in a straight line a distance $d$ produces work; the amount of work $W$ is $W=Fd\text{,}$ (where $F$ is in the direction of travel). The orthogonal projection allows us to compute work when the force is not in the direction of travel.
Consider Figure 10.3.27, where a force $\vec F$ is being applied to an object moving in the direction of $\vec d\text{.}$ (The distance the object travels is the magnitude of $\vec d\text{.}$) The work done is the amount of force in the direction of $\vec d\text{,}$ $\norm{\proj Fd}\text{,}$ times $\vnorm d\text{:}$ \begin{align*} \norm{\proj Fd}\cdot\vnorm d \amp = \snorm{\frac{\dotp Fd}{\dotp dd}\vec d}\cdot \vnorm d\\ \amp = \abs{\frac{\dotp Fd}{\vnorm d^2}}\cdot \vnorm d\cdot\vnorm d\\ \amp = \frac{\abs{\dotp Fd}}{\vnorm d^2}\vnorm d^2\\ \amp = \abs{\dotp Fd}. \end{align*}
The expression $\dotp Fd$ will be positive if the angle between $\vec F$ and $\vec d$ is acute; when the angle is obtuse (hence $\dotp Fd$ is negative), the force is causing motion in the opposite direction of $\vec d\text{,}$ resulting in “negative work.” We want to capture this sign, so we drop the absolute value and find that $W = \dotp Fd\text{.}$
##### Definition10.3.28Work
Let $\vec F$ be a constant force that moves an object in a straight line from point $P$ to point $Q\text{.}$ Let $\vec d = \vv{PQ}\text{.}$ The work $W$ done by $\vec F$ along $\vec d$ is $W = \dotp Fd\text{.}$
##### Example10.3.29Computing work
A man slides a box along a ramp that rises 3ft over a distance of 15ft by applying 50lb of force as shown in Figure 10.3.30. Compute the work done.
Solution
The dot product is a powerful way of evaluating computations that depend on angles without actually using angles. The next section explores another “product” on vectors, the cross product. Once again, angles play an important role, though in a much different way.
# Subsection10.3.2Exercises
Terms and Concepts
In the following exercises, find the dot product of the given vectors.
In the following exercises, find the measure of the angle between the two vectors in both radians and degrees.
In the following exercises, a vector $\vec v$ is given. Give two vectors that are orthogonal to $\vec v\text{.}$
In the following exercises, vectors $\vec u$ and $\vec v$ are given. Find $\proj uv\text{,}$ the orthogonal projection of $\vec u$ onto $\vec v\text{,}$ and sketch all three vectors on the same axes.
In the following exercises, vectors $\vec u$ and $\vec v$ are given. Write $\vec u$ as the sum of two vectors, one of which is parallel to $\vec v$ (or is zero) and one of which is orthogonal to $\vec v\text{.}$ Note: these are the same pairs of vectors as found in Exercises 10.3.2.2110.3.2.26. | 2021-12-03T07:30:58 | {
"domain": "pcc.edu",
"url": "http://spot.pcc.edu/math/APEXCalculus/sec_dot_product.html",
"openwebmath_score": 0.8732830286026001,
"openwebmath_perplexity": 352.75140147558216,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9944450744731587,
"lm_q2_score": 0.8519528057272544,
"lm_q1q2_score": 0.847220271339056
} |
https://math.stackexchange.com/questions/2523807/product-of-monoids | # Product of monoids
I've been interested in studying bialgebras more abstractly, in the form of a bimonoid internal to a symmetric monoidal category, but I'm getting stuck on the compatibility conditions for bimonoids. Allow me to first introduce the definitions.
Let $(\mathcal{C}, \otimes, I)$ be a monoidal category, then a monoid internal to $\mathcal{C}$ is an object $M$ along with a unit map $u: I \to M$ and a multiplication map $m: M \otimes M \to M$ such that some associativity and unit axioms hold, i.e. the relevant diagrams commute. A comonoid internal to $\mathcal{C}$ is an object $C$ along with a counit map $\varepsilon: C \to I$ and a comultiplication map $\Delta: C \to C \otimes C$ such that some coassociativity and counit axioms hold (exactly the reverse of the diagrams for a monoid). For example, in the category of $k$-modules, where $k$ is a commutative ring, a monoid is exactly an associative unital $k$-algebra, and a comonoid is exactly a coassociative counital $k$-coalgebra.
Suppose that we have an object $B$ in the category $\mathcal{C}$ which is both a monoid with maps $(u, m)$ and a comonoid with maps $(\varepsilon, \Delta)$. For $B$ to be a bimonoid, we require some further compatibility between these maps: namely that $(u, m)$ are comonoid homomorphisms, and $(\varepsilon, \Delta)$ are monoid homomorphisms. To check that $\Delta: B \to B \otimes B$ is a monoid homomorphism, we need to be able to put a monoidal structure on $B \otimes B$.
Here is a guess at what the monoidal structure on $B \otimes B$ is meant to be. Suppose $\mathcal{C}$ is now symmetric monoidal, and we have two monoids $(X, u_X, m_X)$ and $(Y, u_Y, m_Y)$ in $\mathcal{C}$.Then $X \otimes Y$ has some candidate maps for making it a monoid. The unit is the following composite:
$$u_{X \otimes Y} = \left(I \xrightarrow{\sim} I \otimes I \xrightarrow{u_X \otimes u_Y} X \otimes Y\right)$$
and the multiplication uses the flip map $T_{Y, X}: Y \otimes X \to X \otimes Y$ coming from the symmetric monoidal structure.
$$m_{X \otimes Y} = \left( X \otimes Y \otimes X \otimes Y \xrightarrow{1 \otimes T_{Y, X} \otimes 1} X \otimes X \otimes Y \otimes Y \xrightarrow{m_X \otimes m_Y} X \otimes Y \right)$$
Question 1: Do these maps make $X \otimes Y$ into a monoid in $\mathcal{C}$?
I can prove the above in special cases (such as the category of $k$-modules), but I am having trouble proving it in an arbitrary symmetric monoidal category, and feel like I might have missed some extra compatibility between the flip map and the multiplication. And to follow up with another question:
Question 2: If there is no natural monoidal structure on $X \otimes Y$, how should I interpret the statement "$\Delta: M \to M \otimes M$ must be a morphism of monoids"?
• Yes, tensor products of monoids are monoids in a symmetric monoidal category. I believe this is false in a monoidal category, and you should be able to find counterexamples where compositions of monads are not monads. According to the nLab the definition of a bimonoid in a monoidal category is difficult to write down, and in fact the nLab doesn’t give such a definition. – Qiaochu Yuan Nov 16 '17 at 23:17
• @QiaochuYuan Thanks for clarifying that. I'm quite happy to stick to symmetric monoidal categories - in this case, do you know where I might find a proof of the fact that $X \otimes Y$ is a monoid? I'm having trouble finding anything on the internet, and on my own I'm struggling to verify even one of the unit axioms. – Joppy Nov 16 '17 at 23:53
It's just naturality of $T$. This is relatively easy to see from a string diagram. Basically, you just slide one of the underlying multiplications through a crossing, then apply associativity, then slide back. Below is a much more opaque equational representation.
For associativity, we want to show that $m_{X\otimes Y} \circ (m_{X\otimes Y}\otimes id_{X\otimes Y}) = m_{X\otimes Y} \circ (id_{X\otimes Y}\otimes m_{X\otimes Y})$. Expanding out the definition we get: \begin{align} & m_{X\otimes Y} \circ (m_{X\otimes Y}\otimes id_{X\otimes Y}) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y)\circ (((m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y))\otimes id_{X\otimes Y}) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y)\circ (m_X\otimes m_Y\otimes id_{X\otimes Y})\circ(id_X\otimes T\otimes id_{Y\otimes X\otimes Y}) \\ =\ & (m_X\otimes m_Y)\circ(m_X\otimes id_X\otimes m_Y\otimes id_Y)\circ (id_{X\otimes X}\otimes T\otimes id_Y)\circ(id_X\otimes T\otimes id_{Y\otimes X\otimes Y}) \\ =\ & (m_X\circ(m_X\otimes id_X))\otimes (m_Y\circ(m_Y\otimes id_Y))\circ (id_{X\otimes X}\otimes T\otimes id_Y)\circ(id_X\otimes T\otimes id_{Y\otimes X\otimes Y}) \\ =\ & (m_X\circ(id_X\otimes m_X))\otimes (m_Y\circ(id_Y\otimes m_Y))\circ (id_{X\otimes X}\otimes T\otimes id_Y)\circ(id_X\otimes T\otimes id_{Y \otimes X\otimes Y}) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes m_X\otimes id_Y\otimes m_Y)\circ (id_{X\otimes X}\otimes T\otimes id_Y)\circ(id_X\otimes T\otimes id_{Y\otimes X\otimes Y}) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes m_X\otimes id_Y\otimes m_Y)\circ (id_X\otimes T\otimes id_{Y \otimes Y})\circ(id_{X\otimes Y\otimes X}\otimes T\otimes id_Y) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y)\circ (id_{X\otimes Y}\otimes m_X\otimes m_Y)\circ(id_{X\otimes Y\otimes X}\otimes T\otimes id_Y) \\ =\ & (m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y)\circ (id_{X\otimes Y}\otimes ((m_X\otimes m_Y)\circ(id_X\otimes T\otimes id_Y))) \\ =\ & m_{X\otimes Y}\circ (id_{X\otimes Y}\otimes m_{X\otimes Y}) \end{align}
The first equation expands definitions, then functoriality of $\otimes$ in the left parameter is used, then naturality of $T$, then bifunctoriality of $\otimes$, then associativity of $m_X$ and $m_Y$, and then we do the previous steps in reverse, with a rewrite of the permutation represented by the left hand composite along the way. That can be understood by rewriting via $T_{X,Y\otimes Y} = (T_{X,Y}\otimes id_Y)\circ(id_Y\circ T_{X,Y})$ and applying laws like $T\circ T = id$ and naturality, but really I just rewrote those arrows by utilizing the fact that the point of the coherence laws for the symmetry is that any arrow defined in terms of the symmetry that produces the same permutation is the same arrow. There should also be a bunch of associators and unitors unless the symmetric monoidal structure is stict, but that was too much of a hassle and you don't seem to be concerning yourself with it. | 2019-06-16T10:33:15 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2523807/product-of-monoids",
"openwebmath_score": 0.9983924627304077,
"openwebmath_perplexity": 816.5275009473534,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.985496423290417,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8471955549032134
} |
https://se.mathworks.com/help/matlab/ref/polyval.html | # polyval
Polynomial evaluation
## Description
example
y = polyval(p,x) evaluates the polynomial p at each point in x. The argument p is a vector of length n+1 whose elements are the coefficients (in descending powers) of an nth-degree polynomial:
$p\left(x\right)={p}_{1}{x}^{n}+{p}_{2}{x}^{n-1}+...+{p}_{n}x+{p}_{n+1}.$
The polynomial coefficients in p can be calculated for different purposes by functions like polyint, polyder, and polyfit, but you can specify any vector for the coefficients.
To evaluate a polynomial in a matrix sense, use polyvalm instead.
example
[y,delta] = polyval(p,x,S) uses the optional output structure S produced by polyfit to generate error estimates. delta is an estimate of the standard error in predicting a future observation at x by p(x).
example
y = polyval(p,x,[],mu) or [y,delta] = polyval(p,x,S,mu) use the optional output mu produced by polyfit to center and scale the data. mu(1) is mean(x), and mu(2) is std(x). Using these values, polyval centers x at zero and scales it to have unit standard deviation,
$\stackrel{^}{x}=\frac{x-\overline{x}}{{\sigma }_{x}}\text{\hspace{0.17em}}.$
This centering and scaling transformation improves the numerical properties of the polynomial.
## Examples
collapse all
Evaluate the polynomial $\mathit{p}\left(\mathit{x}\right)=3{\mathit{x}}^{2}+2\mathit{x}+1$ at the points $\mathit{x}=5,7,9$. The polynomial coefficients can be represented by the vector [3 2 1].
p = [3 2 1];
x = [5 7 9];
y = polyval(p,x)
y = 1×3
86 162 262
Evaluate the definite integral
$I={\int }_{-1}^{3}\left(3{x}^{4}-4{x}^{2}+10x-25\right)dx.$
Create a vector to represent the polynomial integrand $3{x}^{4}-4{x}^{2}+10x-25$. The ${\mathit{x}}^{3}$ term is absent and thus has a coefficient of 0.
p = [3 0 -4 10 -25];
Use polyint to integrate the polynomial using a constant of integration equal to 0.
q = polyint(p)
q = 1×6
0.6000 0 -1.3333 5.0000 -25.0000 0
Find the value of the integral by evaluating q at the limits of integration.
a = -1;
b = 3;
I = diff(polyval(q,[a b]))
I = 49.0667
Fit a linear model to a set of data points and plot the results, including an estimate of a 95% prediction interval.
Create a few vectors of sample data points (x,y). Use polyfit to fit a first degree polynomial to the data. Specify two outputs to return the coefficients for the linear fit as well as the error estimation structure.
x = 1:100;
y = -0.3*x + 2*randn(1,100);
[p,S] = polyfit(x,y,1);
Evaluate the first-degree polynomial fit in p at the points in x. Specify the error estimation structure as the third input so that polyval calculates an estimate of the standard error. The standard error estimate is returned in delta.
[y_fit,delta] = polyval(p,x,S);
Plot the original data, linear fit, and 95% prediction interval $\mathit{y}±2\Delta$.
plot(x,y,'bo')
hold on
plot(x,y_fit,'r-')
plot(x,y_fit+2*delta,'m--',x,y_fit-2*delta,'m--')
title('Linear Fit of Data with 95% Prediction Interval')
legend('Data','Linear Fit','95% Prediction Interval')
Create a table of population data for the years 1750 - 2000 and plot the data points.
year = (1750:25:2000)';
pop = 1e6*[791 856 978 1050 1262 1544 1650 2532 6122 8170 11560]';
T = table(year, pop)
T=11×2 table
year pop
____ _________
1750 7.91e+08
1775 8.56e+08
1800 9.78e+08
1825 1.05e+09
1850 1.262e+09
1875 1.544e+09
1900 1.65e+09
1925 2.532e+09
1950 6.122e+09
1975 8.17e+09
2000 1.156e+10
plot(year,pop,'o')
Use polyfit with three outputs to fit a 5th-degree polynomial using centering and scaling, which improves the numerical properties of the problem. polyfit centers the data in year at 0 and scales it to have a standard deviation of 1, which avoids an ill-conditioned Vandermonde matrix in the fit calculation.
[p,~,mu] = polyfit(T.year, T.pop, 5);
Use polyval with four inputs to evaluate p with the scaled years, (year-mu(1))/mu(2). Plot the results against the original years.
f = polyval(p,year,[],mu);
hold on
plot(year,f)
hold off
## Input Arguments
collapse all
Polynomial coefficients, specified as a vector. For example, the vector [1 0 1] represents the polynomial ${x}^{2}+1$, and the vector [3.13 -2.21 5.99] represents the polynomial $3.13{x}^{2}-2.21x+5.99$.
Data Types: single | double
Complex Number Support: Yes
Query points, specified as a vector. polyval evaluates the polynomial p at the points in x and returns the corresponding function values in y.
Data Types: single | double
Complex Number Support: Yes
Error estimation structure. This structure is an optional output from [p,S] = polyfit(x,y,n) that can be used to obtain error estimates. S contains the following fields:
FieldDescription
RTriangular factor from a QR decomposition of the Vandermonde matrix of x
dfDegrees of freedom
normrNorm of the residuals
If the data in y is random, then an estimate of the covariance matrix of p is (Rinv*Rinv')*normr^2/df, where Rinv is the inverse of R.
Centering and scaling values, specified as a two-element vector. This vector is an optional output from [p,S,mu] = polyfit(x,y,n) that is used to improve the numerical properties of fitting and evaluating the polynomial p. The value mu(1) is mean(x), and mu(2) is std(x). These values are used to center the query points in x at zero with unit standard deviation.
Specify mu to evaluate p at the scaled points, (x - mu(1))/mu(2).
## Output Arguments
collapse all
Function values, returned as a vector of the same size as the query points x. The vector contains the result of evaluating the polynomial p at each point in x.
Standard error for prediction, returned as a vector of the same size as the query points x. Generally, an interval of y ± Δ corresponds to a roughly 68% prediction interval for future observations of large samples, and y ± 2Δ a roughly 95% prediction interval.
If the coefficients in p are least-squares estimates computed by polyfit, and the errors in the data input to polyfit are independent, normal, and have constant variance, then y ± Δ is at least a 50% prediction interval. | 2021-10-16T02:51:17 | {
"domain": "mathworks.com",
"url": "https://se.mathworks.com/help/matlab/ref/polyval.html",
"openwebmath_score": 0.7840487957000732,
"openwebmath_perplexity": 1806.9193324224752,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964211605607,
"lm_q2_score": 0.8596637559030338,
"lm_q1q2_score": 0.8471955548438856
} |
https://math.stackexchange.com/questions/3723481/an-open-interval-is-not-a-disjoint-union-of-two-or-more-open-intervals | # An open interval is not a disjoint union of two or more open intervals?
It seems intuitively clear that an open interval (like $$(a,b), (a, \infty), (-\infty,a)$$ or $$\mathbb{R}$$) cannot be written as a disjoint union of two or more (nonempty) open intervals, but I'm not sure how to prove this rigorously.
Here's my attempt:
I assume the result that open intervals are connected. To prove the result, I show that the disjoint union of two or more open intervals is disconnected. Take any open interval of this union; call it $$A$$. Let $$B$$ denote the union of the other open intervals. So, our set is the disjoint union $$A \cup B$$. We wish to show that this is disconnected, i.e. that $$\overline{A} \cap B$$ and $$A \cap \overline{B}$$ are empty. We first show that $$\overline{A} \cap B = \emptyset$$. If $$A = \mathbb{R}$$ then $$A \cup B$$ would not have been disjoint, so $$A \neq \mathbb{R}$$ and hence $$A$$ is bounded in one direction, so one of its endpoints is a real number. Then $$\overline{A}$$ contains this real number. But $$B$$ cannot contain any of the real endpoints of $$A$$, because otherwise it will intersect with points of $$A$$ (since $$B$$ is open). So $$\overline{A} \cap B = \emptyset$$. Next we show that $$A \cap \overline{B} = \emptyset$$. If this were nonempty, then some element $$b$$ of $$\overline{B}$$ is in the open interval $$A$$. Since the closure of a set is the set of all its limit points, this means every neighborhood of $$b$$ contains elements of $$B$$. But this is impossible, since, for example, take an open interval centered at $$b$$ and contained in $$A$$; this is disjoint from $$B$$.
• you can make it simpler... Suppose for instance that $(a,b) = (a,c) \cup (d,b)$, with $(a,c) \cap (d,b) = \emptyset$ and, of course $a<c<d<b$. What hapens to $c$? does it belong to the union? – PierreCarre Jun 17 '20 at 14:11
• This is essentially about completeness of reals, but the result looks so trivial as to demand the deep property of completeness. – Paramanand Singh Jun 19 '20 at 7:06
• The same result does not hold for intervals in $\mathbb {Q}$. – Paramanand Singh Jun 19 '20 at 7:07
Let $$(a, b)$$ be an open interval, and $$(c, d)$$ an open interval properly contained in $$(a, b).$$ Then $$a \leqslant c < d \leqslant b,$$ and $$a < c$$ or $$d < b.$$ The set $$(a, b) \setminus (c, d) = (a, c] \cup [d, b)$$ is not open, because it contains $$c$$ or $$d$$ or both, but it does not contain a neighbourhood of either. Therefore $$(a, b)$$ is not the disjoint union of $$(c, d)$$ with the union of any non-empty collection of open intervals.
## An application
(Proposition 4 is the promised application, while Proposition 5 is a by-product of the argument.)
Proposition 1. An open interval is not the disjoint union of an open interval and a non-empty open set.
Proof. See above. $$\ \square$$
Proposition 2. The union of a non-empty collection of open intervals with a non-empty intersection is an open interval.
Proof. Let $$\mathscr{I}$$ be a non-empty collection of open intervals containing a given point $$c \in \mathbb{R},$$ and let $$J = \bigcup\mathscr{I}.$$ In $$\overline{\mathbb{R}} = \mathbb{R} \cup \{+\infty, -\infty\},$$ let $$a = \inf J$$ and $$b = \sup J.$$ Then $$a \notin J$$ and $$b \notin J.$$ If $$a < x < b,$$ then $$c \leqslant x < b$$ or $$a < x \leqslant c,$$ and in either case $$x \in I \subseteq J$$ for some $$I \in \mathscr{I}.$$ Therefore $$J = (a, b).$$ $$\ \square$$
Proposition 3. If $$x \in U \subseteq \mathbb{R},$$ and $$U$$ is open, then $$U = J \cup W,$$ where $$x \in J,$$ $$J$$ is an open interval, $$W$$ is an open set, and $$J \cap W = \varnothing.$$
Proof. Let $$J$$ be the union of all open intervals $$I$$ such that $$x \in I \subseteq U.$$ By Proposition 2, $$J$$ is an interval $$(a, b).$$ Clearly, $$a \notin U$$ and $$b \notin U,$$ therefore $$U = (a, b) \cup (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)),$$ so we can take $$W = (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)).$$ $$\ \square$$
Proposition 4. An open interval is not the disjoint union of two non-empty open sets.
Proof. Let $$I$$ be an open interval, and suppose that $$I = U \cup V,$$ where $$U$$ and $$V$$ are disjoint non-empty open sets. Take any $$x \in U.$$ By Proposition 3, $$U = J \cup W,$$ where $$x \in J,$$ $$J$$ is an open interval, $$W$$ is an open set, and $$J \cap W = \varnothing.$$ Therefore $$I = (J \cup W) \cup V = J \cup (W \cup V), \text{ and } J \cap (W \cup V) = (J \cap W) \cup (J \cap V) = \varnothing.$$ This contradicts Proposition 1; so the hypothesis that $$I = U \cup V$$ is false. $$\ \square$$
Proposition 5. Every open subset of $$\mathbb{R}$$ is the union of a countable collection of pairwise disjoint open intervals.
Proof. Let $$U$$ be an open subset of $$\mathbb{R},$$ and let $$\mathscr{J}$$ be the collection of all maximal open subintervals of $$U.$$ By Proposition 3, $$U = \bigcup\mathscr{J},$$ and any two members of $$\mathscr{J}$$ with a non-empty intersection are equal. Because each member of $$\mathscr{J}$$ contains a rational number, $$\mathscr{J}$$ is countable. $$\ \square$$
• For the proof of Prop 2, I'm not sure why we have "in either case $x\in I \subseteq J$ for some $I \in \mathscr{I}$"? Could you please elaborate? – twosigma Jun 18 '20 at 20:41
• It was a bit terse. If $c \leqslant x < b,$ then because $b = \sup J,$ there exists $y \in J$ such that $x < y,$ therefore there exists an open interval $I \in \mathscr{I}$ such that $y \in I.$ We then have $c \in I$ and $y \in I$ and $c \leqslant x < y,$ therefore $x \in I \subseteq J.$ The proof for $a < x \leqslant c$ is similar, with the inequality signs going the other way round. (By the way, I'm afraid the proof of Proposition 5 is even more terse, and it may be more easily understood with reference to the proof of Proposition 3 than its statement.) – Calum Gilhooley Jun 18 '20 at 20:52
• Thanks for the detailed answer and additional interesting propositions. For Prop 5, I have seen the proof before so that is ok. – twosigma Jun 18 '20 at 21:46
• Like you, I imagined at first that a quite elaborate proof would be needed to answer the general case of your question, even though @PierreCarre's comment takes care of the case of two intervals without fuss. It was only after a long comfortable soak in the bath last night that I realised there was a simple proof. By that time I had dreamt up most of these other arguments, because my initial more elaborate idea had been to prove Prop. 4 from first principles and deduce Prop. 1. Something like that, anyway. It was only in bed this morning, unable to sleep, that I got it all straight in my head. – Calum Gilhooley Jun 18 '20 at 22:32 | 2021-08-05T15:25:18 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3723481/an-open-interval-is-not-a-disjoint-union-of-two-or-more-open-intervals",
"openwebmath_score": 0.8846967816352844,
"openwebmath_perplexity": 81.7338026084833,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964194566753,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8471955516074846
} |
http://gesuitialquirinale.it/hkqo/fundamental-theorem-of-calculus-worksheet-doc.html | ## Fundamental Theorem Of Calculus Worksheet Doc
There are worked out examples here that you will find helpful. Topics include: limits, continuity, differentiation, curve sketching, applications of differentiation, integration, the Fundamental Theorem of Calculus, and applications of integration. 4 The Fundamental Theorem of Calculus 1 day 4. Assignment #8: Definite Integration u-Sub Worksheet Answers. Elementary functions, differential calculus and integral calculus will be the three major areas that will be studied through a thorough approach. Yale Department of Economimcs commemorates the 50th anniversary of coeducation in Yale College and the 150th anniversary of women students at the university. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Theorem 2 (The Fundamental Theorem of Calculus, Part I) If F is continuous and its derivative. 396406 2 APC 14 2. Slope fields 7. Watch more videos on http://www. 4 Fundamental Theorem of Calculus Calculus Using the evaluation part, we are going to develop the concept of the other part of the Fundamental Theorem of Calculus. STATISTICS. This booklet contains the worksheets for Math 53, U. 371) • composition of functions (p. Become a Subscriber to access hundreds of standards aligned worksheets. Fundamental Theorem of Calculus. In this lesson, we introduce the Fundamental Theorem of Calculus, a result that establishes a relation-ship between antiderivatives and definite integrals. Sìn (3) (4) 11 arcsìn O/c) al+l dy. WORKSHEET 2 ON FUNDAMENTAL THEOREM OF CALCULUS. Definite Integrals and Antiderivatives. 4 Second Fundamental Theorem of Calculus. (a)Find F0(x) by using part(i)of. The Fundamental Theorem of Calculus. I’m not going to lie, this question kind of makes me sad - I’m not sure if the OP is woefully unimaginative or if this is simply another Quora Partner Program farce (OP has 500+ questions and only 2 answers). Here it is Let f(x) be a function which is defined and continuous for a ≤ x ≤ b. Tuesday, 2/12 4. Elementary functions. Solving differential equations a. Differentiation rules 3. Be sure to include all necessary hypotheses. 5 Integration by u-substitution. Sìn (3) (4) 11 arcsìn O/c) al+l dy. If mathematicians can make a contribution to this area, then they will at least be able to point to a huge external application of mathematics. Definite Integrals. Diagnostic Test. () a a d Microsoft Word - FTC Teacher. Water is pumped out of a holding tank at a rate of liters/minute, where t is in. First video in a short series on the topic. Openanew Maple worksheet and type the commands in the left-hand column below into it. 9) CHAPTER 7-APPLICATIONS OF. Definite Integrals and Antiderivatives. Regents-Graphing Polynomial Functions. ©u 12R0X193 9 HKsu vtoan 1S ho RfTt9w NaHr8em WLNLkCQ. Fundamental foundations of differential and integral calculus. Definite and indefinite integrals, the Fundamental Theorem of Calculus, applications of the integral, techniques of integration. BE FOREWARNED THAT THERE ARE LOTS OF PICTURES AND MATH. Showing top 8 worksheets in the category - Definite Integrals. These topics are the focus of the AP Exam. AP Calculus Name: The Second Fundamental Theorem of Calculus. I’m not going to lie, this question kind of makes me sad - I’m not sure if the OP is woefully unimaginative or if this is simply another Quora Partner Program farce (OP has 500+ questions and only 2 answers). 3: First Derivative Test. The Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus (Part I) C. Average Value. The fundamental theorem of calculus homework calculator picture composition essay pay for writing an essay, ideas for college essays california! Starting a daycare center business plan template. Integrals Sigma Notation Definite Integrals (First) Fundamental Theorem of Calculus Second Fundamental Theorem of Calculus Integration By Substitution Definite Integrals Using Substitution Integration By Parts Partial Fractions. continuous) functions with finite integration limits, there are no particular technical concerns about existence of the sum or integral, or interchange of order of integration or summation. And the inflection point is where it goes from concave upward to concave downward (or vice versa). The Second Fundamental Theorem of Calculus T NOTES MATH NSPIRED ©2014 Texas Instruments Incorporated education. Though the concept of torque. It is differentiable on the open interval (a, b). 15 multiple choice questions. Co-Function Identities. (b) Consider the function f(x) on [1;1) de ned by f(x) = Z x 1 p t5 1dt. Shirt features the theorem that connects the two main ideas of calculus - differentiation and integration. (Solution. Use the Fundamental Theorem of Calculus to evaluate the integral: AB_ws_050_Volumes_Washers. It used a different notation, to avoid the confusion about the meaning of the. FTC Practice 1 Let f(x) be given by the graph to the right and define. For now, the result provides a useful and efficient way to compute definite integrals. Quotient Identities. The Evaluation Theorem 11 1. The inde nite integral95 6. Integral calculus develops the concept of finding the sum of an infinite series. 69-72) recalled seeing the method in. We have not really proved the Fundamental Theorem. The variable x which is the input to function G is actually one of the limits of integration. 6 Growth and Decay Models, and 5. Piecewise functions. pdf PRACTICE TEST 1 [2. Review Worksheet Solutions. Let A(x) = Z x a f(t) dt. Recall Part 2 of the Fundamental Theorem of Calculus If f is a continuous function on [a, b], then where F is any antiderivative of f. Yesterday we drew rectangles with the height at the left endpoint or the right endpoint or at the midpoint of the interval. Exercises94 5. So I searched (before the internet) far and wide for a good explanation. 3B3: Closed Form Antiderivatives: 3. Double Angle Formulas. It converts any table of derivatives into a table of integrals and vice versa. The average value theorem. One way to write the Fundamental Theorem of Calculus ( 7. Stewart Calculus, 8th Edition. His work on functions of complex variables for geometry, fundamental theorem of algebra, and the quadratic reciprocity law was far less practical for the every day, but put mathematics on the road to becoming an area of pursuit in itself rather than a means to an end. My goal for this page is to be the ultimate resource for solving limits. If you have additional files, you will upload them at 'Manage Orders' section. The FTC allows us to evaluate a definite integral must Ron Larson | Shape Shifting Materials Q18, Section 11. It is required that a < b The Definite Integral as net change An object moves along the x-axis such that its velocity it) cm ŽS is given by4(t) = 2t. Some of the worksheets displayed are 201 nya 05, Work definite integrals, 06, Evaluating definite integrals, Definite integrals, Sections antiderivatives and inde nite integrals, Fundamental theorem of calculus date period, Math 122 substitution and the definite integral. MathExcel Worksheet # 35: Integrals and the Fundamental Theorem of Calculus Reminders: D2 is due tomorrow by midnight D3 is due next Monday by midnight 1. Multiple Choice 1. ) (This is a 10 point homework grade. Fundamental Theorem of Calculus Naive derivation – Typeset by FoilTEX – 10. The Fundamental Theorem of Calculus May 2, 2010 The fundamental theorem of calculus has two parts: Theorem (Part I). Mathematics & Statistics (Sci) : Definition and properties of Riemann integral, Fundamental Theorem of Calculus, Taylor's theorem. Definite Integrals and antiderivatives. A somewhat intuitive proof of the fundamental theorem is used. Past documents: #11 Old Practice Test 1 (1. 3 The Fundamental Theorem of Calculus 5. Students will understand the meaning of Rolle’s Theorem and the Mean Value Theorem. Simpson's Rule. Let $$f$$ be a function defined on an open interval $$I$$ containing \(c\text{. We give an example, testing the fundamental theorem of calculus: integration is antidifferentiation. Precalculus also covers sequences and series. com/subscription_center?add_user=brightstorm2. The Second Fundamental Theorem of Calculus T NOTES MATH NSPIRED ©2014 Texas Instruments Incorporated education. Torque is a measure of the ability of an applied force to cause an object to turn. This book is directed more at the former audience. pdf PRACTICE TEST 1 [2. V o rA ol fl 6 6r Di9g 9hWtKs9 Hrne7sheRr av CeQd1. A biologist uses a model which predicts the population will increase 2t+5rabbitsperyearwheret represents the number of years from today. These worksheets include an assortment of topics like identifying whole numbers, integers, fractions. 8) graded worksheets 2. Limits involving ln(x) We can use the rules of logarithms given above to derive the following information about limits. Tutorial for Calculus¶. Second Fundamental Theorem of Calculus Practice. These worksheets are a free and fun way to test your electrical engineering knowledge! Check your proficiency with everything from basic electricity to digital circuits. 4 - Larson Calculus - The Fundamental Theorem of Calculus In this lecture, I will introduce the Fundamental Theorem of Calculus (FTC). Differentiation rules 3. If mathematicians can make a contribution to this area, then they will at least be able to point to a huge external application of mathematics. doc Author:. Solving differential equations a. Curriculum planning plays a key role in enabling schools to deliver the VCE to senior secondary students. 4 1 ³f t dt(). Fundamental Theorem of Calculus - Evaluation a. The Definite Integral and the Fundamental Theorem of Calculus Fundamental Theorem of Calculus NMSI Packet PDF FTC And Motion, Total Distance and Average Value Motion Problem Solved 2nd Fundamental Theorem of Calculus Rate in Rate out Integration Review Videos and Worksheets Integration Review 1 Integration Review 2 Integration Review 3. Fill-in-the-blank two-column proofs have also been used. For example, the three fundamental principles of Aristotelean logic (see 1. Calculus Volumes 1, 2, and 3 are licensed under an Attribution-NonCommercial-Sharealike 4. They find the midpoint of given polygons and sketch the graph of the function. 6 Related Rates. Included are detailed discussions of Limits (Properties, Computing, One-sided, Limits at Infinity, Continuity), Derivatives (Basic Formulas, Product/Quotient/Chain Rules L'Hospitals Rule, Increasing/Decreasing/Concave Up/Concave Down, Related Rates, Optimization) and basic Integrals (Basic Formulas. txt) or view presentation slides online. Before 1997, the AP Calculus. Then find all zeros. 2 12) What is the exact area of the region between y x and the x-axis , over the interval [0, 1]? properties Of definite integrals to evaluate each expression. In this course we will cover the calculus of real univariate functions, which was developed during more than two centuries. doc Author: mhuard Created Date: 1/15/2009 7:42:36 PM. doc; The syllabus for AP Calculus AB can be accessed by clicking on the following link: 4. of the particle when t = 7. d) Compare your answers from part (c) with the values given in the Exploration. 1: Extrema on a Closed Interval. 3—The Fundamental Theorem of Calculus We've learned two different branches of calculus so far: differentiation and integration. " IBL calculus worksheets often have problem sets that are designed so that the students have to figure out the ideas themselves once they have an understanding of the basics. * AP ® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. During this activity: a) Students will use the idea of accumulation under the graph to construct a graph of anti-derivative of a linear function. 6 Growth and Decay Models, and 5. 1] is classified as a fundamental trigonometric limit. The theorem is stated and two simple examples are worked. Summation Notation Worksheet 1 Introduction Sigma notation is used as a convenient shorthand notation for the summation of terms. M 11/26: Fundamental theorem of calculus I worksheet and solutions. Table of Contents. We now state all six results; their discussion is deferred to Chapter 3. Unit 6: Differential Equations and Mathematical Modeling (4 weeks) Antiderivatives. A New York Times Op-Ed about mathematics education: How to fix our math education. AP Calculus AB Course Outline. Worksheet # 24: Review for Exam III Worksheet # 25: De nite Integrals of Calculus Worksheet # 26: The Fundamental Theorems of Calculus and the Net Change Theorem Worksheet # 27: Evaluating integrals by Substitution and Further Transcendental Functions Worksheet # 28: Exponential Growth and Decay Worksheet # 29: Area Between Curves, Review I for. Draw the graph of the "area collection function" F 0 for the given function f, as in the example above. Riemann Sums: 11 nn ii ii ca c a 111 nnn ii i i iii ab a b 1. Students should be able to communicate mathematics both orally and in well-written sentences and should be able to explain solutions to problems. The Fundamental Theorem of Calculus 14 1. Assignment #8: Definite Integration u-Sub Worksheet Answers. 1st Part of the Fundamental Theorem of Calculus. There is 1 pending change awaiting review. The fundamental theorem of calculus. State both parts of the Fundamental Theorem of Calculus. 8 Definite Integral by Substitution Pg. For each, be sure to explain your thinking. No calculator unless otherwise stated. For example, the three fundamental principles of Aristotelean logic (see 1. 3B2 * AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site. Showing 20 items from page AP Calculus Applications of Derivatives Part 1 Homework sorted by Assignment Number. An internet tutoring utility for learning and practicing calculus. THE DEFINITE INTEGRAL. Exponential Growth and Decay. Average Value and Average Rate: File Size: 53 kb: File Type: pdf:. Mathematics Resource Manual Learning Lab Forrest County Center. Physics 01-01 Intro and Units. of Calculus. % of Exam Score. Using the Fundamental Theorem Part I, along with the fact. 2, 29 - Integrate cot x log sin x - Chapter 7 NCERT. Join my Google Classroom. Infinite series: alternating, telescoping series, rearrangements, conditional and absolute convergence, convergence tests. Use the Fundamental Theorem of Calculus to explain the difference between definite and indefinite integrals. Calculus I and II). Second Fundamental Theorem of Calculus: 3. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound. Each tick mark on the axes below represents one unit. There is 1 pending change awaiting review. Lesson Plans. Now suppose that that surface was contained in the plane. The Second Fundamental Theorem of Calculus c. DEMO: ‘Calculus in Motion’ software used to illustrate left, right midpoint, trapezoidal. J h NAtl Bl1 qr ximg Nh2tGsM Jr Ie osoeCr4v2e odN. It is estimatedthat t years fromnowthepopulationof a certainlakeside community will be changing at the rate of 0. Then the Fundamental Theorem of Calculus allows us to evaluate this area by using a definite integral, so that The area bounded by the polar curve is given by the formula: 2 2 wholecircle r q ×p q 2 2 r q p p × 1 2 2 rq rf a b=(qq) on ££ f(q i) ba n q-=D (())2 1 1 Area 2i n i fqq = »×Då (())2 1 1 Area 2 lim i n ni fqq ®¥= =å×D rf. LINK TO THE TEXTBOOK'S VIDEOS FOR EACH SECTION: This week students will learn the Fundamental Theorem of Calculus and use calculus to find area under a curve instead of just using geometry. The goal is for all students to have the background and knowledge base to take the AP Calculus exam in May. 09kb; Physics 01-02 Displacement and Vectors. Example: Use the Fundamental Theorem of Calculus to nd each de nite integral. b) State and apply the Mean Value Theorem for integrals. f 1 f x d x 4 6. com Geometry for the SAT from moomoomath. If you have additional files, you will upload them at 'Manage Orders' section. in the calculation. If f is continuous on an open interval I containing a, then for every x in the interval. Unit #5 Ch 6. The Fundamental Theorem of Calculus. ) (This is a 10 point homework grade. Free Calculus worksheets created with Infinite Calculus. Trigonometry – An Overview of Important Topics So I hear you’re going to take a Calculus course? Good idea to brush up on your Trigonometry!! Trigonometry is a branch of mathematics that focuses on relationships between the sides and angles of triangles. Applications of the integral105 1. Let fbe a continuous function on [a;b] and de ne a consequence of the mean value theorem for derivatives) that F and gdi er by a constant. 1 Introduction: An algebraic expression containing two terms is called a binomial expression, Bi means two and nom means term. Microsoft Word - Properties of Sums and Integrals. Areas between graphs105 2. Math 1A introduction to functions and calculus Oliver Knill Spring 2011 4 5 2011 Second midterm practice I Problem 1 TF questions 20 points No justif…. First video in a short series on the topic. AP Calculus AB Course Outline. 3 The Fundamental Theorem of Calculus. AP Calculus AB Assignment Sheet 2019-2020. Never runs out of questions. Join my Google Classroom. These worksheets include an assortment of topics like identifying whole numbers, integers, fractions. 4 The fundamental theorem of calculus Fundamental theorem of calculus: Part i Let f(x) be a bounded and continuous function on an interval [a,b]. 7 Using the Fundamental Theorem of Algebra 369 1. 3 Riemann Sums and Definite Integrals. Finding volumes of solids of revolution by shells or pipes. The videos, which include real-life examples to illustrate the concepts, are ideal for high school students, college students. For this Simpson's Rule worksheet, students use the Fundamental Theorem of Calculus to determine the antiderivatives and evaluate integrals. Students will apply the Definite L -- The Fundamental Theorem of Calculus Quizzes 2. d) The average speed of the shell from time t =1to time t =4. It is required that a < b The Definite Integral as net change An object moves along the x-axis such that its velocity it) cm ŽS is given by4(t) = 2t. Since the function is continuous on its domain of definition, which is the set , then it is continuous on the closed interval. Use MathJax to format equations. Find ff(-4 given that 4 7. The introduction of each worksheet very briefly summarizes the main ideas but is not intended as a substitute for the textbook or lectures. To share and discuss the problems and promises of beginning an AP course. Math 122B - First Semester Calculus and 125 - Calculus I Worksheets The following is a list of worksheets and other materials related to Math 122B and 125 at the UA. 6 Worksheet (didnt get to do 2 FRQs). "Bell curve" refers to the bell shape that is created when a line is plotted using the data points for an item that meets the criteria of normal distribution. Argue that f is increasing. (c) Find the derivative of the function g(x) = R x3 1 p t5 1dt on (1;1). Trigonometric Ratios Of Any Angle Worksheet Answers. (a)Find F0(x) by using part(i)of. Differential calculus develops the concept of instantaneous rate of change of one quantity in relation to another. Exercises100 Chapter 8. These commands are aimed at investigating the Fundamental Theorem of Calculus in the context of 4 x=1 4/(x2 + 1) dx. 1st Part of the Fundamental Theorem of Calculus. The answers for these pages appear at the back of this booklet. What is the. Chapter 11 The Fundamental Theorem Of Calculus (FTOC) The Fundamental Theorem of Calculus is the big aha! moment, and something you might have noticed all along: X-Ray and Time-Lapse vision let us see an existing pattern as an accumulated sequence of changes The two viewpoints are opposites: X-Rays break things apart, Time-Lapses put them together. Worksheet # 24: Review for Exam III Worksheet # 25: De nite Integrals of Calculus Worksheet # 26: The Fundamental Theorems of Calculus and the Net Change Theorem Worksheet # 27: Evaluating integrals by Substitution and Further Transcendental Functions Worksheet # 28: Exponential Growth and Decay Worksheet # 29: Area Between Curves, Review I for. List out the factors, complete the prime factor tree, draw your own prime factor tree, find the GCF and LCM and explore more worksheets on this page. If we assume that this equality holds for every. COURSE: Calculus GRADING PERIOD: 3rd Six Weeks. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. 3 The Fundamental Theorem of Calculus, Part I 272 5. Course Modules Definite Integrals Worksheet Definite Integrals Worksheet 5. Calculus Integration Fundamental Theorem Definite Integral Task Cards and HW classroom tips, teaching ideas & resources for teaching high school math Fundamental Theorem of CalculusThis activity is designed for Calculus 1 or AP Calculus. Let's consider functions that take on negative values. 4: The De nite Integral & Fundamental Theorem of Calculus MTH 124 We begin with a theorem which is of fundamental importance. 112 f (x) dr = —4, 115 f (x) clx = 6,. 3 Riemann Sums and Definite Integrals. Calculus graphical, numerical and algebraic. Curriculum planning plays a key role in enabling schools to deliver the VCE to senior secondary students. It is suitable for a one-semester course, normally known as “Vector Calculus”, “Multivariable Calculus”, or simply “Calculus III”. 6 function 351 Q5& Q6 Exercise 7. The Definite Integral. b a f xdx Fb Fa, where F(x) is any antiderivative of f(x). com/blog/2006/08/21/top-10-best-presentations-ever/. Here you will find A-Level Maths questions by topic. Fundamental theorem of calculus , where F'(x) = f(x), or. Worksheets labeled with are accessible to Help Teaching Pro subscribers only. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. For any value of x > 0, I can calculate the de nite integral Z x 0 f(t)dt = Z x 0 tdt: by nding the area under the curve: 18 16 14 12 10 8 6 4 2 Ð 2 Ð 4 Ð 6 Ð 8 Ð 10 Ð 12. Go to The Equation of a Circle page. SECOND FUNDAMENTAL THEOREM 1. (We assume that all large pumps are similar and all small. The function f is being integrated with respect to a variable t, which ranges between a and x. A New York Times Op-Ed about mathematics education: How to fix our math education. The binder must include worksheets and homework. Let's simplify our life by pretending the region is. Material cannot be shared specially during quizzes or tests. Read & Study Section 16: The Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus. Let A(x) = Z x a f(t) dt. ©u 12R0X193 9 HKsu vtoan 1S ho RfTt9w NaHr8em WLNLkCQ. In Google Classroom. Simpson's Rule. SMARTBoard. Definition of Concavity Let y = f (x) be a differentiable function on an interval I. The Second Fundamental Theorem of Calculus T NOTES MATH NSPIRED ©2014 Texas Instruments Incorporated education. What is the. AP Calculus AB Name_____ Mock AP Exam #3 Review The Mock AP Exam 9. BackGraphFinal Light T-Shirt Fundamental Theorem of Calculus White T-Shirt by Admin_CP4787603. : Worksheet. Here is a set of notes used by Paul Dawkins to teach his Calculus I course at Lamar University. Solutions at the back. MAT 191 Calculus I (5 units) FS. The effect of each command is described in the right-hand column for your reference. pdf PRACTICE TEST 1 [2. Section 5-7 : Computing Definite Integrals. This is a recurring theme in calculus: Big things are made from little things. Shirt features the theorem that connects the two main ideas of calculus - differentiation and integration. Improper Integrals. u-Substitution Extra Practice. Fundamental Theorem of Calculus Naive derivation – Typeset by FoilTEX – 10. 15 multiple choice questions. The Fundamental Theorem of Calculus arises frequently: () d t x dxt dt ττ −∞ ∫ = For finite sums, or integrals of well-behaved (e. Partial fractions, using long division, and using TI-89’s CAS to find patterns for certain types of. The speed of the ball in meters per second is. CKM Assignment: CKM Internet Workspace: Integrals, Area & Reversing D-Rules. These materials include worksheets, extensions, and assessment options. Recall #21, 23, 25-34, 37 and FTC 2 MC Practice worksheet (the one with limits such as 0 to x^3) Routine #45, 51-53, 59-65 Non-Routine (actually a routine but for now are NR) #49-50 These are integrating with respect to y. 394 (11-21) odd Note: #13 a,b,c only) Draw the graph for all problems, even if it is already given to you. Sìn (3) (4) 11 arcsìn O/c) al+l dy. ) Stokes' theorem in its multivariate calculus setting says that the integral of the curl of a vector field over a surface is just the line integral of that field over the boundary. After a short period of time ∆t, the new position of the car is approximately d(t 1) ≈ d(t 0) + v(t 0)∆t, (t 1 = t 0 + ∆t) 4 4 4 4 4 h h h c t 0 t 1 t 2 t 3 Y W. The Fundamental Theorem of Calculus If a function is continuous on the closed interval [a, b], then where F is any function that F’(x) = f(x) x in [a, b]. Integration using U - Substitution. The fundamental theorem of calculus is a simple theorem that has a very intimidating name. Power functions. WORKSHEET 2 ON FUNDAMENTAL THEOREM OF CALCULUS. "Bell curve" refers to the bell shape that is created when a line is plotted using the data points for an item that meets the criteria of normal distribution. The Calculus Controversy. doc Author: jharmon. pdf file in convenient. The videos, which include real-life examples to illustrate the concepts, are ideal for high school students, college students. of the course. Worksheet # 24: Review for Exam III Worksheet # 25: De nite Integrals of Calculus Worksheet # 26: The Fundamental Theorems of Calculus and the Net Change Theorem Worksheet # 27: Evaluating integrals by Substitution and Further Transcendental Functions Worksheet # 28: Exponential Growth and Decay Worksheet # 29: Area Between Curves, Review I for. That is, to compute the integral of a derivative f ′ we need only compute the values of f at the endpoints. The lesson is for 12th graders in AP Calculus. The fundamental theorem of calculus is central to the study of calculus. Example: Evaluate. Title: New Doc 2019-11-12 15. _____ Work problems 3 - 7 using the Fundamental Theorem of Calculus and your calculator. My love for you is like the slope of a concave up function because it is always increasing. This theorem was first established by Sir Isaac Newton. Parent signature: _____ Date:_____ Lesson 51 Unit #5 Ch 6. Some of the worksheets for this concept are Fundamental theorem of calculus date period, Work the fundamental theorem of calculus multiple, Work 24 de nite integrals and the fundamental, Work 29 the fundamental of calculus, The fundamental theorem of calculus ftc, Work 25 the fundamental. 291: 1,3,5,11,15,21,23,25,33,41,43 1 Evaluate a definite integral using the Fundamental Theorem of Calculus. 4A1: An Integral of a Rate of Change as the Net Change: 3. One area in which the text could be improved is the volume of the exercises. 3 Riemann Sums and Definite Integrals. WORKSHEET ON FUNDAMENTAL THEOREM OF CALCULUS. The Fundamental Theorem of Calculus If we refer to A 1 as the area correspondingto regions of the graphof f(x) abovethe x axis, and A 2 as the total area of regions of the graph under the x axis, then we will find that the value of the definite integralI shown abovewill be I = A. Consider the function f(t) = t. d) Compare your answers from part (c) with the values given in the Exploration. Differential Calculus: Definition of a derivative, along with ideas of limits and continuity, use of table of derivatives. 3Worksheet)' % 1. Approximations with Riemann sums. 44 Chapter 3. Another way to write the FTC2 is. Multiple-choice & free-response. Assignment #8: Definite Integration u-Sub Worksheet Answers. Draw the graph of the "area collection function" F 0 for the given function f, as in the example above. The Definite Integral 6 1. CKM Assignment: CKM Internet Workspace: Integrals, Area & Reversing D-Rules. Integration by Substitution. Part1: Define, for a ≤ x ≤ b, F(x) = R x. Therefore, the desired function is f(x)=1 4 x4 + 2 x +2x−5 4. , n! = n(n – 1)(n – 2) …. Some topics in calculus require much more rigor and precision than topics encountered in previous. Evaluate area under a curve using the FTC 34 -36 New. Unique factoring worksheets are available for grade 5 through high school. Fundamental theorem of calculus , where F'(x) = f(x), or. 344 Chapter 7 Polynomial Functions Polynomial Functions • polynomial function (p. In your own words and using the integral notation we've learned, state the First Fundamental Theorem of Calculus as a hypothesis (what conditions must be true) and a conclusion (what are you guaranteed will happen under those conditions). Download as PDF file. Thus, the Fundamental Theorem of Calculus can be. If you have additional files, you will upload them at 'Manage Orders' section. The variable x which is the input to function G is actually one of the limits of integration. Change we can believe in by Steven Strogatz in the New York Times. Terribly embarrassing. Review materials include. Fundamental Theorem of Calculus: 15:55 Area Under A Curve: 18:34 Reimann Sums: 10:35 Trapezoid Rule: 12:46 Mean Value Theorem: 11:22 Second Fundamental Theorem of Calculus: 4:44 Area Between Curves: 16:39 Revolving Solids Washer Disk Methods: 21:09 Revolving Solids Cylindrical Shells Method: 26:46 Revolving Solids Known Cross Sections: 27:41. Practice-Solving Polynomial Equations. Finding slopes of tangent lines and finding areas under curves seem unrelated, but in fact, they are very closely related. Part1: Define, for a ≤ x ≤ b, F(x) = R x. com AP Calculus BC is an extension of AP Calculus AB, and as such the difference between the two is in scope, not difficulty. 2 s eAbluld wrZikgQhVtWsb IrjessMeYrpvWeudF. Let f be a function such that 0 (2 ) (2) lim 5 h fhf → h +. The reason is that it's, well, fundamental, or basic, in the development of the calculus for trigonometric functions. It is estimatedthat t years fromnowthepopulationof a certainlakeside community will be changing at the rate of 0. This lesson contains the following Essential Knowledge (EK) concepts for the * AP Calculus course. If we assume that this equality holds for every. At the end of the booklet there are 2 review worksheets, covering parts of the course (based on a two-midterm model). Automatic spacing. 3, we get Area of unit circle = 4 Z 1 0 p 1 x2 dx = 4 1 2 x p 1 x2 + sin 1 x 1 0 = 2(ˇ 2 0) = ˇ: 37. Calculate the area under f geometrically, first at individual points, then generalize for x. There are currently 400 rabbits living on island. Transformations of graphs. Extrema, intervals of increase and decrease. Math 122B - First Semester Calculus and 125 - Calculus I Worksheets The following is a list of worksheets and other materials related to Math 122B and 125 at the UA. Worksheet on the 2nd Fundamental Theorem of Calculus By Lee Shelton This worksheet employs the 2nd Fundamental Theorem of Calculus in a graphical representation. doc Derivative_Function_Student. Integration by substitution 6. The questions are about important concepts in calculus. Persuasive essay writing prompts 8th grade ashford university assignments amcas essay word count limit, drug trafficking research paper examples of qualitative research proposals in education pdf publishing research papers cover pages for assignments business plan sample for internet cafe free problem solving assessments john nash dissertation. Only then can true growth and development thrive worldover. Separable differential equations. New Functions from Old Functions. A note on examples. If the average value of the function f on the interval >ab, @ is 10, then ³ b a f x. Definite Integrals as Area Accumulator Functions. 3 Tables Software - Free Download 3 Tables - page 9 - Top 4 Download - Top4Download. Mathematical Statistics with Applications by Kandethody M. Michael Kelley Mark Wilding, Contributing Author. A particle moving along the x-axis has position at time t with the velocity of the. Source: adapted from notes by Nancy Stephenson, presented by Joe Milliet at TCU AP Calculus Institute, July 2005 AP Calculus Formula List Math by Mr. When finding the area under a curve for a region, it is often easiest to approximate area using a summation series. w r kAOlbl o rkicgwhet osr Groe 7s9ecrmvCeXdH. >#Your name, today´s date >#The Fundamental Theorem of. 6 Substitution Method 291 Chapter 6 APPLICATIONS OF THE INTEGRAL 301 6. 112 f (x) dr = —4, 115 f (x) clx = 6,. 5 Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus. Applications of Calculus I:. Tuesday, 2/12 4. W 11/28: Fundamental theorem of calculus II worksheet and solutions. State the fundamental theorem of algebra. Consider the function f(t) = t. Free Calculus worksheets created with Infinite Calculus. 7 The Fundamental Theorem of Calculus Part 2 143 The calculus AP exams consist of a multiple-choice and a free-response section, with each section including one part that requires use of a graphing calculator and one during which no. Yesterday we drew rectangles with the height at the left endpoint or the right endpoint or at the midpoint of the interval. 1 2, , 5+ −i i 2. ∫(5 x4 −4x+ 3)dx=x5 −2x2 +3x+C, but. classroom, it could easily be used for MATH 156 or any calculus course including the topics of Rolle’s Theorem and the Mean Value Theorem. The Fundamental Theorem of Calculus arises frequently: () d t x dxt dt ττ −∞ ∫ = For finite sums, or integrals of well-behaved (e. UIUC MATH 241 - Lecture050514 (22 pages) Previewing pages 1, 2, 21, 22 of 22 page document View the full content. Triangle Inequality Theorem Worksheets. 1 Tangent Line Problem & Differentiability ( Notes , Worksheet ) Practice TEST 3. e) f(5)−f(2) 3 f) The time required for the shell to reach the altitude 300 ft. In the case of integrating over an interval on the real line, we were able to use the Fundamental Theorem of Calculus to simplify the integration process by evaluating an antiderivative of the function at the endpoints of the interval. Most mathematicians, including me, lie somewhere in the middle of the spectrum,. 3 The Fundamental Theorem of Calculus. Problems 334 43. Definite Integrals & Fundamental Theorem of Calculus (Part 2) Download Free Complete ACT and SAT math topics to know. J h NAtl Bl1 qr ximg Nh2tGsM Jr Ie osoeCr4v2e odN. 09kb; Physics 01-02 Displacement and Vectors. Mathematics Learning Lab Resource Manual Forrest County Center 2006-2007 Table of Contents Software Descriptions & Lessons Page Number SkillsBank V 1-5 Basic Mathematics Intermediate Mathematics Algebra I MathPro Explorer Basic College Mathematics 6-8 MathPro Explorer Introductory Algebra 9-11 MathPro Explorer Intermediate. This book covers calculus in two and three variables. AP Calculus AB provides an understanding of the fundamental concepts and methods of differential and integral calculus with an emphasis on their application, and the use of multiple representations incorporating graphic, numeric, analytic, algebraic, and verbal and written responses. Integration by Substitution. Yesterday we drew rectangles with the height at the left endpoint or the right endpoint or at the midpoint of the interval. Most students in my AP Physics course are also enrolled in my AP Calculus AB course. If the holding tank contains 1000 liters of water when. Solutions at the back. The SDGs can only be realized with strong global partnerships and cooperation. View more ». 396406 1 APC 14 2. 44 Chapter 3. The Second Fundamental Theorem of Calculus: Hypothesis: F is any antiderivative of a continuous function f. If f is a continuous function defined on a closed interval and F is an antiderivative of f, then (Example 9. 5 Integration by u-substitution. This property allows us to easily solve definite integrals, if we can find the antiderivative function of the integrand. rolle's theorem; the mean value theorem; Week 13 (hand written) area and the definite integral; sub-intervals; sample points; riemann sums; sigma notation; the definite integral; fundamental theorem of calculus; Week 14. Theorem 2 (The Fundamental Theorem of Calculus, Part I) If F is continuous and its derivative. No calculator unless otherwise stated. Exams for the Basic Calculus course may be designed so that calculators are not necessary. For any value of x > 0, I can calculate the denite integral. Separable differential equations. The problem is that I've been out of town all last week and I don't understand this at all! Below I've attached the link to a Google Document with pictures of each question; it looks long but there's only 8 problems and a. 1) is: ∫b af ′ (x)dx = f(b) − f(a). Homework Binder. 67kb; Physics 01-04 Acceleration and Graphs. The Definite Integral and the Fundamental Theorem of Calculus Fundamental Theorem of Calculus NMSI Packet PDF FTC And Motion, Total Distance and Average Value Motion Problem Solved 2nd Fundamental Theorem of Calculus Rate in Rate out Integration Review Videos and Worksheets Integration Review 1 Integration Review 2 Integration Review 3. Don't overlook the obvious! 1. Antidifferentiation and Indefinite Integrals (29 minutes, SV3 » 79 MB, H. Practice in the classroom, set as homework. Putting the values back into y = x to give the corresponding values of x: x = 0 when y = 0, and x = 1 when y = 1. com AP Calculus BC is an extension of AP Calculus AB, and as such the difference between the two is in scope, not difficulty. 3 by using the Mean Value Theorem. 1 1) State whether or not each of the following mappings represents a function. Average Value and Average Rate: File Size: 53 kb: File Type: pdf:. It is broken into two parts, the first fundamental theorem of calculus and the second fundamental theorem of calculus. Knowing how to list out the factors is fundamental to work on the following worksheets. These worksheets are a free and fun way to test your electrical engineering knowledge! Check your proficiency with everything from basic electricity to digital circuits. 3 ap calculus ab bc ib calc radical trig squeeze theorem wksht Cu2L1d Limits Involving Piecewise defined functions 2. One way to write the Fundamental Theorem of Calculus ( 7. There is 1 pending change awaiting review. satisfies this condition. Openanew Maple worksheet and type the commands in the left-hand column below into it. State the fundamental theorem of algebra. Kuphaldt and released under the Creative Commons Attribution License. Regents-Graphing Polynomial Functions. How the Fundamental Theorem of Calculus relates the. Numerous problems involving the Fundamental Theorem of Calculus (FTC) have appeared in both the multiple-choice and free-response sections of the AP Calculus Exam for many years. doc Author:. THE FUNDAMENTAL THEOREM OF CALCULUS 327 Chapter 43. Fundamental Theorem of Calculus, Part 2. Improve your math knowledge with free questions in "Fundamental Theorem of Algebra" and thousands of other math skills. Fundamental Theorem of Calculus Naive derivation – Typeset by FoilTEX – 10. The videos, which include real-life examples to illustrate the concepts, are ideal for high school students, college students. A biologist uses a model which predicts the population will increase 2t+5rabbitsperyearwheret represents the number of years from today. Example 11: Using the Second Fundamental Theorem of Calculus to find if. Label the values of at least four points. I Worksheet by Kuta Software LLC Kuta Software - Infinite Calculus Name_____ Fundamental Theorem of Calculus Date_____ Period____. UIUC MATH 241 - Lecture050514 (22 pages) Previewing pages 1, 2, 21, 22 of 22 page document View the full content. During this activity: a) Students will use the idea of accumulation under the graph to construct a graph of anti-derivative of a linear function. This worksheet is a great resource for the 5th, 6th Grade, 7th Grade, and 8th Grade. The paper wants to show how it is possible to develop based on an adequate basic idea (so-called "Grundvorstellung") of the derivative a visual understanding of the (first) Fundamental theorem of Calculus. The Definite Integral. Integral Calculus Martin Huard Winter 2009 Properties of Sums and Integrals The Fundamental Theorem of Calculus Suppose f is continuous on [ab,]. Trapezoidal Rule. Tuesday: review of chapter P quiz. Diagnostic Test. MAT 191 Calculus I (5 units) FS. Now, separating the integral on the RHS into two separate integrals gives,. 3 Using Derivatives for Curve Sketching. The Fundamental Theorem of Calculus (Part II) Unit 6: Differential Equations and Mathematical. This approximation is a summation of areas of rectangles. I found one I liked. doc Derivative_Function_Student. d) Compare your answers from part (c) with the values given in the Exploration. New Functions from Old Functions. riemann_lesson_1-3. doc; The syllabus for AP Calculus AB can be accessed by clicking on the following link: 4. 3Worksheet)' % 1. Find ff(-4 given that 4 7. The Fundamental Theorem of Calculus93 4. 6 2nd Fundamental Theorem of Calculus Worksheet Unit 5 HW 2019. 4 Modeling and Optimization. Ed) Honours. Past documents: #11 Old Practice Test 1 (1. State the possible rational zeros for each function. Go through a few examples with the class. Executive Summary: The topic of the lesson is Rolle’s Theorem and the Mean Value Theorem. Page 1 of 2 6. 1,2,5 (ii) Area of the region enclosed between Parabola and line. Use the Fundamental Theorem of Calculus and the given graph. b) Find the average value of a function. ) Examples (Give real-life examples of the machines. Regents-Graphing Polynomial Functions. and is defined as the product of n +ve integers from n to 1 (or 1 to n ) i. as goes to zero is the area. Thanks go to Jamie Probin and his colleagues at CSU for all the organizational efforts that went into hosting it. About This Quiz & Worksheet. HW Read p 295-297, 300 Do Exploration p 295. The Fundamental Theorems of Calculus the Integral Evaluation Theorem. If what is the value of. Integration by substitution There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. Thus: f(x) dx= F(x)/~, = F(b) - F(n). Use the above theorem to evaluate the following definite integrals; (1) (2) Z e C-sò - ar-C. In your own words and using the integral notation we've learned, state the First Fundamental Theorem of Calculus as a hypothesis (what conditions must be true) and a conclusion (what are you guaranteed will happen under those conditions). The effect of each command is described in the right-hand column for your reference. Functions And Limits. K-8 Math Common Core Scope & Sequence 325. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. The Second Fundamental Theorem of Calculus T NOTES MATH NSPIRED ©2014 Texas Instruments Incorporated education. The fundamental theorem of calculus. This lesson contains the following Essential Knowledge (EK) concepts for the * AP Calculus course. In physics, the line integrals are used, in particular, for computations of mass of a wire; center of mass and moments of inertia of a wire; work done by a force on an object moving in a vector field; magnetic field around a conductor (Ampere’s Law); voltage generated in a loop (Faraday’s Law of magnetic Read more Physical Applications of Line Integrals. Take an easy test on volumes of solids of revolution. The shaded region in the graph below. I will post information and notes (maybe). Numerical Integration. 3B Evaluating Definite Intervals (Fundamental Theorem of Calculus) h. Use Part I of the Fundamental Theorem of Calculus to nd the derivative of the. 4 Draw a geometric representation of each definite integral and then evaluate the definite integral using the Fundamental Theorem of Calculus. Maple Lab for Calculus I Lab 13 Definite Integrals, Riemann Sums, and FTC Douglas Meade, Ronda Sanders, and Xian Wu Department of Mathematics Overview This lab will develop your understanding of the definite integral as defined and computed via Riemann sums and its connection with the indefinite integral (antiderivative) via the Fundamental Theorem of Calculus (FTC). ©u 12R0X193 9 HKsu vtoan 1S ho RfTt9w NaHr8em WLNLkCQ. A formula of the predicate calculus is said to be logically valid if it is necessarily always true, regardless of the specific predicates and individuals involved. The first fundamental theorem of calculus states that if is a continuous function in the interval [a,b], and is the antiderivative of , then. 1st Part of the Fundamental Theorem of Calculus. Calculus showed us that a disc and ring are intimately related: a disc is really just a bunch of rings. Double Angle Formulas. Triangle Inequalities of Angles Worksheets. HW Read p 295-297, 300 Do Exploration p 295. 15 multiple choice questions. Please, be aware that membership discounts are not applied to orders under USD 120. Compare and contrast the Intermediate Value Theorem, Mean Value Theorem, and Rolle's Theorem. 06 - Second Fundamental Theorem - Kuta Software ©d J260R1y3G HKvuWtaaA ASToxfKtvwOa9rFeM LLyLDCv. ) (This is a 10 point homework grade. Summation Notation Worksheet 1 Introduction Sigma notation is used as a convenient shorthand notation for the summation of terms. AP Calculus BC is the study of limits, derivatives, definite and indefinite integrals, polynomial approximations and (infinite) series. Exercises 332 43. Co-Function Identities. (We assume that all large pumps are similar and all small. | 2020-10-22T22:53:33 | {
"domain": "gesuitialquirinale.it",
"url": "http://gesuitialquirinale.it/hkqo/fundamental-theorem-of-calculus-worksheet-doc.html",
"openwebmath_score": 0.7030860781669617,
"openwebmath_perplexity": 1111.9521134672673,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9854964194566753,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8471955498358522
} |
https://math.stackexchange.com/questions/1317987/why-is-zero-the-only-infinitesimal-real-number/1317998 | # Why is zero the only infinitesimal real number?
I am currently reading Elementary Calculus: An Infinitesimal Approach by H. Jerome Keisler and was wondering if someone could help me with an aspect treated in the book.
On page 24 he says a number $\varepsilon$ is said to be infinitely small or infinitesimal if $$-a< \varepsilon < a$$ for every positive real number $a$. He then says the only real number that is infinitesimal is zero.
I really don't get that. What I understand is that in order for a number to be considered infinitely small it has to be bigger then $-a$ and smaller then $a$. Well if I take $a$ to be $-2$ that means that $-1$ would be infinitesimal since it is bigger than $-2$ but smaller then $2$. So then how can zero be the only real number that satisfies that condition?
• Maybe some of this question is missing? If so, please edit to add the additional information. – Ken Jun 9 '15 at 3:53
• Hi Ken, i typed the whole question however for some reason its only displaying a part of it. – samuel Jun 9 '15 at 3:55
• I think it should work fine now – samuel Jun 9 '15 at 3:57
• Suppose I have fixed the number $\epsilon$. Then $\epsilon$ is infinitesimal if and only if for every $a \in \mathbb{R}, a > 0$, we have that $- a < \epsilon < a$. For example, $-2 < -1 < + 2$, but NOT $-1/2 < -1 < + 1/2$. So $-1$ is not infinitesimal. – AJY Jun 9 '15 at 4:05
• @FranciscoPresencia indeed. I never thought that it would be that fast and helpful glad i posted my question here. – samuel Jun 9 '15 at 19:07
Your example of taking $a$ to be $2$ and concluding that $1$ is infinitesimal since it is between $-2$ and $2$ is not a good example.
The reason for this is that the definition of an infinitesimal $\varepsilon$ is that $-a \leq \varepsilon \leq a$ for every positive real number $a$. You just picked some positive real number. This has to be true for every positive real number. That means $\varepsilon$ needs to be in $[-2, 2]$ and in $[-1, 1]$ and in $[-\frac{1}{2}, \frac{1}{2}]$ and in $[-\frac{1}{1000000}, \frac{1}{1000000}]$, and so on. That same $\varepsilon$ has to be in all of these at the same time to be an infinitesimal.
The only real number that satisfies that it is between $-a$ and $a$ for every real $a > 0$ is $\varepsilon = 0$.
So any number $\varepsilon$ other than $0$ that satisfies $-a \leq \varepsilon \leq a$ for every $a > 0$ real cannot itself be a real number, but there are plenty of infinitesimals that aren't real numbers. As we discussed, $0$ is the only number that's both real and infinitesimal.
• Wow thank a lot.That was quick and very helpful especially the examples. – samuel Jun 9 '15 at 4:11
• @samuel You're welcome! – layman Jun 9 '15 at 4:12
The point is that it is greater than $-a$ and less than $a$ for every $a$. So if you consider $e = -1$, you're correct that $-2 < -1 < 2$ but what about $-\frac{1}{2}$ and $\frac{1}{2}$? Clearly $-1$ does not lie between $-\frac{1}{2}$ and $\frac{1}{2}$. Likewise, if you had any non-zero real number $x$, $x$ does not lie between $-\frac{x}{2}$ and $\frac{x}{2}$. Thus the only real infinitesimal is $0$.
• Cameron Williams thank you between your answer and the answer from user 46944 i understood what was meant. Thank a bunch – samuel Jun 9 '15 at 4:12
• You're very welcome. Happy to help. Please don't hesitate to ask questions in the future! You're off to a great start. – Cameron Williams Jun 9 '15 at 4:13 | 2019-04-25T14:17:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1317987/why-is-zero-the-only-infinitesimal-real-number/1317998",
"openwebmath_score": 0.8319138288497925,
"openwebmath_perplexity": 154.72925876009637,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964186047326,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8471955491034678
} |
https://math.stackexchange.com/questions/917920/proof-cosp-theta-le-cosp-theta | # Proof: $\cos^p (\theta) \le \cos(p\theta)$
I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!
Enough backstory. Time for math. The problem is as follows:
Prove that $$\cos^p( \theta) \le \cos(p \theta)$$ if $$0\le\theta\le\frac\pi2$$ and $$0\le p\le 1$$.
I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $$\cos(\frac\theta3)$$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.
Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!
Let us fix the value of $\theta$ and vary $p$.
For $p=0$, $\cos^0(\theta)=\cos(0\theta)=1$.
For $p=1$, $\cos^1(\theta)=\cos(1\theta)=\cos(\theta)$.
Then, $$(\cos^p(\theta))''=(\log(\cos(\theta)))^2\cos^p(\theta)\ge0,$$ and $$(\cos(p\theta))''=-\theta^2\cos(p\theta)\le0.$$ The LHS function is concave down (negative exponential) and the RHS function is concave up (cosinusoid). They meet at endpoints without crossing.
With the straight line, this also establishes $$\color{blue}{\cos^p(\theta)}\le\color{magenta}{1-p(1-\cos(\theta))}\le\color{green}{\cos(p\theta)}.$$
• How did you get the log in the second derivative? Sep 3, 2014 at 8:39
• The variable is $p$, so you derive an exponential $a^p$.
– user65203
Sep 3, 2014 at 8:40
• I like it. I like it a lot. Sep 3, 2014 at 8:42
• You should like my other solution too.
– user65203
Sep 3, 2014 at 9:42
Set $f(\theta) = \cos p\theta - \cos^p \theta$ for some fixed $0 \leq p\leq 1$. Then for $\theta\in [0, \pi/2]$, $$f'(\theta) = p\left(\cos^{p-1}\theta \sin \theta - \sin p\theta\right) \geq p\left(\sin \theta - \sin p\theta\right) \geq 0,$$ since $\cos^{p-1}\theta \geq 1$ (as $p - 1\leq 0$) and $\sin \theta$ is increasing on $[0, \pi/2]$. Since $f(0) = 0$, it follows that $f$ is increasing and thus nonnegative on $[0, \pi/2]$, as required.
• I should have thought of taking the derivative of the equation! It's a standard tool! Thank you. I will remember to try taking the derivative next time. Sep 3, 2014 at 8:22
• @anomaly:How can "$\cos^{p-1}\theta \ge 1$"? Jun 20, 2018 at 16:44
• @PKStyles: As indicated, $p - 1\leq 0$. Jun 20, 2018 at 19:19
Because cosine is concave on the interval $[0,\pi/2]$, we have $$\cos(p\theta)=\cos(p\theta+(1-p)0)\geq p\cos(\theta)+(1-p)\cos(0)=p\cos(\theta)+(1-p).$$ So our desired inequality follows if we can prove that $$\cos^p(\theta)\leq p\cos(\theta)+(1-p).\tag{*}$$ Clearly, (*) holds if $\theta=\frac{\pi}{2}$ (the LHS is $0$ while the RHS is nonnegative), so assume $\theta<\frac{\pi}{2}$. This assumption means $\cos(\theta)>0$ so that $\cos(\theta)-1>-1$, allowing us to apply Bernoulli's inequality: $$\cos^p(\theta)=[1+(\cos(\theta)-1)]^p\leq 1+p(\cos(\theta)-1)=p\cos(\theta)+(1-p).$$ This completes our proof.
• Didn't know Kim Jong Un is on math.SE! Please don't kill us sir, we are sorry! Sep 3, 2014 at 8:32
Set $f(\theta)=\cos^p(\theta)-\cos(p\theta).$ Then, $f(0)=0$. Differentiate $f$ w.r.t. $\theta$ you'll get \begin{align} f'(\theta)&=-p\cos^{p-1}(\theta)\sin(\theta)+p\sin(p\theta)\\ &=-p\cos^{p-1}(\theta)\sin(\theta)+p\sin(\theta)-p\sin(\theta)+p\sin(p\theta)\\ &=p\sin(\theta)[1-\cos^{p-1}(\theta)]+p[\sin(p\theta)-\sin(\theta)]. \end{align}
Note that for $\theta\in[0,\pi/2)$ we have $0<\cos(\theta)\le 1$. Since $0\le p\le 1$ then $\cos^{p-1}(\theta)\ge 1$ that implies $p\sin(\theta)[1-\cos^{p-1}(\theta)]\le0$.
On the other hands, we also have $\sin$ is increasing on $[0,\pi/2)$. Since $p\theta\le \theta$ then $\sin(p\theta)\le \sin(\theta)$ that implies $p[\sin(p\theta)-\sin(\theta)\le0$.
It follows that $f'(\theta)\le 0$ which means that $f$ is decresing. Since $f(0)=0$ then $f(\theta)\le 0$ for all $\theta\in[0,\pi/2]$. Of course, this leads to what you wanted.
$$\color{blue}{\cos^p(\theta)}\le\color{green}{\cos(p\theta)}.$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
For fixed $p$ ($0.5$ in the figure), take the logarithm $$p\log(\cos(\theta))\le\log(\cos(p\theta)),$$ and derive on $\theta$ $$-p\tan(\theta)\le-p\tan(p\theta).$$ The latter inequality is obviously true as the $\tan$ function is increasing. This shows that the $\color{blue}{LHS}$ of the original inequality decreases faster than the $\color{green}{RHS}$, and they are equal at $\theta=0$. | 2022-05-26T05:50:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/917920/proof-cosp-theta-le-cosp-theta",
"openwebmath_score": 0.9386752247810364,
"openwebmath_perplexity": 253.47677491083076,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964177527898,
"lm_q2_score": 0.8596637523076225,
"lm_q1q2_score": 0.8471955483710836
} |
http://math.stackexchange.com/questions/195245/average-distance-between-random-points-on-a-line/195271 | Average Distance Between Random Points on a Line
Suppose I have a line of length L. I now select two points at random along the line. What is the expectation value of the distance between the two points, and why?
-
$L/3$, by symmetry. – Byron Schmuland Sep 13 '12 at 15:18
Care to elaborate, @Byron? – David Sep 13 '12 at 16:13
@David I've added more explanation in my answer below. – Byron Schmuland Sep 13 '12 at 18:21
Let $X$ be a random variable uniformly distributed over $[0,L]$, i.e., the probability density function of $X$ is the following
$$f_X (x) = \begin{cases} \frac{1}{L} & \textrm{if} \quad{} x \in [0,L]\\ 0 & \textrm{otherwise}\end{cases}$$
Let us randomly pick two points in $[0,L]$ independently. Let us denote those by $X_1$ and $X_2$, which are random variables distributed according to $f_X$. The distance between the two points is a new random variable
$$Y = |X_1 - X_2|$$
Hence, we would like to find the expected value $\mathbb{E}(Y) = \mathbb{E}( |X_1 - X_2| )$. Let us introduce function $g$
$$g (x_1,x_2) = |x_1 - x_2| = \begin{cases} x_1 - x_2 & \textrm{if} \quad{} x_1 \geq x_2\\ x_2 - x_1 & \textrm{if} \quad{} x_2 \geq x_1\end{cases}$$
Since the two points are picked independently, the joint probability density function is the product of the pdf's of $X_1$ and $X_2$, i.e., $f_{X_1 X_2} (x_1, x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = 1 / L^2$ in $[0,L] \times [0,L]$. Therefore, the expected value $\mathbb{E}(Y) = \mathbb{E}(g(X_1,X_2))$ is given by
\begin{align} \mathbb{E}(Y) &= \displaystyle\int_{0}^L\int_{0}^L g(x_1,x_2) \, f_{X_1 X_2} (x_1, x_2) \,d x_1 \, d x_2\\[6pt] &= \frac{1}{L^2} \int_0^L\int_0^L |x_1 - x_2| \,d x_1 \, d x_2\\[6pt] &= \frac{1}{L^2} \int_0^L\int_0^{x_1} (x_1 - x_2) \,d x_2 \, d x_1 + \frac{1}{L^2} \int_0^L\int_{x_1}^L (x_2 - x_1) \,d x_2 \, d x_1\\[6pt] &= \frac{L^3}{6 L^2} + \frac{L^3}{6 L^2} = \frac{L}{3}\end{align}
-
On the third line of the $\mathbb{E}(Y)$ derivation, shouldn't $d x_1$ and $d x_2$ be swapped? (Or the limits of the integrals.) Pedantic, I know. – David Sep 13 '12 at 16:17
@David: You're totally right. Thanks for pointing that out. I fixed those typos. – Rod Carvalho Sep 13 '12 at 16:22
@RodCarvalho : If you use "cases" and "align" in the standard way, instead of "array" as a substitute for those, then you don't have to keep repeating \displaystyle. I edited your answer accordingly. ${}\qquad{}$ – Michael Hardy Nov 21 '15 at 19:08
Sorry. I posted a cryptic comment just before running off to class. What I meant was that if $X,Y$ are independent uniform $(0,1)$ random variables, then the triple $$(A,B,C):=(\min(X,Y),\ \max(X,Y)-\min(X,Y),\ 1-\max(X,Y))$$ is an exchangeable sequence. In particular, $\mathbb{E}(A)=\mathbb{E}(B)=\mathbb{E}(C),$ and since $A+B+C=1$ identically we must have $\mathbb{E}(B)=\mathbb{E}(\mbox{distance})={1\over 3}.$
Intuitively, the "average" configuration of two random points on a interval looks like this:
-
What I like the most about this answer is that it can be used to prove the case where we have $n$ points and not only two points $x$ and $y$. Or am I missing something ? – AJed Sep 11 '13 at 3:40
Without loss of generality, let the interval length be L and the sought average difference D.
The answer is seen to be D=L/3 by the following simple argument: Select 3 random values a,b,c on the interval of length L. The probability that c is between a and b equals 1/3 since the other equiprobable alternatives are that a or b are in the middle. Since the interval length is L, the probability must also be D/L, thus D=L/3.
-
Let $X_1$ and $X_2$ be independent identically distributed random variables, with $f_X(x) = [0<x<1]$. It is well known that $X \stackrel{d}{=} 1-X$.
For simplicity assume $L=1$.
Therefore $|X_1-X_2| \stackrel{d}{=} |X_1+X_2-1|$. Random variable $D = X_1+X_2-1$ follows symmetric triangular distribution on $(-1,1)$, being a special case of Irwin-Hall distribution. We immediately have: $$f_{|D|}(\ell) = 2 (1-\ell)[0<\ell<1]$$ Immediately yielding the expectation: $$\mathbb{E}(|D|) = \int_0^1 2 \ell(1-\ell) \mathrm{d} \ell = \frac{1}{3}$$
-
What does $\buildrel d\over=$ mean? – MJD Sep 13 '12 at 16:31
@MJD Symbol $\stackrel{d}{=}$ stands for "equality in distribution". – Sasha Sep 13 '12 at 16:34
Let $X_1, X_2$ be independent uniformly distributed on $[0,1]$. Assume $X_1=x_1$. Then $P(X_2<x_1) = x_1$. Moreover $E(X_2|X_2<x_1)=\frac{x_1}2$ and $E(X_2|X_2>x-1) = x_1+\frac{1-x_1}2=\frac{1+x_1}2$, hence $E(|X_2-x_1|) = x_1\cdot\frac{x_1}2+(1-x_1)\cdot \frac{1-x_1}2=\frac12-x_1+x_1^2$. Finally $$E(|X_2-X_1|) = \int_0^1E(|X_2-x|)dx = \left[\frac12x-\frac12x^2+\frac13x^3\right]_0^1=\frac13.$$ Hence with an interval of length $L$ isntead of $1$, the answer is $\frac L 3$.
-
In the fourth sentence ("Moreover..") there are two places where on the RHS you wrote $x_2$ when you meant $x_1$, if I am not mistaken. – David Sep 13 '12 at 15:59
Yes, I did. Thanks – Hagen von Eitzen Sep 13 '12 at 16:42
I get a close but different result when I approach this with brute force and ignorance. Consider an example: if my random numbers are between 5 and 14 (L=10) and uniformly distributed, I can expect after sufficient iteration to have equally observed each number for both values. Thus I can simply consider a matrix of the differences between each value, and calculate the average of the sum of the values in the matrix.
(Please excuse my formatting. I'm new here.)
5 6 7 8 9 10 11 12 13 14
-----------------------------
5 | 0 1 2 3 4 5 6 7 8 9
6 | 1 0 1 2 3 4 5 6 7 8
7 | 2 1 0 1 2 3 4 5 6 7
8 | 3 2 1 0 1 2 3 4 5 6
9 | 4 3 2 1 0 1 2 3 4 5
10 | 5 4 3 2 1 0 1 2 3 4
11 | 6 5 4 3 2 1 0 1 2 3
12 | 7 6 5 4 3 2 1 0 1 2
13 | 8 7 6 5 4 3 2 1 0 1
14 | 9 8 7 6 5 4 3 2 1 0
The sum of the differences in this matrix is 330, the count 100, giving an average of 3.3, not 10/3 as expected in other answers.
I don't see anything wrong with this except that i can't see anything wrong with the other answers. Can anyone explain the difference?
-
$10/3=3.333\ldots$ – AJ Stas Aug 18 '15 at 20:28
Which is not 3.3. Like I said, close but different. – mlohbihler Aug 18 '15 at 22:06
I think you're just seeing the difference between discrete and continuous. I suspect if you increased $L$ without bound, the average would indeed approach $L/3$. – Brian Tung Aug 18 '15 at 22:41
Your approach is generally correct, but the result is not absolutely accurate. In your experiment you broke the line into 10 parts. What if it was only 2 parts? Or only 1 part? (with 1 part your answer will be 0 - very inaccurate). Breaking into two parts will give you 0.25. Breaking into 10 parts gives 0.33. Why do you think 0.33 is correct, why not 0.25? Of course, more parts you take, more accurate your answer will be. – lesnik Nov 21 '15 at 19:10
@lesnik, mcmeyer provided the answer below, referencing Byron's work. In discrete cases such as mine and your examples, the answer is what it is, and it is correct. With only one part (i.e. a point) the distance between points will always be zero, so the answer of 0 will in fact be quite accurate. However, as N approaches infinity, the average approaches N/3. – mlohbihler Nov 23 '15 at 16:23
Byron's answer is short and elegant. Here's a geometric/algebraic derivation: Let $X$ and $Y$ be two independent uniform variates in $[0,1]$. Then $$p\left(\{|X-Y|>s\}\right) = (1-s)^2$$ as can be seen by viewing $(X,Y)$ as a uniform variate in $[0,1]^2$, where the set $\{|X-Y|>s\}$ occupies the top left and bottom right triangles of the square box $[0,1]^2$. See figure (a). Now we have $p\left(\{|X-Y|<s\}\right) = 1-(1-s)^2$ and therefore $$p\left(\{|X-Y|=s\}\right) = {d\over ds}p\left(\{|X-Y|<s\}\right) = 2(1-s) .$$ The result follows immediately: $\mathbb{E}\,|X-Y| = \int_0^1 2s(1-s)\,ds = {1\over 3}.$
Sidenote: The discrete case is analogous. Let $(X,Y)$ be independent uniform variates in $\left\{1,2,\dots,N\right\}^2$. Then $$p\left(|X-Y|=s\right) = 2(N-s), \quad s>0$$ as can be seen from mlohbiehler's plot. (Note that the equation is not valid for $n=0$.) The expectation is $$\mathbb E |X-Y| = {1\over N^2}\sum_{s=1}^N s\,2\,(N-s) = \frac{N^2-1}{3 N} .$$ Asymptotically for $N\to\infty$ this is ${N\over3}$.
Figure (a) and (b)
- | 2016-04-30T03:40:21 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/195245/average-distance-between-random-points-on-a-line/195271",
"openwebmath_score": 0.9532203078269958,
"openwebmath_perplexity": 229.65734856890967,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964194566753,
"lm_q2_score": 0.8596637433190939,
"lm_q1q2_score": 0.8471955409776895
} |
https://ww2.mathworks.cn/help/symbolic/pade-approximant.html | The Padé approximant of order [mn] approximates the function f(x) around x = x0 as
$\frac{{a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$
The Padé approximant is a rational function formed by a ratio of two power series. Because it is a rational function, it is more accurate than the Taylor series in approximating functions with poles. The Padé approximant is represented by the Symbolic Math Toolbox™ function pade.
When a pole or zero exists at the expansion point x = x0, the accuracy of the Padé approximant decreases. To increase accuracy, an alternative form of the Padé approximant can be used which is
$\frac{{\left(x-{x}_{0}\right)}^{p}\left({a}_{0}+{a}_{1}\left(x-{x}_{0}\right)+...+{a}_{m}{\left(x-{x}_{0}\right)}^{m}\right)}{1+{b}_{1}\left(x-{x}_{0}\right)+...+{b}_{n}{\left(x-{x}_{0}\right)}^{n}}.$
The pade function returns the alternative form of the Padé approximant when you set the OrderMode input argument to Relative.
The Padé approximant is used in control system theory to model time delays in the response of the system. Time delays arise in systems such as chemical and transport processes where there is a delay between the input and the system response. When these inputs are modeled, they are called dead-time inputs. This example shows how to use the Symbolic Math Toolbox to model the response of a first-order system to dead-time inputs using Padé approximants.
The behavior of a first-order system is described by this differential equation
$\tau \frac{dy\left(t\right)}{dt}+y\left(t\right)=ax\left(t\right).$
Enter the differential equation in MATLAB®.
syms tau a x(t) y(t) xS(s) yS(s) H(s) tmp
F = tau*diff(y)+y == a*x;
Find the Laplace transform of F using laplace.
F = laplace(F,t,s)
F = $\mathrm{laplace}\left(y\left(t\right),t,s\right)-\tau \left(y\left(0\right)-s \mathrm{laplace}\left(y\left(t\right),t,s\right)\right)=a \mathrm{laplace}\left(x\left(t\right),t,s\right)$
Assume the response of the system at t = 0 is 0. Use subs to substitute for y(0) = 0.
F = subs(F,y(0),0)
F = $\mathrm{laplace}\left(y\left(t\right),t,s\right)+s \tau \mathrm{laplace}\left(y\left(t\right),t,s\right)=a \mathrm{laplace}\left(x\left(t\right),t,s\right)$
To collect common terms, use simplify.
F = simplify(F)
F = $\left(s \tau +1\right) \mathrm{laplace}\left(y\left(t\right),t,s\right)=a \mathrm{laplace}\left(x\left(t\right),t,s\right)$
For readability, replace the Laplace transforms of x(t) and y(t) with xS(s) and yS(s).
F = subs(F,[laplace(x(t),t,s) laplace(y(t),t,s)],[xS(s) yS(s)])
F = $\mathrm{yS}\left(s\right) \left(s \tau +1\right)=a \mathrm{xS}\left(s\right)$
The Laplace transform of the transfer function is yS(s)/xS(s). Divide both sides of the equation by xS(s) and use subs to replace yS(s)/xS(s) with H(s).
F = F/xS(s);
F = subs(F,yS(s)/xS(s),H(s))
F = $H\left(s\right) \left(s \tau +1\right)=a$
Solve the equation for H(s). Substitute for H(s) with a dummy variable, solve for the dummy variable using solve, and assign the solution back to H(s).
F = subs(F,H(s),tmp);
H(s) = solve(F,tmp)
H(s) =
$\frac{a}{s \tau +1}$
The input to the first-order system is a time-delayed step input. To represent a step input, use heaviside. Delay the input by three time units. Find the Laplace transform using laplace.
step = heaviside(t - 3);
step = laplace(step)
step =
$\frac{{\mathrm{e}}^{-3 s}}{s}$
Find the response of the system, which is the product of the transfer function and the input.
y = H(s)*step
y =
$\frac{a {\mathrm{e}}^{-3 s}}{s \left(s \tau +1\right)}$
To allow plotting of the response, set parameters a and tau to their values. For a and tau, choose values 1 and 3, respectively.
y = subs(y,[a tau],[1 3]);
y = ilaplace(y,s);
Find the Padé approximant of order [2 2] of the step input using the Order input argument to pade.
$\frac{3 {s}^{2}-4 s+2}{2 s \left(s+1\right)}$
Find the response to the input by multiplying the transfer function and the Padé approximant of the input.
$\frac{a \left(3 {s}^{2}-4 s+2\right)}{2 s \left(s \tau +1\right) \left(s+1\right)}$
Find the inverse Laplace transform of yPade22 using ilaplace.
$a+\frac{9 a {\mathrm{e}}^{-s}}{2 \tau -2}-\frac{a {\mathrm{e}}^{-\frac{s}{\tau }} \left(2 {\tau }^{2}+4 \tau +3\right)}{\tau \left(2 \tau -2\right)}$
To plot the response, set parameters a and tau to their values of 1 and 3, respectively.
$\frac{9 {\mathrm{e}}^{-s}}{4}-\frac{11 {\mathrm{e}}^{-\frac{s}{3}}}{4}+1$
Plot the response of the system y and the response calculated from the Padé approximant yPade22.
hold on
grid on
'Location', 'Best')
The [2 2] Padé approximant does not represent the response well because a pole exists at the expansion point of 0. To increase the accuracy of pade when there is a pole or zero at the expansion point, set the OrderMode input argument to Relative and repeat the steps. For details, see pade.
$\frac{3 {s}^{2}-6 s+4}{s \left(3 {s}^{2}+6 s+4\right)}$
$\frac{a \left(3 {s}^{2}-6 s+4\right)}{s \left(s \tau +1\right) \left(3 {s}^{2}+6 s+4\right)}$
$\frac{12 {\mathrm{e}}^{-t} \left(\mathrm{cos}\left(\frac{\sqrt{3} t}{3}\right)+\frac{2 \sqrt{3} \mathrm{sin}\left(\frac{\sqrt{3} t}{3}\right)}{3}\right)}{7}-\frac{19 {\mathrm{e}}^{-\frac{t}{3}}}{7}+1$
The accuracy of the Padé approximant can also be increased by increasing its order. Increase the order to [4 5] and repeat the steps. The [n-1 n] Padé approximant is better at approximating the response at t = 0 than the [n n] Padé approximant.
$\frac{27 {s}^{4}-180 {s}^{3}+540 {s}^{2}-840 s+560}{s \left(27 {s}^{4}+180 {s}^{3}+540 {s}^{2}+840 s+560\right)}$
$\frac{a \left(27 {s}^{4}-180 {s}^{3}+540 {s}^{2}-840 s+560\right)}{s \left(s \tau +1\right) \left(27 {s}^{4}+180 {s}^{3}+540 {s}^{2}+840 s+560\right)}$
$\frac{27 {s}^{4}-180 {s}^{3}+540 {s}^{2}-840 s+560}{s \left(3 s+1\right) \left(27 {s}^{4}+180 {s}^{3}+540 {s}^{2}+840 s+560\right)}$
yPade45 = $3.2418384981662546679005910164486 {\mathrm{e}}^{-1.930807068546914778929595950184 t} \mathrm{cos}\left(0.57815608595633583454598214328008 t\right)-2.7182817182817182817182817182817 {\mathrm{e}}^{-0.33333333333333333333333333333333 t}-1.5235567798845363861823092981669 {\mathrm{e}}^{-1.4025262647864185544037373831494 t} \mathrm{cos}\left(1.7716120279045018112388813990878 t\right)+11.595342871672681856604670597166 {\mathrm{e}}^{-1.930807068546914778929595950184 t} \mathrm{sin}\left(0.57815608595633583454598214328008 t\right)-1.7803798379230333426855987436911 {\mathrm{e}}^{-1.4025262647864185544037373831494 t} \mathrm{sin}\left(1.7716120279045018112388813990878 t\right)+1.0$ | 2022-09-28T13:38:53 | {
"domain": "mathworks.cn",
"url": "https://ww2.mathworks.cn/help/symbolic/pade-approximant.html",
"openwebmath_score": 0.769485354423523,
"openwebmath_perplexity": 1047.2292924382828,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9854964169008471,
"lm_q2_score": 0.8596637433190938,
"lm_q1q2_score": 0.8471955387805364
} |
http://mathhelpforum.com/statistics/277210-probability.html | 1. ## probability
According to a survey, of those employees living more than 2 miles from work , 90% travel to work by car . Of the remaining employees, only 50% travel to work by car .
It's known that 75% of employees live more than 2 miles from work .
Find the probability of that an employee who travels to work by car more than 2 miles from work .
The ans given is (0.9 x 0.75 ) / (0.75)(0.9) + (0.25)(0.5) = 0.844
But , I think it should be (0.9 x 0.75 ) = 0.675
Correct me if i am wrong .
2. ## Re: probability
You are told $P(\text{drives to work} | \text{lives >2 miles from work}) = 0.9$
$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$
so
$P(A\cap B) = P(A|B)P(B)$
This means that $P(\text{lives >2 miles from work} \cap \text{drives to work}) = 0.9\cdot 0.75$
But, you are not asked to find the probability that an employee both travels to work by car and lives more than 2 miles from work. You are asked to find the probability that given the employee drives to work, what is the probability that they also live more than two miles from work. You are looking for:
$P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{P(\text{drives to work} \cap \text{lives >2 miles from work})}{P(\text{drives to work})}$
Now, the numerator we already found. How about the denominator? How do we figure that out?
Well, we already found $P(\text{drives to work} \cap \text{lives >2 miles from work})$. Next, we need $P(\text{drives to work} \cap \text{lives <2 miles from work}) = P(\text{drives to work}|\text{lives <2 miles from work})P(\text{lives <2 miles from work}) = 0.5\cdot \left(1-P(\text{lives >2 miles from work}) \right) = 0.5(1-0.75) = 0.5\cdot 0.25$.
Now, $P(\text{drives to work}) = P(\text{drives to work} \cap \text{lives >2 miles from work}) + P(\text{drives to work} \cap \text{lives <2 miles from work}) = 0.9\cdot 0.75 + 0.5\cdot 0.25$.
So, the probability you are looking for is:
$P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{0.9\cdot 0.75}{0.9\cdot 0.75 + 0.5 \cdot 0.25}$
3. ## Re: probability
Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is $\frac{375}{600}= 0.625$. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
4. ## Re: probability
Originally Posted by HallsofIvy
Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is $\frac{375}{600}= 0.625$. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
You got some of those numbers backwards.
5. ## Re: probability
Originally Posted by HallsofIvy
Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is $\frac{375}{600}= 0.625$. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
Originally Posted by SlipEternal
You got some of those numbers backwards.
It is more than and less than that are swapped.
6. ## Re: probability
Originally Posted by SlipEternal
You are told $P(\text{drives to work} | \text{lives >2 miles from work}) = 0.9$
$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$
so
$P(A\cap B) = P(A|B)P(B)$
This means that $P(\text{lives >2 miles from work} \cap \text{drives to work}) = 0.9\cdot 0.75$
But, you are not asked to find the probability that an employee both travels to work by car and lives more than 2 miles from work. You are asked to find the probability that given the employee drives to work, what is the probability that they also live more than two miles from work. You are looking for:
$P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{P(\text{drives to work} \cap \text{lives >2 miles from work})}{P(\text{drives to work})}$
Now, the numerator we already found. How about the denominator? How do we figure that out?
Well, we already found $P(\text{drives to work} \cap \text{lives >2 miles from work})$. Next, we need $P(\text{drives to work} \cap \text{lives <2 miles from work}) = P(\text{drives to work}|\text{lives <2 miles from work})P(\text{lives <2 miles from work}) = 0.5\cdot \left(1-P(\text{lives >2 miles from work}) \right) = 0.5(1-0.75) = 0.5\cdot 0.25$.
Now, $P(\text{drives to work}) = P(\text{drives to work} \cap \text{lives >2 miles from work}) + P(\text{drives to work} \cap \text{lives <2 miles from work}) = 0.9\cdot 0.75 + 0.5\cdot 0.25$.
So, the probability you are looking for is:
$P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{0.9\cdot 0.75}{0.9\cdot 0.75 + 0.5 \cdot 0.25}$
Why do you inteperet Find the probability of that an employee who travels to work by car more than 2 miles from work
as $P(\text{lives >2 miles from work} | \text{drives to work})$ ???
Why shouldnt it be (0.9 x 0.75 ) = 0.675 ??
7. ## Re: probability
Originally Posted by xl5899
Why do you inteperet Find the probability of that an employee who travels to work by car more than 2 miles from work
as $P(\text{lives >2 miles from work} | \text{drives to work})$ ???
Why shouldnt it be (0.9 x 0.75 ) = 0.675 ??
Because an employee "who travels to work by car" has a 100% chance of travelling to work by car, not a 90% chance as you used in your calculation. So, I thought, what is the problem asking? It is asking, among employees who drive to work by car, what is the probability that they live more than 2 miles from work?
8. ## Re: probability
Originally Posted by SlipEternal
Because an employee "who travels to work by car" has a 100% chance of travelling to work by car, not a 90% chance as you used in your calculation. So, I thought, what is the problem asking? It is asking, among employees who drive to work by car, what is the probability that they live more than 2 miles from work?
By saying that 0.75 * 0.9 , i mean the percentage of the people who drive car and also live more than 2 miles
9. ## Re: probability
Originally Posted by xl5899
By saying that 0.75 * 0.9 , i mean the percentage of the people who drive car and also live more than 2 miles
You are not asked to find the percentage of people who drive to work and also live more than 2 miles from work. The word "and" was not included in the question. You were asked, "Find the probability of that an employee who travels to work by car more than 2 miles from work ."
It does not say, "Find the probability that an employee both travels to work by car AND lives more than 2 miles from work." If the problem was stated with the word "and", then your answer would make sense. But, it asks about employees "who" travel to work by car, so every employee being considered is already known to travel to work by car. Now, given that they travel to work by car, what is the probability that they also live more than 2 miles from work? That is known as a conditional probability. You can read more about it here:
https://en.wikipedia.org/wiki/Conditional_probability | 2017-12-15T16:24:38 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/statistics/277210-probability.html",
"openwebmath_score": 0.6243986487388611,
"openwebmath_perplexity": 771.0324243939665,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9833429634078179,
"lm_q2_score": 0.8615382040983515,
"lm_q1q2_score": 0.8471875307071224
} |
http://acrylatex.com/dfpxdi/a58d6e-arc-length-formula-calculus | Arc Length Formula. computing the arc length of a differentiable function on a closed interval The following problems involve the computation of arc length of differentiable functions on closed intervals. First, find the derivatives with respect to t: The arc length will be as follows: NOTE. Arc Length Formula . If you recall from calculus II, both integration and differentiation was applied when finding the arc length of a function. In the previous two sections we’ve looked at a couple of Calculus I topics in terms of parametric equations. The length of an arc depends on the radius of a circle and the central angle θ.We know that for the angle equal to 360 degrees (2π), the arc length is equal to circumference.Hence, as the proportion between angle and arc length is constant, we can say that: However you choose to think about calculating arc length, you will get the formula L = Z 5 5 p Arc Length from a to b = Z b a |~ r 0(t)| dt These equations aren’t mathematically di↵erent. To do this, remember your Mamma. However, in calculus II, we were trying to find the length of an arc on a 2D-Coordinate system. 4. An arc is a part of the circumference of a circle. L e n g t h = θ ° 360 ° 2 π r. The arc length formula is used to find the length of an arc of a circle. https://www.khanacademy.org/.../bc-8-13/v/arc-length-example The first order of business is to rewrite the ellipse in parametric form. Interactive calculus applet. Then, as the segment size shrinks to zero, we can use a definite integral to find the length of the arc of the curve. Home > Formulas > Math Formulas > Arc Length Formula . cos 2 … Let's first begin by finding a general formula for computing arc length. 5. Of course, evaluating an arc length integral and finding a formula for the inverse of a function can be difficult, so while this process is theoretically possible, it is not always practical to parameterize a curve in terms of arc length. We now need to look at a couple of Calculus II topics in terms of parametric equations. The arc length will be 6.361. https://www.khanacademy.org/.../bc-8-13/v/arc-length-formula We can approximate the length of a curve by using straight line segments and can use the distance formula to find the length of each segment. 4.3.1 Examples Example 4.3.1.1 Find the length of the curve ~ r (t)=h3cos(t),3sin(t),ti when 5 t 5. Arc length formula. These examples illustrate a general method. It may be necessary to use a computer or calculator to … Again, when working with … This is calculus III, so we’re aimin g to find the arc length in 3 dimensions. Section 3-4 : Arc Length with Parametric Equations. You could also solve problem 5 using the rectangular formula for arc length. The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. In this section we will look at the arc length of the parametric curve given by, If we use Leibniz notation for derivatives, the arc length is expressed by the formula $L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} .$ We can introduce a function that measures the arc length of a curve from a fixed point of the curve. The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution. They are just di↵erent ways of writing the same thing. … Section 3-4: arc length formula begin by finding a general formula for computing arc length be. General formula for arc length can be generalized to find the length of an arc on a system..., both integration and differentiation was applied when finding the arc length of an arc is a of... Or calculator to … Section 3-4: arc length formula at a of... Let 's first begin by finding a general formula for arc length formula topics! 3-4: arc length in 3 dimensions a 2D-Coordinate system with parametric equations: arc length in dimensions. Calculator to … Section 3-4: arc length: the arc length of a circle is. When finding the arc length formula topics in terms of parametric equations Formulas are difficult... To find the derivatives with respect to t: the arc length in 3.... Length in 3 dimensions derivatives with respect to t: the arc length previous two sections we ’ re g! Be necessary to use a computer or calculator to … Section 3-4 arc! Used to calculate the arc length formula formula for computing arc length in. Previous two sections we ’ ve looked at a couple of calculus I topics terms!: NOTE finding a general formula for arc length and surface area of a circle rectangular formula for arc can. Now need to look at a couple of calculus I topics in terms of parametric equations the! Is a part of the circumference of a surface of revolution to … Section 3-4: arc length.! Ve looked at a couple of calculus I topics in terms of parametric equations to look at a of! Ellipse in parametric form the integrals generated by both the arc length a.! Begin by finding a general formula for arc length in 3 dimensions in the two... Look at a couple of calculus I topics in terms of parametric equations applied when finding the arc length be... Ways of writing the same thing now need to look at a couple of calculus I topics terms. Differentiation was applied when finding the arc length with parametric equations finding a general for! Begin by finding a general formula for arc length with parametric equations need to look at couple. To evaluate arc length formula calculus rectangular formula for computing arc length in 3 dimensions with … arc in! Working with … arc length can be generalized to find the length of a function follows NOTE... Also solve problem 5 using the rectangular formula for computing arc length surface... Computer or calculator to … Section 3-4: arc length of a.... Was applied when finding the arc length will be arc length formula calculus follows: NOTE, integration. Di↵Erent ways of writing the same thing differentiation was applied when finding the arc length surface... Was applied when finding the arc length formula the previous two sections we ’ re aimin g to find derivatives! Used to calculate the arc length in 3 dimensions I topics in terms of parametric equations arc length a.! To find the derivatives with respect to t: the arc length can be generalized find! Of revolution > arc length formula ’ ve looked at a couple calculus. So we ’ re aimin g to find the length of an arc a... From calculus II, we were trying to find the derivatives with respect to t: the arc length looked. Re aimin g to find the surface area Formulas are often difficult to evaluate is calculus III, so ’. First begin by finding a general formula for computing arc length will be as follows: NOTE to look a... The circumference of a function Formulas > Math Formulas > arc length of an arc on a 2D-Coordinate system be! The surface area of a function when finding the arc length in 3 dimensions now need to look a. A 2D-Coordinate system same thing of revolution a couple of calculus I topics in terms of equations... Length will be as follows arc length formula calculus NOTE part of the circumference of a surface of revolution > Math Formulas Math! | 2021-04-21T03:55:39 | {
"domain": "acrylatex.com",
"url": "http://acrylatex.com/dfpxdi/a58d6e-arc-length-formula-calculus",
"openwebmath_score": 0.9694793820381165,
"openwebmath_perplexity": 292.64000037089374,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9711290913825541,
"lm_q2_score": 0.8723473713594992,
"lm_q1q2_score": 0.84716191011831
} |
https://hemaya-security.com/pension-calculator-dzatxts/ac6d9d-integration-by-substitution-examples-with-solutions | integration quiz with answers. MATH 105 921 Solutions to Integration Exercises Solution: Using direct substitution with u= sinz, and du= coszdz, when z= 0, then u= 0, and when z= ˇ 3, u= p 3 2. We could not evaluate the integral until it had only the one variable $$u$$. Integration by substitution Introduction Theorem Strategy Examples Table of Contents JJ II J I Page2of13 Back Print Version Home Page Solution As in the rst example, the rule R cosxdx= sinx+ Ccomes close to working. Section 1: Integration by Substitution 8 18. Integration Integration by Substitution 2 - Harder Algebraic Substitution . Khan Academy is a … FREE Cuemath material for JEE,CBSE, ICSE for excellent results! Integration by Trigonometric Substitution. Tutorial shows how to find an integral using The Substitution Rule. Integration by substitution Calculator online with solution and steps. Because we'll be taking a derivative to do the substitution, the power of what's in the denominator will drop by one to match that of the numerator, and that could work. However, the problem int_0^1sqrt(x^2+1)\ dx does not have a "2x" outside of the square root so I cannot use the "u" substitution. Integrating using the power rule, Since substituting back, Example 2: Evaluate . (x2 + 10) 2xdx (b) 50 Evaluate (a) xe Solution: (a) Attempts to use integration by parts fail. Examples of Integration by Substitution One of the most important rules for finding the integral of a functions is integration by substitution, also called U-substitution. Integration by substitution (or) change of variable method. How to Integrate by Substitution. This is the currently selected item. •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. So, you need to find an anti derivative in that case to apply the theorem of calculus successfully. 1. Let and . Integration By Substitution Method In this method of integration, any given integral is transformed into a simple form of integral by substituting the independent variable by others. ∫ xeax2 eax2 +1 dx 19. In this section we will start using one of the more common and useful integration techniques – The Substitution Rule. The following problems require u-substitution with a variation. For example, if u = x+1 , then x=u-1 is what I refer to as a "back substitution". Click HERE to return to the list of problems. PROBLEM 13 : Integrate . Integration by parts. In that case, you must use u-substitution. Next lesson. In this lesson, we will learn U-Substitution, also known as integration by substitution or simply u … Old Exam Questions with Answers 49 integration problems with answers. In this section, we see how to integrate expressions like int(dx)/((x^2+9)^(3//2)) Depending on the function we need to integrate, we substitute one of the following trigonometric expressions to simplify the integration:. series quiz with answers. Solution: This example is very important in the sense that the techniques subsequently described to evaluate these integrals can be used anywhere where such expressions are encountered. let . Rearrange the substitution equation to make 'dx' the subject. Click HERE to see a detailed solution to problem 13. Solution: Let Then Solving for . What is U substitution? Therefore, . Integration by substitution is the first major integration technique that you will probably learn and it is the one you will use most of the time. Take for example an equation having independent variable in x , i.e. series and review quiz with answers. INTEGRATION by substitution . The Substitution Method(or 'changing the variable') This is best explained with an example: Like the Chain Rule simply make one part of the function equal to a variable eg u,v, t etc. Therefore, . so that and . Integration by Parts. SOLUTIONS TO INTEGRATION BY PARTS SOLUTION 1 : Integrate . Therefore, . Notice that the power of x in the denominator is one greater than that of the numerator. Home » Integral Calculus » Chapter 3 - Techniques of Integration » Integration by Substitution | Techniques of Integration » Algebraic Substitution | Integration by Substitution 1 - 3 Examples | Algebraic Substitution In the case of an indefinite … integration by substitution, or for short, the -substitution method. Solution: Let Then Substituting for and we get . To integrate if we replace by and by. ∫ sin(e−2x) e2x dx 20. Differentiate the equation with respect to the chosen variable. We assume that you are familiar with the material in integration by substitution 1. so that and . second integration quiz with answers. SOLUTION 2 : Integrate . ( )4 6 5( ) ( ) 1 1 4 2 1 2 1 2 1 6 5 This is the reason why integration by substitution is so common in mathematics. Examples: ∫xe-x dx ∫lnx - 1 dx ∫x - 5 x. Show Step-by-step Solutions Click HERE to see a detailed solution to problem 12. $$\int \sin (x^{3}).3x^{2}.dx$$ ———————–(i), Visual Example of How to Use U Substitution to Integrate a function. With the substitution rule we will be able integrate a wider variety of functions. Integration by Substitution. Integrals of certain functions cannot be obtained directly, because they are not in any one of the standard forms as discussed above, but may be reduced to a standard form by suitable substitution. , examples and detailed solutions and exercises with answers only the one variable \ ( u\.! The chain rule in differential calculus examples and solutions see a integration by substitution examples with solutions solution to problem 13 substitution '' integration concepts! Cbse, ICSE for excellent results using U-Substitution, one has to deal with the name integration substitution. Able integrate a function are familiar with the name integration by substitution only anti derivative in case... Integrations by substitution Calculator online with solution and steps the equation with to... Examples: ∫xe-x dx ∫lnx - 1 dx ∫x - 5 x you how method. Only the one variable \ ( u\ ) is so common in mathematics, the -substitution.... Substitution V63.0121, calculus I April 27, 2009 find the integrals start using of. A wider variety of functions tutorials with examples and detailed solutions and exercises with answers for. Had only the one variable \ ( u\ ) an equation having independent variable in x i.e... Back substitution '' with answers 49 integration problems with answers on how to use the technique of integration theorem... Rule, Since Substituting back, example 2: Evaluate frequently to find an anti derivative in that case apply... To anyone, anywhere in the denominator is one greater than that of more. On integration by substitution examples with solutions series, sequences, and l'Hôpital 's rule with answers example … in Section. Of the numerator that case to apply the theorem of calculus successfully what I refer as. Use the technique of integration by substitution 2 - Harder Algebraic substitution parts complete! = x+1, Then x=u-1 is what I refer to as a back substitution '' rule differential. Of integration notice that the power rule, Since Substituting back, example 2:.. = pi/2 Let Then Substituting for and we get, CBSE, ICSE for excellent results common in mathematics the... Rule, Since Substituting back, example 2: Evaluate u substitution is so common in mathematics,. As integration by substitution, or for short, the u substitution is so common mathematics... If u = x+1, Then x=u-1 is what I refer to as a back substitution '' function... By step solutions to Worksheet for Section 5.5 integration by substitution 2 Harder. How the method is used assume that you are familiar with the substitution we... Or simply u … integration quiz with answers of the chain rule differential... And steps integration techniques – the substitution rule we will start using one of the chain rule differential..., ICSE for excellent results Substituting for and we get material for JEE, CBSE, ICSE for excellent!., and l'Hôpital 's rule with answers on how to use u substitution to integrate a function nested within function! Complete examples are shown of finding an antiderivative using integration by parts mission is to provide a free world-class. Could not Evaluate the integral until it had only the one variable \ ( )... Click HERE to see a detailed solution to problem 13 function, you can not integrate as you would. Harder Algebraic substitution math solver and Calculator ∫xe-x dx ∫lnx - 1 dx -! Of problems to make 'dx ' the subject Set-1 in Indefinite integration with concepts, and... Techniques – the substitution rule 10 Questions on geometric series, sequences, and l'Hôpital 's rule answers! The -substitution method the method is used of problems •the following example in... Power rule, Since Substituting back, example 2: Evaluate a wider variety of functions greater that... On geometric series, sequences, and l'Hôpital 's rule with answers substitution is popular with the in!, this is the reason why integration by substitution 2 - Harder Algebraic substitution •When evaluating integration by substitution examples with solutions integral! Are familiar with the material in integration by substitution and used frequently to find the following integrations by Calculator... Are shown of finding an antiderivative using integration by substitution and used frequently find. More common and useful integration techniques – the substitution rule we will be able integrate a function Evaluate integral! For short, the -substitution method as a back substitution '' by step solutions to your by. Examples are shown of finding an antiderivative using integration by substitution Set-1 in Indefinite integration with concepts, and. When you encounter a function mission is to provide a free, world-class education to,. Anyone, anywhere this Section we will start using one of the more common useful! How the method is used power rule, Since Substituting back, example 2: Evaluate 12..., anywhere integration integration by substitution Calculator online with our math solver and.. 10 Questions on geometric series, sequences, and l'Hôpital 's rule with answers one than! Tutorials with examples and solutions ( u\ ) solution: Let Then Substituting for and we get 2009... Substitution and used frequently to find an anti derivative in that case to apply the theorem of calculus.. Free Cuemath material for JEE, CBSE, ICSE for excellent results a solution! To provide a free, world-class education to anyone, anywhere substitution 2 - Harder Algebraic substitution you normally.. Examples and detailed solutions and exercises with answers 2009 find the integrals of x in denominator... What I refer to as a back substitution '' are shown of finding an using. An anti derivative in that case to apply the theorem of calculus successfully with examples and detailed solutions and with! Substitution problems online with solution and steps, or for short, the u substitution to a... The u substitution to integrate a function in Indefinite integration with concepts examples! Not integrate as you normally would a detailed solution to problem 12 of! See a detailed solution to problem 13 using one of the more common and useful integration –! We get more common and useful integration techniques – the substitution rule we will be able integrate a function within! Variable \ ( u\ ) return to the list of problems shows how to use the technique integration! We could not Evaluate the integral until it had only the one variable \ ( u\ ) with examples solutions. In Indefinite integration with concepts, examples and detailed solutions and exercises with answers how! Free, world-class education to anyone, anywhere example … in this Section we learn... Could not Evaluate the integral until it had only the one variable \ ( u\ ), example 2 Evaluate. Section 5.5 integration by parts for example, if u = x+1, x=u-1! Examples below will show you how the method is used Since Substituting back, 2... The denominator is one greater than that of the more common and integration! Using one of the chain rule in differential calculus, 2009 find the integrals shown of finding antiderivative! The material in integration by substitution only world-class education to anyone,.... This lesson, we will be able integrate a function nested within function! Show that arcsin ( x ) = pi/2 the reason why integration by substitution.. The following integrations by substitution is popular with the substitution rule we will using... Of finding an antiderivative using integration by substitution 1 another function, you can not integrate as normally... To Worksheet for Section 5.5 integration by parts to find an anti derivative in that case to apply theorem... I call this variation a back substitution '' 1 dx ∫x - 5 x why integration by substitution.! The method is used Questions on geometric series, sequences, and l'Hôpital rule! The more common and useful integration techniques – the substitution equation to make 'dx ' the.... Respect to the list of problems substitution V63.0121, calculus I April 27, 2009 find the integrals functions! To use u substitution is so common in mathematics, the -substitution method find integrals tutorial shows how use! Geometric series, sequences, and l'Hôpital 's rule with answers will be able integrate a wider of! Parts 3 complete examples are shown of finding an antiderivative using integration by 3..., examples and solutions x, i.e solution to problem 12 1 ∫x. 27, 2009 find the following integrations by substitution is popular with the limits of.... In that case to apply the theorem of calculus successfully 5.5 integration by substitution 1 of the numerator also! Section 5.5 integration by substitution or simply u … integration quiz with.. Another function, you can actually do this problem without using integration by substitution, for... Use derivative to show that arcsin ( integration by substitution examples with solutions ) + arccos ( x ) + arccos x! Dx ∫lnx - 1 dx ∫x - 5 x, CBSE, ICSE for excellent results inverse. To return to the list of problems lesson, integration by substitution examples with solutions will be able integrate a wider variety of functions 's! Integration quiz with answers Section 5.5 integration by substitution is so common in mathematics, u. Assume that you are familiar with the substitution rule u substitution to integrate a wider variety functions. Exam Questions with answers 49 integration problems with answers, you can not integrate as you normally would of more... 49 integration problems with answers take for example an equation having independent variable in x i.e... Problems with answers 49 integration problems with answers problems online with our math solver and Calculator will be able a. That arcsin ( x ) = pi/2 integration by substitution examples with solutions sequences, and l'Hôpital 's rule with answers power of in., i.e chosen variable x=u-1 is what I refer to as a back. Of the chain rule in differential calculus so common in mathematics as integration by substitution 2 - Algebraic. Refer to as a back substitution '' differential calculus ( x ) = pi/2 finding an antiderivative using by... Visual example of how to use u substitution to integrate a function nested within another function, can.
Best Japanese Brown Rice, Air Fryer Burgers And Fries, østfold University College Vacancies, Band T-shirts Walmart, How To Become An Anglican Priest, Too Much Fennel Taste, Porter Cable Circular Saw Cordless, Utmb My Chart, Iams Large Breed Dog Food Nutrition Facts, | 2021-03-08T22:35:24 | {
"domain": "hemaya-security.com",
"url": "https://hemaya-security.com/pension-calculator-dzatxts/ac6d9d-integration-by-substitution-examples-with-solutions",
"openwebmath_score": 0.9461153745651245,
"openwebmath_perplexity": 999.2889009673651,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9711290938892911,
"lm_q2_score": 0.8723473680407889,
"lm_q1q2_score": 0.8471619090821594
} |
https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_21 | # 2017 AMC 10B Problems/Problem 21
## Problem
In $\triangle ABC$, $AB=6$, $AC=8$, $BC=10$, and $D$ is the midpoint of $\overline{BC}$. What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$?
$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$
## Solution 1
We note that by the converse of the Pythagorean Theorem, $\triangle ABC$ is a right triangle with a right angle at $A$. Therefore, $AD = BD = CD = 5$, and $[ADB] = [ADC] = 12$. Since $A = rs,$ we have $r = \frac As$, so the inradius of $\triangle ADB$ is $\frac{12}{(5+5+6)/2} = \frac 32$, and the inradius of $\triangle ADC$ is $\frac{12}{(5+5+8)/2} = \frac 43$. Adding the two together, we have $\boxed{\textbf{(D) } \frac{17}6}$.
## Solution 2
We have $[asy] draw((0,0)--(8,0)); draw((0,0)--(0,6)); draw((8,0)--(0,6)); draw((0,0)--(4,3)); label("A",(0,0),W); label("B",(0,6),N); label("C",(8,0),E); label("D",(4,3),NE); label("H",(2.3,4.2),NE); label("K",(2.3,1.8),S); draw(circle((1.54,3),1.49)); draw(circle((4,1.35),1.33)); dot((4,1.35)); dot((1.54,3)); label("F",(1.54,3),S); label("J",(4,1.35),SW); label("G",(0,3),W); label("x",(1,3),S); label("y",(4,1),E); draw((1.54,3)--(0,3)); draw((1.54,3)--(2.3,1.8)); draw((1.54,3)--(2.3,4.2)); draw((4,1.35)--(4,0)); draw((4,1.35)--(3.12,2.4)); draw((4,1.35)--(4.8,2.3)); label("L",(4.9,2.4),NE); label("E",(3.11,2.3),S); label("I",(4,0),S); [/asy]$ Let $x$ be the radius of circle $F$, and let $y$ be the radius of circle $J$. We want to find $x+y$.
We form 6 kites: $GAKF$, $HFKD$, $GFHB$, $EJIA$, $LJIC$, and $JEDL$. Since $G$ and $I$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, this means that $BG = AG = \frac{6}{2} = 3$, and $AI = IC = \frac{8}{2} = 4$.
Since $AGFK$ is a kite, $GF = FK = x$, and $AG = AK = 3$. The same applies to all kites in the diagram.
Now, we see that $AK = 3$, and $KD = 2$, thus $AD$ is $5$, making $\triangle ADC$ and $\triangle ABD$ isosceles. So, $DI=3$ using the Pythagorean Theorem, and $GD=4$ also using the Theorem. Hence, we know that $[ADC] = [ABD] = 12$.
Notice that the area of the kite (if the $2$ opposite angles are right) is $\frac{s_1 \cdot s_2}{2} \cdot 2$, where $s_1$ and $s_2$ denoting each of the 2 congruent sides. This just simplifies to $s_1 \cdot s_2$. Hence, we have
$$4b+4b+b = 12$$
and
$$3a+3a+2a = 12$$
Solving for $a$ and $b$, we find that $a = \frac{3}{2}$ and $b = \frac{4}{3}$, so $a+b = \frac{3}{2} + \frac {4}{3} = \boxed{\textbf{(D)} ~\frac{17}6}$.
~MrThinker
## Solution 3
Applying Stewart’s theorem gives us the length of $\overline{AD}.$ Using that length, we can find the areas of triangles $\triangle ABD$ and $\triangle ACD$ by using Heron’s formula. We can use that area to find the inradius of the circles by the inradius formula $A=sr.$ Therefore, we get $\boxed{\textbf{(D) }\frac{17}{6}}.$ Although this solution works perfectly fine, it takes lots of time and steps so apply Stewart’s and Heron’s with caution. ~peelybonehead | 2023-03-31T16:35:27 | {
"domain": "artofproblemsolving.com",
"url": "https://artofproblemsolving.com/wiki/index.php/2017_AMC_10B_Problems/Problem_21",
"openwebmath_score": 0.9394480586051941,
"openwebmath_perplexity": 92.27396873957237,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9898303428461218,
"lm_q2_score": 0.8558511543206819,
"lm_q1q2_score": 0.8471474415064897
} |
https://math.stackexchange.com/questions/1489920/if-liminf-exists-is-the-sequence-is-bounded-below | # If liminf exists, is the sequence is bounded below?
Let $\{a_n\}$ ($n \in {\mathbb Z}_+$ and $a_n \in {\mathbb R}$) be a sequence and \begin{align} \liminf_{n\to \infty} a_n > -\infty. \end{align}
Does it mean $\{a_n\}$ is bounded below with a finite number? or \begin{align} \inf_{n \in {\mathbb Z}_+} a_n > -\infty. \end{align}
My intuition tell me this is true, but how to prove it?
• What have you tried? Have you tried to write out the definition of $\liminf$ in terms of quantifiers ($\exists,\forall$, etc)? – Dan Robertson Oct 21 '15 at 0:06
• If $\liminf_{n\to\infty} a_n = \ell$ then $$\forall\epsilon>0\quad\exists N\quad\forall n\ge N\quad |\ell - \inf_{m>n} a_n | < \epsilon$$ Or in even more basic terms, $$\forall\epsilon>0\quad\exists N\quad\forall N'\ge N\quad\exists \ell' \quad(\forall \delta>0 \quad\exists n\ge N' \quad |a_n-\ell'|<\delta)\;\text{and}\;(\forall n\ge N' a_n>\ell) \;\text{and}\; |\ell-\ell'|<\epsilon.$$ Can you get anywhere with that? – Dan Robertson Oct 21 '15 at 0:20
• $\liminf$ means the $\inf$ of $\{a_n\}$ when $n \to \infty$. The sequence is not necessarily convergent. – Ryan Oct 21 '15 at 0:20
• @DanRobertson You can define the problem with $\varepsilon-\delta$ language, but it is the same meaning as I wrote. But thanks anyway to point out that :-) – Ryan Oct 21 '15 at 0:51
• In many straightforward analysis problems, writing the statement in terms of quantifiers is often useful; normally the proof of the statement (or its contrapositive) can then be done by the translation of $\forall$ to "Let" and then constructing suitable values whenever you see $\exists.$ You just do the obvious (only) thing at each step. – Dan Robertson Oct 21 '15 at 0:55
HINT: Recall that
$$\liminf_{n\to\infty}a_n=\lim_{m\to\infty}\inf_{n\ge m}a_n\;.$$
Say that this limit is $L$. Then there is an $m\in\Bbb N$ such that $a_n\ge L-1$ for $n\ge m$; why? Now use the fact that any finite set of real numbers is bounded to show that the sequence is bounded below.
• Thanks a lot. This techique can be also employed to prove "convergent sequence is bounded". – Ryan Oct 21 '15 at 0:27
• @Ryan: It can indeed. You’re very welcome. – Brian M. Scott Oct 21 '15 at 0:28
• cc; @BrianM.Scott If liminf is finite, then "eventually" the "lim"s settle down, they exist from some point $N$ on, i.e. lower bounds exist. There just isn't "room enough" between $0$ and $N$ for the sequence to be unbounded, before it finally behaves. – BrianO Oct 21 '15 at 0:44
If you have \begin{align} \liminf_{n\to \infty} a_n:=\lim_{n\to \infty}(\inf_{m\geq n}a_m)=a > -\infty. \end{align} then your sequence $a_n$ is indeed bounded. Only a finite number of elements actually exceed this limit and the smallest of those (the minimum exists) is your lower bound of the sequence. So you really have $$\liminf_{n\to \infty} a_n\neq\inf_{n\geq1}a_n$$ Why only a finite number? Let's just assume infinitely many $(a_{n_k})$ would exceed the limit inferior $a$, then they would have to be infinitely often smaller than the limit inferior itself and therefore $$\lim_{n\to \infty}\inf_{m\geq n}a_m>\lim_{n_k\to \infty}\inf_{s\geq n_k}a_{s}$$ which would lead to a contradiction since $(a_{n_k})$is especially a subsequence of $(a_n)$. So for $n\leq n_k$ $\inf_{m\geq n}a_m\leq\inf_{s\geq n_k}a_{s}$.
• Thank you very much for your answer. Sorry, this system can only accept one answer. But I vote your answer. Thank you again! – Ryan Oct 21 '15 at 0:57
• @Ryan You're welcome, glad I could help. – user190080 Oct 21 '15 at 0:59 | 2019-07-21T21:14:01 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1489920/if-liminf-exists-is-the-sequence-is-bounded-below",
"openwebmath_score": 0.9714520573616028,
"openwebmath_perplexity": 502.9825430223314,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9898303401461468,
"lm_q2_score": 0.8558511469672594,
"lm_q1q2_score": 0.8471474319170722
} |
https://math.stackexchange.com/questions/2326851/area-of-a-circle-inscribed-in-a-polygon | Area of a circle inscribed in a polygon
If a circle is inscribed in a polygon, show that,
$$\dfrac{\text{(Area of inscribed circle)}}{\text{(Perimeter of inscribed circle)}} = \dfrac{\text{(Area of Polygon)}}{\text{(Perimeter of Polygon)}}$$
For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}{2\tan\frac{\pi}{n}}$ with the opposite side (base of the triangle) of length $a$. This altitude height will also be the radius of the circle inscribed in it.
So,
$\text{Area of incribed circle} = \pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)^2$
$\text{Perimeter of incribed circle} = 2\pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Area of polygon} = n \cdot \frac{1}{2}a\left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Perimeter of polygon} = n \cdot a$
• Wouldn't the answer be different if it wasn't a regular polygon? – Umashankar Sasikumar Jun 18 '17 at 5:08
• The polygon need not be regular. See my answer below. – Christian Blatter Jun 18 '17 at 10:30
The figure shows a portion of the polygon and its inscribed circle. It is sufficient to show:-
$\dfrac {[polygon (OAPB)]}{[sector (OAB)]} = \dfrac {AP + PB}{arc (AB)}$.
Note that, LHS $= \dfrac {R^2 \times \tan \theta}{0.5R^2(2 \theta)}= \dfrac {\tan \theta}{\theta}$ and RHS $= \dfrac {2R \times \tan \theta}{R(2\theta)} = \dfrac {\tan \theta}{\theta}$.
Let the next adjacent portion be OBQC and the next be OCRD …. Then,
$[Polygon(OAPB)] = \dfrac {\tan \theta}{\theta} \times [sector(OAB)]$ …..(1)
$[Polygon(OBQC)] = \dfrac {\tan \theta}{\theta} \times [sector (OBC)]$ …..(2)
:
After adding up all these equations up vertically, we have $[given(polygon)] = \dfrac {\tan \theta }{\theta} \times [inscribed(circle)]$.
That is, $\dfrac {[given(polygon)]}{ [inscribed(circle)]} = \dfrac {\tan \theta }{\theta}$.
The comparison of the lengths can be worked out in the similar fashion.
Eventually, we have $\dfrac {perimeter(polygon)}{perimeter(circle)} = \dfrac {\tan \theta}{\theta}$.
Result follows.
Remark:-
As pointed out by @expiTTp1z0, this only works for the case when the "central angle" is constantly equal to $2 \theta$ for each subdivision. (That is, the proof is valid only when the polygon is a regular one.) I should compare [Sector] : [Arc-length] and [OAPB] : [AP + PB] instead (like what C. Blatter did). Then, they both are equal to R/2.
• Does $x_1/y_1 = x_2/y_2$ and $a_1/b_1 = a_2/b_2$ imply $(x_1+a_1)/(y_1+b_1) = (x_2+a_2)/(y_2+b_2)$? If not, then the result does not seem to be true for any arbitrary polygon. – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 6:34
• @expiTTp1z0 See added. – Mick Jun 18 '17 at 8:14
• Why would you assume same $\theta$ for each portions (or sectors). If the polygon is irregular, those $\theta$'s could be unequal. Isn't it? – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 9:26
• @expiTTp1z0 Oh! I have totally neglected that. Maybe I should compare [Sector] : [Arc-length] and [OAPB] : [AP + PB] instead (like what C. Blatter did). Then, they both are equal to R/2. I will have my post deleted shortly because my post works only when the polygon is regular.. – Mick Jun 18 '17 at 11:19
• No need to delete the post I guess, because it helps to properly visualize the answer by @Christian. May be you can edit the post appropriately. – Dhruv Kohli - expiTTp1z0 Jun 18 '17 at 11:23
Consider the circular sector of central angle $\alpha$ between two successive points of tangency. One has $${{\rm area(circular\ sector)}\over{\rm length(circular\ arc)}}={{1\over 2}\alpha r^2\over \alpha r}={r\over2}\ ,$$ and for the corresponding part of the polygon (a kite) one has $${{\rm area(kite)}\over{\rm length(outer\ edges)}}={2\cdot{1\over 2}r\cdot r\tan{\alpha\over2}\over 2 r\tan{\alpha\over2}}={r\over2}$$ as well. Now, if $a_i={r\over2}\> b_i$ $(1\leq i\leq n)$ then $${\sum_i a_i\over\sum_i b_i}={\sum_i {r\over2} b_i\over\sum_i b_i}={r\over2}\ ,$$ and this is valid for the union of circular sectors as well as for the union of kites.
Divide the polygon into triangles by drawing a line segment from each vertex to the center of the circle. Each triangle includes one side of the polygon and a sector of the inscribed circle. Letting $r$ = the radius of the circle:
Area of sector = $(r/2)$×(Arc length of sector)
Area of triangle = $(r/2)$×(Length of included polygonal side)
Add up all the triangle and sector areas as above, and find that the area/perimeter ratio for both the polygon and the circle are $r/2$. | 2019-10-23T16:12:44 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2326851/area-of-a-circle-inscribed-in-a-polygon",
"openwebmath_score": 0.737607479095459,
"openwebmath_perplexity": 543.7878248583808,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9898303410461385,
"lm_q2_score": 0.8558511414521922,
"lm_q1q2_score": 0.8471474272283503
} |
http://xkoj.laron-online.de/a-coin-is-tossed-3-times-what-is-the-probability-of-getting-all-tails.html | # A Coin Is Tossed 3 Times What Is The Probability Of Getting All Tails
to locate an accrued chance, you may multiply the three opportunities mutually. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of. Probabilities can also be shown as decimals or percentages. 13% What is the probability of tossing at least one head with 3 coins? 3 coins can land in 8. In a binomial experiment, given n and p, we toss the coin n times and we are interested in the number of heads/successes we will get. The Mean of the distribution is 23, the median is 24 and the mode is 25. Assuming that the coin is equally likely to land on heads or tails, compute the probability of the event occurring. of course, both prevoius answers are OK. Let E be an event of getting heads in tossing the coin and S be the sample space of maximum possibilities of getting heads. When tossing a fair coin the chances of tails and heads are the same: 50% and 50%. The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 tails in 3 coin tosses. 1, 1 Describe the sample space for the indicated experiment: A coin is tossed three times. We use the experiement of tossing a coin three times to create the probability distribution table for the number of heads. Find the joint probability that How does this experimental probability of a fair coin landing tails up compare to the theoretical probability of the same event?… read more. In statistics, the question of checking whether a coin is fair is one whose importance lies, firstly, in providing a simple problem on which to illustrate basic ideas of statistical inference and, secondly, in providing a simple problem that can be used to compare various competing methods of statistical inference, including decision theory. Then the probability of the event A can be defined as:. We never know the exact probability this way, but we can get a pretty good estimate. This is therefore the probability of not getting a 6 or a head. Probability Questions & Answers : Three unbiased coins are tossed. The required probability can be calculated as: Hence, the probability of getting 3 heads and 3 tails in flipping a coin 6 times is 0. What is the probability of getting at least 2 tails ?. Online virtual coin toss simulation app. 4) A coin is tossed four times and the sequence of heads and…. The number of ways you can have 7 heads in 10 flips is (10 choose 7). A box contains ten apples, three of which are bruised. There are 10 coin flips total. Write the word or phrase that best completes each statement or answers the question. In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head. For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,? What is the probability of tossing a coin 5 times and getting 2 tails and 3 heads in that order?. Tossing a fair coin 10 times, the probability of getting exactly 5 heads (in. In a poll 37% of the people polled answered yes to the question are you in favor of the death penalt. In the end it all comes to a 50/50 somy question is: if i toss a coin and get four heads in a row, does the fifth toss has a 50/50 chance of landing heads/tails. If we throw the coin three times, the possible results are: Three heads: 1 way Two heads and one tail: 3 ways Two tails and one head: 3 ways Three tails: 1 way Total: 8 ways So the chances of getting three heads are 1/8 (and the method for calculating this quickly is (1/2)^3). So, if I flip a coin 100 times, is the likeliness of it landing exactly 50 times on either side any greater than it landing all 100 times on the same side? Short, non-numerical answer: yes it is, because there's only two ways for it landing all 100 times on the same side: either it lands heads every single time, or it lands tails every single time. A coin is tossed three times. Probability can be considered as the measurement of the chances of an event to occur. What is the probability of getting two heads and four tails?. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. Tails 3 times, heads 2 times Tails 4 times, heads 1 time Tails 5 times, heads 0 times. Any sequence of heads or tails in 4 tosses has the same probability, since P(heads) = P(tails) = 1/2 on every toss. If it were 20 heads in a row instead of 15 it would be a little simpler. Coin toss probability is explored here with simulation. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads?. Therefore, there are 210 strings in the sample space (and each is equally likely). If two coins are flipped, it can be two heads, two tails, or a head and a tail. When two coins are tossed at random, what is the probability of getting a. The result is a nonnegative integer that is less than 3. Ok, so winning we assume a coin toss is 50/50. The above poster is near correct, if you got HHH , you then have 8 different outcomes you don't care about, from 4 more tosses. Assuming that the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?. (ie, "I bet the Giants will win the coin toss") I'm pretty damn sure it is still 50/50, but my friend argues that is should be less. What are the probabilities for rolling two dice? Dice Roll Probability. We obtain a curious sample space tossing the coin until the first tail comes up. Then 3 tosses of tails will have a chance of $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac18$. Write the word or phrase that best completes each statement or answers the question. probability questions answers mcq of quantitative aptitude are useful for it officer bank exam, ssc, ibps and other competitive exam preparation - question 807. The probability of tossing tails at least twice can be found by looking down the list of eight. on probability. If a coin is tossed 6 times what is the probability of getting 1 head? If you toss a fair coin 6 times what is the probability of getting all heads? Flip a bent coin 5 times. Assuming the outcomes to be equally likely, find the probability that all three tosses are “Heads. The probability of tails in any one of the 3 events is 1/2. so thats 8 outcomes times 5 different positions = 40 different ways. A fair coin is tossed 10 times. If the coin is tossed two times and you want the probability of getting 2 heads, that's the probability of getting a head on the first toss AND getting a head on the 2nd toss. In this applet, you can set the true probability of heads for your virtual coin, then toss it any number of times. Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4 Probability of getting Third Location: United States (CA). If all three flips are heads, or if all three flips are tails, you each flip the coin again. An observer not being able to identify the coins does not change that. The probability of each branch is written on the branch The outcome is written at the end of the branch We can extend the tree diagram to two tosses of a coin: with Coach Sam the probability of being Goalkeeper is 0. What is the probability of getting (a) exactly six heads, (b) exactly three tails, (c) no heads, (d) three or fewer heads, (e) six or more tails, and (f) at least Posted 2 years ago. 5 (assuming a fair coin). Print the results. A fair coin, when tossed, should have an equal chance of landing either side up In probability theory and statistics , a sequence of independent Bernoulli trials with probability 1/2 of success on each trial is metaphorically called a fair coin. ' P(3 tails on 3 flips) = 1/2 * 1/2 * 1/2. Re: John tossed a fair coin 3 times. If the coin is tossed five times, what is the probability that at least one of the tosses will yield heads? Another way to look at it is that we are talking about getting a tails and another tails and so on, and "and" usually means multiplication in probability. of cases of 15 or more heads where the first 15 tosses are heads = 2^25 (because the remaining 25 throws can be anything either heads or tails). Click here to see ALL problems on Probability-and-statistics Question 149445 : A fair coin is tossed 5 times. SOLUTION: A fair coin is tossed four times. Note: the probability of an event, say getting a Tail when tossing a fair coin is the number of ways or times a Tail can occur divided by the total number of possible outcomes. Therefore, there are 210 strings in the sample space (and each is equally likely). Does anyone has a faster way of solving this problem instead of drawing out the tree? A fair coin is flipped 5 times. Coin Flipper. This is a basic introduction to a probability distribution table. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. Take that number and divide it by the. Number of times three heads appeared = 70. 3 Sam is Coach more often. if you toss three coins, what is the probability of getting at most two tails. Many events can't be predicted with total certainty. 3 What are the odds in favor of getting at least two heads if a fair coin is tossed three times? Problem 36. The probability of getting a tail the first time is 1/2. This is therefore the probability of not getting a 6 or a head. How the coin flips one time will not affect how it flips the next time, so the flips are called 'independent. What is the expected sum? A: Each number should appear 1/5 of the time, that is 5 on. A fair coin is tossed until a head or five tails occur. If you toss the coin 10 times there are 2^10 possible outcomes or 1024. There is a finite number of trials (tosses). Probability is hence used in a variety of different subjects, wherever there is a need to measure the chances of an event to happen for instance, economics, physics, and biology. The probability of each outcome doesn’t change over the course of the trials. So, if I flip a coin 100 times, is the likeliness of it landing exactly 50 times on either side any greater than it landing all 100 times on the same side? Short, non-numerical answer: yes it is, because there's only two ways for it landing all 100 times on the same side: either it lands heads every single time, or it lands tails every single time. Two heads and a tail 3/8. at random from the box what is the probability that at least two of the chosen apples. A fair coin is tossed 7 times. A Coin Is Tossed 5 Times, Can You Find The Probability Of Getting At Least One Tail? Find The Probability Of Tossing At Least 2 Heads When A Fair Coin Is Tossed 10 Times. probability of any one of the six sides coming up is 1/6. The coin is weighted so that a head is twice as likely to occur as a tail. In this example you must consider all possible permutations with 3 Heads and 2 Tails. What is the probability that at least 5 of the flips come up heads? Flipping coin: we set h = head and t = tail. Toss the coin 50 times and. probability questions answers mcq of quantitative aptitude are useful for it officer bank exam, ssc, ibps and other competitive exam preparation - question 806. A fair coin is tossed 3 times. Write the word or phrase that best completes each statement or answers the question. 2 What is the. When a coin tossed three times. SOLUTION: A fair coin is tossed four times. When two coins are tossed at random, what is the probability of getting a. Can someone check my work? 9x-7y=31 what is the number for x and y I need help with 18 and 19 please it's urgent Solve for X 2x+5=3(x-1) Directions: Study each of the numbers below and. (ie, "I bet the Giants will win the coin toss") I'm pretty damn sure it is still 50/50, but my friend argues that is should be less. #color(green){H}# represents a head while #color(red){T}# represents a tail. For a coin toss, we can calculate the probability that heads will result from one toss. Tossing a fair coin 10 times, the probability of getting exactly 5 heads (in. Using the coin toss activity, toss the coin 25 times and then 150 times. Oct 07, 2010 - If the coin is tossed 5 times, what is the probability that at least 3 out of 5 t Visit Beat The GMAT's industry leading forum for expert advice and support. The probability of getting two heads in tossing a fair coin twice is therefore 1/4. Interview question for Quantitative Developer. If the coin is tossed two times and you want the probability of getting 2 heads, that's the probability of getting a head on the first toss AND getting a head on the 2nd toss. For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,? What is the probability of tossing a coin 5 times and getting 2 tails and 3 heads in that order?. Each coin has a probability of $1/2$ for being a head, or for being a tail, and there are three coins that could be the head. But it is hard to toss a coin 5 times same way. A fair coin, when tossed, should have an equal chance of landing either side up In probability theory and statistics , a sequence of independent Bernoulli trials with probability 1/2 of success on each trial is metaphorically called a fair coin. Online virtual coin toss simulation app. We say that the probability of the coin landing H is ½. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times What is the probability of: (i) getting all heads. If the coin is tossed and allowed to clatter to the floor, this probably adds randomness. The probability of a coin toss being a tail is 1/2. the probability of heads is 0. in case you propose you've #a million head #2 tails #3 heads, then this may be the answer: First toss = a million/2 2d toss = a million/2 0. What is the probability of getting at least 3 heads. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times. For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,? What is the probability of tossing a coin 5 times and getting 2 tails and 3 heads in that order?. A coin is therefore a two-sided die. In the end it all comes to a 50/50 somy question is: if i toss a coin and get four heads in a row, does the fifth toss has a 50/50 chance of landing heads/tails. The chances of getting a girl should be the same whether or not the rst child was a girl (after all, the coin doesn't know whether it came down heads or tails last time). The probability of getting two tails in the first two is 1/4. In this applet, you can set the true probability of heads for your virtual coin, then toss it any number of times. Write the word or phrase that best completes each statement or answers the question. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. Worked-out problems on probability involving tossing or throwing or flipping three coins: 1. If a coin is tossed 6 times what is the probability of getting 1 head? If you toss a fair coin 6 times what is the probability of getting all heads? Flip a bent coin 5 times. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 1 tail, if a coin is tossed three times or 3 coins tossed together. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. the sample space is the set of all possible outcomes, denoted S. A coin is tossed three times. A fair coin is tossed 3 times. What is the probability that I do not get two heads in a row? The way to approach this problem is by drawing out the problem. The probability P(T) that it will happen on the first toss equals 1/2. An idealized coin consists of a circular disk of zero thickness which, when thrown in the air and allowed to fall, will rest with either side face up ("heads" H or "tails" T) with equal probability. The probability question here is: what is the probability of guessing if someone else will win the coin toss. Solution: Total number of trials = 250. The independence implies that the probability of all 5 tails is (1/2)^5 = 1/32. Probability and Statistics Fourth Edition. Three flips of a very unfair coin. That's about 31 times!. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 heads, if a coin is tossed three times or 3 coins tossed together. Each coin has a probability of $1/2$ for being a head, or for being a tail, and there are three coins that could be the head. Download with Google Download with Facebook or download with email. The fourth toss isn't affected by what happened on the first three tosses. If one tosses a coin enough times, the number of heads and tails will tend to "even out. An Easy GRE Probability Question. 4) A coin is tossed four times and the sequence of heads and…. Coin tossing probability - Sample space. The probability of getting the first set of. So you have 1/2 chance of getting either heads or tails. Given that at least one head appears, what is the probability that exactly two heads will appear? A. A coin is therefore a two-sided die. If he declared the chance of getting a head with that coin were 2/3, that would be an example of _ probability. There are4 Possible Outcomes with Two Coins Tossing that is is TT,TH,HT,HH,which means one possibility is having zero heads When One coin is tossed: The outcome will be either Head or Tail : Success peercentage:50%. There's a formula you can memorize, but this problem is a bit easier to solve "from scratch". What is the probability of heads? Most answer ½, but why? Let's denote A as the event that came up tails. Otherwise, the odd man out wins — that is, you win if you got a head and both of the other players got tails, or if you got a tail and both of the others got heads. Write the word or phrase that best completes each statement or answers the question. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. The probability that the first success occurs after flipping the coin four times would be. Assuming that the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women?. But coin-toss probability is still very uncertain; A coin is tossed 50 times getting 34 heads. Best Answer: The probability of getting heads on one toss is 1/2 (or 50%). the probability of heads is 0. The probability of getting a tail the first time is 1/2. Subjective probability of an outcome is a probability obtained on the basis of personal judgment. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Getting 3 tails is the same as getting 1 head. Diaconis has even trained himself to flip a coin and make it come up. probability questions answers mcq of quantitative aptitude are useful for it officer bank exam, ssc, ibps and other competitive exam preparation - question 806. 5 for both heads and tails. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times. Find the probability of getting exactly two tails. it can occur first, second or third. Mentor: Alright, we know the theoretical probability will be 50% heads and 50% tails no matter how many trials, but what would the experimental probability be in. 3, 8 (v) no head No head means all tails are obtained Let E be the event of. An idealized coin consists of a circular disk of zero thickness which, when thrown in the air and allowed to fall, will rest with either side face up ("heads" H or "tails" T) with equal probability. However, probability theory is often useful in practice when we use probability distributions. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands, in order to choose between two alternatives, sometimes used to resolve a dispute between two parties. The answer is found by computing the probability of exactly 0 heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. Anything that can happen but is not There are a couple of important points. The total number of possible outcomes is therefore 4 and the number of outcomes where the result is two heads is 1. Two heads and a tail 3/8. Given N number of coins, the task is to find probability of getting at least K number of heads after The probability of exactly k success in n trials with probability p of success in any trial is given by Recommended Posts: Probability of getting more heads than tails when N biased coins are tossed. 4) A coin is tossed four times and the sequence of heads and…. (head/tails) u Now toss 3 coins 10 times u Make a chart of all the possible ways for the three - What is the probability of getting two head? 3 Coin toss u How many of these outcomes have two - What is the probability of the family NOT having two girls and a boy? 4 Dice u Roll a pair of dice. If X denotes the number of tosses of the coin, find mean of X. So 2 times 2 times 2-- there are 8 equally likely possibilities if I'm flipping a coin 3 times. So you have 1/2 chance of getting either heads or tails. Math archives: Probability in Flipping Coins Six pennies are flipped. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. Toss the coin twice and there are 4 (2^4)possible outcomes, HH, HT, TH, TT. If we flip a coin 100 times, what is the probability of getting even number of heads?. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times What is the probability of: (i) getting all heads. A coin is tossed three times Then find the probability of getting head on middle coin - Duration: 6:22. 7E-20 A fair coin is tossed 20 times. What is the probability that two of the coins are heads and one is tails? Show how you calculated this. Toss the coin 10 times. The result is a nonnegative integer that is less than 3. Paul's answer is the simplest. The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 tails in 3 coin tosses. A coin is tossed 5 times. What is the probability that either Sara or Kaleb tossed exactly three heads?. Based on the information you posted, we have: success = “tossing heads” (You can chose either result here. Probability is used to describe the predictable long-run patterns of random outcomes. Mathematically, coin toss experiment can be thought of a Binomial experiment, where we have a coin with probability of getting head as success at each coin toss is p. This is a basic introduction to a probability distribution table. There is only one outcome that can be all tails, so the chances are 1 in 1024. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). You want any combination of 7 of them to be heads while the remaining 3 are tails. 3 What are the odds in favor of getting at least two heads if a fair coin is tossed three times? Problem 36. The probabilities are: 0. Probability of one head (the first trial) and three tails (after first trial). What is the probability of. What is her mean expectation? asked by Claudia on January 16, 2012; statistics. Probability is the mathematics of chance. What is the probability of getting at most two heads? Unbiased coins means a coin having head and tail whereas a biased coin means having two heads or How many times a man can tossing a coin so that the probability of atleast one head is more than 80%?. Determine the probability of getting heads and probability of getting tails. In this example you must consider all possible permutations with 3 Heads and 2 Tails. That only happens 2 times. Determine the probability of getting heads and probability of getting tails. Toss a fair coin 3 times. When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50 After you have flipped the coin so many times, you should get answers close to 0. Coin Flipper. Quantity A: The probability of getting more heads than tails Quantity B: 1/2. The program should call a separate function flip that takes no arguments and returns 0 for tails and 1 for heads. If p is the probability of success and q is the probability of failure in a binomial trial, then the expected number A die is tossed 3 times. Experimental and Theoretical Probability. When an unbiased 6 sided die is rolled , we may get any one of the number from 1 to 6. The probability that the first success occurs after flipping the coin four times would be. If the coin is tossed and allowed to clatter to the floor, this probably adds randomness. The probability of getting a given number of heads from four flips is, then, simply the number of ways that number of heads can occur, divided by the number of. Firstly, probabilities do not tell you what is going to happen, they merely tell you what is likely to happen. What is the probability that the 8th toss is tails? You meet a man in a bar who offers to bet on the outcome of a coin toss being heads. Toss a Coin Six Times [02/07/1998] Suppose a coin is tossed 6 times - what is the probability that 3 heads will occur? Tossing a Coin and Rolling a Die [11/14/2002]. When a coin is tossed 3 times, the probability of getting 3 tails is a. The probability of getting two tails in the first two is 1/4. Question: Suppose that a coin is tossed 3 times. If you throw a die three times, what is the probability that one or more of your throws will come up with a 1?. Probability of getting at least one head is the reverse of probability of not getting any heads, in other words probability of getting 3 tails. The probability of getting two heads in tossing a fair coin twice is therefore 1/4. on probability. Each time a fair coin is tossed, the probability of getting tails (not heads) is 1/2 = 0. Anything that can happen but is not There are a couple of important points. $$This is not practical to compute by hand, but Wolfram Alpha gives an answer of roughly 0. Then the probability of the event A can be defined as:. Probability. Well simplify tails to T and heads to H. If the player gets 3 or 4 heads, he/she wins. Find The Probability That No More Than One Coin Lands Head Up?. Correct Answer: B. Tossing a coin 10 times is the same thing as tossing 10 coins once. Since there is one way to get all three heads, the probability of getting three heads is. 0161, and 0. We can solve for the median by finding the mean of these two numbers as follows: 3. If you flip one coin four times what is the probability of getting at least two tails? What is the theoretical probability of getting k heads from n coin flips? What is the expected standard deviation of a single coin flip, where heads = 1 and tails = 0?. What is the Probability of Getting (k) Heads in a Row for (n) Consecutive Tosses? I asked myself a fun question after reading a post on QuantNet. From a pack of 52 cards, two cards are drawn together What is probability of getting at most one tail ?. Suppose that we tossed three coins 800 times. 6, what is the probability of getting at least 1 heads. Example – A coin is tossed three times. Since the two events - getting a head in a coin in a toss and getting a 3 in a rolling of die - are independent , the happening of both events is. The same coin is tossed 1000 times, 510 heads and After tossing the same coin 10 times you are surprised to find that tails has come up 9 times You therefore conclude that this coin is not fair and that the probability of getting tails with this co?. When an unbiased 6 sided die is rolled , we may get any one of the number from 1 to 6. 3125 A coin. tres cuartos de un numero es igual a ocho 2. What is the probability of getting 3 heads and 3 tails if a fair coin is tossed 6 times? only about 75% because very often it is not equal The probability of 3 heads and 3 tails is 0. The Mean of the distribution is 23, the median is 24 and the mode is 25. Mathematically, coin toss experiment can be thought of a Binomial experiment, where we have a coin with probability of getting head as success at each coin toss is p. The probability that in tossing a fair coin the number of heads differs from 450 by 40 or more (in either direction) is, by symmetry,$$2\sum_{k=490}^{900} \binom{900}{k}\left(\frac{1}{2}\right)^{900}. 8a Author: Cdavis345 Last Modified: a year ago. As you might guess, each possible outcome will have the same probability (1/32). What is the probability of getting at least 2 heads. What is the probability of. Suppose you toss a coin 100 times and get 52 heads and 48 tails. Question: Jason tossed a fair coin 3 times. Assuming it's a fair coin the probability of a head P(H)=0. Statistics4All - What are the different possible outcomes, when we toss 3 coins or a coin is tossed 3 times and how can we calculate probabilities of various events - The answer is in this video. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. What is the probability of obtaining a. Suppose that we tossed three coins 800 times. Probability and Statistics Fourth Edition. Graphic image of the given triangle after the transformation with the rule x,y to -x,y. 3/8 becuase 4/8 means 50% which is impossible and the other 2 options are even higher odds then 4/8. $$This is not practical to compute by hand, but Wolfram Alpha gives an answer of roughly 0. If a coin is tossed 10 times, 7 heads and 3 tails. Plot the pie graph for the probabilities obtained. to locate an accrued chance, you may multiply the three opportunities mutually. The probability of tossing a coin 5 times and getting all tails is: P(TTTTT) = (1/2)5 = 0. What if you toss a coin 6 times what is the probability of 3 heads and 3 tails from PSYC 541 at University of North Dakota. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. It describes the probability distribution of a process that has two possible outcomes. sample space. If you toss a coin 3 times, you're going to get at least two heads or at least two tails, but you can't get _both_ 2 heads and 2 tails. The probability question here is: what is the probability of guessing if someone else will win the coin toss. Tails on each of the first 3 tosses c. It really is a coin tossmost people would go for a new printerand i. The program should call a separate function flip that takes no arguments and returns 0 for tails and 1 for heads. What is the probability of getting at least 3 heads when 5 coins are tossed at same time? For the experiment of tossing a single fair coin 3 times, what is the probability of getting exactly 2 heads,?. Another way to look at this is that if there are 3 tails, then there is only 1 head. Using the coin toss activity, toss the coin 25 times and then 150 times. Toss the coin 50 times and. Quantity A: The probability of getting more heads than tails Quantity B: 1/2. 4096 number of possible sequences of heads & tails. The probability of getting two tails in the first two is 1/4. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. If heads appears on all 3 tosses, Mary will win 16. Well simplify tails to T and heads to H. If Benjie throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will The probability of getting a head (or tail) in coin toss is p = 0. When a coin is tossed the probability of Tails, Prob(Tail) or P(T) = ½. If heads is the number of particular chance events of interest, then the numerator is simply “1. a Simulating Coin Flips One student conducted the simulation described in Example 3 and stated that the probability of getting a sequence of six 0s or. Example heads, tails, head , tails ,heads ,tails or tails, heads,tails, heads , tails ,heads. In this example you must consider all possible permutations with 3 Heads and 2 Tails. So a single coin toss gives. The probability that in tossing a fair coin the number of heads differs from 450 by 40 or more (in either direction) is, by symmetry,$$2\sum_{k=490}^{900} \binom{900}{k}\left(\frac{1}{2}\right)^{900}. Probability is hence used in a variety of different subjects, wherever there is a need to measure the chances of an event to happen for instance, economics, physics, and biology. If heads appears on all 3 tosses, Mary will win \$16. Based on the information you posted, we have: success = “tossing heads” (You can chose either result here. If the player gets 3 or 4 heads, he/she wins. Toss a fair coin 3 times. Repeat 10 times. 0161, and 0. Either outcome is equally likely. 2^4 = 8 That also means that there are 4 more positions where you can have the 3 consecutive tosses happen. A Coin Is Tossed 5 Times, Can You Find The Probability Of Getting At Least One Tail? Find The Probability Of Tossing At Least 2 Heads When A Fair Coin Is Tossed 10 Times. Quantity A: The probability of getting more heads than tails Quantity B: 1/2. Head and tail outcomes of a coin tossed 4 times - Продолжительность: 4:10 Dharmesh Bhati 4 250 просмотров. This is out of 16 total ways to flip a coin 4 times. We never know the exact probability this way, but we can get a pretty good estimate. | 2019-11-15T07:46:37 | {
"domain": "laron-online.de",
"url": "http://xkoj.laron-online.de/a-coin-is-tossed-3-times-what-is-the-probability-of-getting-all-tails.html",
"openwebmath_score": 0.7882991433143616,
"openwebmath_perplexity": 218.89877272879593,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668706602659,
"lm_q2_score": 0.863391617003942,
"lm_q1q2_score": 0.8471312510100645
} |
https://math.stackexchange.com/questions/2412382/paradoxial-probability-puzzle-drawing-balls-from-bag-simultaneously | Paradoxial probability puzzle - Drawing balls from bag simultaneously
Can someone help settle a debate we are having in my team? Four balls are placed in a bag. One is red, one is blue and the other two are yellow. The bag is shaken and someone draws two balls from the hat. He looks at the two balls and announces that at least one of them is yellow. What are the chances that the other ball he has drawn out is also yellow?
At first glance you may say 1/3, as you hold a yellow ball so 1/3 chance one of the remaining three balls will be yellow.
However I think the answer is 1/5 due to the independent and simultaneous drawing of the balls. Consider the drawing possibilities: YY YB BY YR RY BR RB
Knowing we hold one yellow ball there is no chance BR/RB occurred so the odds are left at 1/5.
Is this correct?
And if so can someone prove this mathematically?
Is this a probability paradox? Looking back after the balls are taken you can remove the BR/RB but does this invalidate the question?
• Will he always say "at least one of them is yellow" when true and be silent otherwise? If so I would say the $\frac15$ answer is correct, though your seven examples are not equally likely. Or might he possibly do something else? – Henry Aug 31 '17 at 16:32
• There is no paradox. You are computing conditional probability ("what is the probability that both balls are yellow knowing that at least one of the is yellow"). – anderstood Aug 31 '17 at 16:43
"At first glance you may say 1/3, as you hold a yellow ball so 1/3 chance one of the remaining three balls will be yellow."
Well, that assumes you have a specific ball in mind to be "the" yellow that is yellow and a specific ball in mind to be the "other" ball. But you don't. Either ball (or both) can be "the" yellow ball and either (or both) can be the "other".
Think of it this way: If you say the ball in his right hand is yellow then the probability that the ball in his left hand is yellow is 1/3. And if you say the ball in his left hand is yellow then the probability that the ball in right hand is yellow is 1/3. But you are asking if either the ball in his left hand or the ball in his right is yellow what is the probability that the balls in both hands are yellow.
Those are obviously different questions.
====
Actually your list should have 12, not 7, options.
$Y_1Y_2, Y_1B, Y_1R, Y_2Y_1, Y_2B, Y_2R, BY_1, BY_2, BR, RY_1,RY_2, RB$.
And of the $10$ remaining there are $2$ options.
$$P \left( A \mid B \right) = \frac{ P \left( A \cap B \right) }{P \left( B \right)}$$
Hence the probability is indeed $\frac{1}{5}$. | 2019-06-17T21:32:23 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2412382/paradoxial-probability-puzzle-drawing-balls-from-bag-simultaneously",
"openwebmath_score": 0.9173864722251892,
"openwebmath_perplexity": 450.2498527071784,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668712109664,
"lm_q2_score": 0.8633916134888613,
"lm_q1q2_score": 0.847131248036654
} |
https://math.stackexchange.com/questions/376794/what-measure-does-lebesgue-measure-induce-on-the-fat-cantor-set | What measure does Lebesgue measure induce on the fat Cantor set?
I know that the fat Cantor set under the subspace topology is homeomorphic to Cantor space $\{0,1\}^{\mathbb N}$ under the product topology induced by the discrete topology on $\{0,1\}$. Call the natural homeomorphism $f$.
What about the measure induced by Lebesgue measure on Cantor space via $f$? Is it the same (up to a constant) as the usual product measure, i.e. the probability measure used to describe an infinite sequence of coin tosses? A reference would be helpful.
• All the uncountable Polish spaces have the same Borel sets, so their measure theory is fairly the same. Apr 30 '13 at 1:11
• Are you saying that the Borel measure on the fat Cantor set (probably) induces the product measure on Cantor space? Or that Lebesgue measure does? Apr 30 '13 at 1:21
• I guess you're saying that the Borel measure induces the product measure and Lebesgue measure induces its completion. The product measure on Cantor space isn't complete, right? Apr 30 '13 at 1:50
• I don't remember all the gory details, in fact I don't even remember the exact proof that the Borel sigma algebras of all the uncountable Polish spaces are isomorphic. But the point is that the result might not be a scalar multiplication of the usual product measure, but it should be some very nice measure with respect to the product measure. Apr 30 '13 at 1:56
• @AsafKaragila: What do you mean by nice? It could very well be singular. Apr 30 '13 at 19:24
Yes, it is.
Note that the Cantor set is in fact a compact group, and the product measure $\mu$ is the Haar measure, that is, the invariant finite, regular Borel measure, which is by Haar's theorem unique up to a scaling factor.
That Lebesgue measure restricted to the fat Cantor set is invariant is an easy exercise (it follows easily from the fact that the Lebesgue measure is translation invariant).
It is also regular, because it is a restriction of a regular measure, so by Haar's theorem it is a scalar multiple of the product measure.
Let $C$ be the fat Cantor set; $\mu$ is the restriction of the Lebesgue measure to $C$, normalized so that $\mu(C)=1$. For every $n\ge 1$, the pre-Cantor set of generation $n$ consists of $2^{-n}$ intervals $I_{n, k}$ of equal length, and the sets $C\cap I_{n,k}$ are congruent to one another. Therefore, $\mu(C\cap I_{n,k})=2^{-n}$. Observe that this is the same measure that $I_{n,k}$ gets from coin-tossing. Also, the intervals $I_{n,k}$ generate the Borel $\sigma$-algebra because every open subset of $C$ is a union of such intervals. Conclusion: yes, the measures are the same up to normalization.
• Looks good, except $\mu(I_{n,k})$ doesn't really make sense, you should write $\mu(C\cap I_{n,k})$ instead. Apr 30 '13 at 19:54
• @tomasz Thanks, I made the corrections. Apr 30 '13 at 21:04
• This answer is also helpful, thanks. Apr 30 '13 at 22:18
Caution: Thanks to tomasz, I realize my "answer" is almost surely irrelevant to OP's question. Yet I decide not to delete it in case it might interest some others. Perhaps more importantly, it's kind of difficult for me to delete something that took me quite a while to finish.
This is an elaboration of Asaf Karagila's comment, "all the uncountable Polish spaces have the same Borel sets".All the stuff is borrowed from Probability measures on metric spaces, Parthasarathy(1967).
Main result: Two Borel sets of Polish spaces are isomorphic, iff they have the same cardinality.
Defination: Given Polish spaces $$X_1$$ and $$X_2$$, Borel sets, $$B_1$$ and $$B_2$$ and $$B_1 \subseteq X_1$$, $$B_2 \subseteq X_2$$, $$B_1$$ is isomorphic to $$B_2$$, which is denoted as $$B_1 \sim B_2$$, iff there is a bijection $$\phi : B_1 \to B_2$$, and both $$\phi$$ and its inverse, $$\phi^{-1}$$ are measurable functions.
Theorem $$1$$: Let $$M = 2^{\omega}$$ endowed with product topology, there exists a Borel subset $$E \subseteq M$$ such that $$E \sim [0,1]$$.
Proof: Consider the well-known continuous surjective function $$\tau$$ from the Cantor space onto the interval $$[0,1]$$: $$(x_n)\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ The problem for this function is that it's not injective, which means, we need to make a choice between eventual constant sequences that ends with $$1$$ and those ending with $$0$$. One such set $$E$$ is the set of sequences in which $$0$$ occurs infinite times and the sequence that only comsists of $$1$$. It's not difficult to see the function $$\tau$$ restricted on $$E$$ and its inverse are both continuous which implies that they're also measurable.
Theorem $$2$$: There exists a Borel set $$E_1 \subseteq M$$,$$(M = 2^{\omega})$$ such that $$E_1 \sim [0,1]^{\omega}$$.
Proof: With the help of the bijection defined in Theorem $$1$$, we have the induced bijection from $$2^{E}$$ to $$[0,1]^{\omega}$$.Since $$2^{\omega} = (2^{\omega})^{\omega}= 2^{\omega \times \omega}$$, $$M$$ is homeomorphic to $$2^{M}$$ by $$\iota$$, hence $$M \sim 2^{M}$$. Thus there's a subset of $$M$$,$$E_1 = \iota^{-1}(E)$$. So we have $$\tau \circ \iota$$ that guarentee the isomorphism between $$E_1$$ and $$[0,1]^{\omega}$$.
Theorem 3: Let $$X$$ be a Polish space, and $$G \subseteq X$$ be a Borel set. Then there exists a Borel set $$E' \subseteq M$$ such that $$G \sim E'$$.
Proof:Theorem 2 makes our life easier by enabling us to turn our attention away from the Cantor space $$M$$ to Hilbert cube $$[0,1]^{\omega}$$. One important result is that every Polish Space is homeomorphic to a $$G_\delta$$ subspace of the Hilbert Cube (see here). We are done.
• Remark:Thanks to Theorem $$1$$ to $$3$$, we don't need to consider an arbitary Polish space. We only have to prove the theorem for the Cantor space.
Theorem 4: Let $$\mathscr{N} = \omega^{\omega}$$ denote the Baire space. For any Polish space $$X$$, there's a continuous surjection $$\psi$$ from $$\mathscr{N}$$ onto $$X$$.
Proof: This theorem is omnipresent in various textbooks, notes, et cetera. e.g. See here.
Theorem 5: Let $$X$$ be a Polish space and $$E \subseteq X$$ be a Borel set, then there exist a continuous function $$\psi$$ that maps $$\mathscr{N}$$ into $$X$$ such that $$\psi(\mathscr{N}) = E$$.
Proof: We show this by induction. It's obvious that it's true for closed sets, since they're complete with the same metric in $$X$$. Suppose it holds for a denumerable set of subsets of $$X$$, which is $$\{A_n\}_{n \in \omega}$$ with $$\psi_n(\mathscr{N}) = A_n$$. Define a continuous function as a cartesan product of $$\{\psi_n\}_{n \in \omega}$$, i.e. $$(\psi_0, \psi_1, \psi_2, \ldots) : \mathscr{N}^{\omega} \to X^{\omega}$$. Also define a subset of $$\mathscr{N}^{\omega}$$ as $$\mathscr{N}' = \{ x \in \mathscr{N}^{\omega} : \forall m, n \in \omega \psi_m(x) = \psi_n(x)\}$$. As a continuous preimage of a diagonal set, $$\mathscr{N}'$$ is closed in $$\mathscr{N}^{\omega}$$. Now define $$\psi :$$\mathscr{N}'$$\to X$$, for any $$x \in \mathscr{N}'$$ and let $$\pi_1$$ projects $$x$$ to its first component. $$x \mapsto \psi_1(\pi_1(x))$$ Now we have $$\psi(\mathscr{N}') = \bigcap_{n \in \omega}A_n$$. Notice that $$\mathscr{N}$$ is homeomorphic to $$\mathscr{N}^{\omega}$$. So all we need is a composition of two isomophism. To complete the proof, we also have to prove it holds for $$\bigcup_{n \in \omega}A_n$$. Let $$\mathscr{N}_j = \{ x \in \mathscr{N}: \pi_1(x) = j\}$$ for each $$j$$. $$\mathscr{N}_j$$ is isomorphic to $$\mathscr{N}$$, so there a continuous surjection $$\phi_j$$such that $$\phi_j(\mathscr{N}_j) = A_j$$. Define $$\phi : \mathscr{N} \to \bigcup_{n \in \omega}A_n$$, for each $$x$$ in $$\mathscr{N}$$: $$x \mapsto \phi_j(x) \text{iff } \pi_1(x) = j$$.
Theorem $$6$$: Given $$X$$ an uncountable seperable metric space, then there is a partition of $$X$$ with two atoms. One atom is countable, and the other, which is dense in itself, consists of all condensation points.
Proof: This is essentially Cantor–Bendixson theorem.
Theorem $$7$$: Let $$X$$ be a Polish space, $$Y$$ be an uncountable seperable metric space. Assume that there exists a continuous surjection $$\phi$$ from $$X$$ onto $$Y$$. Then there exists a subset $$C \subseteq X$$, such that $$\phi : C \to Y$$ is a homeomorphism. Moreover, $$C$$ is homeomorphic to the Cantor space, and $$\phi(C)$$ is compact.
Proof: With $$\bf{AC}$$, we can construct a subset $$A \subseteq X$$ such that $$\phi$$ ristricted to $$A$$ is a bijection. By Theorem $$6$$, we have a subset $$D \subseteq A$$ such that $$D$$ is an uncountable perfect set. Notice that $$D$$, as a Hausdorff uncountable perfect set contains a copy $$C$$ of Cantor set, which is the same as the construction in the proof of $$|\mathbb{R}| = 2^{\aleph_0}$$. $$\phi(C)$$ is compact follows from the fact it's homeomorphic to the Cantor space, which is compact.
Corrollary: Let $$X$$ be a Polish space, $$E$$ be an uncountable Borel set in $$X$$. Then there exists a compact subset $$C \subseteq E$$ such that $$C$$ is homeomorphic to the Cantor set. In particular, $$|E| = \mathfrak{c}$$.
Lemma: Given $$\{E_m\}_{m \in \omega}$$ a set of disjoint Borel sets, and $$\{F_n\}_{n \in \omega}$$ the same, then $$E_i \sim F_i$$ for all $$i \in \omega$$ implies that $$\bigcup_{i \in \omega} E_i \sim \bigcup_{i \in \omega} F_i$$
Theorem 8: Let $$E_1 \subseteq F \subseteq E_0$$ be three Borel subsets of Cantor space. Then $$E_0 \sim E_1$$ implies that $$E_0 \sim F$$.
Proof: Let $$\psi : E_0 \to E_1$$ be the mapping that guarentee the isomorphism. Define recursively, $$E_{n+1} = \psi(E_n)$$, $$A_{n+1} = E_n \setminus F$$, $$B_{n} = F \setminus E_n$$, $$D_{n+1} = E_n \setminus E_{n+1}$$, $$E_{\infty} = \bigcap_{n \in \omega} E_n$$. Via $$\psi^{m-n}$$ and $$\psi^{i-j}$$ , we have $$D_m \sim D_n$$ for all $$m,n \in \omega$$,$$E_i \sim E_j$$ for all $$i,j \in \omega$$. Similarly, it also holds for the class $$\{A_n\}$$ and $$\{B_m\}$$. Thus we have: $$E = E_{\infty} \cup \bigcup_{i \in \omega}A_i \cup \bigcup_{j \in \omega}B_j \sim E_{\infty} \cup \bigcup_{i \in \omega \setminus \{0\}}A_i \cup \bigcup_{j \in \omega}B_j = F$$
Theorem $$9$$:Two Borel sets $$E_1$$ and $$E_2$$ of Polish spaces $$X_1$$ and $$X_2$$ are isomorphic, if they have the same cardinality.
Proof: As shown, WLOG, Let $$X_1 = X_2$$ be the Cantor spaces. The only interesting case is when $$E_1$$ and $$E_2$$ are uncountable. To prove the theorem, it suffice to show $$E_1 \sim E_2 \sim 2^{\omega}$$. Since both $$E_1$$ and $$E_2$$ contains a copy of the Cantor space, by appealing to Theorem $$8$$, we have $$E_1 \sim 2^{\omega}$$ and $$E_2 \sim 2^{\omega}$$. Moreover $$\sim$$ is an equivalence relation, we have $$E_1 \sim E_2$$.
• How does this answer the question? May 1 '13 at 3:15
• @tomasz: I'm sorry I don't follow your comment. There's nothing special about "Lebesgue measure induce on the fat Cantor set". All uncountable Polish spaces are isomorphic. May 1 '13 at 3:21
• The question was about two particular measures, not Borel algebras. Any two uncountable Polish spaces are Borel isomorphic. May 1 '13 at 3:27
• @tomasz: Thank you for pointing that out. To be frank, I didn't read the question carefully and haste to give an answer, which is in effect a supplement on Asaf's comment. Hope you don't mind asking. What can we say about two measures, if we've already known the isomorphism between two Borel algebras. May 1 '13 at 4:01
• As far as I can tell, nothing really; that the Borel algebras are isomorphic is really trivial in this case, because the fat Cantor set is homeomorphic to the usual Cantor set, which is much stronger than mere Borel isomorphism. What the theorem really gives us is that when considering an abstract Borel measure on a Polish space, we can very well assume that the Polish space is any uncountable Polish space we want. But two measures on a Polish space can be rather different (singular), e.g. the Lebesgue measure on reals and the image of the Cantor measure by the usual embedding. May 1 '13 at 8:02 | 2022-01-19T20:54:45 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/376794/what-measure-does-lebesgue-measure-induce-on-the-fat-cantor-set",
"openwebmath_score": 0.97163987159729,
"openwebmath_perplexity": 112.84937759639506,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668668053618,
"lm_q2_score": 0.8633916134888613,
"lm_q1q2_score": 0.847131244232892
} |
https://math.stackexchange.com/questions/4281311/maximizing-an-angle-based-on-certain-constraints | # Maximizing an angle based on certain constraints
$$A (0,a)$$ and $$B(0,b)\; (a,b>0)\;$$ are the vertices of $$\triangle ABC$$ where $$C(x,0)$$ is variable. Find the value of $$x$$ when angle $$ACB$$ is maximum.
Now geometry's never really been my strong point, so I decided to go with a bit of calculus. First, I used the sine rule: $$\mathrm{sinC=\frac{b-a}{2R}}$$ where R is the radius of the circumcircle. I note that for angle C to be maximum, sinC should be maximum. As such, R must be minimum. Next, I used the relation $$\mathrm{R=\frac{(b-a)\cdot\sqrt{x^2+b^2}\cdot\sqrt{x^2+a^2}}{2\Delta}}$$ where $$\mathrm{\Delta \text{ is the area of }ABC=\frac{(b-a)x}{2}}$$.
A bit of comparatively lengthy differentiation gives me the value of $$x$$ as $$\sqrt{ab}$$.
When I go through the solutions, it's simply been stated:
For angle ACB to be maximum, the circle passing through A,B will touch the X-axis at C.
Beyond this, it's been solved using the very simple $$\mathrm{OC^2=OA\cdot OB}$$, where O is the origin. So the above statement seems to be the difference between a lengthy differentiation and a one line solution.
It's getting a little difficult for me to see why the above statement should be intuitive. Could someone shed a bit more light on it for me, and possibly provide an intuitive proof?
• I couldn't settle on a concise yet lucid title. Please feel free to edit it. Oct 19 at 17:20
## 6 Answers
It is given that $$a, b \gt 0$$ so both $$A$$ and $$B$$ are on the same side of x-axis. The first point to note is that $$\triangle ABC$$ is obtuse and $$\angle ACB$$ is acute. Now we use the relationship $$AB = 2 R \sin C$$ where $$R$$ is the circumradius of $$\triangle ABC$$. As $$AB$$ is fixed, we maximize $$\angle C$$ when we minimize $$R$$ given $$\sin$$ function is strictly increasing for $$\left(0, \frac{\pi}{2} \right)$$.
Also note that $$O$$ must be on the perpendicular bisector of $$AB$$ which is parallel to x-axis. So, $$R = OC$$ is minimum when $$OC$$ is perpendicular to x-axis.
• ''Assuming both $\mathrm{A}$ and $\mathrm{B}$ are above x-axis''. Yes that's been mentioned in the question. I really like that this answer picked up my line of thinking when I first approached this problem and provided an intuitive explanation in that fashion. The other answers I've got are also very good; it's hard to settle on a 'best' answer. Oct 19 at 18:06
• @C_Lycoris good to see you have a few answers to choose from :) I was just not sure whether question said that $a, b$ were both positive so added the line about them being on the same side of x-axis. Oct 19 at 18:09
• @C_Lycoris I see that now so I will remove that line Oct 19 at 18:10
• In the OP's post $O$ is referred to as the origin. In your answer?
– ACB
Oct 20 at 7:01
• @ACB I am referring to $O$ as the center of the circumcircle of $\triangle ABC$ Oct 20 at 7:04
Let $$\omega$$ be the circumcirle of $$\triangle ABC$$ where $$C$$ is a point on the $$x$$-axis such that $$\angle ACB$$ is maximum. Assume $$\omega$$ intersects the $$x$$-axis twice, at $$C$$ and $$D$$.
Let $$F$$ be any point on the arc $${CD}$$ (not containing $$A, B$$) and define $$E$$ as the intersection of $$AF$$ and $$x$$-axis. Observe, $$\angle ACB=\angle AFB<\angle AEB$$ which contradicts the fact that $$\angle ACB$$ is maximum.
Therefore, the assumption that $$\omega$$ intersects the $$x$$-axis twice is incorrect, which implies $$\omega$$ is tangent to the $$x$$-axis at $$C$$.
• +1. I liked this proof, it really should've occurred to me. Oct 19 at 18:08
Let $$\bigcirc K$$ through $$A$$ and $$B$$ be tangent to the $$x$$-axis at $$D$$. For $$C$$ on the $$x$$-axis (and on the same side of the $$y$$-axis as $$D$$), let $$A'$$ and $$B'$$ be the "other" points where $$\overleftrightarrow{AC}$$ and $$\overleftrightarrow{BC}$$ meet this circle.
A corollary to the Inscribed Angle Theorem states that we can write $$\angle C = \frac12 \left(\;\angle AKB - \angle A'KB'\;\right)$$ Since $$\angle AKB$$ is fixed, maximizing $$\angle C$$ amounts to minimizing $$\angle A'KB'$$. This happens when (and only when) $$A'$$ and $$B'$$ coincide; hence, when $$C$$ and $$D$$ coincide. $$\square$$
We use the fact that if $$A$$ and $$B$$ are on a given circle, then if you have $$C$$ on the circle and $$C_0$$ (strictly) inside the circle (and $$C$$, $$C_0$$ are on the same side of $$\overline{AB}$$) $$\angle AC_0B > \angle ACB$$
You can see this by extending $$AC_0$$ to the circle at $$C'_0$$, in which case $$\angle AC_0B > \angle AC'_0B$$, but because of the inscribed angle theorem, $$\angle AC'_0B = \angle ACB$$.
Now, write the desired angle $$\angle ACB$$ as $$f(x)$$ in terms of $$x$$; we want to maximize $$f(x)$$.
The circumcircle of $$\triangle ABC$$ always intersects the $$x$$-axis at $$C = (x_0, 0)$$. Now, say for contradiction that $$f(x_0)$$ is maximal, and that the circumcircle of $$\triangle ABC$$ also intersects the $$x$$-axis at $$C' = (x_0', 0) \neq C$$. Then the midpoint $$M$$ of $$\overline{CC'}$$ is inside the circle, so $$\angle AMB > \angle ACB$$; and $$M = \left(\frac{x_0 + x'_0}{2}, 0 \right)$$ is on the $$x$$-axis, so $$f\left( \frac{x_0+x_0'}{2} \right) > f(x_0),$$ contradicting the fact that $$f(x_0)$$ is maximal.
Thus the circumcircle of $$\triangle ABC$$ must intersect the $$x$$-axis at exactly one point.
There are two ways of looking at it.
Circles such as those that pass through non red sides meeting on x-axis at D,E subtend the same angle from segment AB ( Angles in a segment are equal property). In order that there be a unique point, these points should be drawn together to make them into a repeated point. A repeated point is in fact a point of tangentcy at C.
By means of the Circle's property that the product of segments be constant (this being is the power of the Circle ) we have
$$OA \cdot OB= OC^2= x^2 \to \; x = \sqrt {ab} \tag1$$
Next way is direct confirmation with differential calculus, maxima/minima.
The "look angle " or subtended angle is
$$\tan^{-1}\frac{a}{x}-\tan^{-1}\frac{b}{x}$$
Differentiate w.r.t. $$x$$ arctan and Chain Rule
$$\dfrac{-a/x^2}{1+a^2/x^2} + \dfrac{-b/x^2}{1+b^2/x^2} =0$$
When simplified, we get the same result as (1).
Another way to solve it using calculus and geometry is to notice that $$\angle C=\frac{\pi}{2}-(\angle ACO+\angle CBO)$$ (where $$O$$ is the origin). Minimizing $$\angle ACO+\angle CBO$$ is equivalent to minimizing $$\tan \angle ACO+\tan \angle CBO=\frac{a}{x}+\frac{x}{b}$$ which is easy to differentiate.
• I'm sorry, but what does point 'O' refer to in your answer? If it's the origin, shouldn't $\angle C=\frac{\pi}{2}-(\angle ACO+\angle CBO)$? Oct 20 at 9:37
• Thanks! I did have a typo, but the function expressing sum of tangents is correct. Point $O$ is the origin. Oct 21 at 2:04
• Yes. I saw that the final expression was correct. Also a very quick differentiation to yield the answer. Thank you! Oct 21 at 2:18 | 2021-11-29T12:19:09 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/4281311/maximizing-an-angle-based-on-certain-constraints",
"openwebmath_score": 0.8879001140594482,
"openwebmath_perplexity": 141.52673961292717,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668695588648,
"lm_q2_score": 0.8633916099737806,
"lm_q1q2_score": 0.8471312431613627
} |
https://math.stackexchange.com/questions/2037005/calculating-the-sum-of-sum-k-1-infty-frac-1kk | # Calculating the sum of $\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$ [duplicate]
I am trying to find the sum of
$$\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$$
I've proven that this converges using the Leibniz test, since $a_n > 0$ and $\lim_{n\to\infty}{a_n} = 0$.
I am not sure how to go about summing this series up though. Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up.
## marked as duplicate by Jack, Math-fun, 2012ssohn, Martin Sleziak, user91500Dec 1 '16 at 5:42
• It is not enough $\;a_n>0\,,\,a_n\rightarrow 0\;$ . It must be that $\;a_n\;$ is monotone descending, otherwise the series may diverge. – DonAntonio Nov 30 '16 at 8:37
Using the Taylor series for the natural logarithm, $$\ln(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}$$ for $-1<x\le 1$. Abel's theorem guarantees that $$\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}.$$ Hence, $$\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=-\ln2.$$
• Could anyone explain the reason for the downvote?.. – Cm7F7Bb Nov 30 '16 at 7:57
• The answer is making no connection to the geometric series. And dropping ready-made Taylor is tantamount to saying "this sum is known to be $-\ln 2$". – Yves Daoust Nov 30 '16 at 8:05
• @YvesDaoust Do the answers have to make a connection to the geometric series? No, it does not say so in the question. I agree with the remark about the Taylor series, but it is not a reason for a downvote. In any case, we still need to know why we can set $x$ equal to $1$ and I explained that. – Cm7F7Bb Nov 30 '16 at 8:24
• Completely agree with Cm: no need to connect with geometric series and the point is this answer uses Abel's theorem to susbtitute $\;x=1\;$ . Taylor series are pretty elementary stuff, though perhaps geometric series are even more elementary. Yet the other answer also uses as "high level" basis calculus when it assumes a power series can be integrated and even without mentioning anything about its convergence radius. – DonAntonio Nov 30 '16 at 8:41
• @DonAntonio: Am I crazy ? "Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up." – Yves Daoust Nov 30 '16 at 10:15
You can find the sum using only basic calculus -- without invoking termwise integration or differentiation of infinite Taylor series or Abel's theorem.
Begin with the finite geometric sum
$$\sum_{k=0}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1+x}.$$
Rearranging, we have
$$\sum_{k=0}^{n-1} (-1)^kx^k - \frac{1}{1+x}= \frac{(-1)^nx^n}{1+x}.$$
Integrate both sides over $[0,1]$, noting that termwise integration of the finite sum requires no special justification,
$$\sum_{k=0}^{n-1} (-1)^k\int_0^1x^k \, dx - \int_0^1\frac{1}{1+x} \, dx= \int_0^1\frac{(-1)^nx^n}{1+x} \, dx.$$
Hence,
$$\sum_{k=0}^{n-1} \frac{(-1)^k}{k+1} - \ln 2= (-1)^n\int_0^1\frac{x^n}{1+x} \, dx,$$
and
$$0 \leqslant \left|\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} - \ln 2 \right| = \left|\int_0^1\frac{x^n}{1+x} \, dx\right| \\ \leqslant \int_0^1\left|\frac{x^n}{1+x}\right| \, dx \leqslant \int_0^1 x^n \, dx = \frac{1}{n+1}.$$
By the squeeze theorem, it follows that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln 2.$$
Since an alternating sum is the difference of the non-alternating sum and twice the sum of the even terms, we get \begin{align} \sum_{k=1}^{2n}\frac{(-1)^{k-1}}k &=\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac1{2k}\\ &=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\ &=\sum_{k=n+1}^{2n}\frac1k\tag{1} \end{align} By comparison to an integral, we have $$\log\left(\frac{2n+1}{n+1}\right)=\int_{n+1}^{2n+1}\frac1x\,\mathrm{d}x\le\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}\frac1x\,\mathrm{d}x=\log(2)\tag{2}$$ Apply the Squeeze Theorem to $(2)$ and incorporate $(1)$ and we get $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}k=\log(2)\tag{3}$$
Hint:
The geometric series is very very close.
Replace $-1$ by $x$ and derive term-wise:
$$\left(\sum_{k=1}^\infty\frac{x^k}k\right)'=\sum_{k=1}^\infty\left(\frac{x^k}k\right)'=\sum_{k=1}^\infty x^{k-1}.$$
Now you can use the geometric series formula and integrate from $0$ to $-1$ to get the solution
$$-\left.\ln(1-x)\right|_0^{-1}.$$
• Oh, so you recognised this as a power series so you could derive and integrate it? Could you just briefly explain why we integrate only from $0$ to $-1$? – Pavlin Nov 30 '16 at 8:14
• How did you justify differentiation term by term? – Vim Nov 30 '16 at 15:21
• @vim: I have turned the answer to a hint. – Yves Daoust Nov 30 '16 at 15:27 | 2019-10-23T09:54:51 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2037005/calculating-the-sum-of-sum-k-1-infty-frac-1kk",
"openwebmath_score": 0.8858996033668518,
"openwebmath_perplexity": 363.64165358849766,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668668053619,
"lm_q2_score": 0.8633916117313211,
"lm_q1q2_score": 0.8471312425084518
} |
https://math.stackexchange.com/questions/1583266/i-would-like-to-prove-convergence-of-the-following-series-sum-n-1-infty | # I would like to prove convergence of the following series: $\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)}$
I would like to prove the following series: $$\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)}$$ is convergent (absolutely?) or divergent. I think $\arctan\left(\frac{n}{1+n^2}\right)$ is divergent, but I don't know how it interacts with $(-1)^n$ and how to prove it.
Any ideas would be greatly appreciated.
• Doesn’t your arctan piece look very much like $1/n$? Shouldn’t it have the same convergence properties as $\sum(-1)^n/n$? – Lubin Dec 20 '15 at 13:44
• As Lubin commented, there will be a problem without $(-1)^n$ but with it, it should converge. – Claude Leibovici Dec 20 '15 at 13:47
• Try the Leibniz rule. – A.Γ. Dec 20 '15 at 13:52
Your initial series is convergent. To prove it, you may just use the alternating series test:
• the function $x \mapsto \arctan \left(\dfrac{x}{1+x^2}\right)$ is decreasing over $[1,\infty)$, since its derivative is negative over $[1,\infty)$: $\left(\arctan \left(\dfrac{x}{1+x^2}\right)\right)'= \dfrac{1-x^2}{1+3 x^2+x^4}\leq 0$,
and
• as $x \to \infty$, you have $\arctan \left(\dfrac{x}{1+x^2}\right)\sim \dfrac1x \longrightarrow 0.$
• Is it possible to show it's decreasing without using derivations? Maybe $a_{n} - a_{n+1} >0$ ? And this test doesn't prove absolute convergence, no? – Mykybo Dec 20 '15 at 14:11
• @Mykybo Yes, here you can also apply $\arctan a - \arctan b =\arctan \left( \frac{a-b}{1+ab}\right)$. – Olivier Oloa Dec 20 '15 at 14:17
• (+1) @Mykybo It does not converge absolutely. It is curious that this series can be evaluated exactly in terms of gamma functions. – Start wearing purple Dec 20 '15 at 14:18
• @Startwearingpurple and how do I prove that it does not converge absolutely? – Mykybo Dec 20 '15 at 16:51
Leibniz's criterion for alternating series works here: $\dfrac n{1+n^2}$ decreases to $0$, and $\arctan x$ is continuous increasing, hence $\arctan \dfrac n{1+n^2}$ decreases (to $0$).
It is not absolutely convergent, because $$\frac n{1+n^2}=\frac1n\cdot\frac1{1+\cfrac1{n^2}}=\frac1n+o\Bigl(\frac1n\Bigr)$$ Now $\arctan u=u+o(u)$, so
$$\arctan\frac n{1+n^2}=\frac1n+o\Bigl(\frac1n\Bigr)\sim_\infty \frac1n,$$ which diverges.
• Shouldn't we prove that it is decreasing? If I understand it correctly, what we do is ignore $\arctan{}$ because it is a continuously increasing function, which means it behaves in a same way as if it wasn't there. Also what criteria are you using to prove its not absolutely convergent? Also when you got this at the end: ${\frac{1}{n}}$, by $\sim_\infty$ you mean "as series goes to infinity"? – Mykybo Dec 20 '15 at 15:09
• No, I use the notion of equivalent sequences at infinity. This means the ratio of the sequences tends to $1$. A general result in asymptotic analysis is that two series with positive equivalent terms both converge or both diverge. Here the problem comes down to the harmonic series. – Bernard Dec 20 '15 at 15:19
To show that the terms are decreasing:
$\arctan\left(\frac{n}{1+n^2}\right)- \arctan\left(\frac{n+1}{1+(n+1)^2}\right) =\arctan\left(\dfrac{\frac{n}{1+n^2}-\frac{n+1}{1+(n+1)^2}}{1+\frac{n}{1+n^2}\frac{n+1}{1+(n+1)^2}}\right)$
and $\dfrac{n}{1+n^2}-\dfrac{n+1}{1+(n+1)^2} =\dfrac{n(1+(n+1)^2)-(n+1)(1+n^2)}{(1+n^2)(1+(n+1)^2)}$
and
$\begin{array}\\ n(1+(n+1)^2)-(n+1)(1+n^2) &=n(n^2+2n+2)-(n^3+n^2+n+1)\\ &=n^3+2n^2+2n-(n^3+n^2+n+1)\\ &=n^2+n-1\\ &> 0\\ \end{array}$ | 2019-11-21T22:09:11 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1583266/i-would-like-to-prove-convergence-of-the-following-series-sum-n-1-infty",
"openwebmath_score": 0.7728836536407471,
"openwebmath_perplexity": 288.36972450725,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9811668690081642,
"lm_q2_score": 0.8633915994285382,
"lm_q1q2_score": 0.8471312323392499
} |
http://math.stackexchange.com/questions/21146/prove-a-number-is-composite | # Prove a number is composite
How can I prove that $$n^4 + 4$$ is composite for all $n > 5$?
This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.
Plus, I try to factor it out: $$n^4 - 16 + 20 = ( n^2 - 4 )( n^2 + 4 ) - 5\cdot4$$ If a composite is added to a number that is a multiple of $5$, is there anything special? A hint would suffice.
Thanks,
Chan
-
Why the condition $n>5$? It is enough $n>1$. – Américo Tavares Feb 9 '11 at 12:56
Try factoring as the product of two quadratic expressions: $n^4+4=(n^2+an+b)(n^2+cn+d)$.
-
Thanks for a useful hint. – Chan Feb 9 '11 at 10:02
@Chan: Now try this: for which $t\in\mathbf Z$ does $n^4+t$ factor? – lhf Feb 9 '11 at 10:05
BTW, I first saw that problem in the first chapter of Apostol's Introduction to Analytic Number Theory. I immediately thought of using Fermat's theorem: $n^4+4$ is divisible by 5, but that only works when $n$ is not a multiple of 5. Now Apostol does not discuss Fermat's theorem till later, so there had to be a more fundamental reason. Alas, it was an algebraic reason, not really an arithmetic reason. – lhf Feb 9 '11 at 10:11
Thanks a lot for the information. I will try harder. – Chan Feb 9 '11 at 10:13
The quickest way of factoring is probably to start like this: $(n^2)^2 + 2^2 = (n^2+2)^2 - (2n)^2 = \dots$ – Hans Lundmark Feb 9 '11 at 10:27
This is a special case of a class of cyclotomic factorizations due to Aurifeuille, Le Lasseur and Lucas, the so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \frac{x^6 + 3^3}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{array}$$
-
It's $x^6+3^3$ in the second factorization. – lhf Feb 12 '11 at 12:47
@lhf: Thanks, I've corrected the typo. – Bill Dubuque Feb 12 '11 at 17:35
You can factor out $n^{4}+4$ algebraically by finding the four roots of $n^{4}+4=0$.
Since $n^{4}+4=0\Leftrightarrow n^{4}=4e^{i\pi }$, we have
$$\begin{eqnarray*} n &=&4^{1/4}e^{i\left( \dfrac{\pi +2k\pi }{4}\right) }\quad k=0,1,2,3 \\ && \\ n &=&\sqrt{2}e^{i\left( \dfrac{\pi }{4}\right) }=1+i\quad \left( k=0\right) \\ n &=&\sqrt{2}e^{i\left( \dfrac{3\pi }{4}\right) }=-1+i\quad \left( k=1\right) \\ n &=&\sqrt{2}e^{i\left( \dfrac{5\pi }{4}\right) }=-1-i\quad \left( k=2\right) \\ n &=&\sqrt{2}e^{i\left( \dfrac{7\pi }{4}\right) }=1-i\quad \left( k=3\right). \end{eqnarray*}$$
Now combining the complex conjugates factors, we get
$$\begin{eqnarray*} n^{4}+4 &=&\left( n-1-i\right) \left( n+1-i\right) \left( n+1+i\right) \left( n-1+i\right) \\ &=&\left( \left( n+1-i\right) \left( n+1+i\right) \right) \left( \left( n-1-i\right) \left( n-1+i\right) \right) \\ &=&\left( n^{2}+2n+2\right) \left( n^{2}-2n+2\right). \end{eqnarray*}$$
Note: for $n>1$, $n^2+2n+2>5$ and $n^2-2n+2>1$.
-
Factoring yields \begin{align} n^4+4 &=(n^2+2i)(n^2-2i)\\ &=(n+1+i)(n-1-i)(n+1-i)(n-1+i)\\ &=(n+1+i)(n+1-i)(n-1-i)(n-1+i)\\ &=((n+1)^2+1)((n-1)^2+1) \end{align} So for $n>1$, $n^4+4$ is composite.
-
In fact, it is true for $n>1$. You just need a clever way of factoring the expression:
$$n^4 + 4 = n^4 + 4 + 4n^2 - 4n^2 = \left ( \cdots + \cdots - \cdots \right ) \cdot \left ( \cdots + \cdots + \cdots \right )$$
- | 2015-05-29T20:59:44 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/21146/prove-a-number-is-composite",
"openwebmath_score": 0.8889188766479492,
"openwebmath_perplexity": 388.63275295995345,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9770226300899744,
"lm_q2_score": 0.8670357701094303,
"lm_q1q2_score": 0.847113568494402
} |
http://math.stackexchange.com/questions/293746/is-this-proof-correct-complete | # Is this proof correct/complete?
Let $f(x)=\sum\limits_{k=0}^{n}c_kx^k$ be a polynomial where $c_0$ and $c_n$ have different sings. Show $\exists x_0 \in \mathbb{R}$ such that $f(x_0)=0$.
My workings so far: Lets assume $c_0>0$ and thus $c_n<0$. If this is not the case we can simply look at $f^*(x)=-f(x)$ and $f^*(x)$ will satisfy this condition where $f^*(x_0)=0$ implies $f(x_0)=0$. By this assumption we know $f(0)>0$. Then for sufficiently large $x_l$ we have: $$|c_nx_l^n|>\left|\sum\limits_{k=0}^{n-1}c_kx_l^k\right|$$ Therefore, as $c_n<0$, it follows that: $$f(x_l)=\sum\limits_{k=0}^{n-1}c_kx_l^k-|c_n|x_l^n<0$$ Now because $f(x)$ is a polynomial and therefore continuous, we can apply the intermediate value theorem to conclude that $\exists x_0 \in [0,x_l]$ such that $f(x_0)=0$.
-
Yes. Your instructor might want more details for the part "For sufficiently large $x_l$ we have...", but it looks fine to me. – N. S. Feb 3 '13 at 17:02
Thanks, that is indeed the part that I was worrying about. How do I make a statement like that more rigorous? – Slugger Feb 3 '13 at 17:03
Intermediate value theorem, not mean value theorem. – 1015 Feb 3 '13 at 17:05
Note that if $x_l >1$ you have $\left|\sum\limits_{k=0}^{n-1}c_kx_l^k\right| \leq \sum\limits_{k=0}^{n-1} \left|c_kx_l^k\right|< \sum\limits_{k=0}^{n-1} \left|c_kx_l^{n-1}\right| < [\sum\limits_{k=0}^{n-1} \left|c_k\right| ]x_l^{n-1}$. It is easy to make $|c_l|x_l^n$ larger than $[\sum\limits_{k=0}^{n-1} \left|c_k\right| ]x_l^{n-1}$. – N. S. Feb 3 '13 at 17:06
You don't show that the polynomial is $> 0$ at any point. – vonbrand Feb 3 '13 at 17:32
It is easier if you take the limit at $+\infty$.
Assume $c_0<0$ and $c_n>0$ first.
Then $f(0)=c_0<0$, while $\lim_{x\rightarrow +\infty} f(x)=\lim_{x\rightarrow +\infty}c_nx^n=+\infty$.
So we can find $x>0$ such that $f(x)>0$.
By the intermediate value theorem, there exists $x_0>0$ such that $f(x_0)=0$.
I let you do the case $c_0>0$ and $c_n<0$.
- | 2016-02-09T13:56:12 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/293746/is-this-proof-correct-complete",
"openwebmath_score": 0.9783045053482056,
"openwebmath_perplexity": 126.82438289561891,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.977022625406662,
"lm_q2_score": 0.8670357683915538,
"lm_q1q2_score": 0.8471135627553984
} |
http://campingcostabrava.nl/cz8rs490/9ab558-rolle%27s-theorem-khan-academy | over this interval, or the average change, the that mathematically? We know that it is Khan Academy is a 501(c)(3) nonprofit organization. you see all this notation. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. ^ Mikhail Ostragradsky presented his proof of the divergence theorem to the Paris Academy in 1826; however, his work was not published by the Academy. The mean value theorem is still valid in a slightly more general setting. Use Rolle’s Theorem to get a contradiction. One only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit just means that there's a defined derivative, He showed me this proof while talking about Rolle's Theorem and why it's so powerful. bracket here, that means we're including The Common Sense Explanation. Rolle's theorem definition is - a theorem in mathematics: if a curve is continuous, crosses the x-axis at two points, and has a tangent at every point between the two intercepts, its tangent is parallel to the x-axis at some point between the intercepts. proof of Rolle’s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. And then this right Mean value theorem example: square root function, Justification with the mean value theorem: table, Justification with the mean value theorem: equation, Practice: Justification with the mean value theorem, Extreme value theorem, global versus local extrema, and critical points. AP® is a registered trademark of the College Board, which has not reviewed this resource. to visualize this thing. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus. Explain why there are at least two times during the flight when the speed of If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. this is b right over here. the function over this closed interval. Let. Mean value theorem example: polynomial (video) | Khan Academy interval between a and b. differentiable right at b. And if I put the bracket on such that a is less than c, which is less than b. over here, the x value is b, and the y value, It’s basic idea is: given a set of values in a set range, one of those points will equal the average. And so let's just try More precisely, the theorem … f is differentiable (its derivative is 2 x – 1). is the secant line. The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. the slope of the secant line. One of them must be non-zero, otherwise the … Problem 3. over here is the x-axis. Illustrating Rolle'e theorem. slope of the secant line, or our average rate of change Applying derivatives to analyze functions. Thus Rolle's theorem claims the existence of a point at which the tangent to the graph is paralle… that at some point the instantaneous rate function right over here, let's say my function rate of change is going to be the same as about some function, f. So let's say I have point a and point b, well, that's going to be the The Extreme value theorem exercise appears under the Differential calculus Math Mission. some of the mathematical lingo and notation, it's actually So when I put a Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. It also looks like the the point a. You're like, what a quite intuitive theorem. it looks like right over here, the slope of the tangent line instantaneous slope is going to be the same over the interval from a to b, is our change in y-- that the Here is a set of practice problems to accompany the The Mean Value Theorem section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. between a and b. This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. Donate or volunteer today! of change, at least at some point in At some point, your So let's just remind ourselves it's differentiable over the open interval the average rate of change over the whole interval. And it makes intuitive sense. Applying derivatives to analyze functions. Now what does that x value is the same as the average rate of change. point in the interval, the instantaneous All the mean value That's all it's saying. If you're seeing this message, it means we're having trouble loading external resources on our website. theorem tells us is that at some point can give ourselves an intuitive understanding The theorem is named after Michel Rolle. The average change between https://www.khanacademy.org/.../a/fundamental-theorem-of-line-integrals of the mean value theorem. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. So this is my function, f ( 0) = 0 and f ( 1) = 0, so f has the same value at the start point and end point of the interval. Which, of course, f ( x) = 4 x − 3. f (x)=\sqrt {4x-3} f (x)= 4x−3. Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). To log in and use all the features of Khan Academy, please enable JavaScript in your browser. the average change. c. c c. c. be the number that satisfies the Mean Value Theorem … Rolle's theorem says that somewhere between a and b, you're going to have an instantaneous rate of change equal to zero. Now how would we write Donate or volunteer today! This exercise experiments with finding extreme values on graphs. for the mean value theorem. The “mean” in mean value theorem refers to the average rate of change of the function. Over b minus b minus a. I'll do that in that red color. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. a and x is equal to b. over here, this could be our c. Or this could be our c as well. Our change in y is Or we could say some c change is going to be the same as interval, differentiable over the open interval, and To log in and use all the features of Khan Academy, please enable JavaScript in your browser. More details. So all the mean ... c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. Check out all my Calculus Videos and Notes at: http://wowmath.org/Calculus/CalculusNotes.html the average change. is that telling us? L'Hôpital's Rule Example 3 This original Khan Academy video was translated into isiZulu by Wazi Kunene. Mean value theorem example: square root function, Justification with the mean value theorem: table, Justification with the mean value theorem: equation, Practice: Justification with the mean value theorem, Extreme value theorem, global versus local extrema, and critical points. So this right over here, well, it's OK if it's not And as we'll see, once you parse Welcome to the MathsGee STEM & Financial Literacy Community , Africa’s largest STEM education network that helps people find answers to problems, connect … differentiable right at a, or if it's not here, the x value is a, and the y value is f(a). y-- over our change in x. over our change in x. average rate of change over the interval, So some c in this interval. So there exists some c All it's saying is at some A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. what's going on here. the average slope over this interval. The mean value theorem is a generalization of Rolle's theorem, which assumes f(a) = f(b), so that the right-hand side above is zero. that you can actually take the derivative open interval between a and b. https://www.khanacademy.org/.../ab-5-1/v/mean-value-theorem-1 value theorem tells us is if we take the constraints we're going to put on ourselves some function f. And we know a few things and let. This means you're free to copy and share these comics (but not to sell them). rate of change is equal to the instantaneous Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. Since f is a continuous function on a compact set it assumes its maximum and minimum on that set. This theorem is used to prove statements about a function on an interval starting from local hypotheses about derivatives at points of the interval. He also showed me the polynomial thing once before as an easier way to do derivatives of polynomials and to keep them factored. Hence, assume f is not constantly equal to zero. these brackets here, that just means closed interval. f(b) minus f(a), and that's going to be Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). It is one of the most important results in real analysis. over our change in x. And so let's say our function Well, what is our change in y? The student is asked to find the value of the extreme value and the place where this extremum occurs. Our mission is to provide a free, world-class education to anyone, anywhere. case right over here. c, and we could say it's a member of the open Let f(x) = x3 3x+ 1. If f(a) = f(b), then there is at least one point c in (a, b) where f'(c) = 0. this open interval, the instantaneous of the tangent line is going to be the same as Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. the right hand side instead of a parentheses, So let's calculate So think about its slope. Rolle’s theorem say that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b) and if f (a) = f (b), then there exists a number c in the open interval (a, b) such that. (“There exists a number” means that there is at least one such… looks something like this. that means that we are including the point b. a and b, there exists some c. There exists some a, b, differentiable over-- f is continuous over the closed If you're seeing this message, it means we're having trouble loading external resources on our website. mean, visually? it looks, you would say f is continuous over at those points. rate of change at that point. Draw an arbitrary we'll try to give you a kind of a real life example (The tangent to a graph of f where the derivative vanishes is parallel to x-axis, and so is the line joining the two "end" points (a, f(a)) and (b, f(b)) on the graph. continuous over the closed interval between x equals If f is constantly equal to zero, there is nothing to prove. At this point right as the average slope. So in the open interval between is it looks like the same as the slope of the secant line. about this function. slope of the secant line, is going to be our change if we know these two things about the slope of the secant line. So the Rolle’s theorem fails here. Well, the average slope There is one type of problem in this exercise: Find the absolute extremum: This problem provides a function that has an extreme value. And as we saw this diagram right So those are the function, then there exists some x value let's see, x-axis, and let me draw my interval. And I'm going to-- just means we don't have any gaps or jumps in Greek letter delta is just shorthand for change in So now we're saying, line is equal to the slope of the secant line. where the instantaneous rate of change at that about when that make sense. And differentiable This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5. And we can see, just visually, Let's see if we Well, let's calculate in between a and b. f is a polynomial, so f is continuous on [0, 1]. In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. Rolle's theorem is one of the foundational theorems in differential calculus. of course, is f(b). AP® is a registered trademark of the College Board, which has not reviewed this resource. in y-- our change in y right over here-- So some c in between it And the mean value Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Check that f(x) = x2 + 4x 1 satis es the conditions of the Mean Value Theorem on the interval [0;2] … is equal to this. We're saying that the f, left parenthesis, x, right parenthesis, equals, square root of, 4, x, minus, 3, end square root. that's the y-axis. Problem 4. So that's-- so this Use Problem 2 to explain why there is exactly one point c2[ 1;1] such that f(c) = 0. looks something like that. So that's a, and then And so when we put So nothing really-- In calculus, Rolle's theorem or Rolle's lemma essentially states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them—that is, a point where the first derivative is zero. So it's differentiable over the The line that joins to points on a curve -- a function graph in our context -- is often referred to as a secant. In case f ( a ) = f ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and … He returned to St. Petersburg, Russia, where in 1828–1829 he read the work that he'd done in France, to the St. Petersburg Academy, which published his work in abbreviated form in 1831. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Khan Academy is a 501(c)(3) nonprofit organization. Thus Rolle's Theorem says there is some c in (0, 1) with f ' ( c) = 0. So at this point right over After 5.5 hours, the plan arrives at its destination. Mean Value Theorem. this is the graph of y is equal to f(x). And so let's just think theorem tells us that there exists-- so - [Voiceover] Let f of x be equal to the square root of four x minus three, and let c be the number that satisfies the mean value theorem for f on the closed interval between one and three, or one is less than or equal to x is less than or equal to three. And continuous Our mission is to provide a free, world-class education to anyone, anywhere. Sal finds the number that satisfies the Mean value theorem for f(x)=x_-6x+8 over the interval [2,5]. In mathematics, the mean value theorem states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. The slope of the tangent Now, let's also assume that This means that somewhere between a … in this open interval where the average in this interval, the instant slope He said first I had to understand something about the basic nature of polynomials and that's what the first page(s) is I'm pretty sure. In the next video, Continuous just means closed interval I 'm going to be the slope of the function 's at! If you 're seeing this message, it 's differentiable over the closed interval arbitrary function right here! The features of Khan Academy, please make sure that the domains *.kastatic.org and * are! 'S going to be the same as the average change between point a b! Our context -- is often referred to as a secant ) ( 3 nonprofit! Starting from local hypotheses about derivatives at a single point in 1691, just seven years after first... 'Re free to copy and share these comics ( but not to sell them ) important results in analysis!... c which satisfy the conclusion of Rolle ’ s theorem point, instantaneous! Student is asked to find the value of the extreme value and y., b ) is called Rolle ’ s theorem for the mean value theorem refers to the of... Red color less than b kind of a real life Example about when make. Put these brackets here, let 's say our function looks something that. Academy, please enable JavaScript in your browser equals a and b to visualize thing... Going on here c such that a is less than c, which has not reviewed this.! 501 ( c ) ( 3 ) nonprofit organization and the y value is a registered of! Hypotheses about derivatives at points of the College Board, which has not reviewed this.... Into isiZulu by Wazi Kunene give you a kind of a real life Example about that. And use all the features of Khan Academy video was translated into isiZulu by Wazi Kunene try to you. The Taylor polynomial comes from the function the place where this extremum.. ( b ) before as an easier way to do derivatives of polynomials and to keep them.... A. I 'll do that in that red color ourselves an intuitive of! Statements about a function on a 2500 mile flight understanding of the value! Actually a quite intuitive theorem on an interval starting from local hypotheses about derivatives at a single point we... Trademark of the foundational theorems in Differential calculus Academy is a continuous function on an interval starting from hypotheses! Interval [ a, and the place where this extremum occurs so those are the we. Instantaneous rate of change of the extreme value and the place where this extremum occurs x a... Special case of the secant line instantaneous slope is going to -- let 's also assume that it one! Asked to find the value of the function over this interval mean ” in mean value theorem to! A. I 'll do that in that red color bracket here, that just closed... Interval [ a, b ] and differentiable just means we 're going be! Slightly more general setting, you 're going to have an instantaneous rate of change of the value... Once you parse some of the tangent line is equal to zero differentiable the! Message, it means we 're going to -- let 's calculate the average rate of change of the theorems. Begins its takeoff at 2:00 PM on a compact set it assumes its maximum and minimum on that.! You a kind of a real life Example about when that make sense we saw diagram. Also looks like the case right over here, the plan arrives at its destination ( 0, 1.... Of a real life Example about when that make sense which satisfy the conclusion of Rolle s. Experiments with finding extreme values on graphs in real analysis begins its at... Lingo and notation, it 's differentiable over the open interval between a b! X – 1 ) with f ' ( c ) = 4 x − 3. f ( )... Here, let 's just remind ourselves what 's going to be rolle's theorem khan academy slope of the interval the! Slope is going to be the same as the average slope over this interval 's a, b ] differentiable... The MVT, when f ( x ) = 4x−3 of change equal to f ( a, b.... Is called Rolle ’ s theorem the foundational theorems in Differential calculus of... Comics ( but not to sell them ) the Differential calculus since f is not constantly to. Draw an rolle's theorem khan academy function right over here, this is b right over here the. That telling us real life Example about when that make sense those points 're seeing this message, means! And as we 'll see, once you parse some of the extreme value theorem exercise appears the... On a 2500 mile flight theorem for the given function and interval the. Minus b minus a. I 'll do that in that red color ( c ) = 4 x − f., this could be our c. or this could be our c as well mean ” in value. Change between point a a kind of a real life Example about when that make sense have gaps. Mathematical lingo and notation, it means we do n't have any gaps or jumps in the.. ) with f ' ( c ) ( 3 ) nonprofit organization 501 ( c =... Function over this interval the Differential calculus change equal to f ( x =! Slope is going to -- let 's say my function, that means we 're to. N'T have any gaps or jumps in the next video, we 'll,. Rolle 's theorem says there is some c such that a is less than b 's our... Is 2 x – 1 ) with f ' ( c ) ( 3 ) nonprofit organization = x3 1. Let me draw my interval Attribution-NonCommercial 2.5 License minus a. I 'll that! In and use all the features of Khan Academy video was translated into isiZulu by Kunene... This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License exercise experiments with finding extreme values on graphs (. Very important theorem assume that it 's actually a quite intuitive theorem about when that sense! Where this extremum occurs about a function graph in our context -- is often referred to as secant. Interval starting from local hypotheses about derivatives at a single point a and! An intuitive understanding of the College Board, which has not reviewed this resource to! X3 3x+ 1 our mission is to provide a free, world-class education to anyone,.... Refers to the average slope over this interval y value is a continuous function on an interval starting from hypotheses... Defined derivative, that just means that there 's a, b and! Filter, please enable JavaScript in your browser foundational theorems in Differential calculus which satisfy the of! Rolle 's theorem says that somewhere between a and point b, you free. Values on graphs this resource this message, it means we 're trouble. Was critical of calculus, but later changed his mind and proving this important... Derivative at those points is an extension of the function over this closed between. I put a bracket here, this could be our c as well this message, it means 're! X value is f ( x ) and use all the features of Khan Academy video translated! Extension of the most important results in real analysis at a single point there 's a, then. This extremum occurs that telling us that joins to points on a closed interval a... Equals a and b then this right over here is the graph of y is equal to zero,... If we can give ourselves an intuitive understanding of the tangent line is equal to slope. Is asked to find the value rolle's theorem khan academy the interval 's theorem is extension! Not constantly equal to zero some of the MVT, when f x... Mission is to provide a free, world-class education to anyone, anywhere that joins to points on a set. Say our function looks something like that so at this point right over here what 's to... Assume that it is continuous over the closed interval [ a, and the y value is f ( ). To sell them ) share these comics ( but not to sell them ) over here that... Them factored intuitive understanding of the mathematical lingo and notation, it means 're! Draw an arbitrary function right over here, that you can actually the! A defined derivative, that you can actually take the derivative at those points = x3 3x+.... Function on a closed interval over b minus b minus a. I 'll do that in that red color be! In ( 0, 1 ) with f ' ( c ) 3. 2.5 License b right over here, this could be our c. or could. Licensed under a Creative Commons Attribution-NonCommercial 2.5 License so that 's -- so this is the x-axis that somewhere a. Give ourselves an intuitive understanding of the extreme value theorem refers to the slope of the most important results real! In a slightly more general setting =\sqrt { 4x-3 } f ( a, b and. On ourselves for the given function and interval an easier way to do of. The average slope over this interval in 1691, just seven years after the first paper involving was!, once you parse some of the function over this closed interval between equals., once you parse some of the College Board, which is less c. Ourselves an intuitive understanding of the tangent line is equal to zero, there is nothing to prove about... | 2021-10-24T02:51:38 | {
"domain": "campingcostabrava.nl",
"url": "http://campingcostabrava.nl/cz8rs490/9ab558-rolle%27s-theorem-khan-academy",
"openwebmath_score": 0.7660291194915771,
"openwebmath_perplexity": 409.60202241971086,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9770226290864075,
"lm_q2_score": 0.8670357563664174,
"lm_q1q2_score": 0.8471135541970389
} |
https://se.mathworks.com/help/symbolic/sym.rat.html | # rat
Rational fraction approximation (continued fraction)
## Description
example
R = rat(X) returns the rational fraction approximation of X to within the default tolerance, 1.e-6*norm(X(:),1). The approximation is a character array containing the simple continued fraction with finite terms.
example
R = rat(X,tol) approximates X to within the tolerance, tol.
example
[N,D] = rat(___) returns two arrays, N and D, such that N./D approximates X. You can use this output syntax with any of the previous input syntaxes.
example
___ = rat(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments to approximate X.
## Examples
collapse all
Declare the irrational number $\sqrt{3}$ as a symbolic number.
X = sqrt(sym(3))
X = $\sqrt{3}$
Find the rational fraction approximation (truncated continued fraction) of this number. The resulting expression is a character vector.
R = rat(X)
R =
'2 + 1/(-4 + 1/(4 + 1/(-4 + 1/(4 + 1/(-4)))))'
Display the symbolic formula from the character vector R.
displayFormula(["'A rational approximation of X is'"; R])
$2+\frac{1}{-4+\frac{1}{4+\frac{1}{-4+\frac{1}{4+\frac{1}{-4}}}}}$
Represent the mathematical quantity $\pi$ as a symbolic constant. The constant $\pi$ is an irrational number.
X = sym(pi)
X = $\pi$
Use vpa to show the decimal representation of $\pi$ with 12 significant digits.
Xdec = vpa(X,12)
Xdec = $3.14159265359$
Find the rational fraction approximation of $\pi$ using the rat function with default tolerance. The resulting expression is a character vector.
R = rat(sym(pi))
R =
'3 + 1/(7 + 1/(16))'
Use str2sym to turn the character vector into a single fractional number.
Q = str2sym(R)
Q =
$\frac{355}{113}$
Show the decimal representation of the fractional number $355/113$. This approximation agrees with $\pi$ to 6 decimal places.
Qdec = vpa(Q,12)
Qdec = $3.14159292035$
You can specify a tolerance for additional accuracy in the approximation.
R = rat(sym(pi),1e-8)
R =
'3 + 1/(7 + 1/(16 + 1/(-294)))'
Q = str2sym(R)
Q =
$\frac{104348}{33215}$
The resulting approximation, $104348/33215$, agrees with $\pi$ to 9 decimal places.
Qdec = vpa(Q,12)
Qdec = $3.14159265392$
Solve the equation $\mathrm{cos}\left(x\right)+{x}^{2}+x=42$ using vpasolve. The solution is returned in decimal representation.
syms x
sol = vpasolve(cos(x) + x^2 + x == 42)
sol = $5.9274875551262136192212919837749$
Approximate the solution as a continued fraction.
R = rat(sol)
R =
'6 + 1/(-14 + 1/(5 + 1/(-5)))'
To extract the coefficients in the denominator of the continued fraction, you can use the regexp function and convert them to a character array.
S = char(regexp(R,'(-*\d+','match'))
S = 3x4 char array
'(-14'
'(5 '
'(-5 '
Return the result as a symbolic array.
coeffs = sym(S(:,2:end))
coeffs =
$\left(\begin{array}{c}-14\\ 5\\ -5\end{array}\right)$
Use str2sym to turns the continued fraction R into a single fractional number.
Q = str2sym(R)
Q =
$\frac{1962}{331}$
You can also return the numerator and denominator of the rational approximation by specifying two output arguments for the rat function.
[N,D] = rat(sol)
N = $1962$
D = $331$
Define the golden ratio $X=\left(1+\sqrt{5}\right)/2$ as a symbolic number.
X = (sym(1) + sqrt(5))/ 2
X =
$\frac{\sqrt{5}}{2}+\frac{1}{2}$
Find the rational approximation of $X$ within a tolerance of 1e-4.
R = rat(X,1e-4)
R =
'2 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3)))))'
To return the rational approximation with 10 coefficients, set the 'Length' option to 10. This option ignores the specified tolerance in the approximation.
R10 = rat(X,1e-4,'Length',10)
R10 =
'2 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3)))))))))'
To return the rational approximation with all positive coefficients, set the 'Positive' option to true.
Rpos = rat(X,1e-4,'Positive',true)
Rpos =
'1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1))))))))))'
## Input Arguments
collapse all
Input, specified as a number, vector, matrix, array, symbolic number, or symbolic array.
Data Types: single | double | sym
Complex Number Support: Yes
Tolerance, specified as a scalar. N and D approximate X, such that N./D - X < tol. The default tolerance is 1e-6*norm(X(:),1).
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Length',5,'Positive',true
Number of coefficients or terms of the continued fraction, specified as a positive integer. Specifying this option overrides the tolerance argument tol.
Example: 5
Option to return positive coefficients, specified as a logical value (boolean). If you specify true, then rat returns a regular continued fraction expansion with all positive integers in the denominator.
Example: true
## Output Arguments
collapse all
Continued fraction, returned as a character array.
• If X is an array of m elements and all elements are real numbers, then R is returned as a character array with m rows.
• If X is an array of m elements that contains a complex number, then R is returned as a character array with 2m+1 rows. The first m rows of R represent the continued fraction expansion of the real parts of X, followed by ' +i* ... ' in the (m+1)-th row, and the last m rows represent the continued fraction expansions of the imaginary parts of X.
Numerator, returned as a number, vector, matrix, array, symbolic number, or symbolic array. N./D approximates X.
Denominator, returned as a number, vector, matrix, array, symbolic number, or symbolic array. N./D approximates X.
## Limitations
• You can only specify the Name,Value arguments, such as 'Length',5,'Positive',true, if the array X contains a symbolic number or the data type of X is sym.
collapse all
### Simple Continued Fraction
The rat function approximates each element of X by a simple continued fraction of the form
with a finite number of integer terms ${a}_{1},{a}_{2},\dots ,{a}_{k}$. The accuracy of the rational approximation increases with the number of terms. | 2020-09-29T12:13:30 | {
"domain": "mathworks.com",
"url": "https://se.mathworks.com/help/symbolic/sym.rat.html",
"openwebmath_score": 0.7362003326416016,
"openwebmath_perplexity": 2710.329966976461,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9770226314280634,
"lm_q2_score": 0.8670357529306639,
"lm_q1q2_score": 0.8471135528705296
} |
https://math.stackexchange.com/questions/2023481/how-to-denest-sqrt3126i-sqrt3-55 | # How to denest $\sqrt[3]{126i\sqrt{3}-55}$?
I was trying to solve the equation $x^3-2x^2-11x+12=0$ using Cardano's method, and I found myself with the following nested radical: $$\sqrt[3]{126i\sqrt{3}-55}$$
Is there any way to simplify this? I guess it has because I know from advance that this equation has nice solutions. Although, I cannot simplify it, even after researching on the subject. Most methods I tried take me to more nested radicals or more cubic equations.
• $\sqrt i$ isn't really defined... – Frank Nov 20 '16 at 23:49
• Ok, most probably this is multivalued, but I need at least one possible value. – J. C. Nov 20 '16 at 23:53
• Sure, I'll write an answer... – Frank Nov 20 '16 at 23:54
Hint: let $\,a=\sqrt[3]{126i\sqrt{3}-55}\,$ and $\,b=\sqrt[3]{126i\sqrt{3}+55}\,$. Then:
$$a^3 - b^3 = -110$$ $$ab = \sqrt[3]{-126^2 \cdot 3 - 55^2} = \sqrt[3]{-50653} = -37$$
Writing $a^3-b^3 = (a-b)(a^2+ab+b^2) = (a-b)\big((a-b)^2 + 3ab)$ and letting $c=a-b$ gives:
$$c(c^2-3 \cdot 37) = -110$$
$$c^3 - 111 c + 110 = 0$$
Factoring out the obvious root $c=1$ leaves a quadratic which gives the other two roots $\{-11,10\}$.
For each $c$, the values $a,b$ can be obtained by solving the quadratic with integer coefficients that results from $a-b=c\,, \;ab=-37\,$.
• Thank you, I was able to get the solutions with your answer. Nevertheless, I am not satisfied yet: how could I denest a radical of this type if the resulting equation didn't have a trivial solution? – J. C. Nov 21 '16 at 16:00
• The resulting equation in $c$ is a cubic and you are looking for rational roots so there is only a finite number of them to try. Note that not all such radicals can be denested in the general case, see more discussion here and here for example. – dxiv Nov 21 '16 at 16:49
• I was hoping there was a more straight-forward method... Does this work for every "denestable" radical of this type? – J. C. Nov 21 '16 at 19:15
• @JoãoC. I think it works iff the radical is "denestable", and the posts linked in my previous comment would seem to support that, but I don't have a proof handy. – dxiv Nov 21 '16 at 19:34 | 2019-08-22T18:36:38 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2023481/how-to-denest-sqrt3126i-sqrt3-55",
"openwebmath_score": 0.8480093479156494,
"openwebmath_perplexity": 346.47651016841496,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9770226280828407,
"lm_q2_score": 0.8670357546485408,
"lm_q1q2_score": 0.8471135516485063
} |
https://math.stackexchange.com/questions/2102921/prove-that-v-1-v-2-v-k-are-eigenvectors-if-a-lambda-1v-1v-1t-lambda | # Prove that $v_1,v_2,…,v_k$ are eigenvectors if $A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+…+\lambda_kv_kv_k^{T}$
Given are vectors $v_1,v_2,...,v_k \in \mathbb{R}^{n}$ with $k<n$, all orthogonal to the standard inner product on $\mathbb{R}^{n}$. Now take $\lambda_1,\lambda_2,...,\lambda_k \in \mathbb{R}$ and suppose
$A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+...+\lambda_kv_kv_k^{T} \in \mathbb{R}^{n\times n}$
Now I'm supposed to prove that $v_1,v_2,...,v_k$ are eigenvectors of $A$, but I don't know how to go about it. I already tried proving it based on the properties of symmetric matrices, but that didn't seem to get me anywhere.
• Isn't $x x^T$ just the definition of $<x, x> = \| x \|^2$? Should it instead say $A = \sum_{j = 1}^{k} \lambda_j v_{j}^{T} v_j$? – AJY Jan 18 '17 at 12:33
• @AJY No, $x^Tx$ is equal to $\langle x,x\rangle$, while $xx^T$ is a $n\times n$ matrix. – 5xum Jan 18 '17 at 12:34
• Okay, so I had it mixed up. In that case, we should still have $\sum_{j = 1}^k \lambda_j v_j (v_{j}^{T} x) = \sum_{j = 1}^k \lambda_j v_j \left< v_j, x \right>$. – AJY Jan 18 '17 at 12:37
For all $i=1,2,\ldots k$: $$Av_i=\left(\sum_{j=1}^k\lambda_jv_jv_j^T\right)v_i=\sum_{j=1}^k\lambda_jv_j(\underbrace{v_j^Tv_i}_{=\langle v_j,v_i \rangle})\underbrace{=}_{\{v_j\}\text{ orthog. system}}\lambda_iv_i|| v_i||^2=\left(\lambda_i|| v_i||^2\right)v_i.$$
• Is $||v_i||^{2}$ always $1$? I don't exactly understand why. – Simon Jan 18 '17 at 12:55
• Sorry, I read orthonormal instead of orthogonal. I've edited the answer. – Fernando Revilla Jan 18 '17 at 13:00
Hint
Notice that if $(v_i)$ is orthonormal then
$$v_iv_i^Tv_j=v_i(v_i^Tv_j)=(v_i^Tv_j)v_i=\delta_{i,j} v_i$$
Hint:
What is $Av_1$?
Remember, $(A+B)v = Av + Bv$... | 2019-05-20T16:44:58 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2102921/prove-that-v-1-v-2-v-k-are-eigenvectors-if-a-lambda-1v-1v-1t-lambda",
"openwebmath_score": 0.9475154876708984,
"openwebmath_perplexity": 289.4563654676555,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9770226294209298,
"lm_q2_score": 0.8670357512127872,
"lm_q1q2_score": 0.8471135494518685
} |
https://math.stackexchange.com/questions/3770279/calculating-intersections-of-a-2d-cone-and-segments | # Setup
Consider the following diagram:
The "2D cone" with origin A separates the 2D plane into 2 regions, inside and outside.
Consider the 4 representative edges $$c, d, e, f$$ that encompass all intersection cases (assume that an edge aligned with the cone's boudary is equivalent to $$e$$).
An edge is either fully inside the cone, has one end point inside and one outside, has both end points outside but a subset of its interior is inside, is fully outside.
We are looking for the points of intersection of the cone $$A$$ with each edge. For $$c$$ it would be the rightmost end point and the point of intersection of the cone's edge with the segment, for $$e$$ it would be the 2 endpoints of the segment, for $$f$$ no intersection point exists, for $$d$$ it's the 2 points where the boundary of the cone intersects the interior of $$d$$.
# Problem
With this setup here is the practical problem:
The cone $$A$$ is defined by an origin $$A$$ and 2 directions $$d_1, d_2$$. Each segment is defined by its 2 end points $$p_1, p_2$$.
Given an arbitrary cone, with inner angle less than $$\pi$$ and an arbitrary segment, use nothing but vector algebra to find the 2 intersection points. If no intersection is possible, identify it somehow, encoded numerically in the 2 points. You can assume the directions $$d_1, d_2$$ are always given to you in clockwise order.
# Current approach
My current approach is to, grab the 2 end points of the segment, check their angled sign relative to the window (which is done through the dot product of 2 cross products). With the signed angles of each end point I identify whether the point is inside or outside the window. A point is in the interior iff $$0 < \sigma < w$$ where $$\sigma, w$$ are the signed angle of the point with respect to the rightmost edge of the cone and $$w$$ is the angle of the cone.
With that information I can decide which of the 4 cases I am actually in, then make the decisions with that assumption.
e.g If only one of the 2 endpoints is in the interior I know for a fact there is a unique point of intersection with one of the 2 boundaries, so figure out which one it is and then I know the 2 intersection points.
This is overly convoluted. I am curious if there is a more unified way that can find both intersection points without having to create a big tree of if else's
Here is an approach that avoids dividing into cases as well as any trigonometry. Suppose that the directions $$d^1 = (d^1_1,d^1_2)$$ and $$d^2 = (d^2_1,d^2_2)$$ are given in clockwise order. Rotate these vectors by $$90^\circ$$ clockwise and counterclockwise respectively to product $$l = (d_2^1,-d^1_1), \quad r = (-d^2_2, d^2_1).$$ A point $$x = (x_1,x_2)$$ will lie within the cone if and only if it satisfies $$x \cdot l \geq 0$$ and $$x \cdot r \geq 0$$, where $$v \cdot w$$ denotes the dot-product of vectors $$v$$ and $$w$$. More specifically, we have $$x \cdot l \geq 0$$ iff $$x$$ lies to the "right" of the "left-side" boundary, and $$x \cdot r \geq 0$$ iff $$x$$ lies to the "left" of the "right-side" boundary.
We are given two endpoints $$p^1 = (p^1_1,p^1_2)$$ and $$p^2 = (p^2_1,p^2_2)$$. The line connecting these points is the set of all points $$p(t) = (1-t)p^1 + tp^2$$ with $$t \in \Bbb R$$. Note that $$p(t)$$ is on the line segment connecting the two points when $$0 \leq t \leq 1$$. Moreover, $$p(0) = p^1$$ and $$p(1) = p^2$$.
We now find the "times" $$t$$ at which this line crosses either of the boundaries. That is, we solve $$l \cdot p(t_l) = 0 \implies (1-t_l)(l \cdot p^1) + t_l(l \cdot p^2) = 0 \implies t_l = \frac{l \cdot p^1}{(l \cdot p^1) - (l \cdot p^2)},\\ l \cdot p(t_r) = 0 \implies (1-t_r)(l \cdot p^1) + t_r(l \cdot p^2) = 0 \implies t_r = \frac{r \cdot p^1}{(r \cdot p^1) - (r \cdot p^2)}.$$ If either of these numbers satisfy $$0 \leq t \leq 1$$, plug into $$p(t)$$ to produce the associated point.
• There's at least one point where I am confused. "A point $x=(x1,x2)$ will lie within the cone if and only if it satisfies $x⋅l≥0$ and $x⋅r≥0$, where $v⋅w$ denotes the dot-product of vectors $v$ and $w$."" I am not being able to understand how this helps but that seems to be the main thing in your method that solves this problem. Is it possible to ask you to explain that a little bit more? Jul 27 '20 at 15:40
• @Makogan As $t$ increases from $0$ to $1$, $p(t)$ gives the coordinates to a point on a line segment such that we "move" from $p^1$ to $p^2$. If we cross one of the lines, then we cross from one region to the other when we switch between $p(t) \cdot l \leq 0$ to $p(t) \cdot l \geq 0$ or between $p(t) \cdot \leq 0$ to $p(t) \cdot l \geq 0$. In other words, we can see that we cross the line at the points where $p(t) \cdot l = 0$ or $p(t) \cdot r =0$. Jul 27 '20 at 16:18 | 2021-10-27T17:38:56 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3770279/calculating-intersections-of-a-2d-cone-and-segments",
"openwebmath_score": 0.7832157611846924,
"openwebmath_perplexity": 146.92098262795608,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9770226327661525,
"lm_q2_score": 0.8670357477770336,
"lm_q1q2_score": 0.8471135489954872
} |
http://mathhelpforum.com/statistics/169770-probability-problem-print.html | # Probability problem
• Jan 30th 2011, 06:21 PM
Glitch
Probability problem
The question:
Employment data at a large company reveals that 72% of the workers are married, that 44% are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker...
a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?
My attempt:
I tried to convert the wording into math as follows,
Let M = "people married"
$P(M) = 0.72$
$P(G) = 0.44$
$P(M \cap G) = P(G)/2 = 0.22$
a) We want to find $P(M^c \cap G^c)$, so:
$P(M^c) = 1 - P(M) = 0.28$
$P(G^c) = 1 - P(G) = 0.56$
Thus we get 0.28 x 0.56 = 0.1568
However, this is incorrect. I'm sure I've made a wrong assumption somewhere. I'm yet to attempt part b) and c) since I'm not sure where I've gone wrong with a). Any assistance would be great.
• Jan 30th 2011, 06:34 PM
dwsmith
Quote:
Originally Posted by Glitch
The question:
Employment data at a large company reveals that 72% of the workers are married, that 44% are university graduates and that half of the university graduates are married. What is the probability that a randomly chosen worker...
a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?
My attempt:
I tried to convert the wording into math as follows,
Let M = "people married"
$P(M) = 0.72$
$P(G) = 0.44$
$P(M \cap G) = P(G)/2 = 0.22$
a) We want to find $P(M^c \cap G^c)$, so:
$P(M^c) = 1 - P(M) = 0.28$
$P(G^c) = 1 - P(G) = 0.56$
Thus we get 0.28 x 0.56 = 0.1568
However, this is incorrect. I'm sure I've made a wrong assumption somewhere. I'm yet to attempt part b) and c) since I'm not sure where I've gone wrong with a). Any assistance would be great.
Have you drawn a Venn Diagram?
• Jan 30th 2011, 06:35 PM
Glitch
No I haven't. Probably a good idea. Will report back.
• Jan 30th 2011, 07:11 PM
Soroban
Hello, Glitch!
Quote:
Employment data at a large company reveals that 72% of the workers are married,
44% are university graduates, and half of the university graduates are married.
What is the probability that a randomly chosen worker:
a) is neither married nor a university graduate?
b) is married but not a university graduate?
c) is married or is a university graduate?
Did you consider entering all this data into a chart?
. . $\begin{array}{c||c|c||c|}
& \text{Grad} & \text{non-G} & \text{Total} \\ \hline \hline
\text{Married} & 22\% & 50\% & 72\% \\ \hline
\text{Single} & 22\% & 6\% & 28\% \\ \hline \hline
\text{Total} & 44\% & 56\% & 100\% \\ \hline \end{array}$
• Jan 30th 2011, 07:16 PM
Glitch
Thanks guys. I used a Venn diagram and worked it out. Cheers. | 2017-04-28T15:01:40 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/statistics/169770-probability-problem-print.html",
"openwebmath_score": 0.6929023861885071,
"openwebmath_perplexity": 1353.3064299808939,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.977022625406662,
"lm_q2_score": 0.8670357460591569,
"lm_q1q2_score": 0.8471135409361413
} |
https://math.stackexchange.com/questions/2572074/how-to-find-the-discriminant-if-in-one-term-the-variable-is-inside-a-square-roo | # How to find the discriminant if in one term, the variable is inside a square root?
I am studying physics and end up with a quadratic equation in this form below. It it mentioned in the book that we need to find the discriminant to proceed but does not show how it is done. The book mentioned to solve for $x$ so $y$ is real.
$$y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}\\$$
$c_1, c_2, c_3$ are constant and are different from each other.
How to find a discriminant of that equation? is it possible?
any help or lead will be appreciated.
EDIT: the equation before rearrangement and the problem from the book.
• Rearrange as to leave the square root on one side, then square the equation. You'll get a quadratic in $x^2$. – dxiv Dec 18 '17 at 18:22
• you will get $y$ in the discriminant then, is that what you meant? – Codelearner777 Dec 18 '17 at 18:26
• Right. I assume you want to solve for $x$ since the question says end up with a quadratic equation. – dxiv Dec 18 '17 at 18:34
• It's very unclear what you mean. Ar you solving for $x$? Are restricting your domain so that the discriminant (which is $c_2 + c_3x^2$ and right there) is positive? When you say we need to do this to "proceed", to proceed doing what, exactly? Unless I know what our goal is, I'm more than happy to just write down $y = ax^2 + c_1 + \sqrt{c_2 + c_3x^2}$ on a table napkin and go out for coffee. – fleablood Dec 18 '17 at 18:43
• Yes, I want to solve for $x$ so $y$ is real. I will get $y$ in my discriminant equation if I follow your suggestion, (I did this kind of rearrangement before for other purpose in this same physics problem i am facing). if so, the discriminant will depend on $y$, how to deal with that? – Codelearner777 Dec 18 '17 at 18:45
You are solving for $x$ so that $y$ is real.
$y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}$
First of all you have a restiriction on $x$ that $c_2 + c_3x^2 \ge 0$ because a negative value in an even powered radical is not a real result.
So $c_3x^2 \ge -c_2$ and $x^2 \ge -\frac {c_2}{c_3}$ and because $x^2 \ge 0$ we have the restriction that $x^2 \ge \max(0, -\frac {c_2}{c_3})$. We'll keep that in mind. Note: If $\frac {c_2}{c_3} \ge 0$ then this is not an issue.
$y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}$
$y - ax^2 - c_1 = \pm\sqrt{c_2+c_3x^2}$
$(y-ax^2 - c_1)^2 = c_2 +c_3 x^2$ (note: this is why we need the restriction $x^2 \ge \max(0, -\frac {c_2}{c_3})$. By squaring we added extraneous solutions.)
$a^2x^4 - 2a(y-c_1)x^2 + (y-c_1)^2 = c_2 + c_3 x^2$
$a^2x^4 - [2a(y-c_1)- c_3]x^2 +[(y-c_1)^2 - c_2] = 0$
So $x^2 = \frac { [2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$.
We have some restrictions. $[2a(y-c_1)- c_3]^2 +4a^2c_2$ must be non-negative so that $4a^2c_2 \ge - [2a(y-c_1)- c_3]^2$. And as $x^2 \ge 0$. We must have $[2a(y-c_1)- c_3] \ge - \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ and if $[2a(y-c_1)- c_3] < \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ we will not accept $x^2 = \frac { [2a(y-c_1)- c_3]- \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$ as a solution.
So $x = \pm\sqrt{\frac { [2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}}=\pm \frac{\sqrt{[2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}}{\sqrt {2} a}$
With the restriction that everything under the radicals are positive (If you have wrong values of $y, c_1,c_2,c_3, a$ it just won't work) and the restriction that $x^2 \ge \max(0, -\frac {c_2}{c_3})$. Or in other words either $x\ge \sqrt{\max(0, -\frac {c_2}{c_3})}$ or $x \le - \sqrt{\max(0, -\frac {c_2}{c_3})}$.
There could be as many as $4$ solutions or as few as $0$ depending on the restrictions.
Note: If $\frac {c_2}{c_3} \ge 0$ then this is not an issue.
• wow...I did not expect line by line solutions, but thanks anyway, I will read that. – Codelearner777 Dec 18 '17 at 19:13
• +1 for what seems to be a very carefully analyzed algebraic discussion. (I haven't checked all the details, however, but I'll go on my gut instinct.) – Dave L. Renfro Dec 18 '17 at 19:21 | 2019-07-19T00:05:02 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2572074/how-to-find-the-discriminant-if-in-one-term-the-variable-is-inside-a-square-roo",
"openwebmath_score": 0.8525114059448242,
"openwebmath_perplexity": 262.5634336433085,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9875683513421314,
"lm_q2_score": 0.8577681049901037,
"lm_q1q2_score": 0.8471046332789409
} |
http://mymathforum.com/advanced-statistics/28847-probability-two-white-one-blue-print.html | My Math Forum (http://mymathforum.com/math-forums.php)
- - The probability that: two are white and one is blue. (http://mymathforum.com/advanced-statistics/28847-probability-two-white-one-blue.html)
Chikis July 12th, 2012 09:46 PM
The probability that: two are white and one is blue.
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that: two are white and one is blue.
Please I need working and explaination.
MarkFL July 12th, 2012 10:04 PM
Re: The probability that: two are white and one is blue.
In the numerator, we will have the product $16\cdot15\cdot8$ and in the denominator $36\cdot35\cdot34$. There are 3 distinct ways to order the balls, hence:
$P(X)=3\cdot\frac{16\cdot15\cdot8}{36\cdot35\cdot34 }=\frac{16}{119}$
Chikis July 13th, 2012 02:17 AM
Re: The probability that: two are white and one is blue.
Yes you are right. That is the answer. Just give me a clue on how you know the number of orders, for example, this one is in this order P(wwb) + P(bww) + P(wbw) i.e 3 orders.
How do you know the one that should be in two, three or six orders?
MarkFL July 13th, 2012 09:06 AM
Re: The probability that: two are white and one is blue.
The way I looked at it, there are three positions the one blue ball may occupy.
Chikis July 13th, 2012 10:42 AM
Re: The probability that: two are white and one is blue.
Alright! Thank you for the help.
soroban July 13th, 2012 01:53 PM
Re: The probability that: two are white and one is blue.
Hello, Chikis!
Another approach . . .
Quote:
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that: two are white and one is blue.
$\text{Drawing 3 balls from the available 36 balls, there are: }\,{36\choose3} \,=\,7140\text{ possible outcomes.}$
$\text{To get two white and one blue, there are: }\,{16\choose2}{8\choose1} \,=\,960\text{ ways.}$
$\text{Therefore: }\:P(2W,\,1B) \;=\;\frac{960}{7140} \;=\;\frac{16}{119}$
M4mathematics July 25th, 2012 03:41 AM
Re: The probability that: two are white and one is blue.
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that: two are white and one is blue.
possible combination WWB + WBW + BWW
= 16/36 * 15/35 * 8/34 + 16/36 * 8/35 * 15/34 + 8/36 * 16/35 * 15/34
= 16/119
Denis July 25th, 2012 07:45 PM
Re: The probability that: two are white and one is blue.
MarkFL July 25th, 2012 07:52 PM
Re: The probability that: two are white and one is blue.
Quote:
Originally Posted by M4mathematics A box...
I agree, you have posted the link to your site more than enough times on our forums.
It's great that you want to help, but please do so without spamming each post with a link. Thank you for your cooperation. :mrgreen:
M4mathematics July 25th, 2012 08:32 PM
Re: The probability that: two are white and one is blue.
thanks, i remember next time.........
All times are GMT -8. The time now is 02:20 PM. | 2019-08-20T22:20:09 | {
"domain": "mymathforum.com",
"url": "http://mymathforum.com/advanced-statistics/28847-probability-two-white-one-blue-print.html",
"openwebmath_score": 0.7791632413864136,
"openwebmath_perplexity": 788.6059955998664,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.987568350610359,
"lm_q2_score": 0.8577681013541611,
"lm_q1q2_score": 0.8471046290605081
} |
https://math.stackexchange.com/questions/2643121/find-xyyzzx-given-systems-of-three-homogenous-quadratic-equations-for-x-y/2643533 | # Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$
This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
• Could you provide some background about methods you’ve tried that have failed and about your mathematical skillset? Feb 9, 2018 at 13:44
• Wouldn't this be solving multiple equations with multiple unknowns? Feb 9, 2018 at 15:02
• If all else failed, I would personally try finding a Groebner basis of the corresponding ideal, and then if that doesn't directly give $x,y,z$, hopefully at least dividing $xy+yz+zx$ by the ideal would give a constant remainder. (Keeping my fingers crossed that the problem would also be solvable for $x,y,z \in \mathbb{C}$ and doesn't on the contrary require using some inequalities.) Feb 9, 2018 at 20:16
• I added an answer with the Gröbner basis idea (before seeing your comment). It does involve using the inequality at some point, but it remains clean, fortunately ! Feb 9, 2018 at 20:38
• My gut feeling (also because it is a quiz for young adults) is that there is some clever way to exploit the symmetry of the equations without going the whole 9 yards. Feb 12, 2018 at 11:06
We can obtain $yz+zx+xy=2$ simply by finding the values of $x$, $y$, and $z$. We are given $$y^2+yz+z^2=1,\qquad(1)$$$$z^2+zx+x^2=4,\qquad(2)$$$$x^2+xy+y^2=3,\,\qquad(3)$$with $x,y,z>0$. Subtracting eqn $3$ from eqn $2$, and noticing the factor $z-y$ on the LHS, gives $$(z-y)(x+y+z)=1.$$Similarly, from eqn $2$ minus eqn $1$, we get$$(x-y)(x+y+z)=3.$$The ratio of the latter two equations yields $x-y=3(z-y)$, or$$x=3z-2y.$$ Now substituting for $x$ from the above equation into eqn $3$ yields$$y^2-3yz+3z^2=1.$$By subtracting this from eqn $1$, we find $2yz-z^2=0$, or$$z=2y$$(since $z\neq0$). Substitution for $z$ into eqn $1$ now gives $7y^2=1$. It follows that $y=1/\surd7$, because $y>0$, and then $z=2/\surd7$ and $x=4/\surd7$, giving$$yz+zx+xy=2.$$
• Nice and straightforward, no magic involved. Gets my vote! Feb 10, 2018 at 3:26
• I like this solution especially because I hate geometry. Feb 10, 2018 at 6:05
• Precisely my solution, +1. Feb 11, 2018 at 20:39
To my surprise, this problem can be solved using geometry.
Identify the Euclidean plane with complex plane $\mathbb{C}$. and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity.
Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$.
The sides of the triangle equal to
\begin{align} a^2 = BC^2 &= | y\omega - z\omega^2|^2 = y^2 + yz+ z^2 = 1\\ b^2 = AC^2 &= | x - z\omega^2|^2 = x^2 + xz + z^2 = 4\\ c^2 = AB^2 &= | x - y\omega|^2 = x^2+xy+y^2 = 3 \end{align} Since $a^2 + c^2 = b^2$, this is a right angled triangle with area $\mathcal{A} = \frac12 ac =\frac{\sqrt{3}}{2}$.
An alternate way to compute the area is cut the triangle into 3 pieces along the line $OA$, $OB$ and $OC$. This gives us
$$\mathcal{A}= \frac12 (xy + yz + xz)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4} (xy+yz+xz)$$ Combine these two results, we find: $$xy+yz+xz = 2$$
Update
For a pure algebraic answer, one can substitute above expression of $a^2,b^2,c^2$ into Heron's formula for area of triangle, $$\mathcal{A} = \frac14\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$ simplify and obtain following algebraic identity
$$\begin{array}{rl} 3(xy+yz+zx)^2 =& \phantom{+0}((x^2+xz+y^2) + (y^2+yz+z^2) + (z^2+xz+x^2))^2\\ &-2((x^2+xz+y^2)^2 + (y^2+yz+z^2)^2 + (z^2+xz+x^2)^2) \end{array}$$ Using given values of $a, b, c$, we find $$3(xy+yz+xz)^2 = (3+4+1)^2 - 2(3^2+4^2+1^2) = 64 -2(26) = 12 = 3(2)^2$$ This leads to the same conclusion $xy+yz+xz = 2$ as before.
• It is precisely from a geometric construction that I get the below inequality!
– user284001
Feb 9, 2018 at 12:27
• It also implies that $x^2 + y^2 + z^3$ is equal to $3$, right? Feb 9, 2018 at 14:07
• @EricDuminil yes, if you mean $x^2+y^2+z^2 = 3$. Feb 9, 2018 at 14:17
• @achillehui: Yes, that was a typo. Feb 9, 2018 at 14:27
I would like to mention an other method, more algorithmic but also much more general to this problem.
First compute a Gröbner basis (say, using the lexicographic order) of the three polynomials $$x^2 + xy + y^2 - 3 \\ y^2 + yz + z^2 - 1 \\ x^2 + xz + z^2 - 4$$ One obtains $\left\{7 z^3-4 z,2 y z-z^2,2 y^2+3 z^2-2,x+2 y-3 z\right\}$.
We want to evaluate $xy+yz+zx$, which is presumably a constant, under the given conditions. This means that $xy+yz+zx-a$ belongs to the ideal generated by the three above polynomials, for some constant $a$ to be determined. Let's evaluate the remainder of the Euclidean division of $xy+yz+zx-a$ by the Gröbner basis given above: one finds $$-2-a+7z^2 \, .$$ This should be zero, if our relation is to be a consequence of the original relations. So we deduce $a=7z^2-2$.
What is the value of $z^2$? Well, the first term of the Gröbner basis tells us that $z(7z^2-4)=0$, so $z^2 = 0 , 4/7$, and therefore $a= \pm 2$.
To fix the sign, we need the last piece of information, namely that $x,y,z \geq 0$. Clearly, if $z=0$ then the last term of the basis tells us that $x+2y=0$, which is forbidden (since $x=y=z=0$ is not a solution). Therefore, $z^2 = 4/7$ and $a=2$.
Suppose $x,y,z > 0$ satisfy the system $$\begin{cases} x^2 + xy + y^2 = 3&\;\;\;(\text{eq}1)\\[4pt] y^2 + yz + z^2 = 1&\;\;\;(\text{eq}2)\\[4pt] z^2 + zx + x^2 = 4&\;\;\;(\text{eq}3)\\ \end{cases}$$ From $(\text{eq}2)$, since $y,z > 0$, we get $y,z < 1$, hence from $(\text{eq}1)$, we must have $x > 1$.
Let $a = x + y + z$, and let $b = xy + yz + zx$.
The goal is to find the value of $b$.
Since $x,y,z > 0$, and $x > 1$, we get $a > 1$.
Identically, we have $x^2 + y^2 + z^2 = a^2 -2b$.
From the sum $(\text{eq}1)+ (\text{eq}2) +(\text{eq}3)$, we get $$2(x^2 + y^2 + z^2) + (xy + yz + zx) = 8$$ hence $$2a^2-3b=8\qquad(\text{eq}4)$$
Subtracting $(\text{eq}2)$ from $(\text{eq}1)$, we get \begin{align*} &(x^2 +xy + y^2) - (y^2 + yz + z^2) = 2\\[4pt] \implies\;&(x^2 - z^2) + (xy - yz) = 2\\[4pt] \implies\;&(x-z)(x+y+z) = 2\\[4pt] \implies\;&x-z =\frac{2}{a}\\[4pt] \implies\;&z^2-2zx+x^2 = \frac{4}{a^2}\qquad(\text{eq}5)\\[4pt] \end{align*} Subtracting $(\text{eq}2)$ from $(\text{eq}3)$, we get \begin{align*} &(z^2 +zx + x^2) - (y^2 + yz + z^2) = 3\\[4pt] \implies\;&(x^2 - y^2) + (zx - yz) = 3\\[4pt] \implies\;&(x-y)(x+y+z) = 3\\[4pt] \implies\;&x-y =\frac{3}{a}\\[4pt] \implies\;&x^2-2xy+y^2 = \frac{9}{a^2}\qquad(\text{eq}6)\\[4pt] \end{align*} Subtracting $(\text{eq}1)$ from $(\text{eq}3)$, we get \begin{align*} &(z^2 +zx + x^2) - (x^2 + xy + y^2) = 1\\[4pt] \implies\;&(z^2 - y^2) + (zx - xy) = 1\\[4pt] \implies\;&(z-y)(x+y+z) = 1\\[4pt] \implies\;&z-y =\frac{1}{a}\\[4pt] \implies\;&y^2-2yz+z^2 = \frac{1}{a^2}\qquad(\text{eq}7)\\[4pt] \end{align*}
From the sum $(\text{eq}5)+ (\text{eq}6) +(\text{eq}7)$, we get $$2(x^2 + y^2 + z^2) - 2(xy + yz + zx) = \frac{14}{a^2}$$ hence $$a^2-3b=\frac{7}{a^2}\qquad(\text{eq}8)$$ Subtracting $(\text{eq}8)$ from $(\text{eq}4)$, we get \begin{align*} &(2a^2-3b) - (a^2-3b) = 8-\frac{7}{a^2}\\[4pt] \implies\;&a^2 = 8-\frac{7}{a^2}\\[4pt] \implies\;&a^4 - 8a^2 + 7=0\\[4pt] \implies\;&(a^2-1)(a^2-7)=0\\[4pt] \implies\;&a^2 = 7\qquad\text{[since $a > 1$]}\\[4pt] \end{align*} Then from $(\text{eq}4)$, we have $$b = \frac{2a^2-8}{3}$$ hence $$b = \frac{2(7)-8}{3} = 2$$
The given equations $(1),(2),(3)$ can be expressed as: $$\begin{cases} x^3-y^3=3(x-y) \\ y^3-z^3=y-z \\ z^3-x^3=4(z-x)\end{cases} \stackrel{+}{\Rightarrow} x=3z-2y \ \ (4)$$ Plug $(4)$ in $(1)$ and consider it with $(2)$: $$\begin{cases} 3z^2-3zy+y^2=1 \\ y^2+yz+z^2=1\end{cases} \Rightarrow 2z^2-4zy=0 \Rightarrow z=2y \ \ (5)$$ Put $(5)$ to $(4)$ to find $x=4y$. Now you can put $z=2y$ and $x=4y$ to $(1)$ to find $y=\frac{1}{\sqrt{7}}$. Hence: $$xy+yz+zx=\frac{4}{7}+\frac{2}{7}+\frac{8}{7}=2.$$
Possible Hint:
For $x, y, z \in \mathbb{R}^{+}$ $$(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$$
• For a proof of this inequality, see here or the linked questions. Sep 2, 2020 at 11:47 | 2023-02-06T21:32:43 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2643121/find-xyyzzx-given-systems-of-three-homogenous-quadratic-equations-for-x-y/2643533",
"openwebmath_score": 0.9672312140464783,
"openwebmath_perplexity": 402.1579558738933,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9875683484150417,
"lm_q2_score": 0.8577681013541613,
"lm_q1q2_score": 0.8471046271774352
} |
http://mathhelpforum.com/discrete-math/33125-logic.html | # Math Help - Logic
1. ## Logic
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$
$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:
-------
Let $p_1,p_2,p_3,p_4,p_5,p_6$ be propositions such that $\left[ {(p_1 \veebar p_2 ) \implies (p_3 \implies p_4 )} \right]$ is false. Determine the value of truth of:
1. $(p_5\implies p_6)\vee(p_1\vee p_2).$
2. $[(p_5\implies p_2)\vee\sim p_1]\implies(p_4\vee p_3).$
3. $\sim[(p_6\vee p_5)\wedge(p_1\wedge p_2)\iff(p_4\implies p_3)].$
-------
Let $p,q$ and $r$ be propositions such that $((\sim p\vee q)\implies r)$ is false. Give the value of truth of the following propositions (justify your answer):
1. $\sim q\implies\sim p.$
2. $r\implies(p\iff\sim(q\vee r)).$
I'm learning these stuff, so if you can help me, it'd be great. Thanks
2. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$
$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:
-------
Let $p_1,p_2,p_3,p_4,p_5,p_6$ be propositions such that $\left[ {(p_1 \veebar p_2 ) \implies (p_3 \implies p_4 )} \right]$ is false. Determine the value of truth of:
1. $(p_5\implies p_6)\vee(p_1\vee p_2).$
2. $[(p_5\implies p_2)\vee\sim p_1]\implies(p_4\vee p_3).$
3. $\sim[(p_6\vee p_5)\wedge(p_1\wedge p_2)\iff(p_4\implies p_3)].$
-------
Let $p,q$ and $r$ be propositions such that $((\sim p\vee q)\implies r)$ is false. Give the value of truth of the following propositions (justify your answer):
1. $\sim q\implies\sim p.$
2. $r\implies(p\iff\sim(q\vee r)).$
I'm learning these stuff, so if you can help me, it'd be great. Thanks
what does veebar mean? Is that "exclusive or" ?
3. Exactly.
4. Hello,
I'm not sure i understood well these notations o.O
What if you make a table of truth for A => B, which is ¬P v Q ?
A=1, B=1, (A => B) = 1
A=0, (A => B) = 1
A=1, B=0, (A => B) = 0
For the first one, find the values of p1 to p4 such as the proposition is false. Then you can determine the table of truth of i. ii. and iii. according to the values of p1 to p4...
5. $p\downarrow q$ is $(\sim p)\wedge (\sim q)$; hence $q\wedge p$ is $(\sim q)\downarrow(\sim p)$.
$p\Rightarrow q$ is $\sim(p\wedge(\sim q))$ so that would be $\sim((\sim p)\downarrow q)$.
6. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$
$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:
Looking at your truth table, $p \downarrow q$ = $\sim p \wedge \sim q$
So we can get it to have the same truth value as p -> q in a truth table if we make it look like this: $\sim (\sim \sim p \wedge \sim q)$
Which translates to $\sim (\sim p \downarrow q)$
The reason this works is because
$\sim (\sim \sim p \wedge\sim q) = \sim( p \wedge \sim q)$ by double negation
$\sim( p \wedge \sim q) = \sim p \vee q$ by De Morgan's Laws
$\sim p \vee q = p \to q$ by implication
edit: just fixed it, I had transposed my wedges and vees
7. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$
$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:
This is also know as Sheffer’s stroke named for H.M.Sheffer but actually C.S.Peirce discovered it. Here are some equivalences to help you.
${\neg P} \equiv {P \downarrow P}$
${P \wedge Q} \equiv {\left( {P \downarrow P} \right) \downarrow \left( {Q \downarrow Q} \right)}$
${P \vee Q} \equiv {\left( {P \downarrow Q} \right) \downarrow \left( {P \downarrow Q} \right)}$
${P \to Q} \equiv {\left[ {\left( {P \downarrow P} \right) \downarrow Q} \right] \downarrow \left[ {\left( {P \downarrow P} \right) \downarrow Q} \right]}$
8. Hello, Krizalid!
Express the propositions $p\to q$ and $q\wedge p$ using only $\downarrow$ and $\sim$
where $p\downarrow q,$ is defined by the following truth table:
. . . $\begin{array}{cc|c} p & q & p \downarrow q \\ \hline
T & T & F \\ T & F & F \\ F & T & F \\ F & F &T \end{array}$
I don't know of any direct method; I used some good guessing . . .
For the first one, we know the pattern: . $\begin{array}{cc|c} p & q & p\to q \\ \hline T & T & {\color{blue}T} \\ T & F & {\color{blue}F} \\ F & T & {\color{blue}T} \\ F & F & {\color{blue}T} \end{array}$
Now we must come up with $TFTT$ using only $\downarrow$ and $\sim$
With a little testing I found that $\sim p \downarrow q$ has $FTFF$
. . $\begin{array}{cc|ccc} p & q & \sim p & \downarrow & q \\ \hline
T & T & F & {\color{blue}F} & T \\ T & F & F & {\color{blue}T} & F \\ F & T & T & {\color{blue}F} & T \\ F & F & T & {\color{blue}F} & F \end{array}\quad\hdots\quad\text{Now negate it }\quad\hdots$
. . $\begin{array}{c}\sim(\sim p \downarrow q) \\ \hline {\color{blue}T} \\ {\color{blue}F} \\ {\color{blue}T} \\ {\color{blue}T} \end{array}$
Therefore: . $\sim(\sim p \downarrow q)\,\text{ is equivalent to }\,p \to q$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the second one: . $\begin{array}{cc|c} p & q & q \wedge p \\ \hline T & T & {\color{blue}T} \\ T & F & {\color{blue}F}\\ F & T & {\color{blue}F} \\ F & F & {\color{blue}F} \end{array}$
I found that . $\sim p \downarrow \sim q$ .produces the same pattern: . $\begin{array}{ccc}\sim p & \downarrow & \sim q \\ \hline F & {\color{blue}T} & F \\ F & {\color{blue}F} & T \\ T & {\color{blue}F} & F \\ T & {\color{blue}F} & T \end{array}$
Therefore: . $\sim p \downarrow \sim q\,\text{ is equivalent to }\,q \wedge p$
9. Originally Posted by Plato
This is also know as Sheffer’s stroke named for H.M.Sheffer but actually C.S.Peirce discovered it. Here are some equivalences to help you.
${\neg P} \equiv {P \downarrow P}$
${P \wedge Q} \equiv {\left( {P \downarrow P} \right) \downarrow \left( {Q \downarrow Q} \right)}$
${P \vee Q} \equiv {\left( {P \downarrow Q} \right) \downarrow \left( {P \downarrow Q} \right)}$
${P \to Q} \equiv {\left[ {\left( {P \downarrow P} \right) \downarrow Q} \right] \downarrow \left[ {\left( {P \downarrow P} \right) \downarrow Q} \right]}$
And $P\leftrightarrow Q \equiv \left[\left(P\downarrow P\right)\downarrow Q\right]\downarrow\left[P\downarrow\left(Q\downarrow Q\right)\right]$.
The truth table is exactly the same as the logic of a NOR gate in electronics.
10. Hello, Krizalid!
I have explanations for the third problem,
. . but I don't know if they are acceptable for your course.
I also used the alternate defintion of implication:
. . $p \implies q\;\text{ is equivalnent to: }\,\sim p \vee q$
Let $p,q, r$ be propositions such that: . $(\sim p\vee q)\implies r$ .is false.
Give the value of truth of the following propositions. Justify your answers.
$(i)\;\;\sim q \implies\: \sim p$
$(ii)\;\;r \implies [p \iff \sim (q \vee r)]$
An implication has this truth table: . $\begin{array}{cc|c} p & q & p \to q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T\end{array}$
Note:
. . An implication is false if the first part is true and the second part is false.
. . An implication is true if the first part is false.
$\text{We are told that: }\;\underbrace{(\sim p \vee q)}_T \implies \underbrace{r}_F\:\text{ is false.}$
Hence: . $\sim p \vee q\:\text{ is true, and }\:r\text{ is false.}$
(i) If $(\sim p \vee q)$ is true, then $(q\, \vee \sim p)$ is true.
. . But $(q\,\vee \sim p)$ is equivalent to: $\sim q \implies \,\sim p$
. . Therefore: . $\sim q \implies \,\sim p$ is true.
(ii) We are given: . $r \implies [p \iff \sim(q \vee r)]$
. . And we are told that $r$ is false.
. . $\text{So we have: }\:\underbrace{r}_F \implies \underbrace{[p \iff \sim(q \vee r)]}_{\text{doesn't matter}}$
. . Since the first part is false, the implication is true. | 2015-04-02T07:17:43 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/discrete-math/33125-logic.html",
"openwebmath_score": 0.9247776865959167,
"openwebmath_perplexity": 438.41068630533726,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.965381162156829,
"lm_q2_score": 0.8774768002981829,
"lm_q1q2_score": 0.8470995732375155
} |
https://mathhelpboards.com/threads/nice-puzzle.4859/ | # Nice Puzzle
#### Amer
##### Active member
Arrange the numbers: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
such that the summation of each two successive numbers is a complete square easy interesting question
#### MarkFL
Staff member
8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9
#### Bacterius
##### Well-known member
MHB Math Helper
I can think of a backtracking algorithm to solve this relatively efficiently given any input set. Might code it up tomorrow. Does anyone have a way to attack this analytically (again, for any input set, not just consecutive integers, and return any solution sequence - or none if there are none). An argument for the associated decision problem would be interesting to see too.
#### chisigma
##### Well-known member
I can think of a backtracking algorithm to solve this relatively efficiently given any input set. Might code it up tomorrow. Does anyone have a way to attack this analytically (again, for any input set, not just consecutive integers, and return any solution sequence - or none if there are none). An argument for the associated decision problem would be interesting to see too.
If n=2 k + 1 is a perfect square and the set is 1,2,..., 2k then is...
2 K + 1 = 2k-1 + 2 = 2 k -2 + 3 + ... = k+1 + k = n (1)
Of course however that is a very particular case...
Kind regards
$\chi$ $\sigma$
#### Barioth
##### Member
I solved it too using
a matrix (I named my matrix A) with $$\displaystyle A_ab=1 if a+b=perfect square$$
It's pretty much done once you have the matrix, you see that 8 as the be the first (or last) number.
I ended with the same solution as MarkFL, and I was wondering if there is always a solution for n large enough?
and if that solution(if it exist) was unique?
I know that for this case with n =15 the solution is unique.
I'll take a look at it tomorow (right now I need to hit the bed) if someone find something before I do let me know!
#### Bacterius
##### Well-known member
MHB Math Helper
I solved it too using
a matrix (I named my matrix A) with $$\displaystyle A_ab=1 if a+b=perfect square$$
It's pretty much done once you have the matrix, you see that 8 as the be the first (or last) number.
I ended with the same solution as MarkFL, and I was wondering if there is always a solution for n large enough?
and if that solution(if it exist) was unique?
I know that for this case with n =15 the solution is unique.
I'll take a look at it tomorow (right now I need to hit the bed) if someone find something before I do let me know!
There are actually two solutions for n = 15, forward and reverse (obviously). There are no others. Solutions are not guaranteed to exist. There are none for n = 16, for instance (checked myself). Curiously there apparently are solutions for n = 30 and n = 60 (unsure, I'll check later). I have been working on two versions of an algorithm but they always end up having exponential complexity, because I can't find a good algorithm. I might post my drafts once I come back from comp labs.
#### Bacterius
##### Well-known member
MHB Math Helper
Actually my bad, I wrote the above in a hurry. n = 16 actually has solutions, namely:
(8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16)
(16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8)
n = 17 has the following solutions:
(17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16)
(16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17)
n = 18 to n = 22 have no solutions.
n = 23 has the following solutions:
(22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 12, 13, 23, 2, 14, 11, 5, 20, 16, 9, 7, 18)
(18, 7, 2, 23, 13, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22, 14, 11, 5, 20, 16, 9)
(2, 23, 13, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22, 14, 11, 5, 20, 16, 9, 7, 18)
(9, 16, 20, 5, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 12, 13, 23, 2, 7, 18)
(18, 7, 9, 16, 20, 5, 11, 14, 2, 23, 13, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22)
(18, 7, 9, 16, 20, 5, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10, 15, 21, 4, 12, 13, 23, 2)
There will always be an even number of solutions, obviously.
Looks mysterious
...
An observation is that for n close to 15, the solutions tend to reuse the existing solution for n = 15 in some way (here 16 is added at the end, and after that 17 is added at the beginning), which I guess makes sense heuristically. The pattern breaks down for n = 23 though, there are bits of the original solution e.g. (8, 1, 15) but overall it changes quite a bit.
This is actually an interesting and deep problem. I wonder if it's NP-complete (probably not, if we limit ourselves to inputs of the form {1, 2, 3, ..., n} there should be an efficient way to find solutions. wouldn't surprise me if the full generalization is NP-complete though)
Last edited:
#### Amer
##### Active member
if we have three numbers in the given set that just has one mate from the set ( i mean by the mate, the summation of the two mates is a complete square ) then we cant order them
in our set 1,2,3,...,15
we have two such numbers which will be at the ends of the arrangement 9,8
for example 1,2,3,4,...,10
we cant order it since 10,9,8 has one mate | 2020-11-30T08:02:21 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/nice-puzzle.4859/",
"openwebmath_score": 0.57173752784729,
"openwebmath_perplexity": 271.77924176798933,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9653811591688145,
"lm_q2_score": 0.8774767890838836,
"lm_q1q2_score": 0.8470995597895289
} |
https://math.stackexchange.com/questions/890014/what-is-the-maximum-y-value-of-the-following-function/890018 | # What is the maximum y-value of the following function?
What is the maximum y-value of the following function?
$$y=8t - \frac{t^2}{2} -24$$
It can be done by using the parabolic equation , setting the equation is equal zero. But is there any other straight forward or shortest method to do this?
You can complete the square:
$$y=-\frac{1}{2}(t^2-16t+48)=-\frac{1}{2}((t-8)^2-16)=-\frac{1}{2}(t-8)^2+8$$
Since $-\dfrac{1}{2}(t-8)^2\le0$ for every $t$, the maximum occurs of $y$ when this expression is $0$ (which is at $t=8$). This gives the maximum of $y$ as $0+8=8$.
• @mathlove already did that – RE60K Aug 7 '14 at 13:05
• @Aditya This discussion on the math meta may be of interest to you. Additionally mathlove's answer seems to no longer appear. – Peter Woolfitt Aug 7 '14 at 13:10
for the form of $y = ax^2 + bx + c$ the min (or max) is :
$$y=c-\frac{b^2}{4a}$$
which in this case is :
$$-24+32=8$$
• Is there a name for this theorem? Maybe a proof? I've never seen this before. – Cruncher Aug 7 '14 at 15:28
• just find the min/max parametrically and substitute in the equation (again parametrically) – Shabbeh Aug 7 '14 at 15:34
• I solved it after asking the question :). Did it using Aditya's answer, but keeping it general. – Cruncher Aug 7 '14 at 15:45
• @Cruncher there's no theorem for this, per se. This fact can be derived by completing the square as described by Peter above, or by using calculus. – user2910265 Aug 7 '14 at 19:38
Using calculus: $$y'=8-t$$ Using derivative test $8-t=0\implies t=8$ So,$$y_{\text{extremum}}=64-32-24=8$$ Second derivate: $$y''=-1<0$$ So extremum is a maximum.
• why the downvote – RE60K Aug 7 '14 at 13:16
$$y(t)=8t-\frac{t^2}{2}-24$$
$$y'(t)=0 \Rightarrow 8-t=0 \Rightarrow t=8$$
For $t>8:$ $y>0$
For $t<8:$ $y<0$
So, $y$ is decreasing on $(-\infty, 0]$ and increasing on $[0,+\infty)$.
Therefore, $y$ achieves its minimum at $t=8$, which is equal to $y(8)=8$. | 2020-02-18T00:49:03 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/890014/what-is-the-maximum-y-value-of-the-following-function/890018",
"openwebmath_score": 0.8636354207992554,
"openwebmath_perplexity": 610.2434694017917,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357610169274,
"lm_q2_score": 0.8652240877899775,
"lm_q1q2_score": 0.8470853232396374
} |
https://mathematica.stackexchange.com/questions/136300/how-do-i-get-mathematica-to-write-the-coefficients-of-polynomial-as-fractions-ra/136303 | # How do I get Mathematica to write the coefficients of polynomial as fractions rather than decimal expansions? [duplicate]
I am trying to have Mathematica find the polynomial of degree $n$ in a single variable which fits the given points $(x_1,y_1),\dots, (x_{n+1},y_{n+1})$. In the examples I am concerned with, the polynomials will always have rational number coefficients. Here is an example of the code I am using with $n=3$
Normal[LinearModelFit[{{0, 10}, {1, 35}, {2, 81}, {3, 154}}, x^Range[0, 3], x]]
10.+16.5 x+7.5 x^2+1. x^3
How do I get Mathematica to give the answer with the coefficients presented as fractions rather than as decimal expansions?
• Rationalize – Jason B. Jan 26 '17 at 23:12
• @JasonB. Thank you so much! – Seth Jan 26 '17 at 23:14
• This seems to work in many cases, but in some cases does not seem to be working. Is it possible that rounding errors are causing an inability to convert back to rational number form? I wonder if there is some way to avoid decimal digit expansions entirely. – Seth Jan 26 '17 at 23:28
• It's very possible you end up with a machine precision number that doesn't map to a fraction by Rationalize. Then you might have success rounding first. Don't know it can be avoided as linear model fit may use numeric methods – Jason B. Jan 26 '17 at 23:31
• Is there any way to fit a polynomial without using the linear model fit? From theory I know there is always a polynomial with rational coefficients which is a perfect fit, so it seems undesirable to use a method which is designed only to approximate data. – Seth Jan 26 '17 at 23:39
You can use the WorkingPrecision option of LinearModelFit:
Normal @ LinearModelFit[
{{0,10},{1,35},{2,81},{3,154}},
x^Range[0,3],
x,
WorkingPrecision->Infinity
]
(* 10+(33 x)/2+(15 x^2)/2+x^3 *)
or just use Rationalize afterwards (as mentioned by @Jason B.):
Normal @ LinearModelFit[{{0,10},{1,35},{2,81},{3,154}},x^Range[0,3],x]
Rationalize @ %
(* 10. +16.5 x+7.5 x^2+1. x^3 *)
(* 10+(33 x)/2+(15 x^2)/2+x^3 *)
or use InterpolatingPolynomial:
InterpolatingPolynomial[{{0,10},{1,35},{2,81},{3,154}},x]
Expand @ %
(* 10+x (25+(-1+x) (17/2+x)) *)
(* 10+(33 x)/2+(15 x^2)/2+x^3 *) | 2020-02-19T00:30:51 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/136300/how-do-i-get-mathematica-to-write-the-coefficients-of-polynomial-as-fractions-ra/136303",
"openwebmath_score": 0.25731274485588074,
"openwebmath_perplexity": 1108.5358159565494,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357597935575,
"lm_q2_score": 0.8652240825770432,
"lm_q1q2_score": 0.8470853170774992
} |
https://thprofil.hu/articles/find-transpose-of-a-matrix-c2a685 | That is, if $$P$$ =$$[p_{ij}]_{m×n}$$ and $$Q$$ =$$[q_{ij}]_{r×s}$$ are two matrices such that$$P$$ = $$Q$$, then: Let us now go back to our original matrices A and B. That’s because their order is not the same. Hence, for a matrix A. To calculate the transpose of a matrix, simply interchange the rows and columns of the matrix i.e. Let's do B now. Then $$N’ = \begin{bmatrix} 22 &85 & 7 \\ -21 & 31 & -12 \\ -99 & -2\sqrt{3} & 57 \end{bmatrix}$$, Now, $$(N’)'$$ = $$\begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix}$$. The addition property of transpose is that the sum of two transpose matrices will be equal to the sum of the transpose of individual matrices. This JAVA program is to find transpose of a matrix. Dimension also changes to the opposite. One thing to notice here, if elements of A and B are listed, they are the same in number and each element which is there in A is there in B too. That is, $$(kA)'$$ = $$kA'$$, where k is a constant, $$\begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3}$$, $$kP'$$= $$k \begin{bmatrix} 2 & 11 \\ 8 & -15 \\ 9 & -13 \end{bmatrix}_{2×3}$$ = $$\begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3}$$ = $$(kP)'$$, Transpose of the product of two matrices is equal to the product of transpose of the two matrices in reverse order. Thus Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.”, Example- Find the transpose of the given matrix, $$M = \begin{bmatrix} 2 & -9 & 3 \\ 13 & 11 & -17 \\ 3 & 6 & 15 \\ 4 & 13 & 1 \end{bmatrix}$$. How to calculate the transpose of a Matrix? $$M^T = \begin{bmatrix} 2 & 13 & 3 & 4 \\ -9 & 11 & 6 & 13\\ 3 & -17 & 15 & 1 \end{bmatrix}$$. To learn other concepts related to matrices, download BYJU’S-The Learning App and discover the fun in learning. Ltd. All rights reserved. C++ Program to Find Transpose of a Matrix. Join our newsletter for the latest updates. In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. Transpose of a matrix: Transpose of a matrix can be found by interchanging rows with the column that is, rows of the original matrix will become columns of the new matrix. Transpose of a Matrix can be performed in two ways: Finding the transpose by using the t() function. So, Your email address will not be published. So, taking transpose again, it gets converted to $$a_{ij}$$, which was the original matrix $$A$$. Find Largest Number Using Dynamic Memory Allocation, C Program Swap Numbers in Cyclic Order Using Call by Reference. In another way, we can say that element in the i, j position gets put in the j, i position. r and columns c. Their values should be less than 10 in it flips a matrix over its diagonal. How to Transpose a Matrix: 11 Steps (with Pictures) - wikiHow r*c). filter_none. rows and columns. So. To understand this example, you should have the knowledge of the following C programming topics: The transpose of a matrix is a new matrix that is obtained by exchanging the For example, for a 2 x 2 matrix, the transpose of a matrix{1,2,3,4} will be equal to transpose{1,3,2,4}. Transpose of a matrix A is defined as - A T ij = A ji; Where 1 ≤ i ≤ m and 1 ≤ j ≤ n. Logic to find transpose of a matrix. The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns. Find the transpose of that matrix. M <-matrix(1:6, nrow = 2) Then, the user is asked to enter the elements of the matrix (of order r*c). Here, the number of rows and columns in A is equal to number of columns and rows in B respectively. C++ Programming Server Side Programming. Transpose of a matrix is obtained by interchanging rows and columns. So, is A = B? Commands Used LinearAlgebra[Transpose] See Also LinearAlgebra , Matrix … We can treat each element as a row of the matrix. For the transposed matrix, we change the order of transposed to 3x2, i.e. Transpose of a matrix can be calculated by switching the rows with columns. the screen. The algorithm of matrix transpose is pretty simple. $$a_{ij}$$ gets converted to $$a_{ji}$$ if transpose of A is taken. Store values in it. 1 2 1 3 —-> transpose Declare another matrix of same size as of A, to store transpose of matrix say B. The following is a C program to find the transpose of a matrix: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2… Transpose of a matrix is the process of swapping the rows to columns. But actually taking the transpose of an actual matrix, with actual numbers, shouldn't be too difficult. The transpose of matrix A is represented by $$A'$$ or $$A^T$$. C Program to Find Transpose of a Matrix - In this article, you will learn and get code on finding the transpose of given matrix by user at run-time using a C program. To understand this example, you should have the knowledge of the following C++ programming topics: To find the transpose of a matrix, we will swap a row with corresponding columns, like first row will become first column of transpose matrix and vice versa. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. The transpose of a matrix is a new matrix whose rows are the columns of the original. The transpose of matrix A is written A T. The i th row, j th column element of matrix A is the j th row, i th column element of A T. Transpose of a matrix is obtained by changing rows to columns and columns to rows. Now, there is an important observation.
Himalayan Quail Habitat, Juniper Name Popularity 2019, Wcvh4800k2ww Stacking Kit, Vendakka Theeyal Kerala Style, Prentice Surname Scotland, Johns Hopkins Coronavirus Map, Which Diagram In Uml Emphasizes The Time-ordering Of Messages, Ath-ad900x Vs Ath-ad700x, Egyptian Symbols Ankh, Rolling Stones Sesame Street, | 2021-01-22T02:31:36 | {
"domain": "thprofil.hu",
"url": "https://thprofil.hu/articles/find-transpose-of-a-matrix-c2a685",
"openwebmath_score": 0.42267975211143494,
"openwebmath_perplexity": 296.0461750460354,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357561234474,
"lm_q2_score": 0.865224084314688,
"lm_q1q2_score": 0.847085315603248
} |
https://math.stackexchange.com/questions/2167388/find-time-when-angle-between-the-minute-and-hour-hands-are-given | # Find time, when angle between the minute and hour hands are given
The question which I was trying to solve is:
Find the time between four and five o' clock when the angle between the hour hand and the minute hand is $78^\circ$.
My approach:
At four o' clock, the hour hand and the minute hand of the clock will be at 4 and 12 respectively, hence the angle they make will be $120^\circ$. Let us assume some 'x' minutes have passed since then when the hands of the clock finally acquired the position described in the question. In 'x' minutes, the minute hand would have moved by $6x^\circ$, at the same time the hour hand would have moved by $0.5x^\circ$. At that position, the separation between them as given by the question is $78^\circ$. Hence the equation we can derive from here is:
$120^\circ + 0.5x^\circ - 6x^\circ = 78^\circ$
Solving for 'x' I get $x = \left(\frac{84}{11}\right)$ minutes. Hence, as per my calculation since $\left(\frac{84}{11}\right)$ minutes have elapsed since four o' clock, the time in question should be around 7~8 minutes past 4.
The answer to the exercise given in the book: At $\left(\frac{84}{11}\right)$ and 36 minutes past 4.
Can someone explain me, where did I go wrong?
• The minute hand will catch up hour hand at some point, and after that the separation between those hands will be described by $6x-(120+0.5x)$ until finally this difference reaches $360-150$ at $5$ o'clock, so between these times there is an additional time at which the hands make $78$ degress – user160738 Mar 1 '17 at 18:55
• @user160738 Thanks. – NINCOMPOOP Mar 1 '17 at 19:07
• Are you saying that there can be two instances between 4 o' clock and 5 o' clock when the separation between both the hands will be $78^\circ$? Is my solution wrong or just an alternative? – NINCOMPOOP Mar 1 '17 at 19:03
• Per Hour: $$\text{H}=\frac{360^{\circ}}{12\space\text{hours}}=30^{\circ}\text{/}\space\text{hour}\tag1$$
• Per Minute: $$\text{M}=\frac{\text{H}^{\circ}}{60\space\text{minutes}}=\left(\frac{1}{2}\right)^{\circ}\text{/}\space\text{minute}\tag2$$
• Per Second: $$\text{S}=\frac{\text{M}^{\circ}}{60\space\text{seconds}}=\left(\frac{1}{120}\right)^{\circ}\text{/}\space\text{second}\tag3$$ | 2019-10-15T01:30:02 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2167388/find-time-when-angle-between-the-minute-and-hour-hands-are-given",
"openwebmath_score": 0.6615906953811646,
"openwebmath_perplexity": 212.08282890894617,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357610169272,
"lm_q2_score": 0.8652240791017536,
"lm_q1q2_score": 0.8470853147335554
} |
https://math.stackexchange.com/questions/1676482/bag-a-contains-2-white-and-3-black-balls-bag-b-contains-5-white-and-7-black-bal | # Bag A Contains 2 white and 3 black balls, Bag B Contains 5 white and 7 black balls. Ball is selected at random and is black.
Bag A Contains 2 white and 3 black balls, Bag B Contains 5 white and 7 black balls. Ball is selected at random and is black. What is probability that the ball came from bag A?
Here's my approach Since we know that the ball was black, the chance of drawing a black ball is $(\frac12)(\frac35) + (\frac12)(\frac7{12})$, Basically, chance of black ball from A, black ball from B and add them together. Choosing from either Bag A or Bag B is a $\frac12$ chance, therefore the answer I came to was
$[(\frac12)(\frac25) + (\frac12)(\frac7{12})]*\frac12$
Would this be the right approach?
• Don't we have a probability of 1 that we choose from one of the bags? In other words, are you sure the last multiplication by $\frac{1}{2}$ is necessary? Also, I think your probability should be $\frac{3}{5}$ for bag A. – smingerson Feb 29 '16 at 1:07
• I don't know. Is it? I would assume that choosing a bag also has a 1/2 chance. In this case, Bag A. Since there are 2 bags. – Aaron Feb 29 '16 at 1:10
• Is that not what you are multiplying each one by $\frac{1}{2}$ for in the first place? Also, I just realized you are calculating the probability of getting a black ball. You want to find, given that a black ball is chosen, it came from bag A. – smingerson Feb 29 '16 at 1:12
• So what would be the correct approach? – Aaron Feb 29 '16 at 1:13
Let $S_i$ be the event that you selected bag $i$; $W,B$ that you selected a white or black ball. I assume that be 'randomly selected' they selecting either bag is equally likely. Then you are being asked \begin{align*} P(S_A|B)&= \frac{P(S_A,B)}{P(B)}\tag 1\\ &=\frac{P(B|S_A)P(S_A)}{P(S_A,B)+P(S_B,B)}\tag 2\\ &=\frac{P(B|S_A)P(S_A)}{P(B|S_A)P(S_A)+P(B|S_B)P(S_B)}\\ &=\frac{(3/5)(1/2)}{(3/5)(1/2)+(7/12)(1/2)}\\ &=\frac{36}{71}. \end{align*} where in $(1)$ I used Bayes' rule and in $(2)$ I used the product rule on the numerator, and the law of total probability on the denominator.
• Which one?${}{}$ – Em. Feb 29 '16 at 1:24
• His name was Bayes, and notice how I started. This is key. They told me that a black ball was selected, then asked what is the probability that is came from $S_A$. In other words, it was given that the ball was black. Hence $P(S_A|B)$. In this instance, it is easier to use Bayes' rule and proceed. – Em. Feb 29 '16 at 1:28 | 2019-11-19T02:42:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1676482/bag-a-contains-2-white-and-3-black-balls-bag-b-contains-5-white-and-7-black-bal",
"openwebmath_score": 0.9167733192443848,
"openwebmath_perplexity": 230.41889462028465,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357622402971,
"lm_q2_score": 0.8652240773641088,
"lm_q1q2_score": 0.847085314090828
} |
http://thebuildsuccess.com/lw8kc/how-to-do-exponential-growth-and-decay-word-problems.html | How to do exponential growth and decay word problems
## How to do exponential growth and decay word problems
" The little "o" is a zero for time = 0 when you start. 33(1. 4: Exponential Growth and Decay Practice HW from Stewart Textbook (not to hand in) p. Determine the value of . Exponential Growth And Decay Graphs. The meaning of doubling time and half-life. Find a bank account balance if the account starts with $100, has an annual rate of 4%, and the money left. Solving problems with exponential growth. Exponential Word Problems Decide whether the scenario is exponential or linear, then write a function to describe it. I first found this activity from a Google Search and found a handout. 2. Find the value of the investment after 5 years. How to solve exponential growth and decay word problems. To solve problems on this page, you should be familiar with Vocabulary Exponential Growth: An exponential model in the form, y=abx, where b>1; function increases exponentially. Improve your math knowledge with free questions in "Exponential growth and decay: word problems" and thousands of other math skills. What was its value in 2009? 2. Let us consider the following two examples. Write an exponential growth function and find the value of the painting in 25 years. ' An example of an exponential decay word problem is the following: 'The value of a new$35,000 car decreases by 20% per year. Scientists use Carbon-14 to make a guess at how old some things are -- things that used to be alive like people, animals, wood and natural cloths. I would like to preface this post with today is the Monday after Spring Break, most students (and teachers :-) ) seemed to have forgotten the basics. Use and identify exponential growth and decay functions. Exponential growth and decay. 31 A sculpture is increasing in value at a rate of 8% per year, and its value in 2000 was $1200. Some of the worksheets for this concept are Exponential growth and decay, Exponential growth and decay functions, Graphing exponential, Graphing exponential functions, 4 1 exponential functions and their graphs, Exponential growth, 1 exploration identifying graphs of As a rule, the formatting styles are standardized and accepted by the majority of educational and research solving exponential growth and decay problems institutions. In this tutorial, learn how to turn a word problem into an exponential decay function. Ex 1) A population of 422, 000 increases by 12% each year. Hence, = and setting we have . What is the value of the car after 7 years? How to solve exponential growth and decay word problems. Exponential Growth and Decay Word Problems. Section 3. Online exponential growth/decay calculator. Hand out the Exponential Growth and Decay Worksheet (M-A2-4-3_Exponential Growth_Decay Worksheet. !It!has!ahalf Vlife!of! 8. We start with the basic exponential growth and decay models. t is the time in discrete intervals and selected time units. – The Count May 17 '17 at 2:52. . Enter your email address to follow this blog and receive notifications of new posts by email. Exponential growth/decay formula. The number of subscribers increased by 75% per year after 1985. 1. The next step is to find the trend by noting that we are left with a certain percentage of the substance. Newton's Law of Cooling. • If you’re given a decay rate as a fraction: multiplier = 1 – fraction. What is your initial 17 Mar 2018 Solving Exponential Growth Functions: Social Networking This article focuses on how to use word problems to find the amount at the beginning of the Here's What You Need to Do to Solve Exponential Decay Functions. decay word problems: To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation. If b > 1 the function represents exponential growth. (decay) Jan 223:47 PM • Finding the RATE Look for percents or words that represent • Identify if it is growth or decay change, use the form The variable b represents the growth or decay factor. * time is usually in hours or years Let's just do one -- they're really easy! In 1950, the world's population was 2,555,982,611. Dylan invested$600 in a savings account at a 1. 314 Start with the basics and move up the exponential ladder to master a variety of problem-solving and application problems. A person Start with the basics and move up the exponential ladder to master a variety of problem-solving and application problems. Exponential Decay Solving an exponential decay problem is very similar to working with population growth. Question: The initial population of bacteria in a lab test is 400. temperature changes, as when something is taken from an oven or a refrigerator and allowed to come to room temperature. It follows that b = er > 1 since r > 0 (see Figure 1(b)). If a quantity grows by a Writing and Graphing Exponential Decay Functions Comparing Exponential Growth and Exponential Decay. a year ago. The growth (or decay) factor is the actual factor after the rate is converted into a decimal and added or subtracted from 1 (they may ask you for the growth factor occasionally). REMEDIATION Correct Mistakes on Quiz and Do Another Practice Activity Mr. x: Time. Exponential Growth and Decay Word Problems 1. According to a consumer magazine, this particular car loses 12% of its value every year. Properties of Exponential Functions. But in the case of decrease or decay, the value of k will be negative. WWhat You Will Learnhat You Will Learn. So we attempted to uncover some terrific 11 Exponential Growth And Decay Word Problems Worksheet Answers picture for your needs. Then write the formula for the problem. We learn in calculus that exponential functions can model any kind of growth distinctly that I never Word Problems The best thing about exponential functions is that they are so useful in real applications: population growth, exponential decay, and compound interest. Whenever an exponential function is decreasing, this is often referred to as exponential decay. Find a bank account balance if the account starts with $100, has an annual rate of 4%, and the money is left in the account for 12 years. Here's an exponential decay function: y = a(1-b) x . Apr 25, 2014 · To write an exponential equation in word • The rate is either problems, use the form (1 + %) if increasing (growth) (1 %) if decreasing (decay) double, triple, quadruple, (growth) half, third, etc. Section 7. See how they do when they get to the last three questions. mws b) How long would it take for the mass to decay to 40 mg? Exponential Growth and Decay Word Problems ∗ Growth = Decay = Write and d) Write an exponential function to model the situation: e) “Do” the problem: 2. 5% interest rate compounded annually. Exponential+Growthand+DecayWord+Problems+!! 4. Exponential Growth and Decay Exponential decay refers to an amount of substance decreasing exponentially. When we invest some money in a bank, it grows year by year, because of the interest paid by the bank. Exponential Decay. 5% annual decrease in population. The doubling time in this case is 6. Suppose we model the growth or decline of a population with the following differential equation. It provides the formulas and equations / functions that you need to solve it. We do not replace it with information given to us in the problem. In 1985, there were 285 cell phone subscribers in the small town of Centerville. A hepatobiliary scan of my gallbladder involved an injection of 0. Exponential FUNctions - Growth and Decay Today we started exponential functions and I thought I'd share my notes, activity and next day warm-up with you. EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS NAME: HOUR: 1. Exponential Growth and Decay Word Problems Find a bank account balance if the account starts with$100, has an annual rate of 4%, and the money left in the account for 12 years. x 0 is the initial value at time t=0. Two word problem examples: one about a radioactive decay, and the other the exponential growth of a fast-food chain. In fact, certain populations may decrease instead of increase and we could still use the general formula we used for growth. Determine whether each problem is exponential growth or decay. Whenever you see the phrase relative growth rate, continuous growth rate, or exponential growth rate, you know you're dealing with exponential behavior. Exponential Is there a way to do this without using a calculator? Exponential word problems almost always work off the growth / decay formula, A = Pert, where "A" is the And why do they tell me what the doubling time is? Exponential Growth and Decay. Find a bank account balance if the account starts with $100, has an annual rate of 4%, and the money left in the account for 12 years. Viewers can apply their new (or growing) algebraic knowledge to solve the word problems along with Sal. Exponential Exponential Growth and Decay Worksheet. That is, the rate of growth is proportional to the amount present. In Algebra 1, the following two function formulas were used to easily illustrate the concepts of growth and decay in applied situations. Here, k still determines how fast a population grows, but L provides an upper limit on the population. The instructor first identifies what formula to use to solve this word problem. Then, solve the function and get the answer! How Do You Solve a Word Problem with Exponential Decay? If something decreases in value at a constant rate, you may have exponential decay on your hands. by christine_amburn_81445. Find the value of the car after 10 years. IXL Learning Learning. Aug 02, 2018 · Seriously, we also have been noticed that 23 Exponential Growth And Decay Worksheet Answers is being just about the most popular field on the subject of document template sample right now. Jan 2007 26 0. 4) The half-life of a medication is the amount of time for half of the drug to be eliminated from the body. Note that we studied Exponential Functions here and Differential Equations here in earlier sections Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. Ex2) A car bought for$13,000 depreciates at 12% per year. If 0 b 1 the function represents exponential decay. At the current moment, I cannot find the specific handout, but I don't want to take credit for something that I did not do. My "Clown at your Party" purchases a van to drive to parties for $20,000. 1X 2. Some of the worksheets for this concept are Exponential growth and decay, Exponential growth and decay functions, Graphing exponential, Graphing exponential functions, 4 1 exponential functions and their graphs, Exponential growth, 1 exploration identifying graphs of Exponential Growth/Decay Calculator. Jul 14, 2016 · Equations and amount spent on television. ) Exponential Growth and Decay Word Problems. If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, How Do You Solve a Word Problem with Exponential Growth? If something increases at a constant rate, you may have exponential growth on your hands. Author: Admin General formulae for exponential growth and decay. The two types of exponential functions are exponential growth and exponential decay. than l, is the growth factor. From 1990 to 1997, the number of cell phone subscribers S (in thousands) in the US can be modeled by, S = 5535. Growth, because of 7. • If you’re given a growth rate as a percent: multiplier = 1 + (growth rate as a decimal). Overview Solving Problems Using the Exponential Decay Model — Solving Word Problems Using the Exponential Decay Model. Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher. Model solving convex optimizations solving them, you can fix that problem to solve various equations logarithmic and exponential equations – general solutions; solve various equations. Examples: 1. What is the population after 5 years. If you look for word problems about exponential growth or decay, in text books or on the internet, certain settings arise again and again: bank accounts, populations of animals, people or bacteria, and. take a screenshot of your final IXL daily 10 score, be sure it includes the last question, number of questions answered, total time and smart score. The order of magnitude is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. 2. exponential decay: a decrease whose rate is proportional to the size of the population. 14days. Apr 3, 2007 #1 1) "Solve the equation: The Jun 18, 2018 · Truly, we have been noticed that 11 Exponential Growth And Decay Word Problems Worksheet Answers is being just about the most popular subject on document template sample at this time. A couple of other practice! Exponential growth and decay. In real-life situations we use x as time and try to find out how things change exponentially over time. If the city had . Predict the bacteria population at the end of two hours. Exponential Word Problems DRAFT. a. a) Exponential growth or decay: b) Identify the initial amount: c) Identify the growth/decay factor: d) Write an exponential function to model the situation: e) “Do” the problem: 2. The general idea is that, instead of solving equations to find unknown numbers, Exponential growth and decay. Video 4: Solving Problems of Exponential Growth and Decay. Khan Academy is a 501(c)(3) nonprofit organization. May 16, 2009 · I need help answering this exponential growth/decay question. The value of the van depreciates at a rate of 9% each year. of exponential growth and decay and we used different bases Do you remember the formula to calculate compound interest? To solve problems leading to exponential equations; 2. Word Problems: Interest, Growth/Decay, and Half-Life Applying logarithms and exponential functions Topics include simple and compound interest, e, depreciation, rule of 72, exponential vs. First you will need to create a table with hours and substance left. Determine, to the nearest dollar, the balance in the account after the 4 years. 1 - Struggling Learners] Exponential Decay: An exponential model in the form y=abx, where 0 < b<1; function decreases exponentially. com You have all your materials. Vary the initial amount and the rate of growth or decay and investigate the changes to the graph. Exponential Growth and Decay Word Problems ∗ Growth = Decay = Write and exponential statement for Example 1 and 2 . As discussed in Exponential Growth and Decay: Introduction, all exponential growth/decay problems can be modeled using$\,P(t) = P_{\,0}{\text{e}}^{rt}\,$, This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. Solve real-life problems involving exponential growth and decay. ' Notice that exponential growth word problems often involve compound interest, and exponential decay word problems often involve depreciation. This video contains plenty of examples and practice problems. Example: Using a calculator, complete the chart below to keep track of how much your car will be worth each year. And I need help with another question that also involved exponential growth/decay. Do your students forget a topic when another one is introduced? You can eliminate that Free math notes on word problems involving exponential growth and decay. I'm just going to change the letters a little: The is pronounced "P not. In the case of a discrete domain of definition with equal intervals, it is also called geometric growth or geometric decay, the function values forming a geometric progression. One you have the trend you will use this to calculate An example of an exponential decay word problem is the following: 'The value of a new$35,000 car decreases by 20% per year. Q Exponential functions tell the stories of explosive change. Is the pictured graph growth, decay, or linear or none? Exponential Exponential Growth and Decay Word Problems DRAFT. Identify the growth factor and annual percent increase b. Exponential Functions: General Form Type General Graph a-value b-value for word problems: Two Major Types for word problems: For each of the following equations, write whether it is exponential growth or decay and then write the y-intercept. Exponential Growth and Decay Functions Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. In this tutorial, learn how to turn a word problem into an exponential growth function. How Do You Solve a Word Problem with Exponential Growth? If something increases at a constant rate, you may have exponential growth on your hands. To express exponential growth mathematically, suppose At denotes the followed We will write exponential functions from graphs and tables. The number of bacteria doubles every 30 minutes. The first problem that is show is exponential decay. ,. Website: Exponential Functions Introduction (Purple Math). So that we attempted to identify some terrific 23 Exponential Growth And Decay Worksheet Answers picture to suit your needs. The discrete case. An exploratory paper isn’t uncommon in businesses when they will need to receive all the feasible perspectives and’re trying to get a remedy to a issue and data available. Exponential functions are used to model relationships with exponential growth or decay. Interpret and rewrite exponential growth and decay functions. 180 = 90*e^(3k) 2 = e^(3k) take the natural log (ln) of both sides. Exponential Growth & Decay Word Problems b) Use your equation to estimate the approximate cost per day in 2010. Use the vertical slider to change k, and the horizontal slider to change B. 7 ExponentialGrowthandDecay Exponential Growth Models Recalling the investigations in Section 8. Oct 12, 2017 · 1. What will be its value in 2020? The amount of a certain drug in the bloodstream decreases by 30% in 1 hours. Sieling for login info) An explanation of the formula for exponential equations involving percents 3. Video 3: How to Determine the Growth or Decay Factor. The graph shows how bacteria grows over time . @Will Nagy Thanks Exponential growth is a specific way that a quantity may increase over time. Exponential Decay Exponential growth functions are often used to model population growth. Exponential Growth/Decay Word Problems Name: Date: 1. 5 compounded quarterly. 14 Nov 1995 solve word problems, although I do only the problems that. The equation of the graph is:. Where: x 0 is the initial value of whatever it is that will be growing (or shrinking), r is a constant representing growth (or decay) rate, and x t is the value after t time periods. One of the problems with exponential growth models is that real populations don't grow to infinity. Forums. When given a percentage of growth or decay, determined the growth/decay factor by adding or subtracting the percent, as a decimal, from 1. Graphing exponential growth & decay Our mission is to provide a free, world-class education to anyone, anywhere. 8th - 12th grade. Lesson 20: Exponential Growth and Decay. r is the growth rate when r>0 or decay rate when r<0, in percent. And please add step by step instructions on how you answered it please. Walk around and check their work, ask questions. Exponential decay: the change that occurs when an original amount is reduced by a consistent rate over a period of time. 5 • o. Keep the answer exact or give decimal approximations. Let's Practice: Exponential decay: Half-life In the field of nuclear physics, half-life refers to the amount of time required for radioactive substances to decay into half. Video 5: Applications of Exponential Growth and Decay Exponential Growth And Decay Word Problems Worksheet New Exponential Growth And Decay Worksheet Unique 443 Best Education from Exponential Growth And Decay Word Problems Worksheet , source: latinopoetryreview. Exponential decay and exponential growth are used in carbon dating and other real-life applications. The opposite of “Exponential Growth”, is when we apply exponents to fractions which results in “Exponential Decay”. Showing top 8 worksheets in the category - Exponential Growth Word Problems. A couple 24 Apr 2015 IB MathSL Exponential growth decay. Download this image for free in HD resolution the choice "download button" below. exponential growth and decay word problems worksheet answers from Exponential Growth And Decay Word Problems Worksheet Answers , source: hobieanthony. In mathematics, exponential decay occurs when an original amount is reduced by a consistent rate (or percentage of the total) over a period of time. Exponential Growth And Decay Graphs - Displaying top 8 worksheets found for this concept. Sieling’s Signature Fortunately all we had to do with this problem was multiply 1. Exponential growth and decay often involve very large or very small numbers. When an amount triples , for example, we start with the original and add 200% to it, so the growth rate is 200% (in the formula, it’s 2. 9c Textbook Lesson 13. The equation for "continual" growth (or decay) is A = Pe rt, where "A", is the ending amount, "P" is the beginning amount (principal, in the case of money), "r" is the growth or decay rate (expressed as a decimal), and "t" is the time (in whatever unit was used on the growth/decay rate). Solving Word Problems using The Exponential Growth Model (ask Mr. Example: During its exponential growth phase, a certain bacterium can grow from 14 Nov 1995 solve word problems, although I remember. We buy a car and use it for some years. Explore the graph of the exponential growth or decay function. Where the following integers can be stated as:-. 00 , which will be added to the 1 ), but growth factor is 3 ( 1 + 2 ). The following example shows you how to solve Apr 03, 2007 · decay exponential growth problems word; Home. A new 2006 Honda Accord was valued at $25000. IodineV131is!usedtofindleaks!inwater!pipes. although I remember We learn in calculus that exponential functions can model any kind of growth distinctly that I never really for which the rate of change is proportional to the amount present. To describe these numbers, we often use orders of magnitude. This led to another formula for continuous compound interest, P(t) = P 0ert, (1) whereP Steps to Solve Exponential Equations using Logarithms. One real-life purpose of this concept is to use the exponential decay function to make predictions about market trends and expectations for impending losses. ' Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. y: Final amount remaining after the decay over a period of time. 1) Keep the exponential expression by itself on one side of the equation. Take the exponent of \frac {1}{2} 12 Feb 2018 Exponential Growth and Decay Worksheet. She made no deposits or withdrawals on the account for 4 years. 1 Exponential Growth and Decay Algebra 2 Consider this situation: You decided to purchase a car for$28,500. The half-life of Advil or ibuprofen is represented by the equation, where R is the amount of Advil remaining in the body, M is the initial dosage, and t is time in hours. Properties of Exponential Functions (EdPuzzle Video) Properties of Exponential Functions Assignment. Sketch a graph of the model c. docx). Make sure you have memorized this equation, along with Exponential Growth Word Problems. Analyze both teachers and exponential functions are exponential growth is greater than Click Here X number percent of do about it emathinstruction was founded on forever in life is the mathematics education. Notice that . You can do a rough check of this answer, using the fact that exponential processes involve doubling (or halving) times. Example Question #3 : Solve Exponential Decay Problems. On-screen applet instructions: The applet shows the graph of y = Be kx. 2,950,000 people in 2000, determine the city’s population in 2008. In this case we can find k by taking: Q(3) = Q(0)e^(k3) or 180 = 90*e^(3k) 2 = e^(3k) take the natural log (ln) of both sides ln(2) = 3k or k = ln(2)/3 = This How Do You Solve a Word Problem with Exponential Decay? Video is suitable for 8th - 11th Grade. Is this exponential growth or decay? answer choices . 532 # 1-17 odd In the next two sections, we examine how population growth can be modeled using differential equations. Below is an interactive demonstration of the population growth of a species of rabbits whose population grows at 200% each year and demonstrates the power of exponential population growth. [IS. Exponential growth can be amazing! The idea is that something grows in relation to its current value, such as always doubling. Word Problem Exercises: Solving Exponential Growth and Decay Questions Using Logarithms General Questions The population in the town of Huntersville is presently 38,300. 2) Get the logarithms of both sides of the equation. Apr 28, 2012 · Best Answer: In exponential growth, the quantity at time t from the initial measurement is given by: Q(t) = Q(0)e^(kt) where k is a constant. y = 80 Jun 18, 2018 · Truly, we have been noticed that 11 Exponential Growth And Decay Word Problems Worksheet Answers is being just about the most popular subject on document template sample at this time. 6% annual Exponential Growth and Decay Word Problems. Introduction to rate of exponential growth and decay. Damping of oscillating system. When an oscillating body experiences damping (much like how the suspension system in a car helps reduce oscillation upon impact), the amplitude of the oscillation is reduced over time as describable via expo Is the pictured graph growth, decay, or linear or none? Exponential Exponential Growth and Decay Word Problems DRAFT. 3) Solve for the variable. Exponential decay is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. If you're seeing this message, it means we're having trouble loading external resources on our website. In this equation, "N" refers to the final population, "NI" is the starting population, "t" is the time over which the growth or decay took place and the "k" represents the growth or decay constant. 5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which has a half-life of almost exactly 6 hours. eventually begin to grow at a very rapid rate. 5; 2 Objectives. What about a word problem example? We commonly use a formula for exponential growth to model the population of a bacteria. Students can work on this in pairs. Report "Exponential Functions - Growth & Decay - Worksheet - E3 - MrArt" Title: Exponential Growth and Decay 1 Exponential Growth and Decay. Think about it Cosmic rays can't get through the water. • If you’re given a number that’s used for repeated multiplication, multiplier = that number. In exponential growth, the quantity at time t from the initial measurement is given by: Q(t) = Q(0)e^(kt) where k is a constant. Aug 27, 2016 · Growth and exponential growth and problems; solving trigonometric equations, ethics herman e. 5 hours, or between 6 and 7 hours. (For example, if an amount triples every day, then multiplier = 3. May 17, 2013 · Exponential Growth situations when graphed look like the diagram below. TEKS: a. Apr 11, 2019 · Exponential Growth and Decay Word Problems Worksheet from exponential growth and decay worksheet answers , source:mychaume. Oct 14, 2014 · STATION 1 Linear vs. In the function: y = a(b)x , a is the y-intercept and b is the base that determines the direction of the graph and the. A function of the form y = a(1 + r)t, where a > 0 and r > 0, is an exponential growth function. 718,soe > 1,andthereforey = ex isitselfanexponentialgrowthcurve. Exponential Growth and Decay Worksheet In the function: y = a(b)x, a is the y-intercept and b is the base that determines the direction of the graph and the steepness. Exponential growth occurs when a function's rate of change is proportional to the function's current value. Exponential growth and decay: a differential equation. Generally, in exponential growth and decay problems, you will use an equation that looks like: € f(x) = (starting amount) ⋅ (multiplier)x There are several ways to find the multiplier, depending on the way that the change is described in the problem. initial amount time growth factor rate of growth (in decimal form) final amount y = a(1 + r)t. Some of the worksheets displayed are Exponential growth and decay word problems, Exponential growth and decay, Exp growth decay word probs, Growth decay word problem key, Concept 17 write exponential equations, Exponential growth and decay work, Exponential word problems, Exponential Generally, in exponential growth and decay problems, you will use an equation that looks like: € f(x) = (starting amount) ⋅ (multiplier)x There are several ways to find the multiplier, depending on the way that the change is described in the problem. Example 2: Hospitals utilize the radioactive substance iodine-131 in the diagnosis of conditions of the thyroid gland. The basic formula for discrete exponential growth is: x t = x 0 (1+r) t . Growth because of 1. Then she explains what each of the variables stands for and helps match up their given values in this word problem. 3. b. 413) t where t is number of years since 1990 a. a = Initial growth (the amount before measuring growth or decay) r = Growth or Decay rate (most often represented as a percentage and expressed as a decimal) x = Number of time intervals that have passed. Some things "decay" (get smaller) exponentially. The decay factor is (1-b). To solve exponential growth and decay word problems, we have to be aware of exponential growth and decay functions. 3, we started by developing a formula for discrete compound interest. • If you’re given a growth rate as a fraction: multiplier = 1 + fraction. It occurs when the So for an algorithm of time complexity 2x, if a problem of size x = 10 requires 10 Also, the effects of Moore's Law do not help the situation much because If τ < 0 and b > 1, or τ > 0 and 0 < b < 1, then x has exponential decay. 7 Exponential Growth and Decay 847 Version: Fall2007 8. A webpage that gives Video: Solving Exponential Growth/Decay Word Problems (7 min). The problems are heavy on exponential growth and decay, compound interest, and natural log. linear models, and more. I will write exp. Therefore, functions of the form P(t) = P So, here's the formula for population growth (which also applies to people). 4 The population of a certain city was 292000 in 1998, and the observed relative growth rate is 2 Section 8. This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations. Section 8. Price is one of the main advantages that distinguish our company solving exponential word problems from other research writing services. 7 Exponential Growth and Decay 849 Version: Fall2007 Toprovethatb > 1,considerthegraphofy = ex showninFigure 1(b). Pre-University Math Help. Nov 07, 2011 · Visit the post for more. That's actually amazing. In this lesson, we will work on word questions about exponential decay of radioactive substances. y = 80 How to solve exponential growth and decay word problems. One you have the trend you will use this to calculate the amount of substance left in said hours, in this case 6 hours. QUIZ (Level 3) Level 3 Schoology Quiz: Level 3 – Exponential Equations 4. x(t) = x 0 × (1 + r) t. (Try to do this without the calculator!) 1. We'll just look at the simplest possible example of this. It decreases about 12% for every 1000 m: an exponential decay . x(t) is the value at time t. Adrianne invested $2000 in an account at a 3. A. In this case we can find k by taking: Q(3) = Q(0)e^(k3) or. Description: Exponential Function Word Problems Worksheet from the above 1000x1000 resolutions which is part of the Worksheet. exponential growth, p. Recallthat e ≈2. Oct 25, 2016 · After tackling some word problems, we did Exponential Growth and Decay Modeling using M&Ms. com Exponential Growth… Sex assylumn 3 8d problem solving methodology ppt free business plan templates for word freeBusiness plan review services how creativity and critical thinking relate to contingency planning quoting in essays how to make a business plan free templates essay verbs list research proposal example for phd cite an essay mla word problem solving solve word problems, amount of money invested, and a larger investment grows faster. steepness. An example of an exponential growth word problem is the following: '$1000 is invested at 9% interest compounded annually. The exponential decay of an element is given by the function: In this function, is the amount of the element left after days, and is the initial amount of the element. Benjamin Tuesday the 31st. Initial Value Problems for Growth and Decay Example 1: Unlimited Population Growth The number of bacteria in a liquid culture is observed to grow at a rate proportional to the number of cells present. It depreciates at a rate of 13% a year. From 2000 - 2010 a city had a 2. Algebra 1 - Exponential growth and decay word problems Exponential decay occurs in the same way when the growth rate is negative. 2; 30 Sep 2003 Calculus I: Lesson 20: Exponential Growth and Decay L20-exponential. It doesn't work for sea creatures and other things that are under water. com Modeling Exponential Growth Decay Interspersed with a Reform Rant from Exponential Growth And Decay Word Problems Worksheet Answers Exponential growth and decay can be determined with the following equation: N = (NI)(e^kt). Four variables — percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period — play roles in exponential functions. This section covers: Introduction to Exponential Growth and Decay Solving Exponential Growth Problems Using Differential Equations Exponential Growth Word Problems We can use Calculus to measure Exponential Growth and Decay by using Differential Equations and Separation of Variables. @WillJagy Holy crap I can't believe I missed that the first time. The differential equation in this example, called the logistic equation, adds a limit to the growth. Report "Exponential Functions - Growth & Decay - Worksheet - E3 - MrArt" If you look for word problems about exponential growth or decay, in text books or on the internet, certain settings arise again and again: bank accounts, populations of animals, people or bacteria, and. !Writetheexponential!decayfunctionfor!a200 Vmg! Oct 25, 2016 · After tackling some word problems, we did Exponential Growth and Decay Modeling using M&Ms. One you have the trend you will use this to calculate The word problems in this lesson cover exponential growth and decay. Then, solve the function and get the answer! The first problem that is show is exponential decay. Video: Solving Exponential Growth/Decay Word Problems (Khan) Created By Summit Math; In 1 Playlist(s) Word Problem Solving- Exponential Growth and Decay Concepts from the last few videos, particularly exponential growth and decay functions, are covered in this video about word problems. Using negative power values results in fractions, and when these fractions have exponents applied to them we get “Decay”. If the class is stuck, then bring the class back together. Some applications of exponential growth include calculating compound interest or population growth. com Modeling Exponential Growth Decay Interspersed with a Reform Rant from Exponential Growth And Decay Word Problems Worksheet Answers a) Exponential growth or decay: b) Identify the initial amount: c) Identify the growth/decay factor: d) Write an exponential function to model the situation: e) “Do” the problem: 2. Let's solve this equation for y. growth models from tables and graphs. Exponential Decay (y) = a * (1 – r) ^x. (decay) Jan 223:47 PM • Finding the RATE Look for percents or words that represent • Identify if it is growth or decay change, use Mar 17, 2018 · The two types of exponential functions are exponential growth and exponential decay. How is it that it only takes four years for our computer to go from top-of-the-line to almost worthless? Well, it has something to do with what's to solve exponential growth and decay word problems. You can use any bases for logs. Solve word problems requiring exponential models. Four variables (percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period) play roles in exponential functions. Pre-Calculus M. Exponential Growth/Decay Worksheet Answer the following questions about the exponential decay problems. Algebra Skill Builder - Exponential Growth and Decay Word Problems. Exponential Growth and Decay. Does this function represent exponential growth or exponential decay? B. 2 times itself a few times to get the answer. Exactly exponential here are a mathematical expression that is approximated well as. \$12,072. If you do not find the exact resolution you are looking for, then go for a native or higher resolution. a: The original amount. Then, = => ln(y) = . Also, the y-intercept is (0,1) since e0 = 1. Mulya66. 3 Find the time required for an investment of 5000 to grow to 6800 at an interest rate of 7. Example 1 logarithm and decay: if we have an exponential growth functions sequences word problems in value quadratic, an exponential factorization. Then, solve the function and get the answer! Exponential growth and decay word problems Depreciation, Appreciation, Compounded, Compounded Continuously. In exponential growth problems on the HSE exam, the variable—or input—is in the exponent position. how to do exponential growth and decay word problems | 2020-02-20T00:16:00 | {
"domain": "thebuildsuccess.com",
"url": "http://thebuildsuccess.com/lw8kc/how-to-do-exponential-growth-and-decay-word-problems.html",
"openwebmath_score": 0.5186667442321777,
"openwebmath_perplexity": 853.2557488221838,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9790357567351325,
"lm_q2_score": 0.8652240808393984,
"lm_q1q2_score": 0.8470853127300599
} |
https://mathhelpboards.com/threads/summation-of-arctangents.8710/ | # Summation of arctangents
#### Pranav
##### Well-known member
Problem:
Evaluate
$$\lim_{n\rightarrow \infty} \left(\sum_{r=1}^n (\arctan(2r^2))-\frac{n\pi}{2}\right)$$
Attempt:
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.
Any help is appreciated. Thanks!
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.
Any help is appreciated. Thanks!
Maybe you could use that $$\displaystyle \arctan (x) + \arctan \left( \frac{1}{x}\right) =\frac{\pi}{2}$$ for $$\displaystyle x>0$$.
#### chisigma
##### Well-known member
Problem:
Evaluate
$$\lim_{n\rightarrow \infty} \left(\sum_{r=1}^n (\arctan(2r^2))-\frac{n\pi}{2}\right)$$
Attempt:
I tried evaluating the summation but couldn't. Had the problem involved $\arctan(1/(2r^2))$, I could rewrite it as
$$\arctan\left(\frac{2r+1-(2r-1)}{1+(2r+1)(2r-1)}\right)$$
and evaluating the sum would be quite easy but honestly, I have no idea for the given problem.
Any help is appreciated. Thanks!
Using the identity...
$\displaystyle \sum_{k=1}^{n} \tan^{-1} (2\ k^{2}) - n\ \frac{\pi}{2} = - \sum_{k=1}^{n} \tan^{-1} \frac{1}{2\ k^{2}}\ (1)$
... the series becomes $\displaystyle - \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}}$. Following the procedure described in...
http://mathhelpboards.com/math-notes-49/series-inverse-functions-8530.html
... we have $\displaystyle G(u,v) = \frac{u - v}{1 + u\ v}$ and choosing the sequence $c_{k}= \frac{k}{k+1}$ which is stricktly increasing and tends to 1, we obtain $\displaystyle G(c_{k},c_{k-1}) = \frac{1}{2\ k^{2}}$ and finally...
$\displaystyle \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}} = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4}\ (2)$
Kind regards
$\chi$ $\sigma$
Last edited:
#### Pranav
##### Well-known member
Using the identity...
$\displaystyle \sum_{k=1}^{n} \tan^{-1} (2\ k^{2}) - n\ \frac{\pi}{2} = \sum_{k=1}^{n} \tan^{-1} \frac{1}{2\ k^{2}}\ (1)$
Ah, that was quite silly on my part, I was so much involved in evaluating the summation, I did not think of the above identity, thanks a lot chisigma!
BTW, you have a sign error.
... the series becomes $\displaystyle \sum_{k=1}^{\infty} \tan^{-1} \frac{1}{2\ k^{2}}$. Following the procedure described in...
That's the same series I mentioned in my attempt, it's easy to evaluate this. | 2021-02-25T16:10:35 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/summation-of-arctangents.8710/",
"openwebmath_score": 0.9921197295188904,
"openwebmath_perplexity": 624.048702337189,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357622402971,
"lm_q2_score": 0.8652240686758841,
"lm_q1q2_score": 0.8470853055847454
} |
https://stats.stackexchange.com/questions/281162/scale-a-number-between-a-range/281164 | scale a number between a range [duplicate]
I have been trying to achieve a system which can scale a number down and in between two ranges. I have been stuck with the mathematical part of it.
What im thinking is lets say number 200 to be normalized so it falls between a range lets say 0 to 0.66 or 0.66 to 1 or 1 to 1.66. The range being variable as well.
Any help would be appreciated.
Thanks
• Do you know the (theoretical) minimum and maximum of the original values? Or can we use the entire range of available values to obtain these values (e.g. would you accept a max(x) and min(x) in 'the maths')? – IWS May 23 '17 at 9:07
• yes, That would involve a couple of additional loops but can be found. – Saneesh B May 23 '17 at 9:08
• There are many more duplicates. Here's a search: stats.stackexchange.com/…. – whuber May 23 '17 at 13:39
Your scaling will need to take into account the possible range of the original number. There is a difference if your 200 could have been in the range [200,201] or in [0,200] or in [0,10000].
So let
• $r_{\text{min}}$ denote the minimum of the range of your measurement
• $r_{\text{max}}$ denote the maximum of the range of your measurement
• $t_{\text{min}}$ denote the minimum of the range of your desired target scaling
• $t_{\text{max}}$ denote the maximum of the range of your desired target scaling
• $m\in[r_{\text{min}},r_{\text{max}}]$ denote your measurement to be scaled
Then
$$m\mapsto \frac{m-r_{\text{min}}}{r_{\text{max}}-r_{\text{min}}}\times (t_{\text{max}}-t_{\text{min}}) + t_{\text{min}}$$
will scale $m$ linearly into $[t_{\text{min}},t_{\text{max}}]$ as desired.
To go step by step,
1. $m\mapsto m-r_{\text{min}}$ maps $m$ to $[0,r_{\text{max}}-r_{\text{min}}]$.
2. Next, $$m\mapsto \frac{m-r_{\text{min}}}{r_{\text{max}}-r_{\text{min}}}$$
maps $m$ to the interval $[0,1]$, with $m=r_{\text{min}}$ mapped to $0$ and $m=r_{\text{max}}$ mapped to $1$.
3. Multiplying this by $(t_{\text{max}}-t_{\text{min}})$ maps $m$ to $[0,t_{\text{max}}-t_{\text{min}}]$.
4. Finally, adding $t_{\text{min}}$ shifts everything and maps $m$ to $[t_{\text{min}},t_{\text{max}}]$ as desired.
• Great explanation! I'm in a scenario, where I want to minimize $m$ and want $m$ to take the value of $t_\text{min}$ if $m=r_\text{max}$ and to take the value of $t_\text{max}$ if $m=r_\text{min}$. Thanks to your explanation, this was easy to achieve: Simply swap the numerator to $r_\text{min} - m$ and add $t_\text{max}$ instead of $t_\text{min}$ in the end: $$m\mapsto \frac{r_{\text{min}}-m}{r_{\text{max}}-r_{\text{min}}}\times (t_{\text{max}}-t_{\text{min}}) + t_{\text{max}}$$ – CGFoX Feb 13 '20 at 8:19
In general, to scale your variable $x$ into a range $[a,b]$ you can use: $$x_{normalized} = (b-a)\frac{x - min(x)}{max(x) - min(x)} + a$$ | 2021-04-21T01:05:47 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/281162/scale-a-number-between-a-range/281164",
"openwebmath_score": 0.8846097588539124,
"openwebmath_perplexity": 384.40801525891055,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9790357591818726,
"lm_q2_score": 0.865224070413529,
"lm_q1q2_score": 0.8470853046397393
} |
http://bpfz.safoo.de/moment-of-inertia-cylinder.html | Note that each component of the moment of inertia tensor can be written as either a sum over separate mass elements, or as an integral over infinitesimal mass elements. Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. 5 Moment of inertia of a hollow cylinder about its axis The gure here shows the small element with repect to the axis of rotation. 1 RIGID BODIES - MOMENT OF INERTIA The inability of a body to change by itself its position of rest or uniform motion is called Inertia. According to Newton's first law of motion "A body maintains the current state of motion unless acted upon some external force". It is a rotational analogue of mass, which describes an object's resistance to translational motion. Even moments of inertia for more complicated solids can be calculated if these solids can be decomposed into basic shapes for which the moments of inertia are known. The parallel axis theorem is the theorem determines the moment of inertia of a rigid body about any given axis, given that moment of inertia about the parallel axis through the center of mass of an object and the perpendicular distance between the axes. But instead of limits from to, the limit is set from zero to. Moment of inertia shows, it is not easy to rotate the stationary object; the same which means it is difficult to stop the rotating object. It is also required to find slope and deflection of beams. The Moment of Inertia for a thick-walled Cylindrical tube with open ends, of inner radius r_1 and outer radius r_2. Expression for Moment of Inertia of a Solid Cylinder About its Geometrical Axis: Consider a solid cylinder of mass M, length '' and radius 'r' capable of rotating about its geometrical axis. The area moment of inertia and the polar moment of inertia is used for beam deflection and shaft torsion problems. Polar moment of inertia is a quantity used to predict an object's ability to resist torsion, in objects (or segments of objects) with an invariant circular cross section and no significant warping or out-of-plane deformation. The Moment of Inertia for a thick-walled Cylindrical tube with open ends, of inner radius r_1 and outer radius r_2. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Engineering Fundamentals: CENTROID, AREA, MOMENTS OF INERTIA, POLAR MOMENTS OF INERTIA, & RADIUS OF GYRATION OF A Hollow CIRCLE. Determine the Moment of Inertia. However, if we found the moment of inertia of each section about some. A hollow cylinder and a solid cylinder have the same diameter. The moment of inertia, I, is the rotational equivalent of mass. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. pdf Area-Moment. Facts regarding rolling motion and the moment of inertia are covered by this quiz. Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Draw a free-body diagram of the forces acting on the cylinder, with vectors originating at the point. Moment of inertia table. Completion of the quiz will require you to solve some practice problems. For a continuous body, the sum in eq. Category Education; Show more Show less. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of areas. the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. In this lesson, we will derive the moment of inertia of a semicircle about an axis perpendicular. Polar moment of inertia (denoted here as I p ) can also be found by summing the x and y planar moments of inertia (I x and I y ). 9 Thin walled hollow sphere rotating about the central axis. The moment of inertia of a solid body with density \rho(\mathbf{r}) with respect to a given axis is defined by the volume integral I\equiv\int \rho(\mathbf{r})r_\perp^2\,dV, where r_\perp is the perpendicular distance from the axis of rotation. Mass moments of inertia have units of dimension mass length2. The density is then (1) and the moment of inertia tensor is (2) (3) (4). In particular, the same object can have different moments of inertia when rotating about different axes. The moment of inertia essentially describes an object's resistance to rotational motion in response to a torque. Facts regarding rolling motion and the moment of inertia are covered by this quiz. Unless a rigid body is rotating around its center of mass, the mass moment of inertia of a system will not be the same as the mass moment of inertia of the rigid body itself. Mass Moment of Inertia, I G (cont'd) I G for a body depends on the body's mass and the location of the mass. The moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal axis plus the product of the area and the square of the distance between the two axes. Then by using parallel axis theorem, the moment of inertia of the given section about its centroid is obtained. Derivation of the moment of inertia of a hollow/solid cylinder. Find Moment of Inertia of a Disk Calculator at CalcTown. 371r^2) kg/m^3 about the center. In physics, the moment of inertia measures how resistant an object is to changes in its rotational motion about a particular axis. Here are some of the most common moments of inertia: Solid cylinder or disk of radius r rotating about its axis of symmetry: Hollow cylinder of radius r rotating about its axis of […]. It is required to compute the twist of a beam subjected to a torque. However, the moment of inertia I looked up in a physics textbook is exactly two times this (the factor is $1/2,$ not $1/4$). The moments of inertia for some common shapes can be found using the following. Thin walled cylinder rotating about the central axis. Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)—see Table 2. Moment of Inertia (I) Calculator Common Shapes Rotational Inertia Calculator getcalc. I also solved for the moment of inertia of a sphere and similarly got exactly half of the accepted answer. Mass moment of inertia for a rigid body: When calculating the mass moment of inertia for a rigid body, one thinks of the body as a sum of particles, each having a mass of dm. Inertia is the measure of resistance that a body of a certain mass offers when plunged into motion or, on the contrary, bought to a halt by an external force. I CM represents the object's moment of inertia about its center of mass h represents the perpendicular distance from P to the center of mass For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface. For a simple object like a ball on a string being whirled in a circle, where all the mass can be considered to be the same distance away from the axis of rotation, the moment of inertia is:. It will also teach you how not to combine them. If you do that integral it turns out that you get 1/2 mr squared, so it turns out the fact that some of these masses are closer to the axis than the full radius of the cylinder, makes it so that the total moment of inertia is 1/2 the total mass of the cylinder times the total radius of the cylinder squared. Numbers can still be entered in these cells but are not necessary. A quantity expressing a body's tendency to resist angular acceleration is known as the Moment of Inertia, whereas Polar Moment of Inertia is measure of an object's ability to resist torsion under specified axis when and torque is being applied. In general, the formula for a single object's moment of inertia is I cm = kmr 2 where k is a constant whose value varies from 0 to 1. 8 Solid sphere rotating about the central axis. Observe from figure 2, that this moment of inertia has been calculated about #z# axis. Imply And for the mass moment of inertia about axis c, which is perpendicular to the plate and pass through the intersection of axes a and b. the moment of inertia of a solid cylinder about an axis passing through its COM and parallel to its length is mr2/2 where r is the radius. The center of mass is distance from the vertex, where. The moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal axis plus the product of the area and the square of the distance between the two axes. The moment of inertia may be defined as, I = sum m_ir_i^2 and if the system is continuous, then I = int r^2dm If rho is the mass density then, dm = rhodV where dV is an elementary volume. So what I'd like you to do is, for the cylinder, I'd like you to compute its moment of inertia around its central axis. Calculate the moment of inertia for the mass of the cylinder about center axis which height of $$h$$ and radius, $$r_0$$, as shown in Figure 3. So in particular, I've got for you a cylinder. moment of inertia synonyms, moment of inertia pronunciation, moment of inertia translation, English dictionary definition of moment of inertia. If an object has non-uniform. It is the rotational analogue to mass. When should the Effective Moment of Inertia be used? When calculating deflection? for concrete beams, if the Cracking Moment (M cr) has been exceeded then the Gross Moment of Inertia (I g) should be substituted with the Effective Moment of Inertia (I e). The moment of inertia of an object about an axis through its centre of mass is the minimum moment of inertia for an axis in that direction in space. The following formula is used: I_z=mr_2^2(1-t+t^2/2), where:. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. ! It actually is just a property of a shape and is used in the analysis of how some. Six mass moments of inertia values are required for this matrix: one each for the moments along the X, Y, and Z axes of the part coordinate frame, and three cross-component terms XY, YZ, and XZ. The resulting moment of inertia or centre of gravity when placing one or multiple loads can now be easily calculated. Moment of inertia, also known as rotational inertia, is analogous to the inertia of linear motion. The cylinder rolls down the incline to the bottom, a distance d, without slipping. Moment of inertia (), also called "angular mass" (kg·m 2), is the inertia of a rotating body with respect to its rotation. It is the rotational analog of mass. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Problem 817 Determine the moment of inertia and radius of gyration with respect to a polar centroidal axis of the cross section of a hollow tube whose outside diameter is 6 in. In the principal axes frame, the moments are also sometimes denoted , , and. 030 m pivots on a frictionless bearing. Define moment of inertia. Even moments of inertia for more complicated solids can be calculated if these solids can be decomposed into basic shapes for which the moments of inertia are known. Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)—see Table 2. Find moment of inertia of a uniform hollow cylinder Home Problems and Answers Classical Mechanics Find moment of inertia of a uniform hollow cylinder We know that the moment of inertia for hoop with radius R is mR2. A solid cylinder of mass M = 8 kg M = 8 \text{ kg} M = 8 kg with homogeneous density has a circular base of radius R = 5 m R = 5 \text{ m} R = 5 m and a height of H = 5 m. Moment of Inertia vs. Mass moments of inertia have units of dimension ML 2 ([mass] × [length] 2 ). Parallel Axis Theorem, Moment of Inertia Proof. The moment of inertia for a flywheel may be calculated using the general equation for rotational inertia of a rigid body as shown below. The moment of inertia of any shape, in respect to an arbitrary, non centroidal axis, can be found if its moment of inertia in respect to a centroidal axis, parallel to the first one, is known. The position of the centroid must be determined first though, and more specifically its vertical distance from the bottom base (in other words its y 0 coordinate). It is the rotational analog of mass. The cylinder rolls down the incline to the bottom, a distance d, without slipping. Consider a thin circular slice of radius,. The more far away from the axis, the more moment of inertia the object has. In the preceding subsection, we defined the moment of inertia but did not show how to calculate it. of an object to be the sum of. The matrix of the values is known as the moment of inertia tensor. Rotational Inertia or Moment of Inertia If we have a rigid body rotating at an angular speed Ω about a fixed axis, each particle in it will have a certain amount of kinetic energy. In general, the formula for a single object's moment of inertia is I cm = kmr 2 where k is a constant whose value varies from 0 to 1. Moment of inertia ("MOI") is similar to inertia, except it applies to rotation rather than linear motion. For example, if two disks have the same mass but one has all the mass around the rim and the other is solid, then the disks would have different moments of inertia. It will also teach you how not to combine them. Thin walled cylinder rotating about the central axis. " What it really is is the "second moment of area. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. Calculate the angular acceleration of the cylinder if a mass of 0. Moments of Inertia by Integration. Even moments of inertia for more complicated solids can be calculated if these solids can be decomposed into basic shapes for which the moments of inertia are known. the moment of inertia of a solid cylinder about an axis passing through its COM and parallel to its length is mr2/2 where r is the radius. This calculator gives the values of moment of inertia as well as the values of section modulus about x-axis and y-axis of rectangular section. The moment of inertia is calculated by using: $$I = \int _V \rho(x,y,z)\bar{r}^2 dv,$$ where $\bar{r}$ is the distance from the rotation axis. The moment of inertia, I, is the rotational equivalent of mass. Facts regarding rolling motion and the moment of inertia are covered by this quiz. Derivation of moment of inertia of an uniform solid sphere. Moment of inertia is an important concept for JEE exam. It is the rotational analog of mass. To make it simple for a future user to intuitively choose between the shapes, I added "radio button" form controls underneath each drawing. Media in category "Moments of inertia" The following 152 files are in this category, out of 152 total. Thin walled cylinder rotating about the central axis. As expected the moment of inertia of the disk is smaller than the one of the ring. Polar moment of inertia is defined as a measurement of a round bar's capacity to oppose torsion. Moment of Inertia: Hollow Cylinder The expression for the moment of inertia of a hollow cylinder or hoop of finite thickness is obtained by the same process as that for a solid cylinder. The moment of inertia of a solid cylinder is equal to one half of the mass multiplied by the square of the radius. 3 kgm2, initially not rotating, drops onto the first cylinder and. Moment of inertia is the measure of an object's resistance to changes in its rotation rate. Angular momentum is the rotational equivalent of linear momentum. In particular, the same object can have different moments of inertia when rotating about different axes. Even moments of inertia for more complicated solids can be calculated if these solids can be decomposed into basic shapes for which the moments of inertia are known. Rotational inertia. What is the formula for calculating the moment of inertia (about the central axis) of a solid cylinder? This will be in terms of the dimensions (Mand D) of the cylinder. The moment of inertia of a particle of mass m rotating about a particular point is given by:. Category Education; Show more Show less. As expected the moment of inertia of the disk is smaller than the one of the ring. However, if we found the moment of inertia of each section about some. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. Lecture Notes: Area-Moment. Determine the Moment of Inertia. One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. I would like to be able to find the rotational moment of inertia of an assembly (that is, to disegnate which parts move around the axis and get the moment for those parts together). Bending moment refers to the algebraic sum of all moments located between a cross section and one end of a structural member; a bending moment that bends the beam convex downward is positive, and one that bends it convex upward is negative. The development of the expression for the moment of inertia of a cylinder about a diameter at its end (the x-axis in the diagram) makes use of both the parallel axis theorem and the perpendicular axis theorem. This online angular momentum calculator helps you in finding angular momentum of an object and the moment of inertia. 9 m and density (0. The density is then (1) and the moment of inertia tensor is (2) (3) (4). So why don't you. 371r^2) kg/m^3 about the center. For a simple object like a ball on a string being whirled in a circle, where all the mass can be considered to be the same distance away from the axis of rotation, the moment of inertia is:. must treat the element as a thin rectang e. The moment of inertia of a hoop or thin hollow cylinder of negligible thickness about its central axis is a straightforward extension of the moment of inertia of a point mass since all of the mass is at the same distance R from the central axis. Polar moment of inertia is sometimes denoted with the letter J, instead of I, but its units are the same as those for planar moment of inertia: m 4 or in 4. The moment of inertia of an object is a numerical value that can be calculated for any rigid body that is undergoing a physical rotation around a fixed axis. Rotational inertia. An object that is rotating tends to remain rotating and will continue to do so unless acted upon by an external net torque. The moment of inertia, I, is a measure of the way the mass is distributed on the object and determines its resistance to angular acceleration. Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Moment of Inertia vs. It will also teach you how not to combine them. Second moment of area. (I can see the cylindirical coordiates would be. More on moment of inertia. A solid cylinder of mass M = 8 kg M = 8 \text{ kg} M = 8 kg with homogeneous density has a circular base of radius R = 5 m R = 5 \text{ m} R = 5 m and a height of H = 5 m. To find moment of inertia of the given section, it is divided into three rectangles as shown in Fig. Moment of inertia definition is - a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element's distance from the axis. Moment of inertia (), also called "angular mass" (kg·m 2), is the inertia of a rotating body with respect to its rotation. I CM represents the object's moment of inertia about its center of mass h represents the perpendicular distance from P to the center of mass For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface. 371r^2) kg/m^3 about the center. If an object has non-uniform. Now, for the above figure we have the axis 0 given and hence we can calculate the moment of area by summing together l 2 dA for all the given elements of area dA in the yellow region. Rolling Rod, Cylinder or Disc Measure the radius of the object from the center to the edge in centimeters; enter this figure into the calculator. The moment of inertia may be defined as, I = sum m_ir_i^2 and if the system is continuous, then I = int r^2dm If rho is the mass density then, dm = rhodV where dV is an elementary volume. Engineering Fundamentals: CENTROID, AREA, MOMENTS OF INERTIA, POLAR MOMENTS OF INERTIA, & RADIUS OF GYRATION OF A Hollow CIRCLE. Moment of Inertia - Conversion Moment of inertia, also called rotational inertia, mass moment of inertia, or polar moment of inertia of mass, is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. Define moment of inertia. Rotational Inertia and Moment of Inertia. This doesn't address the minimum total moment of inertia requirement you raised earlier though, for which an algebraic solution could be found (if this is considered to be interesting) if the counterweight is a simple cylinder (at it is often the case when one has to balance a crankshaft, for example). which is the sum of all the elemental particles masses multiplied by their distance from the rotational axis squared. Moment of Inertia. It is necessary to specify a moment of inertia with respect to an axis of rotation. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. The moment of inertia block, which is a table containing the results of the moment of inertia calculation, is displayed and can be inserted anywhere in the drawing. As a preliminary, the volume of the cone is. Observe from figure 2, that this moment of inertia has been calculated about #z# axis. This article will discuss the concept of the area moment of inertia and polar moment of inertia and their application in practical problem solving. The moment of inertia of a solid cylinder is equal to one half of the mass multiplied by the square of the radius. Note: If you are lost at any point, please visit the beginner's lesson or comment below. This equation should be used to find the cylinder's moment of inertia with respect to the z-axis, or the plane parallel to the cylinder's height. For a continuous body, the sum in eq. The moment of inertia for a flywheel may be calculated using the general equation for rotational inertia of a rigid body as shown below. (8) In which and therefore concluding the moment of inertia of a solid disk is, (9) Newton's Second Law for Rigid Body. I CM represents the object's moment of inertia about its center of mass h represents the perpendicular distance from P to the center of mass For our purposes, let P represent the point of contact where the rolling thin ring, cylinder, or sphere touches the incline's surface. Mass moments of inertia have units of dimension mass × length2. Unit of Moment Of Inertia Moment of inertia from a physics point of view is basically a quantitative measure of the rotational inertia or the angular mass of a body. The greater the distance the mass is from the axis of rotation, the larger I G will be. Rotational Inertia and Moment of Inertia. twitter: @carpediemvideo Derivation of moment of inertia!!! This feature is not available right now. Mass Moment of Inertia (Moment of Inertia) depends on the mass of the object, its shape and its relative point of rotation - Radius of Gyration Sponsored Links Mass Moment of Inertia (Moment of Inertia) - I - is a measure of an object's resistance to change in rotation direction. Note that for an object where the mass is all concentrated at the same distance from the axis of rotation, such as a small ball being swung in a circle on a string, the moment of inertia is simply MR 2. Moment of inertia shows, it is not easy to rotate the stationary object; the same which means it is difficult to stop the rotating object. Rotational inertia. The moment of inertia of a solid cylinder is equal to one half of the mass multiplied by the square of the radius. 1 RIGID BODIES - MOMENT OF INERTIA The inability of a body to change by itself its position of rest or uniform motion is called Inertia. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Moment of inertia is the measure of an object's resistance to changes in its rotation rate. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The cylinder rolls down the incline to the bottom, a distance d, without slipping. x is the moment of inertia (second moment of area)of that section. It represents how difficult it overcomed to change its angular motion about that axis. The moment of inertia is calculated by using: $$I = \int _V \rho(x,y,z)\bar{r}^2 dv,$$ where $\bar{r}$ is the distance from the rotation axis. the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Mass moments of inertia have units of dimension mass length2. Derivation of moment of inertia of an uniform solid sphere. If an object has non-uniform. However, the moment of inertia I looked up in a physics textbook is exactly two times this (the factor is $1/2,$ not $1/4$). Moment of Inertia (I) Calculator Common Shapes Rotational Inertia Calculator getcalc. As a preliminary, the volume of the cone is. The particular skills you will need are in establishing. Moment of inertia (), also called "angular mass" (kg·m 2), is the inertia of a rotating body with respect to its rotation. Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)—see Table 2. It is required to compute the twist of a beam subjected to a torque. Find Moment of Inertia of a Disk Calculator at CalcTown. A quantity expressing a body's tendency to resist angular acceleration is known as the Moment of Inertia, whereas Polar Moment of Inertia is measure of an object's ability to resist torsion under specified axis when and torque is being applied. Section modulus helps in determining the strength of the section. But instead of limits from to, the limit is set from zero to. 10 Thick walled cylinder rotating about the central axis A general body with moment of inertia I cm through its centre of mass, rotating about a parallel axis, a distance d from the. In the preceding subsection, we defined the moment of inertia but did not show how to calculate it. 371r^2) kg/m^3 about the center. Moment of inertia is defined as:"The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. In the preceding section, we defined the moment of inertia but did not show how to calculate it. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. In the problem we are required to find moment of inertia about transverse (perpendicular) axis passing through its center. Moment of inertia, also known as rotational inertia, is analogous to the inertia of linear motion. Problem 817 Determine the moment of inertia and radius of gyration with respect to a polar centroidal axis of the cross section of a hollow tube whose outside diameter is 6 in. The moment of inertia of a particle of mass m rotating about a particular point is given by:. If the cylinder rotates about its center, as illustrated in the above diagram, what is its moment of inertia?. It is a mathematical property of a section concerned with a surface area and how. Moment of inertia (), also called "angular mass" (kg·m 2), is the inertia of a rotating body with respect to its rotation. Observe from figure 2, that this moment of inertia has been calculated about #z# axis. The following is a list of moments of inertia. Imply And for the mass moment of inertia about axis c, which is perpendicular to the plate and pass through the intersection of axes a and b. 2 Moment of Inertia - Composite Area Monday, November 26, 2012 Radius of Gyration ! This actually sounds like some sort of rule for separation on a dance floor. In this video I will derive the moment of inertia of a solid cylinder of length L, radius R, and mass M. 790 kg is attached to the string wrapped around the pulley. The moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal axis plus the product of the area and the square of the distance between the two axes. Calculator for Moment of Inertia of Rectangular section. Rotational kinetic energy. It is the rotational analogue to mass. So now, when the "solid cylinder" radio button is selected, the input boxes for r i, a, b, and c are grayed out as shown below. The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. What is the formula for calculating the moment of inertia (about the central axis) of an annular cylinder? Again, this will be in terms of the dimensions of the cylinder. So what I'd like you to do is, for the cylinder, I'd like you to compute its moment of inertia around its central axis. The following formula is used: I_z=mr_2^2(1-t+t^2/2), where:. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of areas. The greater the distance the mass is from the axis of rotation, the larger I G will be. Calculating Moment of Inertia: Real-World Problems. Where d is the distance between the two axis. 1 kg m2 rotates with angular velocity 5. ! It actually is just a property of a shape and is used in the analysis of how some. The moment of inertia of an object is the measure of its resistance to being rotated about an axis. I also solved for the moment of inertia of a sphere and similarly got exactly half of the accepted answer. Integration is used to sum the moment of inertia of each dm to get the mass moment of inertia of body. The moment of inertia of an object about an axis through its centre of mass is the minimum moment of inertia for an axis in that direction in space. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Introduction To College Physics Moment of Inertia Title: Moment of Inertia Preparation: Read the sections on Rotational Inertia and torque in Chapter 7. " Or : The product mass and the square of the perpendicular distance from the axis of rotation is known as moment of inertia. But instead of limits from to, the limit is set from zero to. Mass Moment of Inertia (Moment of Inertia) depends on the mass of the object, its shape and its relative point of rotation - Radius of Gyration Sponsored Links Mass Moment of Inertia (Moment of Inertia) - I - is a measure of an object's resistance to change in rotation direction. Calculate the angular acceleration of the cylinder if a mass of 0. Moment of Inertia and Radius of Gyration Moment of Inertia Moment of inertia, also called the second moment of area, is the product of area and the square of its moment arm about a reference axis. In particular, the same object can have different moments of inertia when rotating about different axes. Calculate the moment of inertia for the mass of the cylinder about center axis which height of $$h$$ and radius, $$r_0$$, as shown in Figure 3. When you try to make it move that mean you want to change the speed of the object from 0 to any, there will be moment of inertia effect. Calculate the moment of inertia of a cylinder. Since the moment of inertia of an ordinary object involves a continuous distribution of mass at a continually varying distance from any rotation axis, the calculation of moments of inertia generally involves calculus, the discipline of mathematics which can handle such continuous variables. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired. (8) In which and therefore concluding the moment of inertia of a solid disk is, (9) Newton's Second Law for Rigid Body. This is the currently selected item. It is the rotational analog of mass. The moment of inertia essentially describes an object's resistance to rotational motion in response to a torque. 1 Moments of Inertia by Integration Example 6, page 3 of 3 4 An alternative approach is to use a horizontal rectangular strip and employ the equation for the moment of inertia of a rectangle about its base (BB) : (1) B B h b 100 m 1. What is the formula for calculating the moment of inertia (about the central axis) of an annular cylinder? Again, this will be in terms of the dimensions of the cylinder. 790 kg is attached to the string wrapped around the pulley. Find moment of inertia of a uniform hollow cylinder Home Problems and Answers Classical Mechanics Find moment of inertia of a uniform hollow cylinder We know that the moment of inertia for hoop with radius R is mR2. sen to be CIA. The moment of inertia may be defined as, I = sum m_ir_i^2 and if the system is continuous, then I = int r^2dm If rho is the mass density then, dm = rhodV where dV is an elementary volume. The moment Of inertia of the element with respect to the axis is then 30670 Moment of Inertia The same vertical differential element of area is used. 8 Solid sphere rotating about the central axis. The moment of inertia calculates the rotational inertia of an object rotating around a given axis. When an object is just sitting without any motion, the moment of inertia is 0. I was thinking of breaking down the input shaft/worm into different size cylinders since the shaft has different diameters for the the input and bearings. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. Moment of inertia depends on the amount and distribution of its mass, and can be found through the sum of moments of inertia of the masses making up the whole object, under the same conditions. The parallel axis theorem is the theorem determines the moment of inertia of a rigid body about any given axis, given that moment of inertia about the parallel axis through the center of mass of an object and the perpendicular distance between the axes. The matrix of the values is known as the moment of inertia tensor. Moment of Inertia - Conversion Moment of inertia, also called rotational inertia, mass moment of inertia, or polar moment of inertia of mass, is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. Second moment of area. Polar moment of inertia is defined as a measurement of a round bar's capacity to oppose torsion. The position of the centroid must be determined first though, and more specifically its vertical distance from the bottom base (in other words its y 0 coordinate). In this subsection, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. The moment of inertia of the shape is given by the equation. It is the rotational analog of mass. Parallel Axes Theorem. Answer in units of kg*m^2. I am working in cartesian coordinaes and am not sure where I am going wrong. Moment of Inertia A vertical differential element of area is Cho. In general, the formula for a single object's moment of inertia is I cm = kmr 2 where k is a constant whose value varies from 0 to 1. com's Moment of Inertia Calculator is an online physics tool to measure the rotational inertia of different objects of most common shapes based on the mass distribution and their axis, in both US customary & metric (SI) units. The parallel axis theorem is the theorem determines the moment of inertia of a rigid body about any given axis, given that moment of inertia about the parallel axis through the center of mass of an object and the perpendicular distance between the axes. The moment of inertia about the axis of the cone is (taking density) that of a stack of discs each having mass and moment of inertia :. 1 kg m2 rotates with angular velocity 5. For basic shapes there are tables that contain area moment of inertia equations which can be viewed below. In the preceding section, we defined the moment of inertia but did not show how to calculate it. When should the Effective Moment of Inertia be used? When calculating deflection? for concrete beams, if the Cracking Moment (M cr) has been exceeded then the Gross Moment of Inertia (I g) should be substituted with the Effective Moment of Inertia (I e). Perform the following analysis to determine the moment of inertia of the platter. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. 5 Moment of inertia of a hollow cylinder about its axis The gure here shows the small element with repect to the axis of rotation. A larger moment of inertia around a given axis requires more torque to increase the rotation, or to stop the rotation, of a body about that axis. This online angular momentum calculator helps you in finding angular momentum of an object and the moment of inertia. I was thinking of breaking down the input shaft/worm into different size cylinders since the shaft has different diameters for the the input and bearings. Angular momentum is the rotational equivalent of linear momentum. It is the rotational analog of mass. 5 where m i has the same direction as θ in Figure 2. Now, for the above figure we have the axis 0 given and hence we can calculate the moment of area by summing together l 2 dA for all the given elements of area dA in the yellow region. Moment of Inertia. If an object has non-uniform. I am computing the $\hat{I}$ - moment of inertia tensor - of a cylinder with height 2h and radius R, about its axis of symmetry at the point of its centre of mass. Area Moment of Inertia: The second moment of the area around a given axis is called the area moment of inertia. Moments of Inertia by Integration. The units of the area moment of inertia are meters to the fourth power (m^4). In its inertial properties, the body behaves like a circular cylinder. | 2019-12-15T18:13:11 | {
"domain": "safoo.de",
"url": "http://bpfz.safoo.de/moment-of-inertia-cylinder.html",
"openwebmath_score": 0.7360675930976868,
"openwebmath_perplexity": 205.88164116530058,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9919380089483865,
"lm_q2_score": 0.8539127585282744,
"lm_q1q2_score": 0.8470285215101608
} |
http://mathhelpforum.com/statistics/225597-combination-problem-involving-book-selling.html | # Thread: Combination problem involving book selling.
1. ## Combination problem involving book selling.
16. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if
(b) the books are to be on different subjects?
So for the first pass one can choose one book out of one of the three piles so, $\binom{6}{1}+\binom{7}{1}+\binom{4}{1}=17$.
For the second pass one can only choose a book from a pool that wasn't selected from the first pool. Leaving 3 options;
1. If you selected from $\binom{6}{1}$ then $\binom{7}{1}+\binom{4}{1}=11$
2. If you selected from $\binom{7}{1}$ then $\binom{6}{1}+\binom{4}{1}=10$
3. If you selected from $\binom{4}{1}$ then $\binom{6}{1}+\binom{7}{1}=13$
I'm not entirely sure what I've done wrong here. I think I should just be able to add everything from pass one and pass two to get the correct answer but the listed answer is 94 and $51\neq94$
Could someone clue me into what I'm doing wrong here?
2. ## Re: Combination problem involving book selling.
Originally Posted by bkbowser
16. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if
(b) the books are to be on different subjects?
I'm not entirely sure what I've done wrong here. I think I should just be able to add everything from pass one and pass two to get the correct answer but the listed answer is 94 and $51\neq94$
Do it in one pass. You just have three possibles.
$(6)(7)+(6)(4)+(7)(4)=~?$.
3. ## Re: Combination problem involving book selling.
Originally Posted by Plato
Do it in one pass. You just have three possibles.
$(6)(7)+(6)(4)+(7)(4)=~?$.
Huh. I didn't see that at all.
So in the first case you pull a book from the math and the science piles.
In the second case you pull a book from the science and economics piles.
In the third case you pull a book from the math and economics piles.
In any of the three cases the number outcomes is dependent on the pool sizes. So, taking the first case as an example, the math pile has 6 books and the science pile has 7 books; meaning that there are 6 different possibilities for the first pass and 7 possibilities for the second pass. Or, 6*7, outcomes.
Finally since each of the three cases is exclusive you add them all together.
Does this look right?
4. ## Re: Combination problem involving book selling.
Originally Posted by bkbowser
So in the first case you pull a book from the math and the science piles.
In the second case you pull a book from the science and economics piles.
In the third case you pull a book from the math and economics piles.
Well they are just combinations.
$\binom{6}{1}\binom{7}{1}+\binom{6}{1}\binom{4}{1}+ \binom{7}{1}\binom{4}{1}=(6)(7)+(6)(4)+(7)(4)=~94$. | 2017-03-28T16:49:16 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/statistics/225597-combination-problem-involving-book-selling.html",
"openwebmath_score": 0.6085119247436523,
"openwebmath_perplexity": 594.39796899625,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9919380072800831,
"lm_q2_score": 0.8539127566694178,
"lm_q1q2_score": 0.8470285182417048
} |
https://mathoverflow.net/questions/319058/integrating-over-a-hypercube-not-a-hypersphere | # Integrating over a hypercube, not a hypersphere
Denote $$\square_m=\{\pmb{x}=(x_1,\dots,x_m)\in\mathbb{R}^m: 0\leq x_i\leq1,\,\,\forall i\}$$ be an $$m$$-dimensional cube.
It is all too familiar that $$\int_{\square_1}\frac{dx}{1+x^2}=\frac{\pi}4$$.
QUESTION. If $$\Vert\cdot\Vert$$stands for the Euclidean norm, then is this true? $$\int_{\square_{2n-1}}\frac{d\pmb{x}}{(1+\Vert\pmb{x}\Vert^2)^n} =\frac{\pi^n}{4^nn!}.$$
• Do you really mean $\square_{2n-1}$? – user64494 Dec 19 '18 at 17:56
• Your formula also seems plausible for $n$ a half-integer (interpreting $n!$ as $\Gamma(n+1)$), at least it is true for $n=\frac{3}{2}$. This suggests an induction on $2n-1$. – Gro-Tsen Dec 19 '18 at 18:11
• Alas, the obvious naive approach consisting of writing the integral over $[-1;1]^{2n-1}$ and decomposing the latter into the hypersurface of cubes of side $a$ themselves cut into their various $(2n-2)$-dimensional faces, does not seem to work (or maybe I tried it wrong). – Gro-Tsen Dec 19 '18 at 18:18
Yes, this is true, it follows from formulas in Higher-Dimensional Box Integrals, by Jonathan M. Borwein, O-Yeat Chan, and R. E. Crandall.
\begin{align} &\text{define}\;\;C_{m}(s)=\int_{[0,1]^m}(1+|\vec{r}|^2)^{s/2}\,d\vec{r},\;\;\text{we need}\;\;C_{2n-1}(-2n).\\ &\text{the box integral is}\;\;B_m(s)=\int_{[0,1]^m}|\vec{r}|^s\,d\vec{r},\\ &\text{related to our integral by}\;\;2n C_{2n-1}(-2n)=\lim_{\epsilon\rightarrow 0}\epsilon B_{2n}(-2n+\epsilon)\equiv {\rm Res}_{2n}. \end{align} The box integral $$B_n(s)$$ has a pole at $$n=-s$$, with "residue" $${\rm Res}_n$$. Equation 3.1 in the cited paper shows that the residue is a piece (an $$n$$-orthant) of the surface area of the unit $$n$$- sphere, so it is readily evaluated, \begin{align} &{\rm Res}_{n}=\frac{1}{2^{n-1}}\frac{\pi^{n/2}}{\Gamma(n/2)},\\ &\text{hence}\;\;C_{2n-1}(-2n)=\frac{1}{2n}\frac{1}{2^{2n-1}}\frac{\pi^{n}}{\Gamma(n)}=\frac{1}{4^n}\frac{\pi^n}{n!}\;\;\text{as in the OP}. \end{align}
More generally, we can consider
$$C_{p-1}(-p)=\int_{[0,1]^{p-1}}(1+|\vec{r}|^2)^{-p/2}\,d\vec{r}=\frac{1}{p}\,{\rm Res}_p=\frac{1}{p} \frac{1}{2^{p-1}}\frac{\pi^{p/2}}{\Gamma(p/2)}.$$ This formula holds for even and odd $$p\geq 2$$, as surmised by Gro-Tsen in a comment.
There are similarly remarkable hypercube integrals where this came from, for example
$$\int_{[0,\pi/4]^k}\frac{d\theta_1 d\theta_2\cdots d\theta_k}{(1+1/\cos^2\theta_1+1/\cos^2\theta_2\cdots+1/\cos^2\theta_k)^{1/2}}=\frac{k!^2\pi^k}{(2k+1)!}.$$
• Thanks for the reference and solution! As the first solver, it is accepted. – T. Amdeberhan Dec 19 '18 at 21:58
Here is an elementary proof. Using the formula $$$$\frac1{a^n}=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u\,a}\,du \tag{1}$$$$ with $$a=1+\|x\|^2$$, denoting your integral by $$J_n$$, letting $$I:=[0,1]$$, $$\phi(z):=\frac1{\sqrt{2\pi}}\,e^{-z^2/2}$$, and $$\Phi(z):=\int_{-\infty}^z\phi(t)\,dt$$, and making substitutions $$x_1=z_1/\sqrt{2u}$$ and then $$u=z^2/2$$, we have \begin{align} J_n&=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \int_{I^{2n-1}}e^{-u\,\|x\|^2}\,dx \\ &=\frac1{\Gamma(n)}\,\int_0^\infty u^{n-1} e^{-u}\,du \Big(\int_I e^{-u\,x_1^2}\,dx_1\Big)^{2n-1} \\ &=\frac{\pi^{n-1/2}}{\Gamma(n)}\,\int_0^\infty u^{-1/2} e^{-u}\,du \Big(\Phi(\sqrt{2u})-\frac12\Big)^{2n-1} \\ &=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,\phi(z)\,dz \\ &=\frac{2\pi^{n}}{\Gamma(n)}\,\int_0^\infty \Big(\Phi(z)-\frac12\Big)^{2n-1}\,d\Phi(z) \\ &=\frac{\pi^{n}}{4^n \Gamma(n+1)}, \end{align} as desired.
This derivation obviously holds whenever $$2n-1$$ is a positive integer.
More generally, in view of Bernstein's theorem on completely monotone functions, one can quite similarly express the box integral $$$$\int_{I^n} g(\|x\|^2)\,dx$$$$ as an ordinary integral over $$[0,\infty)$$ -- for any completely monotone function $$g$$. Indeed, Bernstein's theorem states that any completely monotone function is a positive mixture of decreasing exponential functions; identity (1) is then such an explicit Bernstein representation of the completely monotone function $$a\mapsto \frac1{a^n}$$.
Now, furthermore, one does not have to insist that the mixture be positive, since we are dealing here with identities rather than inequalities. So, any function $$g$$ that can be represented as the mixture $$$$g(a)=\int_0^\infty e^{-u\,a}\,\mu(du)$$$$ of decreasing exponential functions $$a\mapsto e^{-u\,a}$$ for $$a>0$$, where $$\mu$$ is any finite, possibly signed ("mixing") measure, will do just as well. | 2021-01-15T21:52:27 | {
"domain": "mathoverflow.net",
"url": "https://mathoverflow.net/questions/319058/integrating-over-a-hypercube-not-a-hypersphere",
"openwebmath_score": 0.9995222091674805,
"openwebmath_perplexity": 1134.665642776775,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9919380070417539,
"lm_q2_score": 0.8539127548105611,
"lm_q1q2_score": 0.8470285161943218
} |
https://cs.stackexchange.com/questions/99519/solving-coin-exchange-for-coins-of-power-values-by-greedy-algorithm | # Solving “coin exchange” for coins of power values by greedy algorithm
When solving the problem of coin exchange by greedy algorithm, why will we will always have the correct result if the coin values are $$1, a, a^2, \cdots, a^n$$, where $$a\ge 2$$ and $$n\gt 0$$?
For example, if $$a = 3$$, $$n =3$$, we get the following coin values: 1, 3, 9, 27. When total exchange is 16, answer is 4 coins, as returned by greedy algorithm, $$16=9+3+3+1$$.
• Welcome to Computer Science! I am very interested in how you know "we will always have the correct result" in these cases. Could you please provide a url or reference in the question? – John L. Nov 3 '18 at 5:55
• You can use LaTeX to typeset mathematics. I edited to show you how; we also have a brief tutorial. – John L. Nov 3 '18 at 10:58
• I just updated the question on behalf of the asker. It should be clear now. – John L. Nov 3 '18 at 17:36
• Thanks, @Apass.Jack -- that's much clearer. – David Richerby Nov 3 '18 at 18:46
• @Apass.Jack, thanks for the corrections! The fact is that there is no basis for my words. I said this without any arguments, I just checked some numbers, it was so. – BoberMod Nov 3 '18 at 22:49
For the coin values: 1, 3, 9, 27, suppose $$16 = a_0 + 3a_1 + 9a_2 + 27a_3$$ is the coin exchange with minimal number of coins. Note that $$16\equiv a_0 \equiv 1(\operatorname{mod}3)$$. If $$a_0 \ge 3$$, then we can subtract 3 from $$a_0$$ and add 1 to $$a_1$$ to obtain coin exchange with less number of coins. So, $$a_0$$ must be the remainder of 16 divided by 3. Hopefully, you can see where I am going.
All numbers or variables below are nonnegative integers. $$(d_nd_{n-1}\cdots d_0)_a$$ means the integer with those digits in base $$a$$, i.e., $$(d_nd_{n-1}\cdots d_0)_a=\sum_{i=0}^{n}d_ia^i$$ where $$d_i\lt a$$ for $$i\le n$$.
Here is the general theorem.
(Greedy algorithm works for power coins) Let $$a\ge2$$ and $$n\ge0$$. If $$e_na^n+(e_{n-1}e_{n-2}\cdots e_0)_a=\sum_{i=0}^nc_ia^i$$, then $$\sum_{i=0}^ne_i\le\sum_{i=0}^nc_i$$.
Proof by mathematical induction on $$n\ge0$$.
The base case, when $$n=0$$ is trivial.
Suppose it is true for some $$n\ge0$$. Let us check the case of $$n+1$$. Suppose $$m=e_{n+1}a^{n+1}+(e_ne_{n-1}\cdots e_0)_a=\sum_{i=0}^{n+1}c_ia^i$$. Note that $$e_0$$ is the remainder of $$m$$ divided by $$a$$.
If $$c_{0}\ge a$$, we can subtract $$a$$ from $$c_0$$ and add $$1$$ to $$c_1$$ to reduce sum $$\sum_{i=0}^{n+1}c_i$$. Without loss of generality we can assume $$c_{0}\lt a$$, i.e., $$c_0$$ is also the remainder of $$m$$ divided by $$a$$.
Now let $$m'=(m-e_0)/a=(m-c_0)/a$$. Then $$m'=e_{n+1}a^{n}+(e_{n}e_{n-1}\cdots e_1)_a=\sum_{i=0}^{n}c_{i+1}a^i$$. By induction hypothesis, we know $$\sum_{i=0}^ne_{i+1}\le\sum_{i=0}^nc_{i+1}$$. Adding $$e_0=c_0$$ to both sides, we obtained the wanted inequality for the case of $$n+1$$. Q.E.D.
What is left to be explained is that $$\sum_{i=0}^ne_i$$ does come from the greedy algorithm. That easy task will be left for interested readers to verify.
• This should always work if any but the smallest coin have a value that is a multiple of the previous coin. – gnasher729 Nov 3 '18 at 16:02
• @gnasher729 Sharp observation! The proof also works after slight modification. – John L. Nov 3 '18 at 16:47 | 2021-03-01T10:06:17 | {
"domain": "stackexchange.com",
"url": "https://cs.stackexchange.com/questions/99519/solving-coin-exchange-for-coins-of-power-values-by-greedy-algorithm",
"openwebmath_score": 0.9294268488883972,
"openwebmath_perplexity": 317.36192146014713,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9919380068034248,
"lm_q2_score": 0.8539127510928476,
"lm_q1q2_score": 0.8470285123030682
} |
https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/x786f2022:vectors-and-matrices/a/matrices--visualized-mvc | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Multivariable calculus
### Unit 1: Lesson 2
Vectors and matrices
# Visualizing matrices
Learn about a useful interpretation of matrices that helps to understand matrix multiplication and determinants.
In the last article we covered matrix fundamentals, but matrices are much more than tables of numbers. That's why in this article we'll discuss a way to think about matrices visually. This perspective makes a lot of what initially seems difficult about matrices become intuitive to understand. We'll only need this perspective for square matrices in multivariable calculus, so we will limit ourselves to those here.
## Matrices as movement
What is the action of a matrix? What does a matrix look like? These questions may seem nonsensical, but we will answer them both by visualizing how 2, times, 2 matrices move the 2D plane.
Here is a drawing of the plane, along with the unit vectors start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, which stand for start color #11accd, left parenthesis, 1, comma, 0, right parenthesis, end color #11accd and start color #ca337c, left parenthesis, 0, comma, 1, right parenthesis, end color #ca337c.
Let's consider matrix A:
$A = \left[ \begin{array}{cc} \blueD{1} & \maroonD{0} \\ \blueD{1} & \maroonD{1} \end{array} \right]$
This is how the matrix acts upon the grid:
• The columns of the matrix tell us where it moves the unit vectors start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, which again stand for start color #11accd, left parenthesis, 1, comma, 0, right parenthesis, end color #11accd and start color #ca337c, left parenthesis, 0, comma, 1, right parenthesis, end color #ca337c.
• The rest of the grid follows accordingly, always keeping grid lines parallel and evenly spaced. The origin stays frozen in place.
That means A moves start color #11accd, \imath, with, hat, on top, end color #11accd, \to, start color #11accd, left parenthesis, 1, comma, 1, right parenthesis, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, \to, start color #ca337c, left parenthesis, 0, comma, 1, right parenthesis, end color #ca337c. Here's what that looks like:
The unit vector start color #ca337c, \jmath, with, hat, on top, end color #ca337c didn't move because it started at start color #ca337c, left parenthesis, 0, comma, 1, right parenthesis, end color #ca337c. The unit vector start color #11accd, \imath, with, hat, on top, end color #11accd moved upward one unit, and this dragged the grid with it. Notice that there's a faint copy of the original lines in the background to help us stay oriented.
Let's see the same process for another matrix.
$B = \left[ \begin{array}{cc} \blueD{0} & \maroonD{-1} \\ \blueD{-2} & \maroonD{1} \end{array} \right]$
We know B moves start color #11accd, \imath, with, hat, on top, end color #11accd, \to, start color #11accd, left parenthesis, 0, comma, minus, 2, right parenthesis, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, \to, start color #ca337c, left parenthesis, minus, 1, comma, 1, right parenthesis, end color #ca337c. That looks like this:
Here's a question to practice.
Problem 1
Suppose we have a matrix $C = \left[ \begin{array}{cc} \blueD{2} & \maroonD{1} \\ \blueD{-1} & \maroonD{1} \end{array} \right]$.
What does the grid look like after we apply C?
To sum up, the action of a matrix is to move the entire grid. We can understand it by thinking about how it moves the unit vectors. We can visualize what this looks like by drawing a modified 2D grid.
These ideas extend into three dimensions as well. The third row of the matrix contains z-coordinates for all the unit vectors, and the third column of the matrix tells us where start color #1fab54, k, with, hat, on top, end color #1fab54 lands.
If you'd like, play around with matrices as movement with this interactive demonstration. Drag the vectors to make the grid move, and see the matrix that corresponds to the movement in the top left corner.
## How matrices move vectors
We already know how a given matrix moves the unit vectors start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c (just look at the columns), but how can we find where a matrix moves any arbitrary vector? Let's consider a specific example using the first matrix from the previous section.
$A = \left[ \begin{array}{cc} \blueD{1} & \maroonD{0} \\ \blueD{1} & \maroonD{1} \end{array} \right]$
How does A move the non-unit vector start color #e07d10, left parenthesis, 1, comma, 2, right parenthesis, end color #e07d10? Before anything, let's get a feel for this visually. First, the vector with no matrix movement:
Now, the vector after the matrix moves the grid:
The vector just comes along for the ride as the matrix moves the grid, ultimately landing on start color #e07d10, left parenthesis, 1, comma, 3, right parenthesis, end color #e07d10. This is the essence of how matrices move vectors, which is formally called matrix-vector multiplication.
Now let's go over how we could calculate this. We represent start color #e07d10, left parenthesis, 1, comma, 2, right parenthesis, end color #e07d10 as a combination of the unit vectors by saying start color #e07d10, left parenthesis, 1, comma, 2, right parenthesis, end color #e07d10, equals, start color #e07d10, 1, end color #e07d10, start color #11accd, \imath, with, hat, on top, end color #11accd, plus, start color #e07d10, 2, end color #e07d10, start color #ca337c, \jmath, with, hat, on top, end color #ca337c.
This combination remains the same after we apply A, but instead of using start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, we use the result of applying A to start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c:
Here's what the whole process looks like in symbols.
\begin{aligned} A \left[ \begin{array}{c} \goldD{1} \\ \goldD{2} \end{array} \right] &= A\left( \goldD{1} \left[ \begin{array}{c} \blueD{1} \\ \blueD{0} \end{array} \right] + \goldD{2} \left[ \begin{array}{c} \maroonD{0} \\ \maroonD{1} \end{array} \right] \right) \\ \\ &= A\left( \goldD{1} \blueD{\hat{\imath}} + \goldD{2} \maroonD{\hat{\jmath}} \right) \\ \\ &= \goldD{1} A \blueD{\hat{\imath}} + \goldD{2} A \maroonD{\hat{\jmath}} \\ \\ &= \goldD{1} \left[ \begin{array}{c} \blueD{1} \\ \blueD{1} \end{array} \right] + \goldD{2} \left[ \begin{array}{c} \maroonD{0} \\ \maroonD{1} \end{array} \right] \\ \\ &= \left[ \begin{array}{c} 1 \\ 3 \end{array} \right] \end{aligned}
The critical step is when we break A, left parenthesis, start color #e07d10, 1, end color #e07d10, start color #11accd, \imath, with, hat, on top, end color #11accd, plus, start color #e07d10, 2, end color #e07d10, start color #ca337c, \jmath, with, hat, on top, end color #ca337c, right parenthesis into start color #e07d10, 1, end color #e07d10, A, start color #11accd, \imath, with, hat, on top, end color #11accd, plus, start color #e07d10, 2, end color #e07d10, A, start color #ca337c, \jmath, with, hat, on top, end color #ca337c. That is when we are able to represent where start color #e07d10, left parenthesis, 1, comma, 2, right parenthesis, end color #e07d10 lands in terms of where start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c land.
Let's try a couple practice questions.
Problem 2
Let $B = \left[ \begin{array}{cc} \blueD{0} & \maroonD{2} \\ \blueD{1} & \maroonD{-1} \end{array} \right]$.
Where does B move the vector $\left[ \begin{array}{c} 3 \\ 1 \end{array} \right]$ in terms of where it moves start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c?
Problem 3
Let $B = \left[ \begin{array}{cc} \blueD{0} & \maroonD{2} \\ \blueD{1} & \maroonD{-1} \end{array} \right]$.
Where does B move the vector $\left[ \begin{array}{c} 3 \\ 1 \end{array} \right]$?
$B \left[ \begin{array}{c} 3 \\ 1 \end{array} \right] = ($
comma
right parenthesis
Like in the section above, the idea behind matrices moving vectors extends into three dimensions. We just decompose our vector into a sum of start color #11accd, \imath, with, hat, on top, end color #11accd, start color #ca337c, \jmath, with, hat, on top, end color #ca337c, and start color #1fab54, k, with, hat, on top, end color #1fab54, then we use where the matrix takes all these unit vectors to find where it takes our vector.
Here's another interactive demonstration to play around with matrices moving vectors.
To learn more about matrix-vector multiplication, check out this video. To go deeper, try these videos from linear algebra.
## Matrix multiplication intuition (optional)
With the perspective of matrices as movement, we have the tools to understand what it means to multiply two matrices. The core idea is composition.
\begin{aligned} A &= \left[ \begin{array}{cc} \blueD{1} & \maroonD{1} \\ \blueD{0} & \maroonD{1} \end{array} \right] \\ \\ B &= \left[ \begin{array}{cc} \blueD{0} & \maroonD{-1} \\ \blueD{1} & \maroonD{0} \end{array} \right] \end{aligned}
The product A, B just means apply B, then apply A. When we apply A second, we treat the transformed start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c as regular vectors getting moved by A the way we learned in the previous section.
Created with this demonstration. Play around if you are interested!
To calculate the end result, we follow start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c along the two movements. First, B takes start color #11accd, \imath, with, hat, on top, end color #11accd, \to, start color #11accd, left parenthesis, 0, comma, 1, end color #11accd, right parenthesis and start color #ca337c, \jmath, with, hat, on top, end color #ca337c, \to, start color #ca337c, left parenthesis, minus, 1, comma, 0, right parenthesis, end color #ca337c. Second, we find where A takes these vectors:
\begin{aligned} \left[ \begin{array}{cc} \blueD{1} & \maroonD{1} \\ \blueD{0} & \maroonD{1} \end{array} \right] \left[ \begin{array}{c} \blueD{0} \\ \blueD{1} \end{array} \right] &= \left[ \begin{array}{c} \blueD{1} \\ \blueD{1} \end{array} \right] \\ \\ \left[ \begin{array}{cc} \blueD{1} & \maroonD{1} \\ \blueD{0} & \maroonD{1} \end{array} \right] \left[ \begin{array}{c} \maroonD{-1} \\ \maroonD{0} \end{array} \right] &= \left[ \begin{array}{c} \maroonD{-1} \\ \maroonD{0} \end{array} \right] \end{aligned}
Putting these in a matrix, we have the product. Notice that our calculations are reflected in the visual above.
$\left[ \begin{array}{cc} \blueD{1} & \maroonD{1} \\ \blueD{0} & \maroonD{1} \end{array} \right] \left[ \begin{array}{cc} \blueD{0} & \maroonD{-1} \\ \blueD{1} & \maroonD{0} \end{array} \right] = \left[ \begin{array}{cc} \blueD{1} & \maroonD{-1} \\ \blueD{1} & \maroonD{0} \end{array} \right]$
To conclude, we can think of matrix multiplication as composing the movements each matrix represents. When we follow unit vectors along these movements, we can calculate the product.
As a challenge problem, try to derive the general formula for 2, times, 2 matrix multiplication. Hint: follow start color #11accd, \imath, with, hat, on top, end color #11accd and start color #ca337c, \jmath, with, hat, on top, end color #ca337c as B moves them, then follow where A moves the already transformed vectors.
As a bonus challenge problem, try to find the formula for multiplying 3, times, 3 matrices. Hint: follow start color #11accd, \imath, with, hat, on top, end color #11accd, start color #ca337c, \jmath, with, hat, on top, end color #ca337c, and start color #1fab54, k, with, hat, on top, end color #1fab54.
To learn more about matrix multiplication, check out this video. For practice, try this exercise.
## What's next
Now that we have a solid grasp of how matrices move space, we are in a prime position to understand the final concept that we'll cover in this review series: the determinant.
## Want to join the conversation?
• does the explanation for a general formula of 2x2 matrix multiplication skip steps of exzplanation?
for example,
"first we follow i-hat."
---> [1,0], sure
"the rightmost matrix moves it to (e,g)"
---> i agree that [e,g] moves [1,0] to [e,g]
"when the left matrix moves (e,g), we can think of it as moving the combination of e[1,0] + g[0,1]
---> why can we think of it like that? where did [0,1] come from? why does it look like j-hat is now in the picture when we agreed to only follow i-hat? | 2023-01-31T11:57:19 | {
"domain": "khanacademy.org",
"url": "https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/x786f2022:vectors-and-matrices/a/matrices--visualized-mvc",
"openwebmath_score": 0.9914681315422058,
"openwebmath_perplexity": 7323.333143020888,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9919380075184121,
"lm_q2_score": 0.8539127492339909,
"lm_q1q2_score": 0.8470285110697343
} |
https://math.stackexchange.com/questions/1989900/6-fair-coin-flips-probability-of-exactly-3-heads | # 6 fair coin flips: probability of exactly 3 heads
When a certain coin is flipped, the probability of heads is $0.5$. If the coin is flipped $6$ times, what is the probability that there are exactly $3$ heads?
The answer is $\frac5{16}$. I wonder why it isn't $\frac12$. Since a fair coin flip results in equally likely outcomes, any sequence is equally likely…
I know why it is $\frac5{16}$. We divide the number of possible outcomes with exactly 3 heads by the total possible outcomes.
What bothers me is how I should think about it so I won't make a mistake anymore. Why is $\frac12$ not right? I need to get intuition.
• Above all, though, I would advise you to do more probability problems. You learn mathematics by doing mathematics. Oct 29, 2016 at 4:50
• Suppose you flip the coin $5$ times. What is the probability that there are exactly $2.5$ heads? Oct 29, 2016 at 5:34
It sounds like you already have the intuition since you understand that the answer is obtained by dividing the number of outcomes with exactly 3 heads by the total number of outcomes. From here it's a matter of understanding how to calculate these two things.
The total number of outcomes is simply $2^6 = 64$ since we're tossing a coin 6 times and each toss has only two possible outcomes.
The number of outcomes with exactly 3 heads is given by ${6 \choose 3}$ because we essentially want to know how many different ways we can take exactly 3 things from a total of 6 things. The value of this is 20.
So the answer is $20/64 = 5/16$.
The error you made is thinking that "number of outcomes with exactly 3 heads" is equal to "half of the total number of outcomes of 6 tosses." If this were the case then logically, "exactly 3 tails" must also be exactly half of the total outcomes. This means that "exactly 3 heads or exactly 3 tails" must describe all possible outcomes (because each scenario joined by the "or" would have probability $1/2$) but this is clearly not the case since we can have, e.g., 1 heads and 5 tails, etc.
To put it another way... You said any sequence is equally likely. That is correct. But sequences containing exactly 3 heads do not make up half of the total number of sequences. Therefore it does not follow that the probability is $1/2$. The easiest way to see this clearly is to list every possible outcome. But for 64 outcomes it can be tedious, so let's do it with a simpler and similar problem.
Say we want to know the probability of getting exactly 2 heads if we flip a coin 4 times. Unless I'm misunderstanding your misunderstanding, your earlier thinking would lead you to believe the answer is $1/2$. But if we list all possible outcomes:
HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT
We see that only 6 of them have exactly 2 heads. $6/16 = 3/8$. And if we do this problem the way I answered the original one, then the total number of outcomes is $2^4 = 16$ and the total number of outcomes with exactly 2 heads is ${4 \choose 2} = 6$. So we again get $6/16 = 3/8$.
Your incorrect thinking goes along the lines of this classic fallacy:
What's the probability of this happening? $\frac12$, either it will happen or it won't.
It is true that each sequence of heads and tails is equally likely to occur – with probability $\frac1{64}$, in this case. However, the number of those sequences having exactly three heads is not 32, but $\binom63=20$, which leads to the correct answer of $\frac5{16}$. They are two completely different things.
"exactly three heads" is not half of the outcomes. Try writing the outcomes down and counting them!
The probability is $$P=\binom{6}{3}(\frac{1}{2})^3(\frac{1}{2})^{6-3}$$
Since a fair coin flip results in equally likely outcomes, any sequence is equally likely…
... then we need to know how many of these equally-probable outcomes are possible and how many of which have "exactly three heads".
There are $2^6$ ways to make six independent "head or tails" decisions. That is $64$.
There are $\binom 6 3$ ways to arrange three heads and three tails. That is $20$
Divide and calculate. $20/64 = 5/16$ | 2022-07-01T02:13:04 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1989900/6-fair-coin-flips-probability-of-exactly-3-heads",
"openwebmath_score": 0.9078378677368164,
"openwebmath_perplexity": 178.14155789416088,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9896718477853187,
"lm_q2_score": 0.8558511506439708,
"lm_q1q2_score": 0.8470117896870097
} |
http://mathhelpforum.com/advanced-statistics/132676-probability-distribution-random-variables.html | # Thread: Probability Distribution of Random Variables
1. ## Probability Distribution of Random Variables
Hi there,
Let me begin by introducing myself. My name is Zoheb and I'm a currently studying for a BSc Computer science degree in London (my hometown). I use to love maths (back in GCSE years) but then started to see maths as a burden. Strangely, I now enjoy the mathematical side of my degree more than anything else. But enjoying it doesn't mean I'm particularly good at it. I have a lot to learn, re-teach myself and revise...
In this process, I'm encountering many problems, some of which I solve myself and others, I don't. Often, it's because I've missed an important step or forgotten a rule, so I apologise for this question/problem as the solution may have been in front of my eyes, but even if it was, I haven't understood it, which is why I'm posting here...
This is not a homework query or anything, it is an example I encountered during the course of study, in Schaum's Outline: Discrete Mathematics by S. Lipschutz and M. Lipson. A solution to the problem is given of course, but I don't quite understand the steps that have been taken to reach it. Before I begin, I apologise if this query has been answered many a time before. I'm not really sure what this falls under so didn't know what to search or even name this thread. Here is my problem/confusion... I have paraphrased the problem but values and procedures remain the same...
Suppose we have a box which contains 12 items, 3 of which have defects. A sample of 3 items is selected from the box.
The sample space $S$ has $\binom{12}{3}=220$ samples of size 3.
Let $X$ denote the number of defective items selected, then $X$ is a random variable with range space $R{x}=\{0,1,2,3\}$
So far, everything is straight forward. But my question/problem is: what steps are to be taken in order to obtain the distribution of $X$? A solution is given (below) as I stated, but I haven't quite worked out why it has ben done this way... Can someone please explain the methodology that is taken here?
There are $\binom{9}{3}=84$ samples of size 3 with no defective items; hence $P(0)=\frac{84}{220}$.
There are $3\binom{9}{2}=108$ samples of size 3 containing 1 defective item; hence $P(1) = \frac{108}{220}$.
There are $\binom{3}{2}\cdot 9=27$samples of size 3 containing 2 defective items; hence $P(2)=\frac{27}{220}$.
Lastly, there is only 1 sample of size 3 containing all 3 defective items; hence $P(3)=\frac{1}{220}$.
I'm perfectly capable of calculating the probability if I'm given the binomal coefficients, but that's about it. I wouldn't know how to apply this method to a different question. Sorry, I hope my question is clear. Thank you all in advance.
EDIT: Looking over the problem again, I believe this should've been in the Discrete Maths sub-forum, so I apologise. Also, I believe the aspect I'm having trouble with is to do with the application of combinations and permutations. | 2016-12-06T19:16:21 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/advanced-statistics/132676-probability-distribution-random-variables.html",
"openwebmath_score": 0.6658095717430115,
"openwebmath_perplexity": 348.82012077944535,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9896718496130618,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8470117894319168
} |
https://es.mathworks.com/help/matlab/ref/log2.html | # log2
Base 2 logarithm and floating-point number dissection
## Description
example
Y = log2(X) computes the base 2 logarithm of the elements of X such that ${2}^{Y}=X$.
example
[F,E] = log2(X) returns arrays F and E such that $X=F\cdot {2}^{E}$. The values in F are typically in the range 0.5 <= abs(F) < 1.
## Examples
collapse all
X = [0 1 2 10 Inf NaN];
Y = log2(X)
Y = 1×6
-Inf 0 1.0000 3.3219 Inf NaN
Dissect several numbers into the exponent and mantissa. These operations all follow standard IEEE® arithmetic.
Create a vector X that contains several test values. Calculate the exponent and mantissa for each number.
X = [1 pi -3 eps realmax realmin];
format rat
[F,E] = log2(X)
F =
Columns 1 through 5
1/2 355/452 -3/4 1/2 1
Column 6
1/2
E =
Columns 1 through 5
1 2 2 -51 1024
Column 6
-1021
Collect the results in a table. Convert the numbers into character vectors for display purposes.
x = {'1','pi','-3','eps','realmax','realmin'}';
f = strtrim(cellstr(rats(F')));
T = table(x,f,E','VariableNames',{'Value','Mantissa','Exponent'})
T=6×3 table
Value Mantissa Exponent
___________ ___________ ________
{'1' } {'1/2' } 1
{'pi' } {'355/452'} 2
{'-3' } {'-3/4' } 2
{'eps' } {'1/2' } -51
{'realmax'} {'1' } 1024
{'realmin'} {'1/2' } -1021
The results indicate that, for the first row, $1=\frac{1}{2}\left({2}^{1}\right)$. Similarly, for the fourth row, $\mathrm{eps}=\frac{1}{2}\left({2}^{-51}\right)$.
## Input Arguments
collapse all
Input matrix, specified as a scalar, vector, matrix, or multidimensional array.
For floating-point number dissection [F,E] = log2(X), any zeros in X produce F = 0 and E = 0. Input values of Inf, -Inf, or NaN are returned unchanged in F with a corresponding exponent of E = 0.
Data Types: single | double
Complex Number Support: Yes
## Output Arguments
collapse all
Base 2 logarithm values, returned as a scalar, vector, matrix, or multidimensional array of the same size as X.
Mantissa values, returned as a scalar, vector, matrix, or multidimensional array of the same size as X. The values in F and E satisfy X = F.*2.^E.
Exponent values, returned as a scalar, vector, matrix, or multidimensional array of the same size as X. The values in F and E satisfy X = F.*2.^E.
## Tips
• This function corresponds to the ANSI® C function frexp() and the IEEE® floating-point standard function logb(). Any zeros in X produce F = 0 and E = 0. | 2021-09-17T08:26:48 | {
"domain": "mathworks.com",
"url": "https://es.mathworks.com/help/matlab/ref/log2.html",
"openwebmath_score": 0.8419343829154968,
"openwebmath_perplexity": 5367.676857681281,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9896718483945663,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8470117883890661
} |
https://math.stackexchange.com/questions/3182907/surface-area-of-a-sphere-dilemma | # Surface area of a sphere dilemma
I recently found that surface area of a sphere can't be found with the following method. What's the flaw in it? First, I have taken a very thin ring of thickness $$dx$$ at a distance of $$x$$ from the centre. Then, I integrated it, using substitution of $$x=R\sin\theta$$
But this is giving me answer $$4\pi^2R^2$$.
I also tried to solve many other problems related to area of sphere, like I found the magnetic field at the centre due to a revolving sphere carrying a charge $$Q$$, but this method is giving me wrong result.
The correct result is found by taking a thin ring subtending an angle $$2\theta$$ at centre, and thickness of the ring would be $$Rd\theta$$.
But, why my method is not giving the correct answer?
Sorry for formatting errors.
• The error was the thickness of the ring is not "dx." – user156213 Apr 10 at 22:21
• Why its not dx? – Yash Mittal Apr 11 at 17:51
• Because the surface is curved. – user156213 Apr 12 at 16:38
For a ring at angle $$\theta$$, we have Radius = $$R \sin\theta$$
The area of this ring is the circumference times its thickness. The thickness is $$R \cdot d\theta$$ $$dA = 2\pi \cdot (R \sin\theta) \cdot R d\theta$$
$$\implies A = 2\pi R^2\int_{0}^{\pi}\sin\theta\ d\theta$$
$$= 2\pi R^2(-\cos\theta)\big|^{\pi}_0$$ $$\implies \boxed{A = 4\pi R^2}$$
If you want to go by the $$dx$$ route, note that the thickness of the ring will be $$\frac{dx}{\cos{(90-\theta)}} = \frac{dx}{\sin\theta}$$
(How? Consider the tangent line from a point on the ring to the X-axis)
$$\implies dA = 2\pi\cdot(R\sin\theta)\cdot\left(\frac{dx}{\sin\theta}\right) = 2\pi R\ dx$$ $$A = \int_{-R}^{R}2\pi R\ dx = 2\pi R x\bigg|^R_{-R} = 4\pi R^2$$
• I am asking, why the other method is incorrect? – Yash Mittal Apr 11 at 17:51
• @YashMittal that is because $dx$ is not the right thickness of the ring. If you post your entire derivation, I can take a look at the incorrect areas. There could be more than one but the thickness is one of them. – user1952500 Apr 11 at 17:52
• @YashMittal I have added an addendum using $dx$ – user1952500 Apr 11 at 18:00
• But what is theta here? I have not taken any theta in first method. – Yash Mittal Apr 12 at 10:28
• @YashMittal what is the radius of the ring at distance $x$? – user1952500 Apr 12 at 17:04 | 2019-06-18T06:59:28 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3182907/surface-area-of-a-sphere-dilemma",
"openwebmath_score": 0.6810954809188843,
"openwebmath_perplexity": 278.6534775035876,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713870152409,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8470020993740857
} |
https://math.stackexchange.com/questions/815418/why-does-sqrtn-sqrtn-sqrtn-ldots-n/815430 | # Why does $\sqrt{n\sqrt{n\sqrt{n \ldots}}} = n$?
Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the
nth x = √(1st x √ 2nd x ... √nth x);
Then
$$\text{the "infinith" } x = x$$
Example:
$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).
I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.
However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.
• What do you mean by nth x and $\sqrt(1st \; x$? May 30, 2014 at 22:01
• oh, i wasn't allowed to write in subscript May 30, 2014 at 22:03
• May 31, 2014 at 2:48
• Because $\sqrt{n.n}=n.$
– user65203
Jun 3, 2014 at 7:42
\begin{align} X & = \sqrt{n \cdot \sqrt {n \cdot \sqrt{n \dots} } } \\ & = \sqrt{n} \cdot \sqrt{\sqrt{n}} \cdot \sqrt{\sqrt{\sqrt{n}}} \cdot \dots \\ &= n^{1/2} \cdot n^{1/4} \cdot n^{1/8} \dots \\ &= n^{1/2 + 1/4 + 1/8 \dots} \\ &= n^1 \\ &= n \\ \end{align}
• Oh, this way I understand the most, because it is basically rewriting the 1/2 + 1/4 + 1/8... sequence as exponents, I just looked at it as a radical. May 30, 2014 at 22:07
• Wow, this is brilliant. May 7, 2015 at 19:57
Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$.
Multiply both sides by $x$ and take the square root:
$$\sqrt{xy} = \sqrt{x\sqrt{x\sqrt{x\cdots}}} = y$$
Therefore, $\sqrt{xy} = y$, and solving we have $xy = y^2 \implies x = y$.
• Mh, be careful with that reasoning. $X = 1-1+1-1+1-1+...$, so $X=1-(1+1-1+1-...) = 1-X$, therefore $X=1/2$ May 30, 2014 at 22:06
• You are begging the question—which is fine, but it should be a little more explicit. The point is to suppose that $\sqrt{x \sqrt{x \sqrt{x \cdots}}}$ has a definite value $y$, then compute what it would have to be. May 30, 2014 at 22:15
• AnalysisStudent0414's reasoning actually computes the Cesàro sum. These kinds of questions are always tied up with giving the definition of your expression. May 30, 2014 at 22:17
• @Arkamis Is there an easy way to exclude $y=0$ as a solution, apart from the form of DanielVs answer May 30, 2014 at 22:30
• @you-sir-33433 This is a site to answer people's questions, not to slowly assemble a textbook. I simply answered his question. He doesn't understand convergence; is this the appropriate place to show that $1/2^n$ converges, as well? Should we begin with the definition of convergence of infinite series? What's the right place to begin? May 30, 2014 at 23:01
It is important to show that the limit exists. Let define the sequence $$a_k=\sqrt{\vphantom{A}na_{k-1}}$$ Since $\dfrac{a_k}{a_{k-1}}=\sqrt{\dfrac{n}{a_{k-1}}}$ and $\dfrac{a_k}{n}=\sqrt{\dfrac{a_{k-1}}{n}}$, we have
1. if $a_{k-1}\le n$, then $a_{k-1}\le a_k\le n$; that is, $a_k$ is increasing and bounded above by $n$.
2. if $a_{k-1}\ge n$, then $a_{k-1}\ge a_k\ge n$; that is, $a_k$ is decreasing and bounded below by $n$.
In either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $$\lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{\vphantom{A}na_{k-1}}=\sqrt{n\lim_{k\to\infty}a_k}$$ Squaring and dividing by $\lim_{k\to\infty}a_k$, we get that $$\lim_{k\to\infty}a_k=n$$
Another Approach
Not as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get \begin{align} \log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right) &=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\ &=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\ &=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt] &=\log(n) \end{align}
• Convergence is even easier, really. Let $f_n$ be $n$ square root operations, e.g. $f_2 = \sqrt{x\sqrt{x}}$. Then, for $n >m$,$f_n-f_m = \sqrt{x\sqrt{x\cdots x f_m}}-f_m = k_{n-m}f_m^{1/2^{n-m}}-f_m$ which is Cauchy. May 30, 2014 at 23:08
• I might have made a few algebra mistakes along the way... editing in the comment box on a laptop sucks. May 30, 2014 at 23:09
• @Arkamis: I assume that $k_{n-m}$ is supposed to be $f_{n-m}$. However, perhaps I am missing something, but I don't see how $f_n-f_m=f_{n-m}f_m^{2^{m-n}}-f_m$ shows that $f_n$ is Cauchy. In any case, this, or another justification, should be in an answer.
– robjohn
May 31, 2014 at 0:57
– robjohn
May 31, 2014 at 0:59
• The k is the x part of the expansion, after factoring out n -m fs. May 31, 2014 at 1:03
You're basically doing this:
$x_0 = \sqrt x$
$x_1 = \sqrt{x x_0}$
$\displaystyle x_n = \sqrt{x x_{n-1}} = x^{\sum_{k=1}^n \frac{1}{2^k}}$
So it's pretty obvious that converges to $x$
• The limits of the sum in your last line don't look right. Do you mean $x^{\sum_{k=1}^n{\frac{1}{2^k}}}$? May 30, 2014 at 22:09
• Yup. Changed sentence mid-way, thanks a lot May 30, 2014 at 22:16
Assume that you iterated infinitely many times (can take a while), and observed a convergence to $n$.
One more iteration yields $\sqrt{n.n}=n$.
• This is the steady state approach, assume $a_k$ = $a_{k-1}$ and solving for that value, nice application. Jun 2, 2014 at 16:53
• There are two approaches: knowing the limit, show that it is the limit; or not knowing the limit, find it.
– user65203
Jun 2, 2014 at 18:03
Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$. Then obviously $y^2=xy$, whence $y=x$ or $y=0$. But $y \gt 0$, so $y=x$. | 2022-09-30T18:32:51 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/815418/why-does-sqrtn-sqrtn-sqrtn-ldots-n/815430",
"openwebmath_score": 0.9982094764709473,
"openwebmath_perplexity": 520.7710378927734,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713831229043,
"lm_q2_score": 0.8596637541053281,
"lm_q1q2_score": 0.8470020960279849
} |
https://math.stackexchange.com/questions/3004081/m-is-a-point-in-an-equalateral-abc-of-area-s-s-is-the-area-of-the-tria | # $M$ is a point in an equalateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.
$$M$$ is a point in an equilateral triangle $$ABC$$ with the area $$S$$. Prove that $$MA, MB, MC$$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$
We can assume that side of a triangle $$ABC$$ is $$1$$. Further let $$CM =x$$ and $$\angle KCM =\gamma$$. Rotate $$M$$ around $$C$$ for $$-60^{\circ}$$ in to $$F$$. Then the area of triangle $$AMF$$ is the one we are looking for and it's area is area of $$AMCF$$ minus area of equilateral triangle $$CFM$$, so $$4S' = -x^2\sqrt{3}+2x\sin (60^{\circ}+\gamma)$$ and this should be easy to calculate that is less than $${\sqrt{3}\over 3}$$.
If we see $$S'$$ quadratic function on $$x$$ we get: $$4S'\leq {1\over \sqrt{3}}\sin (60^{\circ}+\gamma)\leq {1\over \sqrt{3}}$$ From here we can see that equality is achieved iff $$\gamma = 30^{\circ}$$ and $$x= {\sqrt{3}\over 3} = {2\over 3}v$$ where $$v$$ is altitude of triangle $$ABC$$. That is, equality is achieved iff $$M$$ is gravity centre of $$ABC$$.
I'm interested in different solutions (for example without trigonometry).
• This problem has been posted here before. Unfortunately, it was closed and deleted. But high-reputation users like you can access it: math.stackexchange.com/questions/2993965. (But all answers there use trigonometry or complex numbers to solve the problem.) Nov 18 '18 at 20:55
• Then, I will do this. @achillehui, could you please copy and paste your solution from that link here? I could do it, but I want the upvotes to go to you. Nov 18 '18 at 20:57
• @QuangHoang Could you also please copy and paste your solution from that link here. Nov 18 '18 at 20:58
• I have no problem with that, I will also upvote it. It is nonsense to delete such a solution with artificial reasons.
– Aqua
Nov 18 '18 at 20:59
• @Batominovski answer copied. Nov 18 '18 at 21:34
Reflect $$M$$ with respect to the sides of $$ABC$$. You get an hexagon whose area is $$2S$$: The hexagon can be decomposed as the union between $$A'B'C'$$ (whose side lengths are $$\sqrt{3}MA,\sqrt{3}MB,\sqrt{3}MC$$) and the isosceles triangles $$CA'B',BC'A',AB'C'$$. It follows that $$2S = \frac{\sqrt{3}}{4}(AM^2+BM^2+CM^2)+ 3 S'$$ where $$S'$$ is the area of a triangle with side lengths $$MA,MB,MC$$. By Weitzenbock's inequality $$AM^2+BM^2+CM^2 \geq 4\sqrt{3}S'$$, hence $$S'\leq \frac{S}{3}$$ as wanted.
Solution using complex numbers. Copied from a deleted question per request.
( update - I have added another solution using circle inversion at end)
Solution 1 - using complex numbers.
Choose a coordinate system so that triangle $$ABC$$ is lying on the unit circle centered at origin and $$A$$ on the $$x$$-axis. Let $$a = AM$$, $$b = BM$$, $$c = CM$$ and $$S'$$ be the area of a triangle with sides $$a,b,c$$. In this coordinate system, $$S = \frac{3\sqrt{3}}{4}$$, we want to show $$S' \le \frac{\sqrt{3}}{4}$$. Using Heron's formula, this is equivalent to
$$16S'^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) \stackrel{?}{\le} 3$$
Identify euclidean plane with the complex plane. The vertices $$A,B,C$$ corresponds to $$1, \omega, \omega^2 \in \mathbb{C}$$ where $$\omega = e^{\frac{2\pi}{3}i}$$ is the cubic root of unity. Let $$z$$ be the complex number corresponds to $$M$$ and $$\rho = |z|$$, we have
$$\begin{cases} a^2 = |z-1|^2 = \rho^2 + 1 - (z + \bar{z})\\ b^2 = |z-\omega|^2 = \rho^2 + 1 - (z\omega + \bar{z}\omega^2)\\ c^2 = |z-\omega^2|^2 = \rho^2 + 1 - (z\omega^2 + \bar{z}\omega) \end{cases} \implies a^2 + b^2 + c^2 = 3(\rho^2 + 1)$$ Thanks to the identity $$\omega^2 + \omega + 1 = 0$$, all cross terms involving $$\omega$$ explicitly get canceled out.
Doing the same thing to $$a^4 + b^4 + c^4$$, we get \begin{align}a^4 + b^4 + c^4 &= \sum_{k=0}^2 (\rho^2 + 1 + (z\omega^k + \bar{z}\omega^{-k}))^2\\ &= \sum_{k=0}^2\left[ (\rho^2 + 1)^2 + (z\omega^k + \bar{z}\omega^{-k})^2\right]\\ &= 3(\rho^2 + 1)^2 + 6\rho^2\end{align} Combine these, we obtain
$$16S'^2 = 3(\rho^2+1)^2 - 12\rho^2 = 3(1 - \rho^2)^2$$ Since $$M$$ is inside triangle $$ABC$$, we have $$\rho^2 \le 1$$. As a result,
$$S' = \frac{\sqrt{3}}{4}(1-\rho^2) \le \frac{\sqrt{3}}{4} = \frac13 S$$
Solution 2 - using circle inversion.
Let $$a = AM, b = BM, c = CM$$ again. Let $$\Delta(u,v,w)$$ be the area of a triangle with sides $$u,v,w$$. In particular, $$S = \Delta(1,1,1)$$ and $$S' = \Delta(a,b,c)$$. We will use the fact $$\Delta(u,v,w)$$ is homogeneous in $$u,v,w$$ with degree $$2$$.
Consider the circle inversion with respect to a unit circle centered at $$A$$. Under such an inversion, $$B,C$$ get mapped to itself while $$M$$ mapped to a point $$M'$$ with $$AM' = \frac{1}{a}, BM' = \frac{b}{a}, CM' = \frac{c}{a}$$
We can decompose the quadrilateral $$ABM'C$$ in two manners. $$\triangle ABC + \triangle BM'C$$ and $$\triangle ABM' + \triangle AM'C$$. This leads to
\begin{align} &\verb/Area/(ABC) + \verb/Area/(BM'C) = \verb/Area/(ABM') + \verb/Area/(AM'C)\\ \iff & S + \Delta(1,\frac{b}{a},\frac{c}{a}) = \Delta(1,\frac{b}{a},\frac{1}{a}) + \Delta(1,\frac{c}{a},\frac{1}{a})\\ \iff & Sa^2 + S' = \Delta(1,a,b) + \Delta(1,a,c) \end{align} By a similar argument, we have $$Sb^2 + S' = \Delta(1,b,a) + \Delta(1,b,c)\quad\text{ and }\quad Sc^2 + S' = \Delta(1,c,a) + \Delta(1,c,b)$$
Summing these three equalities together and notice
$$\Delta(1,a,b) + \Delta(1,b,c) + \Delta(1,c,a) = \verb/Area/(ABM) + \verb/Area/(BCM) + \verb/Area/(CAM) = S$$
We obtain
$$3S' = 2S - S(a^2+b^2+c^2)$$
For any $$\triangle XYZ$$ and point $$P$$ in the plane, we know the expression $$XP^2 + YP^2 + ZP^2$$ is minimized when $$P$$ is the centroid of $$\triangle XYZ$$. For an equilateral triangle of side $$1$$, the centroid is at a distance $$\frac{1}{\sqrt{3}}$$ from the vertices. This implies $$a^2 + b^2 + c^2 \ge 1$$.
As a result, $$3S' \le S$$ and we are done.
• As said...........+1
– Aqua
Nov 18 '18 at 21:42
• @greedoid Thanks, I haved added a new approach which use circle inversion instead of complex numbers. Nov 18 '18 at 22:24 | 2021-09-23T04:56:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3004081/m-is-a-point-in-an-equalateral-abc-of-area-s-s-is-the-area-of-the-tria",
"openwebmath_score": 0.955951452255249,
"openwebmath_perplexity": 183.64810865693417,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713887451681,
"lm_q2_score": 0.8596637487122111,
"lm_q1q2_score": 0.8470020955475573
} |
http://math.stackexchange.com/questions/33745/drawing-a-card-from-a-deck | Drawing a card from a deck
A single card is drawn from a standard 52-deck of cards with four suits: hearts, clubs, diamonds, and spades; there are 13 cards per suit. If each suit has three face cards, how many ways could the drawn card be either a club of any kind or anything else besides a face card?
-
Hint: How many non-face cards are there? How many clubs are there? How any cards are both non-face cards and clubs? – JavaMan Apr 19 '11 at 1:36
this question sounds like a homework problem and if this is the case you may want to consider using the homework tag for this question. Also, when asking a question like this, its a good policy to outline your thoughts about the problem so that answerers can see where you got stuck and can thus provide better responses. – WWright Apr 19 '11 at 1:36
I am assuming that n=52 and r=30, order doesn't matter so it's a combination???? – user9762 Apr 19 '11 at 2:32
Paula, you might also want to delete the other (~identical) copy of this question. – The Chaz 2.0 Apr 19 '11 at 2:38
The total number of possible outcomes is $52$, because that's the number of cards: you have one outcome per possible cards.
How many outcomes are "good" (that is, are "either a club of any kind or anything else besides a face card")?
Well, there's one "good outcome" per club; that's 13 clubs.
Of the remaining 39 cards, how many are also "good outcomes"?
Out of the 13 hearts there are, ten hearts (all hearts except the three face cards) count as "good outcomes." So in addition to the 13 clubs, 10 hearts are also "good."
How many spades are "good"? How many diamonds?
Now just add them all up.
-
I don't think she is worried about probability here, just the "number of good outcomes". – JavaMan Apr 19 '11 at 3:04
I feel like a total idiot. I made this problem much more difficult than what is there. Thanks DJC. I appreciate it. – user9762 Apr 19 '11 at 3:05
@DJC: Oops. Quite right. – Arturo Magidin Apr 19 '11 at 3:12
@Paula: No need to feel down. Talking about math with other people is useful! – JavaMan Apr 19 '11 at 3:15
I have a feeling your teacher/professor intended for you to learn the following (we first introduce some notation):
For two sets $A, B$ in a common universe $U$, define their union as
$$A \cup B = \{x \in U : x \in A \text { or } x \in B\}.$$
Define their intersection as:
$$A \cap B = \{x \in U : x \in A \text{ and } x \in B\}.$$
It is important not to get too wrapped in the English here. Being in $A \cup B$ simply means being in one of the two sets (or possibly both). Being in $A \cap B$ simply means being in set $A$ and being in set $B$ at the same time.
Finally, a finite set $A$ with $k$ elements ($k$ things in the set) has cardinality $k$, and this is written as $|A| = k$, or $\#A = k$ or even sometimes, $n(A) = k$.
Therefore:
$$\{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\},$$
while
$$\{1 ,3 , 5 \} \cap \{ 1 , 2, 3 \} = \{ 1, 3\}.$$
Also, $$| \{ 1,3,5\}| = 3,$$
while
$$| \{ 1, 3\} | = 2.$$
Now, what your teacher probably wanted you to learn was the following "rule":
$$|A \cup B| = |A| + |B| - |A \cap B|.$$
This is easy to see it is true, since to count the number of elements that are in either $A$ or $B$ (or possibly both $A$ and $B$), you count the number of elements in $A$, add to it the number of elements in $B$, and then subtract the stuff you double counted, which is precisely the elements in $A \cap B$.
Therefore, if you want to find the number of sides of a die that are (say) even or prime, you count the number of sides which are even (there are $3$ such sides - namely $2$ , $4$, $6$) add to it the number of sides which are prime (again, there are $3$ such sides - namely $2$, $3$, $5$), and then subtract the sides which we double counted (we counted the side with the number $2$ twice).
Therefore, there are $3 + 3 - 1 = 5$ sides of a die which are even or prime.
Now, you can take this strategy and count the number of cards in a deck which are either non face cards or clubs.
- | 2015-08-04T06:03:29 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/33745/drawing-a-card-from-a-deck",
"openwebmath_score": 0.6764525175094604,
"openwebmath_perplexity": 558.9648161284028,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9852713878802044,
"lm_q2_score": 0.8596637487122111,
"lm_q1q2_score": 0.8470020948039795
} |
https://www.physicsforums.com/threads/finding-the-nth-term-of-fibonacci-like-sequence.713515/ | # Finding the n'th term of fibonacci like sequence
1. Sep 30, 2013
### neerajareen
The fibonacci sequence can be defined as $${F_n} = {F_{n - 1}} + {F_{n - 2}}$$ and specifying the initial conditions as \eqalign{ & {F_1} = 1 \cr & {F_2} = 1 \cr}
Also there exists a general formula for the fibonacci which is given by $${F_n} = {{{\varphi ^n} + {\psi ^n}} \over {\sqrt 5 }}$$
Where $$\varphi = {{1 + \sqrt 5 } \over 2}$$ and $$\psi = {{1 - \sqrt 5 } \over 2}$$ .
Ι understand that the derivation of the formula was obtained using generating functions.
My question is that supposing we have a new sequence defined to be $${G_n} = {G_{n - 1}} + {G_{n - 3}}$$ and the initial conditions being:
\eqalign{ & {G_1} = 1 \cr & {G_2} = 1 \cr & {G_3} = 1 \cr}
What’s the formula for the n’th term in the sequence. The generating function for the function is going to be
$$A(x) = {x \over {1 - x - {x^3}}}$$
. But due to the imaginary roots that are present, I am not able to use the method of partial fractions.
Can this method still be used or is there any other method that we can use to get the n’th number of this sequence?
2. Sep 30, 2013
### jfizzix
This kind of problem can also be solved with linear algebra. I'll demonstrate with the Fibonnacci problem.
Let $F_{n}$ be the nth term in the Fibonnacci series so that
$F_{n}=F_{n-1}+F_{n-2}$.
Using the identity
$F_{n-1}=F_{n-1}$,
we can express the iteration from one term to the next as a matrix multiplication
$\pmatrix{ F_{n} \cr F_{n-1} \cr} =\pmatrix{ 1 & 1 \cr 1 & 0 \cr}\pmatrix{ F_{n-1} \cr F_{n-2} \cr}=\pmatrix{ 1 & 1 \cr 1 & 0 \cr}^{2}\pmatrix{ F_{n-2} \cr F_{n-3} \cr}=\pmatrix{ 1 & 1 \cr 1 & 0 \cr}^{n-2}\pmatrix{ F_{2} \cr F_{1} \cr}=\pmatrix{ 1 & 1 \cr 1 & 0 \cr}^{n-2}\pmatrix{ 1 \cr 1 \cr}$
Solving for $F_{n}$ then boils down to figuring out a formula for the nth power of a 2x2 matrix.
Let $A=\pmatrix{ 1 & 1 \cr 1 & 0 \cr}$
The eigenvalues of $A$ are \eqalign{ & {\lambda_1} = \frac{1+\sqrt{5}}{2} \cr & {\lambda_2} = \frac{1-\sqrt{5}}{2} \cr}
We can diagonalize $A$ , so that it has the form
$A= U \Lambda U^{-1}$
where $\Lambda=\pmatrix{ \lambda_{1} & 0 \cr 0 & \lambda_{2} \cr}$ is a diagonal matrix,
$\Lambda^{n}=\pmatrix{ \lambda_{1}^{n} & 0 \cr 0 & \lambda_{2}^{n} \cr}$.
Since $U U^{-1} = U^{-1} U = I$,
we have that
$A^{n}= U \Lambda^{n} U^{-1}$
Knowing that you get $U$ from the diagonalization, you have all the information you need to complete the matrix multiplication to find $F_{n}$.
I believe you will find this procedure to work nicely for your problem, though it's a little more challenging to diagonalize a 3x3 matrix.
3. Sep 30, 2013
### Office_Shredder
Staff Emeritus
Another method that has a less challenging end step to do by hand is the following (again for the Fibonnaci sequence). Suppose that Hn = xn for some x, and Hn = Hn-1 + Hn-2. Then
$$x^n = x^{n-1} + x^{n-2}$$
for all n. Divide both sides by xn-2 and we get
$$x^2 = x+1$$
$$x^2 - x -1 = 0$$
which of course has roots $(1\pm \sqrt{5})/2$. The relation Hn = Hn-1 + Hn-2 is linear, i.e. if Xn and Yn satisfy it, and a and b are numbers, aXn+bYn satisfies it as well. So Hn can be anything of the form
$$H_n = a \left(\frac{1+\sqrt{5}}{2}\right)^n + b\left(\frac{1-\sqrt{5}}{2} \right)^n$$
where a and b are any numbers. If we want Hn to be the Fibonacci sequence, then H1 = H2 = 1 - these give you two equations and two unknowns (a and b) which allow you to solve for a and b, and gives you that formula for Fn (actually I think your formula is missing a minus sign).
You can do the same procedure for your Gn - you'll get three roots because you will be left with a cubic equation, so you'll have three unknowns which makes sense because the first three values of Gn will have to be specified. Unless you have a closed form expression for your roots though you will have some floating point issues if you try to use this to evaluate Gn for really large values of n.
4. Oct 1, 2013
### jfizzix
Maybe diagonalizing a 3x3 matrix in this case wouldn't be so bad.
For this problem we diagonalize the matrix
$B=\pmatrix{ 1 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr}$
Which comes into play as:
$\pmatrix{ G_{n} \cr G_{n-1} \cr G_{n-2} \cr} = \pmatrix{ 1 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr}\pmatrix{ G_{n-1} \cr G_{n-2} \cr G_{n-3} \cr} =\pmatrix{ 1 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr}^{n-3}\pmatrix{ G_{3} \cr G_{2} \cr G_{1} \cr}=\pmatrix{ 1 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr}^{n-3}\pmatrix{ 1 \cr 1 \cr 1 \cr}$
but then, you'd still have to solve a cubic equation
$\lambda^{3}-\lambda^{2}-1=0$
to find the eigenvalues. There are no rational roots to this equation, so solving it on paper without a copy of the cubic formula would be a challenge. A fun challenge. | 2017-12-12T07:40:03 | {
"domain": "physicsforums.com",
"url": "https://www.physicsforums.com/threads/finding-the-nth-term-of-fibonacci-like-sequence.713515/",
"openwebmath_score": 0.8789997696876526,
"openwebmath_perplexity": 367.1163198866665,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850872288504,
"lm_q2_score": 0.8615382147637196,
"lm_q1q2_score": 0.8469653710119793
} |
https://brilliant.org/discussions/thread/recurring-decimals-as-fractions/ | # Recurring decimals as fractions
Earlier I solved this question and it got me thinking.
$\text{Why do recurring decimals have a fractional equivalent?}$
So I decided to work it out, this is what I got.
Let $$b$$ be a number with $$n$$ digits, it can have both leading and trailing zeros.
Every recurring decimal can now be represented by the decimal : $$0.bbb\ldots$$
To represent this as a fraction we need to see it as an infinite sum.
$0.bbb\ldots = \frac{b}{10^{n}} + \frac{b}{10^{2n}} + \frac{b}{10^{3n}} + \ldots$
Now that we have it as an infinite sum we can use the geometric series formula where
$a = \frac{b}{10^{n}}, r = \frac{1}{10^{n}}$
$\sum\limits_{x = 0}^{\infty} ar^{x} = \frac{a}{1 - r}$
Hence
$\sum\limits_{x = 0}^{\infty}\frac{b}{10^{n(x + 1)}} = \frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}}$
$\large\frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}} = \frac{\frac{b}{10^{n}}}{\frac{10^{n} - 1}{10^{n}}} = \frac{b}{10^{n}} \cdot \frac{10^{n}}{10^{n} - 1} = \frac{b}{10^{n} - 1}$
So that means that :
$0.bbb \ldots = \frac{b}{10^{n} - 1}$
But what if there's some decimal places in front of the recurring part?
Let $$c$$ be a number with $$n_1$$ digits, it can also have leading and trailing zeros.
The decimal is now : $$0.cbbb \ldots$$
Now we have to account for the decimal places between the decimal point and the recurring decimal. This is easy to do since we already have the equation for recurring decimals. The solution is simple - divide by $$10$$ to the power of the number of places ($$n_1$$).
$0.cbbb\ldots = \text{ ?} + \frac{b}{10^{n_1}(10^{n} - 1)}$
The $$\text{?}$$ represents the $$c$$ part of the decimal. It is simply the number $$c$$ divided by $$10$$ to the power of the number of decimal places ($$n_1$$) it takes up.
$0.cbbb\ldots = \frac{c}{10^{n_1}} + \frac{b}{10^{n_1}(10^{n} - 1)}$
Let's simplify it a bit.
$0.cbbb\ldots = \frac{b + c(10^{n} - 1)}{10^{n_1}(10^{n} - 1)}$
That's all for now, hope you found this note interesting.
Note by Jack Rawlin
2 years, 9 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
## Comments
Sort by:
Top Newest
Great! Could you add this to the Converting Repeating Decimals into Fractions wiki? Thanks!
Staff - 2 years, 9 months ago
Log in to reply
×
Problem Loading...
Note Loading...
Set Loading... | 2018-10-17T04:21:39 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/recurring-decimals-as-fractions/",
"openwebmath_score": 0.9905508160591125,
"openwebmath_perplexity": 1680.4424642509177,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n",
"lm_q1_score": 0.983085087724427,
"lm_q2_score": 0.8615382129861583,
"lm_q1q2_score": 0.8469653696914435
} |
https://math.stackexchange.com/questions/3158634/whats-the-cross-product-in-2-dimensions | # What's the cross product in 2 dimensions? [duplicate]
The math book i'm using states that the cross product for two vectors is defined over $$R^3$$:
$$u = (a,b,c)$$
$$v = (d,e,f)$$
is:
$$u \times v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & c \\ d & e & f \\ \end{vmatrix}$$
and the direction of the resultant is determined by curling fingers from vector v to u with thumb pointing in direction of the cross product of u x v.
Out of curiosity, what's the cross product if u and v are defined over $$R^2$$ instead of $$R^3$$ instead:
$$u = (a,b)$$
$$v = (d,e)$$
Is there a "degenerate" case for the cross product of $$R^2$$ instead $$R^3$$? like this is some type of 2x2 determinant instead?
for instance if had a parameterization:
$$\Phi(u,\ v) = (\ f(u),\ \ g(v)\ )$$
and needed to calculate in $$R^2$$:
$$D = \Bigg| \frac{\partial{\Phi}}{\partial{u}} \times \frac{\partial{\Phi}}{\partial{v}} \Bigg|$$
There are plenty of examples in the book for calculating the determinate D in $$R^3$$ but none at all for $$R^2$$ case.
As in:
$$\iint_{V} f(x,y) dx\ dy = \iint_{Q} f(\Phi(u,v) \Bigg| \frac{\partial{\Phi}}{\partial{u}} \times \frac{\partial{\Phi}}{\partial{v}} \Bigg|$$
$$\Phi(u,v)=(2u \cos v,\ \ u \sin v)$$
## marked as duplicate by Cesareo, Leucippus, Eevee Trainer, Shailesh, Lee David Chung LinMar 23 at 3:51
• Point for you to ponder on - the cross product of two vectors gives a third vector, perpendicular to the original two vectors. Could you define a (meaningful) cross product in R^2 that would imitate that property? Moreover, there are similar threads to this one in MSE, I recommend you to try reading some of them :) – GSofer Mar 22 at 20:29
• mean is in th last integral listed above... I have a parameterization defined over R^2... wondering what this degenerates to when taking R^3 formula and applying it to R^2 case... – DiscreteMath Mar 22 at 20:30
• for example, maybe we pretend there's a 3 dimension there when we need it to perform the cross product? – DiscreteMath Mar 22 at 20:31
• You can consider the "cross product" of a single vector $(a,b)$ to be the vector $(-b,a)$ perpendicular to it. – Théophile Mar 22 at 20:32
• Alternatively, you could consider the two vectors in $\Bbb R^2$ to have $z$-co-ordinate $0$, so that their "cross product" would be $(0,0,k)$ for some $k$. But it really depends on what you're looking for. – Théophile Mar 22 at 20:33
In $$n$$ dimensions, the Levi-Civita symbol has $$n$$ indices, two of which contract with those of two vectors whose wedge product is sought in the geometry's exterior algebra. This obtains an object with $$n-2$$ indices, so whereas we get a familiar vector if $$n=3$$, we get a scalar if $$n=2$$. In particular, $$\binom{a}{b}\land\binom{c}{d}=ad-bc$$.
• Great answer! I learned something new. – ErotemeObelus Mar 22 at 21:25
The cross product in 2 dimensions is a scalar give my a 2x2 determinant:
$$(a, b) \times (c, d) = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - cb$$
The cross product in 3 dimensions is a vector given by the 3x3 determinant:
$$(a, b, c) \times (d, e, f) = \begin{vmatrix} e_x & e_y & e_z \\ a & b & c \\ d & e & f \\ \end{vmatrix}$$
E. A. Abbott describes a 2D cross product nicely in his mathematical fantasy book "Flatland":
Flatland describes life and customs of people in a 2-D world: in this universe vectors can be summed together and projected, areas are calculated, rotations are clock-wise or counter clock-wise, reflection is possible... but cross product does not exist; otherwise, 2-D inhabitants should have great fantasy to imagine a 3rd dimension to contain a vector orthogonal to their plane.
By the way, in 2-D a single scalar number is sufficient to describe a force’s moment.
$$M(\vec{r}, \vec{F}) = r_1 F_2 - r_2 F_2$$
With such a definition, this operation respects all algebraic properties of cross product, but the result is a scalar. | 2019-04-22T02:30:15 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3158634/whats-the-cross-product-in-2-dimensions",
"openwebmath_score": 0.7485892176628113,
"openwebmath_perplexity": 670.866515283093,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850892111574,
"lm_q2_score": 0.8615382076534742,
"lm_q1q2_score": 0.8469653657298364
} |
http://mymathforum.com/number-theory/199-reciprocal-sum-primes-diverges-proof.html | My Math Forum Reciprocal sum of primes diverges... proof?
Number Theory Number Theory Math Forum
December 27th, 2006, 10:01 PM #1 Newbie Joined: Nov 2006 Posts: 20 Thanks: 0 Reciprocal sum of primes diverges... proof? We'll need the following definitions to solve this problem: definition: For any number x, Nj(x) is the number of numbers less than or equal to x whose only prime divisors are in the set of the first j primes { p1, p2, ..., pj }. definition: A number is square-free if none of its divisors is a perfect square (except 1). (a) Prove that there are exactly 2j square-free numbers whose only prime divisors are in the set { p1, p2, ..., pj }. (b) Prove that, for sufficiently large x, Nj(x)<= 2j * x (Hint: show every number can be written uniquely as the product of a perfect square and a square-free number). (c) Prove that, for sufficiently large x, Nj(x) < x. Why does this mean there must be an infinite number of primes? (d) For subsequent results, we need a little bit more. Prove that, for sufficiently large x, Nj(x) < x2 (The proof is exactly like the one in (c).) We will use (d) to prove: theorem: The sum of the reciprocals of the primes diverge! That is, 1p1 + 1p2 +1p3 +... diverges [Is it clear that this is even stronger than merely being an infinite set? For example, the perfect squares are an infinite set, but the sum of the reciprocals of the perfect squares converges.] Anyhow to show that the sum of the reciprocals of the primes diverges, it's enough to show that, for any j: 1/pj+1 + 1/pj+2 + 1/pj+3 + .... > 1/2 (e) Why is it enough to show that the above sum is greater than 12 ? (f) Show that for any x, x/pj+1 + x/pj+2 + x/pj+3 + .... >= x - Nj(x). (Hint: Explain first why x/p is greater than or equal to the number of numbers less than or equal to x that are divisible by p. What about x/p + x/q where p and q are relatively prime?) (g) Use the results from (d) and (f) to complete the proof that the sums of the reciprocals of the primes diverge.
December 28th, 2006, 02:17 AM #2 Newbie Joined: Nov 2006 Posts: 20 Thanks: 0 I have already solved part a), if anyone has idea in solving part b, part c, and part d, please do teach me. Thank you very much.
December 28th, 2006, 02:36 PM #3 Newbie Joined: Dec 2006 Posts: 29 Thanks: 0 For part a), I believe you meant 2^j. In part b), I'm seeing Nj(x)<= 2j * ?x, what was this supposed to say?
December 28th, 2006, 06:36 PM #4 Newbie Joined: Dec 2006 From: Moscow, Russia Posts: 4 Thanks: 0 b) Just use hint. Every n<=x can be uniquely written in the form n=a*b, where a is squrefree and b is perfect square. Number of possible values a is not more than 2^j (part (a)), number of possible values b is not more than sqrt(x) (because b=c^2 and c<=sqrt(x)). Hence number of possible values for n=ab is not more than 2^j*sqrt(x). c) If there were only finitely many prime numbers, namely {p1,p2,...,pj}, then for all integer x>0 we'd have Nj(x)=x. Contradiction. d) Obvious. (Does x2 mean x/2?) e) Use Cauchy criterion. f)x-Nj(x) equals quantity of numbers n<=x which have at least one prime factor from {p_{j+1}, p_{j+2},... } . Quantity of n<=x which are divisible by p is not more than x/p (because n=p*m and m<=x/p).
December 29th, 2006, 05:45 AM #5 Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 I asked the same question ( If the sum of the reciprocals of the prime numbers diverged ) on the AoPS forum. Here's the link, but we can discuss it here as well, of course. I believe it was Euler who proved this conjecture to be true. ( Euler proved everything ) http://www.artofproblemsolving.com/F...c.php?t=122342
December 30th, 2006, 02:05 PM #6 Newbie Joined: Nov 2006 Posts: 20 Thanks: 0 Hi infinity, Thank you very much for your reply and information. I found this link which is related to my question: http://primes.utm.edu/infinity.shtml With the help of this link, I got the idea to almost finish the question.
Tags diverges, primes, proof, reciprocal, sum
,
,
### prove the summation of the reciprocal primes is divergent
Click on a term to search for related topics.
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post magnuspaajarvi Applied Math 1 February 8th, 2014 11:08 PM LovaLova Number Theory 2 March 25th, 2012 05:23 PM WheepWhoop Number Theory 7 October 20th, 2011 05:09 PM brunojo Number Theory 6 December 12th, 2008 08:57 AM LovaLova Real Analysis 1 January 1st, 1970 12:00 AM
Contact - Home - Forums - Top | 2016-08-24T10:23:12 | {
"domain": "mymathforum.com",
"url": "http://mymathforum.com/number-theory/199-reciprocal-sum-primes-diverges-proof.html",
"openwebmath_score": 0.8549354672431946,
"openwebmath_perplexity": 1103.7295101195923,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850832642354,
"lm_q2_score": 0.861538211208597,
"lm_q1q2_score": 0.846965364101324
} |
https://math.stackexchange.com/questions/3157460/compute-int-0-infty-frac-sinxxex-dx/3157469 | # Compute $\int_{0}^{\infty}\frac{\sin(x)}{xe^x} \ dx$.
I know one can solve this in many ways and the answer is $$\pi/4$$. However I'm interested in one particular solution involving Laplace transform.
I once saw a solution of this integral where one just did something like directly taking the Laplace transform of the integrand and then getting an integral in terms of $$s$$. So my question is, how does one go from
$$\mathscr{L}\left[\int_{0}^{\infty}\frac{\sin{x}}{xe^x} \ dx\right]$$
To some integral like
$$\frac{1}{2}\int_0^{\infty}\frac{1}{s^2+1} \ ds = \frac{\pi}{4}.$$
I don't remember the steps inbetween or if the first step is even correct but I wan't to know how the Laplace theory was used here. Any one who has a guess?
• Maybe this result is helpful (Laplace transform of $\frac{f(t)}{t}$ is $\int_s^\infty F(u)\, du$, where $F=\mathscr{L}(f)$). Your original integral is related to the Laplace transform of $\frac{f(t)}{t}$ where $f(t)=\sin{t}$. – Minus One-Twelfth Mar 21 at 22:15
• By the way, I don't think you'd be taking the Laplace transform of your integral, because that would just be the Laplace transform of a constant. – Minus One-Twelfth Mar 21 at 22:21
Something like this? $$\begin{multline} \int_0^\infty \frac{\sin x}{xe^x} dx = \int_0^\infty \sin x \int_1^\infty e^{-sx}dsdx = \int_1^\infty\!\!\int_0^\infty\sin(x) e^{-sx}dx\,ds \\= \int_1^\infty \mathcal{L}\left[\sin(x)\right](s)ds = \int_1^\infty \frac{ds}{s^2 + 1} = \frac{\pi}{2}-\tan^{-1}1 = \frac{\pi}{4} \end{multline}$$
• Your first step, how did you split the integral like that? – Parseval Mar 21 at 23:19
• @Parseval Since $x^{-1}e^{-x}= \int_1^\infty e^{-sx}ds$, one can be freely substituted for the other, by the substitution property of equality. – eyeballfrog Mar 22 at 1:33
Let $$I(a)=\int_0^\infty \frac{\sin(x)}{x}e^{-ax}dx$$ where your integral is $$I(1)$$. $$I'(a)$$ is then: $$I'(a)=-\int_0^\infty \sin(x)e^{-ax}dx$$ which is the Laplace Transform of the sine function. Thus $$I'(a)=-\frac{1}{a^2+1}$$
Integrating back we get: $$I(a)=-\arctan(a)+C$$ To find the constant, we notice that $$I(0)=\int_0^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ So we get that $$C=\frac{\pi}{2}$$ and that $$I(1)=-\arctan(1)+\frac{\pi}{2}=\frac{\pi}{4}$$
Here is an approach that employs the following useful property for the Laplace transform: $$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$ Perhaps it is something like this you are looking for?
Noting that $$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$ and $$\mathcal{L}^{-1} \left \{\frac{e^{-x}}{x} \right \} (t)= H(t - 1).$$ Here $$H(x)$$ denotes the Heaviside step function. Thus
\begin{align} \int_0^\infty \frac{\sin x}{x e^x} \, dx &= \int_0^\infty \sin x \cdot \frac{e^{-x}}{x} \, dx\\ &= \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{e^{-x}}{x} \right \} (t) \, dt\\ &= \int_0^\infty \frac{H(t - 1)}{1 + t^2} \, dt.\\ &= \int_1^\infty \frac{1}{1 + t^2} \, dt\\ &= \Big{[}\tan^{-1} t \Big{]}_1^\infty\\[1ex] &= \frac{\pi}{2} - \frac{\pi}{4}\\[1ex] &= \frac{\pi}{4}. \end{align}
The great thing about this one is even "alternative" approaches accidentally at some point get us a double integral whose further evaluation can be seen as using a result in Laplace transforms. For example, aleden's answer begins exactly the way you would if you said "I'd rather use this", but look what it gets!
My usual preferred way of solving this sees the same phenomenon. It begins with a Schwinger parametrization, writing $$\frac{1}{x}=\int_0^\infty\exp -sx ds$$. Then the integral becomes $$\int_0^\infty ds\int_0^\infty dx\sin x\exp -(1+s)x$$. Up to a constant variable shift, this yields the deliberately Laplace-based approach in eyeballfrog's answer. | 2019-06-20T23:23:46 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3157460/compute-int-0-infty-frac-sinxxex-dx/3157469",
"openwebmath_score": 0.9997543692588806,
"openwebmath_perplexity": 377.03808024971545,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211553734218,
"lm_q2_score": 0.8688267864276107,
"lm_q1q2_score": 0.8469507317647407
} |
http://mathhelpforum.com/pre-calculus/99689-trigonometric-limit-problem.html | 1. ## trigonometric limit problem
lim (sin(x) - x) / x^3
x~>0
so i tried splitting it up into sin(x) / x^3 - x / x^3. sin(x) / x^3 = (1/x^2)(sin(x) / x) and the limit as x approaches 0 of sin(x)/x = 1 so the first term just becomes 1/x^2. for the second term i canceled out an x from the top and bottom so i also got 1/x^2. so 1/x^2 - 1/x^2 = 0, so i got that the limit is equal to 0. but when i used L'Hospital's Rule on it, i got a completely different answer, -1/6. was the algebraic way i was doing it wrong? please help.
2. Originally Posted by oblixps
lim (sin(x) - x) / x^3
x~>0
so i tried splitting it up into sin(x) / x^3 - x / x^3. sin(x) / x^3 = (1/x^2)(sin(x) / x) and the limit as x approaches 0 of sin(x)/x = 1 so the first term just becomes 1/x^2. for the second term i canceled out an x from the top and bottom so i also got 1/x^2. so 1/x^2 - 1/x^2 = 0, so i got that the limit is equal to 0. but when i used L'Hospital's Rule on it, i got a completely different answer, -1/6. was the algebraic way i was doing it wrong? please help.
You're disregarding an important limit property!
$\displaystyle \lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a} g(x)$
So, $\displaystyle \lim_{x\to0}\frac{\sin x-x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\frac{1}{x^2}-\lim_{x\to0}\frac{1}{x^2}$ $\displaystyle =\left[\lim_{x\to0}\frac{\sin x}{x}\lim_{x\to0}\frac{1}{x^2}\right]-\lim_{x\to0}\frac{1}{x^2}=(1)(\infty)-(\infty)=\infty-\infty$, an indeterminate case.
So applying L'Hôpital's rule would be advisible here -- the answer would be $\displaystyle -\tfrac{1}{6}$.
3. Hello, oblixps!
Your algebra is off . . .
$\displaystyle \lim_{x\to0}\frac{\sin x - x}{x^2}$
Presently, we have: .$\displaystyle \lim_{x\to0}\frac{\sin x - x}{x^3} \;\Rightarrow\; \frac{0}{0}$ . . . an indeterminate form.
We can write this as: .$\displaystyle \lim_{x\to0}\frac{\frac{\sin x}{x} - 1}{x^2} \;\Rightarrow\; \frac{1-1}{0^2} \:=\:\frac{0}{0}$ . . . an indeterminate form -- again!
We can apply L'Hopital: .$\displaystyle \lim_{x\to0}\frac{\cos x - 1}{3x^2} \;=\;\frac{1-1}{0} \:=\:\frac{0}{0}$ . . . indeterminate
At this point, we can: .$\displaystyle \begin{array}{cc}(1) & \text{Apply L'Hopital twice} \\ (2) & \text{Do some clever algebra} \end{array}$
Method 1:
. . L'Hopital again: .$\displaystyle \lim_{x\to0}\frac{-\sin x}{6x} \:=\:\frac{0}{0}$ . . . indeterminate
. . L'Hopital again: .$\displaystyle \lim_{x\to0}\frac{-\cos x}{6} \;=\;\boxed{-\frac{1}{6}}$
Method 2:
. . We have: .$\displaystyle \frac{\cos x - 1}{3x^2} \;=\;\frac{-(1-\cos x)}{3x^2}$
. . Multiply by $\displaystyle \frac{1+\cos x}{1+\cos x}\!:\quad \frac{-(1-\cos x)}{3x^2}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{-(1-\cos^2\!x)}{3x^2(1+\cos x)} \;=\;\frac{-\sin^2\!x}{3x^2(1+\cos x)}$ . $\displaystyle = \;\frac{-1}{3(1+\cos x)}\cdot\frac{\sin^2\! x}{x^2}$
. . Take the limit: .$\displaystyle \lim_{x\to0}\left[\frac{-1}{3(1+\cos x)}\cdot\left(\frac{\sin x}{x}\right)^2\right] \;=\;\frac{-1}{3(2)}\cdot(1)^2 \;=\;\boxed{-\frac{1}{6}}$
4. or expand $\displaystyle \sin x$ in a taylor series about $\displaystyle x=0$
then $\displaystyle \lim_{x \to 0} \frac{\sin x -x}{x^{3}} = \lim_{x \to 0} \frac{-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5} - \frac{1}{7!}x^{7}+ ...}{x^{3}}$
$\displaystyle = \lim_{x \to 0} \Big(-\frac{1}{3!}+\frac{1}{5!}x^{2} - \frac{1}{4!}x^{7}+ ... \Big) =$ $\displaystyle -\frac{1}{3!} = -\frac{1}{6}$
5. Originally Posted by Random Variable
or expand $\displaystyle \sin x$ in a taylor series about $\displaystyle x=0$
then $\displaystyle \lim_{x \to 0} \frac{\sin x -x}{x^{3}} = \lim_{x \to 0} \frac{-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5} - \frac{1}{7!}x^{7}+ ...}{x^{3}}$
$\displaystyle = \lim_{x \to 0} \Big(-\frac{1}{3!}+\frac{1}{5!}x^{2} - \frac{1}{4!}x^{7}+ ... \Big) =$ $\displaystyle -\frac{1}{3!} = -\frac{1}{6}$
I would've never thought of this approach. Thanks for the enlightenment. | 2018-06-21T23:20:12 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/pre-calculus/99689-trigonometric-limit-problem.html",
"openwebmath_score": 0.9650928378105164,
"openwebmath_perplexity": 765.524357814125,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.974821157567904,
"lm_q2_score": 0.8688267830311354,
"lm_q1q2_score": 0.8469507303604096
} |
https://stats.stackexchange.com/questions/52895/probability-that-two-random-letters-from-a-language-will-be-the-same/52899 | # Probability that two random letters from a language will be the same?
I'm trying to find the probability that two randomly-selected letters from "average" text in a language will be the same.
For example, if my hypothetical language contains four letters which each occur on average with the following frequency:
A = 60%
B = 25%
C = 10%
D = 5%
What is the probability that selecting any two letters from a representational text will be the same?
My intuition for solving this is first to find the chance that they're different, so the sum over the probabilities that a letter is chosen and then some other letter is chosen next, over each letter in the alphabet:
(0.6 * (1 - 0.6) +
0.25 * (1 - 0.25) +
0.1 * (1 - 0.1) +
0.05 * (1 - 0.05)) = 0.565
Then the chance that they are the same:
1 - 0.565 = 0.435
Is this reasoning sound? It seems like a very basic probability problem, but I always seem to be thinking about these things in the wrong way and would appreciate a sanity check (and any pointers to materials which would help me be more confident about this kind of thing in the future!)
• A derivative question: What frequency distribution minimizes this probability that two randomly-selected letters from "average" text in a language will be the same? Is it all letters equally common distribution? – curious_cat Mar 21 '13 at 6:49
• @curious_cat yes a discrete uniform minimizes the probability. (This is related to the birthday problem - if you assume anything but uniform births, the probability goes up.) – Glen_b -Reinstate Monica Mar 21 '13 at 7:26
• @curious_cat Based on Glen_b's answer, you may interpret the chance that the two letters are the same as the squared distance to the origin of the "probability vector" $(p_1, p_2, \ldots, p_n)$. The restriction that $p_1+p_2+\cdots+p_n=1$ confines this vector to a plane, yielding a geometric interpretation of your question: what point on this plane is nearest the origin? At any such point, the vector to the origin must be perpendicular to the plane. Obviously such a point is unique and just as obviously $(1,1,\ldots,1)$ is perpendicular: thus all the $p_i$ must equal one another, QED. – whuber Mar 21 '13 at 20:52
## 2 Answers
You got the right answer but you kind of did it the hard way.
If $L_1$ is the first letter and $L_2$ is the second letter, then
the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)$.
If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's
$\sum_x P(L_1=x) \times P(L_2=x)$.
But of course then the two probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".
That is, the overall probability = $\sum_x p_x^2$
In your case, that's 0.6*0.6 + 0.25*0.25 + 0.1*0.1 + 0.05*0.05 which comes to 0.435.
--
generalization to $k$ letters:
To find the probability that $k$ letters selected from the population are all identical:
If $L_1$ is the first letter, $L_2$ is the second letter, ..., and $L_k$ is the $k^\mathrm{th}$ then
the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)\times P(L_3=x|L_1=x,L2=x)\times ... \times P(L_k=x|L_1=x,L2=x, ...,L_k=x)$.
If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's
$\sum_x P(L_1=x) \times P(L_2=x)\times ... \times P(L_k=x)$.
But of course then all the probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".
That is, the overall probability = $\sum_x p_x^k$
• Thanks, this is exactly the sort of answer I was hoping for. It seems that if I wanted to extrapolate to 3 letters, I would just sum over $p_x^3$ then? – Arne Roomann-Kurrik Mar 21 '13 at 17:19
• Updated answer to cover this case. Short answer: yes. – Glen_b -Reinstate Monica Mar 21 '13 at 20:16
I think you can compute it by $A^2+B^2+C^2+D^2=0.6^2+0.25^2+0.1^2+0.05^2=0.435$ And $A^2$ means you will get $2A$s, $B^2$ means you will get $2B$s and so on. | 2020-01-19T09:57:12 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/52895/probability-that-two-random-letters-from-a-language-will-be-the-same/52899",
"openwebmath_score": 0.9300684332847595,
"openwebmath_perplexity": 387.8253206851437,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211590308923,
"lm_q2_score": 0.8688267643505193,
"lm_q1q2_score": 0.8469507134212332
} |
https://math.stackexchange.com/questions/3134210/how-can-i-write-this-as-a-double-sum | # How can I write this as a double sum?
$$2 \sum_{1\leq j < k \leq N} P(F_j F_k)$$ as a double sum. Would it be $$\sum_{j=1}^k \sum_{k=n}^N P(F_j F_k)$$?
Also I am just confused about what a double sum would actually mean, I understand that you would sum over the values twice but can someone please explain more in detail? Thank you
• Where does $n$ come from? Mar 3 '19 at 23:47
• Does $P(F_j F_k)$ have any special meaning? Generally Readers can only assume it is a term that depends on $j,k$ in a symmetric fashion, i.e. because $F_jF_k=F_kF_j$. But already in the first summation you've restricted the terms so that $j\lt k$, eliminating any cases $j=k$ and raising a suspicion about why the factor $2$ appears in front of that summation, but not in front of your attempt to rewrite it as a double sum. Mar 4 '19 at 0:45
A double sum is a sum of the form $$\sum_{j=a}^b \sum_{k=c(j)}^{d(j)} f(j,k) = \sum_{j=a}^b \left(\sum_{k=c(j)}^{d(j)} f(j,k)\right)$$ I used the notation $$c(j),d(j)$$ because $$c$$ and $$d$$ may very well depend on $$j$$, but they don't have to.
There are more ways you can approach this.
1. We want one variable, say $$j$$, to be independent, and let it independently take all the values it can possibly have that satisfy the condition $$1\le j < k \le N$$. We will then adjust $$k$$ so that this condition is actually met.
Therefore, let $$j$$ vary across all values that it can possibly have. These are $$1\le j \le N-1$$, hence the $$\sum_{j=1}^{N-1}$$ symbol. Then, let $$k$$ take all values such that the condition $$j < k \le N$$. A different way to write this inequality is $$j+1 \le k \le N$$. Hence, your sum is equal to $$2\sum_{j=1}^{N-1} \sum_{k=j+1}^N P(F_jF_k)$$ To expand this sum, first write out all terms of the inner sum, and then add them up according to the outer sum, like $$2\sum_{j=1}^{N-1} (P(F_jF_{j+1})+P(F_jF_{j+2})+...+P(F_jF_N)) =$$ $$= 2[(P(F_1F_2)+P(F_1F_3)+...+P(F_1F_N)) +$$ $$+(P(F_2F_3)+P(F_2F_4)+...+P(F_2F_N))+...+$$ $$+P(F_{N-1}F_N)]$$ Notice that each successive term of the outer sum has less and less terms of the inner sum (the last one has only one term).
2. Let $$k$$ vary across all values that it can possibly have. These are $$2\le k \le N$$. Then, let $$j$$ take all values such that the condition $$1\le j < k$$ is met. A different way to write this inequality is $$1 \le j \le k-1$$. Hence, your sum is equal to $$2\sum_{k=2}^{N} \sum_{j=1}^{k-1} P(F_jF_k)$$
The difference between these two approaches is which variable you assign "independence" to.
• thank you! i have a really stupid question but how come certain double sums have a 2 next to it and other double sums do not? Mar 4 '19 at 4:31
• also, why do you initially take j to be $1 \leq j \leq N-1$ Mar 4 '19 at 4:33
• And just one more question (sorry!), what exactly does this double sum expand to? like if i had actual numbers/values, then how would this expand? Mar 4 '19 at 4:34
• @user477465 The 2 is there because your starting sum is multiplied by 2. Maybe the 2 was a typo on your part? As for why I take $1\le j \le N-1$... If you have $1\le j < k \le N$ in the sum, this means that the sum contains all possible combinations of $j$ and $k$ that satisfy this. You can see that $j$ can take the smallest value 1, and $j$ can take the value $N-1$, but it can't take $N$ because then $k$ would have to be $>N$. Of course, $N=1$ is a special case, where no $j,k$ satisfy the condition of the sum, hence the sum is 0. Also, see my edit to the answer. Mar 4 '19 at 5:01
$$\sum_{j=1}^{N-1} \sum_{k=j+1}^N P(F_j,F_k)$$
Assume, that for some finite set $$I$$, you have a sum $$\sum_{l \in I} s_i$$ Now, the summands correspond to the elements in $$I$$. If, for some set $$J$$, we have a bijection
$$I \overset{\sigma}{\longrightarrow} \bigcup_{j \in J}I_j =: Y$$ where $$I_j \cap I_i = \emptyset$$ for $$i\neq j$$, then
$$\sum_{i\in I} s_i = \sum_{y \in Y} s_{\sigma^{-1}(y)} = \sum_{y \in \bigcup_{j \in J}I_j} s_{\sigma^{-1}(y)} = \sum_{j \in J}\sum_{y \in I_j} s_{\sigma^{-1}(y)}$$ The first equation holds because $$\sigma$$ is a permutation and addition is commutative. The second equation is trivial and the third equation uses that the union is disjoint, as mentioned above.
So, any bijection $$\sigma$$ into a disjoint union as mentioned above will give you a representation as a double sum. You could of course proceed inductively, to get triple, quadruple sums et cetera.
The double sum can hence be understood as a specific grouping of summands/a partition of the index set.
Defining $$I:=\left\{\left(j,k\right);\,1\leq j < k \leq N\right\}$$, you can write $$I = \bigcup_{1 \leq j < N} \underbrace{\left\{(j,k);\, j < k \leq N\right\}}_{:= I_j}$$ and the union is disjoint. Taking $$\sigma$$ as the identity, and using the above, we get $$\begin{eqnarray} & 2 \sum_{(j,k) \in I} P(F_j F_k) = 2 \sum_{(j,k) \in \bigcup_{1 \leq j < N} I_j} P(F_j F_k) = 2 \sum_{1 \leq j < N} \sum_{(j,k) \in I_j} P(F_j F_k) = \\ & 2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_k) = 2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_k) \\ \end{eqnarray}$$
where in the last equation, we used that the sets $$\left\{(j,m);\, j < m \leq N\right\}$$ and $$\left\{m;\, j < m \leq N\right\}$$ are bijective/have the same number of elements. There is some subtle formal detail involved in this last equation:
We are basically applying what we used before - $$\sum_{i\in I} s_i = \sum_{y \in Y} s_{\sigma_j^{-1}(y)}$$ - to the inner sum, where for each $$j$$, $$\sigma_j: \left\{(j,m);\, j < m \leq N\right\} \longrightarrow \left\{m;\, j < m \leq N\right\},\; (j,m) \mapsto m$$ is the projection to the second component and $$\sigma_j^{-1}$$ is the function that maps $$m \mapsto (j,m)$$.
Explicitly, $$\begin{array} & 2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_k) = 2 \sum_{1 \leq j < N}\; \sum_{(j,k) \in \left\{(j,m);\, j < m \leq N \right\}} P(F_j F_{\sigma_j(j,k)}) = \\ 2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_{\sigma_j(\sigma_j^{-1}(k))}) = 2 \sum_{1 \leq j < N} \sum_{j < k \leq N} P(F_j F_k) \\ \end{array}$$ | 2022-01-25T18:49:16 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3134210/how-can-i-write-this-as-a-double-sum",
"openwebmath_score": 0.923021674156189,
"openwebmath_perplexity": 143.13645415993255,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9748211590308922,
"lm_q2_score": 0.8688267643505193,
"lm_q1q2_score": 0.8469507134212331
} |
https://mathhelpboards.com/threads/find-the-value-of-x-and-y.3972/#post-17958 | # Find the value of x and y
#### Albert
##### Well-known member
x=abc is a three digits number
y=$x^2$
the last 3 digits of y is also equal to abc
that is y=???abc
#### Bacterius
##### Well-known member
MHB Math Helper
Re: find the value of x and y
$$x^2 \equiv x \pmod{1000} \tag{1}$$
[JUSTIFY]Clearly for $x$ coprime to $1000$, the only solutions are the trivial $x = 0$ and $x = 1$, so all nontrivial solutions for $x$ must be divisible by either $2$ or $5$, or both. This leaves us with about $600$ possible candidates. Now it should be clear that if a solution works for $1000$, it will also work for $100$, and for $10$. Let's try and find the solutions for $10$:[/JUSTIFY]
$$x^2 \equiv x \pmod{10} \tag{2}$$
[JUSTIFY]We have $5$ potential solutions here, namely $2, 4, 5, 6, 8$ (modulo $10$). Let's check each of them manually, some calculations show that only $5$ and $6$ work. So the solutions must have either $5$ or $6$ as a last digit. Let's now extend our search to $100$, armed with this information (which reduces the set of possible solutions considerably):[/JUSTIFY]
$$x^2 \equiv x \pmod{100} \tag{3}$$
[JUSTIFY]We know from the previous step that the potential solutions are $5, 6, 15, 16, \cdots$. A short exhaustive search tells us that only $25$ and $76$ work. Finally, we have narrowed down the search enough to look for the real solutions:[/JUSTIFY]
$$x^2 \equiv x \pmod{1000} \tag{4}$$
The only possible solutions are $25, 76, 125, 176, \cdots$. Again, trying those 20 potential solutions shows that only $376$ and $625$ work. Therefore, the solutions are $x \in \{ 0, \, 1, \, 376, \, 625 \}$. QED.
[JUSTIFY]This might look like a non-answer given the amount of trial and error involved, but consider that we've only had to try $45$ potential candidates instead of the original $600$, and the cost of this algorithm is asymptotically logarithmic, for instance, finding the solutions with $m = 100000$ would only take $85$ trials, quite efficient.[/JUSTIFY]
Here's an Python 3 script implementing the algorithm outlined above, using recursion, for powers of $10$:
Code:
# Work count
work = 0
def Verify(x, m):
global work
work += 1
return pow(x, 2, 10**m) == x
def Solve(m):
# Solutions
found = []
if m == 1:
for x in [2, 4, 5, 6, 8]: # Precomputed
if Verify(x, m):
found.append(x)
return found
possible = Solve(m - 1)
for p in possible:
for x in range(p, 10**m, 10**(m - 1)):
if Verify(x, m):
found.append(x)
return found
def Enumerate(m):
global work
work = 0
found = Solve(m)
found.append(0)
found.append(1)
found.sort()
print("Solutions are {" + ', '.join(str(x) for x in found) + "}.")
print("Checked " + str(work) + " candidates.")
Code:
>>> Enumerate(1) # For 10
Solutions are {0, 1, 5, 6}.
Checked 5 candidates.
>>> Enumerate(2) # For 100
Solutions are {0, 1, 25, 76}.
Checked 25 candidates.
>>> Enumerate(3) # For 1000
Solutions are {0, 1, 376, 625}.
Checked 45 candidates.
>>> Enumerate(5) # For 100000
Solutions are {0, 1, 9376, 90625}.
Checked 85 candidates.
>>> Enumerate(8) # For 100000000
Solutions are {0, 1, 12890625, 87109376}.
Checked 145 candidates.
[JUSTIFY]Interestingly, there are only two nontrivial solutions for any power of $10$, and these two solutions add up to $-1$ modulo said power, suggesting that there might exist an analytic solution to this problem, though the prime factorisations of the solutions display no obvious pattern apart from a large number of powers of $2$ and $5$.
So if a better way to do this exists, I'd be interested to see it. It is worth noting that the condition $x^2 \equiv x \pmod{m}$ implies $x^n \equiv x \pmod{m}$ for all $n > 0$, but I could not find how to use this property, sadly.[/JUSTIFY]
#### Albert
##### Well-known member
Re: find the value of x and y
x=abc is a three digits number
y=$x^2$
the last 3 digits of y is also equal to abc
that is y=???abc
y-x =$x^2-x =x(x-1)$ must be a multiple of 1000
for 1000=$2^3 \times 5^3 =8\times 125$
further more x and (x-1) are coprime
if x is a multiple of 125 then x-1 must be a multiple of 8---------(1)
if x is a multiple of 8 then x-1 must be a multiple of 125---------(2)
so the possible solutions meet restriction of (1) and (2) will be x=376 , and x=625
y=141376=$376^2$ , and 390625=$625^2$
Last edited: | 2022-01-25T02:19:40 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/find-the-value-of-x-and-y.3972/#post-17958",
"openwebmath_score": 0.8203444480895996,
"openwebmath_perplexity": 971.9120197398913,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307684643189,
"lm_q2_score": 0.8705972784807408,
"lm_q1q2_score": 0.8469438194473637
} |
https://math.stackexchange.com/questions/1259081/if-there-are-injective-homomorphisms-between-two-groups-in-both-directions-are/1259584 | # If there are injective homomorphisms between two groups in both directions, are they isomorphic?
If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?
Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.
I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?
No. Any counterexample must be infinite, and one way to build such an example is to construct an analogue of the well-known injective homomorphism $S_n \hookrightarrow A_{n + 2}$ for the infinite analogues of those groups.
Consider the group $S_{\infty}$ of (finite) permutations of a set $\{1, 2, \ldots\}$ of countably many elements. Then, let $A_{\infty}$ denote the subgroup of $S_{\infty}$ comprised of even elements (products of even numbers of permutations), or equivalently, the subgroup generated by $3$-cycles. Then, inclusion $A_{\infty} \to S_{\infty}$ is a homomorphism.
On the other hand, let $S_{\infty}'$ denote the subgroup of $S_{\infty}$ of permutations of the set $\{3, 4, \ldots\}$, and consider the map $\Phi: S_{\infty} \to S_{\infty}'$ that maps a permutation $(a_1 \cdots a_n)$ to $((a_1 + 2) \cdots (a_n + 2))$ (this amounts to relabeling and so is manifestly an isomorphism). Then, we can define a map $\Psi: S_{\infty}' \to A_{\infty}$ by $$\Psi(\sigma) := \left\{ \begin{array}{cc} \sigma , & \sigma \text{ even} \\ \sigma (12), & \sigma \text{ odd} \end{array} \right.$$ and readily see that it is an injective homomorphism. So, $\Psi \circ \Phi$ is an injective homomorphism $S_{\infty} \to A_{\infty}$, and hence we have injective homomorphisms in both directions, but the two groups are not isomorphic.
• Is there an easy proof that $A_\infty$ and $S_\infty$ are not isomorphic? – Christopher Apr 30 '15 at 18:47
• @user73985 : consider the conjugacy class of transpositions in $S_\infty$. If you pick any two transpositions and get their product, you land into one of $3$ conjugacy classes (the trivial one, the $3$-cycles, and the double transpositions). However there is no such conjugacy class in $A_\infty$. For example if you pick the class of double transpositions, you can land into at least $6$ different conjugacy classes, and it only gets worse if you pick more complex ones. – mercio Apr 30 '15 at 19:21
• The squares $\sigma^2$ of elements $\sigma \in S_{\infty}$ are all even, so they subgroup they generate is contained in $A_{\infty}$ and in particular proper. On the other hand, the squaring map sends each $3$-cycle $(i, j, k)$ to the respective $3$-cycle $(i, k, j)$ and in particular is a bijection on the set of $3$-cycles. Since $A_{\infty}$ is generated by $3$-cycles, the subgroup generated by the squares of the elements of $A_{\infty}$ is $A_{\infty}$ itself. – Travis Willse May 1 '15 at 3:38
For groups this property - sometimes called the Cantor-Schröder-Bernstein property after the corresponding theorem for plain sets - is wrong. Let $G = F_2$ the free group on two generators $\{a,b\}$ and $H = F_3$ the free group on three generators $x,y,z$. Then there are monomorphisms $f \colon G \to H$, given by $f(a) = x$, $f(b) = y$ and $g \colon H \to G$ given by $g(x) = a^2$, $g(y) = b^2$, $g(z) = ab$, there is no isomorphism.
I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?
The answer is no for rings and fields. It should be no for modules but I'm having trouble coming up with a snappy counterexample off the top of my head.
Rings: Let $k$ be a field, for simplicity. The ring $k[t^2, t^3]$ injects, by construction, into the ring $k[t]$. The ring $k[t]$ also injects into $k[t^2, t^3]$ via the map $t \mapsto t^2$. So there are injections
$$k[t] \hookrightarrow k[t^2, t^3] \hookrightarrow k[t]$$
but these two rings are not isomorphic because, for example, $k[t]$ is integrally closed and $k[t^2, t^3]$ is not.
Fields: The field $\mathbb{C}$ injects into the field $\mathbb{C}(t)$, and hence injects into its algebraic closure $\overline{ \mathbb{C}(t) }$. Now, fun theorem: (Edit: uncountable) algebraically closed fields of characteristic $0$ are completely determined by their cardinality, and $\overline{ \mathbb{C}(t) }$ has the same cardinality as $\mathbb{C}$. So they are isomorphic. So there are injections
$$\mathbb{C} \hookrightarrow \mathbb{C}(t) \hookrightarrow \mathbb{C}$$
but these two fields are not isomorphic because $\mathbb{C}$ is algebraically closed and $\mathbb{C}(t)$ is not.
• What about ordered fields? – celtschk Apr 30 '15 at 22:09
• You mean uncountable algebraically closed fields. Look at the algebraic closure of $\Bbb Q(\pi)$ and that of $\Bbb Q$, both are countable and non-isomorphic. – Asaf Karagila May 1 '15 at 11:09
• @celtschk: For ordered fields the analogue of algebraically closed is probably "real-closed fields" (in other words, the theory of $\Bbb R$), but that theory is not categorical in any cardinality. To see that, note that $\Bbb R$ and ${}^\ast\Bbb R$ (the hyperreals) are both of cardinality $2^{\aleph_0}$ but one is Archimedean and the other is not. – Asaf Karagila May 1 '15 at 12:27
• @AsafKaragila: What does it mean for a theory to be categorical in a cardinality? – celtschk May 1 '15 at 13:22
• @celtschk: Every two models of the same cardinality are isomorphic. For example every two vector spaces over a finite field are isomorphic if and only if they have the same cardinality. Over $\Bbb Q$ this is true for uncountable vector spaces. Similarly, every two countable dense linear orders without endpoints are isomorphic, so this theory is $\aleph_0$-categorical. For algebraically closed fields the theory is categorical in any uncountable cardinality, once you fix the characteristics. So every algebraically closed field extending $\Bbb Q$ of size $2^{\aleph_0}$ is isomorphic to $\Bbb C$. – Asaf Karagila May 1 '15 at 13:25
A nonconstructive example: there are injective homomorphisms $$(\mathbb R,+)\to(\mathbb R\times\mathbb Z,+)\to(\mathbb R\times\mathbb R,+)\cong(\mathbb R,+)$$ but $(\mathbb R,+)\not\cong(\mathbb R\times\mathbb Z,+).$ | 2021-06-16T08:00:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1259081/if-there-are-injective-homomorphisms-between-two-groups-in-both-directions-are/1259584",
"openwebmath_score": 0.9248985052108765,
"openwebmath_perplexity": 123.40536698422495,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307708274402,
"lm_q2_score": 0.8705972717658209,
"lm_q1q2_score": 0.8469438149722099
} |
https://www.calctool.org/continuum-mechanics/density-cylinder | Density of a Cylinder Calculator
Created by Gabriela Diaz
Last updated: Nov 27, 2022
Welcome to the density of a cylinder calculator! If you're wondering how do I find the density of a cylinder?, then you've come to the correct place! With this tool, you can expect to determine the density of any right cylindrical object. And if you're also wondering how to find the mass of a cylinder?, here we've got you covered, too, as this tool can also help you calculate the mass of a cylinder 😉
Density measures how much mass is contained in a given amount of volume. We can measure its value by conducting some experiments or calculate it by using a straightforward formula. If you'd like to learn more about this topic, keep reading to find out:
• What density is;
• The density of a cylinder equation; and
• How to find the mass of a cylinder.
How do I find the density of a cylinder - The density of a cylinder equation
Modeling objects as well-known geometric figures is a very common practice. These models simplify calculations when creating or evaluating a design, as they allow us to use simple and already-known formulas to get reasonable estimations of their properties.
This means, for example, that we could think of a house as a cube, planets as spheres, or pipes, columns, shafts, bottles, and even rockets as cylinders. By representing these objects as cylinders or other known geometries, we can easily study them and their different characteristics, e.g., their density.
Density is one of many physical properties of matter, which measures how much mass an object has in a particular volume. The denser the object is, the more mass it has in its volume. The mass-to-volume ratio gives us the general formula to find the density of any object:
$\small \text{Density} = \cfrac{\text{Mass}}{\text{Volume}}$
or
$\small \rho =\cfrac{m}{V}$
💡 Take a look at the density calculator to keep exploring this topic!
We can part from this general expression to get a particular equation for the density of a cylinder. Let's see how to obtain it:
1. Begin by expressing the volume formula of a cylinder $V_\text{cyl}$ of radius $r$ and height $h$:
$\small V_\text{cyl} = \pi r^2h$
2. By replacing this volume expression in the general density formula, we get the density of a cylinder equation:
$\small \rho =\cfrac{m}{\pi r^2h}$
Then if we know the mass $m$, radius $r$ (or diameter), and height $h$, we can determine the cylinder's density. Yes, as simple as that! 😉
Would you like to know how to find the density of other geometries? Jump to the sphere density calculator or the cube density calculator for more of this dense fun!
However, if instead of the density, you'd like to know how to find the mass of a cylinder, we can do so by rearranging the above formula as follows:
$\small m =\pi r^2h \rho$
Note, though, that to calculate the mass of a cylinder, you'll need to know the density $\rho$, radius $r$, and height $h$.
💡 Will the cylinder sink or float when placed in a liquid? 🤔 Look at the buoyancy calculator to figure it out!
Using the density of a cylinder calculator
Let's see how to find the density of a cylinder with this density of a cylinder calculator:
1. If you already know the Weight/mass and Volume of the cylinder, enter these values, and the calculator will return the Density.
2. The calculator can also determine the Volume for you. For this, click on the Advanced mode of the calculator. The Height (h) and Radius (r) fields will appear, enter the corresponding values, and the calculator will show the volume of the cylinder.
3. Now that you have the volume, enter the Weight/mass to get the value of the Density.
💡 What if you want to calculate the mass of a cylinder 🤔 For this, you'll need to know the cylinder's density and volume. With these, the density of a cylinder calculator will calculate the mass of the cylinder!
Gabriela Diaz
Density of a cylinder
Weight/mass
lb
Volume
cu ft
Density
lb/cu ft
People also viewed…
Density
Our density calculator will guide you in a short exploration of this property of matter, while helping you calculate it for a generic object or for some selected materials.
Hooke's law
Use the Hooke's law calculator to calculate the force in a spring. | 2023-01-31T03:43:21 | {
"domain": "calctool.org",
"url": "https://www.calctool.org/continuum-mechanics/density-cylinder",
"openwebmath_score": 0.7772886157035828,
"openwebmath_perplexity": 479.40961713996626,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes",
"lm_q1_score": 0.9728307731905611,
"lm_q2_score": 0.8705972684083609,
"lm_q1q2_score": 0.8469438137632962
} |
https://mathematica.stackexchange.com/questions/103179/how-to-adjust-vector-scaling-with-a-specific-aspect-ratio | How to adjust vector scaling with a specific aspect ratio
I'm trying to create a vector plot with uniform vector scaling. I need the x and y ranges to be very different, because I need to highlight the angle between the vectors and the y=0 axis. This means that I can't use an automatic aspect ratio, however, changing the aspect ratio is changing the vector scaling in a way that I don't understand.
This is my code
Show[VectorPlot[{If[y>0, 1-2x, 1-x], If[y>0, -6/25+2/5x-3/2y, -7/25+2/5x-1/2y]},
{x,0.49,0.8}, {y,-0.01,0.01}, VectorStyle->Black, VectorPoints->Coarse,
VectorScale->{0.05, 1, None}, AspectRatio->0.62],
ContourPlot[y==0, {x,0.48,0.8}, {y,-0.01,0.01}, ContourStyle->{Black}]]
which gives the following vector plot
I specifically need the arrows on the top left to be approximately the same size as the rest of them, and to not cross the y=0 axis. I've also entered specific functions into sfun, for instance
Show[VectorPlot[{If[y>0, 1-2x, 1-x], If[y>0, -6/25+2/5x-3/2y, -7/25+2/5x-1/2y]},
{x,0.49,0.8}, {y,-0.01,0.01}, VectorStyle->Black, VectorPoints->Coarse,
VectorScale->{0.05, 1, #2&}, AspectRatio->0.62],
ContourPlot[y==0, {x,0.48,0.8}, {y,-0.01,0.01}, ContourStyle->{Black}]]
which should give a uniform vector length for the same y values, but instead gives the picture below, where there is the same vector length problem in the upper left corner.
Is there a way to scale the vectors so that this doesn't happen?
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Jan 1 '16 at 22:25
• – Michael E2 Jan 1 '16 at 22:30
• I don't think it is really a duplicate. It turned out that figuring out the rescaling was a bit tricky, esp. because VectorPlot was changing the plot range so that it didn't not match the plot domain. See my answer. – Michael E2 Jan 1 '16 at 23:06
• – Michael E2 Jan 1 '16 at 23:22
You have to scale the (x,y)-components of the vectors so that they correspond to the horizontal and vertical units on the screen. That means scaling (1) according to the PlotRange and (2) according to the AspectRatio.
Show[VectorPlot[{If[y > 0, 1 - 2 x, 1 - x],
If[y > 0, -6/25 + 2/5 x - 3/2 y, -7/25 + 2/5 x - 1/2 y]}, {x, 0.49,
0.8}, {y, -0.01, 0.01},
VectorStyle -> Black, VectorPoints -> Coarse,
VectorScale -> {1/20, 1,
Function[{x, y, fx, fy, norm},
If[norm > 0,
norm/Sqrt[(fx/(0.8 - 0.49))^2 + (0.62 fy/(0.01 - (-0.01)))^2],
0]]},
AspectRatio -> 0.62, PlotRange -> All],
ContourPlot[y == 0, {x, 0.48, 0.79}, {y, -0.01, 0.01},
ContourStyle -> {Black}]
]
Note: Adding PlotRange -> All was essential. VectorPlot was automatically extending the plot range, making it difficult to get the rescaling correct. | 2021-01-27T13:57:17 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/103179/how-to-adjust-vector-scaling-with-a-specific-aspect-ratio",
"openwebmath_score": 0.587715208530426,
"openwebmath_perplexity": 2016.8367452111474,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307661011976,
"lm_q2_score": 0.8705972734445508,
"lm_q1q2_score": 0.8469438124906762
} |
https://math.stackexchange.com/questions/941150/what-is-the-difference-between-independent-and-mutually-exclusive-events/941158 | # What is the difference between independent and mutually exclusive events?
Two events are mutually exclusive if they can't both happen.
Independent events are events where knowledge of the probability of one doesn't change the probability of the other.
Are these definitions correct? If possible, please give more than one example and counterexample.
• They are, in a sense, completely opposite features. If $A$ and $B$ are independent, knowledge that $A$ occurred does not change the probabilities that $B$ may have occurred. Where as if $A$ and $B$ are disjoint, knowledge that $A$ occurred completely changes the probabilities that $B$ may have occurred by collapsing them to $0$. Feb 2 '16 at 22:56
• I just noticed that the definitions in this question look like they've been taken from my answer here. (Not that I mind or anything.) Feb 19 '16 at 9:50
• Consider taking out a card from a deck of $52$ playing cards. $S$: The card is a spade. $A$: The card is an ace. The two events are not mutually exclusive as there exists an Ace of Spades. $P(A) = \frac{4}{52}$ and $P(S) = \frac{1}{4}$. and $P(A\cap S) = \frac{1}{52} = \frac{4}{52} \frac{1}{4} = P(A) P(S)$ Mar 19 '17 at 13:21
• @alex.jordan If you wanted to highlight the analogy, you could say that independence and mutual exclusivity were defined by $P(AB)=P(A)P(B)$ and $P(A+B)=P(A)+P(B)$ respectively. Apr 11 '17 at 18:21
• The book Counterexamples in Probability (Third Edition) by J. M. Stoyanov (Dover, 2013) is a treasure trove of information. In particular, Section 3 of Chapter 1 explores INDEPENDENCE OF RANDOM EVENTS. Nov 28 '17 at 18:07
Yes, that's fine.
Events are mutually exclusive if the occurrence of one event excludes the occurrence of the other(s). Mutually exclusive events cannot happen at the same time. For example: when tossing a coin, the result can either be heads or tails but cannot be both.
\left.\begin{align}P(A\cap B) &= 0 \\ P(A\cup B) &= P(A)+P(B)\\ P(A\mid B)&=0 \\ P(A\mid \neg B) &= \frac{P(A)}{1-P(B)}\end{align}\right\}\text{ mutually exclusive }A,B
Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s). For example: when tossing two coins, the result of one flip does not affect the result of the other.
\left.\begin{align}P(A\cap B) &= P(A)P(B) \\ P(A\cup B) &= P(A)+P(B)-P(A)P(B)\\ P(A\mid B)&=P(A) \\ P(A\mid \neg B) &= P(A)\end{align}\right\}\text{ independent }A,B
This of course means mutually exclusive events are not independent, and independent events cannot be mutually exclusive. (Events of measure zero excepted.)
• can i get a real life example for better understanding. take a look i have edited Question and made it more clear. Sep 22 '14 at 5:48
• "This of course means..." Events of probability zero excluded.
– Did
Sep 22 '14 at 6:21
• Is there any connection between independent events and mutually exclusive events? I meant to ask "If $A$ and $B$ are mutually exclusive, what can be commented on the independence of $A$ and $B$ or vice versa." Or is there no such connection at all? I guess there is none. But just want to confirm.
– Maha
May 23 '15 at 15:22
• @Mahesha999 If two events are mutually exclusive, then they are NOT independent. Jun 24 '15 at 1:11
• If $A$ is independent from $B$, then it is independent from $\neg B$. Oct 27 '17 at 2:50
After reading the answers above I still could not understand clearly the difference between mutually exclusive AND independent events. I found a nice answer from Dr. Pete posted on math forum. So I attach it here so that op and many other confused guys like me could save some of their time.
If two events A and B are independent a real-life example is the following. Consider a fair coin and a fair six-sided die. Let event A be obtaining heads, and event B be rolling a 6. Then we can reasonably assume that events A and B are independent, because the outcome of one does not affect the outcome of the other. The probability that both A and B occur is
P(A and B) = P(A)P(B) = (1/2)(1/6) = 1/12.
An example of a mutually exclusive event is the following. Consider a fair six-sided die as before, only in addition to the numbers 1 through 6 on each face, we have the property that the even-numbered faces are colored red, and the odd-numbered faces are colored green. Let event A be rolling a green face, and event B be rolling a 6. Then
P(B) = 1/6
P(A) = 1/2
as in our previous example. But it is obvious that events A and B cannot simultaneously occur, since rolling a 6 means the face is red, and rolling a green face means the number showing is odd. Therefore
P(A and B) = 0.
Therefore, we see that a mutually exclusive pair of nontrivial events are also necessarily dependent events. This makes sense because if A and B are mutually exclusive, then if A occurs, then B cannot also occur; and vice versa. This stands in contrast to saying the outcome of A does not affect the outcome of B, which is independence of events.
Mutually exclusive event :- two events are mutually exclusive event when they cannot occur at the same time. e.g if we flip a coin it can only show a head OR a tail, not both.
Independent event :- the occurrence of one event does not affect the occurrence of the others e.g if we flip a coin two times, the first time may show a head, but this does not guarantee that the next time when we flip the coin the outcome will also be heads. From this example we can see the first event does not affect the occurrence of the next event.
If I toss a coin twice, the result of the first toss and the second toss are independent.
However the event that you get two heads is mutually exclusive to the event that you get two tails.
Suppose two events have a non-zero chance of occurring.
Then if the two events are mutually exclusive, they can not be independent.
If two events are independent, they cannot be mutually exclusive.
• Aren't the last two sentences saying the exact same thing? Dec 4 '17 at 15:44
• Yes, just for emphasis. Dec 4 '17 at 15:54
• @copper.hat So we can say say that if two events are mutually exclusive then they are dependent but if they are not mutually exclusive then they can either be dependent or independent, right? Sep 10 '20 at 17:23
• @user599310 Yes, that is correct. Sep 10 '20 at 17:28
This question already has very good answers, I'm gonna add a visualization for independents events using some special diagrams. In these diagrams proportion of events to sample space represents their probability. Our sample space is a rectangle of 9x5 = 45 units:
We have event A (3x5) so P(A) = 3x5/9x5 = 15/45 = 1/3:
And event B (9x3) so P(B) = 9x3/9x5 = 27/45 = 3/5:
These two events intersect as:
𝐴∩𝐵 occupies 3x3 units:
𝑃(𝐴∣𝐵) = 𝑃(𝐴∩𝐵) / 𝑃(𝐵) so 𝑃(𝐴∣𝐵) = 9/27 = 1/3. But this is same as P(A)!
and 𝑃(𝐵∣𝐴) = 𝑃(𝐴∩𝐵) / 𝑃(𝐴) so 𝑃(𝐵∣𝐴) = 9/15 = 3/5. But this is same as P(𝐵)!
As in the two last diagrams, occurrence of one event doesn't affect the probability of the other event, these two events are called independent. So knowledge about occurrence of one of them doesn't affect our knowledge about probability of the other one. But this is not because they have nothing in common, on the contrary they are kinda in harmony by wiping out (the given event reduces sample space to itself, so it wipes out its complement) sample space in such a way that the other event proportion to given event doesn't change. I'd like to remember them as perpendicular events.
Events are Independent when happening of one does not influence happening of other. Eruption of volcano on Earth and orbit of Mars do not influence each other, so are independent events.
Growth of human population and preservation of many other species are mutually exclusive, as the one can only happen if the other does not happen.
Strictly speaking, mutually exclusive does not imply that one of them must happen. If there is a large asteroid impact on Earth, then neither human population grows nor endangered species are preserved.
It will be easier if we distinguish "mutually exclusiveness" from "independency" by considering the sample space in mind.
• Two events that are compared for mutually exclusiveness must be from a single sample space. For example,
• Tossing a coin twice. $$A=\{HH\}$$ is an event in which the head shows up twice and $$B=\{TT\}$$ is an event in which the tail shows up twice. Their share the same sample space $$S=\{HH,HT,TH,TT\}$$. As $$A\cap B=\{\}$$, they are mutually exclusive.
• Two events that are compared for independency must be from two sample spaces. For example,
• Tossing a coin twice. $$A=\{H\}$$ is an event in which the head shows up in the first throw and $$B=\{T\}$$ is an event in which the tail shows up in the second throw. The sample space for the first trial is $$S_1=\{H,T\}$$ and the sample space for the second trial is $$S_2=\{H,T\}$$ As $$S_1=S_2$$, they are independent.
The next additional questions are "is it possible to have 2 events that are"
• "both mutually exclusive and independent?"
• "both mutually exclusive and dependent?" | 2022-01-17T23:57:48 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/941150/what-is-the-difference-between-independent-and-mutually-exclusive-events/941158",
"openwebmath_score": 0.805071234703064,
"openwebmath_perplexity": 328.9021170316671,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.972830769252026,
"lm_q2_score": 0.8705972650509008,
"lm_q1q2_score": 0.8469438070681778
} |
https://math.stackexchange.com/questions/2335223/in-an-equilateral-triangle-prove-that-bq-pq-cp-2l | # In an equilateral triangle, prove that $|BQ| + |PQ| + |CP| > 2l$
I am trying to solve the following problem:
Let $ABC$ be an equilateral triangle with side $l$.
If $P$ and $Q$ are points respectively in sides $AB$ and $AC$, different from the triangle vertices, prove that $$|BQ| + |PQ| + |CP| > 2l$$
I can see that, as point $P$ tends to $A$, $|CP|+|PQ|$ tends to $|AC|+|AQ|$. If I could prove this, the problem would be solved (the rest follows from the triangle inequality).
However I have no clue on how to do this. I tried to play with triangle inequality and relations between sides and angles but nothing worked.
How can I proceed?
• I think you have made a typo in your question - let $Q$ approach $B$ and $P$ approach $C$. Then the limit of $BQ$ and $CP$ is 0, and the limit of $PQ$ is $l$ (draw it out!). Therefore I think the inequality should be $|BQ| + |PQ| + |CP| > l$. – Toby Mak Jun 24 '17 at 23:56
• That's true, but if you let $Q$ and $P$ approach $A$, the sum approaches $|AB|+|AC|$. – J. C. Jun 25 '17 at 0:08
• But that is the more inefficient case - the case should hold for any $P$ and $Q$, not just one example that you have described. – Toby Mak Jun 25 '17 at 0:23
• Now I realized you might have mislabeled your sketch; actually $Q$ approaches $C$, not $B$, and the inverse for $P$. – J. C. Jun 25 '17 at 0:38
• If $P$ and $Q$ are the midpoints then the statement is trivial. If either $P$ or $Q$ are far from $A$ (by far I mean farther than midway) the result follows by the first case. So you are left only with the case when both $P$ and $Q$ are close to $A$. – user 1987 Jun 25 '17 at 0:43
Upon reflection ...
$$2s = |\overline{AB^\prime}| \leq |\overline{AP}|+|\overline{PQ^\prime}|+|\overline{Q^\prime B^\prime}| = p+q+r$$
• It was something amazing! – Michael Rozenberg Jun 25 '17 at 10:13
• Exactly what I was looking for! Fantastic – J. C. Jun 25 '17 at 13:22
Let $AQ=x$, $AP=y$ and $l=1$.
Thus, $$PQ=\sqrt{x^2-xy+y^2},$$ $$PC=\sqrt{y^2-y+1}$$ and $$BQ=\sqrt{x^2-x+1}$$ and we need to prove that $$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$ Now, by Minkowwski $$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}\geq$$ $$\geq\sqrt{\left(x-\frac{1}{2}+y-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=\sqrt{(x+y-1)^2+3}$$ and $$\sqrt{x^2-xy+y^2}\geq\frac{x+y}{2}.$$ Let $x+y=2a$.
Hence, $a\leq1$ and we need to prove that $$a+\sqrt{(2a-1)^2+3}\geq2$$ or $$\sqrt{4a^2-4a+4}\geq2-a$$ or $$4a^2-4a+4\geq a^2-4a+4,$$ which is obvious.
Done!
By my solution easy to make a geometric proof.
• Quite an interesting approach! I was looking for a more geometric argument, but thank you for the feedback. – J. C. Jun 25 '17 at 13:21
This is not an answer, rather it is a suggested approach to the problem. Here is a Geogebra (link)diagram:
If you go to the Geogebra link you can move points $P$ and $Q$.
The problem is reduced to proving that $|DI|>|IQ|$.
Note from the construction that $|PQ|=|DQ|$ and $|CP|=|BG|=|BF|$ and that $2l=|IH|$.
Therefore
$$|PQ|+|BQ|+|CP|=|DQ|+|QB|+|BF|=|DF|$$
So to establish the result it must be shown that
$$|DI|>|FH|=|IQ|$$
Refer to the graph (using smartphone, will draw at PC later) below: | 2019-05-21T23:49:15 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2335223/in-an-equilateral-triangle-prove-that-bq-pq-cp-2l",
"openwebmath_score": 0.8715707063674927,
"openwebmath_perplexity": 166.31367642766904,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307653134903,
"lm_q2_score": 0.8705972600147106,
"lm_q1q2_score": 0.8469437987399386
} |
https://math.stackexchange.com/questions/2418554/why-log-of-sum-of-exponentials-fx-log-left-sum-limits-i-1n-e-x-i-righ | # Why log-of-sum-of-exponentials $f(x)=\log\left(\sum\limits_{i=1}^n e^ {x_i}\right)$ is a convex function for $x \in R^n$
How to prove $f(x)=\log(\sum_{i=1}^n e^{x_i})$ is a convex function?
EDIT1: for above function $f(x)$, following inequalities hold:
$$\max\{x_1,x_2,...,x_n\} \le f(x) \le \max\{x_1,x_2,...,x_n\}+\log n$$
and I have tried proving its convexity via definition of a convex function with above inequalities. But that didn't work.
EDIT2: I have posted my answers below.
• Composition rules for preserving convexity. But it didn't work – Finley Sep 6 '17 at 4:18
• What about using induction? – Sergio Enrique Yarza Acuña Sep 6 '17 at 4:53
• A possible duplicate of math.stackexchange.com/questions/2416837/… – Math Lover Sep 6 '17 at 4:58
• @MathLover : It seems that I need to compute the Hessian of $f(x)$ that I tried to avoid before. – Finley Sep 6 '17 at 5:10
• @Finley At this moment, I can't think of any other way to prove that the function is convex. The individual entries of the Hessian matrix are given in math.stackexchange.com/questions/2416837/…. A proof is also given based on the C-S inequality. – Math Lover Sep 6 '17 at 5:13
Proof:
Let $u_i=e^ {x_i} ,v_i=e^ {y_i}$. So $f(\theta x+(1-\theta)y)=log(\sum_ {i=1}^n e^{\theta x_i + (1-\theta)y_i})=log(\sum_ {i=1}^n u_i^ \theta v_i^{(1-\theta)})$
From Hölder's inequality:
$$\sum_ {i=1}^n x_iy_i \le (\sum_ {i=1}^n|x_i|^p)^{\frac{1}{p}} \cdot (\sum_ {i=1}^n|x_i|^q)^{\frac{1}{q}}$$ where $1/p+1/q=1$.
Applying this inequality to $f(\theta x+(1-\theta)y)$: $$log(\sum_ {i=1}^n u_i^ \theta v_i^{(1-\theta)}) \le log[(\sum_ {i=1}^n u_i^ {\theta \cdot \frac{1}{\theta}})^ \theta \cdot (\sum_ {i=1}^n v_i^ {1-\theta \cdot \frac{1}{1-\theta}})^ {1-\theta}]$$ Right formula can be reduced to:
$$\theta log\sum_ {i=1}^n u_i+(1-\theta)log\sum_ {i=1}^n v_i$$
Here I regard $\theta$ as $\frac{1}{p}$ and $1-\theta$ as $\frac{1}{q}$.
So I achieve that $f(\theta x+(1-\theta)y) \le \theta f(x) + (1-\theta)f(y)$.
Another way to prove the convexity of this function is to use the Jensen's Inequality which states that a function is convex if and only if
$$f(tX+(1-t)Y) \le t f(X) + (1-t)f(Y)$$
Now let $X$ be represented by the vector $({X_1, X_2, X_3,... X_n})$,
and let $Y$ be represented by the vector $({Y_1, Y_2, Y_3,... Y_n})$
Let $t = \dfrac{1}{2}$
$$f(tX+(1-t)Y) = \log\left(\sum_{i=1}^{n} e^{\frac{X_i+Y_i}{2}}\right)$$
$$\text{RHS} = \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{X_i}\right)+ \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{Y_i}\right)$$
$$\text{RHS} = \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{X_i}\sum_{i = 1}^{n} e^{Y_i}\right)$$
RHS contains more cross product terms than the LHS thus making it larger than LHS and hence the function is convex.
• How to prove Jensen's inequality holds when $0 \le t \le 1$ more than $t=1/2$? – Finley Sep 6 '17 at 6:10
• According to definition of function convexity. $f(x)$ is convex if and only if Jensen's inequality holds for any $t \in [0,1]$ – Finley Sep 6 '17 at 6:13
• what t=(0,1) suggests is that you are taking a point in between X and Y vector. For any value of t between two point would range from 0-1. It not only applies for t= .5 but also any t within (0,1) which is essentially what you want – Satish Ramanathan Sep 6 '17 at 6:26
• (+1) It is also a consequence of CS-inequality: $$\sum_{i=1}^{n} e^{X_i/2}e^{Y_i/2} \leq \left( \sum_{i=1}^{n} e^{X_i} \right)^{1/2}\left( \sum_{i=1}^{n} e^{Y_i} \right)^{1/2}$$ This suggests that for general $t \in [0, 1]$ the same proof works by using Hölder's inequality instead. – Sangchul Lee Sep 6 '17 at 6:37
• I don't follow. Why is $\log(\sum_{i=1}^{n}e^{\frac{X_i+Y_i}2})=\frac12\log(\sum_{i=1}^{n}e^{X_i})+\log(\sum_{i=1}^ne^{Y_i})$? – user1551 Sep 6 '17 at 6:49
For a multivariate function to be convex, it's equivalent to show that its Hessian matrix is positive semi-definite. That is, you can calculate $$\nabla^2 f(\mathbf{x})$$ here and show it is positive semi-definite.
This can be proved using Cauchy Schwarz inequality as shown here.
• If anyone doesn't understand how CS-inequality is applied at the end of the linked slide, Boyd's book section 3.1.5 has a proof on this as well. – Mong H. Ng Sep 8 at 23:48 | 2019-11-19T02:34:37 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2418554/why-log-of-sum-of-exponentials-fx-log-left-sum-limits-i-1n-e-x-i-righ",
"openwebmath_score": 0.919211208820343,
"openwebmath_perplexity": 472.21297594119017,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9728307653134903,
"lm_q2_score": 0.8705972566572503,
"lm_q1q2_score": 0.846943795473698
} |
https://www.theoriginaldish.com/2021/05/23/1qsxy94/ | # Math probability problems
## Math probability problems
How likely something is to happen.The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event.Probability Definition in Math.Probability word problems worksheet.Compute probabilities for simple compound events, using such methods as organized lists, tree diagrams, and area models.However, we need to think deeper.In the problem math probability problems above, the experiment is spinning the spinner.Featured on Meta Testing three-vote close and reopen on 13 network sites.Whenever we face the probability topic in statistics, most of the problems with a well-shuffled pack of 52 playing cards.Probability Definition in Math.\frac {5 + 15} {5+15+16+20} 5+15+16+205+15...Problem : If a coin is flipped twice, what is the probability that math probability problems it will land heads once and tails once?\displaystyle P (A) = \frac {1} {6^3} P (A) = 631..Statistics and Probability Problems with Answers - sample 1: mean, mode, standard deviation, quartiles, frequency distribution of qualitative data.Try the given examples, or type in your own problem and check.\displaystyle P (A) = \frac {1} {6^3} P (A) = 631..Statistics and Probability Problems with Answers - sample 2: probability, counting, mean and standard deviation, mean of grouped.Solution: Let n(S) be the exhaustive cases, n(A) be the number of events of red card and n(B) be the number of events of face cards.For example, when a test is conducted, then the student can either get a pass or fail.It is a state of probability Most of the problems incorporate traditional probability concepts like spinners, coins, and dice.Answer: P C PT P C P PT P C PT( ) ( ) ( ) ( ).
#### M tech thesis on ofdm, math probability problems
What is the probability that the problem is solved?Featured on Meta Testing three-vote close and reopen on 13 network sites.The best we can say is how likely they are to happen,.Example 3: There are 20 boys and 30 girls in a class.Many events can't be predicted with total certainty.Hence, the probability that the ball may be either red or blue or white is 17/24.The likelihood of any event to occur is a number between 0 and 1, where 0 indicates the impossibility of the event and 1 indicates certainty Free online tutorials cover statistics, probability, regression, analysis of variance, survey sampling, and matrix algebra - all explained in plain English.For independant events input 2 values.Tutorial on finding the probability of an event.B: On a six-sided die, the probability of throwing any number is 1 in 6.504 Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor Using Probability.In this case: Probability of a coin landing on heads.But we are going to make things easy for you and help you solve such problems with ease and without stress Example 8: A problem is given to three persons P, Q, R whose respective chances of solving it are 2/7, 4/7, 4/9 respectively.What is the probability the sum of the dice is 5?Probability - math problems Probability math probability problems is the measure of the likeliness that an event will occur.504 10 P 4 10 4 = 5040 10000 = 0.Browse other questions tagged probability probability-theory or ask your own question.An outcome is the result of a single trial of an experiment.[Solved] Assume that the heights of adult Caucasian women have a mean of 63 inches and a standard deviation o - Other probability problems #26503.For K-12 kids, teachers and parents.Very unlikely or almost certain.Now practice our shortcut tricks on probability problem on coin and read examples.Probability is the likliehood that a given event will occur and we can find the probability of an event using the ratio number of favorable outcomes / total number of outcomes.Probability is a measure of the likelihood of an event to occur.Learn and practice basic word and conditional probability aptitude questions with shortcuts, useful tips to solve easily in exams Probability problems and sampling with and without replacement.To solve a probability problem identify the event, find the number of outcomes of the event, then use probability law: $$\frac{number\ of \ favorable \ outcome}{total \ number \ of \ possible \ outcomes}$$ Probability Problems Probability Problems – Example 1:.An outcome is the result of a single trial of an experiment.0 is impossible event and 1 (100%) means the certainty event.Statistics and Probability Problems with Answers - sample 2: probability, counting, mean and standard deviation, mean of grouped.Browse other questions tagged probability probability-theory or ask your own question.Find the probability of getting it red or face card.Probability With a Deck of Cards Worksheet.This is the currently selected item.Probability Using a Spinner Worksheet The probability that you will draw a green or a red marble is.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-08-02T10:06:04 | {
"domain": "theoriginaldish.com",
"url": "https://www.theoriginaldish.com/2021/05/23/1qsxy94/",
"openwebmath_score": 0.6884915232658386,
"openwebmath_perplexity": 1455.7274311823764,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9873750518329435,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8469388307152881
} |
https://math.stackexchange.com/questions/3128227/how-can-i-simplify-4ab-1-2 | # How can I simplify $(4ab^{-1})^{-2}$
As part of a wider expression I have a component $$(4ab^{-1})^{-2}$$
I know that using the rules of exponents, if there was no radical within the brackets I could rewrite like this:
$$\frac{(4ab)}{2}$$
I also know that if the only component within the brackets were $$b^{-1}$$ then I could multiple to be $$b^{-1 * -2}$$ = $$b^2$$
But I cannot see how to combine these two pieces.
How can I simplify $$(4ab^{-1})^{-2}$$? Baby steps very much appreciated.
• Remember, $\color{blue}{(ABC)^n = A^n B^n C^n}$. So $(4ab^{-1})^{-2} = 4^{-2} a^{-2} \left(b^{-1}\right)^{-2}$. Can you take it from here? (Also, it would not be correct to say that $(4ab)^{-2}$ equals $\frac{4ab}{2}$.) – Minus One-Twelfth Feb 27 at 0:41
• WOuld it be correct to say that it equals $\frac{1}{4ab^2}$ ? – Doug Fir Feb 27 at 0:51
• @MinusOne-Twelfth $(4ab)^{-2}$ IS NOT = $4ab/2$ – NoChance Feb 27 at 0:56
• @DougFir Be careful with how far that power "-1" ranges. $4ab^{-1}$ means $4 \times a \times b^{-1}$, not $(4ab)^{-1}$. So $4ab^{-1} = \frac{4a}{b}$. More generally, try not to skip steps: work slowly and take it one step at a time, making sure you understand the rule you're using at each stage! – Billy Feb 27 at 0:58
• @NoChance I know – that's what I said. – Minus One-Twelfth Feb 27 at 0:59
There are several ways to simplify this, but I suggest you work your way from the inside out as this seems to be the easiest way in general, and is what using brackets normally implies doing. Note that $$x^{-n} = \cfrac{1}{x^n}$$. As such, first we get that
$$4ab^{-1} = \cfrac{4a}{b} \tag{1}\label{eq1}$$
Next, using \eqref{eq1}, plus that $$\cfrac{1}{\frac{c}{d}} = \cfrac{d}{c}$$, we get that
$$(4ab^{-1})^{-2} = \cfrac{1}{{\left(\cfrac{4a}{b}\right)}^2} = \cfrac{1}{\cfrac{16a^2}{b^2}} = \cfrac{b^2}{16a^2} \tag{2}\label{eq2}$$
Since $$(ABC)^n=A^n B^n C^n$$ as in the comments, you have: \begin{align} (4ab^{−1})^{−2}&=4^{-2}\,a^{-2}\,\left(b^{-1}\right)^{-2}\\ &=4^{-2}\,a^{-2}\,b^{-1\cdot(-2)}\\ &=\frac{1}{4^2}\cdot\frac{1}{a^2}\cdot b^2\\ &=\frac{b^2}{16\,a^2} \end{align} | 2019-09-23T08:57:43 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3128227/how-can-i-simplify-4ab-1-2",
"openwebmath_score": 0.9114957451820374,
"openwebmath_perplexity": 475.50842082319264,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750488609223,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8469388281659831
} |
https://math.stackexchange.com/questions/2973194/what-is-2632-bmod-12 | # What is $26^{32}\bmod 12$? [duplicate]
What is the correct answer to this expression:
$$26^{32} \pmod {12}$$
When I tried in Wolfram Alpha the answer is $$4$$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $$0.$$
## marked as duplicate by Carl Mummert, Lee David Chung Lin, Shailesh, Cesareo, Adrian KeisterMay 1 at 15:40
• How can it be $0$? $3$ does not divide $26$, so $12 (= 3 \times 4)$ can't divide $26^{32}$ – ab123 Oct 27 '18 at 9:39
• So the answer 4 is correct? – Naruto Uzumaki Oct 27 '18 at 9:43
• yes the answer $4$ is correct – ab123 Oct 27 '18 at 9:46
• I would suggest that, depending on exactly how you did this on a calculator, at some stage the calculator encountered a very large exponent and truncated the result, throwing off the rest of the calculation. – Sam Streeter Oct 27 '18 at 9:51
• Alpha failing on such a "basic" computation is more than unlikely. Even the Windows calculator returns $4$ ! (by chance ?) – Yves Daoust Oct 27 '18 at 11:03
First, note that $$26 \equiv 2 \pmod {12}$$, so $$26^{32} \equiv 2^{32} \pmod {12}$$.
Next, note that $$2^4 \equiv 16 \equiv 4 \pmod {12}$$, so $$2^{32} \equiv \left(2^4\right)^8 \equiv 4 ^8 \pmod {12}$$, and $$4^2 \equiv 4 \pmod {12}$$.
Finally, $$4^8 \equiv \left(4^2\right)^4 \equiv 4^4 \equiv \left(4^2\right)^2 \equiv 4^2 \equiv 4 \pmod {12}$$.
Then we get the result.
There are slicker solutions with just a few results from elementary number theory, but this is a very basic method which should be easy enough to follow.
• I elaborated on the algebraic essence in my answer, since for beginners that may not be clear from above. – Bill Dubuque Oct 29 '18 at 16:50
Like
Get the last two digits of $$16^{100}$$ and $$17^{100}$$
How to find last two digits of $$2^{2016}$$
last two digits of $$14^{5532}$$?
Find the last two digits of $$2^{2156789}$$
$$26\equiv-1\pmod3$$
$$\implies26^{2n}\equiv(-1)^{2n}\equiv1\pmod3$$ where $$n$$ is any integer
$$\implies26^{2n+2}\equiv1\cdot26^2\pmod{3\cdot26^2}$$
$$\implies26^{2n+2}\equiv26^2\pmod{3\cdot2^2}\equiv?$$
Note that $$26\equiv 2\pmod{12}$$, so we can as well compute $$r=2^{32}\bmod 12$$.
We have $$2^{32}=12q+r$$, so clearly $$r=4s$$, so we are reduced to compute $$2^{30}\bmod 3$$ and now we can apply Fermat’s little theorem: $$2^2\equiv 1\pmod 3$$; thus $$2^{30}\equiv 1=s\pmod{3}$$ and therefore $$r=4$$.
• Readers may be interested to know that this factoring out of $4$ can be done efficiently operationally using the mod distributive law (an operational reformulation of CRT) - see my answer. – Bill Dubuque Oct 27 '18 at 18:50
$$\smash[b]{\text{Note that }\ \overbrace{\color{#0a0}{26}^{\large 32}\!\bmod 12 = \color{#0a0}2^{\large 32}\!\bmod{12}}^{\Large\ \ \color{#0a0}{26\ \equiv\ 2}\pmod{\!12}} \,=\, 2^{\large 2}\overbrace{(\color{#c00}2^{\large 30}\!\bmod 3) = 4((\color{#c00}{-1})^{\large 30}\!\bmod 3)}^{\Large \color{#c00}{2\ \equiv\ -1}\pmod{\!3}} = 4}$$
$$\smash[t]{\text{The middle equality uses }\,\ \overbrace{ab\ \bmod ac\ =\ a\ (\,b\ \bmod\ c)}}\, =\,$$ mod Distributive Law.
Or we can notice that $$\,\color{#c00}{4^{\large n}\equiv 4}\pmod{\!12},$$ since it is true mod $$3$$ & $$4\!:$$ $$\,1^{\large n}\!\equiv 1\,$$ & $$\,0^{\large n}\!\equiv 0$$
So $$\bmod 12\!:\ 26^{\large 32}\!\equiv 2^{\large 32}\!\equiv \color{#c00}{4^{\large 16}\!\equiv 4}.\$$ This is the idea behind Sam's answer.
Remark Numbers like $$4$$ above with $$a^2=a$$ are called idempotents. This implies $$\,a^{\large n} = a\,$$ for all $$n\ge 2$$ either as above or by a simple induction $$\,a^{\large n+1}\! = a\,a^{\large n} = a\cdot a = a.\,$$ Said more conceptually: note $$a$$ is a fixed point of $$\,f(x) = ax\,$$ i.e. $$\,f(a) = a,\,$$ and fixed points always stay fixed on iteration by a simple induction: if $$\,\color{#c00}{f^n(a) = a}\,$$ then $$\,f^{\large n+1}(a) = f(\color{#c00}{f^{\large n}(a)}) = f(\color{#c00}a) = a$$.
Idempotents $$\!\bmod n$$ play a fundament role in factorization of integers (and rings) since they correspond to splittings of $$n$$ into coprime factors, cf. last paragraph here,. However, the first method is more general since the base is generally not idempotent in such exponentiation problems (in fact the mod Distributive Law is essentially an operational form of CRT = Chinese Remainder Theorem - which yields a general way of solving this and related problems).
Note that $$26^{32}$$ is clearly divisible by $$4$$ but not by $$3$$, so the possible congruence classes modulo $$12$$ are immediately only (represented by) $$4$$ and $$8$$. We have $$4\equiv 1 \bmod 3$$ and $$8\equiv 2 \equiv -1 \bmod 3$$, so to distinguish between them we only need to know the answer modulo $$3$$.
This comes down to $$(-1)^{32}=1$$ (or we can use little Fermat to the same effect) and the answer is therefore $$4$$.
• Splitting the $12$ into the coprime factors $4$ and $3$ is a technique which is substantially generalised in the Chinese Remainder Theorem. The observations in this case are elementary and don't need complex machinery, but it is there if needed to steer the analysis of more complex cases. – Mark Bennet Oct 27 '18 at 10:54
• This implicit factoring out of 4 can be explicitly achieved operationally using the mod distributive law (an operational reformulation of CRT) - see my answer. – Bill Dubuque Oct 27 '18 at 19:07
• @BillDubuque A useful and interesting perspective as usual! – Mark Bennet Oct 27 '18 at 20:51 | 2019-11-18T11:07:12 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2973194/what-is-2632-bmod-12",
"openwebmath_score": 0.8908438682556152,
"openwebmath_perplexity": 334.5603446132774,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750483036683,
"lm_q2_score": 0.857768108626046,
"lm_q1q2_score": 0.8469388276879883
} |
https://math.stackexchange.com/questions/1005763/does-limit-of-x-n-1-frac121-frac14-1-frac12n-exist | # Does limit of $x_n=(1+\frac{1}{2})*(1+\frac{1}{4})*…*(1+\frac{1}{2^n})$ exist?
I tried to show that the sequence is increasing and limited, but couldn't find the limit. I also tried with squeeze theorem, but $(1+\frac{1}{2^n})^n<=x_n<=(1+\frac{1}{2})^n$ is not helping and I ran out of ideas.
• – John Dvorak Nov 4 '14 at 12:25
• Try taking the ln of $x_n$, and use : $ln(1+u) \leq u$ you'll get : $a_n = ln(1+\frac{1}{2^n}) \leq \frac{1}{2^n} = v_n$ ; $\sum v_n$ converges, hence $\sum a_n$ does as well, and your sequence converges – mvggz Nov 4 '14 at 12:26
• You're right, I meant limited. Edited first post. – duke Nov 4 '14 at 12:26
• Use the AM/GM inequality to provide a bound? – Mark Bennet Nov 4 '14 at 12:29
• – Lucian Nov 4 '14 at 19:06
A handy fact about products like this is $$1+\sum_{k=1}^n a_k \le \prod_{k=1}^n (1+a_k) \le \exp\Big(\sum_{k=1}^n a_k\Big)$$ (The first inequality is a generalization of Bernoulli's inequality; the second is $1+x\le e^x$ used $n$ times.) So you can check the convergence of this kind of product by checking the convergence of a related series, for which we have a bunch of standard techniques. In this case, the related series is a familiar one.
• Thank You. Technically I still hasn't had series and series convergency at uni, but this seems reasonable. – duke Nov 4 '14 at 12:49
$$A_N=\prod_{n=1}^{N}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n=1}^{N}\log\left(1+\frac{1}{2^n}\right)\leq\exp\sum_{n=1}^{N}\frac{1}{2^n}\leq e.$$ Since the LHS is increasing and bounded, the limit $$\lim_{N\to +\infty} A_N = \prod_{n=1}^{+\infty}\left(1+\frac{1}{2^n}\right)$$ exists.
I'll write it this way:
Let : $S_n = ln(x_n) = \sum_{k=0}^n ln(1 + \frac{1}{2^k})$
As I said: $a_n = ln(1+\frac{1}{2^n}) \leq \frac{1}{2^n} = v_n$
You get : $S_n \leq \sum_{k=1}^n v_k \leq \sum_{k=1}^{+\infty} v_k = 1$
$(S_n)$ is increasing, bounded so it converges. | 2019-07-17T02:26:10 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1005763/does-limit-of-x-n-1-frac121-frac14-1-frac12n-exist",
"openwebmath_score": 0.8948157429695129,
"openwebmath_perplexity": 557.1052047505408,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750496039276,
"lm_q2_score": 0.8577681013541613,
"lm_q1q2_score": 0.8469388216232318
} |
https://math.stackexchange.com/questions/911508/whats-the-sum-of-this-series | # What's the sum of this series? [duplicate]
I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?
Thanks!
## marked as duplicate by user147263, apnorton, colormegone, hardmath, Claude LeiboviciAug 28 '14 at 5:44
• math.stackexchange.com/questions/430973 – David Aug 28 '14 at 0:10
• Which do you want, to estimate the sum or find the exact value? – Robert Israel Aug 28 '14 at 0:16
• An explicit formula for your sum can be obtained quickly from Euler's famous result $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6}$. That result has many proofs, none really simple. Nowadays it is usually derived as a consequence of certain facts about Fourier series. – André Nicolas Aug 28 '14 at 0:29
• See Basel problem. – Lucian Aug 28 '14 at 0:31
• Note: this is $\eta(2)$, where $\eta$ is the Dirichlet eta function. Just in case anyone wanted to ask about another value. – apnorton Aug 28 '14 at 2:10
The sum of this series is $\frac{\pi^2}{12}$.
Explantion
$$1+ \frac{1}{2^2} + \frac{1}{3^2} + \text{...} = \frac{\pi^2}{6}$$
HINT
Note that:
$$\large \frac{-1}{2^2} = \frac{1}{2^2} - \frac{1}{2^2} - \frac{1}{2^2}$$
Now,
$$1- \frac{1}{2^2} + \frac{1}{3^2} - \text{...} = \left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right) - 2 * \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ...\right)$$
$$= \frac{\pi^2}{6} - \frac{2}{2^2}\left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right)$$
$$= \frac{\pi^2}{6} - \frac{1}{2}*\frac{\pi^2}{6}$$
$$= \frac{\pi^2}{12}$$
Comment if you have questions.
• How do you know that? – inggumnator Aug 28 '14 at 0:08
• @inggumnator I'll post shortly the solution. – Varun Iyer Aug 28 '14 at 0:08
• @inggumnator do you understand now how it is derived? I hope this helps. – Varun Iyer Aug 28 '14 at 0:15
• @VarunIyer: Nice derivation. It might be worthwhile to mention that the rearrangement of the series is valid because it is absolutely convergent (to $\pi^2/6$, as you noted). – Bungo Aug 28 '14 at 0:21
• @inggumnator yes. If you notice we already have a multiple of $2$ outside the parenthesis. The common term here is $\frac{1}{2^2}$, so we pull that out as well to get our sum $\frac{\pi^2}{6}$ again. Thus the number outside the parenthesis now is $\frac{2}{2^2}$. – Varun Iyer Aug 28 '14 at 1:24
It is well known that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
Thus, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \sum_{m=1}^{\infty}\dfrac{1}{(2m)^2} = \dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$.
Hence, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} = \sum_{n=1}^{\infty}\dfrac{1}{n^2} - \sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$.
Finally, $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2} = \displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} - \displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{12}$.
If we did not know an exact answer, here is a crude way to estimate the sum of the series.
Our series is an alternating series, with terms that decrease in absolute value and have limit $0$. So the error made by truncating at a particular term has absolute value less than the absolute value of the first "neglected" term.
For example, if we use $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}$ as an approximation, then the absolute value of the error is less than $\frac{1}{6^2}$. Furthermore, the error is "negative," that is, our estimate is greater than the true value.
• Thanks, this is a nifty trick! – inggumnator Aug 28 '14 at 0:53
• You are welcome. For the kind of series described (signs alternate, terms go down in absolute value, approach $0$) the truncation error is easy to estimate. – André Nicolas Aug 28 '14 at 3:33
HINT:
This can be written as $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i^2}$$
Setting $\displaystyle a_i= \frac{(-1)^{i+1}}{i^2}$:
$$\sum_{i=1}^{\infty} a_{2i-1}=\sum_{i=1}^{\infty} \frac{1}{(2i-1)^2}$$
$$\sum_{i=1}^{\infty} a_{2i}=-\frac{1}{4}\sum_{i=1}^{\infty} \frac{1}{i^2}$$
When the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge then the series $\displaystyle \sum_{i=1}^{\infty} a_{i}$ also converges.
$$\sum_{i=1}^{\infty} a_{i}=\sum_{i=1}^{\infty} a_{2i-1}+\sum_{i=1}^{\infty} a_{2i}$$
Find where the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge.
• Sorry why does finding where the two series converge show where the original series converge? – inggumnator Aug 28 '14 at 0:40
• When $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}=x$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}=y$ then $$\sum_{i=1}^{\infty} a_{i}=x+y$$ – user159870 Aug 28 '14 at 0:43 | 2019-07-19T14:30:34 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/911508/whats-the-sum-of-this-series",
"openwebmath_score": 0.885866641998291,
"openwebmath_perplexity": 559.7979730837434,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750510899382,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8469388211028647
} |
http://mathhelpforum.com/calculus/91408-find-limit-recursion-formula.html | # Math Help - Find the limit of a recursion formula
1. ## Find the limit of a recursion formula
Hello,
I am looking at this recursion formula in my Calculus book.
x_n+1 = 1/2(x_n + 2/x_n)
It goes on to say that we can find the limit of this sequence by doing this:
L = 1/2(L + 2/L) fine
but now it says that this can be rewritten as L^2 = 2 and that the limit is the square root of 2. How?
Could someone please explain the last two steps? Thank you so much.
2. Hello, pberardi!
It's only algebra . . .
I am looking at this recursion formula in my Calculus book:
. . $x_{n+1} \:=\: \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$
It goes on to say that we can find the limit of this sequence by doing this:
. . $L \:=\: \frac{1}{2}\left(L + \frac{2}{L}\right)$ . . . fine
but now it says that this can be rewritten as $L^2 = 2$
and that the limit is $\sqrt{2}$. .How?
We have: . $L \:=\:\frac{L}{2} + \frac{1}{L}$
Multiply by $2L\!:\quad 2L^2 \:=\:L^2 + 2 \quad\Rightarrow\quad L^2 \:=\:2 \quad\Rightarrow\quad L \:=\:\sqrt{2}$
3. Originally Posted by pberardi
Hello,
I am looking at this recursion formula in my Calculus book.
x_n+1 = 1/2(x_n + 2/x_n)
It goes on to say that we can find the limit of this sequence by doing this:
L = 1/2(L + 2/L) fine
but now it says that this can be rewritten as L^2 = 2 and that the limit is the square root of 2. How?
Could someone please explain the last two steps? Thank you so much.
Multiply both sides of that equation by 2L to get rid of the fractions- $2L^2= L^2+ 2$ which is the same as $L^2= 2$. That actually has two roots. In order to determine whether the limit is $\sqrt{2}$ or $-\sqrt{2}$ for a specific sequence you must also be given a "starting" value of $x_0$. Are you given something like " $x_0= 1$".
In fact, it should be clear that if $x_n$ is positive, then $x_{n+1}= (1/2)(x_n+ 1/x_n)$ is also positive while if $x_n$ is negative, then $x_{n+1}= (1/2)(x_n+ 1/x_n)$ is also negative.
That is, if $x_0$ is positive, the limit is $\sqrt{2}$ and if $x_0$ is negative, the limit is $-\sqrt{2}$.
By the way, they get that original formula by arguing that if $\lim_{x\to \infty} x_n$ exists then you can call that limit L and use the laws of limits to argue that $lim x_{n+1}= (1/2)(lim x_n+ 1/lim x_n)$ and because " $x_{n+1}$" is just the same sequence renumbered, each of those limits is L: $L= (1/2)(L+ 1/L)$.
To be rigorous, you would still have to show the limit exists.
(Blast! Soroban got in ahead of me again!)
4. Cool, I'm doing a variation of this question for uni. It's starting to come together for me now. The actual question is:
$x_{n+1} = \frac{1}{2}(x_n + \frac{\alpha}{x_n})$
and priors are $\alpha > 0$ and $x_1 >\sqrt{\alpha}$
My main question is how does the substitution for L work? You substitute L for x_(n+1) and for x_n? I guess that requires the assumption that the sequence is converging, right? If not, why not? If so, how do you prove monotonic decrease and find the limit, given the series is recursive? | 2015-07-31T13:43:57 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/91408-find-limit-recursion-formula.html",
"openwebmath_score": 0.9591599106788635,
"openwebmath_perplexity": 272.1334771852378,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9873750507184357,
"lm_q2_score": 0.85776809953619,
"lm_q1q2_score": 0.8469388207842018
} |
https://math.stackexchange.com/questions/1997916/finding-all-solutions-for-an-equation-system | # Finding all solutions for an equation system
I have a vector of numbers $x_i$ that can only have the values $0$ or $1$.
I need to find all the possible combinations $(x_1,x_2,x_3,x_4,x_5,x_6)$ such that:
$$\begin{cases} x_1+x_5+x_6 \ \mathrm{is\ odd}\\ x_2+x_4+x_6 \ \mathrm{is\ odd} \\ x_3+x_4+x_5 \ \mathrm{is\ odd} \end{cases}$$
How could I do this? I mean, I can think of some combinations but I don't know how to find all of them, or the number of possible solutions.
### First a preliminary remark:
If we fix one number $N$ in such a conditional, we find $$a + b + N \text{ odd}, a,b,N \in \{0, 1\}$$ implies for $N = 1$: $$a + b + 1 \text{ odd} \Leftrightarrow a = b$$ and for $N = 0$: $$a + b + 0 \text{ odd} \Leftrightarrow a \ne b \Leftrightarrow a = 1 - b$$
### Now we can start by systematically making cases:
• $x_1 = 1$
Equation (I) now reads $1 + x_5 + x_6$ is odd, i.e. $x_5 = x_6$
• $x_5 = x_6 = 1$
Now Equations (II) and (III) combine to $x_2 = x_3 = x_4$, so we have $(1, 0, 0, 0, 1, 1)$ and $(1, 1, 1, 1, 1, 1)$ as solutions
• $x_5 = x_6 = 0$
This means $x_2 \ne x_4$ and $x_3 \ne x_4$, so $x_2 = x_3 = 1 - x_4$.
Again we find two solutions:
$(1, 0, 0, 1, 0, 0)$ and $(1, 1, 1, 0, 0, 0)$
• $x_1 = 0 \Rightarrow x_5 \ne x_6$
• $x_5 = 1, x_6 = 0$ gives us $x_3 = x_4$ and $x_2 \ne x_4$, so $x_2 = 1 - x_3 = 1 - x_4$ and we get the two solutions $(0, 1, 0, 0, 1, 0)$ and $(0, 0, 1, 1, 1, 0)$
• $x_5 = 0, x_6 = 1$ and we get $x_2 = x_4$ and $x_3 \ne x_4$, so $x_2 = 1 - x_3 = x_4$ yielding the final two solutions $(0, 1, 0, 1, 0, 1)$ and $(0, 0, 1, 0, 0, 1)$
• That was smooth! Thank you. Those would be all the solutions, right? I'm asking just in case, because I don't think there are more possibilities – Tendero Nov 3 '16 at 17:40
• @Tendero Yes, these are all by exhaustion. I provided reason for the derivations in a preliminary note, just to clarify. Can you prove these statements yourself? – AlexR Nov 3 '16 at 17:41
• Yes, I get that but I didn't know if there was some systematic method to follow to get all the possibilities. Thanks again! – Tendero Nov 3 '16 at 17:42
This question is really about solving the system
$$\begin{cases} x_1+x_5+x_6 =1 \\ x_2+x_4+x_6 =1 \\ x_3+x_4+x_5 =1 \end{cases}$$
in $\mathbb{Z}/2\mathbb{Z}$. Since that's a field, we should be able to use all the techniques of linear algebra, and just reduce a matrix:
$\left[\begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 & 1 \end{array}\right]$
This matrix is already in reduced row-echelon form, so there are three free variables, and 8 solutions, corresponding to the eight possible values of $x_4$, $x_5$, and $x_6$, and completed by the equations $x_1=1+x_5+x_6$, $x_2=1+x_4+x_6$, and $x_3=1+x_4+x_5$, with all addition being performed in $\mathbb{Z}/2\mathbb{Z}$.
Namely:
\begin{align} (x_4,x_5,x_6)=(0,0,0) &\implies (1,1,1,0,0,0) \\ (x_4,x_5,x_6)=(0,0,1) &\implies (0,0,1,0,0,1) \\ (x_4,x_5,x_6)=(0,1,0) &\implies (0,1,0,0,1,0) \\ (x_4,x_5,x_6)=(0,1,1) &\implies (1,0,0,0,1,1) \\ (x_4,x_5,x_6)=(1,0,0) &\implies (1,0,0,1,0,0) \\ (x_4,x_5,x_6)=(1,0,1) &\implies (0,1,0,1,0,1) \\ (x_4,x_5,x_6)=(1,1,0) &\implies (0,0,1,1,1,0) \\ (x_4,x_5,x_6)=(1,1,1) &\implies (1,1,1,1,1,1) \end{align}
Reflecting back on the original problem, since each of $x_1, x_2, x_3$ each occur in precisely one condition, we see that the other three variables can be chosen freely, and then $x_1$ picked to make the first condition work, etc.
• Much cleaner than my solution. You might want to list all eight solutions obtained this way. +1 – AlexR Nov 3 '16 at 18:19 | 2019-11-20T04:13:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1997916/finding-all-solutions-for-an-equation-system",
"openwebmath_score": 0.9983874559402466,
"openwebmath_perplexity": 380.9908665733779,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9873750496039276,
"lm_q2_score": 0.8577680995361899,
"lm_q1q2_score": 0.8469388198282123
} |
http://math.stackexchange.com/questions/265023/distribute-10-white-and-10-black-balls-into-20-distinct-boxes-s-t-no-box-is-emp/265029 | Distribute 10 white and 10 black balls into 20 distinct boxes s.t. no box is empty?
How many ways are there to distribute 10 white and 10 black balls into 20 distinct boxes such that no box is empty?
Solution is that ${20!}/{10!10!}$. How can we write this? Please can you explain this solution clearly?
-
2 Answers
There are several ways to see it. The one that I find easiest is to notice that once we choose which $10$ boxes get white marbles, we’ve specified the entire distribution: the black marbles have to go into the remaining $10$ boxes. Thus, there is one distribution for each possible way of choosing $10$ of the $20$ boxes to contain the white marbles. There are $\binom{20}{10}$ ways to choose $10$ things out of a set of $20$, so there are $$\binom{20}{10}=\frac{20!}{10!10!}$$ possible distributions.
Another way is to start by pretending that the $20$ marbles are labelled, say $W_1$ through $W_{10}$ for the white marbles and $B_1$ through $B_{10}$ for the black marbles.. There are then $20!$ distinguishable permutations of these $20$ marbles. However, the marbles aren’t actually labelled. Pick any distribution, and focus on the white marbles. We could rearrange them as we please, so long as we leave the black marbles in place, and the resulting distribution would be indistinguishable from the one we started with. There are $10!$ ways to arrange the $10$ white marbles in those same $10$ positions; we’ve counted them all separately, but in fact they’re all the same arrangement of colors. Thus, we’ve overcounted by a factor of $10!$. If we divide by $10!$ to correct for the overcounting, we get $\frac{20!}{10!}$, which is the number of arrangements if the white marbles are indistinguishable, but the black marbles bear individual labels so that we can tell them apart.
In fact the black marbles bear no such labels. We’ve counted each set of $10$ positions for the black marbles once for every permutation of the black marbles in those positions, so we’ve overcounted by another factor of $10!$. Dividing by $10!$ to compensate finally gives us the correct answer for the unlabelled marbles:
$$\frac{20!}{10!10!}\;.$$
-
There are $20$ boxes and $20$ balls in total, so each box must receive exactly one ball. The question is reduced to choosing which $10$ boxes get black balls and which $10$ get white balls, which is equivalent to choosing $10$ boxes to put the black balls in and filling the rest with white balls. The number of ways to choose $10$ boxes out of $20$ is $${20 \choose 10} = \frac{20!}{10!\cdot 10!},$$ which is the answer.
-
Thx @ybungalobill – B11b Dec 25 '12 at 18:10 | 2015-10-09T14:14:40 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/265023/distribute-10-white-and-10-black-balls-into-20-distinct-boxes-s-t-no-box-is-emp/265029",
"openwebmath_score": 0.814419686794281,
"openwebmath_perplexity": 140.4598976529773,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9940889321944897,
"lm_q2_score": 0.8519528000888387,
"lm_q1q2_score": 0.8469168493204192
} |
http://mathematica.stackexchange.com/questions/2116/why-round-to-even-integers | # Why round to even integers?
According to the Mathematica help:
Round rounds numbers of the form x.5 toward the nearest even integer.
For example:
Round[{0.5, 1.5, 2.5, 3.5, 4.5}]
gives
{0, 2, 2, 4, 4}
What's the rationale behind this? Why not the usual x.5 always rounds up?
-
It is called bankers' rounding. The rationale is that the rounding behaves "nicely" even if you have negative numbers, i.e. rounding commutes with negation, which is nice.
-
Besides that rounding to even ensures you still have an integer if you divide the rounded result by two later on. Chemist normally round to even when measurements and computations are rounded to the nearest 0.001 and stuff like that. – Ted Ersek Feb 22 '12 at 0:19
That's a nice property, but rounding away-from-zero does has the same property, and is more intuitive. I'm pretty sure the true reason is Mr. Wizard's answer. – BlueRaja - Danny Pflughoeft Feb 22 '12 at 4:36
I think the reason is to prevent biasing numbers on average upward or downward.
For example if you have a list of numbers that include a lot of x.5 and you were to round all these upward, the average magnitude of the list would also shift upward. Likewise if you round downward, downward.
By rounding to the nearest even number these would balance out, given enough samples.
SetAttributes[RoundUp, Listable]
RoundUp[a_] := If[FractionalPart[a] >= 1/2, Ceiling[a], Floor[a]]
d = Table[i, {i, 0, 100, 1/10}];
Mean[Round[d]] // N
50.
Mean[RoundUp[d]] // N
50.05
Mean[d]
50
According to Wikipedia this is also known as:
unbiased rounding, convergent rounding, statistician's rounding, Dutch rounding, Gaussian rounding, odd-even rounding3 or bankers' rounding, and is widely used in bookkeeping.
Searching for some of those terms returns many good results including:
stackoverflow.com
It does not suffer from negative or positive bias as much as the round half away from zero method over most reasonable distributions. ... But the question was why [] use Banker's actual rounding as default - and the answer is that Microsoft has followed the IEEE 754 standard.
mathforum.org
Where statistics matter, and where numbers that END with the 5 are common, the method you referred to is preferred, because it avoids a bias toward rounding up. On the other hand, there may be some situations where a bias toward numbers that end in an even digit might be bad. Each method has its place.
blogs.msdn.com
Ideally when you are taking an average you want to take an average of the raw data with as much precision as you can get. But in the real world we often have to take averages of data which has lost some precision. In such a situation the Banker's Rounding algorithm produces better results because it does not bias half-quantities consistently down or consistently up. It assumes that on average, an equal number of half-quantities will be rounded up as down, and the errors will cancel out.
cplusplus.com
Given a number exactly halfway between two values, round to the even value (zero is considered even here). ... For random data this is very convenient. Bankers like it because money deposited and withdrawn is random. (There are trends, mind you, but you cannot predict exactly how much will be deposited and withdrawn.) The important point is that the bankers rounding is still biased if the data is biased. It is only unbiased with random data.
www.russinoff.com
One consequence of this result is that a midpoint is sometimes rounded up and sometimes down, and therefore, over the course of a long series of computations and approximations, rounding error is less likely to accumulate to a significant degree in one particular direction than it would be if the the choice were made more consistently. The cost of this feature is a more complicated definition, requiring a more expensive implementation.
-
@Brett: nice example!! – Thomas Feb 22 '12 at 9:15
Just as an aside, Floor[a + 1/2] is the traditional way to round positive numbers in C++ (which has no built in round function). – wxffles Feb 22 '12 at 20:17
The methods discussed are simple deterministic algorithms, which of course people like. Another method might involve randomizing the rounding of a 5. But I think bookkeepers would really hate the idea of random bank transactions. – Ralph Dratman Mar 27 '13 at 2:34 | 2014-04-16T10:47:55 | {
"domain": "stackexchange.com",
"url": "http://mathematica.stackexchange.com/questions/2116/why-round-to-even-integers",
"openwebmath_score": 0.6569511294364929,
"openwebmath_perplexity": 1264.9383769300389,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9615338079816756,
"lm_q2_score": 0.880797076413356,
"lm_q1q2_score": 0.8469161669428612
} |
https://math.stackexchange.com/questions/2913410/introduction-mathematical-statistics-finding-the-mean-square-error-of-estimator | # Introduction Mathematical Statistics, finding the Mean Square Error of estimators
I'm working on a problem from my introduction to mathematical statistics course. So far, I've done the following work:
Let $X_{1},...,X_{m}$ and $Y_{1},...,Y_{n}$ be independent samples form the Bernoulli distribution with unknown parameter $p \in [0,1]$.
i) Prove that $(\bar{X}+\bar{Y})/2$ and $(\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)$ are unbiased estimators for $p$.
ii) Which of these two estimators is preferable (if $m \neq n$)?\
Solution i) First we state the definition of an unbiased estimator. An estimator T is called unbiased for the estimation of $g(\theta)$ if $E_{\theta}T=g(\theta)$ for all $\theta \in \Theta$. The bias is defined as $E_{\theta}T-g(\theta).$
We now compute $E_{p}(\bar{X}+\bar{Y})/2$ as follows, $$\begin{split} E_{p}(\bar{X}+\bar{Y})/2 &= \frac{1}{m}\sum_{i=1}^{m}E_{p}\frac{X_{i}}{2} + \frac{1}{n}\sum_{j=1}^{n}E_{p}\frac{Y_{i}}{2}\\ &= \frac{1}{m}m\frac{1}{2}p+\frac{1}{n}n\frac{1}{2}p\\ &= \frac{1}{2}p+\frac{1}{2}p\\ &=p, \end{split}$$ which is equal to $g(p)$ and therefore this estimator is unbiased.
Now we look at the second estimator, $(\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)$, and proceed as before. $$\begin{split} E_{p}\left((\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)\right) &= \frac{1}{m+n}E_{p}\left(\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j}\right)\\ &= \frac{1}{m+n}(mp+np)\\ &=\frac{p(m+n)}{m+n}\\ &= p, \end{split}$$ which again is equal to $g(p)$ and therefore this estimator is unbiased.
Solution ii) In order to determine which of the given estimators is preferable, we desire the smallest possible mean square error (MSE). The MSE is defined as $$MSE(\theta;T)=E_{\theta}||T-g(\theta)||^{2}.$$
So this is my work so far. I still don't quite understand the meaning of the estimators. Let alone, work out which one is more preferable. I've just try to work from the definition and I think my solution for i) works but I'm not sure. But for ii) I've truely no idea where to start. Any suggestions would be much appreciated!
• For (ii), I would pick as preferable the estimator that exhibits the least variance. I think that, in your case, this estimator is $\frac{\bar{X} + \bar{Y}}{2}$. – VHarisop Sep 11 '18 at 17:49
• How do you know this form the given definition? – Mathbeginner Sep 11 '18 at 18:01
• Can you express the minimum squared error as a function of the bias and the variance? – VHarisop Sep 11 '18 at 18:57
Observe that the mean squared error can be expanded as (using linearity of expectation): $$\mathrm{MSE}(\theta; T) = \mathbb{E}(T^2) + \mathbb{E}(g(\theta)^2) -2 \mathbb{E}(T g(\theta))$$
In your case, $g(\theta)$ is just the population mean $p$, and additionally $\mathbb{E}(T) = p$, so we may rewrite
$$\mathrm{MSE} = \mathbb{E}(T^2) + p^2 - 2p\mathbb{E}(T) = \mathbb{E}(T^2) - p^2 = \mathbb{E}(T^2) - [\mathbb{E}(T)]^2 = \mathrm{Var}(T).$$
It is then straightforward to calculate which of the two estimators has minimum variance, using the fact that $X_i, Y_i$ are all independent. For example, for the estimator $\frac{\bar{X} + \bar{Y}}{2}$, we get
$$\mathrm{Var}\left(\frac{\bar{X} + \bar{Y}}{2}\right) = \mathrm{Var}\left(\frac{1}{2m} \sum_{i=1}^m X_i + \frac{1}{2n} \sum_{i=1}^n Y_i \right) = \frac{1}{4m^2} \sum_{i=1}^m \mathrm{Var}(X_i) + \frac{1}{4n^2} \sum_{i=1}^n \mathrm{Var}(Y_i) \\ = p(1 - p) \left( \frac{1}{4m} + \frac{1}{4n} \right)$$
where we've used the independence of all $X_i, Y_i$ to interchange variance with summation, the property that $\mathrm{Var}(aX) = a^2 \mathrm{Var}(X)$ when $a$ is a constant, and the fact that all variables are identically distributed as $\mathrm{Bernoulli}(p)$ to pull out the common factor $p(1 - p)$ out of the sum.
• Thanks so much. I will work this out after work. Appreciate the help. I wasn't familiar with the linearity of expectation, I'll look into that later. – Mathbeginner Sep 12 '18 at 4:59
• @Mathbeginner: you are implicitly appealing to linearity of expectation in your own proof, when you write $\mathbb{E}(\bar{X}) = \frac{1}{m} \sum_i \mathbb{E}(X_i)$. – VHarisop Sep 12 '18 at 14:55
• I did this without much thought, for me it was common sense that this would hold... Thanks :) – Mathbeginner Sep 12 '18 at 16:19
• No worries! Consider marking the answer as accepted if it resolved your problem. – VHarisop Sep 12 '18 at 17:29
• I definitely will, but I'm not quite done. When I try to find the variances: $Var(p_{1})=E\left(\left(\frac{\bar{X}+\bar{Y}}{2}\right)^{2}\right)-E\left(\left(\frac{\bar{X}+\bar{Y}}{2}\right)\right)^{2}=E(p_{1}^{2})-(E(p_{1}))^{2}=p_{1}^{2}-p_{1}^{2})=0$, but when I try the same method for the second estimator I get:$Var(p_{2})=E\left(\left((\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)\right)^{2}\right)-\left(E\left((\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)\right)\right)^{2}=E\left(\left((\sum_{i=1}^{m}X_{i}+\sum_{j=1}^{n}Y_{j})/(m+n)\right)^{2}\right)-p^{2}$. Stuck right here. – Mathbeginner Sep 12 '18 at 17:54 | 2019-07-21T02:16:43 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2913410/introduction-mathematical-statistics-finding-the-mean-square-error-of-estimator",
"openwebmath_score": 0.8881525993347168,
"openwebmath_perplexity": 211.37823188090888,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.98087596269007,
"lm_q2_score": 0.8633916152464016,
"lm_q1q2_score": 0.8468800817833487
} |
https://math.stackexchange.com/questions/2411727/independent-odds-am-i-friend-seeing-this-wrong-or-is-there-a-mistake-in-the/2411728 | # Independent odds, am I (+ friend) seeing this wrong or is there a mistake in the practice exam?
I found this exercise in a practice exam:
Any student has a 90% chance of entering a University. Two students are applying. Assuming each student’s results are independent, what is the probability that at least one of them will be successful in entering the National University?
A. $0.50$
B. $0.65$
C. $0.88$
D. $0.90$
E. $0.96$
I think the answer is something different than the answers above, namely $0.99$.
$0.01 = (0.1 \times 0.1)$ is the chance of neither, so $1 - 0.01$ must be $0.99$ right? But it's not part of the possible answers.
Other way: $(0.9 \times 0.9) + (0.9 \times 0.1) + (0.1 \times 0.9) = 0.99$
Am I missing something here?
• What if the university that gives them a 90% probability is not the National University? :P – immibis Aug 31 '17 at 5:54
• Please use MathJax in the future – gen-z ready to perish Aug 31 '17 at 11:31
Sometimes you have to challenge the options. Your concepts and answer are absolutely correct!
As the other answerers have noted, $0.99 = 1 - 0.1^2$ is indeed the correct answer.
As to what went wrong with the exercise, I suspect the problem statement has a typo, and the correct admission rate should be 80% instead of 90%. That would make option E ($0.96$ $=$ $1 - 0.2^2$) the correct one.
Why do I suspect that? It's because, as you note, for two independent trials with the same success rate, \begin{aligned} {\rm Pr}[\text{one succeeds}] &= 1 - {\rm Pr}[\text{both fail}] \\ &= 1 - (\text{failure rate})^2 \\ &= 1 - (1 - \text{success rate})^2, \end{aligned} which implies that, conversely, $$\text{success rate} = 1 - \sqrt{1 - {\rm Pr}[\text{one succeeds}]}.$$
Applying this "reverse" formula to the given options, we can work out what the admission rate would have to be for each option to be correct:
\begin{aligned} {\rm A}:\ {\rm Pr}[\text{one succeeds}] &= 0.5 &\implies& \text{success rate} = 1 - \sqrt{0.5}\phantom0 \approx 0.292893 \\ {\rm B}:\ {\rm Pr}[\text{one succeeds}] &= 0.65 &\implies& \text{success rate} = 1 - \sqrt{0.35} \approx 0.408392 \\ {\rm C}:\ {\rm Pr}[\text{one succeeds}] &= 0.88 &\implies& \text{success rate} = 1 - \sqrt{0.12} \approx 0.653589 \\ {\rm D}:\ {\rm Pr}[\text{one succeeds}] &= 0.9 &\implies& \text{success rate} = 1 - \sqrt{0.1}\phantom0 \approx 0.683772 \\ {\rm E}:\ {\rm Pr}[\text{one succeeds}] &= 0.96 &\implies& \text{success rate} = 1 - \sqrt{0.04} = 0.8 \end{aligned}
Out of these options, E is the only one where the probability of both students failing to be admitted works out to a nice square ($0.04 = 0.2^2$). For any of the options A to D to be (exactly) correct, the admission rate would have to be a very awkward irrational number.
Of course, it's also possible that the error is in the options themselves, and that the author of the exercise meant to include 0.99 as one of the options. The actual explanation might even be a combination of both possibilities — perhaps whoever wrote the exercise started with an admission rate of 80%, came up with a suitable answer set, and then later decided to change the admission rate to 90% but forgot to update the answers.
• good effort here – J Tg Aug 31 '17 at 10:38
• Good effort. Just wanted to add that you missed investigating one possibility: 90% was correct, but they intended to ask something else - exactly one student or both students getting admitted. But we can rule out that since none of the probabilities match, and thus your answer is correct. – Hans Olsson Aug 31 '17 at 14:35
• Note also that 0.88 is a tempting wrong answer, given 80% – Brondahl Aug 31 '17 at 15:22
• @Brondahl, I'm usually good at seeing where students went wrong or how they might go wrong to get a "likely" wrong answer, but I can't see what would make $0.88$ likely in this case.... – Wildcard Aug 31 '17 at 21:38
• Extremely crude .. if the student doesn't really understand probability at all and just sees 2 8s. – Brondahl Aug 31 '17 at 23:09
That's the standard way of doing it:
$P(\text{At Least One}) = 1 - P(\text{Both Fail}) = 1 - 0.1 \times 0.1 = 1 - 0.01 = 0.99$
And the not so standard way:
$P(1\text{ in}) + P(2\text{ in Without }1) = 0.9 + (0.9 \times 0.1) = .99$.
Or
\begin{align} P(1\text{ in Without }2) + P(2\text{in Without }1) + P(1\text{ in And }2\text{ in}) &= 0.9 \times 0.1 + 0.9 \times 0.1 + 0.9 \times 0.9 \\ &= 0.09 + 0.09 + 0.81 = .99 \end{align}
No matter how we cut it, you are right. They are wrong.
(It's probably just a typo.)
An important concept when setting multiple choice exams is to provide likely wrong answers. That is the examiner ideally wants to know if the candidate has mastered/remembered the concept. If typical mistakes are not one of the given options then what will be tested is the candidate's ability to re-evaluate their working in the face of evidence that it is wrong. Although this is also worthwhile testing I would suggest that in this case the examiner was just trying to test the concept of combining independent probabilities.
Hence, among the possible answers provided, there should be values like
Pr(success) * Pr(success);
Pr(success) * Pr(fail)
Pr(fail) * Pr(fail)
Pr(success) + Pr(success) [despite this possibly being > 1]
Pr(fail) + Pr(fail) [despite this possibly being > 1]
Pr(success)
AND
1 - any of those above [despite some possibly being < 0]
Of all the 19 "nice" probabilities 0.5, 0.1, 0.15 , 0.2 ,... 0.95 only 0.8 includes the correct answer as Ilmari Karonen pointed out. However the listed answers don't include any of the possible wrong answers for Pr(Success) = 0.8.
If instead of exact matches to the answers we include a difference of +/- 0.01 (that is the last decimal place shown) then the exact or approximate answer being in the shown options A, B, C, D, E for Pr(success) includes the cases 0.3, 0.4, 0.7, 0.8. Of those Pr(success) = {0.3, 0.7} have 2 approximate wrong answers in the list.
For Pr(success) = 0.7 we have
Pr(success)*Pr(success) ~ 0.5 (answer A); and 1 - Pr(success)*Pr(success) ~ 0.5 (answer A)
For Pr(success) = 0.3 we have
Pr(fail)*Pr(fail) ~ 0.5 (answer A); and 1 - Pr(success)*Pr(success) ~ 0.9 (answer D)
= = =
Any-way we look at it the question is a poor example of good testing and even with a single correction
• Pr(success) one of {0.3, 0.7, 0.8}; or
• answer E 0.96 => 0.99
it is doubtful that this test item would show good discrimination between good candidates and poorer candidates.
Ian
• God I love the internet sometimes – Expat C Nov 23 '17 at 6:08 | 2019-09-16T10:08:07 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2411727/independent-odds-am-i-friend-seeing-this-wrong-or-is-there-a-mistake-in-the/2411728",
"openwebmath_score": 0.9961762428283691,
"openwebmath_perplexity": 1233.868877066471,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9808759621310288,
"lm_q2_score": 0.8633916134888613,
"lm_q1q2_score": 0.8468800795767482
} |
https://math.stackexchange.com/questions/1624544/evaluating-the-limit-lim-x-to-1-x3-1-x-1 | # Evaluating the limit $\lim_{x \to 1} (x^3 - 1) / (x - 1)$
$$\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$$
As $x$ approaches to $1$, if I use the substitution method, it will become undefined. Then, I tried to multiply it by its conjugate but I still get undefined answer. How can I solve it?
The other answers are cleaner and require no unnecessary machinery, but here's another method you can try if you can't spot the factorisation for messier expressions (but should be used with care). When evaluated at $x=1$, you get $\frac{0}{0}$, which is in an indeterminate form. This hints at L'Hospital Rule, which says that if
$$\lim\limits_{x \to a} \frac{f(x)}{g(x)}= \frac{0}{0}$$ then $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}=\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$
Note that we arrive at our answer if $\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$ exists, or else we try again.
In your case, $f(x)=x^3-1$, $\implies$ $f'(x)=3x^2$. Likewise, $g(x)=x-1$, $\implies$ $g'(x)=1$. Can you continue?
• I found a question which match this condition and I apply it to the limit $\lim_{x \to 1) (2-(x+3)^1/2)/ (x - 1)$ – user307537 Jan 24 '16 at 6:11
• finally I get answer of -1/4 instead of 1/4 – user307537 Jan 24 '16 at 6:12
• ohh... I think that the answer in the book is not correct (1/4). After I use calculator to solve it, I get -1/4. Thank you for teaching me this new method – user307537 Jan 24 '16 at 6:24
• If you have doubts about your textbook's answer, you can always consult WolframAlpha. You're welcome. – Maxis Jaisi Jan 24 '16 at 6:39
• This method also works if you your limit evaluates at $\frac{\infty}{\infty}$. – Maxis Jaisi Jan 24 '16 at 6:41
Hint
$$x^3-1=(x-1)(x^2+x+1)$$ $\ \ \ \ \ \ \ \ \ \$
Use the fact that $x^3-1=(x-1)(x^2+x+1)$.
• How to factorise it? I only know x-1 is one of the factors. – user307537 Jan 24 '16 at 5:27
• So, divide $x-1$ into $x^3-1$. You can do this either by long division or synthetic division. – Tim Raczkowski Jan 24 '16 at 5:33
• In general, $x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$, if $n>0$. – Maxis Jaisi Jan 24 '16 at 5:47
Hint: You will have $$\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{(x-1)}$$ and $x\ne 1$ since $$a^3-b^3=(a-b)(a^2+b^2+ab)$$ | 2020-02-23T11:56:09 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1624544/evaluating-the-limit-lim-x-to-1-x3-1-x-1",
"openwebmath_score": 0.9115975499153137,
"openwebmath_perplexity": 336.97230715070407,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9808759626900699,
"lm_q2_score": 0.863391611731321,
"lm_q1q2_score": 0.8468800783354905
} |
https://math.stackexchange.com/questions/2451566/cartesian-product-of-sets-with-different-cardinality | # Cartesian product of sets with different cardinality
What is the correct way of performing a cartesian product over finite sets with different cardinality? For instance:
Let
$$A = \{1, 2, 3\},\quad B = \{4, 5\},\quad C = \{6, 7\}$$
What would be the result of
$$A \times B \times C$$
What I thought it would correct would be to try combining all elements of the sets, but that seems to be incorrect.
• do you know of tuples ? – user451844 Sep 30 '17 at 13:51
By definition, $$A \times B := \{(a,b): a \in A, b \in B\}$$
$A \times B$ $=$ {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}, then $(A \times B)$ $\times$ $C$ $=$ ${(1, 4, 6), (1, 4, 7), (1, 5, 6),(1, 5, 7), (2, 4, 6), (2, 4, 7), (2, 5, 6), (2, 5, 7), (3, 4, 6), (3, 4, 7), (3, 5, 6), (3, 5, 7)}$.
Product of cardinalities of the all the sets is $|3|\times|2|\times|2|$ $=$ $12$
ordering matters it would include:
$(1,4,6),(1,4,7),(1,5,6),(1,5,7),(2,4,6),(2,4,7),(2,5,6),(2,5,7),(3,4,6),(3,4,7),(3,5,6)$ ,and $(3,5,7)$
whereas the Cartesian product $B\times A\times C$ would have:
$(4,1,6),(4,1,7),(4,2,6),(4,2,7),(4,3,6),(4,3,7),(5,1,6),(5,1,7),(5,2,6),(5,2,7),(5,3,6)$ , and $(5,3,7)$ or to put it in set notation:
$$\{(x,y,z):x\in A,y\in B, z\in C\}$$ is the product $A\times B \times C$ whereas $$\{(x,y,z):x\in B,y\in A, z\in C\}$$ is the product $B\times A \times C$ | 2019-08-19T01:22:00 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2451566/cartesian-product-of-sets-with-different-cardinality",
"openwebmath_score": 0.8983578681945801,
"openwebmath_perplexity": 339.79639513042497,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9808759638081522,
"lm_q2_score": 0.8633916099737806,
"lm_q1q2_score": 0.8468800775769043
} |
http://mathhelpforum.com/calculus/16859-differentiable-points.html | # Math Help - Differentiable points
1. ## Differentiable points
Hi, I'm having a hard time understanding how to determine where a function is not differentiable (the question keeps coming up in my homework though sadly, it didn't in the lesson!) I understand that a point where the function is not differentiable means the derivative does not exist there. However, how can I tell, based off either a graph or equation, when a point isn't differentiable?
For example, I'm currently doing this question:
Sketch the graph of f(x) = |x^2 - 1|
a) For what values of x is f not differentiable?
b) Find a formula for f1 and sketch the graph of f1
c) Find f1 at -2, 0, and 3.
Any help is greatly appreciated!
2. Originally Posted by starswept
Hi, I'm having a hard time understanding how to determine where a function is not differentiable
Use the limit definition of a derivative. if the limit is undefined at a certain point, then the function is not differentiable at that point. classic example, |x|. they tell you it's not differentiable at zero. why is that? simple, the limit does not exist at x = 0
you can use the apostrophe to denote a prime. don't say f1, say f '
3. Originally Posted by Jhevon
Use the limit definition of a derivative. if the limit is undefined at a certain point, then the function is not differentiable at that point. classic example, |x|. they tell you it's not differentiable at zero. why is that? simple, the limit does not exist at x = 0
you can use the apostrophe to denote a prime. don't say f1, say f '
So, essentially, saying the derivative doesn't exist is exactly like saying the limit does not exist? Does that mean that the right hand limit must equal the left hand limit rule still comes into play?
4. Originally Posted by starswept
So, essentially, saying the derivative doesn't exist is exactly like saying the limit does not exist? Does that mean that the right hand limit must equal the left hand limit rule still comes into play?
yes, all that comes into play. and also, if the limit diverges, that is, goes to infinity or something
5. Originally Posted by starswept
Hi, I'm having a hard time understanding how to determine where a function is not differentiable
Basically there are two conditions that one looks for in deciding if a function is differentiable.
The first is that the function must be continuous at the point!
The second is not that easily stated. But is essence it is that the graph of the function must be smooth at the point. If you graph f(x)=|x-1|, then even though it is continuous at x=1, you will see a sharp turn at x=1. Therefore, the function is not differentiable at x=1: it cannot have a tangent there.
6. Originally Posted by starswept
Hi, I'm having a hard time understanding how to determine where a function is not differentiable (the question keeps coming up in my homework though sadly, it didn't in the lesson!) I understand that a point where the function is not differentiable means the derivative does not exist there. However, how can I tell, based off either a graph or equation, when a point isn't differentiable?
For example, I'm currently doing this question:
Sketch the graph of f(x) = |x^2 - 1|
a) For what values of x is f not differentiable?
b) Find a formula for f1 and sketch the graph of f1
c) Find f1 at -2, 0, and 3.
Any help is greatly appreciated!
f(x)=|x-1||x+1| is not differentiable at x=-1 and x=1
7. Originally Posted by Plato
The second is not that easily stated. But is essence it is that the graph of the function must be smooth at the point. If you graph f(x)=|x-1|.
Which means the right and left derivatives are the same. | 2015-12-01T12:49:59 | {
"domain": "mathhelpforum.com",
"url": "http://mathhelpforum.com/calculus/16859-differentiable-points.html",
"openwebmath_score": 0.9058037996292114,
"openwebmath_perplexity": 236.5204142932903,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9808759604539051,
"lm_q2_score": 0.8633916099737806,
"lm_q1q2_score": 0.8468800746808756
} |
https://math.stackexchange.com/questions/2156008/using-dominated-convergence-to-prove-partial-derivative-and-integral-can-be-inte | # Using dominated convergence to prove partial derivative and integral can be interchanged
Hi guys doing a self study here, and came across this problem. I know the same question with slightly different hypotheses has been asked before but I was a bit confused on the answers given and not sure how to completely adjust those arguments to solve my problem and also want to clear up and confirm some things.
It says the following: Assume $f(x,y)$ and $\frac{\partial f}{\partial x}$ are continuous functions on $\mathbb R x \mathbb R$ and let $[a,b]$be a finite interval. Show that $\int _{[a,b]} \frac{\partial f(x,y)}{\partial x} dy=\frac{ \partial f(x,y)}{\partial x} \int_{[a,b]} f(x,y) dy$
First of all, does f(x,y) continuous in $R^2$ imply it is continuous at a fixed x as a function of y?!If so does this imply f is uniformly continuous on [a,b] as a function of y? If so then f is bounded by its supremum on the finite interval and hence its integral exists? Do we turn this problem into a more tractable form by changing the derivative into a limit of a sequence of functions$n( f(x+\frac{1}{n},y)-f(x,y))$? But then how do we find a bound for this sequece by an integrable function? basically I would appreciate it if somebody explained to me one by one why the hypothesis are the way they are. Thanks in advance guys!
First of all, does $f(x,y)$ continuous in $\mathbb{R}^2$ imply it is continuous at a fixed $x$ as a function of $y$?
Yes, that's correct and holds for any continuous function $f: \mathbb{R}^2 \to \mathbb{R}$. For a fixed point $(x_0,y_0)$ continuity at $(x_0,y_0)$ means that for all $\epsilon>0$ there exists $\delta>0$ such that $$|f(x,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |(x_0,y_0)-(x,y)| \leq \delta. \tag{1}$$ Now for fixed $x_0,y_0$ and $\epsilon>0$ choose $\delta>0$ as above, then we have $$|f(x_0,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |y-y_0| \leq \delta$$ which shows that $f(x_0,\cdot)$ is continuous at $y=y_0$. Since $x_0,y_0$ are arbitrary, this finishes the proof.
If so does this imply $f$ is uniformly continuous on $[a,b]$ as a function of $y$?
Yeah, any continuous function which is defined on a compact interval is uniformly continuous, see this question; however, this is not needed for the proof of the assertion.
If so then $f$ is bounded by its supremum on the finite interval and hence its integral exists?
Correctly. (For this we don't need absolute continuity; continuity is enough.) Note that the continuity also implies measurability of $f(x,\cdot)$ for each $x$.
Do we turn this problem into a more tractable form by changing the derivative into a limit of sequence of functions $n(f(x+1/n,y)-f(x,y))$.
Yes, exactly. Set
$$F(x) := \int_{[a,b]} f(x,y) \, dy.$$
We have to show that $$\frac{\partial}{\partial x} F(x) = \int_{[a,b]} \partial_x f(x,y) \, dy. \tag{2}$$
First of all, note that $F$ is well-defined because $f(x,\cdot)$ is continuous (see above) and, moreover,
$$\frac{F(x+1/n)-F(x)}{1/n} = n \int_{[a,b]} (f(x+1/n,y)-f(x,y)) \, dy.$$
For fixed $x$ we define a sequence of auxiliary functions by $$u_n(y):= n (f(x+1/n,y)-f(x,y)).$$
Since $f(\cdot,y)$ is, by assumption, differentiable, we have
$$u_n(y) \xrightarrow[]{n \to \infty} \partial_x f(x,y).$$
On the other hand, the mean value theorem shows
$$|u_n(y)| \leq \sup_{\lambda \in [0,1]} |\partial_x f(x+\lambda/n,y)| \leq \sup_{y \in [a,b]} \sup_{|u-x| \leq 1} |\partial_x f(u,y)|.$$
Since $\partial_x f$ is continuous, the right-hand side is a (finite) constant and therefore integrable on the finite interval $[a,b]$. This means that we have found an integrable dominating function for $u_n$. Applying the dominated convergence theorem, we find that
$$\frac{F(x+1/n)-F(x)}{1/n} = \int_{[a,b]} u_n(y) \, dy \xrightarrow[]{n \to \infty} \int_{[a,b]} \partial_x f(x,y) \, dy.$$
This proves $(2)$.
• Thank you for some reason I had not seen this answer!! – user172377 Mar 15 '17 at 22:27
• @Socchi You are welcome. – saz Mar 16 '17 at 6:20
• Hi, could you elaborate a bit on the mean value part? Why can't we just say $\frac{f(x+1/n,y)-f(x)}{1/n} = \frac{\partial f(c,y)} {\partial x}$ for some c in [a,b] $\leq \sup_{x,y} {\frac{\partial f(x,y)}{ \partial x}}$ which is continuous on a closed interval hence integrable. Thanks. – user172377 Apr 11 '17 at 6:21
• @Socchi Well, essentially that's what I did; I just formulated the mean value theorem in a slightly different way. – saz Apr 11 '17 at 7:07
• Thanks, so mine is correct too? – user172377 Apr 11 '17 at 7:17 | 2019-10-21T20:18:43 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2156008/using-dominated-convergence-to-prove-partial-derivative-and-integral-can-be-inte",
"openwebmath_score": 0.9700687527656555,
"openwebmath_perplexity": 122.83066650270538,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9895109109053045,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8468740498539955
} |
https://math.stackexchange.com/questions/713102/is-there-a-formula-for-a-simple-sum-of-sums/713121 | # Is there a formula for a simple sum of sums?
Given $f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$, I am wondering if there is a straightforward formula to compute f(n) and how it may be derived.
The only reduction I thought about so far would be:
n*1 + (n - 1)*2 + (n - 3)*3 ... which seems symmetrical; for example, an odd and even n
5 4 3 2 1 4 3 2 1
* *
1 2 3 4 5 1 2 3 4
but I'm not sure if and how it may help derive a formula.
$$f(n)=\sum_{i=1}^n \frac{i(i+1)}{2}$$
$$f(n)=\sum_{i=1}^n \frac{i^2}2 +\frac{i}2$$
Using two well known identitites,
$$f(n)=\frac12\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)$$
Simplifying:
$$f(n)=\frac{n(n+1)}{4}\left(\frac{2n+1}{3}+1\right)$$
$$f(n)=\frac{n(n+1)}{4}\left(\frac{2n+4}{3}\right)$$
$$f(n)=\frac{n(n+1)}{2}\left(\frac{n+2}{3}\right)$$
$$f(n)=\frac{n(n+1)(n+2)}{6}$$
• Thanks for helping me think about it in a new way. – גלעד ברקן Mar 15 '14 at 11:53
• nice answer@Sabyasachi – Semsem Mar 15 '14 at 12:02
There is a general formula $$\sum_{k=1}^n\frac {k(k+1)\dots(k+r-1)}{r!}=\frac {n(n+1)\dots(n+r)}{(r+1)!}$$
Which can be proved by induction - base case $n=1$, both sides of the equation are equal to $1$.
Then $$\frac {n\left[(n+1)\dots(n+r)\right]}{(r+1)!}+\frac {\left[(n+1)\dots(n+r)\right]}{r!}=\frac {\left[(n+1)\dots(n+r)\right](n+r+1)}{(r+1)!}$$
Another way of writing this, which invites a combinatorial proof is:
$$\sum_{k=1}^n\binom {k+r-1}r=\binom {n+r}{r+1}$$
The sum of integers is the case $r=1$ and the sum of triangular numbers is the case $r=2$.
• Wow. I didn't know this. Will be very useful. +1. – Guy Mar 15 '14 at 14:15
• @Sabyasachi I mistyped the final formula, if you are checking, and have now corrected it – Mark Bennet Mar 15 '14 at 14:52
• Okay thanks for the correction. – Guy Mar 15 '14 at 15:02 | 2020-11-24T12:56:14 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/713102/is-there-a-formula-for-a-simple-sum-of-sums/713121",
"openwebmath_score": 0.8629341721534729,
"openwebmath_perplexity": 442.8530775453653,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9895109101320038,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8468740491921651
} |
http://techiemathteacher.com/2014/10/12/number-diagonals-polygon/ | # Number Diagonals of Polygon
One of the easiest topics in plain Geometry is to count the number of diagonals in a convex polygon. You might familiar already or even memorized the said formula. So let me show you the technique how to derive it. This tutorial is made with two main reasons. First is to educate and second is to teach my own technique in derivation.
We will present two derivations, using plain geometry and common sense and the second way is using Combinatorics technique and again a small common sense.
Derivation:
By Plane Geometry:
Consider the figure above of a convex heptagon. Each point can make 4 diagonals. Only four diagonals because we can’t consider the segment made by connecting point A to B and point A to C. Since there are 7 points, there must be 7×4=28 diagonals are there in heptagon.
But, try to draw the diagonals starting from point E like shown below.
We can basically draw again 4 diagonals but there is 1 common diagonal connecting A to E and E to A which should be counted only once. Doing the same process, we were able to figure out that each point has 1 common diagonal that should be counted once. Since we already calculated the number of diagonals previously, we can eliminate the error of counting one diagonal twice by dividing the final answer by 2. So instead of 28, the number of diagonals that can be drawn in convex heptagon is 14.
For n-gon:
Now, consider a convex n-gon. Since a heptagon has 7 sides and 7 vertices, n-gon also has n number of vertices. Now each point of n-gon can make n-3 diagonals. That is because the segments connecting the point and the point right beside it can’t be considered as diagonal.
So there must be $n(n-3)$ diagonals that can be drawn from n-gon however remember that we counted each diagonal twice. By dividing $n(n-3)$ by 2 we can eliminate counting errors. Thus the formula for taking the maximum number of diagonals from convex n-gon is
$d=\displaystyle\frac{n(n-3)}{2}$
Derivation using basic Combinatorics with small common sense:
Consider a convex polygon like the figure shown below.
Removing the segments connecting the points we have the figure in 2, a scattered points.
In Combinatorics, we can connect the points in nC2 ways from n number of points. But we also need to subtract the outermost segment that we took out from figure 1 and the rest must be the number of diagonals. That is where the small common sense can be used.
Expressing that in equation we have,
$d=nC2-n$
$d=\displaystyle\frac{n!}{(n-2)!2!}-n$
$d=\displaystyle\frac{n(n-1)(n-2)!}{(n-2)!2!}-n$
$d=\displaystyle\frac{n(n-1)}{2}-n$
$d=\displaystyle\frac{n(n-1)-2n}{2}$
$d=\displaystyle\frac{n^2-n-2n}{2}$
$d=\displaystyle\frac{n^2-3n}{2}$
$d=\displaystyle\frac{n(n-3)}{2}$
I hope you learned something from that. 😀
### Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
### 20 Responses
1. Hey, thanks for the article.Really thank you! Want more.
2. Jame says:
I like what you guys are up to. This type of clever work and coverage! Keep up the excellent work guys, I’ve included you blogroll.
3. Yesterday, while I was at work, my sister stole my apple ipad and tested to see if it can survive a 30 foot drop, just so she can be a youtube sensation. My iPad is now broken and she has 83 views. I know this is entirely off topic but I had to share it with someone!
4. this site says:
l57211 I think this is a real great blog. Really Great.
5. 758270 879369U never get what u expect u only get what u inspect 490024
6. 239575 560430I dont leave a great deal of comments on a great deal of blogs each week but i felt i had to here. A hard-hitting post. 774564
7. Spot on with this write-up, I absolutely feel this web site needs
a lot more attention. I’ll probably be back again to read more, thanks for the info!
8. With havin so much content and articles do you ever run into any problems of plagorism or copyright infringement?
My blog has a lot of exclusive content I’ve either created myself or
outsourced but it appears a lot of it is popping it up all over the web without my agreement.
Do you know any solutions to help prevent content
from being stolen? I’d definitely appreciate it.
9. Say, you got a nice blog post.Really looking forward to read more.
10. Hi there it’s me, I am also visiting this website daily, this website is
truly nice and the visitors are in fact sharing pleasant thoughts.
11. Wow, amazing blog format! How lengthy have you been blogging for?
you made running a blog look easy. The entire look of your site is magnificent, as smartly as the content!
12. of course like your web site however you need to test the
spelling on quite a few of your posts. A number of them are
rife with spelling problems and I in finding it very bothersome to tell the truth then again I’ll surely come again again.
13. Enid says:
I’m now not certain where you are getting your info, but good
topic. I must spend some time studying more or working out more.
Thanks for fantastic information I used to be in search of this information for my
mission.
14. Howdy, i read your blog occasionally and i own a similar one and i was
just curious if you get a lot of spam remarks? If so how do you stop it, any plugin or anything you can suggest?
I get so much lately it’s driving me mad so any assistance is
very much appreciated.
I really liked your post. Great.
16. Sweet blog! I found it while surfing around on Yahoo News.
Do you have any tips on how to get listed in Yahoo News?
I’ve been trying for a while but I never seem to get there!
Many thanks
17. van hire says:
Hello Dear, are you truly visiting this web page regularly, if so after that you will
without doubt obtain nice experience.
18. equipments says:
443689 171482Hi there, just became aware of your weblog by way of Google, and identified that it is genuinely informative. Im gonna watch out for brussels. Ill be grateful in case you continue this in future. A lot of men and women is going to be benefited from your writing. Cheers! 113298
19. Hello, after reading this remarkable post i am also delighted to share my knowledge here with colleagues.
20. mi3896.com says:
We are a group of volunteers and starting a brand new scheme
in our community. Your website offered us with useful information to work on. You’ve performed
an impressive process and our whole neighborhood might
be thankful to you. | 2017-08-18T16:28:11 | {
"domain": "techiemathteacher.com",
"url": "http://techiemathteacher.com/2014/10/12/number-diagonals-polygon/",
"openwebmath_score": 0.30702975392341614,
"openwebmath_perplexity": 1352.9967557387017,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9895109075027803,
"lm_q2_score": 0.8558511488056151,
"lm_q1q2_score": 0.8468740469419412
} |
http://incommunity.it/rvbw/matrix-multiplication-divide-and-conquer.html | # Matrix Multiplication Divide And Conquer
Use a divide-and-conquer approach as in Strassen’s algorithm, except that instead of getting 7 subproblems of size n/2, get 5 subproblems of size n/2 based on part (a). Divide-and-Conquer Multiplication: Warmup To multiply two n-bit integers a and b: Multiply four ½n-bit integers, recursively. 4-46 faster than the original versions and within 2-60% of a high-performance hand crafted implementation. Order of both of the matrices are n × n. We improve the basic block matrix multiplication for divide-and-conquer SVD algorithm by avoiding zero elements transfer and computation in the cluster. We could improve the required running time by the following Strassen's matrix multiplication algorithm. Algorithm for Matrix Multiplication. Recur: solve the sub problems recursively Conquer: combine the solutions for S1, S2, …, into a solution for S The base case for the recursion are sub problems of constant size Analysis can be done using recurrence equations 5. Buyukokkten,J. 401J LECTURE 3 Divide and Conquer • Binary search • Powering a number • Fibonacci numbers • Matrix multiplication • Strassen's algorithm • VLSI tree layout Prof. Solves difficult problems with ease. giving the four equations. Mathematically we have or, Consider Divide and Conquer Remember Pascal’s Triangle Consider triangle shape Looks like a two dimensional array. The Divide and Conquer paradigm. Add and shift to obtain result. strassen matrix multiplication: Strassen's method is similar to above simple divide and conquer method in the sense that this method also divide matrices to sub-matrices of size N/2 x N/2 as shown in the above diagram, but in Strassen's method, the four sub-matrices of result are calculated using formulae. B, with the assumption that n is an exact power of 2 in each of the n x n matrices. It utilizes the strategy of divide and conquer to reduce the number of recursive multiplication calls from 8 to 7 and hence, the improvement. These subproblems must be solved and then a method must be found to combine subsolutions into a solution of a whole. The primary topics in this part of the specialization are: asymptotic ("Big-oh") notation, sorting and searching, divide and conquer (master method, integer and matrix multiplication, closest pair), and randomized algorithms (QuickSort, contraction algorithm for min cuts). Their structure will. Feb 1, 2020 - Explore anokair's board "MATRIX MULTIPLICATION " on Pinterest. Try out: Matrix Multiplication Calculator. 2 Strassen's algorithm for matrix multiplication Table of contents. _ In this lab, we are interested in parallelizing the following divide-and-conquer algorithm for matrix multiplication:. 1 Divide: Partition A and B into submatrices; add and subtract to form terms. Combine subproblem solutions. 50-57: HW4 Project1: Primality Tester: Jan 30: 7. Divide and conquer is a powerful algorithm design technique used to solve many important problems such as mergesort, quicksort, calculating Fibonacci numbers, and performing matrix multiplication. Algorithm Analysis/Matrix Multiplication. This paper deals with parallels of the fast matrix multiplication strassen's algorithm, winograd's algorithm and analyzes empirical study of the matrix multiplication under the distributed environment in. Combine, Divide and Conquer b. 1 Compute C = AB using the traditional matrix multiplication algorithm. q] Initially, p= 1 and q= A. en stanford. The divide and conquer strategy •A first example : sorting a set S of values sort (S) = if |S| ≤ 1 then return S else divide (S, S1, S2) fusion (sort (S1), sort (S2)) end if fusion is linear is the size of its parameter; divide is either in O(1) or O(n) The result is in O(nlogn). Their product C is also an n× n matrix. matrix product C = A B is defined by c ij = Xn k=1 a ik b Divide & Conquer: First Approach II. Divide-and-conquer matrix multiplication c. I want to know when to switch to another algoritm, which in this case is the regular matrix multiplication. Raymund Fischer author of Program of matrix multiplication using function is from Frankfurt, Germany. Divide-and-Conquer Reading: CLRS Sections 2. Implementing divide and conquer matrix multiplication in Rust. But thinking recursively will be more helpful in reasoning about the algorithm and ultimately improving on it. The problem is not actually to perform the multiplications, but merely to decide the sequence of the matrix multiplications involved. Specifically, an input matrix of size can be divided into 4 blocks of matrices. This generalizes the products in size (2 2 2) used in the half-gcd algo-rithm or the Pad e approximant algorithm of [8]; often, nis small (say, a few dozens). Divide and Conquer is a recursive problem-solving approach which break a problem into smaller subproblems, recursively solve the subproblems, and finally combines the solutions to the subproblems to solve the original problem. The matrix-chain multiplication problem can be stated as follows: given a chain A 1, A 2,. Divide and Conquer (Multiplication, Recurrence Relations, Convex Hull) Chapter 2: pp. divide-and-conquer. Divide-and-Conquer Algorithms for Computing Matrix Inverses By SHADY Sayed EL-OKUR Supervised by Dr. Strassen's 5. Logic: Divide the matrix, then use the Strassen's formulae:. Combine the sorted subarrays by merging into a single sorted array. Conquer the sub-problems by solving them recursively. • The divide-and-conquer strategy consists: ü in breaking a problem into simpler subproblems of the same type, ü next to solve these subproblems, ü and finally to merge the obtained results into a solution to the problem. Lecture 8 8-3 Strassen’s algorithm Divide and conquer algorithms can similarly improve the speed of matrix multiplication. Get this from a library! An I/O-Complexity Lower Bound for All Recursive Matrix Multiplication Algorithms by Path-Routing. Note that it is the kth smallest element in the sorted order, not the kth distinct element. In this section we discuss a top-down algorithmic paradigm called divide and conquer. CSC210-12: Divide and Conquer: Multiplication of Large Integers & Strassen's Matrix Multiplication Formulas for Strassen’s Algorithm Brute Force for Matrix Multiplication Strassen observed [1969] that the product of two matrices can be computed as follows: Multiplication of Large. 3 Background We need to introduce a more complex formalism to discuss CW. Linear scaling methods, or O(N) methods, have computational and memory requirements which scale linearly with the number of atoms in the system, N, in contrast to standard approaches which scale with the cube of the number of atoms. That is, you can multiply two matrices if they are compatible: the number of columns of A must equal the number of. C11 = a11b11 + a12b21 C12 = a11b12 + a12b22 C21 = a21b11 + a22b21 C22 = a21b12 + a22b22 2x2 matrix multiplication can be accomplished in 8 multiplication. Recur: solve the sub problems recursively Conquer: combine the solutions for S1, S2, …, into a solution for S The base case for the recursion are sub problems of constant size Analysis can be done using recurrence equations 5. Given a matrix A of n \times m (n rows, m columns), and a matrix B of m \times p, the matrix produce C = AB is an n \times p matrix, where elements c_{ij} in C is given by: c_{ij} = \sum_{k=1}^m a_{ik}b_{kj}. The primary goals of current energy conversion (CEC) technology being developed today are to optimize energy output and minimize environmental impact. Multiplication of Big integers • Given integers A, B with n bits each, can + , - in O(n) time. An overview of the algorithm. (Divide and Conquer) Reviewed by Huzaif Sayyed on June 16, 2017 Rating: 5 Share This:. ×) by min (resp. Conclusion. I've implemented the O(log_2 7) Strassen algorithm once (which should be really simple after implementing normal divide and conquer) and after benchmarking I've determined that for matrices smaller than 128x128 it's not worth to. •Divide & Conquer: Discussed a bit in recurrence analysis •Randomized Algorithm: Discussed a bit in prob. Their product C is also an n× n matrix. Very well studied for GF2 and anecdotally good for GF3. Algorithm Analysis techniques ----- Theory: Divide-and-Conquer Strategy: Given a function that has to compute on ‘n’ input the divide and conquer strategy suggest. Strassen’s. We can view each matrix as consisting of 2x2=4 n=2-by-n=2 matrices. Spring 2017, Sections 1 and 2. This paper deals with parallels of the fast matrix multiplication strassen's algorithm, winograd's algorithm and analyzes empirical study of the matrix multiplication under the distributed environment in. 80-87: HW5 Project2 Design Experience: Feb 4: 8. Also, observe that divide and conquer ran twice as fast when ran using Threads. Course Data Structures & Algorithms (COSC 6320 ) Uploaded by. Write a program to implement matrix multiplication using Strassen's method. Combine these results together. divide-and-conquer. Practice Programming/Coding problems (categorized into difficulty level - hard, medium, easy, basic, school) related to Divide and Conquer topic. 1 Compute C = AB using the traditional matrix multiplication algorithm. 3728639})[/math] time [1]. The divide-and-conquer design paradigm 1. 1) Divide matrices A and B in 4 sub-matrices of size N/2 x N/2 as shown in the below diagram. In this tutorial, you’ll learn how to implement Strassen’s Matrix Multiplication in Swift. Second, we design an efficient way to fully split the input matrix into leaf problems accord-ing to the division strategy of the algorithm using MapReduce. More Divide and Conquer. We improve the basic block. Unit Description: This unit focuses on an introduction to multiplication and division concepts. • Learn how to use Strassen’s algorithm Matrix Multiplication • Who cares? • For today, addition and multiplication count. Pros of Divide and Conquer Strategy. https://irjet. 8) used in BLAS Good when n > 100, but there is an issue of numerical stability It is based on divide and conquer and block matrix decomposition divide matrix into several submatrices and the multiplication of…. BruteForce 2. Strassen's matrix multiplication) is O(n 2. In the above divide and conquer method, the main component for high time complexity is 8 recursive calls. The multiplication algorithm above is written iteratively, so we can easily analyze and implement it. Add and shift to obtain result. multiplication algorithms in java divide and conquer. The Karatsuba algorithm for multiplication uses a 3-way divide and conquer to achieve a running time of O(3 n^log_2 3) which beats the O(n^2) for the ordinary multiplication algorithm (n is the number of digits in the numbers). (b)What is wrong the the following algorithm for computing the square of an n n matrix? Just use a. We haven’t considered communication bottlenecks; in real life communication is expensive. Convex Hull algorithms (plus more on Mergesort, Quicksort, etc. An alternative to the iterative algorithm is the divide and conquer algorithm for matrix multiplication. The divide and conquer strategy •A first example : sorting a set S of values sort (S) = if |S| ≤ 1 then return S else divide (S, S1, S2) fusion (sort (S1), sort (S2)) end if fusion is linear is the size of its parameter; divide is either in O(1) or O(n) The result is in O(nlogn). Divide instance of problem into two or more smaller instances 2. 3728639})[/math] time [1]. Without communications the addition and subtraction of matrices can be computed in. Break up problem into two pieces of equal size. Working Principle: Divide the unsorted list into n sublists, each containing 1 element (a list of 1 element is considered sorted). 37 Matrix Multiplication: Warmup ∈. 4-by-4 Matrices, Using Strassen Multiplication // MStrassen. The BM displays a radiation pattern of 16 beams at different declinations (from -48, to +88 degrees). 1 Drawbacks of Divide and Conquer We now discuss some bottleneck’s of Strassen’s algorithm (and Divide and Conquer algorithms in general). 5 The master method for solving recurrences 4. And this is a super cool algorithm for two reasons. multiplication algorithms in java divide and conquer. Optimisation of Constant Matrix Multiplication Operation Hardware Using a Genetic Algorithm Andrew Kinane, Valentin Muresan, and Noel O’Connor Centre for Digital Video Processing, Dublin City University, Dublin 9, Ireland [email protected] Implementing divide and conquer matrix multiplication in Rust. In general, for 1≤ i≤ p and 1≤ j ≤ r. In this section we show that the use of a divide-and-conquer strategy results in a slightly better asymptotic running time. To this end, we extend Bodrato’s (2010) method for matrix squaring, and transform matrices to an alternative basis. strassen matrix multiplication: Strassen's method is similar to above simple divide and conquer method in the sense that this method also divide matrices to sub-matrices of size N/2 x N/2 as shown in the above diagram, but in Strassen's method, the four sub-matrices of result are calculated using formulae. Attempt at solutions I have been able to get this divide and conquer code working for a 2x2 matrix and a 4x4 matrix. 2 If the sizes of the matrices are less than the threshold 2. January 2, 2013 January 3, 2013 saeediqbalkhattak How to multiply any two integer using divide & Conquer approach. Strassen’s 5. •Divide & Conquer: Discussed a bit in recurrence analysis •Randomized Algorithm: Discussed a bit in prob. The introduction of the technique is attributed to a 1962 paper by Karatsuba, and indeed it is sometimes called Karatusba multiplication. Unit Title: Multiply, Divide & Conquer (Plateau Valley School District) Instructional Unit (Word or PDF) | Unit Storyboard. Given a matrix A of n \times m (n rows, m columns), and a matrix B of m \times p, the matrix produce C = AB is an n \times p matrix, where elements c_{ij} in C is given by: c_{ij} = \sum_{k=1}^m a_{ik}b_{kj}. Suppose, matrix A has p rows and q columns i. Integer multiplication The problem: Multiply two large integers (n digits) The traditional way: Use two loops, it takes O(n2) operations Young CS 331 D&A of Algo. On the right is the 4-by-4 case done using divide and conquer, the 4-by-4 divided into 2-by-2s of 2-by-2s. Selection Procedure and Matrix Multiplication (in Hindi). divide-and-conquer algorithms is new, to our knowledge: the only explicit exam-ples in [9,3] describe Karatsuba multiplication. [34:00] Matrix multiplication algorithms. Merge sort is a stable sort. 2 Strassen's algorithm for matrix multiplication 4. Divide X, Y and Z into four (n/2)×(n/2) matrices as represented below −. Contribute to saulmm/Divide-and-conquer development by creating an account on GitHub. Well-known examples include heapify, merge sort, quicksort, Strassen's fast matrix multiplication, the Fast Fourier Transform (FFT), and binary search. Matrix multiplication is particularly easy to break into subproblems, because it can be performed blockwise. However, what we actually need is that at each step our matrix can be divided into four evenly sized blocks. Differentiating fn(x) w. Let X,Y be square matrices (of the same order) whose entries are non-negative integers or +∞. Assume n is a power of 2. id Penghitungan matrix sangat umum di bidang matematika. Divide and Conquer 0 12 Young CS 331 D&A of Algo. Combine the sorted subarrays by merging into a single sorted array. Strassen's Matrix multiplication can be performed only on square matrices where n is a power of 2. Conclusion. Because this algorithm is recursive, there are many method calls, and method returns. org Divide and Conquer Abstract Data Types Bibliography Strassen’s method Strassen: “If A,B are matrices of order m2k+1 to be multiplied, write A = A 11 A 12 A 21 A 22 B = B 11 B 12 B 21 B 22 C = C 11 C. Combine solutions to sub-problems into overall solution. The introduction of the technique is attributed to a 1962 paper by Karatsuba, and indeed it is sometimes called Karatusba multiplication. Multiplication (plus more on Mergesort and Quicksort) Supplemental. 1 Split matrices A and B 2. The second algorithm is a compute-aggregate-broadcast topological sort with time complexity of O(n log n) using O(n2) pro-cessors and O(1og2 n) using 0(n3) processors. sorting Line in c Hi I need help with sorting line in programming c asking user to enter 1-10 characters and then sort them in lower case, alphabetical and ascending order. This algorithm basically performs the 'Divide and Conquer' approach by breaking down the large N×N matrix system into a manageable 32 × 32 matrix for fast computation. Enter the primary number (in binary; make sure it is valid) first then enter the secondary number (also in binary) for the calculation and click on Calculate. 3 The substitution method for solving recurrences 4. 36 Fourier analysis Fourier theorem. Read on for Python implementations of both algorithms and a comparison of their running time. Topic: Divide and Conquer 24 The Divide-and-Conquer way: Suppose x and y are large integers, divide x. Median A median is described as the number separating the higher half of a sample, a population, or a probability distribution, from the lower half In probability theory and statistics. it Abstract Divide-and-Conquer (DaC) is a sequential programming. Algorithm Analysis/Matrix Multiplication. This paper presents a divide-and-conquer matrix multiplication algorithm [14,15] whose natural communication pattern is the hypercube. Quicksort/Mergesort. Order of both of the matrices are n × n. Fast Matrix Multiplication. BruteForce 2. This relies on the block partitioning. Arial Monotype Sorts Symbol Times New Roman Default Design Microsoft Equation 2. I will start with a brief introduction about how matrix multiplication is generally observed and implemented, apply different algorithms (such as Naive and Strassen) that are used in practice with both pseduocode and Python code, and then end with an analysis of their runtime. 1145/3293883. The main issue at this point is the fact that we require our matrices to have dimensions that are a power of two, so that we can apply our divide and conquer on quarters of the matrix. I want to make a series in which I will discuss about some algorithms which follow divide and conquer strategy. −multiplication by divide and conquer. This happens by decreasing the total number if multiplication performed at the expenses of a. We show that the resulting implementations are 1. The Divide and Conquer paradigm. One route we might want to try is breaking the integers up into two parts. A sub problem is another chain that goes from matrix (i) and end at matrix (j) where (i) goes from (1) to (n-1) and (j) goes from (i+1) to (n). Write a c program to find out transport of a matrix. com/bePatron?u=20475192 U. Greedy Algorithms Idea: Find solution by always making the choice that looks. Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen's algorithm. Probably most well-known technique in Computer Science. Matrix multiplication chains, dynamic. If your matrix is 3 x 3 or larger, finding the determinant takes a bit more work: 3 x 3 matrix: Choose any element and cross out the row and column it belongs to. But the algorithm is not very practical, so I recommend either naive multiplication, which runs in $\mathcal{O}(n^3)$, or S. COEN 279/AMTH 377 Design and Analysis of Algorithms Department of Computer Engineering Santa Clara University then the two points that define Divide and Conquer divide the problem into disjoint subproblems (hopefully split the problem in half)and solve them recursively, and the solution is then formed form. Algorithm for Matrix Multiplication. (b)What is wrong the the following algorithm for computing the square of an n n matrix? Just use a. Selection Procedure and Matrix Multiplication (in Hindi). 3 The D & C approach 2. This relies on the block partitioning. Third, matrix multiplication is the most expensive step of the whole divide-and-conquer SVD algorithm. Jordana, b a Department of Electrical Engineering and Computer Science, UC Berkeley b Department of Statistics, UC Berkeley Abstract This work introduces Divide-Factor-Combine (DFC), a parallel divide-and-conquer framework for noisy matrix factorization. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen's algorithm. Binary Multiplication Rules. For instance, if the array. Strassen in 1969 which gives an overview that how we can find the multiplication of two 2*2 dimension matrix by the brute-force algorithm. To see what this means, carve into four,. Break up polynomial into low and high powers. This paradigm, divide-and-conquer, breaks a problem into subproblems that are similar to the original problem, recursively solves the subproblems, and finally combines the solutions to the subproblems to solve the original problem. This algorithm basically performs the 'Divide and Conquer' approach by breaking down the large N×N matrix system into a manageable 32 × 32 matrix for fast computation. Dynamic programming, like the divide-and-conquer method, solves problems by combining the solutions to subproblems. Divide-and-conquer suggests that we divide the subarray into two subarrays of as equal size as possible. 50 Divide-and-conquer Decimation in frequency. Week 7: Divide and Conquer Example 3: Matrix multiplication: •Assume we are given two n ×n matrix X and Y to multiply. Goal Implementing a large matrix-matrix multiplication on FPGA Approach Using divide and conquer techniques to describe the matrix multiplication algorithm and then using SDSoC for high-level synthesis Benefits High-performance implementation, short time-to-market design Credit This work has been done under the ENPOWER project (funded by EPSRC) at the University of Bristol. 1 maximum subarray 4. The divide-and-conquer design paradigm 1. MATRIX MULTIPLICATION in Python. Divide-and-Conquer Let us investigate this recursive version of the matrix multiplication. View Show abstract. (This is how matrix multiplication got down under O(n^3): Strassen figured out how to multiply 2 2x2 matrices using 7 multiplications instead of 8. n /2 × n/2 matrices and then adds their. a triply nested loop as the base case of the recursion, 2. For example, we can model a graph with an adjacency matrix and. Divide and conquer is applied to many problems sorting matrix multiplication from CIS 502 at National Tsing Hua University, China. We aim to improve the sub-matrix multiplication procedure by addressing issues P 1 and P 2. De nition 4 (bilinear map). Applications. Strassen's Matrix Multiplication algorithm is the first algorithm to prove that matrix multiplication can be done at a time faster than O(N^3). Given two square matrices A and B of size n x n each, find their multiplication matrix. Strassen's algorithm : about O(n^2. Y1 - 2016/10/1. Who Should Enroll Learners with at least a little bit of programming experience who want to learn the essentials of. Binary Multiplication Rules. Divide-and-Conquer Examples Sorting: mergesort and quicksort Binary tree traversals Multiplication of large integers Matrix multiplication: Strassen’s algorithm Closest-pair and convex-hull algorithms Binary search: decrease-by-half (or degenerate divide&conq. ) Matrix Multiplication 29. Because this algorithm is recursive, there are many method calls, and method returns. Quicksort/Mergesort. Divide and Conquer Following is simple Divide and Conquer method to multiply two square matrices. 1 Drawbacks of Divide and Conquer We now discuss some bottleneck’s of Strassen’s algorithm (and Divide and Conquer algorithms in general). – The above naturally leads to divide-and-conquer solution: ∗ Divide X and Y into 8 sub-matrices A, B, C, and D. 1 Create a new thread to compute Mi = A'i B'i 2. We report the first IPS observations with the instrument, employing a Butler Matrix (BM) of 16×16 ports, fed by 16 east - west lines of 64 dipoles (1/4 of the total array). C program to find determinant of a matrix 12. Where the idea came from is unclear, however the goal was to reduce the number of multiplications needed to complete the algorithm. 2 Strassen's algorithm for matrix multiplication 4. DIVIDE AND CONQUER II ‣ master theorem ‣ integer multiplication ‣ matrix multiplication ‣ convolution and FFTSECTION 5. For example; given that matrix A is a 3 x 3 matrix, for matrix multiplication AB to be possible, matrix B must have size 3 x m where m can be any number of columns. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. ・n / bi = size of subproblem at level i. Divide and Conquer: The Karatsuba algorithm (multiplication of large integers) Instructor: L aszl o Babai Updated 01-13-2020 NOTATION. 4 The recursion-tree method for solving recurrences 4. DECREASE-AND-CONQUER & TRANSFORM-AND-CONQUER: DECREASE-AND-CONQUER Insertion Sort, Topological Sorting, Decrease-by-ConstantFactor Algorithms fake-coin problem,. it leaves open the possibility that you can first divide a problem and then apply a different algorithm to those smaller problems. Combine, Divide and Conquer b. The divide and conquer approach is an algorithm design paradigm which can be used to perform matrix multiplication. It enables us to reduce O(n^3) time complexity to O(n^2. The multiplication algorithm above is written iteratively, so we can easily analyze and implement it. To solve a given problem, it is subdivided into one or more subproblems each of which is similar to the given problem. • His method uses. Strassen's Matrix Multiplication algorithm is the first algorithm to prove that matrix multiplication can be done at a time faster than O(N^3). Strassen’s algorithm is a pretty smart algorithm which performs the same operation. Tree data structures. GENERAL METHOD: Given a function to compute on n inputs the divide-and-conquer strategy suggests splitting the inputs into k distinct subsets, 1Divide & Conquer approach: Strassen’s algorithm 1. Matrix multiplication is particularly easy to break into subproblems, because it can be performed blockwise. Winner of the Standing Ovation Award for "Best PowerPoint Templates" from Presentations Magazine. 2 Strassen's algorithm for matrix multiplication Table of contents. divide-and-conquer. Strassen’s algorithm is a pretty smart algorithm which performs the same operation. 2 Strassen's algorithm for matrix multiplication 4. The combine() method combines each of the multiplied matrice and combines the four parts to get the output matrice of that multiplication. Divide-and-Conquer Let us investigate this recursive version of the matrix multiplication. [Fourier, Dirichlet, Riemann] Any (sufficiently smooth) via fast matrix multiplication. Therefore the matrix-chain multiplication problem is given a chain of matrices to multiply, we need to find the grouping which minimizes the total number of scalar multiplications In most cases the time we spend to find this special grouping is more than made up by the savings in the effort to actually compute the solution. As a result of multiplication you will get a new matrix that has the same quantity of rows as the 1st one has and the same quantity of columns as the 2nd one. Here is below a schematic diagram for the matrix chain that we have. Divide and Conquer Ming-Hwa Wang, Ph. Times New Roman Monotype Sorts Symbol brknbars Equation Matrix Multiplication Winograd Multiplication Generalization Analysis Two By Two Multiplication 2x2 Works for Matrices Divide and Conquer? Strassen's Equations Using The Equations Analysis of Strassen. CSE 6331 Algorithms. org/rec/journals/corr/abs-1802-00003 URL. 1 Binary search 2. If the subproblems are relatively…. This package of Matlab functions computes the Takagi factorization of a complex-symmetric matrix. Multiplication: Karatsuba algorithm; Closest-Pair; Goddard: Part A - Divide and Conquer / Sorting and Searching. How quickly can you multiply a kn n matrix by an n kn matrix, using Strassen's algorithm as a subroutine? Answer the same question with the order of the input. Topic: Divide and Conquer 23 3. That is, we want to compute the product of A and A itself. Before stating the divide and conquer formulation, we introduce a few notations. Write a program to implement matrix multiplication using Strassen's method. However, there are more efficient algorithms for matrix multiplication than the naive approach. length, but these values. The second algorithm is a compute-aggregate-broadcast topological sort with time complexity of O(n log n) using O(n2) pro-cessors and O(1og2 n) using 0(n3) processors. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The Chain Matrix Multiplication Problem. Divide And Conquer Algorithm sample code - Build a C++ Program with C++ Code Examples - Learn C++ Programming. From here, I want to differentiate each column of the matrix with respect to g(x). Combine these results together. BibTeX @INPROCEEDINGS{Pauca97architecture-efficientstrassen's, author = {Paul Pauca and Xiaobai Sun and Siddhartha Chatterjee and Alvin Lebeck}, title = {Architecture-efficient Strassen's Matrix Multiplication: A Case Study of Divide-and-Conquer Algorithms}, booktitle = {In International Linear Algebra Society(ILAS) Symposium on Algorithms for Control, Signals and Image Processing}, year = {1997}}. Hardest if towards the middle - median (?) How might we approach?. Divide and Conquer. • Learn how to use Strassen’s algorithm Matrix Multiplication • Who cares? • For today, addition and multiplication count. Divide the problem into smaller problems. Instead of changing the size at the beginning, we. Easy if it is the smallest item or largest. Solve two parts recursively. 1 of Introduction to Algorithms introduces Merge sort algorithm, Chapter 4 "Divide and conquer" introduces The maximum-subarray problem and Strassen's algorithm for matrix multiplication. Divide and Conquer Following is simple Divide and Conquer method to multiply two square matrices. Strassen's method of matrix multiplication is a typical divide and conquer algorithm. It consists of rows and columns. 4 The recursion-tree method for solving recurrences 4. CS 56101 Section 002. Breaking down a problem into multiple independent subproblems, solving the subproblems (recursively), and combining those solutions into a solution for the original problem. Here the dimensions of matrices must be a power of 2. ; Roberts, J. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. Divide and Conquer † A general paradigm for algorithm design; inspired by emperors and colonizers. net/archives/V5/i3/IRJET-V5I3362. This paper discusses and compares several parallelization strategies for tree-structured computations. C program to find inverse of a matrix 8. We have implemented a simple formula for you to find the Strassen's matrix multiplication of the 4x4 matrix. BruteForce 2. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Classical matrix multiplication b. Most common usage. The three main steps of divide and conquer algorithm paradigm are: Divide the problem into subproblems; Conquer the subproblems by solving them recursively or directly. Conclusion. The following computer algorithms are based on divide-and-conquer programming approach − Merge Sort; Quick Sort; Binary Search; Strassen's Matrix Multiplication; Closest pair (points) There are various ways available to solve any computer problem, but the mentioned are a good example of divide and conquer approach. Divide and Conquer | Set 4 (Karatsuba algorithm for fast multiplication) Given two binary strings that represent value of two integers, find the product of two strings. Introduction to Algorithms 6. 4-by-4 Matrices, Using Strassen Multiplication // MStrassen. Veldhuizen tveldhui@acm. Reading: Chapter 18. Example-Matrix Multiplication Consider the problem of computing the product of two matrices. Polynomial time algorithms, and Divide and Conquer: Counting Inversions KT Chapter 5 TR Video (Week 1) 16/1 : Counting Inversions: Running time and Correctness proofs. ×) by min (resp. Collect and combine the solutions into the overall solution In contrast to the partitioning strategy, divide and conquer uses recursive partitioning with concurrent. For better under understanding lets see one more sorting technique called quick sort. A computer hacker learns from mysterious rebels about the true nature of his reality and his role in the war against its controllers. Introduction For any symmetric matrix A, there exists a special form of SVD, called Takagi factorization: where Q is unitary and is the diagonal singular value matrix. •These are huge matrices, say n ≈50,000. We show that the resulting implementations are 1. Divide and Conquer to Multiply and Order. The best known algorithm for matrix multiplication runs in time. https://irjet. Let X,Y be square matrices (of the same order) whose entries are non-negative integers or +∞. 2 Strassen's algorithm for matrix multiplication Table of contents. Divide-And-Conquer Approach. Divide and Conquer. ・n / bi = size of subproblem at level i. No longer only TWO subproblems Conquer: Solve each subproblem (directly or recursively), and Combine: Combine the solutions of the subproblems into a global solution. ! Formula for doing this is ! This runs in O(n3) time ! In fact, multiplying an n x m by an m x q takes nmq operations kj n k Z ij!X ikY " = = 1 0. 3 Master method T (n)=aT. ECE750-TXB Lecture 5: Veni, Divisi, Vici Todd L. to devise a better algorithm that would not be possible without Divide-and-Conquer. 5 The master method for solving recurrences 4. Soumyottam Chatterjee. De nition 4 (bilinear map). Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Compare it with the running time for Strassen's algorithm. Divide-and-ConquerMatrix Factorization Lester Mackey aAmeet Talwalkar Michael I. Both merge sort and quicksort employ a common algorithmic paradigm based on recursion. 1) before explaining the master theorem (Section 2. However, there are more efficient algorithms for matrix multiplication than the naive approach. University of Houston. Integer multiplication The problem: Multiply two large integers (n digits) The traditional way: Use two loops, it takes O(n2) operations Young CS 331 D&A of Algo. •The native algorithm will have to multiply one row of X by one column of Z (i. Solves difficult problems with ease. Such systems are able to support a large volume of parallel communication of various patterns in constant time. A divide and conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly. Divide And Conquer! - Matrix Multiplication; Understanding Divide And Conquer; K-merge Operation Of Sorted Arrays Using Divide And Conquer Method; Learning Divide And Conquer Algorithms; Divide And Conquer Closest-Pair Algorithm; Partitioning A Matrix; Math. Running Time: T(n) = ( nlg7) = O(n2:81) Matrix Product Checking Your friend gives you a program binary that multiplies matricesor so he claims. Breaking down a problem into multiple independent subproblems, solving the subproblems (recursively), and combining those solutions into a solution for the original problem. 7 Determining thresholds 2. Start studying Divide and Conquer. Sparse Matrix Multiplication 311. O(n) multiplication) to find out only one entry of the result Z •Total time will be O(n3). Algorithm Analysis techniques ----- Theory: Divide-and-Conquer Strategy: Given a function that has to compute on ‘n’ input the divide and conquer strategy suggest. Veldhuizen tveldhui@acm. #include int main(){. Conquer by solving these problems. 6 Proof of the master theorem. De nition 4 (bilinear map). Apr 28, 2020 - Lecture 13 : Recurrences and Divide and Conquer - PPT, Algorithms Notes | EduRev is made by best teachers of. Dynamic programming, like the divide-and-conquer method, solves problems by combining the solutions to subproblems. I will start with a brief introduction about how matrix multiplication is generally observed and implemented, apply different algorithms (such as Naive and Strassen) that are used in practice with both pseduocode and Python code, and then end with an analysis of their runtime. strassen matrix multiplication: Strassen's method is similar to above simple divide and conquer method in the sense that this method also divide matrices to sub-matrices of size N/2 x N/2 as shown in the above diagram, but in Strassen's method, the four sub-matrices of result are calculated using formulae. Square Matrix Multiplication using Divide and Conq Example of File. Simple Matrix Multiplication Method Divide and Conquer Method Strassen's Matrix Multiplication Method PATREON : https://www. The inner most Recursive call of multiplyMatrix() is to iterate k (col1 or row2). •These are huge matrices, say n ≈50,000. Algorithm Analysis/Matrix Multiplication. Nipun Vats. Divide and Conquer Following is simple Divide and Conquer method to multiply two square matrices. Strassen's method is similar to above simple divide and conquer method in the sense that this method also divide matrices to sub-matrices of size N/2 x N/2 as shown in the above diagram, but in Strassen. Applying a divide and conquer strategy recursively (view A i;j, B i;j and C i;j as matrices instead of scalars) allows matrix multiplication over n = 2N size matrices to be performed using only 7N = 7log 2 n= nlog 2 7 = O(n2:81) multiplications. The matrix product is now. Combine: 7 products into 4 terms using 8 matrix additions. Divide X, Y and Z into four (n/2)×(n/2) matrices as represented below −. Dynamic Programming vs. Our discussion will be based on the following problems: 1 Sorting (a review of merge sort) 2 Counting inversions 3 Dominance counting 4 Matrix multiplication We will focus on the ideas most relevant to illustrating. Jordana, b a Department of Electrical Engineering and Computer Science, UC Berkeley b Department of Statistics, UC Berkeley Abstract This work introduces Divide-Factor-Combine (DFC), a parallel divide-and-conquer framework for noisy matrix factorization. Numerous individual/specialized algorithms. 4 Divide-and-Conquer 4 Divide-and-Conquer 4. ・a ≥ 1 is the number of subproblems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Application of Divide and Conquer (in Hindi) Lesson 3 of 6 • 1 upvotes • 14:55 mins. Divide-and-Conquer Reading: CLRS Sections 2. BruteForce 2. Upper triangular matrix in c 10. divide and conquer (algorithmic technique) Definition: Solve a problem, either directly because solving that instance is easy (typically, because the instance is small) or by dividing it into two or more smaller instances. Yufei Tao Divide and Conquer. Break up polynomial into low and high powers. There has been a significant amount of work in recent years in the field of matrix multiplication algorithms as it has found its application in many areas. ! Formula for doing this is ! This runs in O(n3) time ! In fact, multiplying an n x m by an m x q takes nmq operations kj n k Z ij!X ikY " = = 1 0. Divide & Conquer Review >Integer multiplication >Matrix multiplication >Fast Fourier Transform >Integer multiplication again Outline for Today >Processor provides ability to multiply small (<= 64 bit) numbers >Multiplying arbitrary-size integers is a classic problem. Abstract | PDF (237 KB) (2009) A fast method for solving both the time-dependent Schrödinger equation in angular coordinates and its associated "m-mixing" problem. 1 Divide: Partition A and B into submatrices; add and subtract to form terms. Probably most well-known technique in Computer Science. Brief review of the tridiagonal DC method. Buyukokkten,J. 8) used in BLAS Good when n > 100, but there is an issue of numerical stability It is based on divide and conquer and block matrix decomposition divide matrix into several submatrices and the multiplication of…. 1 Strassen’s matrix multiplication 4. 3 Master method T (n)=aT. Recall that when multiplying two matrices, A =aij and B =bjk, the resulting matrix C =cik is given by. ・k = log b n levels. We improve the basic block. *; class Matrix{ public static void main Divide and Conquer (4). Matrix Chain Multiplication Dynamic Programming solves problems by combining the solutions to subproblems just like the divide and conquer method. Their structure will. Quicksort/Mergesort. e If we want to compute 2^8 we actually.
054xjbwnft3, z74mjcpqaa, nby4mooh97bx, m5e68v0fwmdbgib, vpqh0zc6mrlof, mpwzhbgk9sivjjp, rr9ebvdhg37yv, ir3gc98cjwint, wrk2k9us1wnyqgi, av2di6ip918p, i0ugm39xk5, unmqvrq3t6, 01szm0271o, x5bgxlzn8tfxev, 5akvnzvzwg61, mfpo4ld3een, 6w202cqfdix, mwmru667ebzx, duswt55y8sa, dpb7l7gkbqd, aq088c7n8jd, gr9n25mvibmgv, uprdfvrpzki, upqde8hjxz, bzkl4qxev3gjkwh, ribqkzowtjt7e, kadv63j943, oxlbdg50ql5gv, 9zedeblcfxelxy, j29c6hptmsnjh | 2020-05-25T18:09:39 | {
"domain": "incommunity.it",
"url": "http://incommunity.it/rvbw/matrix-multiplication-divide-and-conquer.html",
"openwebmath_score": 0.6675629019737244,
"openwebmath_perplexity": 1410.6004866033325,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9895109112146249,
"lm_q2_score": 0.8558511451289037,
"lm_q1q2_score": 0.8468740464805816
} |
https://mathhelpboards.com/analysis-50/reason-why-integral-zero-6647.html?s=9d017fe07dcf5abaf0820d527a904d47 | # Thread: Reason why Integral is Zero
1. Hello i would like to know why this integral is Zero:
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$
Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$
But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?
2.
3. Hi shen07!
Originally Posted by shen07
But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?
I believe you already said it: the "pole" lies outside the closed curve.
In that case the circular integral is zero.
If the curve were around a pole, it would be $2\pi i$.
4. Originally Posted by shen07
Hello i would like to know why this integral is Zero:
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$
Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$
But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?
Cauchy's Integral Theorem states that for any complex function which is closed and holomorphic everywhere in and on the boundary, its contour integral is equal to 0. As your function only has a singular point at z = -2, which is not in the boundary, your function satisfies the conditions and so Cauchy's Integral Theorem applies.
5. Originally Posted by shen07
Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$
By using $\displaystyle \gamma(a;r)$ you are making sure that the pole at $\displaystyle z=a$ is included in your curve by letting it the center of the disk since the radius is nonzero .
What if i use Cauchy's Theorem, since $$f(z)=\frac{1}{z+2}$$ is holomorphic on and inside $$\gamma(0;1)$$, using Cauchy's Theorem
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z=0$$
7. Originally Posted by shen07
What if i use Cauchy's Theorem, since $$f(z)=\frac{1}{z+2}$$ is holomorphic on and inside $$\gamma(0;1)$$, using Cauchy's Theorem
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z=0$$
Yes , since the only pole is at $z=-2$ which is out of the circular curve .
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 2020-02-25T21:11:34 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/analysis-50/reason-why-integral-zero-6647.html?s=9d017fe07dcf5abaf0820d527a904d47",
"openwebmath_score": 0.942291796207428,
"openwebmath_perplexity": 321.5124243991886,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.989510908740062,
"lm_q2_score": 0.855851143290548,
"lm_q1q2_score": 0.8468740425436512
} |
https://math.stackexchange.com/questions/1706333/why-yc-1x-iff-yc-2x-implies-c-1-c-2-c-i-is-a-relation | # Why $yC_1x \iff yC_2x$ implies $C_1 = C_2$? $C_i$ is a relation.
Here is the text from the book Topology by Munkres:
Studying equivalence relations on a set $A$ and studying partitions of $A$ are really the same thing. Given any partition $\scr D$ of $A$, there is exactly one equivalence relation on $A$ from which it is derived.
The proof is not difficult. To show that the partition $\scr D$ comes from some equivalence relation, let us define a relation $C$ on $A$ by setting $xCy$ if $x$ and $y$ belong to the same element of $\scr D$. Symmetry of $C$ is obvious; reflexivity follows from the fact that the union of the elements of $\scr D$ equals all of $A$; transitivity follows from the fact that distinct elements of $\scr D$ are disjoint. It is simple to check that the collection of equivalence classes determined by $C$ is precisely the collection $\mathscr{D}$.
To show there is only one such equivalence relation, suppose that $C_1$ and $C_2$ are two equivalence relations on $A$ that give rise to the same collection of equivalence classes $\mathscr{D}$. Given $x\in A$, we show that $yC_1 x$ if and only if $yC_2 x$, from which we conclude that $C_1=C_2$. Let $E_1$ be the equivalence class determined by $x$ relative to the relation $C_1$; let $E_2$ be the equivalence class determined by $x$ relative to the relation $C_2$. Then $E_1$ is an element of $\scr D$, so that it must equal the unique element of $D$ of $\scr D$ that contains $x$. Similarly, $E_2$ must equal $D$. Now by definition, $E_1$ consists of all $y$ such that $yC_1x$; and $E_2$ consists of all $y$ such that $yC_2x$. Since $E_1=D=E_2$, our result is proved.
The text of course proves that $yC_1x \iff yC_2x$, but why it implies $C_1 = C_2$? For example suppose elements are humans, so we can define $C_1$ for "person x and person y are in relation $C_1$ if each of them has two hands"; and, $C_2$ for "person x and person y are in relation $C_2$ if each of them has two foots". $yC_1x \iff yC_2x$ holds but $C_1 \ne C_2$?
Edit - PS - we ignore the set of people with two foots and less than two hand and with two hands and less than two foots.
• A relation is a set of ordered pairs. The statement $yC_1x\iff yC_2x$ means precisely that the ordered pairs in the first relation are precisely the ordered pairs of the second relation. Hence the sets are the same. – symplectomorphic Mar 20 '16 at 21:44
• As there are people with one hand and two feet, the two equivalence relations are not the same. – Doug M Mar 20 '16 at 22:06
• @DougM - otherwise they have prosthetic hand. – Liebe Mar 20 '16 at 22:43
There are persons with two foots and less than two hands; so there is no $\iff$ relationship between the two
The source of the confusion here is what $C_1=C_2$ means; since there is no "intrinsic way" to know what does it mean for two completely abstract relationships to be equal, we call them equal if they act in the same way.
• I just tried to understand why $yC_1x \iff yC_2x$ implies $C_1 = C_2$? My counterexample oops was bad. How to prove the implication? – Liebe Mar 20 '16 at 21:39
• By the way, what if we ignore the set of people with two foots and less than two hand and with two hands and less than two foots? – Liebe Mar 20 '16 at 21:42
• @Liebe then they would be equal. The source of the confusion here is what $C_1 = C_2$ means; since there is no "intrinsic way" to know what does it mean for two completely abstract relationships to be equal, we call them equal if they act in the same way. – Ant Mar 20 '16 at 21:49
• Thank you very much. Your comment is the answer for what I asked! If you please add it to the answer section so I can 'accept' it for this question not remained unaccepted-answered. – Liebe Mar 20 '16 at 21:59
• I will be disappointed if no effort is made to make sure that we respect the common usage for the definition of a relation. Actually, I don't care that much, but the definition used should then be clarified in the question. Having a question that becomes difficult because the definition is unclear and a lot of discussions based on this lack of a precise definition should have no place in mathematics. – Dominic108 Mar 20 '16 at 22:06
The point is that a relation on a set $S$ is defined as a subset of $S^2$. We write $x C y$ to mean that the pair $(x,y) \in C$. So if two relations are the same as subsets, they are the same.
• I thought that a relation means the reason for 'connectivity'. What that is called, the reason for 'connectivity'? I don't know how to it more clear, sorry my maths is weak. – Liebe Mar 20 '16 at 21:49
• In Enderton's book A Mathematical Introduction to Logic, for example, a relation is defined as a set of ordered pairs. – Dominic108 Mar 20 '16 at 21:51
• It's the same situation with functions. For functions $f_1$ and $f_2$ to be the same, what we need is $f_1(x) = y$ if and only if $f_2(x) = y$. The same function might have different descriptions (what Patrick Stevens is calling "definitions in intension"), but the description is not an intrinsic part of the function. – Robert Israel Mar 21 '16 at 0:23
This is a fairly subtle point.
Define $C_1$ to be the relation on $\mathbb{N}$ given by $x C_1 y$ if and only if $x-y$ is even.
Define $C_2$ to be the relation on $\mathbb{N}$ given by $x C_2 y$ if and only if $x-y+1$ is odd.
These are not verbatim the same relation, but actually the collection of all pairs related by $C_1$ is the same as the collection of all pairs related by $C_2$.
We say the relations have different definitions in intension but the same definition in extension: their descriptions are different but their ultimate outcome is the same. $C_1$ will view $x$ as being related to $y$ if and only if $C_2$ does.
If we have "$y C_1 x$ if and only if $y C_2 x$", then we have that $C_1$ and $C_2$ are the same in extension. We usually write $C_1 = C_2$ for "$C_1$ and $C_2$ are the same in extension", and we will (perhaps slightly sloppily, depending on your point of view) call them equal. Most people do not actually distinguish between equality and equality-in-extension; when they say "equal" they mean "equal in extension".
• Your answer assumes that the definition of a relation is a formula with two variables, but is this definition really the common usage? Do you have any standard textbook that defines a relation in this way? – Dominic108 Mar 20 '16 at 21:58
• I've been deliberately agnostic about how a relation is defined here. We all know what a relation is: it's a thing that relates things. If you want to implement a relation in set theory, you'll (usually) end up with a collection of ordered pairs, but I prefer not to give that as the definition of a relation because it's… not really what a relation is. Just as a natural number isn't really a finite ordinal, but we implement them in that way in set theory. Does that make sense? – Patrick Stevens Mar 21 '16 at 8:01
• Well, as long as the answer to the question depends on the specific definition chosen, I think it makes no sense not to discuss definitions. But, I understand that we want to validate the OP reasoning as much as possible and start from there. So, I understand that you we should not jump to fast into a specific definition. The OP was considering a general concept of relation based on sentences in English. So, we have to acknowledge that as much as possible. But, the answer to the question requires that we consider the standard math definition, so we have to bring the OP there. – Dominic108 Mar 21 '16 at 16:04
A relation on $A$ is a subset of $A\times A$, and $xC_1y$ is shorthand for $(x,y)\in C_1$, where $(x,y)$ is an ordered pair.
Thus a relation is determined by the pairs of elements that are in relation with each other, if you have $xC_1y\iff xC_2y$ the subsets $C_1$ and $C_2$ of $A\times A$ are the same set, even if you originally specified the $2$ relations in (apparently) different ways
• I thought that a relation means the reason for 'connectivity'. What that is called, the reason for 'connectivity'? I don't know how to it more clear, sorry my maths is weak. – Liebe Mar 20 '16 at 21:51
• When we informally say something like "let $xRy$ if $2|x-y$, where $x,y\in\mathbb{N}$ what's meant is "let $R\subseteq\mathbb{N}\times\mathbb{N}=\{(x,y):2|x-y\}$" because that's how relations on a set are defined. Of course in most situations the definition is just cumbersome and unneeded, but it's relevant in this case, because it follows from it that 2 relations are the same if they are the same as sets, regardless of the way we describe them in more or less formal language – Alessandro Codenotti Mar 20 '16 at 21:57
It depends on how we define a relation. Usually a relation on $A$ is a subset of $A \times A$. In this case, the implication is clear. However, it would also be natural to define a relation as a formula on two variables. It is not the usual definition, but it is natural. It is also closer to a sentence in English, so perhaps more intuitive. In this case, indeed there is no implication.
A set is determined by the elements in it: two sets $X, Y$ are equal provided $x\in X$ iff $x\in Y$. A relation on $A$ is just a certain subset of $A\times A$, and you've shown that a point $(x, y)\in A\times A$ lies in $C_1$ iff it lies in $C_2$. | 2019-11-12T13:03:55 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1706333/why-yc-1x-iff-yc-2x-implies-c-1-c-2-c-i-is-a-relation",
"openwebmath_score": 0.8726533651351929,
"openwebmath_perplexity": 167.6635965507014,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9688561730622295,
"lm_q2_score": 0.8740772351648677,
"lm_q1q2_score": 0.8468551250226481
} |
http://divorcedocuments.ca/5t7pxs/85a968-argument-of-complex-number | Follow 722 views (last 30 days) bsd on 30 Jun 2011. The modulus and argument of a Complex numbers are defined algebraically and interpreted geometrically. 8. 0. The modulus and argument are fairly simple to calculate using trigonometry. You can also determine the real and imaginary parts of complex numbers and compute other common values such as phase and angle. The principal amplitude of (sin 4 0 ∘ + i cos 4 0 ∘) 5 is. For example, 3+2i, -2+i√3 are complex numbers. The modulus of a complex number z, also called the complex norm, is denoted |z| and defined by |x+iy|=sqrt(x^2+y^2). 6. Argument of a Complex Number Description Determine the argument of a complex number . Finding the complex square roots of a complex number without a calculator. See also. The argument of the complex number 0 is not defined. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full … Python complex number can be created either using direct assignment statement or by using complex function. 7. If I use the function angle(x) it shows the following warning "??? Complex Numbers and the Complex Exponential 1. Subscript indices must either be real positive integers or logicals." Modulus of a complex number, argument of a vector In the case of a complex number, r represents the absolute value or modulus and the angle θ is called the argument of the complex number. The modulus of z is the length of the line OQ which we can find using Pythagoras’ theorem. Yes, the argument of a complex number can be negative, such as for -5+3i. Please reply as soon as possible, since this is very much needed for my project. Find the argument of the complex number, z 1 = 5 + 5i. In the Argand's plane, the locus of z ( = 1) such that a r g {2 3 (3 z 2 − z − 2 2 z 2 − 5 z + 3 )} = 3 2 π is. Conversion and Promotion are defined so that operations on any combination of predefined numeric types, whether primitive or composite, behave as expected.. Complex Numbers The argument of z is the angle formed between the line joining the point to the origin and the positive real axis. Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer Normally, we would find the argument of a complex number by using trigonometry. Calculate with cart. The square |z|^2 of |z| is sometimes called the absolute square. 1 How can you find a complex number when you only know its argument? value transfers the cartesian number into the second calculator. Thanking you, BSD 0 Comments. how to find argument or angle of a complex number in matlab? Vote. As result for argument i got 1.25 rad. (4.1) on p. 49 of Boas, we write: z = x+iy = r(cosθ +isinθ) = rei θ, (1) where x = Re z and y = Im z are real numbers. Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. Argument of a Complex Number Description Determine the argument of a complex number . Identify the argument of the complex number 1 + i Solve a sample argument equation State how to find the real measurement of the argument in a given example Skills Practiced. the complex number, z. We can define the argument of a complex number also as any value of the θ which satisfies the system of equations $\displaystyle cos\theta = \frac{x}{\sqrt{x^2 + y^2 }}$ $\displaystyle sin\theta = \frac{y}{\sqrt{x^2 + y^2 }}$ The argument of a complex number is not unique. It's interesting to trace the evolution of the mathematician opinions on complex number problems. The argument of a complex number is the angle formed by the vector of a complex number and the positive real axis. For instance, an electric circuit which is defined by voltage(V) and current(C) are used in geometry, scientific calculations and calculus. For a complex number in polar form r(cos θ + isin θ) the argument is θ. 0. Example #4 - Argument of a Complex Number in Radians - Exact Measurement. Given a quadratic equation: x2 + 1 = 0 or ( x2 = -1 ) has no solution in the set of real numbers, as there does not exist any real number whose square is -1. Argument in the roots of a complex number. Click hereto get an answer to your question ️ The argument of the complex number sin 6pi5 + i ( 1 + cos 6pi5 ) is Functions. Dear sir/madam, How do we find the argument of a complex number in matlab? It has been represented by the point Q which has coordinates (4,3). An alternative option for coordinates in the complex plane is the polar coordinate system that uses the distance of the point z from the origin (O), and the angle subtended between the positive real axis and the line segment Oz in a counterclockwise sense. What can I say about the two complex numbers when divided have a complex number of constant argument? Either undefined, or any real number is an argument of 0 . Looking forward for your reply. What is the argument of Z? The angle between the vector and the real axis is defined as the argument or phase of a Complex Number. The argument of a complex number In these notes, we examine the argument of a non-zero complex number z, sometimes called angle of z or the phase of z. 7. (1) If z is expressed as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1. Example.Find the modulus and argument of z =4+3i. The argument is measured in radians as an angle in standard position. Lernen Sie die Übersetzung für 'argument complex number of a' in LEOs Englisch ⇔ Deutsch Wörterbuch. This is the angle between the line joining z to the origin and the positive Real direction. What is the argument of 0? Trouble with argument in a complex number. It is denoted by $$\arg \left( z \right)$$. Hot Network Questions To what extent is the students' perspective on the lecturer credible? That means we can use inverse tangent to figure out the measurement in degrees, then convert that to radians. Does magnitude and modulus mean the same? Complex and Rational Numbers. Following eq. If I use the function angle(x) it shows the following warning "??? View solution ∣ z 1 + z 2 ∣ = ∣ z 1 ∣ + ∣ z 2 ∣ is possible if View solution. 1 A- LEVEL – MATHEMATICS P 3 Complex Numbers (NOTES) 1. The argument of z is denoted by θ, which is measured in radians. abs: Absolute value and complex magnitude: angle: Phase angle: complex: Create complex array: conj : Complex conjugate: cplxpair: Sort complex numbers into complex conjugate pairs: i: … Consider the complex number $$z = - 2 + 2\sqrt 3 i$$, and determine its magnitude and argument. Solution.The complex number z = 4+3i is shown in Figure 2. Modulus and argument. Note Since the above trigonometric equation has an infinite number of solutions (since $$\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\theta$$ and let us call them conventions 1 and 2 for simplicity. Complex Number Vector. For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted … Misc 13 Find the modulus and argument of the complex number ( 1 + 2i)/(1 − 3i) . and the argument of the complex number $$Z$$ is angle $$\theta$$ in standard position. Then, the argument of our complex number will be the angle that this ray makes with the positive real axis. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. 0 ⋮ Vote. What I want to do is first plot this number in blue on the complex plane, and then figure out what it is raised to the 20th power and then try to plot that. A complex number is a number of the form a+bi, where a,b — real numbers, and i — imaginary unit is a solution of the equation: i 2 =-1.. We can represent a complex number as a vector consisting of two components in a plane consisting of the real and imaginary axes. Phase of complex number. View solution. Phase (Argument) of a Complex Number. I am using the matlab version MATLAB 7.10.0(R2010a). Complex numbers which are mostly used where we are using two real numbers. Let us discuss another example. This leads to the polar form of complex numbers. I'm struggling with the transformation of rad in degrees of the complex argument. We can note that the complex number, 5 + 5i, is in Quadrant I (I'll let you sketch this one out). Argument of Complex Numbers. Geometrically, the phase of a complex number is the angle between the positive real axis and the vector representing complex number.This is also known as argument of complex number.Phase is returned using phase(), which takes complex number as argument.The range of phase lies from-pi to +pi. The magnitude is also called the modulus. I want to transform rad in degrees by calculation argument*(180/PI). The angle φ is in rad, here you can convert angle units. i.e from -3.14 to +3.14. Commented: Seungho Kim on 3 Dec 2018 Accepted Answer: Sean de Wolski. Here we introduce a number (symbol ) i = √-1 or i2 = -1 and we may deduce i3 = -i i4 = 1 (2) The complex modulus is implemented in the Wolfram Language as Abs[z], or as Norm[z]. In spite of this it turns out to be very useful to assume that there is a number ifor which one has (1) i2 = −1. Argument of z. Julia includes predefined types for both complex and rational numbers, and supports all the standard Mathematical Operations and Elementary Functions on them. I have the complex number cosine of two pi over three, or two thirds pi, plus i sine of two thirds pi and I'm going to raise that to the 20th power. Instead, it’s the angle between two of our axes, so we know this is a right angle. a = ρ * cos(φ) b = ρ * sin(φ) Therefore, the two components of the vector are it’s real part and it’s imaginary part. But as result, I got 0.00 degree and I have no idea why the calculation failed. View solution. However, in this case, we can see that our argument is not the angle in a triangle. Starting from the 16th-century, mathematicians faced the special numbers' necessity, also known nowadays as complex numbers. Complex Numbers Conversion of the forms of complex numbers, cartesian, to polar and exponentiation with →, the other was with ←. The argument of the complex number sin 5 6 π + i (1 + cos 5 6 π ) is. How do we find the argument of a complex number in matlab? We note that z … Solution for find the modulus and argument of the complex number (2+i/3-i)^2 You can use them to create complex numbers such as 2i+5. Examples with detailed solutions are included. We know this is a right angle joining z to the polar form r ( θ. Form of complex numbers ( NOTES ) 1 i want to transform in... Or any real number is the angle in standard position mostly used where we are using two numbers... Kim on 3 Dec 2018 Accepted Answer: Sean de Wolski but as result, i got 0.00 and., here you can also determine the real and imaginary parts of complex numbers \ ) in position! De Wolski numbers, and determine its magnitude and argument are fairly simple to using... As soon as possible, since this is very much needed for my project or as [... The forms of complex numbers are defined as the argument of 0 joining z to origin. Sometimes called the absolute square real positive integers or logicals. you know. To the origin and the argument of a complex number \ ( z \right ) \ is! Of ( sin 4 0 ∘ ) 5 is by using argument of complex number absolute.... Vector consisting of the mathematician opinions on complex number in radians the evolution of the complex number radians... Number sin 5 6 π ) is angle \ ( \theta \.... To polar and exponentiation with →, the other was with ← standard! Square roots of a complex exponential ( i.e., a phasor ), then |re^ ( )! ( 180/PI ) the standard Mathematical Operations and Elementary Functions on them see that our argument is.... You find a complex number and the positive real direction it is denoted |z| and defined by |x+iy|=sqrt ( )! S imaginary part use inverse tangent to figure out the Measurement in degrees, then convert that to radians line! 1 + z 2 ∣ is possible if view solution the line OQ which we can that... The form x+iy, where x and y are real numbers and i no!, -2+i√3 are complex numbers, and supports all the standard Mathematical Operations and Elementary Functions on them 5 π! How can you find a complex number in matlab numbers ( NOTES ) 1 the length of the x+iy... Übersetzung für 'argument complex number is the students ' perspective on the lecturer credible axes, we! Starting from the 16th-century, mathematicians faced the special numbers ' necessity, also called the complex norm, denoted... Struggling with the positive real direction for my project in degrees of the axis. The 16th-century, mathematicians faced the special numbers ' necessity, also called the square. As result, i got 0.00 degree and i = √-1 -2+i√3 complex! Needed for my project cartesian, to polar and exponentiation with → the... Complex exponential ( i.e., a phasor ), then |re^ ( iphi ) |=|r| ' necessity, also the. Number z = 4+3i is shown in figure 2 on 30 Jun 2011 inverse tangent figure! Note that z … the argument of a complex number problems to create complex numbers are defined numbers... Absolute square [ z ], or as norm [ z ] Questions to extent... About the two components in a triangle possible if view solution ∣ z 1 = 5 + 5i \arg... Argument * ( 180/PI ) + 2\sqrt 3 i\ ), and supports all the standard Operations! What extent argument of complex number the angle formed between the line joining the point to the polar form (... Which is measured in radians as an angle in a plane consisting of the form x+iy, x. Are real numbers and i have no idea why the calculation failed which has coordinates ( 4,3 ) not angle! + isin θ ) the argument is not defined a calculator which are mostly where. Of constant argument using the matlab version matlab 7.10.0 ( R2010a ) square roots a. Hot Network Questions to what extent is the angle between two of our axes, so know... Norm, is denoted by θ, which is measured in radians,! Inverse tangent to figure out the Measurement in degrees of the mathematician opinions complex... Angle of a ' in LEOs Englisch ⇔ Deutsch Wörterbuch if i use the angle. Represented by the point Q which has coordinates ( 4,3 ) common values such as phase and angle \right \... In figure 2 algebraically and interpreted geometrically expressed as a vector consisting of the opinions... Was with ← How can you find a complex number when you know. Is shown in figure 2 nowadays as complex numbers Jun 2011 has support! Mathematicians faced the special numbers ' necessity, also known nowadays as complex numbers are defined and... Common values such as phase and angle a calculator that our argument is θ origin... Vector are it ’ s the angle formed between the line joining the point to the origin the. 'M struggling with the transformation of rad in degrees by calculation argument * ( 180/PI.. Cos θ + isin θ ) the argument of a complex number sin 5 π! That means we can use inverse tangent to figure out the Measurement in degrees calculation. Here you can use them to create complex numbers when divided have complex... \ ) in standard position and y are real numbers – MATHEMATICS P 3 complex numbers of. Mathematician opinions on complex number without a calculator as an angle in standard.! The line OQ which we can use inverse tangent to figure out the Measurement in degrees of the mathematician on! |Z| and defined by |x+iy|=sqrt ( x^2+y^2 ) complex variables was with ← ( R2010a ) complex number a! Joining z to the origin and the positive real axis is defined as the is... What can i say about the two components of the mathematician opinions on complex number z. Z ], or any real number is an argument of a number. Convert that to radians numbers of the vector and the positive real axis is as...????????????????????! X ) it shows the following warning ???????????! Between two of our complex number is the angle between the line the. De Wolski two components in a triangle ) \ ) in standard.. Can i say about the two complex numbers and argument of complex number other common values such as phase and angle (. Can use inverse tangent to figure out the Measurement in degrees, then convert to... Cos θ + isin θ ) the argument of the form x+iy, where and. Angle in a triangle Measurement in degrees of the line joining the point to the polar r!, so we know this is the angle in standard position the square. Function angle ( x ) it shows the following warning ??. X and y are real numbers and symbolic complex variables rad in degrees of the complex norm is. Elementary Functions on them but as result argument of complex number i got 0.00 degree and i have idea... Be real positive integers or logicals. exponential ( i.e., a phasor ), convert! Of rad in degrees by calculation argument * ( 180/PI ) when divided a... An angle in a triangle 1 ) if z is the angle formed between the vector are it ’ imaginary. Number problems determine the real and imaginary parts of complex numbers such as phase and angle i have idea! Interesting to trace the evolution of the mathematician opinions on complex number \ ( z )! 1 ) if z is denoted by \ ( z \ ) called the complex modulus implemented. Types for both explicit complex numbers are defined algebraically and interpreted geometrically simple... | 2022-05-24T19:18:28 | {
"domain": "divorcedocuments.ca",
"url": "http://divorcedocuments.ca/5t7pxs/85a968-argument-of-complex-number",
"openwebmath_score": 0.9001246690750122,
"openwebmath_perplexity": 582.8222687533695,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_q1_score": 0.9688561712637257,
"lm_q2_score": 0.8740772302445241,
"lm_q1q2_score": 0.8468551186835116
} |
http://math.stackexchange.com/questions/115176/a-probability-question-using-percentages/115178 | # a probability question using percentages
This question is confusing me as I am not used to seeing percentages in a possibility question.
in a large insurance agency
- 60% of the customers have automobile insurance
- 40% of the customers have homeowners insurance
- 75% of the customers have on type or the other or both
a) find the proportion of customers with both types of insurance.
b) find the probability that a customer has homeowner insurance
given that he has automobile insurance
c) find the probability that a customer has automobile insurance given that
he has home insurance
I am not asking for someone to solve this, I am just wondering if it would be the same logic as a normal "rolling the dice n times question"?
so far I have thought of this :
$P(A) = \frac{6}{10}$ customers have auto insurance
$P(B) = \frac{4}{10}$ have homeowner insurance
$P(C) = \frac{7.5}{10}$ have one type or the other or both
a) so we have to do $A \land B$ and since they are independent events we can just multiply $P(A) \times P(B)$ : $$(6/10) * (4/10) = 24/100$$
b) $$P(B|A) = P(B \land A)/P(A) = (24/100)/6/10 = 4/10$$
now thats as far as I have done but I started thinking that I did not use $C$ at all so I think I am doing something wrong and I should take $C$ into account but not sure how. could some one tell me if I am doing anything wrong.
-
The meaning of, say, $17$% is $\frac{17}{100}$, so translation from percent language to ordinary number language is automatic: $17$% **is** $0.17$. – André Nicolas Mar 1 '12 at 6:29
Yea got that :) ... another quick question which I thought I shouldnt post a whole new topic for : the statement : "Only one tenth of 1% of the individuals in a certain population have a particular disease (an incident rate of .001), does that mean only 0.001 percent are affected ? – Ahoura Ghotbi Mar 1 '12 at 6:36
$1$% is $1/100$, which is $0.01$. One tenth of that is $0.001$. But what you quoted says that, it says the indidence rate is $0.001$. It is definitely not $0.001$ percent, which is $0.00001$. – André Nicolas Mar 1 '12 at 6:43
You need to use the following formula, which is undoubtedly part of your course material: $$P(A\cup B)=P(A)+P(B)-P(A\cap B). \qquad(\ast)$$
At least informally, this is fairly easy to see by drawing a diagram. Draw two intersecting circles, and label them $A$ and $B$. The probability of $A\cup B$ is kind of the weight of $A\cup B$. If we add $P(A)$ and $P(B)$, we have counted the probability of $A\cap B$ twice, so we must subtract it.
Let $A$ be the event "has auto insurance" and let $B$ be the event "has home insurance."
For Question a), we want $P(A\cap B)$. Using the formula $(\ast)$, we find that $0.75=0.60+0.40-P(A\cap B)$, and therefore $P(A\cap B)=0.25$.
For Question b), you want $P(B|A)$. Here you use the fact that $$P(A\cap B)=P(B|A)P(A).$$ From a), you know $P(A\cap B)$. And you certainly know $P(A)$. Now you can find $P(B|A)$.
Question c) is answered very much like Question b).
Remark: Note that $A$ and $B$ are not independent. If they were, we would have $P(A\cap B)=P(A)P(B)$. But we saw that $P(A\cap B)=0.25$. Note that $P(A)P(B)=0.24$. Not equal!
However, interestingly enough, the two numbers $0.25$ and $0.24$ are quite close to each other. Informally, although $A$ and $B$ are not independent, they are fairly close to being independent.
-
yea it is part of the course, just saw it. but how about the third statement, statement C? how would I include that in my calculations ?? it cant be there for no reason right? :S – Ahoura Ghotbi Mar 1 '12 at 6:07
and also part a mentions the customers who have BOTH types, so is what I am doing correct ? – Ahoura Ghotbi Mar 1 '12 at 6:09
Thanks man, are you by any chance located in Canada, and available for tutoring? – Ahoura Ghotbi Mar 1 '12 at 6:17
Yes, $P(A) = 6/10$ and $P(B) = 4/10$, but no, $A$ and $B$ are not independent. The third piece of information is telling you $P(A \cup B) = 75/100$.
-
could you elaborate a bit more on how I can account for P(A union B) in my calculations? – Ahoura Ghotbi Mar 1 '12 at 5:54
See Andre's answer – Robert Israel Mar 1 '12 at 17:20 | 2014-08-30T16:36:12 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/115176/a-probability-question-using-percentages/115178",
"openwebmath_score": 0.7598447799682617,
"openwebmath_perplexity": 421.2650021016579,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9688561658682131,
"lm_q2_score": 0.8740772318846386,
"lm_q1q2_score": 0.846855115556452
} |
https://brilliant.org/discussions/thread/easy-d/ | # Try to solve it 2
$\large \lim_{x\to1} \frac{x+x^2+x^3+\ldots +x^n-n}{\sqrt x-1} = \ ?$
Note by Abdulrahman El Shafei
5 years, 5 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Without L'Hôpital's rule
Rationalize the denominator by multiplying the expression with $\frac{\sqrt{x} + 1} {\sqrt{x} + 1}$. Distribute the $n$ and rewrite the expression as
$\lim _{ x \to 1 } \frac{ (x - 1) + (x^{2} - 1) + (x^{3} - 1) + \cdots + (x^{n} - 1) }{x -1} \cdot (\sqrt{x} + 1)$
$= \lim _{ x \to 1 } \left( \frac{ x - 1 } { x -1 } + \frac{ x^{2} - 1 } { x -1 } + \frac{ x^{3} - 1 } { x -1 } + \cdots + \frac{ x^{n} - 1 } { x -1 }\right) \cdot (\sqrt{x} + 1)$
$= \lim _{ x \to 1 } \left( (1) + (1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + \cdots + (1 + x + x^{2} + x^{3} + \cdots + x^{n-1}) \right) \cdot (\sqrt{x} + 1)$
Now solve by substitution:
$= ( 1 + 2 + 3 + \cdots + n) \cdot (1 + 1) = \boxed{n(n+1)}$
- 5 years, 5 months ago
Amazing solution! :)
- 5 years, 1 month ago
$n^{2}+n$???
- 5 years, 5 months ago
Did you use L.Hospital's Rule ?
- 5 years, 5 months ago
Yes. Can u prove the L.hospital rule?
- 5 years, 5 months ago
n(n+1)
- 5 years, 5 months ago
Did you use L.Hospital's Rule ?
- 5 years, 5 months ago
Yes. However I cannot post my solution. My Latex skills s*cks. Sorry.
- 5 years, 5 months ago
n(n+1) May be thats the answer. If it is correct please notify.
- 5 years, 5 months ago
Did you use L.Hospital's Rule ?
- 5 years, 5 months ago
n(n+1)
- 5 years, 5 months ago
Did you use L.Hospital's Rule ?
- 5 years, 5 months ago
n(n+1)
- 4 years, 8 months ago
Since this is an indeterminate form of the $\dfrac{0}{0}$ form, we can simply use the L.Hospital's Rule to evaluate the given limit, which is as you follow:
$\begin{array}{} & \lim_{x\to1} \dfrac{x+x^2+x^3+\ldots +x^n-n}{\sqrt x-1} \\ & = \lim_{x \to 1} \dfrac{1+2x+3x^2+\ldots +nx^{n-1}}{\dfrac{1}{2 \sqrt x}} \\ & = n(n+1) \end{array}$
- 5 years, 5 months ago
There is an answer without using L.Hospital's Rule ,....can u find it ?
- 5 years, 5 months ago
There will be no difference in the answer whether you solve it using L.Hospital's rule or not. But yeah, there obviously exists a way to evaluate it without using L.Hospital's rule, as Pranshu did. And notice that the answer is still the same.
- 5 years, 5 months ago
We could just use L'Hopital's rule but sometimes we can search about another answer ...I was looking for an approach that does not use this rule and I am sorry I didn't mention , but your solution still elegant :))
- 5 years, 5 months ago
n(n+1)
- 2 years, 9 months ago | 2021-03-01T04:05:07 | {
"domain": "brilliant.org",
"url": "https://brilliant.org/discussions/thread/easy-d/",
"openwebmath_score": 0.9967843890190125,
"openwebmath_perplexity": 4514.20299437972,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9688561658682131,
"lm_q2_score": 0.874077222043951,
"lm_q1q2_score": 0.8468551060222411
} |
https://math.stackexchange.com/questions/3017393/fermats-little-theorem-and-carmichael-numbers | # Fermat's Little Theorem and Carmichael Numbers
Fermat's little theorem states that if $$p$$ is a prime number and $$a$$ is a positive integer, then $$p|a^p-a$$.
However, the converse is false, that is, for integers $$a$$ and $$p$$, if $$p|a^p-a$$, then $$a$$ is a prime number, is a false statement. For instant, $$561|a^{561}-a$$ for some integer $$a$$, but $$561$$ is actually a composite number, and such numbers are called "Carmichael numbers".
In other words, a Carmichael number is a composite integer, say $$k$$, such that $$k|a^k-a$$ for all integers $$a$$.
This is what I know, am I right or I misunderstand something?
and
Do we have a way to find Carmichael numbers?
• Search the web for find Carmicael numbers maths.lancs.ac.uk/jameson/carfind.pdf Nov 28 '18 at 17:04
• There is a nice criterion for a squarefree odd composite $N>1$ : $N$ is a Carmichael number if and only if $p-1\mid N-1$ holds for every prime $p\mid N$. If $N$ is not squarefree odd and composite, it cannot be a Carmichael number. Moreover, it can be shown that $N$ must have at least three prime factors. Nov 28 '18 at 19:00
Numbers of the form $$(6k+1)(12k+1)(18k+1)$$ are Carmichael numbers if each of the three factors is prime. This gives some examples - already $$k=1$$ works. Actually, the sequence A046025 gives more values, e.g.,
$$k=1, 6, 35, 45, 51, 55, 56, 100, 121, 195, 206, 216, 255, 276, 370, 380, 426, 506, 510, 511, 710, 741, 800, 825, 871, 930, 975, 1025, 1060, 1115, 1140, 1161, 1270, 1280, 1281, 1311, 1336, 1361, 1365, 1381, 1420, 1421, 1441, 1490, 1515, 1696, 1805, 1875, 1885$$ | 2021-10-20T23:12:21 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3017393/fermats-little-theorem-and-carmichael-numbers",
"openwebmath_score": 0.8360247611999512,
"openwebmath_perplexity": 116.94078070335249,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9939964029584999,
"lm_q2_score": 0.8519528000888386,
"lm_q1q2_score": 0.8468380187787274
} |
http://math.stackexchange.com/questions/199258/is-the-vector-in-the-space-of-3-other-vectors/199261 | # Is the vector in the space of 3 other vectors
I have a set of 3 vectors $$IE = {[1, 1, -3]; [2, -1, 3]; [-6, 3, -9]}$$
I want to know if the vector [1, 4, -12] , belongs (or is in the span?) to my previous set.
So here's what I did.
$$\begin{matrix} 1 & 2 & -6 & [c1]\\ 1 & -1 & 3 & [c2]\\ -3 & 3 & -9 & [c3]\\ \end{matrix} = \begin{matrix} 1 \\ 4 \\ -12 \\ \end{matrix}$$
Now I start Gauss
L2-L1
$$\begin{matrix} 1 & 2 & -6 \\ 0 & -3 & 9 \\ -3 & 3 & -9 \\ \end{matrix} = \begin{matrix} 1 \\ 3 \\ -12 \\ \end{matrix}$$
L3+3L1
$$\begin{matrix} 1 & 2 & -6 \\ 0 & -3 & 9 \\ 0 & 9 & -27 \\ \end{matrix} = \begin{matrix} 1 \\ 3 \\ -9 \\ \end{matrix}$$
L2 / 3
and then L3-9L2
$$\begin{matrix} 1 & 2 & -6 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ \end{matrix} = \begin{matrix} 1 \\ -1 \\ 0 \\ \end{matrix}$$
1. The variable C3 (3rd column) has no pivot. which means it is a free variable.
2. There is no contradiction , so the vector $$[1, 4, -12]$$ is in the space of my Set.
I have 2 questions :
1. does the fact that I have a free variable change anything ?
2. A contradiction happens when I have a line full of zeros that equals a non zero value, right ?
Like
$$\begin{matrix} 0 & 0 & 0 \\ \end{matrix} = \begin{matrix} 2 \\ \end{matrix}$$
Is there any other way to have a contradiction ?
Sorry if the title isn't clear, I have trouble traducing the question..
Thanks
-
Note that you also multiplied L2 by -1 at some point at the end to reach your conclusion - what you have written is correct, I'm just pointing out that you did one more row-reduction step. – Michael Boratko Sep 19 '12 at 18:04
We could also continue row reduction to find out just what values of $c_1,c_2,c_3$ give the linear combination we are interested in:
$$\left[ \begin{array}{ccc|c} 1 & 2 & -6 &1 \\ 0 & 1 & -3 &-1 \\ 0 & 0 & 0 &0\end{array} \right] \sim \left[ \begin{array}{ccc|c} 1 & 0 & 0 &3 \\ 0 & 1 & -3 &-1 \\ 0 & 0 & 0 &0\end{array} \right]$$
So $c_1=3$, and $c_2=3c_3 - 1$. Let's see if this works in general: $$3\left[ \begin{array}{c} 1 \\ 1 \\ -3\end{array} \right] + (3c_3-1)\left[ \begin{array}{c} 2 \\ -1 \\ 3\end{array} \right]+ c_3\left[ \begin{array}{c} -6 \\ 3 \\ -9\end{array} \right] = \left[ \begin{array}{c} 3+2(3c_3-1)-6c_3 \\ 3-(3c_3-1)+3c_3 \\ -9+3(3c_3-1)-9c_3\end{array} \right]=\left[ \begin{array}{c} 1 \\ 4 \\ -12\end{array} \right]$$ So this means you can actually find infinitely many linear combinations by taking different values for $c_3$.
2. If you do row reduction, that is essentially the only way to have a contradiction. The reason it is a contradiction is because if you remember that the matrix is "shorthand" for a system of linear equations, the linear equation corresponding to a row of zeros on the left and a number like 2 on the right is of the form$$0c_1 + 0c_2 + 0c_3 = 2 \implies 0 = 2$$ which is why it is a contradiction. Any other time, if you reduce to row echelon form and you don't have a row of zeros on the left corresponding to a nonzero number in the augmented column, you essentially are using "back-substitution" to solve the system of equations. | 2014-04-19T17:59:30 | {
"domain": "stackexchange.com",
"url": "http://math.stackexchange.com/questions/199258/is-the-vector-in-the-space-of-3-other-vectors/199261",
"openwebmath_score": 0.8561221361160278,
"openwebmath_perplexity": 143.10545169085555,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9579122756889437,
"lm_q2_score": 0.8840392710530071,
"lm_q1q2_score": 0.8468320699327809
} |
https://rstudio-pubs-static.s3.amazonaws.com/780087_a50acc9bce4142ab94d6d94adaa40b53.html | ## Chapter 7 - Ulysses’ Compass
The chapter began with the problem of overfitting, a universal phenomenon by which models with more parameters fit a sample better, even when the additional parameters are meaningless. Two common tools were introduced to address overfitting: regularizing priors and estimates of out-of-sample accuracy (WAIC and PSIS). Regularizing priors reduce overfitting during estimation, and WAIC and PSIS help estimate the degree of overfitting. Practical functions compare in the rethinking package were introduced to help analyze collections of models fit to the same data. If you are after causal estimates, then these tools will mislead you. So models must be designed through some other method, not selected on the basis of out-of-sample predictive accuracy. But any causal estimate will still overfit the sample. So you always have to worry about overfitting, measuring it with WAIC/PSIS and reducing it with regularization.
Place each answer inside the code chunk (grey box). The code chunks should contain a text response or a code that completes/answers the question or activity requested. Make sure to include plots if the question requests them. Problems are labeled Easy (E), Medium (M), and Hard(H).
Finally, upon completion, name your final output .html file as: YourName_ANLY505-Year-Semester.html and publish the assignment to your R Pubs account and submit the link to Canvas. Each question is worth 5 points.
## Questions
7E1. State the three motivating criteria that define information entropy. Try to express each in your own words.
#1. First motivating criteria is the information uncertainity must be continuous that way the possible outcomes can be seen in a pattern
#2. Second motivating criteria is that the uncertainity must be increasing with number of possible outcomes so that various uncertainities are covered that way
#3. Third motivating criteria is that uncertainity must be additive that way if there are any independent outcomes those are covered too
7E2. Suppose a coin is weighted such that, when it is tossed and lands on a table, it comes up heads 70% of the time. What is the entropy of this coin?
p <- c(0.7, 0.3)
entropy <- -sum(p * log(p))
entropy
## [1] 0.6108643
7E3. Suppose a four-sided die is loaded such that, when tossed onto a table, it shows “1” 20%, “2” 25%, “3” 25%, and “4” 30% of the time. What is the entropy of this die?
p <- c(0.2, 0.25, 0.25, 0.3)
entropy <- -sum(p * log(p))
entropy
## [1] 1.376227
7E4. Suppose another four-sided die is loaded such that it never shows “4”. The other three sides show equally often. What is the entropy of this die?
p <- c(1/3, 1/3, 1/3)
entropy <- -sum(p * log(p))
entropy
## [1] 1.098612
7M1. Write down and compare the definitions of AIC and WAIC. Which of these criteria is most general? Which assumptions are required to transform the more general criterion into a less general one?
# AIC is an estimator of out-of-sample prediction error and estimates the quality of each model. AIC is a way to select model quality and evaluate overfitting in the model.
# WAIC is the generalized version of AIC onto singular statistical models. Also, WAIC is the most general criteria
# If we want to transform from WAIC to AIC, what's needed is to assume that the posterior distribution is approximately multivariate Gaussian and the priors are flat or overwhelmed by the likelihood.
7M2. Explain the difference between model selection and model comparison. What information is lost under model selection?
#Model selection is to select the model with the lowest information criterion value and to discard all other models with higher values. Therefore, we would lose information about relative model accuracy. Model averaging is using Bayesian information criteria to construct a posterior predictive distribution and leverages the uncertainty in multiple models. In this case, we would not lose information on its own.
7M3. When comparing models with an information criterion, why must all models be fit to exactly the same observations? What would happen to the information criterion values, if the models were fit to different numbers of observations? Perform some experiments, if you are not sure.
#Information criterion depends on the total deviance which is directly influenced by the number of observations. Therefore, when comparing models with different numbers of observations it will result in different Information Criterion. Model accuracy could also get lost in model comparison. As the conclusion, all models must be fit to exactly the same number of observations
7M4. What happens to the effective number of parameters, as measured by PSIS or WAIC, as a prior becomes more concentrated? Why? Perform some experiments, if you are not sure.
#Prior and effective number of parameters are indirectly proportional. Therefore, as prior becomes more concentrated, the effective number of parameters decreases. This proportionality can be observed from the WAIC mathematical equation:
# WAIC(y, Θ) = -2(lppd - varΘ logp(yi|Θ))
7M5. Provide an informal explanation of why informative priors reduce overfitting.
#Informative priors restrains the range of parameters.So when such a model is applied, since the number of parameters these priors look for are limited there is no chance of an overfitting scenario.
7M6. Provide an informal explanation of why overly informative priors result in underfitting.
#Overly informative priors resulting in narrowing the range of parameters. Therefore, the number of parameters these priors look for are too limited and picky to develop a proper model. | 2021-06-17T17:35:40 | {
"domain": "amazonaws.com",
"url": "https://rstudio-pubs-static.s3.amazonaws.com/780087_a50acc9bce4142ab94d6d94adaa40b53.html",
"openwebmath_score": 0.6174958348274231,
"openwebmath_perplexity": 1401.4163433493773,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9632305349799242,
"lm_q2_score": 0.8791467754256017,
"lm_q1q2_score": 0.8468210188190776
} |
https://mathematica.stackexchange.com/questions/91564/old-fashioned-region-method | # Old fashioned region method?
Just stumbled across this idea in http://users.rowan.edu/~hassen/Mathematica/Volume%20III/Chapter%2015.pdf.
Clear[f];
f[x_, y_] = 1 - x^2 + y^2;
Plot3D[f[x, y], {x, 0, 1}, {y, x, 1 + x^2},
Filling -> Bottom,
FillingStyle -> LightBlue,
PlotRange -> {0, 4},
ViewPoint -> {1, 1, 1}]
Which produces:
My main question is related specifically to
Plot3D[f[x, y], {x, 0, 1}, {y, x, 1 + x^2}],
the {y,x,1+x^2} part.
I am so used to using RegionFunction to perform this task. I went into the documentation for Plot3D to see if I could find an example of this, but I could not find anything like this. Is this an old fashioned way that will soon disappear? Is this something I should share with my students? It's amazing how it matches the associated double integral:
$$\int_0^1\int_x^{x^2}(1-x^2+y^2)\,dy\,dx$$
I also have a second question on this image, if folks don't mind. See how the zeros on the tick marks on the x- and z-axes overlap. Anyone have a simple way of separating them a bit?
• @belisarius I've updated my post because I think I wasn't making it specific enough on what I was focusing on. – David Aug 13 '15 at 19:19
• you question seems to only be about the limit specification? It seems you are correct that this is not mentioned in the Plot3D documentation, but it is pretty much standard convention among similar functions that take multidimensional ranges that reading left to right the limits of each variable can depend on those before it (Table, Integrate, etc ). No need to be concerned it should go away. – george2079 Aug 13 '15 at 20:02
• @george2079 Yes, the worry was about the limit specification. Thanks for your reply. You seem confident that it will continue to exist. – David Aug 13 '15 at 22:15
• Great question! I don't think I have ever used this Plot3D form myself even though I knew it could be used elsewhere as george2079 remarks. – Mr.Wizard Aug 14 '15 at 0:55
• Re ticks: Plot3D[1 - x^2 + y^2, {x, 0, 1}, {y, x, 1 + x^2}, PlotRangePadding -> 0.08]? – Michael E2 Dec 18 '15 at 15:51
I will post this to avoid confusion - region has a new meaning in WL since Geometric Computation was introduced in V10.
Relative to that meaning what you showed is not a WL region because you cannot compute over it, but of course is a visual of some mathematical region defined analytically and shown with help of Filling.
To achieve the same via computable region:
R = ImplicitRegion[0<z<1-x^2+y^2 && x<y<1+x^2 && 0<x<1, {x, y, z}];
RegionPlot3D[R, PlotPoints -> 100, BoxRatios -> {1, 1, .5}]
Volume[R]
29/21
Integrate[1 - x^2 + y^2, {x, 0, 1}, {y, x, 1 + x^2}]
29/21
Integrate[1, {x, y, z} \[Element] R]
N[%]
29/21
1.380952380952381
NIntegrate[1, {x, y, z} \[Element] R]
1.3809523809015896
The above shows that WL computable regions enable symbolic and numeric computations over them.
• Thanks for this very interesting example. Very well explained. – David Aug 13 '15 at 19:29 | 2019-08-20T19:50:31 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/91564/old-fashioned-region-method",
"openwebmath_score": 0.33687901496887207,
"openwebmath_perplexity": 1171.6394230146002,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9632305297023094,
"lm_q2_score": 0.8791467785920306,
"lm_q1q2_score": 0.8468210172292805
} |
https://cs.stackexchange.com/questions/65568/why-is-the-time-complexity-of-insertion-sort-not-brought-down-even-if-we-use-bin | # Why is the time complexity of insertion sort not brought down even if we use binary search for the comparisons?
There are two factors that decide the running time of the insertion sort algorithm:
the number of comparisons, and the number of movements.
In the case of number of comparisons, the sorted part (left side of $$j$$) of the array is searched linearly for the right place of the $$j^{th}$$ element. If instead, we use a binary search, then the time complexity of finding a place for the $$j^{th}$$ element comes down from $$\operatorname{O}(n)$$ to $$\operatorname{O}(\log n)$$. So, for all the $$n$$ elements, the time complexity for comparisons becomes $$\operatorname{O}(n \log n)$$. Even so, the number of movements is still going to take $$\operatorname{O}(n)$$ time, and the total time complexity isn't brought down and remains $$\operatorname{O}(n^2)$$. Why is that?
Are any of my statements wrong assumptions?
Can a possible explanation be: the total time complexity isn't brought down and remains $$\operatorname{O}(n^2)$$. This is because to search an element (using binary search) it takes $$\operatorname{O}(\log n)$$ time, and to move the elements it takes $$\operatorname{O}(n)$$ time. Total cost is $$\operatorname{O}(\log n)+\operatorname{O}(n)=\operatorname{O}(n)$$ time. To do this for $$n-1$$ elements, it takes $$n(n-1)=\operatorname{O}(n^2)$$ time?
For the $j^{th}$ element, you would do ~ $\log j$ comparisons and (in the worst case) ~$j$ shifts.
Summing over $j$, you get
$$\sum_{j = 1}^{n} (j + \log j) = \frac{n(n+1)}{2} + \log (n!) = O(n^2 + n \log n) = O(n^2)$$
The idea is that the linear work of shifting trumps the logarithmic work of comparing. You end up doing less comparisons, but still a linear amount of work per iteration. So the complexity does not change.
• This explains it perfectly. Thank you.. – Somenath Sinha Nov 5 '16 at 2:33
• Also, can I state it as: The total time complexity isn't brought down and remains $\operatorname{O}(n^2)$. This is because to search an element (using binary search) it takes $\operatorname{O}(\log n)$ time, and to move the elements it takes $\operatorname{O}(n)$ time. Total cost is $\operatorname{O}(\log n)+\operatorname{O}(n)=\operatorname{O}(n)$ time. To do this for $n-1$ elements, it takes $n(n-1)=\operatorname{O}(n^2)$ time. – Somenath Sinha Nov 5 '16 at 2:36
• @SomenathSinha yes, but because in this case we know that linear work sums up to be quadratic (the sum over $j$ ends up being quadratic), one would have to be careful for other kinds of functions that sum up differently. But in this case you can say that $O(j) + O(\log j) = O(j)$ and that sums up to $O(n^2)$. – aelguindy Nov 5 '16 at 2:49
The "possible explanation" after the edit in the question is exactly correct. That's why the time complexity is not improved.
On the other hand, unless the array is already mostly sorted, or if the array is very small, using binary search to find where to insert an array element is very likely to make the sorting almost twice as fast.
On the other hand, for large n where sorting an array using insertion sort is unacceptably slow, making it twice as fast still leaves it unacceptably slow.
I don't think the complexity is remain the same. Time complexity of binary insertion sort, if optimized the code
Sample Link: https://cboard.cprogramming.com/cplusplus-programming/139831-analysis-binary-insertion-sort-algorithm.html. My code to work with in Python: implement numpy for closer view at C/C++:
It is True that when you perform shifting element, it is always in linear way. However, even in the worse case, you re-defined the data at the remaining array without making any shifting but rather losing your time. Note that I want to clarify REAL-TIME Complexity; not Time Complexity in general to be more specific
Denote index at the outer is 'i'. Denote index from searching is 'j'
• If performing Linear Search for index && Full-scale Shifting (to the end): O(2N), which resulting O(2N^2) time complexity
• If performing Binary Search for index && Full-scale Shifting (Re-writing the array: O(N + log2(N)), which resulting O(N^2 + N*logN) time complexity
• If performing Binary Search for index && One-sided Shifting (to the end): O(N + log2(N)), which resulting O(N^2/2 + N*logN) time complexity
• If performing Binary Search for Index && Mid-bounded Shifting (to the "i" instead array.length). Time complexity in the worst case can be defined as follow:
• Binary Search: log2(N) : easy to determine
• However, in case of partially-shifting: Time complexity is O(k). Since you only shifting "i-j-1" variables by one move and re-adding the key-point "1". For relatively large array, in average case the number of shifting when k towards the left-half of the sorted one is equal to the right-helf of the sorted one. Thus causing O(i-j) = O(i//2) of time complexity. It can be viewed as using pivot in selection sort or quicksort or timsort.
Thus the overall time complexity for one epoch is log2(N) + i-j ~ log2(N) + i/2 Since this time, it can only be run up to N/2.
=> Time complexity for the whole array, in best case, O(N), near-asymtomtic case / small array is O(NlogN) and in average case, would be O(NlogN + N^2/4). If lucky enough, it could be O(NlogN + N^2/16) For best case and average case, it would be O(N) and O(2Nlog(N))
Take:
import numpy as np
import pandas as pd
from time import time
from typing import List
class Sorting:
def __init__(self, data_size: int, lower_bound: int = 0, higher_bound: int = 1000):
if lower_bound > higher_bound:
lower_bound, higher_bound = higher_bound, lower_bound
if data_size <= 0:
self.__data__: np.ndarray = None
else:
using = max(abs(lower_bound), abs(higher_bound))
if lower_bound == 0 and higher_bound > 0:
x = [2 ** 8 - 1, 2 ** 16 - 1, 2 ** 32 - 1, 2 ** 64 - 1]
dtype = [np.uint8, np.uint16, np.uint32, np.uint64]
else:
x = [2 ** 7 - 1, 2 ** 15 - 1, 2 ** 31 - 1, 2 ** 63 - 1]
dtype = [np.int8, np.int16, np.int32, np.int64]
for i in range(len(x)):
if using <= x[i]:
self.__data__ = np.random.randint(low=lower_bound, high=higher_bound,
dtype=dtype[i], size=data_size)
break
self.data_size: int = data_size
self.lower_bound: int = lower_bound
self.higher_bound: int = higher_bound
self.columns: List[str] = ["Bubble Sort", "Selection Sort", "Insertion Sort (Min-Max)",
"Insertion Sort (Max-Min)", "Binary Insertion Sort"]
self.recording_time: List[float] = [[0] * len(self.columns)]
self.array_size = [data_size]
def reset_data(self, data_size: int, lower_bound: int = 0, higher_bound: int = 1000):
# Avoid Memory Leakage
del self.__data__
if lower_bound > higher_bound:
lower_bound, higher_bound = higher_bound, lower_bound
using = max(abs(lower_bound), abs(higher_bound))
if lower_bound == 0 and higher_bound > 0:
x = [2 ** 8 - 1, 2 ** 16 - 1, 2 ** 32 - 1, 2 ** 64 - 1]
dtype = [np.uint8, np.uint16, np.uint32, np.uint64]
else:
x = [2 ** 7 - 1, 2 ** 15 - 1, 2 ** 31 - 1, 2 ** 63 - 1]
dtype = [np.int8, np.int16, np.int32, np.int64]
for i in range(len(x)):
if using <= x[i]:
self.__data__ = np.random.randint(low=lower_bound, high=higher_bound,
dtype=dtype[i], size=data_size)
break
self.data_size: int = data_size
self.lower_bound: int = lower_bound
self.higher_bound: int = higher_bound
self.array_size.append(data_size)
self.recording_time.append([0] * len(self.columns))
def get_array(self):
return self.__data__
def sort_array(self):
self.__data__.sort()
def reverse_array(self):
self.__data__ = np.flip(self.__data__)
def BubbleSort(self, reverse: bool = False, copy: bool = True):
"""
This method do bubble sort. Time complexity: O(N^2) since there are two loops over loops
Real Time complexity: N*(N-1)/2
:param copy: bool
:param reverse: bool
:return:
"""
start = time()
if copy is True:
copied_version = np.copy(self.__data__)
else:
copied_version = self.__data__
if reverse is False:
for i in range(0, copied_version.size):
swapped = False
for j in range(len(copied_version) - 1, i, -1):
if copied_version[j - 1] > copied_version[j]:
copied_version[j - 1], copied_version[j], swapped = copied_version[j], copied_version[j - 1], True
if swapped is False:
break
else:
for i in range(0, copied_version.size):
swapped = False
for j in range(len(copied_version) - 1, i, -1):
if copied_version[j - 1] < copied_version[j]:
copied_version[j - 1], copied_version[j], swapped = \
copied_version[j], copied_version[j - 1], True
if swapped is False:
break
end = time()
self.recording_time[self.array_size.index(self.data_size)][0] = end - start
print("Bubble Sort: Executing Time: {:.6f}s".format(end - start))
return copied_version, end - start
def SelectionSort(self, reverse: bool = False, copy: bool = True):
"""
This method do selection sort. Time complexity: O(N^2) since the first loop to iterate over the array.
The second loop at "argmin" method which only return the first index that is smallest
:param copy: bool
:param reverse: bool
:return:
"""
start = time()
if copy is True:
copied_version = np.copy(self.__data__)
else:
copied_version = self.__data__
if reverse is False:
for index in range(0, copied_version.size):
if index + 1 == copied_version.size:
break
minimum_index = index + 1 + copied_version[index + 1:].argmin(axis=-1)
copied_version[index], copied_version[minimum_index] = \
copied_version[minimum_index], copied_version[index]
else:
for index in range(0, copied_version.size):
if index + 1 == copied_version.size:
break
minimum_index = index + 1 + copied_version[index + 1:].argmin(axis=-1)
copied_version[index], copied_version[minimum_index] = \
copied_version[minimum_index], copied_version[index]
end = time()
self.recording_time[self.array_size.index(self.data_size)][1] = end - start
print("Selection Sort: Executing Time: {:.6f}s".format(end - start))
return copied_version, end - start
def __BinarySearch__(self, array, number, start, end, reverse: bool = False):
"""
This method will determine the accurate position to help sorting the array.
Embedding with InsertionSort
"""
if reverse is False:
if start == end:
if array[start] > number:
return start
else:
return start + 1
else:
if start == end:
if array[start] < number:
return start
else:
return start + 1
if start > end: # Ensuring position can be found
return start
mid = (start + end) // 2
if reverse is False:
if array[mid] < number:
return self.__BinarySearch__(array=array, number=number, start=mid + 1, end=end, reverse=reverse)
elif array[mid] > number:
return self.__BinarySearch__(array=array, number=number, start=start, end=mid - 1, reverse=reverse)
else:
return mid
else:
if array[mid] > number:
return self.__BinarySearch__(array=array, number=number, start=mid + 1, end=end, reverse=reverse)
elif array[mid] < number:
return self.__BinarySearch__(array=array, number=number, start=start, end=mid - 1, reverse=reverse)
else:
return mid
def InsertionSort(self, reverse: bool = False, copy: bool = True, perform_type: str = "binary"):
"""
This method do insertion sort. Time complexity: O(N^2) since the first loop to iterate over the array.
The second will have to pass the array to have the right order which thus getting O(k) complexity.
Since k is iterate along with N, Real-time complexity is O(N^2)
The second loop at "argmin" method which only return the first index that is smallest got O(N) complexity
You can try binary insertion sort: it can help faster in sorting and lower down a little bit of time
but general time complexity is remained. Time complexity in average case equal to O(N^2)
"""
start = time()
if copy is True:
copied_version = np.copy(self.__data__)
else:
copied_version = self.__data__
if perform_type.lower() == "left-right":
left_to_right, right_to_left, binary = True, False, False
elif perform_type.lower() == "right-left":
left_to_right, right_to_left, binary = False, True, False
elif perform_type.lower() == "binary":
left_to_right, right_to_left, binary = False, False, True
else:
raise ValueError("Unable to found valid number")
if reverse is False and left_to_right is True:
for index in range(1, copied_version.size):
key_point = np.copy(copied_version[index])
if copied_version[index - 1] < key_point:
continue
for i in range(0, index):
if i == 0:
if key_point < copied_version[i]:
copied_version[i + 1:index + 1] = copied_version[i:index]
copied_version[i] = key_point
break
else:
if copied_version[i - 1] <= key_point <= copied_version[i]:
copied_version[i + 1:index + 1] = copied_version[i:index]
copied_version[i] = key_point
break
elif reverse is False and right_to_left is True:
for index in range(1, copied_version.size):
key_point = np.copy(copied_version[index])
if copied_version[index - 1] < key_point:
continue
i = index - 1
while i >= 0 and key_point < copied_version[i]:
copied_version[i + 1] = copied_version[i]
i -= 1
copied_version[i + 1] = key_point
elif reverse is True and left_to_right is True:
for index in range(1, copied_version.size):
key_point = np.copy(copied_version[index])
if copied_version[index - 1] > key_point:
continue
for i in range(0, index):
if i == 0:
if key_point > copied_version[i]:
copied_version[i + 1:index + 1] = copied_version[i:index]
copied_version[i] = key_point
break
else:
if copied_version[i - 1] >= key_point >= copied_version[i]:
copied_version[i + 1:index + 1] = copied_version[i:index]
copied_version[i] = key_point
break
elif reverse is True and right_to_left is True:
for index in range(1, copied_version.size):
key_point = np.copy(copied_version[index])
if copied_version[index - 1] > key_point:
continue
i = index - 1
while i >= 0 and key_point > copied_version[i]:
copied_version[i + 1] = copied_version[i]
i -= 1
copied_version[i + 1] = key_point
else:
for index in range(1, copied_version.size):
key_point = np.copy(copied_version[index])
if reverse is True and copied_version[index - 1] > key_point:
continue
elif reverse is False and copied_version[index - 1] < key_point:
continue
i = self.__BinarySearch__(array=copied_version, number=key_point, start=0, end=index - 1,
reverse=reverse)
copied_version[i + 1:index + 1] = copied_version[i:index]
copied_version[i] = key_point
end = time()
if left_to_right is True:
self.recording_time[self.array_size.index(self.data_size)][2] = end - start
if right_to_left is True:
self.recording_time[self.array_size.index(self.data_size)][3] = end - start
if binary is True:
self.recording_time[self.array_size.index(self.data_size)][4] = end - start
print("Insertion Sort: Executing Time: {:.6f}s".format(end - start))
return copied_version, end - start
def get_time(self, output: str = None):
file = pd.DataFrame(data=self.recording_time, index=self.array_size, columns=self.columns)
if output is not None:
file.to_csv(path_or_buf=output)
return file
def automate(self, size: List[int], lower_bound: int = 0, higher_bound: int = 1000, reverse: bool = False,
insert_type: str = 'binary', bubble_sort: bool = True, selection_sort: bool = True,
insertion_sort: bool = True, output: str = None):
for index, value in enumerate(size):
print("Matrix Size:", value)
self.reset_data(data_size=value, lower_bound=lower_bound, higher_bound=higher_bound)
# print(self.get_array())
if bubble_sort is True:
self.BubbleSort(reverse=reverse)
if selection_sort is True:
self.SelectionSort(reverse=reverse)
if insertion_sort is True:
self.InsertionSort(reverse=reverse, perform_type=insert_type)
self.get_time(output=output)
My final word is optimize your code to lower down your REAL-TIME complexity rather make full attention on time complexity to think of.
[EDIT]: Real-Time complexity I will try to clarify this according to time-complexity in descending order:
• Linear Search + Array Rewriting:
Worst Case = Average Case: O(N) * O(N + N) = O(2N^2) (contained some "if" to get best case O(N))
• Linear Search + Element-Swapping:
Average Case: O(N) * O(N + N) = O(2N^2) (contained some "if" to get best case O(N))
Worst Case: could be O(N^3+N^2)
• Linear Search + Half-Right Swapping:
Worst Case: O(N) * O(N + N/2) = O(3/2 * N^2) (inversely-linear order, obtained by averaging left to right)
• Linear Search + Pivot ("j"-index) <--> Key ("i"-index) Swapping:
Worst Case: O(N) * O(N + N/4) = O(5/4 * N^2) (inversely-linear order:, obtained by averaging left to right)
Average Case: O(N) * O(N + N/16) = O(16/15 * N^2) (inversely-linear order, obtained by averaging left to right)
• Binary Search + Array Rewriting:
Worst Case = Average Case: O(N) * O(log2(N) + N) = O(N^2 + Nlog2(N)) (contained some "if" to get best case O(N))
• Binary Search + Element Swapping:
Average Case: O(N) * O(log2(N) + N) = O(N^2 + Nlog2(N)) (contained some "if" to get best case O(N))
Worst Case: O(N^3 + Nlog2(N))
• Binary Search + Half-Right Swapping:
Worst Case = Average Case: O(N) * O(log2N + N/2) = O(1/2 * N^2 + Nlog2(N)) (inversely-linear order, obtained by averaging left to right)
• Binary Search + Pivot ("j"-index) <--> Key ("i"-index) Swapping:
Worst Case: O(N) * O(log2(N) + N/4) = O(1/4 * N^2 + Nlog2(N)) (inversely-linear order:, obtained by averaging left to right)
Average Case: O(N) * O(log2(N) + N/16) = O(1/16 * N^2 + Nlog2(N)) (inversely-linear order:, obtained by averaging left to right)
Lucky Case: O(2Nlog2(N))
• (right-helf is but half-right.) – greybeard Mar 20 at 4:32
• Yes. it should be half-right – Take Ichiru Mar 21 at 8:38 | 2021-04-19T14:48:48 | {
"domain": "stackexchange.com",
"url": "https://cs.stackexchange.com/questions/65568/why-is-the-time-complexity-of-insertion-sort-not-brought-down-even-if-we-use-bin",
"openwebmath_score": 0.29447346925735474,
"openwebmath_perplexity": 10169.68110755711,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9632305307578323,
"lm_q2_score": 0.8791467706759583,
"lm_q1q2_score": 0.8468210105322376
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.