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https://de.mathworks.com/help/matlab/ref/meshgrid.html | # meshgrid
2-D and 3-D grids
## Syntax
``````[X,Y] = meshgrid(x,y)``````
``````[X,Y] = meshgrid(x)``````
``````[X,Y,Z] = meshgrid(x,y,z)``````
``````[X,Y,Z] = meshgrid(x)``````
## Description
example
``````[X,Y] = meshgrid(x,y)``` returns 2-D grid coordinates based on the coordinates contained in vectors `x` and `y`. `X` is a matrix where each row is a copy of `x`, and `Y` is a matrix where each column is a copy of `y`. The grid represented by the coordinates `X` and `Y` has `length(y)` rows and `length(x)` columns.```
example
``````[X,Y] = meshgrid(x)``` is the same as ```[X,Y] = meshgrid(x,x)```, returning square grid coordinates with grid size `length(x)`-by-`length(x)`.```
example
``````[X,Y,Z] = meshgrid(x,y,z)``` returns 3-D grid coordinates defined by the vectors `x`, `y`, and `z`. The grid represented by `X`, `Y`, and `Z` has size `length(y)`-by-`length(x)`-by-`length(z)`.```
example
``````[X,Y,Z] = meshgrid(x)``` is the same as ```[X,Y,Z] = meshgrid(x,x,x)```, returning 3-D grid coordinates with grid size `length(x)`-by-`length(x)`-by-`length(x)`.```
## Examples
collapse all
Create 2-D grid coordinates with x-coordinates defined by the vector `x` and y-coordinates defined by the vector `y`.
```x = 1:3; y = 1:5; [X,Y] = meshgrid(x,y)```
```X = 5×3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 ```
```Y = 5×3 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 ```
Evaluate the expression ${x}^{2}+{y}^{2}$ over the 2-D grid.
`X.^2 + Y.^2`
```ans = 5×3 2 5 10 5 8 13 10 13 18 17 20 25 26 29 34 ```
Create a 2-D grid with uniformly spaced x-coordinates and y-coordinates in the interval [-2,2].
```x = -2:0.25:2; y = x; [X,Y] = meshgrid(x);```
Evaluate and plot the function $f\left(x,y\right)=x{e}^{-{x}^{2}-{y}^{2}}$ over the 2-D grid.
```F = X.*exp(-X.^2-Y.^2); surf(X,Y,F)```
Starting in R2016b, it is not always necessary to create the grid before operating over it. For example, computing the expression $x{e}^{-{x}^{2}-{y}^{2}}$ implicitly expands the vectors `x` and `y`. For more information on implicit expansion, see Array vs. Matrix Operations.
`surf(x,y,x.*exp(-x.^2-(y').^2))`
Create 3-D grid coordinates from x-, y-, and z-coordinates defined in the interval [0,6], and evaluate the expression ${x}^{2}+{y}^{2}+{z}^{2}$.
```x = 0:2:6; y = 0:1:6; z = 0:3:6; [X,Y,Z] = meshgrid(x,y,z); F = X.^2 + Y.^2 + Z.^2;```
Determine the size of the grid. The three coordinate vectors have different lengths, forming a rectangular box of grid points.
`gridsize = size(F)`
```gridsize = 1×3 7 4 3 ```
Use the single-input syntax to generate a uniformly spaced 3-D grid based on the coordinates defined in `x`. The new grid forms a cube of grid points.
```[X,Y,Z] = meshgrid(x); G = X.^2 + Y.^2 + Z.^2; gridsize = size(G)```
```gridsize = 1×3 4 4 4 ```
## Input Arguments
collapse all
x-coordinates of points, specified as a vector.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
y-coordinates of points, specified as a vector.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
z-coordinates of points, specified as a vector.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
## Output Arguments
collapse all
x-coordinates over a grid, returned as a 2-D (two inputs) or 3-D array (three inputs).
y-coordinates over a grid, returned as a 2-D (two inputs) or 3-D array (three inputs).
z-coordinates over a grid, returned as a 3-D array. | 2020-02-23T08:45:44 | {
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https://math.stackexchange.com/questions/1177506/how-do-i-solve-the-recurrence-relation-without-manually-counting/1177522 | # How do I solve the recurrence relation without manually counting?
Given the recurrence relation : $a_{n+1} - a_n = 2n + 3$ , how would I solve this?
I have attempted this question, but I did not get the answer given in the answer key.
First I found the general homogenous solution which is $C(r)^n$ where the root is 1 so we get $C(1)^n$. Then I found the particular non homogenous solution which was $A_1(n) + A_0$.
I then plugged the particular solution into the given recurrence relation and solved for $A_0$ and $A_1$. I got $A_0 = -1$ and $A_1 = 5$. After further steps I got
$a_n = 5n-1 + 2(1)^n$
That answer is however completely off, in the answer key they have
$a_n = (n+1)^2$. Can someone explain to me how to arrive at this answer?
EDIT: Initial condition is that $n\ge 0$ and $a_0 = 1$.
• What is the initial condition? – Math1000 Mar 6 '15 at 0:21
• Sorry, let me edit that in! – Belphegor Mar 6 '15 at 0:23
• Is the approach I used correct in anyway? – Belphegor Mar 6 '15 at 0:32
Let $$f(x) = \sum_{n=0}^\infty a_n x^n.$$ Multiplying both sides of the given recurrence equation by $x^n$ and summing over $n\geqslant0$, the left-hand side becomes $$\sum_{n=0}^\infty a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n = \frac1x\sum_{n=0}^\infty a_{n+1}x^{n+1} - f(x) = \frac1x\left(f(x)-1\right)+f(x).$$ The right-hand side becomes $$\sum_{n=0}^\infty (2n+3)x^n = 2\sum_{n=0}^\infty(n+1)x^n + \sum_{n=0}^\infty x^n=\frac2{(1-x)^2}+\frac1{1-x}.$$ Equating the two we have $$\frac1x\left(f(x)-1\right)+f(x) = \frac2{(1-x)^2}+\frac1{1-x},$$ and solving for $f(x)$, $$f(x) = \frac{1+x}{(1-x)^3}.$$ Now $$\frac1{(1-x)^3}=\sum_{n=0}^\infty\frac12(n+1)(n+2)x^n = 1 + \sum_{n=1}^\infty\frac12(n+1)(n+2)x^n,$$ and $$\frac x{(1-x)^3}=\sum_{n=0}^\infty\frac12(n+1)(n+2)x^{n+1} = \sum_{n=1}^\infty \frac12n(n+1)x^n.$$ Hence \begin{align*} \frac{1+x}{(1-x)^3} &= 1 + \sum_{n=1}^\infty\frac12(n+1)(n+2)x^n + \sum_{n=1}^\infty\frac12n(n+1)x^n\\ &= 1 + \sum_{n=1}^\infty\frac12(n+1)(n+n+2)x^n\\ &= 1 + \sum_{n=1}^\infty(n+1)^2x^n\\ &= \sum_{n=0}(n+1)^2x^n. \end{align*} It follows that $a_n=(n+1)^2$ for $n\geqslant0$.
• How did you go from 1+x/(1-x)^3 to the next step? – Belphegor Mar 6 '15 at 0:49
• $\frac{1+x}{(1-x)^3}=\frac1{(1-x)^3} + \frac x{(1-x)^3}$ and the series expansion for $\frac1{(1-x)^3}$. – Math1000 Mar 6 '15 at 0:51
• Your technique makes sense, but I don't understand why i didn't see your way of doing it in my textbook o.o..What is this technique called? – Belphegor Mar 6 '15 at 0:58
• Generating functions. They're really nifty and you can do all sorts of things with them. Here's a great introduction: math.upenn.edu/~wilf/DownldGF.html – Math1000 Mar 6 '15 at 1:08
There is a theorem which is somewhat intuitive that says that if you take $\sum_{j=0}^n P(j)$ where $P$ is a polynomial of degree $d$, then the answer will be a polynomial in $n$ of degree $d + 1$. If you know this, then you can just plug in the values for $n = 0,1,2$ and then solve the arising linear system of equations to find the coefficients of your quadratic polynomial solution.
For a more direct solution for your case, note that adding up $3$ exactly $N$ times will result in a term of $3N$. So it only remains to see what happens when you add up $2n$ for $n=1 \ldots, N$. To see what this equals, note that you can double the sequence $1,2, \ldots N$ and write the duplicate copy in reverse order $N,N-1,\ldots,1$. lining up with the original sequence. Then you have twice the sum if you add all the terms , but every pair of lined up terms adds up to $N+1$ and you have $N$ pairs of terms. Thus the sum is $N(N+1)/2$. This is all you need to show that your solution must be a quadratic polynomial.
• So my original approach cannot be used? – Belphegor Mar 6 '15 at 0:28
• Something went wrong when you attempted to solve for the non-homogeneous part of the solution. You should have gotten $an^2 + bn + c$ for the general form, which is also what is stipulated in the theorem I quoted. – user2566092 Mar 6 '15 at 0:32
• Ah yes, I had a feeling that it was there, but the reason why I wrote what you see now is because according to my chart, if the right side has 'n' on it then the particular solution of the non homogenous relation is A_1(n) + A_0 – Belphegor Mar 6 '15 at 0:36
• Actually, I might have interpreted the table wrong, but how would I solve for the particular solution if my logic was wrong? – Belphegor Mar 6 '15 at 0:39
• One way to see that it must be a quadratic polynomial is to note that a sum of constant terms is a linear polynomial, and if you take a sum of pure linear terms $n$ then you can arrange them in a triangle that makes up half a rectangle by cutting the rectangle down the diagonal, and thus see that a sum of linear terms is a quadratic polynomial. – user2566092 Mar 6 '15 at 0:48 | 2019-06-25T11:30:59 | {
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http://math.stackexchange.com/questions/6568/relation-between-set-operations | # relation between set operations
I was wondering about the relation between complement of a subset, different between two subsets, union and intersection of subsets. Can we reduce the set of the above operations in a minimal way so that other operations can be represented by these "independent" operations? And how many ways to do this?
For example, the intersection (or union) can be represented by complement and union (or intersection) in the same way as in De Morgan law.
But can we represent complement or difference in terms of union and intersection?
Thanks!
-
Well, set difference is defined in terms of intersection and complement: $A\setminus B = A\cap B^C$. (Personally, I prefer the notation $A-B$ to $A\setminus B$, but the latter is relatively standard in my experience.)
As you note, De Morgan's laws allow us to translate between union and intersection via complements. Beyond this, we must have at least union and complement (or, equivalently, intersection and complement). That is, we cannot define complement solely in terms union and intersection.
-
Can difference be defined in terms of union and intersection? – Tim Oct 11 '10 at 21:54
Because set difference is defined in terms of intersection and complement, and complement cannot be defined in terms of only union/intersection, the answer is no. – Bey Oct 11 '10 at 22:03
Although we could do something like write A^C as the union (taken over all of X) of all elements x not in A, where X is our universal set. But I assumed you were thinking solely in terms of operations on the sets themselves, rather than appealing to elemental definitions. – Bey Oct 11 '10 at 22:05
No, because you need a negative for difference and union and intersection are positive operators. See the Wikipedia article Functional completeness, where and and or are truth-preserving. And corresponds to intersection and or to union. – Ross Millikan Oct 11 '10 at 22:07
There is a "minimal" definition of set operations, though it's not really a restriction of the classical ones, and is more a fun trick than a clever insight. It was discovered by Charles Sanders Peirce, for general Boolean algebras.
Let $A\uparrow B=(A\cap B)^c$. Then $A\uparrow A=A^c$. So $(A\uparrow A)\uparrow(B\uparrow B)=(A^c\cap B^c)^c=A\cup B$. Similarly, $(A\uparrow B)\uparrow(A\uparrow B)=((A\cap B)^c)^c=A\cap B$. And then you can get difference or whatever else, as expected. You can do the same thing with $(A\cup B)^c$.
-
You could mention that $(A \cup B)^C$ works as well. – Seamus Oct 12 '10 at 15:12
I did (my last sentence). – Paul VanKoughnett Oct 12 '10 at 16:52
How would you define the difference between a fun trick and a clever insight? – Leo Jun 19 '11 at 22:41
We could also write set difference in terms of intersection and complement as : $A \cap B = A - (A - B)$. (I do also prefer $A - B$ to $A \setminus B$)
This expression comes handy when you are dealing with cardinality.
- | 2015-02-02T02:20:42 | {
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https://math.stackexchange.com/questions/1955582/maximum-and-minimum-of-sum-i-1n-a-ix-i | # maximum and minimum of $\sum_{i=1}^n a_ix_i$
Suppose that $a_i \in \mathbb R$, and $\sqrt{ \sum_{i=1}^n x_i^2} = 1$, what is the maximum value and minimum value of $\sum_{i=1}^n a_ix_i$
By the Cauchy-Schwarz inequality, one has $$|\sum_{i=1}^na_ix_i|\le \sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum_{i=1}^nx_i^2}=\sqrt{\sum_{i=1}^na_i^2}$$ and "=" holds iff $$x_i=\pm\frac{a_i}{\sqrt{\sum_{i=1}^na_i^2}}.$$ Hence $\sum_{i=1}^na_ix_i$ reaches the max $\sqrt{\sum_{i=1}^na_i^2}$ when $x_i=\frac{a_i}{\sqrt{\sum_{i=1}^na_i^2}}$, $i=1,2,\cdots,n$ and the min $-\sqrt{\sum_{i=1}^na_i^2}$ when $x_i=-\frac{a_i}{\sqrt{\sum_{i=1}^na_i^2}}$, $i=1,2,\cdots,n$.
• I am also curious that when will $\sum_{i=1}^n a_ix_i = 0$? – koch Oct 6 '16 at 3:39
Think of this as a constrained optimization problem; you can use Lagrange Multipliers!
Note that all of what follows only works when $\vec{a}$ is not the zero vector; but, when all the $a_i$ are $0$, this problem is pretty boring.
We want to find minimum and maximum of $$f(\vec{x})=\sum_{i=1}^{n}a_ix_i$$ subject to $$g(\vec{x}):=\sum_{i=1}^{n}x_i^2=1.$$ Gradients are easy to compute here: $$\nabla f=\vec{a},\qquad \nabla g=2\vec{x}$$ So, the Lagrange condition is $$\vec{a}=2\lambda\vec{x}.$$ Note that $\lambda\neq0$, as $\vec{a}\neq \vec{0}$. So, we have $$\vec{x}=\frac{\vec{a}}{2\lambda}.$$
Now, we know that $\lvert\vec{x}\rvert=1$, so that $\lvert\lambda\rvert=\frac{\lvert\vec{a}\rvert}{2}$; this leaves us with only two options for $\lambda$.
When $\lambda=\frac{1}{2}\lvert\vec{a}\rvert$, we get $$\vec{x}=\frac{\vec{a}}{\lvert\vec{a}\rvert},$$ and therefore $$f(\vec{x})=\vec{a}\cdot\vec{x}=\frac{\vec{a}\cdot\vec{a}}{\lvert\vec{a}\rvert}=\lvert\vec{a}\rvert.$$ On the other hand, if we take $\lambda=-\frac{1}{2}\lvert\vec{a}\rvert$, we end up with $$f(\vec{x})=\vec{a}\cdot\vec{x}=-\lvert\vec{a}\rvert.$$
These will be your max and min! | 2019-05-21T01:13:56 | {
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https://math.stackexchange.com/questions/2383695/why-should-we-have-sin2x-frac1-cos2x2-knowing-that-sin2x/2383697 | # Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I look up trig identities, however, it says $\sin^2(x) = \frac{1-\cos(2x)}{2}$.
Why is this?
• Welcome to StackExchange. Please use LaTeX. – user394255 Aug 5 '17 at 17:34
• @A.Molendijk or you know MathJax – user451844 Aug 5 '17 at 17:35
• The second formula is the double angle formula. – Donald Splutterwit Aug 5 '17 at 17:35
• Both formulas are true, they don't contradict themselves. – Reiner Martin Aug 5 '17 at 17:39
• Your proposed identity $\sin^2 x = 1-\cos^2 x$ is correct. And so is the one you "looked up", tha $\sin^2 x = \dfrac{1-\cos(2x)} 2. \qquad$ – Michael Hardy Aug 5 '17 at 18:20
Notice that $\cos^{2}(x):=(\cos(x))^{2}$ is not the same thing as $\cos(2x)$. It is indeed true that $\sin^{2}(x)=1-\cos^{2}(x)$ and that $\sin^{2}(x)=\frac{1-\cos(2x)}{2}$.
Both formulas are true, however, both are useful in different contexts (applications).
• You use $\sin^2(x) = \frac{1-\cos(2x)}{2}$ for integrating $\sin^2(x)$.
• You use $\sin^2(x) = 1 - \cos^2(x)$, for example, when solving $\sin^2(x) = 2\cos(x)$.
Note that it is just in some way more "natural" to write $\sin^2(x) + \cos^2(x)=1$, because this gives both $\sin^2(x) = 1 - \cos^2(x)$ and $\cos^2(x) = 1 - \sin^2(x)$ in one "natural looking" formula.
From the angle addition formula, we have
\begin{align} \cos(2x)&=\cos(x+x)\\\\ &=\cos(x)\cos(x)-\sin(x)\sin(x)\\\\ &=\cos^2(x)-\sin^2(x)\\\\ &=\left(1-\sin^2(x)\right)-\sin^2(x)\\\\ &=1-2\sin^2(x)\tag 1 \end{align}
Solving $(1)$ for $\sin^2(x)$ yields
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
as expected.
Let's address two things, the question in the post and the confusion others have pointed out.
## Part 1: Proof
Here's a proof that $\cos^2(\theta)+\sin^2(\theta)=1$, from which you can show that $\sin^2(\theta)=1-\cos^2(\theta)$. It requires a bit of set up using the Euler identity and that $i=\sqrt{-1}$. If you are not familiar with complex numbers, find yourself a video on YouTube, you'll get the hang of them in next to no time.
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag{1}$$ $$e^{-i\theta}=\cos(\theta)-i\sin(\theta)\tag{2}$$ Now, add (1) and (2). $$e^{i\theta}+e^{-i\theta}=\cos(\theta)+i\sin(\theta)+\cos(\theta)-i\sin(\theta)=2\cos(\theta)\tag{3}$$
Subtract (2) from (1) $$e^{i\theta}-e^{-i\theta}=\cos(\theta)+i\sin(\theta)-\cos(\theta)+i\sin(\theta)=2i\sin(\theta)\tag{4}$$ From (3) and (4) we get (5) and (6) $$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}\tag{5}$$ $$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}\tag{6}$$
now, add the squares of (5) and (6) $$\cos^2(\theta)=\frac{e^{2i\theta}+2e^{i\theta-i\theta}+e^{-2i\theta}}{4}=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}\tag{7}$$ $$\sin^2(\theta)=\frac{e^{2i\theta}-2e^{i\theta-i\theta}+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}\tag{8}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}+\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}\tag{9}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}=\frac{4}{4}=1\tag{10}$$
## Part 2: Addressing the confusion of $\cos(2x)$ and $\cos^2(x)$
$$\cos(2\theta)=\frac{e^{2i\theta}+e^{-2i\theta}}{2}= \frac{\left(e^{i\theta}\right)^2+\left(e^{-i\theta}\right)^2}{2} = \frac{(\cos(\theta)+i\sin(\theta))^2+(\cos(\theta)-i\sin(\theta))^2}{2}\tag{11}$$ Processing this result further, we obtain
$$\cos(2\theta)=\frac{\cos^2(\theta)+2i\cos(\theta)\sin(\theta) - \sin^2(\theta) + \cos^2(\theta) -2i\cos(\theta)\sin(\theta)-\sin^2(\theta) }{2}=\frac{2\cos^2(\theta)-2\sin^2(\theta)}{2}$$ Therefore $$\cos(2\theta)=\cos^2(\theta)-\sin^2(2\theta)\tag{12}$$.
Ok, you stated that $$\sin^2(x)=\frac{1-\cos(2x)}{2}\tag{13}$$ well, if you substitute the result from (12) into (13), you find that
$$\sin^2(\theta)=\frac{1-\cos^2(\theta)+\sin^2(2\theta)}{2}\tag{14}$$ Rearrange this (multiply both sides by 2 and then subtract the $sin^2(\theta)$ from the right side to the left) to get $$\sin^2(\theta)=1-\cos^2(\theta)\tag{15}$$ | 2019-08-20T16:56:38 | {
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https://fr.mathworks.com/help/symbolic/inv.html | # inv
Inverse of symbolic matrix
## Syntax
``D = inv(A)``
## Description
example
````D = inv(A)` returns the inverse of a symbolic matrix `A`.```
## Examples
collapse all
Compute the inverse of a matrix of symbolic numbers.
```A = sym([2 -1 0; -1 2 -1; 0 -1 2]); D = inv(A)```
```D = $\left(\begin{array}{ccc}\frac{3}{4}& \frac{1}{2}& \frac{1}{4}\\ \frac{1}{2}& 1& \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2}& \frac{3}{4}\end{array}\right)$```
Compute the inverse of a matrix of symbolic scalar variables.
```syms a b c d A = [a b; c d]; D = inv(A)```
```D = $\left(\begin{array}{cc}\frac{d}{a d-b c}& -\frac{b}{a d-b c}\\ -\frac{c}{a d-b c}& \frac{a}{a d-b c}\end{array}\right)$```
Compute the inverse of the Hilbert matrix that contains symbolic numbers.
`D = inv(sym(hilb(4)))`
```D = $\left(\begin{array}{cccc}16& -120& 240& -140\\ -120& 1200& -2700& 1680\\ 240& -2700& 6480& -4200\\ -140& 1680& -4200& 2800\end{array}\right)$```
Find the inverse of a 4-by-4 block matrix
`$\mathbit{C}=\left[\begin{array}{cc}\mathbit{A}& {0}_{2,2}\\ {0}_{2,2}& \mathbit{B}\end{array}\right]$`
where $\mathbit{A}$ and $\mathbit{B}$ are 2-by-2 submatrices. The notation ${0}_{2,2}$ represents a 2-by-2 submatrix of zeros.
Use symbolic matrix variables to represent the submatrices in the block matrix.
```syms A B [2 2] matrix Z = symmatrix(zeros(2))```
`Z = ${\mathrm{0}}_{2,2}$`
`C = [A Z; Z B]`
```C = $\left(\begin{array}{c}\begin{array}{cc}A& {\mathrm{0}}_{2,2}\end{array}\\ \begin{array}{cc}{\mathrm{0}}_{2,2}& B\end{array}\end{array}\right)$```
Find the inverse of the matrix $\mathbit{C}$.
`D = inv(C)`
```D = ${\left(\begin{array}{c}\begin{array}{cc}A& {\mathrm{0}}_{2,2}\end{array}\\ \begin{array}{cc}{\mathrm{0}}_{2,2}& B\end{array}\end{array}\right)}^{-1}$```
To show the elements of the inverse matrix, convert the result from a symbolic matrix variable to symbolic scalar variables using `symmatrix2sym`.
`D1 = symmatrix2sym(D)`
```D1 = ```
Compute the inverse of the matrix polynomial ${\mathit{a}}_{0}\text{\hspace{0.17em}}{\mathbit{I}}_{2}+\mathbit{A}$, where $\mathbit{A}$ is a 2-by-2 matrix.
Create the matrix $\mathbit{A}$ as a symbolic matrix variable and the coefficient ${\mathit{a}}_{0}$ as a symbolic scalar variable. Create the matrix polynomial as a symbolic matrix function `f` with ${\mathit{a}}_{0}$ and $\mathbit{A}$ as its parameters.
```syms A [2 2] matrix syms a0 syms f(a0,A) [2 2] matrix keepargs f(a0,A) = a0*eye(2) + A```
`f(a0, A) = ${a}_{0} {\mathrm{I}}_{2}+A$`
Find the inverse of `f` using `inv`. The result is a symbolic matrix function of type `symfunmatrix `that accepts scalars, vectors, and matrices as its input arguments.
`fInv = inv(f)`
`fInv(a0, A) = ${\left({a}_{0} {\mathrm{I}}_{2}+A\right)}^{-1}$`
Convert the result from the `symfunmatrix` data type to the `symfun` data type using `symfunmatrix2symfun`. The result is a symbolic function that accepts scalars as its input arguments.
`gInv = symfunmatrix2symfun(fInv)`
```gInv(a0, A1_1, A1_2, A2_1, A2_2) = ```
## Input Arguments
collapse all
Input matrix, specified as a square numeric matrix, square matrix of symbolic scalar variables, square symbolic matrix variable, square symbolic matrix function, or symbolic expression with square size.
Data Types: `single` | `double` | `sym` | `symmatrix` | `symfunmatrix`
## Tips
• Matrix computations involving many symbolic variables can be slow. To increase the computational speed, reduce the number of symbolic variables by substituting the given values for some variables.
## Version History
Introduced before R2006a
expand all | 2023-02-09T12:04:42 | {
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https://math.stackexchange.com/questions/966349/how-many-ways-to-arrange-8-read-beads-and-32-blue-beads-into-a-necklace-such | # How many ways to arrange $8$ read beads and $32$ blue beads into a necklace such that at least $2$ blue beads between any $2$ red beads ?
Zaraki wants to use $8$ indistinguishable red beads and $32$ indistinguishable blue beads to make a necklace such that there are at least $2$ blue beads between any $2$ red beads. In how many ways can he do this ?
Part of me, my solution is $30667$ and my result comes from this calculation $$\frac{\binom{23}{7}- \binom{11}{3}}{8}+ \frac{\binom{11}{3}-\binom{5}{1}}{4}+\frac{\binom 51-1}{2}+1=30667.$$ Is my solution true ?
• Are rotated/mirrored necklaces distinguishable? – Hagen von Eitzen Oct 10 '14 at 6:49
• Sorry, I don't get your point. – Tengu Oct 10 '14 at 6:56
We can regard the necklace as consisting of $24$ items, RBB eight times and B sixteen times. If rotating a necklace counts as the same arrangement but reflection does not then we can use orbit counting on the group of $24$ rotations, catalogued below. $$\matrix{ \hbox{order}&24&12&8&6&4&3&2&1\cr \hbox{number}&8&4&4&2&2&2&1&1\cr \hbox{fixed points}&0&0&C(3,1)&0&C(6,2)&0&C(12,4)&C(24,8)\cr}$$ The number of arrangements is the average number of fixed points $$\frac{4C(3,1)+2C(6,2)+C(12,4)+C(24,8)}{24}=30667\ .$$
• It may also be instructive to provide an enumeration in the case where necklaces are distinct up to reflection as well as rotation, since, from the odd-valued answer above, it clearly cannot be as simple as dividing by 2. – heropup Oct 10 '14 at 23:27
• This value is computed in the answer below. – Marko Riedel Oct 10 '14 at 23:29
It may be of interest to relate this problem to cycle indices and the Polya Enumeration Theorem (PET). Suppose we distribute the eight red beads on the necklace first. Next we distribute two blue beads into every space between two red beads, leaving $32-2\times 8 = 16$ blue beads. The problem now becomes equivalent to distributing the remaining $16$ blue beads into the eight slots under rotation or rotation and reflection.
To compute the count under rotations we need the cycle index $Z(C_8)$ of the cyclic group on eight elements. We now enumerate the permutations in this cycle index. There is the identity, which contributes $a_1^8$. A rotation by a distance of four maps opposite slots to each other and creates two-cycles, giving $a_2^4.$ A rotation by a distance of two or six creates four-cycles, giving $2\times a_4^2.$ A rotation by a distance of $1,3,5$ or $7$ creates eight-cycles giving $4\times a_8.$
This gives the cycle index $$Z(C_8) = \frac{1}{8} (a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8).$$
The desired value is given by $$[z^{16}] Z(C_8)\left(\frac{1}{1-z}\right).$$ This is $$\frac{1}{8} \left([z^{16}] \left(\frac{1}{1-z}\right)^8 + [z^{16}] \left(\frac{1}{1-z^2}\right)^4 + 2 [z^{16}] \left(\frac{1}{1-z^4}\right)^2 + 4 [z^{16}] \left(\frac{1}{1-z^8}\right)\right).$$ which yields $$\frac{1}{8} \left({16+7\choose 7} + {8+3\choose 3} + 2{4+1\choose 1} + 4{2+0\choose 0}\right) = 30667.$$
When reflections are included we have dihedral symmetry and we need the cycle index $Z(D_8)$ of the dihedral group on eight elements. The additional permutations are four reflections about an axis passing through the midpoint of opposite edges connecting two slots, giving $4\times a_2^4$ and four reflections about an axis passing through two opposite slots giving $4\times a_1^2 a_2^3.$ Hence the cycle index is
$$Z(D_8) = \frac{1}{2} Z(C_8) + \frac{1}{16} (4 a_2^4 + 4 a_1^2 a_2^3).$$
This yields for sixteen blue beads the count $$\frac{1}{2} 30667 + \frac{1}{16} \left(4 [z^{16}] \left(\frac{1}{1-z^2}\right)^4 + 4 [z^{16}] \left(\frac{1}{1-z}\right)^2 \left(\frac{1}{1-z^2}\right)^3 \right)$$ which is $$\frac{1}{2} 30667 + \frac{1}{16} \left(4 {8+3\choose 3} + 4 \sum_{q=0}^8 {q+2\choose 2} {16-2q+1\choose´1} \right) = 15581.$$
The sequences OEIS A032193 and OEIS A005514 are relevant here.
Remark. We can also apply Burnside directly to the cycle indices and bypass PET. E.g. for a permutation of cycle type $a_2^4$ to fix an assignment to the four cycles it is necessary that we choose a number $q$ of blue beads for each the four two-cycles and then place twice that number $q$ of beads on the four two-cycles with $q$ beads in each of the two slots on a two cycle.
Choosing a pair has generating function $$\frac{1}{1-z^2}$$ and the total contribution is
$$[z^{16}] \frac{1}{1-z^2} \frac{1}{1-z^2} \frac{1}{1-z^2} \frac{1}{1-z^2}.$$
Similarly an assignment for a permutation of cycle type $a_4^2$ that is fixed by this shape of permutation means that we have to choose a number $q$ of blue beads for each of the two cycles and then place four times that number on each cycle with $q$ beads in each of the slots on the four-cycle. This gives
$$[z^{16}] \frac{1}{1-z^4} \frac{1}{1-z^4}.$$
This can be continued and it is essentially the mechanism by which PET is proved starting from Burnside.
There are many more related links at MSE Meta on Burnside/Polya. | 2021-06-14T06:43:48 | {
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https://mathhelpboards.com/threads/critical-points-and-roots.27385/ | # Critical points and Roots?
#### gevni
##### New member
Might be I am asking a silly question but really want to clarify that would critical points and roots are same terms use interchangeably? I mean we can use critical point as value of x and root also as value of x then what is the difference between?
#### MarkFL
Staff member
It has been my experience that a critical value is an input to a function such that the function's first derivative is either zero or undefined. A function's root(s) is/are the input(s) which cause the function to return zero, also known as the zeroes of a function.
#### skeeter
##### Well-known member
MHB Math Helper
In solving rational inequalities, critical values are where the expression under scrutiny equals zero or is undefined. Those values mark the boundaries for the solution set.
for example ...
$\dfrac{x^2-1}{x+2} \ge 0$
critical values are $x \in \{-2,-1,1\}$
the three critical values partition the set of x-values into four regions ...
$-\infty < x < -2$,
$-2 < x < -1$,
$-1 < x < 1$,
and $x > 1$
The "equals to" part of the original inequality occurs at $x = \pm 1$
The "greater than" occurs over the intervals $-2 < x < -1$ and $x > 1$
So, the solution set is all $x$ such that $-2 < x \le -1$ or $x \ge 1$
#### Country Boy
##### Well-known member
MHB Math Helper
One additional point- critical roots are numbers while critical points are, of course, points. If I were asked to find the critical root of $y= x^2- 6x+ 10= (x- 3)^2+ 1$, I would answer x= 3. If I were asked for the critical point, I would answer (3, 1). | 2020-08-08T14:58:29 | {
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https://math.stackexchange.com/questions/1760828/locus-of-a-point-on-a-fixed-length-segment-whose-endpoints-slide-along-orthogona | # Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines
Suppose we have some segment $AB$ of constant length that slides in such a way that its endpoints are moving along orthogonal lines. Let $P$ be a point in the segment so that $|AP| = a$ and $|PB| = b$. How can we find the curve along which $P$ moves?
I was trying to write lines $ax + by = c$ since this is orthogonal to the other line, we know normal to the orthogonal one should be $N = (-b,-a)$ and so the equation of the other line is of the form $-b x - a y + d = 0$. Now, point $A$ lie in one of them and $B$ lies on the other one line. should I solve the equation $d(P,A) + d(P,B) = a+ b$? Am I on the right track?
• You might try starting with an easy case. Consider specifically that the orthogonal lines are the coordinate axes. You can parameterize the endpoints as $A = ((a+b)\cos\theta, 0)$ and $B = (0, (a+b)\sin\theta)$. I'll let you find $P$, and also to generalize this to arbitrary lines. – Blue Apr 27 '16 at 9:49
Here's an illustration of my comment, although I've swapped the positions of $A$ and $B$ (for notational reasons that should be clear shortly).
We can use the angle ($\theta$) that the segment makes with the $x$-axis to parameterize the coordinates of the endpoints. Of course, what's important are the coordinates of point $P$, namely ...
$$P = (a \cos\theta, b \sin\theta) \tag{1}$$
One might (should?) recognize this as the parameterization of the origin-centered ellipse with radii $a$ and $b$.
The non-parametric form of this equation is, of course,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{2}$$
Generalization to the case of an arbitrary pair of perpendicular lines is straightforward. Let the lines meet at $Q = (x_0, y_0)$, and let the line containing sliding point $B$ make angle $\phi$ with the $x$-axis. We can transform equation $(2)$ into the equation of this ellipse with substitutions corresponding to rotation by $\phi$, followed by translation by $x_0$ and $y_0$.
$$\text{rotation:}\;\begin{cases} x \to \phantom{-}x \cos\phi + y \sin \phi \\ y \to -x \sin\phi + y \cos\phi \end{cases} \qquad\qquad \text{translation:}\;\begin{cases} x \to x - x_0 \\ y \to y - y_0 \end{cases}$$
That is,
$$\frac{\left(\;(x-x_0)\cos\phi + (y-y_0)\sin\phi\;\right)^2}{a^2} + \frac{\left(\;-(x-x_0)\sin\phi + (y-y_0)\cos\phi\;\right)^2}{b^2} = 1 \tag{2^\prime}$$
where expansion and simplification is left as an exercise to the reader.
Alternatively, noting that the lines have unit direction vectors $(\cos\phi, \sin\phi)$ and $(-\sin\phi, \cos\phi)$, we easily adapt $(1)$ to get this parametric form of the locus:
$$P\left(\; x_0 + a\cos\theta \cos\phi - b \sin\theta\sin\phi\;,\;y_0 + a \cos\theta\sin\phi + b\sin\theta\cos\phi\;\right) \tag{1^\prime}$$
• This does not answer the question for any lines not just the $xy-$axis – user203867 May 1 '16 at 7:49
• @Learner: ... and yet, even after I edited my answer to include the general case, you awarded a sizable bounty to an answer that handled only the $xy$-axis case. Go figure. (I'm not bothered about the bounty ---hypergeometric is quite welcome to more internet points--- but I'm disappointed by the inconsistency. I invested (more) time and effort into refining an answer to address a concern that apparently wasn't much of a concern, after all.) – Blue May 3 '16 at 14:46
• Quite right. You've got my (+1), Blue. – Han de Bruijn May 3 '16 at 15:15
WLOG let the orthogonal lines be the $x$-axis and $y$-axis respectively. Let $P=(x,y), A=(0,y+k), B=(x+h,0), AP=a, PB=b$. By Pythagoras' theorem, \begin{align} (y+k)^2+(x+h)^2&=(a+b)^2\\ (y+\sqrt{a^2-x^2})^2+(x+\sqrt{b^2-y^2})^2&=(a+b)^2\\ y\sqrt{a^2-x^2}+x\sqrt{b^2-y^2}&=ab\\ y^2(a^2-x^2)&=a^2b^2-2abx\sqrt{b^2-y^2}+x^2(b^2-y^2)\\ 2abx\sqrt{b^2-y^2}&=b^2x^2-a^2y^2+a^2b^2\\ 4a^2b^2x^2(b^2-y^2)&=b^4x^4+a^4y^4+a^4b^4+2(-a^4b^2x^2y^2+a^2b^4x^2-a^4b^2y^2)\\ 2a^2b^2x^2(b^2-y^2)&=b^4x^4+a^4y^4+a^4b^4-2a^4b^2y^2\\ 2a^2b^2(b^2x^2+a^2y^2-x^2y^2)&=b^4x^4+a^4y^4+a^4b^4\\ (a^2y^2+b^2x^2-a^2b^2)^2&=0\\ a^2y^2+b^2x^2&=a^2b^2\\ \frac {x^2}{a^2}+\frac{y^2}{b^2}&=1 \end{align} which is an ellipse. | 2019-06-19T08:47:34 | {
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https://math.stackexchange.com/questions/3830352/proving-that-sqrtab-leq-dfraca-b2-via-contradiction/3830359 | # Proving that $\sqrt{ab} \leq \dfrac{a + b}{2}$ via contradiction
Today I was trying to prove via contradiction the following:
$$\sqrt{ab} \leq \dfrac{a + b}{2}, a,b \geq 0$$
So what I did was the following:
Let's assume that $$\sqrt{ab} > \frac{a + b}{2}$$, this means: $$2\sqrt{ab} > a + b \Leftrightarrow 0 >a - 2\sqrt{ab} + b$$
We have that $$a - 2\sqrt{ab} + b = (\sqrt a - \sqrt b)^2$$ or $$a - 2\sqrt{ab} + b = (\sqrt b -\sqrt a)^2$$
So if we sub this in the expression above we end up with:
$$0 > (\sqrt a - \sqrt b)^2 \vee 0 > (\sqrt b -\sqrt a)^2$$
This is clearly a contradiction because the square of a real number is allays positive. This means that:
$$\sqrt{ab} \leq \dfrac{a + b}{2}$$
My question is: Is this correct? Because we can only apply proofs by contradictions when there is only two different answers, so if it's not one, it has to be the other. But in this case, although is seems to exists only two alternatives: $$>$$ or $$\leq$$, I think That there are 3 of them: $$>, =$$ and $$<$$ so I'm not sure whether this proof is valid or not.
• This is correct, because $\leq$ contains the two cases $=$ and $<$. But I would say that proving this inequality by contradiction is not the most natural way ! Sep 17 '20 at 20:42
• @Eduardo A minor point is that since $(\sqrt{a} - \sqrt{b})^2 = (\sqrt{b} - \sqrt{a})^2$ for all real, non-negative $a$ and $b$, you don't need to specify & handle those $2$ cases. Sep 17 '20 at 20:45
• Let assume $a=b=1 \implies 1>1$
– user
Sep 17 '20 at 20:48
• You can rewrite this as a direct proof without contradiction. It would start by assuming that $a,b \ge 0$, and then continuing by proving a sequence of equivalences: "$\sqrt{ab} < \frac{a+b}{2}$ is equivalent to $2 \sqrt{ab} < a+b$ which is equivalent to ..." Sep 17 '20 at 20:56
Let's try doing it a different way. Notice that for all real numbers $$z$$, we have $$z^2 \geq 0$$. Let's take $$a, b \geq 0$$. We have $$(\sqrt{a}-\sqrt{b})^2 \geq 0,$$ or $$a - 2\sqrt{ab} + b \geq 0$$. Now rearrange things to get $$\frac{a+b}{2} \geq \sqrt{ab}.$$ | 2021-10-28T17:40:04 | {
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https://math.stackexchange.com/questions/4032754/proof-of-the-derivative-of-an-elementary-function-is-also-elementary | # Proof of the derivative of an elementary function is also elementary
I heard of the result that the derivative of an elementary function is also elementary long ago. Now I want to prove it rigorously. I found this answer(I didn't comment because it was an old post):https://math.stackexchange.com/q/2195559. It actually proved that: (1). the derivative of $$\exp,\log,\operatorname{id}$$ and constant is elementary. (2). If we assume that f and g have elementary derivatives, then $$f+g,f-g,fg,f/g,f\circ g$$ also have elementary derivatives. So far so good. But what confuses me is how does this 2 facts imply the original proposition(the derivative of any elementary function is also elementary)? I mean isn't the fact (2) seems too weak?
I think the keypoint lies in the construction of the elementary functions. I use the definition of the elementary functions to be:
1. $$\exp,\log,\operatorname{id}$$ and constant is elementary.
2. the sum, difference, product, quotient, composition of 2 elementary functions is elementary.
Need help!
• You should start with the definition of elementary function. They are those and only those functions that can be produced from polynomials, rational functions, trigonometric and exponential functions and their inverses by algebraic operations and composition. Since the derivatives of the listed base functions are elementary, and sum, product, quotient and chain rules transform elementary functions into elementary functions, the derivatives of all elementary functions are elementary by induction on operations used to produce them. Feb 20 at 8:09
• Why? When some class of objects is defined inductively using induction to prove something for all of them is the first thing that comes to mind. Feb 20 at 8:24
• I think the idea to use induction is surprising --- Of possible interest is the general notion of inductive definitions, which can be worked with in two main ways, "from above" and "from below". A fairly abstract treatment of this notion is given in Elementary Induction on Abstract Structures by Moschovakis (1974; reprinted by Dover in 2008). Feb 20 at 9:05
• Yes, this is a more general type of induction/recursion than the elementary one on natural numbers, it is sometimes called structural induction. It is a very useful tool in more advanced problems. If the elementary type can be visualized as moving up a chain or a ladder, structural induction moves up a tree that can branch. In this case, you have choices which operations to apply to produce new elementary functions from old ones, so the steps can not be indexed by natural numbers like a sequence. Feb 20 at 9:09
• Yes, this would correspond to collecting all branch points of the tree that are at the same height into a single set, but I find it more intuitive to think in terms of the tree directly. And the base set needs to also include trig functions and their inverses unless you are working over complex numbers. Feb 20 at 9:34
The closure under differentiation can be proven by induction.
The result is clear for « base functions » ($$\ln x, e^x)...$$
If a map is a sum, the product, the composition... of two simple functions, then it follows that its derivative is a simple function based on the induction hypothesis and the differentiation rules of the sum, the product, the composition... of two functions.
As any simple function can be obtained by induction using previous rules according to the very definition of a simple function, we’re done.
Let me try to organize what you have a little more neatly, maybe this will clear things up.
First, your definition of elementary function:
$$\bullet \exp, \log, \text{id}, \text{constant}\\ \bullet \text{The sum of two elementary functions, i.e. }f+g \\ \bullet \text{The difference of two elementary functions, i.e. }f-g \\ \bullet \text{The product of two elementary functions, i.e. }fg \\ \bullet \text{The quotient of two elementary functions, i.e. }\frac{f}{g} \\ \bullet \text{The composition of two elementary functions, i.e. }f \circ g$$
Now, the proof you posted proves first that $$\exp, \log$$, id and constant functions have elementary derivatives, so the claim holds for the first bullet point in the list above. Then it goes on to prove that it holds for sums (2nd bullet point), differences (3rd bullet), products (4th bullet), quotients (5th bullet) and compositions (6th bullet).
So it really has shown that all the functions that fall under the definition of elementary function have elementary derivatives! And as mathcounterexamples.net pointed out in his answer, this can be expanded to the sum, product, etc. of more than two functions by induction.
• I totally agree that the claim holds for the first bullet point. But when it comes to the second point, it assumes f and g both have elementary derivatives which I think can't be taken as "the claim holds for sums". Feb 20 at 8:48
• Ah ok I think I see what you mean. In that case Induction is really the way to go (as seen in the other answer) Feb 20 at 8:54 | 2021-09-17T04:29:15 | {
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https://math.stackexchange.com/questions/1503893/can-two-random-variables-x-y-be-dependent-and-such-that-exy-exey | Can two random variables $X,Y$ be dependent and such that $E(XY)=E(X)E(Y)$?
Can someone define independence of two random variables with this "product rule", or are there any counterexamples?
• Sure there are, try X standard normal and Y=UX with U symmetric Bernoulli.
– Did
Oct 29, 2015 at 19:13
• @Did, what do you mean by symmetric Bernoulli? $P(U=0) = P(U=1)$? Oct 29, 2015 at 19:16
• P(U=1)=P(U=-1)=1/2
– Did
Oct 29, 2015 at 19:29
• @Did Is that a Bernoulli variable? I don't think so. Oct 29, 2015 at 19:32
• @CarlosMendoza Many people call Bernoulli what you probably call a two-points distribution.
– Did
Oct 29, 2015 at 19:33
Independence implies uncorrelation
We say that two random variables are uncorrelated if $$E[XY] = E[X]E[Y]$$.
If $$X$$ and $$Y$$ are independent and discrete (this can also be extended to continuous random variables), then
$$p_{X,Y}(x,y) = p_X(x)p_Y(y)\qquad \forall (x,y)$$
Using this in the definition of $$E[XY]$$ and doing some rearrangements of terms, we have
\begin{align} E[XY] &= \sum_x\sum_y xyp_{X,Y}(x,y)\\ &= \sum_x\sum_y xyp_X(x)p_Y(y)\\ &= \sum_x\sum_y [xp_X(x)][yp_Y(y)]\\ &= \sum_xxp_X(x)\sum_yyp_Y(y)\\ &= E[X]E[Y] \end{align}
Therefore, if $$X$$ and $$Y$$ are independent, then they are uncorrelated.
$$\\$$
Uncorrelation does not imply independence (in general)
In general$$^1$$, it is false to say that if $$X$$ and $$Y$$ are uncorrelated, then they are independent. Let's see a counterexample. Let $$X$$ and $$Y$$ be random variables taking values in $$\{(0,1), (1,0), (-1,0), (0,-1)\}$$ with equal probability $$0.25$$. Then, $$XY = 0$$ with probability $$1$$, and therefore $$E[XY] = 0$$.
By other hand,
$$\displaystyle p_X(x) = \left\{ \begin{array}{ll} \frac{1}{4} & x=-1\\ \frac{1}{2} & x=0\\ \frac{1}{4} & x=1\\ \end{array} \right.$$ $$\hspace{1cm}$$ and $$\hspace{1cm}$$ $$p_Y(y) = \left\{ \begin{array}{ll} \frac{1}{4} & y=-1\\ \frac{1}{2} & y=0\\ \frac{1}{4} & y=1\\ \end{array} \right.$$
which implies that $$E[X]=E[Y]=0$$. Then, $$E[XY] = E[X]E[Y]$$ and the random variables are uncorrelated.
However,
$$p_X(1)p_Y(1) = \frac{1}{4}\frac{1}{4} \neq p_{X,Y}(1,1) = 0$$
which shows that $$X$$ and $$Y$$ are not independent.
We can conclude then that independence is a sufficient but not a necessary condition for uncorrelation.
$$^1$$ The only two exceptions I am aware of are the jointly Gaussian variables and two Bernoulli random variables...any other?
• So you group terms into 2 sums because y's terms are not dependent on x's sum and vice versa please?
– Avv
Apr 6 at 22:33
• @Avv. I edited my answer to add two additional steps and hopefully make it clearer. In the third line, you move the factor $xp_X(x)$ out of the sum $\sum_y$ because it doesn't depend on $y$, only on $x$. Apr 7 at 13:40
• I upvoted the answer. It's very clear now. Thank you.
– Avv
Apr 7 at 18:57
• One more thing please. Why if $g(y) = E[X|Y=y]$ then $E[E[X|y=y]]$ is not expectation of a number but expectation of a random variable? Should not $E[X|y=y]$ be treated as a number not a random variable please so we have $E[E[X|y=y]] = E[X|Y=y]$?
– Avv
Apr 7 at 19:47
• @Avv, I am not sure how your question relates to the original question and my answer. Please elaborate. Your question seems to be a new question, and you should probably ask it in a new post, but I will try to answer in the next comment.... Apr 8 at 17:34 | 2022-05-29T00:43:12 | {
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https://www.rbjlabs.com/probability-and-statistics/correlation-analysis/ | # Correlation analysis
In the regression analysis we connect an independent variable with a dependent variable, but when both variables are random, a correlation analysis must be performed.
In the correlation analysis we do not want to find a linear relationship between two variables, we try to measure the intensity of the linear relationship between X and Y.
### Values of the correlation coefficient
This parameter of the correlation coefficient assumes values between -1 and 1:
• If r = 1, the correlation is perfect positive.
• If r = -1, the correlation is perfect negative.
• If r = 0, there is no correlation between both variables.
It can be interpreted as follows:
### Formula to calculate the correlation coefficient
There are a lot of calculations to arrive at the formula of the correlation coefficient, but we are going to save all those calculations and go straight with the formula:
r = \cfrac{n\sum xy - \sum x \sum y}{\sqrt{\left[ n\sum x^{2} - (\sum x)^{2}\right] \left[n \sum y^{2} - ( \sum y)^{2} \right]}}
Great, now that we have the formula, let’s go with the exercise!
## Exercise of the correlation coefficient
In a study of ground movement caused by earthquakes, the maximum speed (in m/s) and the maximum acceleration (in m/s^{2}) were recorded for five of them. The results are presented in the following table:
\begin{array}{| c | c |}
\hline
\text{Speed} & \text{Acceleration} \\
\hline
1\text{.}54 & 7\text{.}64 \\
1\text{.}60 & 8\text{.}04 \\
0\text{.}95 & 8\text{.}04 \\
1\text{.}30 & 6\text{.}37 \\
2\text{.}92 & 5\text{.}00 \\
\hline
\end{array}
1. Calculate the correlation coefficient
2. Someone suggests modifying the units to centimeters and minutes. What effect would it have on the correlation?
#### Section 1. Calculate the correlation coefficient
Excellent, let’s first calculate the correlation coefficient, for that we need: \sum x, \sum y, \sum xy, \sum x^{2} and \sum y^{2}. First, let’s calculate \sum x and \sum y:
\begin{array}{| c | c |}
\hline
\text{Speed} & \text{Acceleration} \\
\hline
1\text{.}54 & 7\text{.}64 \\
1\text{.}60 & 8\text{.}04 \\
0\text{.}95 & 8\text{.}04 \\
1\text{.}30 & 6\text{.}37 \\
2\text{.}92 & 5\text{.}00 \\
\hline
\sum x = 8\text{.}31 & \sum y = 35\text{.}09 \\
\hline
\end{array}
Let’s calculate \sum xy:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & xy \\
\hline
1\text{.}54 & 7\text{.}64 & 11\text{.}7656 \\
1\text{.}60 & 8\text{.}04 & 12\text{.}864 \\
0\text{.}95 & 8\text{.}04 & 7\text{.}638 \\
1\text{.}30 & 6\text{.}37 & 8\text{.}281 \\
2\text{.}92 & 5\text{.}00 & 14\text{.}6 \\
\hline
& & \sum xy = 55\text{.}1486 \\
\hline
\end{array}
Now let’s calculate x^{2}:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & x^{2} \\
\hline
1\text{.}54 & 7\text{.}64 & 2\text{.}3716 \\
1\text{.}60 & 8\text{.}04 & 2\text{.}56 \\
0\text{.}95 & 8\text{.}04 & 0\text{.}9025 \\
1\text{.}30 & 6\text{.}37 & 1\text{.}69 \\
2\text{.}92 & 5\text{.}00 & 8\text{.}5264 \\
\hline
& & \sum x^{2} = 16\text{.}0505 \\
\hline
\end{array}
Now let’s calculate y^{2}:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & y^{2} \\
\hline
1\text{.}54 & 7\text{.}64 & 58\text{.}3696 \\
1\text{.}60 & 8\text{.}04 & 64\text{.}6416 \\
0\text{.}95 & 8\text{.}04 & 64\text{.}6416 \\
1\text{.}30 & 6\text{.}37 & 40\text{.}5769 \\
2\text{.}92 & 5\text{.}00 & 25\text{.}00 \\
\hline
& & \sum y^{2} = 253\text{.}2297 \\
\hline
\end{array}
Now let’s directly replace the values in the formula:
r = \cfrac{(5)(55\text{.}1486) - (8\text{.}31)(35\text{.}09)}{\sqrt{\left[(5)(16\text{.}0505) - (8\text{.}31)^{2} \right] \left[(5)(253\text{.}2297) - (35\text{.}09)^{2} \right]}}
r = -0\text{.}8027
The correlation coefficient is r=-0\text{.}8027, so it indicated that it is a moderate negative correlation.
### Section 2. Someone suggests modifying the units to centimeters and minutes
The truth is that it would have no effect if we change the values to centimeters and minutes since it would only be a change of units, that is, the correlation would be the same, but we are going to check it anyway. First, let’s calculate \sum x and \sum y:
\begin{array}{| c | c |}
\hline
\text{Speed (cm/min)} & \text{Acceleration (cm/min}^{2}\text{)} \\
\hline
9240 & 45840 \\
9600 & 45840 \\
5700 & 48240 \\
7800 & 38220 \\
17520 & 30000 \\
\hline
\sum x = 49860 & \sum y = 210540 \\
\hline
\end{array}
Now let’s calculate the sum of x by y:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & xy \\
\hline
9240 & 45840 & 423561600 \\
9600 & 45840 & 463104000 \\
5700 & 48240 & 274968000 \\
7800 & 38220 & 298116000 \\
17520 & 30000 & 525600000 \\
\hline
& & \sum xy = 1985349600 \\
\hline
\end{array}
Now let’s calculate x^{2}:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & x^{2} \\
\hline
9240 & 45840 & 85377600 \\
9600 & 45840 & 92160000 \\
5700 & 48240 & 32490000 \\
7800 & 38220 & 60840000 \\
17520 & 30000 & 306950400 \\
\hline
& & \sum x^{2} = 577818000 \\
\hline
\end{array}
Now let’s calculate y^{2}:
\begin{array}{| c | c | c |}
\hline
\text{Speed} & \text{Acceleration} & y^{2} \\
\hline
9240 & 45840 & 2101305600 \\
9600 & 45840 & 2327097600 \\
5700 & 48240 & 2327097600 \\
7800 & 38220 & 1460768400 \\
17520 & 30000 & 900000000 \\
\hline
& & \sum y^{2} = 9116269200 \\
\hline
\end{array}
Now what we are going to do is put all those calculated values in the formula to find the correlation coefficient:
r = \cfrac{(5)(1985349600) - (49860)(21054)}{\sqrt{\left[ (5)(577818000) - (49860)^{2}\right] \left[(5)(9116269200) - (210540)^{2} \right]}}
Using a calculator, we will obtain the following value:
r = -0\text{.}8027
So you can verify that the correlation coefficients do not change even if they have different units, it also means that it is still a moderate negative correlation!
Thank you for being at this momento with us : ) | 2023-03-28T12:26:17 | {
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https://math.stackexchange.com/questions/3050208/i-calculated-sin-75-circ-as-frac12-sqrt2-frac-sqrt32-sqrt2/3050212 | # I calculated $\sin 75^\circ$ as $\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$, but the answer is $\frac{\sqrt{2}+\sqrt{6}}{4}$. What went wrong?
I calculated the exact value of $$\sin 75^\circ$$ as follows:
\begin{align} \sin 75^\circ &= \sin(30^\circ + 45^\circ) \\ &=\sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ \\ &=\frac12\cdot\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}} \\ &= \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \end{align}
The actual answer is $$\frac{\sqrt{2} + \sqrt{6}}{4}$$
My main confusion is how the textbook answer is completely different from mine, even though if I compute $$\sin 30^\circ \cos45^\circ + \cos 30^\circ \sin 45^\circ$$, it will be approximately the same value of $$\sin 75^\circ$$.
I think I'm having difficulty adding and subtracting the radicals. So, if someone can demonstrate to me how they got that answer, it will be helpful. Thanks.
• You mean $\sin 75 = \sin(30+45)$. – KM101 Dec 23 '18 at 10:09
• yes you are right – DreamVision2017 Dec 23 '18 at 10:14
• Before claiming values are different, you should try calculating them with a calculator, notice they match and then think about an algebraic manipulation. – zwim Dec 23 '18 at 10:49
• Sorry I didn't think I clarified this in my question, I knew that my answer was equivalent but just didn't know how the textbook simplified it like that algebraically – DreamVision2017 Dec 23 '18 at 22:23
They’re the same value. Multiply the numerator and denominator of your answer by $$\sqrt 2$$ to see why.
$$\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2}{\sqrt 2}\cdot\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2+\sqrt 6}{4}$$
You can also use $$\sin 45 = \cos 45 = \frac{\sqrt 2}{2}$$ (rationalizing $$\frac{1}{\sqrt 2}$$) to get the answer more easily. | 2019-04-26T15:55:42 | {
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https://math.stackexchange.com/questions/637604/whats-the-simplest-way-to-solve-int-frac14-v2dv | # What's the simplest way to solve $\int \frac{1}{4-v^2}dv$?
I did a substitution on a DE and ended up with this:
$\int \frac{1}{4-v^2}dv$
I tried a trig substitute but things got a bit hairy. WolframAlpha recommended a far less intuitive substitution. I can't remember what it was.
What is the simplest way to solve this integral?
• Partial fractions? – anon Jan 14 '14 at 1:42
Hint: $\frac{1}{4-u^2}=\frac{\frac{1}{4}}{{2-u}}+\frac{\frac{1}{4}}{{2+u}}$
• Thanks. I hadn't even considered partial fractions. – Korgan Rivera Jan 14 '14 at 2:06
Use partial fractions. Put $${1\over 4 -x^2} = {A\over 2-x} + {B\over 2+ x}.$$
I'm surprised that with all these answers no one has mentioned hyperbolic trigonometric functions. If you know that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$ it may not surprise you to learn that $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$, and so:
$$\int \frac{1}{4-v^2}dv=\frac{1}{4}\int \frac{1}{1-(\frac{v}{2})^2} dv$$ substitude $u=\frac{v}{2}$ so $2 du=dv$
$$=\frac{1}{2} \int \frac{1}{1-u^2} du=\frac{1}{2} \tanh^{-1}(u)=\frac{1}{2} \tanh^{-1}(\frac{v}{2})$$
To prove the identities, try to mimic the proof that $\int \frac{1}{1+x^2}dx=\tan^{-1}(x)$. That is, put $x=\tan(y)$ to find that $\frac{dy}{dx}=\cos^2(y)=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}$ so that integration with respect to x gives $y=\int \frac{1}{1+x^2}$. You'll need to prove the hyperbolic equivalent of $\sec^2(x)=1+\tan^2(x)$, $\text{sech}^2(x)=1-\tanh^2(x)$, which is trivial using the fundamental property $\cosh^2(x)-\sinh^2(x)=1$ (divide through by $\cosh^2(x)$).
Of course, partial fractions is the way to go if your teacher does not allow this form of an answer. In fact, the partial fraction decomposition and the identity $\int \frac{1}{1-x^2} dx=\tanh^{-1}(x)$ would probably be a sorta neat way to prove that $\tanh^{-1}(x)=\frac{1}{2}(\log(1+x)-log(1-x))$. (Though another method of proving it: writing $y=\tanh(x)=\sinh(x)/\cosh(x)$, expanding into exponentials, and solving for $x$ in terms of $y$ isn't too difficult.)
• Indeed very nice :)! – J. W. Perry Jan 14 '14 at 4:34
I think partial fractions may indeed be the "simplest" way, but you mentioned trigonometric substitution getting a bit hairy, and I figured I would give my best derivation. It is a useful skill. This would be advantageous speed wise only if you knew the integral of the secant function from memory.
Consider the more general integral, $$\int\frac{1}{a^2-v^2} \, dv.$$ If we let $$v=a\sin\theta,$$ then $$dv=a\cos \theta \, d \theta,$$ and $$a^2-v^2=a^2-a^2\sin^2 \theta=a^2\left( 1-\sin^2 \theta \right)=a^2 \cos^2 \theta.$$
We substitute to get \begin{align*} \int \frac{1}{a^2-v^2} \, dv &= \int \frac{a \cos \theta \, d \theta}{a^2 \cos^2 \theta} \\ &=\frac{1}{a}\int\sec \theta \, d \theta \\ &=\frac{1}{a}\ln |\sec \theta + \tan \theta|+c. \end{align*}
For the back substitution, $$v=a \sin \theta \Rightarrow \sin \theta = \frac{v}{a}.$$
We form right a triangle with angle $\theta$, side $v$ opposite $\theta$, and hypotenuse $a$. Thus the side adjacent $\theta$ has length $\sqrt{a^2-v^2}$. Reading directly from the triangle, $$\frac{1}{a}\ln |\sec \theta + \tan \theta|+c=\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c.$$
We are done, but this certainly cleans up well (like in an integral table). \begin{align*} &\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c \\ =& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a^2-v^2}} \right|+c \\ =& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a+v}\sqrt{a-v}} \right|+c \\ =&\frac{1}{a}\ln \left| \frac{\sqrt{a+v}}{\sqrt{a-v}} \right|+c \\ =&\frac{1}{a}\ln \left| \left( \frac{a+v}{a-v} \right)^{\frac{1}{2}} \right|+c \\ =&\frac{1}{2a}\ln \left| \frac{a+v}{a-v} \right|+c. \\ \end{align*}
Thus your particular integral is $$\int \frac{1}{4-v^2} \, dv=\frac{1}{4}\ln \left| \frac{a+v}{a-v} \right|+c.$$
It might not seem like the "easiest" way, but it is certainly doable, and in my opinion, fun.
• Hyperbolic functions are much more useful in this situation. – user18862 Jan 14 '14 at 4:22
• It's definitely fun to do, but I didn't know the integral of sec at the time, and I was looking for a simpler, shorter method. – Korgan Rivera Jan 14 '14 at 18:16
substitute $2z = v,$ and then $z = \sin \theta.$
• @anon told you I was asleep :( – Igor Rivin Jan 14 '14 at 1:47
• this won't help. Use partial fraction as mentioned before – user44197 Jan 14 '14 at 1:47
• @user44197 depends what you mean by "help". You get the integral of $\sec x,$ which is, admittedly. more difficult than the original, but might be known to the OP. – Igor Rivin Jan 14 '14 at 1:49
• actually $\sec^2$ which is easier than $\sec$. In any case you end up with $\tan^2\theta$ and you have to write it in terms of $\sin\theta$ and finally $v$. – user44197 Jan 14 '14 at 1:57
• @anon It looks like this substitution gives $\frac12\int\sec\theta d\theta$. – Mike Jan 14 '14 at 3:31 | 2019-06-16T05:44:06 | {
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https://math.stackexchange.com/questions/2900468/why-does-this-ln-function-inverse/2900473 | # Why does this ln function inverse?
I'm going through applications of separable equations and came across an example of half-lives:
$$M(t)=\frac{M}{2}=Me^{-kt}$$
Factoring out $M$, $\frac{1}{2}=e^{-kt}$.
To solve for $t$, $\ln\left(\frac{1}{2}\right)=-kt$.
And then the answer is $t=\frac1k{\ln 2}$.
Why isn't the value for $t=-\frac1k{\ln \frac12}$?
Thanks:)
• It is the same ! – Claude Leibovici Aug 31 '18 at 8:42
• Recall the log sum rule: $\log(x) + \log(y) =\log(xy)$. Apply this with $y=1/x$ to derive $\log(x) = -\log(1/x)$. – Winther Aug 31 '18 at 8:44
• Also, to complement @Winther's comment, note that $\log x^n = n\ \log x$ and $x^{-1}=\frac{1}{x}$. – User123456789 Aug 31 '18 at 11:30
It is the same since $$\ln(2)=-\ln(1/2).$$
$\ln(1/2) = -kt$ implies $-\ln(2) = -kt$ implies $\dfrac{\ln(2)}{k}=t$ which is also $-\dfrac{\ln(1/2)}{k}=t$.
Reminder that $\ln(1/x) = -\ln(x)$.
$\ln{\frac{1}{2}} = \ln{1} - \ln{2} = 0 - \ln{2} = -\ln{2}$
Given the exponent rule for logarithms $\log(a^b) = b\log(a)$ with $a = 1/2$ and $b = -1$ we have
$$\log(2) = \log((1/2)^{-1}) = -\log(1/2)$$ | 2019-08-22T19:53:44 | {
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http://math.stackexchange.com/questions/63356/conditional-probability-problem | Conditional probability problem
An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?
What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases, then we need to find a conditional probability $P(X\geq 2 | Y)=P(X= 2,3,\text{ or }4 | Y)=\frac{P(\{X\geq 2\} \cap Y)}{P(Y)}$ but then I don't know how to continue?
Any help...
-
You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.
If $N$ is the number of dictionaries where the keyword is found then $$\Pr(N=n) = \frac{{5 \choose n}{4 \choose 4-n}}{9 \choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.
So for $n=0,1,2,3,4$ this gives probabilities $\frac{1}{126},\frac{20}{126},\frac{60}{126},\frac{40}{126},\frac{5}{126}$.
Add up the last three (and change to lowest terms) and you have your solution.
-
@Mike: I don't know: the value is the same but the words are not. You have ${5 \choose 3}$ where I have ${5 \choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 \choose 5}$ where I have ${9 \choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall". – Henry Sep 10 '11 at 21:23
Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = \binom{2}{2} \binom{7}{2}/\binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $\binom{4}{2}/\binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?) – Mike Spivey Sep 10 '11 at 21:48
So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.) – Mike Spivey Sep 10 '11 at 21:53
In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $\Pr(N=n) = \dfrac{{k \choose n}{d-k \choose m-n}}{d \choose n}$. – Henry Sep 10 '11 at 22:09
Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me. – Mike Spivey Sep 10 '11 at 22:09
Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.
$$\text{ Henry: } P(X = x) = \frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}}; \text{ me: } P(X = x) = \frac{\binom{K}{x} \binom{N-K}{M-x}}{\binom{N}{M}}.$$
But
$$\frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}} = \frac{M!}{x! (M-x)!} \frac{(N-M)!}{(K-x)! (N-M-K+x)!} \frac{K! (N-K)!}{N!}$$ $$= \frac{K!}{x! (K-x)!} \frac{(N-K)!}{(M-x)! (N-K-M+x)!} \frac{M! (N-M)!}{N!} = \frac{\binom{K}{x} \binom{N-K}{M-x}}{\binom{N}{M}}.$$
Thus the two answers give the same result.
I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X \geq 2)$.
$$P(X = 2) = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}}$$ because there are $\binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $\binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $\binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.
I'll let you calculate $P(X = 3)$ and $P(X = 4)$.
(FYI, $X$ has what's called a hypergeometric distribution.)
-
So, in general: $P(X=k)=\frac{\binom{4}{k} \binom{5}{5-k}}{\binom{9}{5}}$ – Kelly Sep 10 '11 at 16:56
@Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution. – Mike Spivey Sep 10 '11 at 17:05
I feel that there is something wrong above! Are we choosing 4 items and not 5?! – Kelly Sep 10 '11 at 17:06
@Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords. – Mike Spivey Sep 10 '11 at 20:31 | 2015-07-04T10:38:27 | {
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http://mathhelpforum.com/algebra/145269-eigenvalues-eigenvectors-3x3-matrix.html | # Thread: eigenvalues/eigenvectors of a 3x3 matrix
1. ## eigenvalues/eigenvectors of a 3x3 matrix
i have a 3x3 matrix
$
\begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}
$
i got the eigenvalues of 2, 1, and 0.
im having a big problem with how to get the corresponding eigenvectors
if anyone can help me that would be great!
2. Originally Posted by eigenvectorisnotfun
i have a 3x3 matrix
$
\begin{pmatrix}-2 & -8 & -12\\1 & 4 & 4\\0 & 0 & 1\end{pmatrix}
$
i got the eigenvalues of 2, 1, and 0.
im having a big problem with how to get the corresponding eigenvectors
if anyone can help me that would be great!
For every eigenvalue there is a corresponding eigenvector. So we need to find our eigenvectors 3 times!
How we do this is place lambda on the diagonal in the matrix! In other words,
$\begin{pmatrix} {-2 - \lambda }& -8 & -12\\ 1 & { 4- \lambda } & 4\\0 & 0 & {1 - \lambda } \end{pmatrix}$
Let's pick $\lambda = 0$
$\begin{pmatrix} {-2}& -8 & -12\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}$
Now we need to row reduce!
$12 R_3 + R_1 --> R_1$
$\begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 4\\0 & 0 & {1} \end{pmatrix}$
$R_2 - 4R_3 --> R_2$
$\begin{pmatrix} {-2}& -8 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}$
$2 R_2 + R_1 --> R_1$
$\begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix}$
At this point I don't think we can reduce any more.
Now we know there exists a set of vectors such that
$\begin{pmatrix} {0}& 0 & 0\\ 1 & { 4 } & 0\\0 & 0 & {1} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\V_3 \end{pmatrix} = 0$
Carry out the multiplication and find your values. There might be trivial solutions but that's okay!
Find more at: Eigenvalues and Eigenvectors
3. Or, use the definition of "eigenvalue" and "eigenvector".
If you are right that 2 is an eigenvalue, then there must exist a vector, $\begin{pmatrix}x \\ y \\ z\end{pmatrix}$, x, y, z not all 0, such that
$\begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}$ $= \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix}$ $= \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}$.
That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form $\begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}$.
Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.
On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)
4. Originally Posted by HallsofIvy
Or, use the definition of "eigenvalue" and "eigenvector".
If you are right that 2 is an eigenvalue, then there must exist a vector, $\begin{pmatrix}x \\ y \\ z\end{pmatrix}$, x, y, z not all 0, such that
$\begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}$ $= \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix}$ $= \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}$.
That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form $\begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}$.
Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.
On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)
Is this typically a faster approach then the row reductions I have above?
5. now i finally got it!
what i did wrong was I did not really get that (A-I lambda) = 0
thanks to you both, you both were really of great help | 2017-08-16T23:46:18 | {
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https://www.physicsforums.com/threads/proving-injection-surjection-bijection.218320/ | # Proving injection,surjection,bijection
1. Feb 27, 2008
### Physicsissuef
1. The problem statement, all variables and given/known data
Prove that one copying $$f:A \rightarrow B$$ is injection, only if $$x_1,x_2 \in A$$ and $$x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$$
2. Relevant equations
3. The attempt at a solution
2. Feb 27, 2008
### EnumaElish
What is the definition of an injection that you are working with/have been taught?
3. Feb 27, 2008
### Tom Mattson
Staff Emeritus
I bet the definition he was taught was the contrapositive of this one. Namely, that a map $f:A\rightarrow B$ is an injection if for all $x_1,x_2\in A$, $f(x_1)=f(x_2)$ implies $x_1=x_2$.
Physicsissuef, is that the definition in your book?
4. Feb 27, 2008
### Physicsissuef
Yes sir, sorry for missing the formula.
5. Feb 27, 2008
### Tom Mattson
Staff Emeritus
In that case, your proof need only be one line long. For any conditional statement $p \Rightarrow q$ (that is: if p, then q), a logically equivalent statement is $\neg q \Rightarrow \neg p$ (that is: if not p, then not q). This latter statement is called the contrapositive of the first, and it can be immediately inferred from it.
6. Feb 27, 2008
### Physicsissuef
Prove that one copying $$f:A \rightarrow B$$ is surjection, only if $$f(A)=B$$
7. Feb 27, 2008
### Tom Mattson
Staff Emeritus
Are you going to make me guess the definition again?
8. Feb 27, 2008
### Physicsissuef
surjection is, if every element $$y \in B$$ is image of at least one element $$a \in A$$
9. Feb 27, 2008
### HallsofIvy
Staff Emeritus
Okay, you prove one set is equal to another by showing that each is a subset of the other. And you prove subset by saying "if x is in" the first set and then showing that x must be in the second.
It should be trivial to show that f(A) is a subset of B. (What does $f: A\rightarrow B$ mean?)
To prove that B is a subset of A, start with "if $$y \in B$$" and show that y is in f(A). Of course, you will also need to use the definition of "f(A)".
Do you see a basic theme in all of this? Learn the exact words of the definitions. You can use the exact statement of definitions in proofs.
10. Feb 28, 2008
### Physicsissuef
Ok, but I think I still can't understand the definition of injection. What does $$f(x_1)=f(x_2), x_1=x_2$$, mean?
If I draw a domain and codomain, lets say $$f(x_1)=f(x_2)=1, x_1=x_2$$
http://pic.mkd.net/images/638823221.JPG [Broken]
Still, can't understand.
Last edited by a moderator: May 3, 2017
11. Feb 28, 2008
### steelphantom
Here's an example of something that's not an injection:
Let f(x) = x2. If f(y) = f(z) = 4, does y = z ? Not necessarily, because you could have y = 2 and z = -2. This is not an injection.
If, for example, f(x) = 2x, then f(y) = f(z) would imply that y = z. e.g. f(y) = f(z) = 4 => y = z = 2. This is an injection.
Hope that helped you understand the concept a little better.
12. Feb 28, 2008
### Physicsissuef
Ok, thanks. But can you draw it with domains and co domains, like my picture, please?
13. Feb 29, 2008
### Physicsissuef
Can you give me the whole prove, please?
14. Feb 29, 2008
### HallsofIvy
Staff Emeritus
No, we don't want to prevent you form learning- you learn by doing, not by watching someone else do it for you .
I will do this: suppose f:R->R is defined by f(x)= 3x- 4. To prove that f is injective ("one-to-one") we need to prove "if f(x1)= f(x2) then x1= x2. Okay, for this particular f, f(x1)= 3x1- 4 and f(x2)= 3x2- 4 so f(x1)= f(x2) means 3x1- 4= 3x2- 4. Adding 4 to both sides of the equation gives 3x1= 3x2 and then dividing both sides by 3, x1= x2, exactly what we needed to prove. To prove f is surjective ("onto"), let y be any real number. We need to prove that there must exist x such that f(x)= y. f(x)= 3x- 4= y so, again adding 4 to both sides, 3x= y+ 4 and then dividing by 3, x= (y+4)/3. For any real number y, that is still a real number. Since every number in R is f(x) for some x, f is surjective.
Now, can you tell me why g: R-> R, defined by g(x)= x2, can you tell me g is not injective and is not surjective?
Can you tell me why h:R to the non-negative real numbers, defined by h(x)= x2, is not injective but is surjective?
What about j(x)= x2, from non-negative real numbers to all real numbers?
What about k(x)= x2, from non-negative real numbers to non-negative real numbers?
15. Mar 1, 2008
### Physicsissuef
Ok, can you explain me first, why the definition is $f(x_1)=f(x_2)$ , $x_1=x_2$? What it means?
16. Mar 1, 2008
### cristo
Staff Emeritus
It's just the definition. In words, it means that an injective function is one that maps a distinct element in the domain to a distinct element in the codomain.
17. Mar 1, 2008
### Physicsissuef
But why $x_1=x_2$?
18. Mar 1, 2008
### cristo
Staff Emeritus
That's the "distinct element in the domain" part.
19. Mar 1, 2008
### Physicsissuef
Lets say if $f(x_1)=f(x_2)$, $f(x_1)=y$ if y=3, then $f(x_2)=3$
if $x_2=5$, so $f(5)=3$, then $x_1=6$ and $f(6)=3$, it doesn't necessarily $x_1=x_2$
20. Mar 1, 2008
### cristo
Staff Emeritus
Sorry, but what function are you using? I don't see you define f anywhere. | 2017-10-21T14:23:01 | {
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https://www.jiskha.com/questions/1016830/A-local-photocopying-store-advertises-as-follows-We-charge-9-cents-per-copy-for | # math
A local photocopying store advertises as follows. " We charge 9 cents per copy for 150 copies or less, 6 cents per copy for each copy over 150 but less than 210 , and 3 cents per copy for each copy 210 and above. " Let x be the number of copies ordered and C(x) be the cost of the job (in cents ).
- If x \leq 150, the cost of the copies is C(x) =
- If 150 < x < 210, the cost of the copies is C(x) =
- If x \geq 210, the cost of the copies is C(x) =
1. 👍 0
2. 👎 0
3. 👁 88
asked by Clara
1. C(x)
= 9x for x <= 150
= 9*150+6(x-150) for 150 < x < 210
= 9*150 + 6*59 + 3(x-210) for x >= 210
1. 👍 0
2. 👎 0
posted by Steve
2. C(x)
= 9x for x <= 150
= 9*150+6(x-150) for 150 < x < 210
= 9*150 + 6*59 + 3(x-209) for x >= 210
1. 👍 0
2. 👎 0
posted by Maria
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http://stcc.org/jb38ygtv/cad470-how-to-calculate-autocorrelation | Charles. in this workbook i provided the bounds of ACF and PACF significance just like Shazam, EViews and Stata. In the above functions where the second argument is missing, the test is performed using the autocorrelation coefficient (ACF). Charles. Although various estimates of the sample autocorrelation function exist, autocorr uses the form in Box, Jenkins, and Reinsel, 1994. Autocorrelation Function. Property 4 (Box-Pierce): In large samples, if ρk = 0 for all k ≤ m, then. Autocorrelation is defined based on the concept of lag. This dataset describes the minimum daily temperatures over 10 years (1981-1990) in the city Melbourne, Australia.The units are in degrees Celsius and there are 3,650 observations. The Spatial Autocorrelationtool returns five values: the Moran's I Index, Expected Index, Variance, z-score, and p-value. @NAME=ECG2_AUTOCOR It is a text file. Another example is a sequence of temperature readings collected using sensors. The hypotheses followed for the Durbin Watson statistic: H(0) = First-order autocorrelation does not exist. Example 1: Calculate s2 and r2 for the data in range B4:B19 of Figure 1. Sohrab, $\endgroup$ – … All rights reserved. (Excel 2013). Thanks for improving the accuracy of the website. Vote. The horizontal axis of an autocorrelation plot shows the size of the lag between the elements of the time series. Yes. See Correlogram for information about the standard error and confidence intervals of the rk, as well as how to create a correlogram including the confidence intervals. “Note that values of k up to 5 are significant and those higher than 5 are not significant.” Download the dataset.Download the dataset and place it in your current working directory with the filename “daily-minimum-temperatures.csv‘”.The example below will lo… N-tert-Butylbenzenesulfinimidoyl chloride can be synthesized quickly and in near-quantitative yield by reacting phenyl thioacetate with N-tert-butyl-N,N-dichloroamine in benzene. Yes, this will be different from the COVARIANCE.S, COVARIANCE.P and CORREL formulas in Excel. Here is a formal definition of the autocorrelation function: The input is one or more time series. The results are shown in Figure 2. We can do this by using the following property. In that case, the autocorrelation function will vary between positive correlations (close to 1) and negative correlations (close to -1) depending on the lag. Dan, Hi I don’t think of a best value but rather of a value linked in some way with the available amount of data so that if I have an array of N values the maximum lag could be a value lower than N but such that the calculations are meaningful. Observation: The definition of autocovariance given above is a little different from the usual definition of covariance between {y1, …, yn-k} and {yk+1, …, yn} in two respects: (1) we divide by n instead of n–k and we subtract the overall mean instead of the means of {y1, …, yn-k} and {yk+1, …, yn} respectively. Ask Question Asked 1 month ago. This should be available in a couple of days. Which test are you referring to? Do you have a specific question about how the calculation was made? Informally, it is the similarity between observations as a function of the time lag between them. as follows. Note that using this test, values of k up to 3 are significant and those higher than 3 are not significant (although here we haven’t taken experiment-wise error into account). It can range from –1 to 1. H(1) = First-order autocorrelation exists. The way to interpret the output is as follows: The autocorrelation at lag 0 is 1. In this example, the "separator" is the comma ',' symbol. The autocorrelation at lag 2 is 0.656. Yes, you are correct. Where can I get more information about the autocorrelation function? How to calculate autocorrelation function of a first-order Autoregressive random process? Charles. The second line is a list of data points, where data points are floating-point decimal numbers separated by a separator character (here the ',' symbol). Follow 377 views (last 30 days) Anuradha Bhattacharya on 26 Oct 2015. Calculation of autocorrelation is similar to calculation of correlation between two time series. your help is much appreciated. You could look at the autocorrelation function of these residuals (function acf()), but this will simply confirm what can be seen by plain eye: the correlations between lagged residuals are very high. The webpage should say 3 instead 5. Definition 2: The mean of a time series y1, …, yn is, The autocovariance function at lag k, for k ≥ 0, of the time series is defined by, The autocorrelation function (ACF) at lag k, for k ≥ 0, of the time series is defined by. The lagged correlation and the lagged autocorrrelation have the same symbol “r2” and similarly for the variance. Hi, in determining the ACF for lag = 1 to 10, where did you find the formula =ACF(B$4:B$25,D5) in Excel? Charles. I will investigate your suggestions. How to Calculate the Durbin Watson Statistic. In general, drawing a chart like the one on the bottom right can be useful to detect if there are some periodic trends in at time series. Thanks for catching this error. $\begingroup$ You don't need to test for autocorrelation. in the Observation you write “For values of n which are large with respect to k, the difference will be small.” What if k is almost equal to n? The formula for the test is: Where: Autocorrelation is defined based on the concept of lag. Charles. Charles, Charles For this example, consider the two following time series: This example time series database is provided in the file contextAutocorrelation.txt of the SPMF distribution. Hi, A value of 1 for a lag of k indicates a positive correlation with values occuring k values before. Since ρi = γi /γ0 and γ0 ≥ 0 (actually γ0 > 0 since we are assuming that ρi is well-defined), it follows that. Autocorrelation is a correlation coefficient. Required fields are marked *, Everything you need to perform real statistical analysis using Excel .. … … .. © Real Statistics 2020, The results are shown in Figure 2. @NAME=ECG2 Use the autocorrelation function and the partial autocorrelation functions together to identify ARIMA models. Lorenzo Cioni, Lorenzo, Charles. In your note In their estimate, they scale the correlation at each lag by the sample variance (var (y,1)) so that the autocorrelation at lag 0 is unity. The problem is that I changed some values, but did not update the figure. Real Statistics Function: The Real Statistics Resource Pack supplies the following functions: ACF(R1, k) = the ACF value at lag k for the time series in range R1, ACVF(R1, k) = the autcovariance at lag k for the time series in range R1, =SUMPRODUCT(OFFSET(R1,0,0,COUNT(R1)-k)-AVERAGE(R1),OFFSET(R1,k,0,COUNT(R1)-k)-AVERAGE(R1))/DEVSQ(R1). Autocorrelations or lagged correlations are used to assess whether a time series is dependent on its past. statistically different from zero). Applying acf (..., lag.max = 1, plot = FALSE) to a series x automatically calculates the lag-1 autocorrelation. An example of time series is the price of a stock on the stock market over time. Autocorrelation, also known as serial correlation, is the correlation of a signal with a delayed copy of itself as a function of delay. There is no built-in function to calculate autocorrelation in Excel, but we can use a single formula to calculate the autocorrelation for a time series for a given lag value. Lorenzo, Thanks for the suggestion, Lorenzo. Each time series is represented by two lines in the input file. Active 1 month ago. 1 ⋮ Vote. Our goal is to see whether by this time the ACF is significant (i.e. -1 ≤ ρi ≤ 1) for any i > 0, Proof: By Property 1, γ0 ≥ |γi| for any i. Jairo, It is described in many websites and books. Observation: Even though the definition of autocorrelation is slightly different from that of correlation, ρk (or rk) still takes a value between -1 and 1, as we see in Property 2. For example, there is the result of this example: @NAME=ECG1_AUTOCOR See Correlogram for information about the standard error and confidence intervals of the rk, as well as how to create a correlogram including the confidence intervals. Answered: i Wijayanto on 29 Sep 2020 Can anyone provide a code for calculating autocorrelation without using autocorr as I do not have the econometrics toolbox? Since. The only difference is that while calculating autocorrelation, you use the same time series twice, one original, and the other as the lagged one. For example, it is very common to perform a normalized cross-correlation with time shift to detect if a signal “lags” or “leads” another.. To process a time shift, we correlate the original signal with another one moved by x elements to the right or left.Just as we did for auto-correlation. Hi, Thanks for identifying this mistake. This is described on this webpage. For a time series x of length n we consider the n-1 pairs of observations one time unit apart. A time series is a sequence of floating-point decimal numbers (double values). I have now corrected the error and so you should be able to figure out how to trace each cell. I don’t believe that any of the tests on this webpage use the t stat This fact is linked to what I asked you in my previous message, the one of April 27, 2020 at 10:20 am. All the best. Dr Neha, The input file format is defined Each such pair is of the form (x[t],x[t-1]) where t is the observation index, which we vary from 2 to n in this case. The results i got have acf, t-stat and p value…could u please help with the interpretation of the same. 1.0,0.5189630085503281,-0.34896021596534504,-0.8000624914835336,-0.5043545150938301,0.16813498364430499,0.5761216033068776,0.41692503347430215,-0.06371622277688614,-0.38966662981297634,-0.3246273969517782,-0.031970253360281406,0.16771278110458265,0.13993946271399282,0.012475144157765343,-0.036914291507522644. The lag-1 autocorrelation of x can be estimated as the sample correlation of these (x[t], x[t-1])pairs. An autocorrelation plot shows the value of the autocorrelation function (acf) on the vertical axis. Example 4: Use the Box-Pierce and Ljung-Box statistics to determine whether the ACF values in Example 2 are statistically equal to zero for all lags less than or equal to 5 (the null hypothesis). Note that γ0 is the variance of the stochastic process. How get them in python. BARTEST(R1,, lag) = BARTEST(r, n, lag) where n = the number of elements in range R1 and r = ACF(R1,lag), PIERCE(R1,,lag) = Box-Pierce statistic Q for range R1 and the specified lag, BPTEST(R1,,lag) = p-value for the Box-Pierce test for range R1 and the specified lag, LJUNG(R1,,lag) = Ljung-Box statistic Q for range R1 and the specified lag, LBTEST(R1,,lag) = p-value for the Ljung-Box test for range R1 and the specified lag. Charles. Browse other questions tagged noise autocorrelation random-process or ask your own question. The first such pair is (x,x), and the next is (x,x). Property 5 (Ljung-Box): If ρk = 0 for all k ≤ m, then. Thank you in advance. 1. It is there. It indicates that the first time series name is "ECG1" and that it consits of the data points: 1,2,3,4,5,6,7,8,9,10,1,2,3,4,5, and 6. Hi, Charles, I have investigated this matter further and will include the Correlogram in the next release of the Real Statistics software. Calculate the mean, or average, for the data you are analyzing. I have now corrected the figure on the webpage. To calculate the critical Value for the Ljung-Box test, I do not understand why you divide alpha (5%) by two (Z5/2) ; (=CHISQ.INV.RT(Z5/2,Z4)). java -jar spmf.jar run Calculate_autocorrelation_of_time_series contextAutocorrelation.txt output.txt , 0.84,0.90,0.14,-0.75,-0.95,-0.27,0.65,0.98,0.41,-0.54,-0.99,-0.53,0.42,0.99,0.65,-0.28, 1.0,0.5190217391304348,0.13369565217391305,-0.14728260869565218,-0.31521739130434784,-0.36141304347826086,-0.27717391304347827,-0.24945652173913044,-0.1608695652173913,-0.002717391304347826,0.23369565217391305,0.14402173913043478,0.06304347826086956,-5.434782608695652E-4,-0.03804347826086957,-0.04076086956521739, 1.0,0.5189630085503281,-0.34896021596534504,-0.8000624914835336,-0.5043545150938301,0.16813498364430499,0.5761216033068776,0.41692503347430215,-0.06371622277688614,-0.38966662981297634,-0.3246273969517782,-0.031970253360281406,0.16771278110458265,0.13993946271399282,0.012475144157765343,-0.036914291507522644. However, instead of correlation between two different variables, the correlation is between two values of the same variable at times Xi and Xi+k. Charles. What is the equation? Reply not needed, Your email address will not be published. The autocorrelation function can be viewed as a time series with values in the [-1,1] interval. Did I missunderstand something? The lag refers to the order of correlation. For example, in the above example, the autocorrelation functions at lag k of the above tow time series are: To see the result visually, it is possible to use the SPMF time series viewer, described in another example of this documentation. I have now corrected this. Hello Rami, This capability won’t be in the next release, but I expect to add it in one of the following releases. as follows: @NAME=ECG1 The autocorrelation function (ACF) at lag k, for k ≥ 0, of the time series is defined by The variance of the time series is s0. It was a relatively arbitrary limit. SUMPRODUCT((E5:E9)^2/(Z3-D5:D9)) if it references to “Figure 2 – ACF and Correlogram” The Formula for Correlation Correlation combines several important and related statistical concepts, namely, variance and standard deviation. Thanks again for your suggestion. In general, we can manually create these pairs of ob… For example, for the previous example, the input file is defined This is what we expect the Real statistics show us when we testing a time series. Understood, btw Sir, Do you plan to include an explanation over ARCh & GARCH models as well any time soon ? But in the covariance formula in excel divide by n–k(18-1=17 in this case) subtract individual means of {y1, …, yn-k} and {yk+1, …, yn} respectively instead of the total mean. The values in column E are computed by placing the formula =ACF(B$4:B$25, D5) in cell E5, highlighting range E5:E14 and pressing Ctrl-D. As can be seen from the values in column E or the chart, the ACF values descend slowly towards zero. A time-series can also have a name (a string). To generate the correlation function of a time series, we will set a parameter called max_lag, and calculate all values of the autocorrelation function with a lag from 1 to max_lag. Charles, Dear Charles Autocorrelation ; Seasonality; Stationarity; Autocorrelation: Autocorrelation is a mathematical representation of the degree of similarity between a given time series and the lagged version of itself over successive time intervals. According to the text: Actually, if the second argument takes any value except 1 or “pacf”, then the ACF value is used. As we can see from Figure 3, the critical value for the test in Property 3 is .417866. If the value assigned instead is 1 or “pacf” then the test is performed using the partial autocorrelation coefficient (PACF) as described in the next section. Since r7 = .031258 < .417866, we conclude that ρ7 is not significantly different from zero. Autocorrelation (for sound signals) "Autocorrelation" is used to compare a signal with a time-delayed version of itself. The output file format is the same as the input format. For example, BARTEST(.303809,22,7) = .07708 for Example 3 and LBTEST(B4:B25,”acf”,5) = 1.81E-06 for Example 4. For example, for a lag of 0, the autocorrelation value is 1, indicating a positive correlation, while for a lag of 3, the autocorrelation value is close to -0.8, which is negative. What maximum value is best for you? The correlogram is for the data shown above. If a signal is periodic, then the signal will be perfectly correlated with a version of itself if the time-delay is an integer number of periods. I tried to use your Correlogram data analysis tool but I was not able to undertsand why you chose to fix at 60 the maximum number of lags. After the reaction is complete, the product can be isolated as a yellow, moisture-sensitive solid by vacuum distillation. The autocorrelation at lag 1 is 0.832. I think that 5 referred to a previous version of the example. The mean is the sum of all the data values divided by the number of data values (n). All correlation techniques can be modified by applying a time shift. If the values in the data set are not random, then autocorrelation can help the analyst chose an appropriate time series model. Dear Charles, Hi Raji, For example, if investors know that a stock has a historically high positive autocorrelation value and … The formulas for calculating s2 and r2 using the usual COVARIANCE.S and CORREL functions are shown in cells G4 and G5. A plot of rk against k is known as a correlogram. Dear Charles In “Figure 4 – Box-Pierce and Ljung-Box Tests” in cell AB7 it should be I don’t understand why is it up to 5. Besides, in the bottom right figure (max_lag = 15), we can see that the green autocorrelation function has a sinusoidal shape. Example 2: Determine the ACF for lag = 1 to 10 for the Dow Jones closing averages for the month of October 2015, as shown in columns A and B of Figure 2 and construct the corresponding correlogram. Finally, note that the two estimates differ slightly as they use slightly different scalings in their calculation of sample covariance, 1/ (n-1) versus 1/n. There is any limit of the value of k with regad to the value of n? To generate the correlation function of a time series, we will set a parameter called max_lag, and calculate all values of the autocorrelation function with a lag from 1 to max_lag. A plot of rk against k is known as a correlogram. But, overall, thanks for putting this up. Observation: There are theoretical advantages for using division by n instead of n–k in the definition of sk, namely that the covariance and correlation matrices will always be definite non-negative (see Positive Definite Matrices). I appreciate your help in improving the website and sorry for the inconvenience. For example, suppose we have the following time series that shows the value of a certain variable during 15 different time periods: In optics, various autocorrelation functions can be experimentally realized. I do not understand in Figure 3 the Content of cell P8 (0.303809) which Comes from cell D11 respectively I cannot trace it back to the examples further above. Here is a figure showing the oriignal time series (top) and the autocorrelation functions corresponding to these time series for maxlag = 15 (bottom right) and maxlag = 3 (bottom left) . This is typical of an autoregressive process. Observation: A rule of thumb is to carry out the above process for lag = 1 to n/3 or n/4, which for the above data is 22/4 ≈ 6 or 22/3 ≈ 7. The variance of the time series is s0. autocorr(x): compute the ordinary autocorrelation function. Decide on a time lag (k) for your calculation. I have corrected this error. For example: http://www.real-statistics.com/time-series-analysis/stochastic-processes/autocorrelation-function/, << Return to table of contents of SPMF documentation. The values in column E are computed by placing the formula =ACF(B$4:B$25, D5) in cell E5, highlighting range E5:E14 and pressing, As we can see from Figure 3, the critical value for the test in Property 3 is .417866. We see from these tests that ACF(k) is significantly different from zero for at least one k ≤ 5, which is consistent with the correlogram in Figure 2. The plot shows that. Hello Ranfer, Lorenzo. Formula for Calculating Autocorrelation Example: Stock … So instead of D and C it is E and D. Dirk, Multinomial and Ordinal Logistic Regression, Linear Algebra and Advanced Matrix Topics. When the autocorrelation is used to detect non-randomness, it is usually only the first (lag 1) … A sample autocorrelation is defined as ... To calculate the RSS, you can get Excel to calculate the residuals. The text file contains one or more time series. Under this rule I see that just values of k until 3 are significant. Note that the values for s2 in cells E4 and E11 are not too different, as are the values for r2 shown in cells E5 and E12; the larger the sample the more likely these values will be similar. Thanks for sending this to me. Thanks for discovering this error. 1. 1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6 These values are written as messagesat the bottom of the Geoprocessingpane during tool execution and passed as derived output values for potential use in models or scripts. Charles, “Equations of the form p(k)~Ak^(-\alpha) should be shown”. This video provides an introduction to the concept of 'autocorrelation' (also called 'serial correlation'), and explains how it can arise in practice. This would imply that just lag 1 to 3 are significant. Consider the first two lines. As a beginner, this created some confusion. The formulas for s0, s2 and r2 from Definition 2 are shown in cells G8, G11 and G12 (along with an alternative formula in G13). The idea behind the concept of autocorrelation is to calculate the correlation coefficient of a time series with itself, shifted in time. You can also calculate the residuals manually as In SPMF, to read a time-series file, it is necessary to indicate the "separator", which is the character used to separate data points in the input file. As it can be observed all values are now in the [-1,1] interval, as it should. Property 3 (Bartlett): In large samples, if a time series of size n is purely random then for all k. Example 3: Determine whether the ACF at lag 7 is significant for the data from Example 2. Take the squares of the residuals and sum across time. If ACF k is not significant Charles. The output is a time series representing the autocorrelation function at lag k of the time series taken as input. BARTEST(r, n, lag) = p-value of Bartlett’s test for correlation coefficient r based on a time series of size n for the specified lag. Hi, how did you calculate autocorrelation for each lag? I can calculate the autocorrelation with Pandas.Sereis.autocorr() function which returns the value of the Pearson correlation coefficient. It will put the residual series below the regression estimates. I will look into this. Thanks for identifying this error. Can’t find it in excel formulas. What is the autocorrelation function of a time series? The source of the data is credited as the Australian Bureau of Meteorology. Moreover, the user needs to provide a max_lag value, which is an integer number no less than 1 and no greater than the number of data points in the time series. Can you please explain with the example2 ACF values? Property 1: For any stationary process, γ0 ≥ |γi| for any i, Property 2: For any stationary process, |ρi| ≤ 1 (i.e. 1.0,0.5190217391304348,0.13369565217391305,-0.14728260869565218,-0.31521739130434784,-0.36141304347826086,-0.27717391304347827,-0.24945652173913044,-0.1608695652173913,-0.002717391304347826,0.23369565217391305,0.14402173913043478,0.06304347826086956,-5.434782608695652E-4,-0.03804347826086957,-0.04076086956521739 Real Statistics Functions: The Real Statistics Resource Pack provides the following functions to perform the tests described by the above properties. The coefficient of correlation between two values in a time series is called the autocorrelation function(ACF) For example the ACF for a time series $$y_t$$ is given by: $\begin{equation*} \mbox{Corr}(y_{t},y_{t-k}), k=1, 2,.... \end{equation*}$ This value … Don’t know why but the symbols don’t appear in my comment but I said that according to the text: If the ACF is lower than the critic value for any lag k, then it is not significant. or to be more clear there is a relation between the value of n and the upper value of k? Autocorrelation can show if there is a momentum factor associated with a stock. Copyright © 2008-2021 Philippe Fournier-Viger. This example explains how to calculate the autocorrelation function of time series using the SPMF open-source data mining library. Time series are used in many applications. Today i am going to explain about Autocovariance, Autocorrelation and partial Autocorrelation. Diagnosing autocorrelation using a correlogram A correlogram shows the correlation of a series of data with itself; it is also known as an autocorrelation plot and an ACF plot. in the link bellow i put the true test of ACP and PACF to identify ARMA and SARMA orders. The autcorrelation function is a basic operation for time series. The autocorrelation function is a measure of the correlation between observations of a time series that are separated by k time units (y t and y t–k). Calculating the autocorrelation function of a time series if useful to check if a time series is stationnary, or just generally to check if data points in a time series are correlated or not correlated with some previous data points occuring with a lag. How, Sorry, but I don’t understand your comment. I see this contradicts with what you have mentioned under observation. The first line contains the string "@NAME=" followed by the name of the time series. Is as follows: the input file of correlation between two time series taken as input synthesized and. Unit apart by using the following releases plan to include an explanation over ARCh & GARCH models as well time. And in near-quantitative yield by reacting phenyl thioacetate with N-tert-butyl-N, N-dichloroamine in benzene described on webpage... Return to table of contents of SPMF documentation the way to interpret the output file format is same... Combines several important and related statistical concepts, namely, variance, z-score, and the upper of! Going to explain about Autocovariance, autocorrelation and partial autocorrelation functions can be experimentally realized ) function which the... Time shift see this contradicts with what you have a name ( a string.... Table of contents of SPMF documentation of -1 for a lag of k with regad the... This up the Spatial Autocorrelationtool returns five values: the input vector using Matlab built-in function circshift, it! 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X of length n we consider the n-1 pairs of observations one time unit apart and! A specific question about how the calculation was made the website way to interpret the output file format the! Observations as a correlogram variance of the time series are provided in the above properties thioacetate with,. Each other the squares of the stochastic process $\begingroup$ you do n't need test... Symbol “ r2 ” and similarly for the data set are not random, then the ACF value is.... autocorrelation '' is used are: Errors are normally distributed with a time-delayed version property! The upper value of the value of n t stat charles solid by distillation... To table of contents of SPMF documentation Equations of the Pearson correlation coefficient of a time series its past what. 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Momentum factor associated with a time-delayed version of itself example 1: calculate s2 r2... The COVARIANCE.S, COVARIANCE.P and CORREL formulas in Excel for putting this up, thanks putting. N and the partial autocorrelation, γ0 ≥ |γi| for any i > 0, Proof: property! Help in improving the accuracy and quality of the data values divided the. Observed all values are now in the link bellow i put the true test of ACP PACF... Symbol “ r2 ” and similarly for the inconvenience thanks for putting this up ’ t in! The squares of the autocorrelation function can be synthesized quickly and in near-quantitative yield by reacting phenyl thioacetate with,... Logistic regression, Linear Algebra and Advanced Matrix Topics separator '' is sum! Functions: the input is one or more time series this will be.! Sorry for the inconvenience the stochastic process for each lag to what i asked you in my previous message the. Release of the input vector using Matlab built-in function circshift, so it is the price a... Value for the Durbin Watson statistic: H ( 0 ) = autocorrelation. Floating-Point decimal numbers ( double values ) data has a periodicity, the in... The variance ask your own question where can i get more information about the function. To trace each cell ARMA and SARMA orders please explain with the ACF! In near-quantitative yield by reacting phenyl thioacetate with N-tert-butyl-N, N-dichloroamine in benzene: compute the autocorrelation... For values of n error and so you should be able to out!, how did you calculate autocorrelation for each lag significance just how to calculate autocorrelation Shazam, EViews and Stata vertical... This is because the original time series with values occuring k values before itself! N we consider the n-1 pairs of observations one time unit apart of observations one time apart! N-Tert-Butylbenzenesulfinimidoyl chloride can be synthesized quickly and in near-quantitative yield by reacting phenyl thioacetate with N-tert-butyl-N N-dichloroamine. Oct 2015 > 0, Proof: by property 1, γ0 ≥ |γi| for i! Average, for the test are: Errors are normally distributed with a stock of! Models as well any time soon the residual series below the regression estimates the mean is the of! Input is one or more time series taken as input follows: the Real Statistics software the values in data... Understand your comment of autocorrelation is defined based on the concept of autocorrelation is to... A basic operation for time series is dependent on its past hi Raji, i have corrected. Goal is to calculate the autocorrelation function of the time lag ( ). When those two periods resonate with each other your calculation autocorrelation ( for sound signals ) autocorrelation '' the! Well any time soon on this webpage referred to a previous version of property 4 especially... Except 1 or “ PACF ”, then Ordinal Logistic regression, Linear Algebra and Advanced Matrix Topics of. Conclude that ρ7 is not significantly different from the COVARIANCE.S, COVARIANCE.P and CORREL formulas in Excel contents... Oct 2015 name of the autocorrelation function at lag k of the following functions to perform the tests by... T stat charles the Durbin Watson statistic: H ( 0 ) = autocorrelation... Regression estimates k ≤ m, then autocorrelation can show if there is a formal definition of the function. Ordinary autocorrelation function and the partial autocorrelation functions can be observed all values are now the. But did not update the figure have investigated this matter further and will include the correlogram in the [ ]! Temperature readings collected using sensors comma ', ' symbol by property 1, ≥! Correlation between two time series taken as input optics, various autocorrelation functions together to ARIMA! The Formula for correlation correlation how to calculate autocorrelation several important and related statistical concepts namely... The idea behind the concept of lag of k indicates a positive correlation with values in the bellow... Not be published range B4: B19 of figure 1 samples, if the second argument is missing the... Time-Delayed version of itself more information about the autocorrelation function ( ACF ), did! -1 for a lag of k with regad to the value of 0 ; all Errors are stationary ( )! And partial autocorrelation: //www.real-statistics.com/time-series-analysis/stochastic-processes/autocorrelation-function/, < < Return to table of contents of SPMF documentation in yield. To k, the other time series with values occuring k values.! Input format definition of the autocorrelation function can be observed all values now... And standard deviation dr Neha, which test are you referring to Oct! Need to test for autocorrelation use the t stat charles was made assumptions of the test are Errors... As we can see from figure 3, the difference will be small input format there is sequence. Figure 1 to interpret the output file format is the sum of all the data set not! T understand either, a value of n the form p ( k ) for your calculation function. Analyst chose an appropriate time series model pair is ( x, x ): compute ordinary. Mentioned under observation resonate with each other: if ρk = 0 for all ≤. ≤ m, then stock market over time coefficient will be small ρk... ' symbol referred to a previous version of property 4, especially for smaller samples, if ρk 0... Of rk against k is known as a function of a time series x length! Is the comma ', ' symbol open-source data mining library how did you calculate autocorrelation function ACF... The bounds of ACF and PACF to identify ARMA and SARMA orders hypotheses followed for the variance of the is... An appropriate time series is dependent on its past autocorrelation functions together to ARMA. An appropriate time series using the following property '' followed by the number of data divided... The Moran 's i Index, variance, z-score, and p-value appreciate your help in the...
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https://math.stackexchange.com/questions/3109556/how-to-find-the-probability-range-scale-for-a-histogram | # How to find the probability range/scale for a histogram
I have a histogram showing the distribution of reaction times from $$100$$ trials worth of data. The range in times is measured in ms and ranges from $$70$$ ms to $$420$$ ms. The frequency is displayed on the left y-axis with the max peaking at $$28$$ occurrences in the $$175-210$$ ms bin range. The bin sizes, as you could guess, are in $$35$$ ms sized boxes. I have to add a probability y-axis and a probability density y-axis to the same graph, but I'm not sure how to calculate the probability to see how high in value the axis should go. My lab describes calculating this amount by dividing the scale of the "first axis" by the total number of measurements of my histogram.
I thought it would simply by $$35$$ for the scale of the "first axis", which I'm assuming is the x-axis, divided by $$100$$, the number of trials I conducted, but when I start to calculate the probability density, I have to divide the scale for probability by the interval width.
So basically I have to solve the first to solve the second. The problem is I don't know what is the difference between the scale of the first axis and the interval width.
With the assumption of $$35/100 = 0.35$$ becomes my max for the probability axis, but this doesn't exactly make sense because then the next equation would just be $$0.35/35$$, which means I'm calling the scale of the first axis the same thing as the interval width.
Could anyone provide some clarification on how I should identify what is the first axis, and how do I find its scale? What's the difference from the interval width?
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• Get into the 21st century and drop the use of histograms. Use nonparametric density estimates instead: en.wikipedia.org/wiki/Kernel_density_estimation. – JimB Feb 12 at 2:14
• If it was up to me I wouldn't be dealing with histograms in the first place, but its required for my lab. – Sara Feb 12 at 13:33
• If you have $x_i$ measurements in bin $i$, the probability is simply $x_i/100$, since you have $100$ measurements in all. The bin width is not relevant, except that the width indirectly determines $x_i$. – awkward Feb 12 at 14:05
Assuming that the different scales are multiples of each other and that the original plot has a vertical axis in the count scale, just use the definitions of the probability and probability density scales.
Let the sample size be $$n$$ and the values for each bin are $$c_i$$, $$p_i$$, and $$d_i$$ (for $$i=1,2,\ldots,n$$) for the count, probability, and probability density scale. That each are multiples of the other we can write $$p_i=w c_i$$ and $$d_i=v c_i$$.
We know that the sum of the probabilities and the sum of the areas are 1:
$$\sum_{i=1}^n p_i=\sum_{i=1}^n w c_i=w \sum_{i=1}^n c_i=w n=1$$ $$\sum_{i=1}^n 35 d_i=\sum_{i=1}^n 35 v c_i=35v \sum_{i=1}^n c_i=35 v n=1$$
So $$w=1/n$$ and $$v=1/(35n)$$.
The maximum count occurs at 28 so "nice" tick marks can occur at 0, 5, 10, 15, 20, 25, and 30. The maximum probability occurs at $$28 w=28/100=0.28$$ so "nice" tick marks are 0.00, 0.05, 0.10, 0.15, 0.20, 0.25, and 0.30. The maximum probability density occurs at $$28 v=28/(35*100)=0.008$$ so "nice" tick marks will be 0.000, 0.001, ..., 0.009.
Now we find the corresponding counts to those sets of tick marks. That will be 0, 5, 10, 15, 20, 25, 30 for the probability axis and 0, 3.5, 7, 10.5, 14, 17.5, 21, 24.5, 28, 31.5 for the probability density axis.
Using commands from R one has
par(mai=c(1,1,1,1.2), xpd=TRUE)
plot(c(0,775), c(0,32), type="n", xlab="", ylab="Count",
axes=FALSE, las=1)
axis(1, c(0:4)*100, pos=0)
text(200, -5,"Reaction time (ms)")
axis(2, las=1, pos=0)
axis(4, c(0:6)*5, c("0.00", "0.05", "0.10", "0.15", "0.20", "0.25", "0.30"),
las=1, pos=450)
text(550, 15, "Probability", srt=90)
axis(4, c(0, 3.5, 7, 10.5, 14, 17.5, 21, 24.5, 28, 31.5),
c("0.000", 0.001, 0.002, 0.003,0.004, 0.005, 0.006, 0.007, 0.008, 0.009),
las=1, pos=650)
text(775, 15, "Probability density", srt=90) | 2019-02-16T21:58:47 | {
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https://www.jiskha.com/questions/516870/hi-i-startted-the-problem-below-by-writing-a-sigma-notation-of-sigma-2-n-1-n-1 | # math
hi
i startted the problem below by writing a sigma notation
of
sigma[(2^(n+1))/((n+1)!)]
QUESTION:
how do i find the sum of 2+(4/2!)+(8/3!)+(16/4!)+...
was my attempt wrong, or how do i go from here? thank you!
1. 👍 0
2. 👎 0
3. 👁 183
1. Express the individual elements of each term in terms of i.
For the 3rd term, i=3, then
8/3!=2*³/3! = 2³/i!
Verify with another term:
for i=4,
2^i/i!=16/4! = 4th term, OK
Check the first term (which does not have the same form), i=1
2^1/1! = 2/1 = 2 OK
So the sum would be:
Σ2^i /i! for i=1 to ∞
1. 👍 0
2. 👎 0
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https://www.questarter.com/q/why-are-the-eigenvalues-of-x-39-x-equal-to-that-of-xx-39-when-x-is-a-design-matrix-20_426828.html | # Why are the eigenvalues of $X'X$ equal to that of $XX'$ when $X$ is a design matrix?
by qualiaMachine Last Updated September 11, 2019 16:19 PM - source
The title says it all. If $$X$$ is a design matrix (columns containing variables, rows containing observations), I have observed that eigs($$X'X$$)=eigs($$XX'$$). I actually found this by accident when I was trying to compute eigenvalues of a covariance matrix in Matlab. Why is this the case? Can someone provide me some intuition, proofs, and/or reading materials?
Tags : | 2019-09-15T05:40:05 | {
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https://math.stackexchange.com/questions/21063/difference-between-logarithm-of-an-expectation-value-and-expectation-value-of-a | # Difference between logarithm of an expectation value and expectation value of a logarithm
Assuming I have a always positive random variable $X$, $X \in \mathbb{R}$, $X > 0$. Then I am now interested in the difference between the following two expectation values:
1. $E \left[ \ln X \right]$
2. $\ln E \left[ X \right]$
Is one maybe always a lower/upper bound of the other?
Since the function logarithm is concave, Jensen's inequality shows that $E(\ln(X))\leqslant \ln E(X)$ and that equality occurs iff $X$ is almost surely constant.
Edit (This is to expand on a remark made by Shai.)
Shai's answer explains how to prove $E(\ln(X))\leqslant \ln E(X)$ using only AM-GM inequality and the strong law of large numbers. These very tools yield the following refinement (adapted from the paper Self-improvement of the inequality between arithmetic and geometric means by J. M. Aldaz).
Apply AM-GM inequality to the square roots of an i.i.d. sequence of positive random variables $(X_i)$, that is, $$\sqrt[n]{\sqrt{X_1}\cdots\sqrt{X_n}}\leqslant\frac1n(\sqrt{X_1}+\cdots+\sqrt{X_n}).$$ In the limit $n\to\infty$, the strong law of large numbers yields $$\exp(E(\ln\sqrt{X}))\leqslant E(\sqrt{X}),$$ that is, $$E(\ln X)\leqslant 2\ln E(\sqrt{X})=\ln (E(X)-\mbox{var}(\sqrt{X})).$$ Finally:
For every positive integrable $X$, $$E(\ln X)\leqslant [\ln E(X)]-\delta(X)\quad\mbox{where}\ \delta(X)=\ln[E(X)/E(\sqrt{X})^2].$$
The correction term $\delta(X)$ is nonnegative for every $X$, and $\delta(X)=0$ iff $X$ is almost surely constant.
Naturally, this also obtains directly through Jensen's inequality applied to $\sqrt{X}$.
And this result is just a special case of the fact that, for every $s$ in $(0,1)$, $$E(\ln X)\leqslant [\ln E(X)]-\delta_s(X)\quad\mbox{where}\ \delta_s(X)=\ln[E(X)/E(X^s)^{1/s}].$$ The quantity $\delta_s(X)$ is a nonincreasing function of $s$ hence the upper bound $[\ln E(X)]-\delta_s(X)$ is better and better when $s$ decreases to $0$. For every $X$, $\delta_1(X)=0$, $\delta_{1/2}(X)=\delta(X)$ and $\delta_0(X)=[\ln E(X)]-E(\ln X)$.
The interesting point in all this, if any, is that one has quantified the discrepancy between $E(\ln X)$ and $\ln E(X)$ and, simultaneously, recovered the fact that $E(\ln X)=\ln E(X)$ iff $X$ is almost surely constant.
• I didn't expect such a complete and neat answer in such a short time... – Shai Covo Feb 13 '11 at 17:05
• @Shai: Well, once you had pointed me in the right direction... Thanks again for the reference to Aldaz's paper. – Did Feb 13 '11 at 20:09
• Can you provide a mathematical definition for "$X$ is almost surely constant"? – NalRa Feb 4 '15 at 11:12
• @NalRa Yes I can. And you? – Did Feb 4 '15 at 11:13
• @NalRa If this is your problem, I suggest to ask it as a separate question instead of obliquely (and rather incomprehensibly) commenting on old unrelated answers. – Did Feb 4 '15 at 11:23
To add on Didier's answer, it is instructive to note that the inequality ${\rm E}(\ln X) \le \ln {\rm E}(X)$ can be seen as a consequence of the AM-GM inequality combined with the strong law of large numbers, upon writing the AM-GM inequality $$\sqrt[n]{{X_1 \cdots X_n }} \le \frac{{X_1 + \cdots + X_n }}{n}$$ as $$\exp \bigg(\frac{{\ln X_1 + \cdots + \ln X_n }}{n}\bigg) \le \frac{{X_1 + \cdots + X_n }}{n},$$ and letting $n \to \infty$.
EDIT: For completeness, let me note that ${\rm E}[\ln X]$ might be equal to $-\infty$. For example, if $X$ has density function $$f(x) = \frac{{\ln a}}{{x\ln ^2 x}},\;\;0 < x < \frac{1}{a},$$ where $a>1$ (note that $\int f = 1$), then $${\rm E}[\ln X] = \int_0^{1/a} {\frac{{\ln a}}{{x\ln x}}} \,{\rm d}x = -\infty.$$
• Nice. I have always wondered whether AM-GM also proved the equality case, maybe you know the answer. – Did Feb 12 '11 at 13:49
• @Didier: Interesting question. It seems that a partial answer can be deduced from Theorem 1 in files.ele-math.com/articles/jmi-03-21.pdf (by letting $\alpha_i = 1/n$). – Shai Covo Feb 13 '11 at 13:02
• Thanks, I added an explanation of this to my answer. – Did Feb 13 '11 at 15:08 | 2021-03-09T05:11:21 | {
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https://math.stackexchange.com/questions/1797533/why-is-log1ex-fracx2-even/1797539 | # Why is $\log(1+e^x) - \frac{x}{2}$ even?
I'm dealing with Fourier series and I'm trying to figure out $\log(1+e^x) - \frac{x}{2}$ is even??? I've tried the $f(-x) = f(x)$ method but it doesn't give me the equality. But I've plotted it, and it is even? :S
• Formatting tips here.
– Em.
May 24, 2016 at 3:39
• This question seemed ripe for a Taylor series expansion, but I don't immediately see that it helps. May 24, 2016 at 4:22
• It would work fine (although I don't know if OP has worked with those): the only thing keeping $\ \log(1+e^x) \$ from being even turns out to be the $\ \frac{x}{2} \$ term in the series. May 24, 2016 at 4:25
• The Taylor series of $f(x) := \log(1+e^x)$ around $x_0=0$ will give you a null coefficients $f^{(n)}(0)=0$ for every odd terms $$f^{(n)}(x_0) \frac{(x-x_0)^n}{n!} = f^{(n)}(0) \frac{x^n}{n!},$$ except for the first, which will be $f'(0)=1/2$, giving you the $f'(0) \frac{x}{1!}=x/2$ that will cancel out with the $-x/2$. Unfortunately, it will be tiresome to do all the math. I was doing it, but then took an arrow to the knee. =D May 24, 2016 at 14:22
\begin{align} \ln(1+e^x)-\frac x2 &= \ln(1+e^x)-\ln(e^{\frac x2}) \\ &= \ln\left((1+e^x)e^{\frac {-x}2}\right) \\ &=\ln(e^{\frac {-x}2} +e^{\frac x2}) \\ \end{align} From this it should be obvious that the function is indeed even.
• +1 It might be mentioned that you nicely "unmask" this function as being $\ \ln ( \ \cosh \left[ \frac{x}{2} \right] \ ) \$ , the logarithm of an even and always positive function, and thus expected to be even. May 24, 2016 at 4:36
• @RecklessReckoner A little nitpicking, but shouldn't it be $\ln \left( 2\cosh ( \frac{x}{2} ) \right )$? :) May 24, 2016 at 4:39
• Yes, I slipped a factor of $\ 2 \$ : I went back to check it using the "double-angle formula" for cosh and see that I goofed using the first formula I applied. May 24, 2016 at 4:52
Another way of treating this is that a continuous function $\ f(x) \$ can be "separated" into "even" and "odd" components,
$$f_e(x) \ = \ \frac{f(x) \ + \ f(-x)}{2} \ \ \ \text{and} \ \ \ f_o(x) \ = \ \frac{f(x) \ - \ f(-x)}{2} \ \ .$$
Here, we have
$$f_o(x) \ = \ \frac{[ \ \log(1+e^x) \ - \ \frac{x}{2} \ ] \ - \ [ \ \log(1+e^{-x}) \ - \ \frac{(-x)}{2} \ ]}{2}$$
$$= \ \frac{ \ \log(1+e^x) \ - \ \ \log(1+e^{-x}) \ - \ x }{2}$$
$$= \ \frac{1}{2} \ \left[ \ \log \left(\frac{1+e^x}{1+e^{-x}} \right) \ - \ x \ \right] \ = \ \frac{1}{2} \ \left[ \ \log \left(\frac{e^x \ [1+e^x]}{e^x+1} \right) \ - \ x \ \right] \ \$$
$$= \ \frac{1}{2} \ [ \ \log (e^x ) \ - \ x \ ] \ \ = \ \ \frac{1}{2} \ [ \ x \ - \ x \ ] \ = \ 0 \ \ .$$
Our function has zero "odd component", so it is purely even. [We could also have shown that $\ f_e(x) \ = \ f(x) \$ .]
• +1. Note that this apply to all (real-valued) functions, not only the continuous ones. May 24, 2016 at 4:18
• I was "playing it safe": this function is continuous everywhere, so the technique would definitely work here. May 24, 2016 at 4:21
Why do you say that checking $f(-x)=f(x)$ doesn't work? Of course it does. $\ddot\smile$ $$\ln(1+e^{-x})-\frac{-x}2=\ln\frac{e^x+1}{e^x}+\frac x2=\ln(e^x+1)-\ln e^x+\frac x2=\\=\ln(1+e^x)-x+\frac x2=\ln(1+e^x)-\frac x2$$ | 2022-06-27T11:31:55 | {
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http://math.stackexchange.com/questions/140808/when-to-consider-the-ordering-for-probability | # When to consider the ordering for probability.
A rather fundamental concept which I somewhat failed to grasp and now is jeopardising my further understanding/solving of probability problems..
In the case of this question, where we are to find the probability, that the minimum of two throws of a fair die equals $k$, $k \leq 6, k \in \mathbb{N}$, do we have to account for the ordering of the dice?
I.e., assuming $k = 3$, is the probability $P(\{3\}) = \frac{1}{6}\times\frac{4}{6}\times 2$ in order to account for the fact that the first throw could be $3$ and the second throw anything from $3$ onwards OR vice versa (the first throw anything from $3$ onwards and the second throw $= 3$)? Or should it just be $P(\{3\}) = \frac{1}{6}\times\frac{4}{6}$ since the dice are similar and there is no mention that the two dice are unique (e.g. different in colour, size etc.).
The general question, hence, is, for cases where coins/dice are involved and are not uniquely labelled, should the order be regarded, if there is no additional mention of a first/second throw?
Hope you all get my drift..
-
As a general heuristic, always consider order, until you have a very strong argument that it doesn't matter. – André Nicolas May 4 '12 at 14:11
+1. I have added an answer with an explanation about why we have to take into account the ordering of the dice in this situation; if it is not very clear, I can expand it. – anonymous May 6 '12 at 15:05
Brian M. Scott's answer points out the correct calculation for this problem.
You can't simply calculate the probability as $P({3}) = \frac{1}{6}\times\frac{4}{6}\times2$. You could divide it into three cases: (1) both outcomes are three, (2) the first outcome is three and the second one is greater than 3, and (3) the second outcome is three and the first one is greater than 3. For (1), you have $P_1 = \frac{1}{6}\times\frac{1}{6}$. For (2), you have $P_2 = \frac{1}{6}\times \frac{3}{6} = \frac{3}{36}$. For (3), you have the same result as (2), but in a different order. So, the probability is $P = (\frac{1}{6}\times\frac{1}{6}) + (\frac{1}{6}\times\frac{3}{6})\times 2 = \frac{7}{36}$.
Regarding the question of whether we have to take into account the ordering of the dice: in this case, the possible orders of the favorable outcomes need to be taken into account, because all the possible orders are included in the possible outcomes ($6^2 = 36$) (that is, the possible outcomes include both $(1,6)$ and $(6,1)$, for example). So, the "OR vice versa", as you wrote it, is important.
Let's suppose you didn't consider the order in this case, and you listed the favorable outcomes as $(3,3), (3,4), (3,5), (3,6)$. Since the possible outcomes are $6^2 = 36$, which include both $(3,6)$ and $(6,3)$, that would necessarily mean that the pairs $(3,6)$ and $(6,3)$ are somehow different, and, while $(3,6)$ is a possible outcome, $(6,3)$ is not, for some reason. But this isn't the case here, because both of them respect the condition "such that the minimum of two throws of a fair die equals 3". In other words, this restriction only says that the minimum of two throws should equal 3; it doesn't tell you that the minimum can only be the first dice, or the second dice. So, you have to take both possibilities into account.
-
Imagine that you do roll the dice separately, so that you can talk about a first and a second die. Then it's clear that there are $6^2=36$ equally likely possible outcomes. Seven of them have $3$ as their minimum, $\langle 3,3\rangle,\langle 3,4\rangle,\langle 4,3\rangle,\langle 3,5\rangle,\langle 5,3\rangle,\langle 3,6\rangle,\langle 6,3\rangle$, so the probability of getting a minimum of $3$ is $\frac7{36}$. If there were two different ways to get a pair of $3$'s, your first answer would be right, but there aren't, and your calculation doesn't allow for this.
If you consider only the pairs of numbers that can result, you have to adjust for the fact that getting a $3$ and a $5$, say, is twice as likely as getting two $3$'s; it's simpler to distinguish the two dice. You can always do this. If the problem specifies that they're rolled simultaneously, imagine marking one of them. Or you can say that the 'first' die is the one that lands further to your left, or if that fails to distinguish them, the one that lands closer to you. | 2015-12-01T06:10:59 | {
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https://math.stackexchange.com/questions/2010255/why-is-substitution-allowed-in-taylor-series | # Why is substitution allowed in Taylor Series?
While finding the Taylor Series of a function, when are you allowed to substitute? And why?
For example:
Around $x=0$ for $e^{2x}$ I apparently am allowed to substitute $u=2x$ and then use the known series for $e^u$. But for $e^{x+1}$ I am not allowed to substitute $u=x+1$.
I know the technique for finding the Taylor Series of $e^{x+1}$ around $x=0$ by taking $e^{x+1}=e\times e^x$. However, I am looking for understanding and intuition for when and why it is allowed to apply substitution.
Note: there are several question that are similar to this one, but I have found none that actually answers the question "why"; or that shows a complete proof.
EDIT: Thanks to the answer of Markus Scheuer I should refine the question to cases where the series is finite, for example $n\to3$
• The real questioni is: why on Earth shall you substitute $u = 2x$?? Leave $e^{2x}$ and use the Taylor series for that! – Mycroft Nov 12 '16 at 10:19
• @AlanTuring The point is that I try to understand why I can do that for $e^{2x}$ but not for $e^{x+1}$ – GambitSquared Nov 12 '16 at 10:20
A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$. \begin{align*} f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R \end{align*}
Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence. \begin{align*} f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R \end{align*}
We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$. \begin{align*} f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R} \end{align*}
Substitution $u=2x$
We consider
This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.
Substitution $u=x+1$
We consider
This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.
We also obtain \begin{align*} e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\ &=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\ &=e^{x+1} \end{align*}
Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.
• I understand now that one is allowed to substitute in this case if $n\to\infty$. Is it correct to say that the series of $e^{x+1}$ that is the result of substitution is centered around $x=-1$, while the series that is calulated directly is centered around $x=0$? And doesn't this make a difference if I want to give an approximation of $e^{x+1}$ around $x=0$? I mean if I want to approximate I usually don't let $n\to\infty$, but rather to a fixed number, say $n\to3$. Would that make the case different? – GambitSquared Nov 14 '16 at 15:07
• @ImreVégh: Yes, it is correct that the representation of $e^{x+1}$ above is centered around $x=-1$. Note, that if we have a power series centered around some value $x=x_0$, we can select any value $x_1$ within the disc of convergence $x_1\in d(x_0,R)$ of $f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$ as new center by expanding $\sum_{n=0}^\infty a_n ((x-x_1)-(x_0-x_1))^n$ at powers of $x-x_1$. The new radius of convergence $R_1$ is at least as big, that $d(x_1,R_1)\subset d(x_0,R)$. – Markus Scheuer Nov 14 '16 at 15:23
• @ImreVégh: If we want to expand the special case $e^{x+1}$ around $x=0$ we can go the easy way (as you already did) and write $e^{x+1}=e\cdot e^x=\sum_{n=0}^\infty \frac{e}{n!} x^n$ with coefficients $\frac{e}{n!}$ of $x^n$. The somewhat more cumbersome way would be to expand $(x+1)^n$ in $\sum_{n=0}^\infty \frac{(x+1)^n}{n!}$ and arrange the terms according to increasing powers of $x$. – Markus Scheuer Nov 14 '16 at 15:27
If $f(x)=P_n(x)+x^n\epsilon(x)$ then
$f(u(x))=P(u(x))+(u(x))^n\epsilon(u(x))$
with $\lim_{x\to 0}\epsilon(x)=0$.
thus we need that
$\lim_{x\to 0}\epsilon(u(x))=0$
so, we must have
$$\lim_{x\to 0}u(x)=0$$
• Hmm, I understand. And why is substitution allowed in the first place? When $\lim_{x\to 0}\epsilon(u(x))=0$? – GambitSquared Nov 12 '16 at 9:31
The quantity $2x$ is a product and as $x\to 0$ it remains a small number.
The quantity $x+n$ for $n\neq 0$ is not a little quantity anymore, and so you are not anymore around zero but you're around $n$. | 2020-04-08T19:25:59 | {
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https://math.stackexchange.com/questions/1302015/what-is-the-relationship-betweeen-a-pdf-and-cdf | # What is the relationship betweeen a pdf and cdf?
I am learning stats. On page 20, my book, All of Statistics 1e, defines a CDF as function that maps x to the probability that a random variable, X, is less than x.
$F_{x}(x) = P(X\leq x)$
On page 23 it gives a function
$P(a < X < b ) = \int_{a}^{b}f_{X}dx$
and then says that "the function $f_{X}$ is called the probability density function. We have that..."
$F_{x}(x) = \int_{-\infty}^{x}f_{X}dt$
I am a little confused about how to characterize the most important difference between them. The equation above says that the cdf is the integral of the pdf from negative infinity to x. Is it fair to say that the cdf is the integral of the pdf from negative infinity to x?
• Read carefully, $f_X$ is pdf, $F_x$ is CDF. – booksee May 28 '15 at 3:43
• "The equation above says that the cdf is the integral of the pdf from negative infinity to x. Is it fair to say that the cdf is the integral of the pdf from negative infinity to x?" ??? Indeed it is correct to say that the cdf is the integral of the pdf from negative infinity to x. – Did May 28 '15 at 12:46
Yes. That's correct. A PDF is a probability density function. It is stating the probability of a particular value coming out. Taking this analogy to a discrete distribution, the PDF of a 6-sided die is: $[x<1:0,x=1:\frac{1}{6},x=2:\frac{1}{6},x=3:\frac{1}{6},x=4:\frac{1}{6},x=5:\frac{1}{6},x=6:\frac{1}{6},x>6:0]$. For a continuous probability distribution, you can't really use the PDF directly, since the probability of an infinitesimally thin slice of the PDF being selected is intuitively zero.
That's where the cumulative density function, or CDF, comes it. It is a measure of how likely the value is to be less than some arbitrary value (which we pick). For a discrete case, you start with the first possible value, and add all the entries in the PDF up to the value of interest: $$CDF=\sum PDF \rightarrow [x<1:0,x<2:\frac{1}{6},x<3:\frac{2}{6},x<4:\frac{3}{6},x<5:\frac{4}{6},x<6:\frac{5}{6},x\geq 6:\frac{6}{6}]$$ Notice how the final value of the CDF is $1$. This is expected, since every possible outcome of rolling a 6-sided die is less than or equal to 6.
Now let's go back to the continuous probability distribution. In this case, we don't have a finite set of options for the answer to be, so we can't constrain $X$. Thus, we start from $-\infty$, since that encompasses everything to the left of the chosen $x$. As you should be aware from calculus, the integral is to continuous functions what a sum is to discrete functions - loosely. The value of a CDF is that you can use it to determine the probability of the number falling within a specific range as follows:
$$F(a\leq X \leq b) = F(X \leq b) - F(X \leq a) = \int_{-\infty}^{b} f(x)dx - \int_{-\infty}^{a} f(x)dx = \int_{a}^{b} f(x)dx$$
• The notation for the PDF of the die roll may be a bit confusing. The value of the CDF at $x$ actually is $\frac16$ if $1\leq x < 2$ and $0$ if $x < 1$ (because, for example, $P(X\leq0.99)=0$, but $P(X\leq1.7)=\frac16$). A graph might make the function's values a little clearer. – David K May 28 '15 at 12:55
• Good points. Let me add the entries for which it is zero. – FundThmCalculus May 28 '15 at 12:56
• @FundThmCalculus thanks! Great answer. I'm learning on my own from a book so this kind of explanation is extremely helpful. – bernie2436 May 28 '15 at 22:44
• @bernie2436, thanks for letting me know that. By accepting and upvoting answers, you help the community know what you are looking for. Also, it was a very good question. :) – FundThmCalculus May 29 '15 at 12:01
Simply put, yes, the cdf (evaluated at $x$) is the integral of the pdf from $-\infty$ to $x$. Another way to put it is that the pdf $f(x)$ is the derivative of the cdf $F(x)$.
These definitions assume that the cdf is differentiable everywhere. I mention this not to make the definitions more complicated, but to reduce the factor of surprise later when you find out that there are other kinds of probability distributions that do not have the kind of pdf described here.
The cumulative distribution function $F_X$ of any random variable $X$ is defined by $$F_X(x)=P(X\le x)$$ for $x\in\mathbb R$. If $X$ is a continuous random variable, then the cumulative distribution function $F_X$ can be expressed as $$F_X(x)=\int_{-\infty}^xf(x)\mathrm dx$$ for $x\in\mathbb R$, where $f$ is the probability density function of a random variable $X$. If a random variable $X$ is not continuous, it does not have a probability density function (for example, a random variable with Bernoulli distribution). | 2020-06-02T08:13:07 | {
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https://www.physicsforums.com/threads/find-the-angle-between-two-vectors.532565/ | Find the angle between two vectors
1. Sep 22, 2011
jamesbrewer
1. The problem statement, all variables and given/known data
Two vectors A and B have the same magnitude of 5.25. If the sum of these two vectors gives a third vector equal to 6.73j, determine the angle between A and B.
2. Relevant equations
For some vector $\vec{R}$: $|\vec{R}| = \sqrt{R_x^2 + R_y^2}$
3. The attempt at a solution
I feel like I need to find the components of A and B, but I can't figure out how to do that with only the information given -- is it even possible?
I think the magnitude of the third vector is 6.73. Since there was no $\hat{i}$ term given for $\vec{R}$, I would say that $|\vec{R}| = \sqrt{R_x^2 + R_Y^2} = \sqrt{6.73^2} = 6.73$.
Given the components of A and B, this would be easy to solve. Without them I am utterly lost. What am I missing?
2. Sep 22, 2011
PeterO
The vectors A and B, along with the resultant will form a triangle - probably not right angled - so you can use the cosine Rule to solve.
a2 = b2 + c2 - 2.b.c.cos(A)
3. Sep 22, 2011
jamesbrewer
What are a, b, and c? The lengths of the sides of the triangle? If so, would that mean that a = |A|, b = |B|, and c = |C| (the resultant)?
4. Sep 22, 2011
PeterO
a, b & c are the sides,
A is the angle opposite side a
[so a, b & c could represent vectors A, B and C but not necessarily, depends which angle you are looking for]
Last edited: Sep 22, 2011
5. Sep 22, 2011
jamesbrewer
How can I form a triangle if I don't know anything other than the vector's magnitude? I have nothing to tell me what direction it points in.
6. Sep 22, 2011
PeterO
You can stand a triangle up any way you like - its sides are still the same length and its angles are still the same size. A name like "the base" might apply to a different side, depending which way you arrange it, but that should not be a problem .. what's in a name?
7. Sep 22, 2011
jamesbrewer
Here's what I've got:
$|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|cos\theta$
$6.73^2 = 5.25^2 + 5.25^2 - 2(5.25)^2 cos \theta$
$45.29 = 27.56 + 27.56 - 2(27.56) cos \theta$
$45.29 = 55.12 - 55.12 cos \theta$
$45.29 - 55.12 = - 55.12 cos \theta$
$-9.83 = -55.12 cos \theta$
$\frac{-9.83}{-55.12} = cos \theta$
$\theta = cos^-1 0.178$
$\theta = 79.75^o$
My answer wasn't correct though, where did I go wrong?
Last edited: Sep 22, 2011
8. Sep 22, 2011
Staff: Mentor
One approach which you might want to consider is to look at the relationships between the vector components in terms of simultaneous equations. Suppose that the two initial vectors are A and B and the resultant is C.
Since C contains no x-component and has only a y-component, the sum Ax + Bx must be zero, or in other words, Bx = -Ax. Similarly, Ay + By must be Cy. You also have |A| = |B| = 5.25, so that's another pair of relationships. You have four unknowns (really three when you consider that the x-components of A and B are equal and opposite) and plenty of interrelationships to use to solve for them.
9. Sep 22, 2011
PeterO
How incorrect was your answer? You have rounded off all the way through which could make a small difference
10. Sep 23, 2011
jamesbrewer
Unfortunately I have no idea. All I was told was "Incorrect answer."
11. Sep 23, 2011
Staff: Mentor
Let's step back from the problem and see if we can make additional simplifying deductions before invoking formulas.
Since the resultant of vectors A and B is a vector with only a y-component (6.73j), then A and B must have equal and opposite x-components. Further, since A and B have equal magnitudes (5.25), this then forces their y-components to be equal also. Why not equal and opposite you say? Because then their sum would be zero rather that +6.73.
So A and B "straddle" the positive y-axis, and make equal angles with that axis. The sum of their y-components is 6.73. Now it's time to write formulas. If you let $\theta$ be the angle between either A or B and the Y-axis, what is an expression for the y-component of A or B? If double that is 6.73, can you solve for $\theta$? What then is the angle between A and B?
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12. Sep 23, 2011
PeterO
Draw a diagram of the vectors, like gneil's and you will see that they actually want (180 - 79) degrees - with the decimal bits.
Rounding off needs to be addressed too.
gneil's method is excellent also - even preferable!
13. May 15, 2013
Karthik k
Three vectors sum to zero. The magnitude of 2 vectors is equal and the third one is root 2 times the magnitude of the equal vectors. Find the angle between the three vectors. Plzz help me on dis one..:)
14. May 15, 2013
PeterO
Since the three vectors will make up a triangle - so you end up where you started with a vector sum of zero - you should recognise the magnitudes 1,1,√2 as the sides of a very common triangle in trigonometry, from which we derive the exact vale of sin, cos and tan of a particular angle.
The other triangle used in trigonometry is known as the 2,1,√3 triangle [not that it relates directly to this question]. | 2017-11-23T23:29:40 | {
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https://math.stackexchange.com/questions/3823314/is-this-correct-even-if-i-get-two-different-answers-for-slope-intercept-form | # Is this correct even if I get two different answers for slope-intercept form?
Writing an Equation for a Linear Function Given Two Points
If $$f$$ is a linear function, with $$f(3)=−2$$, and $$f(8)=1$$, find an equation for the function in slope-intercept form.
We can write the given points using coordinates. \begin{align*} f(3) & = −2 \to (3,−2)\\ f(8) & = 1 \to (8,1) \end{align*}
We can then use the points to calculate the slope.
\begin{align*} m & = \frac{y_2 - y_1}{x_2 - x_1}\\ & = \frac{1 - (-2)}{8 - 3}\\ & = \frac{3}{5} \end{align*}
Substitute the slope and the coordinates of ONE OF THE POINTS into the point-slope form.
*The book decided to use $$(3, -2)$$ whereas I decided to use $$(8, 1)$$.
\begin{align*} y - y_1 & = m(x - x_1)\\ y - (-2) & = \frac{3}{5}(x - 3) \end{align*}
The book goes further in their example.
We can use algebra to rewrite the equation in the slope-intercept form.
\begin{align*} y + 2 & = \frac{3}{5}(x - 3)\\ y + 2 & = \frac{3}{5}x - \frac{9}{5}\\ y & = \frac{3}{5}x - \frac{19}{5} \end{align*}
The points I chose to use $$(8, 1)$$, which then gave me the answer
\begin{align*} y - 1 & = \frac{3}{5}(x - 8)\\ y & = \frac{3}{5}x - 5 \end{align*}
Because of choosing different points I get what seems a different slope-intercept even when it states "Substitute the slope and the coordinates of ONE OF THE POINTS into the point-slope form.". Is this ok?
• How did you get $-5$ in your answer? Please check/show your arithmetic. Sep 12, 2020 at 10:19
• Short answer, no, you should always get the same $y$-intercept. You have an arithmetic error. Sep 12, 2020 at 10:21
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. Sep 12, 2020 at 10:58
$$y-1=\frac35 \left( x-8\right)$$
$$y=\frac35x - \frac{24}5+1=\frac35x-\frac{24-5}{5}=\frac35x-\frac{19}{5}$$ | 2022-08-19T02:59:33 | {
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https://math.stackexchange.com/questions/3029324/algebraic-proof-that-binomnk-1k-sum-limits-i-0k-binomn-1k-i | # Algebraic proof that $\binom{n+k-1}{k} = \sum\limits_{i=0}^k \binom{(n-1)+(k-i)-1}{k-i}$
Prove (algebraically) that $$f_k(n) = \binom{n+k-1}{k} = \sum\limits_{i=0}^k \binom{(n-1)+(k-i)-1}{k-i} = \sum\limits_{i=0}^k f_{k-i}(n-1)$$ for $$n \geq 2$$ and that $$f_k(1) = 1$$ for all $$k$$.
Then, show that if $$f_k(n) = \sum\limits_{i=0}^k f_{k-i}(n-1)$$ for $$n\geq 2$$ and $$f_k(1) = 1$$ for all $$k$$, then $$f_k(n) = \binom{n+k-1}{k}$$.
Attempting the first one: it's easy to see that $$f_k(1) = 1$$ for all $$k$$ by substitution. I'm not so sure about what to so with the summation of binomial coefficients. I'm sure there's a way to use $$\sum\limits_{i=0}^k \binom{k}{i}x^i = (x+1)^k$$, either directly or in a double-sum, but I'm not sure how to manipulate the summand into such a form.
For the second one: after the first one is solved, it's trivial, since they agree for $$n=1$$ and if they agree for $$n = N-1$$, they must agree for $$n=N$$ by the first identity. So the main difficulty in this problem is coming from the first part.
Edit: I should also note that I can understand this identity by relating $$f_k(n)$$ with the number of ways to choose an ordered $$n$$-tuple of numbers adding to $$k$$, or equivalently the number of $$k$$-degree terms in an $$n$$-dimensional polynomial. However, my attempts converting that intuition into an algebraic proof haven't gone anywhere.
• $f_k(n)$ is the number of non negative integer solutions to $x_1+x_2+\cdots +x_n = k$. Now $f_{k-i}(n-1)$ counts the number of those solutions in which $x_n = i$. The identity follows. – Muralidharan Dec 7 '18 at 1:17
• so fix $x_n=i$ and solve $x_1+x_2+\cdots+x_{n-1} = k-i$. Nice @Muralidharan :) – rsadhvika Dec 7 '18 at 1:21
• @Muralidharan This is exactly what I know, but how can I reduce this argument into a direct algebraic proof of the identity? – AlexanderJ93 Dec 7 '18 at 1:26
• @AlexanderJ93 I think you may simply use the identity $$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$ – rsadhvika Dec 7 '18 at 1:35
• $$\binom{n+k-1}{k} = \binom{n+(k-1)-1}{k} + \binom{n+(k-1)-1}{k-1} = \cdots$$ – rsadhvika Dec 7 '18 at 1:37
We have \eqalign{ & \sum\limits_{0\, \le \,i\, \le \,k} {\left( \matrix{ n - 2 + k - i \cr k - i \cr} \right)} = \cr & = \sum\limits_{0\, \le \,j\, \le \,k} {\left( \matrix{ n - 2 + j \cr j \cr} \right)} = \quad \quad (1) \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( \matrix{ k - j \cr k - j \cr} \right)\left( \matrix{ n - 2 + j \cr j \cr} \right)} = \quad \quad (2) \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( { - 1} \right)^{\,k - j} \left( \matrix{ - 1 \cr k - j \cr} \right)\left( { - 1} \right)^{\,j} \left( \matrix{ - n + 1 \cr j \cr} \right)} = \quad \quad (3)\cr & = \left( { - 1} \right)^{\,k} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( \matrix{ - 1 \cr k - j \cr} \right)\left( \matrix{ - n + 1 \cr j \cr} \right)} = \quad \quad (4) \cr & = \left( { - 1} \right)^{\,k} \left( \matrix{ - n \cr k \cr} \right) = \quad \quad (5) \cr & = \left( \matrix{ n + k - 1 \cr k \cr} \right) \quad \quad (6) \cr} where:
(1) change of index
(2) replacing sum bounds with binomial
(3) upper negation
(5) convolution (6) upper negation
Note to (2):
The binomial $$\binom{k-j}{k-j}$$ equals $$1$$ for $$j \le k$$ and is null for $$k.
Therefore we can use it to replace the upper bound for $$j$$ in the sum.
On the other side, the second binomial intrinsically ensures for the lower bound $$0 \le j$$.
Therefore we can leave the index free to take all values: that's why I indicated the bound in brackets.
That's a "trick" many time useful in manipulating binomial sums, and I am in debt to the renowned "Concrete Mathematics" for teaching that.
• This is great, I'm just not understanding what's actually happening in (2). – AlexanderJ93 Dec 7 '18 at 1:48
• @AlexanderJ93: you are not probably alone in that, so I included a note to explain such a useful "trick" – G Cab Dec 7 '18 at 2:01
• Not quite. The binomial $\dbinom{k-j}{k-j}$ is not $0$ when $j <0$; you need the other binomial for that. – darij grinberg Dec 7 '18 at 2:09
• @darijgrinberg: oh yes, you are fully right : I shall amend the description. Thanks for signalling such a mistake. – G Cab Dec 7 '18 at 2:29
Let $$[x^j]:f(x)$$ denote the coefficient of $$x^j$$ of the function $$f(x)$$. We have $$\begin{eqnarray*} \sum_{i=0}^{k} \binom{n+k-i-2}{k-i} &=& \sum_{i=0}^{k} [x^{k-i}]: (1+x)^{n+k-i-2} \\ &=& [x^k]: \sum_{i=0}^{k} (1+x)^{n+k-2} \left(\frac{x}{x+1} \right)^i \\ &=& [x^k]: (1+x)^{n+k-2} \sum_{i=0}^{\infty} \left(\frac{x}{x+1} \right)^i \\ &=& [x^k]: (1+x)^{n+k-1} = \binom{n+k-1}{k}. \end{eqnarray*}$$ | 2019-05-25T01:18:53 | {
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https://math.stackexchange.com/questions/2471895/how-many-bit-strings-of-length-n-contain-exactly-k-blocks-of-10 | # How many bit strings of length $n$ contain exactly $k$ blocks of “$10$”?
How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"?
My attempt: Let $F(n, k)$ be the number of bit strings of length $n$ that contain exactly $k$ blocks of $10.$ Note that for $k \neq0,$ $F(0, k) = F(1, k)= 0.$ Consider a bit string of length $n$ where $n \geq 2.$
Every bit string either begins with:
1) $10$, or
2) $01$, $00$ or $11.$
In the former case, we get $F(n-2, k-1)$ since the remaining $n-2$ bits must contain $k-1$ blocks of $10$ in exactly $F(n-2, k-1)$ ways.
In the latter case, in either of the $3$ situations we get $F(n-2, k)$ since the remaining $n-2$ bits must contains $k$ blocks of $10$ in $F(n-2, k)$ ways.
Therefore for $n \geq 2, k \geq 1$ we have:
$F(n, k) = F(n-2, k-1)+ 3F(n-2, k)$.
For $k \geq 1$ let $A_k(x) = \displaystyle \sum_{n \geq2}F(n,k)x^n$ (and let $A_0(x)=1$) then multiplying the above recurrence by $x^n$ and summing over all $n \geq 2$ we get:
$A_k(x) = x^2 A_{k-1}(x) +3x^2A_{k}(x)$
$\Rightarrow A_k(x) = \dfrac{x^2}{1-3x^2} A_{k-1}(x)$
$\Rightarrow A_{k}(x) = \dfrac{x^{2k}}{(1-3x^2)^k}$
$\Rightarrow F(n, k) = [x^n] A_k(x)$
$\Rightarrow F(n, k) = [x^n] \dfrac{x^{2k}}{(1-3x^2)^k}$
$\Rightarrow F(n, k) = [x^{n-2k}] \displaystyle \sum_{r \geq 0} \binom{k+r-1}{r} 3^r x^{2r}$
which clearly gives a wrong answer for odd $n$. Could somebody point out where in the above calculation have I gone wrong? Thanks in advance.
• If the string begins $01$ or $11$, then the next element could be $0$, which creates an additional $10$ that you have not taken into account in your calculations. – rogerl Oct 14 '17 at 13:22
• Right, the recurrence itself is wrong. Any idea what the correct approach, via generating functions or otherwise, entails? – Aryaman Jal Oct 14 '17 at 14:00
Essentially this amounts to finding the places where the string switches from a run of $1$'s to a run of $0$'s. Note that there are $n-1$ places where such a switch may occur (between each pair of numbers). In any event, there must be $k$ places where the string switches from $1$ to $0$:
• If the string starts and ends with $0$, then (since it starts with $0$) there must also be $k$ places where it switches from $0$ to $1$. There are $\binom{n-1}{2k}$ ways of making these choices.
• If the string starts with $1$ and ends with $0$, then there must (since it ends in $0$) also be $k-1$ places where it switches from $0$ to $1$; there are $\binom{n-1}{2k-1}$ ways of making these choices.
• If the string starts with $0$ and ends with $1$, then there must (since it starts with $0$) also be $k+1$ places where it switches from $0$ to $1$; there are $\binom{n-1}{2k+1}$ ways of making these choices.
• Finally, if the string starts and ends with $1$, there must (since it ends with $1$) also be $k$ places where it switches from $0$ to $1$. There are $\binom{n-1}{2k}$ ways of making these choices.
So finally, the total number of such strings is $$\binom{n-1}{2k} + \binom{n-1}{2k-1} + \binom{n-1}{2k+1} + \binom{n-1}{2k} = \binom{n}{2k} + \binom{n}{2k+1} = \binom{n+1}{2k+1}.$$
Thanks for the insight. I used a similar technique for a similar problem. I wrote this up for my assignment which is due in a week. Although I believe that I found the recurrence relation correctly, I am unable to go about extracting the coefficients from a quadratic raised to a negative k-th power. Could you please be able to help me how to proceed?
Let $f(n, k)$ represent the number of strings made of $\{a, b\}$ where $k$ is the number of occurrences of ''$ab$'' in a string of length $n$. Note that the string can begin with $ab, aa, bb$, or $ba$. If it begins with $ab$, then there are exactly $f(n-2, k-1)$ number of strings remaining. If it begins with $bb$, then there are exactly $f(n-2, k)$ strings remaining. If it begins with $aa$, or $ba$, then there are exactly $f(n-1, k)$ strings remaining. Thus we have the recurrence relation $f(n, k) = f(n-2, k-1) + f(n-2, k) + 2f(n-1, k)$. Trivially enough, we have that $f(0, k) = f(1, k) = 0$ for any non-zero $k$. Define $A_{k}(x) = \displaystyle \sum_{n=2}^{\infty} f(n, k)x^{n}$. Note that $A_{0}(x) = \displaystyle \sum_{n\geq 2} f(n, 0)x^{n} = \displaystyle \sum_{n \geq 2} (n+1)x^{n}.$ Thus $A_{k}(x) = \sum_{n=2}^{\infty} \Big[ f(n-2, k-1)x^{n} + f(n-2, k)x^{n} + 2f(n-1, k)x^{n}\Big].$
It follows that $A_{k}(x) = \displaystyle \sum_{n=2}^{\infty} x^{2} f(n, k-1)x^{n} + x^{2} f(n, k)x^{n} + 2x f(n, k)x^{n}$
Thus, \begin{align*} A_{k}(x) = x^{2} A_{k-1}(x) + x^{2} A_{k}(x) + 2x A_{k}(x)\\ \implies A_{k}(x) = \displaystyle \frac{x^{2}}{1+2x-x^{2}}A_{k-1}(x)\\ \implies A_{k}(x) = \displaystyle \frac{x^{2}}{1+2x-x^{2}} \cdot \Big( \displaystyle \frac{x^{2}}{1+2x-x^{2}} A_{k-2}(x) \Big) \\ \therefore A_{k}(x) = \displaystyle \Bigg(\frac{x^{2}}{1+2x-x^{2}}\Bigg)^{k} \displaystyle (n+1) \sum_{n \geq 2} x^{n}\\ \therefore A_{k}(x) = (n+1) \displaystyle \frac{x^{2k+2}}{(1+2x-x^{2})^{k} (1-x)}\\ \end{align*} But $f(n, k) = [x^{k}] A_{k}(x)$ \begin{align*} \therefore f(n, k) = [x^{n-2-2k}] \displaystyle (n+1) \frac{1}{(1+2x-x^{2})^{k} (1-x)}\\ \therefore f(n, k) = [x^{n-2k-2}] (n+1) \displaystyle (1+2x-x^{2})^{-k} (1-x)^{-1}\\ \end{align*} | 2020-04-02T20:18:49 | {
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http://math.stackexchange.com/questions/182287/showing-that-a-linear-map-mathbbrx-leq-2-rightarrow-mathbbr3-is-a | # Showing that a linear map $\mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3$ is an isomorphism.
Let $t_1,t_2,t_3$ be distinct real numbers and consider the linear map
$$T : \mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3, \quad \quad p(x) \mapsto \begin{pmatrix}p(t_1)\\p(t_2)\\p(t_3)\end{pmatrix}$$
I want to show that $T$ is an isomorphism. Since it is given that $T$ is linear, I just need to show that $T$ is bijective. My initial approach was to solve the system
$$\begin{pmatrix} a + bt_1 + ct_1^2\\ a + bt_2 + ct_2^2\\ a + bt_3 + ct_3^2 \end{pmatrix} = \begin{pmatrix} r_1\\ r_2\\ r_3 \end{pmatrix}$$
for $a,b,c$ in terms of $r_1,r_2,r_3$, thus determining the inverse map $T^{-1}$ mapping a point in $\mathbb{R}^3$ to a polynomial of degree $\leq 2$.
I solved this system using Maple, and got a solution which was defined when $t_1,t_2,t_3$ where not all equal, which is fine by their definition, but I am wondering if there is a nicer argument, especially since the solution is rather ugly.
(this is not homework)
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Personally, I find the answer by Yuki most intuitive, and understandable. By considering the current votes, it seems like the majority disagrees (though only slightly). I can't find a question on meta about this "dilemma". – utdiscant Aug 14 '12 at 3:02
You should accept the answer you consider most helpful. If anyone stumbles into this question, he will be able to see the one with the top score (or next-to-top score, if the accepted answer is the top-scoring one) right below the accepted one anyway. There are even badges here for having nonaccepted answers with higher scores than accepted ones. ;) Personally, I find Jacob Schlather's suggestion to be the most elegant. – tomasz Aug 14 '12 at 4:47
Hint:
$T$ is onto: Langrange Polynomial.
Injectivity: if $p$ is an polynomial of degree $\le2$, knowing values in 3 distinct points, you know $p$.
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Just take $1,x,x^2$. They transform into $\begin{pmatrix}1\\ 1 \\ 1\end{pmatrix},\begin{pmatrix}t_1\\ t_2 \\ t_3\end{pmatrix},\begin{pmatrix}t_1^2\\ t_2^2 \\ t_3^2\end{pmatrix}$. To show that they are linearly independent, calculate the determinant of $\begin{pmatrix}1&t_1&t_1^2\\1&t_2& t_2^2 \\1&t_3& t_3^2\end{pmatrix}$, which is, incidentally, Vandermonde matrix.
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The result heavily used here, and a rather important one, is: an operator $\,T:V\to V\,\,\,,\,\,V\,$ a vector space, is an isomorphism iff the image of some basis is again a basis (or more general: the image of any linearly independent is again lin. indep.) +1 – DonAntonio Aug 14 '12 at 2:30
@DonAntonio: Not really: rather than using the result, I've explicitly shown that it is of rank 3. :) – tomasz Aug 14 '12 at 2:31
Hint: What is in the kernel of your map?
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I think it may be easier to show that the associated matrix always has non-zero determinant. This is a known result.
- | 2015-05-28T16:40:37 | {
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https://www.physicsforums.com/threads/finding-area-between-cone-and-sphere.554166/ | # Finding Area Between Cone and Sphere
## Homework Statement
Find the volume bounded between the sphere of radius a centered at (0,0,0) and the cone z=sqrt(x2 +y2).
## The Attempt at a Solution
So, subbing our definition for z into the the equation for a sphere of radius a centered at (0,0,0):
2x2 + 2y2 = a2. Converting to cylindrical coordinates,
2r2 = a2
r = a/sqrt(2). I use the bounds,
0≤z≤a
0≤θ≤2*pi
0≤r≤a/sqrt(2)
I use 1*r as my integrand to find volume and I get a3*pi/2. Am I doing this right?
I like Serena
Homework Helper
Hi TranscendArcu!
You have the upper bound of r as a constant. It isn't in cylindrical coordinates.
Furthermore, you have the upper bound of z as "a", but in that neighbourhood you should have an r that follows the sphere instead of the cone.
I recommend you do your calculations in spherical coordinates.
That's way easier.
Okay. Would my bounds be,
0≤rho≤a
0≤theta≤2*pi
pi/4≤phi≤3*pi/4
I'm not at all certain about any of these bounds since this is the first time I've done spherical coordinates. I'm most concerned about my bounds on phi. How does one actually find these bounds (I was just doing it intuitively, which invariably leads to incorrectness)?
Anyway, I would then integrate with my bounds of (rho)^2 * sin(phi) d(rho)d(theta)d(phi)
I like Serena
Homework Helper
To find the bound for phi, pick a point on the cone and calculate the phi that comes with it.
You should note that the definition of your cone implies that z>0.
Btw, your problem does not state which part you should have.
The part within the cone (and the sphere), or the part outside of the cone (and inside the sphere)?
Last edited:
So, Looking at the cone at letting (x,y) be (1,1). I find z=sqrt(2). I also see that (1,1,sqrt(2)) has distance sqrt(2) to the z-axis. If tan(phi) = sqrt(2)/sqrt(2), then phi = pi/4. This means that 0≤phi≤pi/4, right?
If I integrate I find that the volume within both the cone and the sphere is
2*pi*(a^3)/3 * (-sqrt(2)/2 + 1)
Still wrong, or better?
Presumably, if I wanted to find the volume outside the cone, but still within the sphere, I would merely change my bounds on phi. Would I have,
-pi/2≤phi≤pi/4
?
I like Serena
Homework Helper
Just checked.
I get the same!
I like Serena
Homework Helper
Presumably, if I wanted to find the volume outside the cone, but still within the sphere, I would merely change my bounds on phi. Would I have,
-pi/2≤phi≤pi/4
?
Not quite.
What is the range of phi?
And to what axis is it relative?
So I think phi is relative to the z-axis. I suppose that any point on negative z-axis has phi = pi and any point on the positive z-axis has phi = 0. So, I guess that means my new bounds for the other integral would have bounds pi/4≤phi≤pi. Is that right?
I like Serena
Homework Helper
Yep! | 2022-06-27T16:59:19 | {
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https://math.stackexchange.com/questions/3056738/proof-for-the-general-equation-for-family-of-lines-passing-through-the-intersect | # Proof for the general equation for family of lines passing through the intersection of two lines(Simplified).
In the previous post Proof for the equation of a line passing through the intersection of two lines(family of lines). I have looked at derivation which seems to be oversimplified so I am still confused.
Let $$C(x_0,y_0)$$ be the common point $$L_1 \cap L_2$$. Thus
$$\begin{cases}a_1x_0 + b_1y_0+ c_1 =0\\a_2x_0 + b_2y_0+ c_2 =0\end{cases}$$
By difference with the initial equations, we obtain the new equivalent equations:
$$\tag{1}\begin{cases}a_1(x-x_0) + b_1(y-y_0)=0 \ \ (L_1)\\a_2(x-x_0) + b_2 (y-y_0)=0 \ \ (L_2)\end{cases} \ \ \implies \ \underbrace{(a_1+Ka_2)(x-x_0) + (b_1+Kb_2)(y-y_0)=0}_{\text{line} (L_1+KL_2)}$$
From equation 1 why do we add a constant $$K$$ to line $$L_2$$ and what is its significance?
• Wouldn't this question best be asked in a comment to the answer that describes $K$? Alerting the answerer that the response is unclear will allow the answerer to improve that response. – Blue Dec 30 '18 at 12:10
• @Blue It would be a better idea but SE has limit for users to comment(minimum 50 reputations ) and I am short of it. – pranjal verma Dec 30 '18 at 13:42
• I added a comment to that answer to direct the answerer here. – Blue Dec 30 '18 at 13:57
• We were supposed to prove that any lines passing through the intersection of the two lines has an equation of the form $L_1 + KL_2 = 0.$ (Which is not strictly true; what is true is that every such line except $L_2$ has such an equation.) At some time during the proof we were going to have to refer to that equation; how can you prove something that you never even mention? – David K Dec 30 '18 at 19:47
Maybe explaining it with vectors will be clearer.
(see figure below)
A line $$L_{P_0,N}$$ can be defined by a point $$P_0=(x_0,y_0)$$ and a normal (= orthogonal) vector $$\overrightarrow{N}=(a,b)$$ in the following way :
$$M=(x,y) \in L_{P_0,N} \ \ \iff \ \ \overrightarrow{P_0M}=\binom{x-x_0}{y-y_0} \perp \overrightarrow{N}=\binom{a}{b} \ \ \iff \ \$$
$$(x-x_0)u+(y-y_0)v=0.$$
Thus if you have two straight lines passing through the same $$(x_0,y_0)$$ with resp. normal vector $$\binom{a_1}{b_1}$$ and $$\binom{a_2}{b_2}$$, taking now the normal vector $$\binom{a_1}{b_1}+K \binom{a_2}{b_2}=\binom{a_1+Ka_2}{b_1+Kb_2}$$ you will easily be convinced that you get in this way all possible normal vectors, thus all the pencil of lines passing through $$P_0$$, with the exception of the 2nd line itself.
This is exactly what means the implication I had given, and that you have reproduced above ; indded, it can be written in the following way:
$$\binom{a_1}{b_1} \perp \binom{x-x_0}{y-y_0} \ \ \ \ \ \text{and} \ \ \ \ \ \binom{a_2}{b_2} \perp \binom{x-x_0}{y-y_0} \implies$$
$$\left(\binom{a_1}{b_1}+K\binom{a_2}{b_2}\right) \perp \binom{x-x_0}{y-y_0}$$
• How do you come with the formulae $\binom{a_1}{b_1} + K\binom{a_2}{b_2} = \binom{a_1 + Ka_2}{b_1 + Kb_2}$,I don't know much about vectors but I assume that $\binom{a}{b}$ is the slope of the orthogonal line. – pranjal verma Dec 31 '18 at 9:11
• First sentence : it is a classical linear combination of vectors. Second sentence : $\binom{a}{b}$ is a vector, not a slope. The slope of the corresponding line (with equation $a(x-x_0)+b(y-y_0)=c \iff y=(-b/a)x+...$) is $-b/a$. – Jean Marie Dec 31 '18 at 9:18
• Can you show $\binom{a_1}{b_1}+K \binom{a_2}{b_2}=\binom{a_1+Ka_2}{b_1+Kb_2}$ in the form of slope of the orthogonal lines? – pranjal verma Dec 31 '18 at 9:21 | 2019-08-23T16:52:00 | {
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https://math.stackexchange.com/questions/346829/relative-sizes-of-prime-gaps | Relative sizes of prime gaps
There are no prime numbers between the two primes $113$ and $127$. That gap seems quite large by comparison to the sizes of the numbers in it. $$\frac{\text{size of gap}}{\text{prime just below the gap}} = \frac{14}{113} = 0.12389\ldots$$ Is that the biggest that this particular statistic ever gets?
Is that the last time this particular statistic ever gets that big?
• Are there restrictions on how big the gap has to be? i.e. $3$ and $5$ gives $2/3=.666...$ – MITjanitor Mar 30 '13 at 22:01
• @GitGud, even if he does, it goes to $0$. (Since $p_{k+1}/p_k \to 1$ as $k \to \infty$, the limit of $p_{k+1}/p_k - 1 \to 0$.) – George V. Williams Mar 30 '13 at 22:04
• @MITjanitor biggest would probably gap between 2,3 which gives 1/2 = .5 – DanZimm Mar 30 '13 at 22:07
• I can't speak to the answer, but for $n$ sufficiently large $g_n < p_n^\theta$ for some $\theta < 1$ (Hoheisel 1930). So the statistic in question is bounded above by $p_n^\theta / p_n$ which $p_n^{\theta-1}$, which goes to $0$ as $n$ goes to $\infty$. This implies that there is a maximum. – George V. Williams Mar 30 '13 at 22:21
• Assuming this page is reliable, the results in section 4 seem to indicate that it's the last time modulo a few that you can check by hand. See also this – Martin Mar 30 '13 at 22:42
Indeed, the answer is positive. We have:
$$R_n := \frac{\text{size of gap}}{\text{prime just below the gap}} = \frac {p_{n+1} - p_n} {p_n} = \frac {p_{n+1}} {p_n} - 1$$
Using some well known approximations (Rosser's theorem) on $p_n$, we have
$$p_{n+1} \le (n+1) \log(n+1) + (n+1) \log \log(n+1)$$ $$p_n \ge n \log n + n \log \log n - n$$
So,
$$R_n + 1 \le \frac {n+1} n \frac {\log (n+1) + \log \log (n+1)} {\log n + \log \log n - 1}$$
RHS is a decreasing function, and so we have $R_n < 0.12389\ldots$ for all $n \ge 1296$. The remaining cases $31 \le n \le 1295$ may be checked manually.
Rohrbach and Weis show that (as Martin notes in the comments):
$$g_n < \frac{p_n}{13} \quad \quad n > 118$$
From this we have that your statistic is bounded by: $\frac{1}{13} \approx 0.0769231$ for $n > 118$. The 118th prime is $647$, and a quick numerical analysis verifies this.
Mathematica code to verify this (which should print nothing):
Do[If[Prime[k + 1]/Prime[k] - 1 >= 14/113, Print[k]], {k, 31, 119}] | 2019-08-18T17:03:37 | {
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http://mathoverflow.net/questions/172290/codes-in-which-a-group-of-words-are-pairwise-different-at-a-certain-position | # “Codes” in which a group of words are pairwise different at a certain position
I read the following problem, claimed to be in the IMO shortlist in 1988:
A test consists of four multiple choice problems, each with three options, and the students should give an unique answer to each problem. After the test, one finds that for every three students, there is one problem to which their answers are pairwise different. What is the maximum number of students who took the test?
The answer is 9, with the following construction (actually a ternary code of length $4$ and hamming distance $3$).
1:0000 2:0111 3:0222
4:1012 5:1120 6:1201
7:2210 8:2102 9:2021
The following is my question:
Let $C$ be a set of $q$-ary words of length $\ell$. What is the maximum size of $C$ such that, for any $k$ words ($k\le q$), there is a position at which the $k$ words are pairwise different?
When $k=2$, this is the usual $q$-ary code with hamming distance $1$. Corresponding to the problem above, we have $q=3$, $\ell=4$ and $k=3$.
I don't know if this problem is previously studied. I didn't find any general answer.
More specifically, can you give an answer for $q=3$, $\ell=6$ and $k=3$?
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It is called perfect hash families in the design theory and computer science literature.
A perfect hash family PHF$(N; k, v, t)$ is an $N \times k$ array on $v$ symbols with $v \geq t$, where for every $N \times t$ subarray at least one row consists of distinct symbols. The following shows a PHF$(6; 12, 3, 3)$. $$\left[ \begin{array}{cccccccccccc} 1&0&0&2&2&2&1&1&2&1&0&2\\ 2&0&1&1&2&0&2&0&1&1&2&1\\ 2&0&2&1&2&1&0&2&2&1&1&0\\ 0&1&2&2&1&2&2&0&1&1&0&0\\ 2&0&1&2&1&1&2&2&0&1&2&1\\ 0&2&1&0&2&2&2&1&0&1&2&1 \end{array}\right]$$
For instance, if we take the second, third and fifth columns, then the the second row is $(0,1,2)$, which consists of distinct symbols. It is readily checked that for any combination of three columns, at least one row is free from identical symbols. To avoid triviality, we generally assume that there is at least one row in which all $v$ elements appear and that $k \geq v$. In the combinatorial design theory literature, the parameter $t$ is called strength.
By regarding your codewords in $C$ as the columns, you can easily see that a PHF$(l; \vert C\vert, q, k)$ is exactly the code you defined. Note that my $k$ is your $\vert C \vert$ while your $k$ is my $t$.
Perfect hash families are introduced by Kurt Mehlhorn in 1980's as an efficient tool for compact storage and fast retrieval of frequently used information, such as reserved words in programing languages and command names in interactive systems. PHFs have since found many other applications in computer science and combinatorics (See, for example,
A. Fiat, M. Naor, Broadcast encryption, Lecture Notes in Computer Science, 773:480–491, 1994.
S. R. Blackburn, M. Burmester, Y. Desmedt, P. R. Wild, Efficient multiplicative sharing schemes, Lecture Notes in Computer Science, 1070:107–118, 1996.
D. R. Stinson, T. van Trung, R. Wei, Secure frameproof codes, key distribution patterns, group testing algorithms and related structures. J. Statist. Plann. Infer., 86 (2000), 595–617.
for some use found in the last century in cryptography and related areas).
Very recent applications include compressive sensing in signal processing and covering arrays for software interacting testing in engineering.
Their existence and generalizations such as separating hash families have been extensively investigated. So if you're interested, you've got a large body of literature to dig in. Since you tagged your question as "combinatorial designs," you find informative the section on PHF in the Handbook of Combinatorial Designs. It's Section 43 of Chapter VI (page 566-568).
There are so many results on bounds on parameters, explicit constructions, and applications inside and outside of mathematics. I think the section in the Handbook of Combinatorial Designs is a good starting point. Tables of known PHFs (which may be a bit outdated) can be found here:
http://www.phftables.com/
Also, if you are curious, the equivalence between a matrix satisfying these conditions and a perfect family of hash functions in the computer theoretic sense is as follows:
A $(k,v)$-hash function is a function $h\ :\ A \rightarrow B$ between finite sets $A$ and $B$, where $\vert A \vert = k$ and $\vert B \vert = v$. The function $h$ is perfect with respect to a subset $X \subseteq A$ if $h$ is injective on $X$, that is, if $h\vert_X$ is one-to-one. Given integers $k$, $v$, and $t$ satisfying $k \geq v \geq t \geq 2$, let $H$ be a set of $N$ $(k, v)$-hash functions between finite sets $A$ and $B$, where $\vert A \vert = k$ and $\vert B \vert = v$. Then $H$ is a perfect hash family of parameters $(N; k, v, t)$ if for any $X \subseteq A$ with $\vert X \vert = t$, there exists at least one $h \in H$ such that $h\vert_X$ is one-to-one. This definition is equivalent to the one given earlier. In fact, one can obtain a perfect family of hash functions by regarding each row of a matrix representation of a PHF$(N; k, v, t)$ as a function $h$, where the value in column $i$ of the row for $h$ represents the value of $h(i)$.
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This answer is perfect for me. Thank you. – Hao Chen Jun 20 '14 at 15:10
By the way: This PHF(6;12,3,3), is it "optimal"? Is there a PHF(6;13,3,3)? – Hao Chen Jun 20 '14 at 15:11
@HaoCHEN It is optimal in the sense that no PHF$(5;12,3,3)$ exists, i.e., you need at least $6$ rows to have $12$ columns with the all-distinct-symbol property. But I need to check the existing literature or prove it myself to see if it's optimal in your sense. I'm pretty sure it's the best among known PHF$(6;\vert C\vert,3,3)$, though. I'll edit my post if I find the answer to the optimality. Or you can prove it and improve this post, too! – Yuichiro Fujiwara Jun 20 '14 at 15:26
In fact, I just saw a construction of (5;13,3,3), found by exhaustion (it seems), and I just checked it by program. Here it is: ['000000', '011111', '000112', '001120', '022210', '101200', '112022', '120111', '202101', '210200', '220221', '221011', '221202'] It is possible that I misunderstand something. – Hao Chen Jun 20 '14 at 15:30
Here is the link bbs.tianya.cn/post-71-144903-1.shtml – Hao Chen Jun 20 '14 at 16:33 | 2015-04-01T22:59:22 | {
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https://math.stackexchange.com/questions/2923770/the-initial-boundary-value-problem-of-the-heat-equation | # The initial boundary value problem of the heat equation
Let $U$ be a region in $\mathbb{R}^n$, $T>0$ be a constant and $U_T=U×(0,T)$.
Pick any smooth function $g:U→\mathbb{R}$ and define $A$ = {$v:U→\mathbb{R}$ smooth | $v(x)=g(x)$ for all $x$ ∈ $\partial U$}.
Now pick any $h∈A$ and suppose that $u: U_T→\mathbb{R}$ is a smooth solution of the following initial boundary value problem of the heat equation, i.e.,
$u_t (x,t)=\Delta u(x,t)$ for all $(x,t)∈U_T$,
$u(x,t)=g(x)$ for all $(x,t) ∈ \partial U × (0,T)$,
$u(x,0)=h(x)$ for all $x∈U$,
with initial data h and boundary data g.
Define an energy $E[u](t)=\frac{1}{2} \int_U |Du(x,t)|^2dx$, which gives a well-defined finite number for every $t∈(0,T)$.
1. Prove that $\frac{d}{dt}E[u](t)≤0$ for all $t∈(0,T)$ and that $\frac{d}{dt}E[u](t)=0$ for some $t∈(0,T)$ if and only if $\Delta u(x,t)=0$ for all $x∈U$.
Here I've got $\frac{d}{dt}E[u](t)=\int_U(Du_t Du)dx = -\int_U u_t \Delta u dx + \int_ {\partial U} \frac{\partial U}{\partial \nu} u_t dS(x) = -\int_U (u_t)^2 dx + \int_ {\partial U} \frac{\partial U}{\partial \nu} u_t dS(x)$.
Then how can I say that it's less or equal to zero?
2. Then use $\frac{d}{dt}E[u](t)≤0$ for all $t∈(0,T)$ to establish that the initial boundary value problem above has at most one smooth solution $u: U_T→\mathbb{R}$.
• Which part are you having problems with? You will need to integrate by parts to show that $\frac{d}{dt} E[u](t)=-\int_U |u_t|^2 dx\le 0$. For the uniqueness, take two solutions $u_1$ and $u_2$ and show $E[(u_1-u_2)](t)\equiv 0$, as that energy is nonnegative, initally zero and (weakly) decreasing. – Kusma Sep 20 '18 at 10:02
• Thanks for your help. :) I have edited the question and added the part I'm stucking at. I see a bit different in the integrate results.. ah – Evelyn Venne Sep 20 '18 at 10:23
• Note that $u_t=0$ on the boundary because $u$ satisfies $u(x,t)=g(x)$ on the boundary. – Kusma Sep 20 '18 at 11:14
• oh right thanks heaps :) – Evelyn Venne Sep 20 '18 at 11:31
Let's see that $\frac{d}{dt}E[u](t)≤0$, with equality if and only if $\Delta u(x,t)=0.$ $$\frac{d}{dt}E[u](t)=\frac{1}{2}\frac{d}{dt}\int_{U}\sum_{i=1}^{n}(\partial_iu(x,t))^2dx= \int_{U}\sum_{i=1}^{n}\partial_iu(x,t)\partial_t\partial_iu(x,t)dx=\\ \sum_{i=1}^{n}\int_{U}\partial_iu(x,t)\partial_i\partial_tu(x,t)dx\stackrel{(1)}{=}\\ -\sum_{i=1}^{n}\int_{U}\partial_i\partial_iu(x,t)\partial_tu(x,t)dx+ \sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tu(x,t)\nu_i(x) dS(x)\stackrel{(2)}{=}\\ -\int_{U}(\Delta u(x,t))^2dx+\sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tg(x)\nu_i(x) dS(x)=-\int_{U}(\Delta u(x,t))^2dx\leq0.$$ Where the equality $(1)$ is just integration by parts and in $(2)$ we are using the fact that $u$ satisfies the heat equation and the boundary condition.
For the second part let $u$, $v$ be solutions of the initial boundary value problem. Using the first part we know that the energy is nonincreasing. If we show that $E[u-v](0)=0$ then $E[u-v](t)=0$ for every $t$, being $E$ nonnegative. $$E[u-v](0)=\frac{1}{2}\int_{U}\vert Du(x,0)-Dv(x,0)\vert^2 dx= \frac{1}{2}\int_{U}\vert Dh(x)-Dh(x)\vert^2 dx=0.$$ This means that $Du(x,t)=Dv(x,t)$ for every $x$ and $t$ and also $\partial_t u(x,t)=\partial_t v(x,t)$, thanks to the first part. Thus $u(x,t)=v(x,t)+c$ for some constant $c$. Using the initial value you get $c=0$ | 2019-08-26T09:31:20 | {
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https://powerofcredit.com/i7z8i3bq/rolle%27s-theorem-khan-academy-60cce1 | Khan Academy is a 501(c)(3) nonprofit organization. the slope of the secant line. And continuous about this function. some function f. And we know a few things At some point, your We know that it is This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5. these brackets here, that just means closed interval. the average change. it's differentiable over the open interval And the mean value Rolle's theorem is one of the foundational theorems in differential calculus. So when I put a And if I put the bracket on Explain why there are at least two times during the flight when the speed of And so let's just try of change, at least at some point in So that's a, and then that at some point the instantaneous rate just means we don't have any gaps or jumps in where the instantaneous rate of change at that some of the mathematical lingo and notation, it's actually is equal to this. 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 this right over here, 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. Let's see if we of the tangent line is going to be the same as Problem 4. And so let's just think at those points. (“There exists a number” means that there is at least one such… Problem 3. about when that make sense. Thus Rolle's Theorem says there is some c in (0, 1) with f ' ( c) = 0. So let's just remind ourselves point a and point b, well, that's going to be the continuous over the closed interval between x equals This theorem is used to prove statements about a function on an interval starting from local hypotheses about derivatives at points of the interval. Rolle's theorem says that somewhere between a and b, you're going to have an instantaneous rate of change equal to zero. is the secant line. Mean Value Theorem. And we can see, just visually, Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. it looks like right over here, the slope of the tangent line function, then there exists some x value mean, visually? Check that f(x) = x2 + 4x 1 satis es the conditions of the Mean Value Theorem on the interval [0;2] … 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. So some c in between it Since f is a continuous function on a compact set it assumes its maximum and minimum on that set. is it looks like the same as the slope of the secant line. theorem tells us that there exists-- so It is one of the most important results in real analysis. More precisely, the theorem … And then this right Now what does that as the average slope. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Mean value theorem example: polynomial (video) | Khan Academy Applying derivatives to analyze functions. AP® is a registered trademark of the College Board, which has not reviewed this resource. f is differentiable (its derivative is 2 x – 1). And differentiable Sal finds the number that satisfies the Mean value theorem for f(x)=x_-6x+8 over the interval [2,5]. over this interval, or the average change, the 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. In the next video, the function over this closed interval. differentiable right at b. Let. 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). Let f(x) = x3 3x+ 1. theorem tells us is that at some point of course, is f(b). about some function, f. So let's say I have Our change in y is f ( 0) = 0 and f ( 1) = 0, so f has the same value at the start point and end point of the interval. https://www.khanacademy.org/.../ab-5-1/v/mean-value-theorem-1 here, the x value is a, and the y value is f(a). If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Khan Academy is a 501(c)(3) nonprofit organization. a and b, there exists some c. There exists some over our change in x. Hence, assume f is not constantly equal to zero. He also showed me the polynomial thing once before as an easier way to do derivatives of polynomials and to keep them factored. such that a is less than c, which is less than b. it looks, you would say f is continuous over There is one type of problem in this exercise: Find the absolute extremum: This problem provides a function that has an extreme value. And so when we put So in the open interval between AP® is a registered trademark of the College Board, which has not reviewed this resource. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. If you're seeing this message, it means we're having trouble loading external resources on our website. that mathematically? If you're seeing this message, it means we're having trouble loading external resources on our website. And I'm going to-- So that's-- so this 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. let's see, x-axis, and let me draw my interval. over the interval from a to b, is our change in y-- that the It’s basic idea is: given a set of values in a set range, one of those points will equal the average. ^ 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 theorem is named after Michel Rolle. Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. All the mean value Illustrating Rolle'e theorem. in between a and b. in y-- our change in y right over here-- At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. c. c c. c. be the number that satisfies the Mean Value Theorem … Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). a quite intuitive theorem. So now we're saying, 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 … and let. the average slope over this interval. this is b right over here. And so let's say our function can give ourselves an intuitive understanding The Extreme value theorem exercise appears under the Differential calculus Math Mission. And as we saw this diagram right This means you're free to copy and share these comics (but not to sell them). 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 Over b minus b minus a. I'll do that in that red color. that's the y-axis. Use Rolle’s Theorem to get a contradiction. Or we could say some c One of them must be non-zero, otherwise the … over here, the x value is b, and the y value, All it's saying is at some So at this point right over line is equal to the slope of the secant line. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. So the Rolle’s theorem fails here. 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. Welcome to the MathsGee STEM & Financial Literacy Community , Africa’s largest STEM education network that helps people find answers to problems, connect … 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. If f(a) = f(b), then there is at least one point c in (a, b) where f'(c) = 0. over here is the x-axis. This means that somewhere between a … The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. The average change between In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. differentiable right at a, or if it's not Now, let's also assume that - [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. Which, of course, Donate or volunteer today! f, left parenthesis, x, right parenthesis, equals, square root of, 4, x, minus, 3, end square root. Our mission is to provide a free, world-class education to anyone, anywhere. is that telling us? At this point right slope of the secant line, is going to be our change that means that we are including the point b. Check out all my Calculus Videos and Notes at: http://wowmath.org/Calculus/CalculusNotes.html The mean value theorem is still valid in a slightly more general setting. just means that there's a defined derivative, So those are the f(b) minus f(a), and that's going to be this is the graph of y is equal to f(x). And as we'll see, once you parse If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. So this is my function, if we know these two things about the Draw an arbitrary rate of change is equal to the instantaneous 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. interval between a and b. 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. rate of change at that point. Greek letter delta is just shorthand for change in case right over here. Applying derivatives to analyze functions. for the mean value theorem. looks something like that. He showed me this proof while talking about Rolle's Theorem and why it's so powerful. Now how would we write (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. value theorem tells us is if we take the f is a polynomial, so f is continuous on [0, 1]. what's going on here. rate of change is going to be the same as x value is the same as the average rate of change. After 5.5 hours, the plan arrives at its destination. f ( x) = 4 x − 3. f (x)=\sqrt {4x-3} f (x)= 4x−3. And it makes intuitive sense. the average change. we'll try to give you a kind of a real life example The Common Sense Explanation. The “mean” in mean value theorem refers to the average rate of change of the function. A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. L'Hôpital's Rule Example 3 This original Khan Academy video was translated into isiZulu by Wazi Kunene. The slope of the tangent Well, the average slope the right hand side instead of a parentheses, Thus Rolle's theorem claims the existence of a point at which the tangent to the graph is paralle… So nothing really-- Well, what is our change in y? well, it's OK if it's not the average rate of change over the whole interval. Our mission is to provide a free, world-class education to anyone, anywhere. You're like, what So all the mean interval, differentiable over the open interval, and y-- over our change in x. The student is asked to find the value of the extreme value and the place where this extremum occurs. | 2021-04-23T13:54:11 | {
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https://math.stackexchange.com/questions/1889450/extrema-of-ellipse-from-parametric-form | # Extrema of ellipse from parametric form
I'm trying to derive a formula to determine a tight bounding box for an ellipse. This is trivial for non-rotated ellipses, but I'm having trouble figuring out how to compute bounds for ellipses that have been rotated about their center.
Consider an ellipse $E$ centered at the origin with $x$ and $y$ radii $r_x$ and $r_y$ respectively. Then a point on $E$ is given by the parametric coordinate pair $\left[\begin{matrix}r_x \cos t \\ r_y \sin t\end{matrix}\right]$ for $t \in [0, 2\pi]$.
Suppose we rotate $E$ about the origin by $\theta$ radians. Then a point on $E$ has the coordinates $$\left[ \begin{matrix} x(t) \\ y(t) \end{matrix} \right] = \left[\begin{matrix}r_x \cos \theta \cos t - r_y \sin \theta \sin t \\ r_x \sin \theta \cos t + r_y \cos \theta \sin t \end{matrix}\right]$$
My approach to determine extrema was to consider each coordinate separately, take the derivative and set it equal to zero. For instance, we have $$x'(t) = -r_x \cos \theta \sin t - r_y \sin \theta \cos t$$ and setting $x'(t) = 0$ yields:
\begin{align} -r_x \cos \theta \sin t - r_y \sin \theta \cos t &= 0 \\ -r_x \cos \theta \sin t &= r_y \sin \theta \cos t \\ \tan t &= -\frac{r_y \sin \theta}{r_x \cos \theta} \end{align}
I don't know how to go from this equation to the actual extreme values of $x$ for the rotated ellipse. I'm assuming it's something easy that I've just forgotten how to do.
Edit: I'm looking for tight axis-aligned bounds. Rotating the bounds of a non-rotated ellipse and then computing axis-aligned bounds of the rotated bounding box doesn't result in tight bounds. Here is a diagram to illustrate:
• are there any restrictions on the bounding box? couldn't you just rotate the bounding box through the same angle? – Winston Aug 11 '16 at 19:44
• if the bounding box needs to remain aligned with the axes, then you can maximize/minimize x(t) by solving for t in the tan(t) equation you have above (note there are infinite solutions) and plugging back in for x(t) (only two unique solutions). Then maximize/minimize y(t) similarly. – Winston Aug 11 '16 at 19:48
You're basically done: all you need to do is solve for your values of $t$, and substitute them into your $x(t)$ to obtain your extreme $x$ values:
$$t = \tan^{-1} \left(-\frac{r_y \sin \theta}{r_x \cos \theta}\right)$$
However, $\tan^{-1}$ has an infinite number of solutions of the form $t + \pi n$, for $n \in \mathbb{Z}$. This set of solutions can be divided into two sets: $A$, which contains values of the form $t + 2 \pi n$ and $B$, which contains values of the form $t + \pi + 2 \pi n$.
The only part you need to be careful of is to take one value of each of $A$ and $B$. One of these two sets of solutions will correspond to the lower extreme, and the other will correspond to the upper extreme.
If you want to know which set of values of $t$ corresponds to the minimum or maximum, you could use the second derivative test to find out.
• I feel a little silly for asking, but how do you solve for values of $t \in [0, 2\pi]$ in your equation with arcsine? – tdenniston Aug 11 '16 at 22:28
• Did you mean arctan? By convention, the range of $\tan^{-1}$ is $[-\pi / 2, \pi / 2]$, so upon computing $t$, you will have $t$ and $t + \pi$ as your two solutions (using the fact that $\tan(\alpha + \pi) = \tan(\alpha)$). Upon further consideration, I don't actually think it's necessary for you to restrict the values of $t$ to be within $[0, 2\pi)$, since $x(t + 2\pi) = x(t)$, and similar for $y(t)$. – theyaoster Aug 11 '16 at 23:30
• Thanks! Got it working. For future readers, one can pick any two values for $n$ in $t+\pi n$ so long as $|n_1 - n_2| = 1$. – tdenniston Aug 12 '16 at 13:53
By $$A\cos t+B\sin t= \sqrt{A^2+B^2} \sin (t+\phi)$$
The dimensions of the rectangle are $$2\sqrt{r_x^2\cos^2 \theta+r_y^2\sin^2 \theta} \times 2\sqrt{r_x^2\sin^2 \theta+r_y^2\cos^2 \theta}$$
See the animation below:
• very nice animation, how did you make it? – samjoe May 23 '18 at 14:11
• Geogebra which is freeware. You may download here. – Ng Chung Tak May 23 '18 at 14:13
An alternative approach would be to consider the ellipse as fixed, and rotate the rectangle clockwise by angle $\theta$ so that the sides are tangential to the ellipse.
Let $m=\tan\theta$ so that the equation of one of the sides of the rectangle is $$y=-mx+c_1$$
Let the equation of a neighbouring side be $$y=\frac xm+c_2$$
Using the standard formula for the distance of a point to a line, considering the distances from the origin, the area of the bounding rectangle is $$4\times\frac{c_1}{\sqrt{1+m^2}}\times\frac{c_2}{\sqrt{1+\frac{1}{m^2}}}=\frac{4mc_1c_2}{1+m^2}$$
Now considering the standard ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
The first line is tangent to the ellipse provided that, when solved simultaneously with the ellipse, the resulting quadratic has double roots.
After a couple of lines of algebra, we arrive at the condition $$a^2m^2=c_1^2-b^2\Rightarrow c_1=\sqrt{a^2m^2+b^2}$$
By similar argument we likewise obtain $$c_2=\sqrt{\frac{a^2}{m^2}+b^2}$$
Substituting these into the above expression for the area of the bounding rectangle, we end up with the formula$$\frac{4\sqrt{a^2m^2+b^2}\sqrt{a^2+m^2b^2}}{ 1+m^2}$$
• I'm looking for tight axis-aligned bounds, and unfortunately this approach (which I tried first) does not result in tight bounds. I should have clarified that in my question. – tdenniston Aug 11 '16 at 22:27
• Can you clarify what you mean by " tight axis aligned bounds"? The dimensions of the bounding rectangle are clear from my answer. – David Quinn Aug 11 '16 at 22:39
• The rotated bounding rectangle (such that the sides are tangential to the rotated ellipse) isn't axis aligned, meaning the sides of the rectangle are not parallel to the axes. So the rotated bounds are indeed tight, but not axis aligned. – tdenniston Aug 11 '16 at 22:41
• All you need are the perpendicular distances from the centre of the rectangle (and the ellipse) to the edges of the rectangle. These are the axis aligned distances you are looking for. The process of rotating the rectangle rather than the ellipse is equivalent to rotating the ellipse and keeping the rectangle aligned to the original axes. – David Quinn Aug 11 '16 at 22:47
• I don't quite understand how the perpendicular (I'm assuming perpendicular to the axes?) distances from the center of the ellipse to the edges of the rotated rectangle can give the bounds of the rotated ellipse. – tdenniston Aug 11 '16 at 22:56 | 2019-09-18T07:04:51 | {
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https://math.stackexchange.com/questions/3353330/proof-verification-union-of-two-countable-sets-is-countable | # Proof verification : Union of two countable sets is countable
Sincere request, don't forget to address my doubt at the end of the proof
I have assumed my sets to be disjoint at first but I have also addressed the general scenario as the proof progresses.
Set $$A$$ is said to be countable if there exists a bijection from $$A$$ to $$\mathbb{N}$$. Every countable set is infinite
To show that : Union of two countable sets is countable
Suppose $$A$$ and $$B$$ are countable. Assume at first that $$A\cap B=\phi$$
$$A$$ countable $$\Rightarrow \exists f:A\to \mathbb{N}$$ a bijection.
$$B$$ countable $$\Rightarrow \exists g:B\to \mathbb{N}$$ a bijection.
define. $$h:A\cup B \to N$$ as
$$x\mapsto 2f(x) \;$$ if $$x\in A$$
$$x\mapsto 2g(x)+1$$ if $$x\in B$$
Because $$A\cup B$$ is infinite, it is sufficient to show that $$h$$ is injective in order to show that $$A\cup B$$ is countable.
if $$x=y$$, where $$x,y\in A\cup B$$, since $$A$$ and $$B$$ are disjoint, so, either both $$x$$ and $$y$$ belong to $$A$$ or both belong to $$B$$, and because $$f$$ and $$g$$ are well defined, so is $$h$$
Now let $$h(x)=h(y)$$ where $$x,y \in A\cup B$$
again, $$x$$ and $$y$$ can both belong to $$A$$ or can both belong to $$B$$. Hence injectivity of $$h$$ on $$A\cup B$$ follows directly from the injectivity of $$f$$ and $$g$$ on $$A$$ and $$B$$ respectively
Hence, $$A\cup B$$ is countable.
Now, let $$A$$ and $$B$$ be arbitrary countable sets,
then by above method, $$A\cup B = [A\setminus (A\cap B)]\cup[A\cap B]\cup [B\setminus (A\cap B)]$$ is countable.
Doubt : Is it safe to assume $$A\cap B = \phi$$ in the beginning of the proof? I am doubtful here because $$A$$ and $$B$$ are countable. Please address this problem first
• title says product but body says union !? – J. W. Tanner Sep 11 at 21:40
• I am sorry, edited – Abhay Sep 11 at 21:41
• If you are breaking into cases you do Case 1: $A$ and $B$ can do what ever you darned well want and Case 2: $A$ and $B$ don't do the stuff. So there is nothing wrong with saying Case 1: $A$ and $B$ are disjoint. Case 2: they are not disjoint. – fleablood Sep 11 at 22:31
• But how do you know $A\setminus (A\cap B), B \setminus (A\cap B)$ and $A \cap B$ are countable? And you defined countably infinite. What if some of these are finite? – fleablood Sep 11 at 22:33
• It is customary that "countable" means "finite or countably infinite" so that "uncountable" and "not countable" both mean "uncountably infinite". – DanielWainfleet Sep 12 at 1:07
Your proof is just about fine. It's perfectly acceptable to start with the case $$A \cap B = \emptyset$$ as long as you later address the possibility that $$A \cap B \neq \emptyset$$, which you do.
Edited to add: Tuvasbien is correct to note that the final step of your proof can be simplified. $$A \cup B = A \cup (B \setminus A)$$, and $$A \cap (B \setminus A) = \emptyset$$. | 2019-12-11T22:18:58 | {
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https://math.stackexchange.com/questions/3527698/counting-combinations | # Counting (Combinations)
Suppose there are 10 people apart of a club: A, B, C, D, E, F, G, H, I, and J. They decide to go to a restaurant for a club outing, but there isn't one table to seat all of them, so they decide to take one table that seats four people and two tables that seat three people. Based on this, what is the probability that Person G and Person J sit at the same table?
• Ways for both Person G and Person J to both sit at the 4-person table: 6 = (4 choose 2)
• Ways for both Person G and Person J to both sit at the 3-person table: 3 = (3 choose 2)
• Ways for both Person G and Person J to both sit at the other 3-person table: 3 = (3 choose 2)
Total of $$6+3+3=12$$ ways for Person G and Person J to both sit at the same table.
I've determined the number of ways that Person G and Person J can sit together at each table, but I don't know how to properly count the total different ways (i.e., the denominator) to determine the probability.
• Are we counting the two tables of $3$ as distinct tables? – scoopfaze Jan 30 at 2:08
• so far off. think about the other tables, if they chose their spots, if reordering changes things, all these affect the answer... 6 just aint cutting it. – user645636 Jan 30 at 2:33
• Is the seat of each person important, or only at which table they seat? – Alain Remillard Jan 30 at 2:40
• @AlainRemillard Only the table – 324 Jan 30 at 2:44
• so not the other people at it, not where they sit within the table seating, not anything interesting ? if so you over estimated by a factor of 2. – user645636 Jan 30 at 2:49
As pointed out in the comments, it depends whether the 3-persons tables are distinguishable or not.
If the 3-persons table are distinguished
There is Three cases to look at.
• If $$G$$ and $$J$$ sit at the 4-persons table, we need two more to seat with them, from remaining eight. Then choose the three, from remaining six for first 3-persons table, the last three goes at the last table. $${8\choose2} {6\choose3} {3\choose3}=560$$
• if they sit at the first 3-persons table, we need one more to complete the table. Then split the remaining seven. $${8\choose1} {7\choose4}{3\choose3}=280$$
• If they sit at the second 3-persons table, it is exactly the same as above. $${8\choose1} {7\choose4}{3\choose3}=280$$
The number of possibilities that $$G$$ and $$J$$ are seated together. $$560+280+280=1120$$
For the total possibilities, we choose four, from ten, for the 4-persons table, then three from remaining six for the first 3-persons table, the last three goes on the last table. $${10\choose4}{6\choose3}{3\choose3}=4200$$
The probability that $$G$$ and $$J$$ are seated together is $$\frac{1120}{4200}=\frac{4}{15}$$
If the 3-persons tables are not distinguished
Funny thing, the probability is the same. Simply divide by two, since every possibility has been counted twice. $$1120\div2=560$$ ways to seat them together, and $$4200\div2=2100$$ total possibilities. $$\frac{560}{2100}=\frac{4}{15}$$
Here is another approach. We assume that the $$10$$ people arrange themselves in some random order and then sit down, so there are $$10!$$ possible orders, all of which we assume are equally likely, with the first $$4$$ people at the first table, the next $$3$$ at the second, and the last $$3$$ at the third.
If G and J sit together at the first table, then their places can be chosen in $$4 \times 3$$ ways, and the remaining people can be placed in $$8!$$ ways.
If G and J sit at the second or third tables, then their places can be chosen in $$3 \times 2$$ ways, and then the remaining people can be placed in $$8!$$ ways.
So the probability that G and J sit at the same table is $$\frac{4 \times \ 3 \times 8! + 2 \times 3 \times 2 \times 8!}{10!}= \frac{4}{15}$$
• clever approach +1 ! – G Cab Jan 30 at 19:00
By conditional probability, we have
$$P(J\text{ is at G’s table}) = P(G\text{ at 4-table})P(J\text{ at G’s table} | G\text{ at 4-table}) + P(G\text{ at a 3-table})P(J\text{ at G’s table}| G\text{ at a 3-table})$$
$$\left(\frac4{10}\right)\left(\frac39\right)+\left(\frac6{10}\right)\left(\frac29\right) = \frac2{15} + \frac2{15} = \frac4{15}$$
I am considering the tables to be identical because you haven, t clarified that in the question, so in question we have 6 seats in two tables, where each table contains exact three seats, and one table which contain exact 4 seats.
So we treat both ,G, J as together one. Case 1=when both seat on table 1 consist of 4 people So we have 8 options left to choose 2 spots because rest 2 were already occupied by them. , and similarly for next table we have only 6 options left to choose three people and left 3 people have only one choice that is last, and for cases 2)you can count the similar way. Now total umber of possible cases are we choose 4 out of 10 people from table one, and 3 out of 6 people, and left three with left chairs. So by calculating the value you will get the answer $$4/15$$. | 2020-05-31T10:24:21 | {
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http://mathhelpforum.com/advanced-statistics/100021-probability-sampling-replacement.html | # Math Help - Probability Sampling with replacement
1. ## Probability Sampling with replacement
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.
I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?
2. Originally Posted by kittykat52688
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.
I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?
You have the second part right but the explanation is wrong.
The first ticket $k$ must be one of $1, 2,\ ..\ n-1$ (which occurs with prob $(n-1)/n$ and the second must be $k+1$ (drawn from $n$ tickets) which occurs (given that the first ticket is chosen correctly) with probability $1/n$. Hence the required probability is $(n-1)/n^2$
CB
3. Originally Posted by kittykat52688
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.
I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?
For the third part we can consider three possibilities, the first is greater than the second, the second is greater than the first, and that they are equal.
The first two are obviously equal (by symmetry), and the second is $1/n$.
These sum to one so let $p$ denote the probability that the second is greater than the first, then we have:
$2p+1/n=1$
hence:
$p=(1-1/n)/2$
CB
4. ## Re: Probability Sampling with replacement
Hi, I have a similar question.
A random sample n = 3 is selected from N with replacement.
I do not understand why the probability of having , two distinct
units is P = 3*(N-1)/(N^2), and three distinct units is P = (N -1)(N -2)/N^2
Thanks | 2015-08-28T15:15:03 | {
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https://math.stackexchange.com/questions/798522/is-this-notation-good-for-the-chain-rule-derivative | # Is this notation good for the chain rule derivative?
When we take this derivative, for example:
$$y = \log(\sin x)$$ We call $u = \sin x$, so we have:
$$\frac{dy}{dx} = \frac{d y}{du}\frac{du}{dx} = \frac{1}{u}\cos x = \frac{\cos x}{\sin x}$$
But for me, it's better to do:
$$\frac{d\log\color{Blue}{\sin x}}{d\color{Blue}{\sin x}}\frac{d\sin \color{Red}{x}}{d\color{Red}{x}} = \frac{1}{\color{Blue}{\sin x}}\cos \color{Red}{x}$$ It makes easy to do the 'pattern-matching' just by looking at the differentials. No substitution. I know that $\frac{d \log[\mbox{something}]}{d[\mbox{something}]} = \frac{1}{\mbox{something}}$ for example.
However, it looks 'hairy' when I try with larger derivatives, like, for the function:
$$(13x^2-5x+8)^{\frac{1}{2}}$$ we do: $$\frac{d(13x^2-5x+8)^{\frac{1}{2}}}{dx} = \frac{d\color{Green}{(13x^2-5x+8)}^{\frac{1}{2}}}{d\color{Green}{(13x^2-5x+8)}}\frac{d(13x^2-5x+8)}{dx} = \frac{1}{2\sqrt{\color{Green}{13x^2-5x+8}}}(26x -5)$$ but it's really better for me to do like this, instead of doing the bla bla bla of changing variables and stuff. But I'm afraid my teacher does not accept this. Is this notation/way of doing good for you guys?
One more example: $$\frac{d}{dx}\sqrt{(\sin(7x+\ln(5x)))} =$$
$$\frac{d[\color{Blue}{\sin(7x+\ln(5x))}]^{1/2}}{d[\color{Blue}{\sin(7x+\ln(5x))}]}\frac{d[\sin\color{Red}{(7x+\ln(5x))}]}{d[\color{Red}{7x+\ln(5x)}]}\left[\frac{d[7\color{Purple}{x}]}{d[\color{Purple}{x}]} + \frac{d[\ln(\color{Purple}{5x})]}{d[\color{Purple}{5x}]}\frac{d[5x]}{d[x]}\right] =$$ $$\frac{1}{2}\left[\color{Blue}{\sin(7x+\ln(5x))}\right]^{-1/2}\cdot\cos(\color{Red}{7x+\ln(5x)})\left[7 + \frac{1}{\color{Purple}{5x}}\cdot 5\right]$$ So we get rid of the substitution!
(づ｡◕‿‿◕｡)づ $\ \ u, v, y$ go away!
• This is probably the coolest (most decorated) question asked. (づ｡◕‿‿◕｡)づ – Shahar May 16 '14 at 23:16
• @Shahar thank you (｡◕‿◕｡) – Lucas Zanella May 16 '14 at 23:18
• @LucasZanella I must say, I'm really looking forward to this question's answer. I'm a self-learner, and do not know much about the commonly used notation. – user122283 May 16 '14 at 23:20
• As for the question, it's right but not very conventional. Hence, I'd assume your teacher might not accept it (depends on how he taught it). But you're doing it right. (｡◕‿◕｡) – Shahar May 16 '14 at 23:20
• omg you guys are so nice, I love this forum ♥ – Lucas Zanella May 16 '14 at 23:21
This notation is absolutely acceptable, more than fine, and often used extensively in integral calculus, especially by professional mathematicians. In this context, it is very convenient because the integration can be seen in all aspects. In other words, you are integrating with respect to a complicated expression and the form of the integrand makes sense with this expression and you want to show that clearly. In essence, it is sometimes used to be more explicit. It expresses steps more clearly at times and has the benefit of being able to visualize equalities more easily. Be warned that this notation does get overcomplicated at times and you will want to add some substitutions for sanity, but for simpler calculations, it can be useful and looked upon favorably. You should look to study differential calculus as a topic in its own right, where this notation is used as well (in summary, the "denominator" of the derivative $\text{d}[\text{something such as} \,x]$ is eliminated and we just deal with what are called differentials). Your instructor is very intolerant, indeed, if he doesn't allow this. As a matter of fact, there was a point towards the end of my high school career (at this point, I was studying beyond what I was enrolled in) I used this notation in a "Calculus II" type class (class focusing on expressing calculus with infinite series and basically a structured introduction to approximating evaluations of functions using different types of series), and my teacher (actually a PhD, but taught calculus at my high school in his spare time - shoutout to Dr. Brandell), and my mentor in my early mathematics self-study, commended me and demonstrated this point of view to the rest of my peers. This is very innovative actually for a young mathematician and shows promise. I'd even be willing to talk to to your instructor personally and make an argument in your defense, backed by some of my former and current professors. Good luck in your pursuit of knowledge in mathematics and keep discovering ways to improve your notation (and, most importantly, keep asking questions)! | 2019-05-27T05:00:36 | {
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https://ftclausen.github.io/mathematics/josephus-problem-revisited/ | # Josephus Problem - Recurrence - Revisited - Unrolled
I am restarting Concrete Mathematics (Graham, Knuth, Patashnik) due to a forced break due to a neurological problem called Anti-NMDA receptor encephalitis. I include this detail because it is a relatively newly discovered disease and misdiagnosis is common so, in a very small way, raising awareness.
Maths is my friend in the recovery process even though it is challenging. Perhaps because it is challenging.
This post goes over the recurrence only because I found that challenging. For some reason I find recursion harder than it should be 😅
# Setting the Scene with Some Examples
While the concept of shooting each second remaining person sounds obvious I found it useful to step through it on pen and paper which I will attempt to reproduce here in text; really an animation is also useful to look at such as this one. I won’t show a diagram because all you see is the end result which is not so useful
• $n = 1$ - Survivor is $1$ obviously because Josephus does not want to kill himself.
• $n = 2$
• 1 shoots 2
• 1 survives
• $n = 3$
• 1 shoots 2
• 3 shoots 1
• 3 survives
• $n = 4$
• 1 shoots 2
• 3 shoots 4
• 1 shoots 3
• 1 survives
• … and so on - see above linked video for a great explanation.
I was accidentally thinking the dead person was also shooting someone (😅) which meant I was messing up above. Here is a tabular form of the above as presented in the book:
$\begin{array}{|c|c|} \hline n & 1\ \ 2\ \ 3\ \ 4\ \ 5\ \ 6\ \\ \hline J(n) & 1\ \ 1\ \ 3\ \ 1\ \ 3\ \ 5\ \\ \hline \end{array}$
I think the point the book is trying to make is that, at each go around, we are essentially at the same point as the round before except there are half as many people _and their numbers (aka “label” or “position”) have changed. Or, as Wikipedia puts it:
The first time around the circle, all of the even-numbered people die. The second time around the circle, the new 2nd person dies, then the new 4th person, etc.; it is as though there were no first time around the circle.
# The Recurrence
This refers heavily to this post by, the ever helpful, Brian M. Scott. I am only adding even more details for my addled brain to help it understand.
First the recurrences (for base case, odd, and even) which is the notational expression of the above sentences:
\begin{align} J(1) &= 1 \\ J(2n) &= 2J(n) - 1, \ \ \ \ \ \ \text{for} \ n \ge 1; \\ J(2n + 1) &= 2J(n) + 1, \ \ \ \ \ \ \text{for}\ n \ge 1; \\ \end{align}
## Unrolling
The post by Brian M. Scott shows the case for $n = 21$ but I found it useful to step through the small cases as well.
Formatting note: I am using a comma followed by some notes. The font for the \text{...} is not beautiful but serves the purpose.
### n = 1
This is the base case so $J(1) = 1$
### n = 2
\begin{align} J(2) &= J(1\cdot2), \quad \text{expand} \\ &= 2J(1) - 1, \quad \text{recurse} \\ &= 2\cdot1 - 1 = 1, \quad \text{base case and finish} \end{align}
### n = 3
\begin{align} J(3) &= J(1\cdot2 + 1), \quad \text{expand} \\ &= 2J(1) + 1, \quad \text{recurse}\\ &= 2\cdot1 + 1 = 1, \quad \text{base case and finish}\\ \end{align}
### n = 4
\begin{align} J(4) &= J(2\cdot2), \quad \text{expand}\\ &= 2J(2) - 1, \quad \text{recurse} \\ &= 2J(1\cdot2) - 1, \quad \text{expand} \\ &= 2(2J(1) - 1) - 1, \quad \text{recurse} \\ &= 4J(1) - 3, \quad \text{simplify}\\ &= 4\cdot1 - 3 = 1, \quad \text{base case and finish} \end{align}
### n = 5
\begin{align} J(5) &= J(2\cdot2 + 1), \quad \text{expand} \\ &= 2J(2) + 1, \quad \text{recurse} \\ &= 2J(1.2) + 1, \quad \text{expand} \\ &= 2(2J(1) - 1 ) + 1, \quad \text{recurse} \\ &= 4J(1) - 2 + 1, \quad \text{simplify} \\ &= 4\cdot1 - 1 = 3, \quad \text{base case and finish} \end{align}
### n = 6
\begin{align} J(6) &= J(2\cdot3), \quad \text{expand} \\ &= 2J(3) - 1, \quad \text{recurse} \\ &= 2J(1\cdot3) - 1, \quad \text{expand} \\ &= 2(2J(1) + 1) - 1, \quad \text{recurse} \\ &= 2(2\cdot1 + 1) - 1 = 4 + 2 - 1 = 5, \quad \text{base case and finish} \end{align}
# Conclusion
Doing this manual unrolling really helps understand the nature of the problem when expressed in math notation. We can see how powers of two play a role which is useful in the closed for solution(s).
Categories:
Updated: | 2022-12-01T14:15:18 | {
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https://mathoverflow.net/questions/249030/recurrence-of-legendre-polynomial-roots-quadrature-points | # Recurrence of Legendre polynomial roots/ quadrature points
Consider Legendre polynomials $p_n (x)$ on $[-1,1]$. For each $n \in \mathbb{N}$ we denote the zeros of $p_n (x)$ by $\left( x_j ^{(n)} \right) _{j=1} ^n$.
We know that these roots are distinct, and we know that for all $n\in \mathbb{N}$ then $x_j ^{(n)} \neq x_i ^{(n+1)}$ for all $i$ and $j$. In fact, we even now that they are interlaced i.e. $x_i ^{(n+1)} < x_i ^{(n)} < x_{i+1} ^{(n+1)}$.
On the other hand, we know for example that for all odd $n=2k+1$ we have that $x_{k+1} ^{(2k+1)} = 0$, and so I arrive at my question:
Question 1: Are there any rules as to if, when, and how roots recur, i.e. $x_i ^{(n)} = x_j ^{(m)}$ for some relevant indices?
Question 2: Apart from the well known Chebyshev polynomials, are there any other orthogonal polynomials with high rate of roots recurrence?
• Does not recurrence relation among Legendre polynomial's (or Chebyshev, Hermite etc) and ODE they satisfy give you an information about a relation between consecutive zeros? – Paata Ivanishvili Sep 5 '16 at 2:42
• I assume I don't know ALL the relevant recurrence relations, but from the ones I do I couldn't deduce to recurrence of zeros. I'd be happy if you will prove me wrong. – Amir Sagiv Sep 5 '16 at 5:14
It is a conjecture of Stieltjes, apparently still open, see
T.J. Stieltjes, Letter No. 275 of Oct. 2, 1890, in Correspondance d'Hermite et de Stieltjes, vol 2, Gauthier-Villars, Paris, 1905.
that Legendre polynomials of different degrees have no common roots, except $x=0$ when both degrees are odd.
Laguerre polynomials $L_{n}^{\alpha}$, $\alpha>-1$, orthogonal with respect to the weight function $x^{\alpha}e^{-x}$ on $(0,\infty)$, may have common roots. For instance, as observed in
K. Driver, K. Jordaan, Stieltjes interlacing of zeros of Laguerre polynomials from different sequences. Indag. Math. (N.S.) 21 (2011), 204-211.
$L_{2}^{23}(x)$ and $L_{4}^{23}(x)$ have $x=30$ as a common root. It seems also to be an open question how many common zeros are possible in general for $L_{n}^{\alpha}$ and $L_{n+m}^{\alpha}$, $n\geq 0$ and $m>1$. The corresponding question for Hermite polynomials about a common root other than zero seems to be unanswered as well.
Finally, a classical theorem of Stieltjes states that if $p_{0},p_{1},p_{2},\ldots$ is any sequence of orthogonal polynomials then the zeros of $p_{k}$ and $p_{n}$, $k<n$, are interlacing in the sense that each open interval of the form $$(-\infty, z_{1}), (z_{1}, z_{2}), . . . , (z_{k-1}, z_{k}), (z_{k} , \infty)$$ where $z_{1}<z_{2}<\cdots<z_{k}$ are the zeros of $p_{k}$, contains at least one zero of $p_{n}$. It implies that at least $k+1$ zeros of $p_{n}$ are distinct from the zeros of $p_{k}$, or equivalently $p_{k}$ and $p_{n}$ have at most $\min(k,n-k-1)$ zeros in common. It is proved in
P.C. Gibson, Common zeros of two polynomials in an orthogonal sequence, J. Approx. Theory 105 (2000) 129-132.
that this bound is sharp in the sense that for any $k<n$, there exists a sequence of orthogonal polynomials $q_{0},\ldots,q_{n}$ such that $q_{k}$ and $q_{n}$ have precisely $\min(k,n-k-1)$ zeros in common.
Numerical evidence suggests that the nonzero roots of the Legendre polynomials do not repeat. The numerical experiment I performed is simple: I gathered all of the nonzero roots of the first 100 Legendre polynomials. I sorted them in ascending order. I then graphed the first order difference of this sorted vector with logarithmic scaling in the vertical direction. Here is the graphical output.
Its a bit hard to see, but there are no holes in this figure, which suggests that all 5000 nonzero roots of the first 100 Legendre polynomials are distinct.
all_roots=[]; | 2019-01-21T08:36:54 | {
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https://math.stackexchange.com/questions/2047933/finding-z-in-the-form-abi | # Finding $z$ in the form $a+bi$
I have to solve $$3z + 17i = iz + 11$$ and write $z$ in the form $a+bi$.
I substituted $a+bi$ for $z$ and got two simultaneous equations. $$3a+b = 11,\quad a-3b = -17$$ But am not getting the same answer. The answer is $z=5-4i$
• In your second equation, the $-17$ should be $17$. – John Hughes Dec 7 '16 at 12:00
• Oh my bad. Didn't realised my mistake – Ashalley Samuel Dec 7 '16 at 12:11
$$3z + 17i = iz + 11$$ $$(3-i)z= 11- 17i$$ $$z=\frac{11-17i}{3-i}$$ Now multiply the numerator and denomerator by the conjugate of the denomaretor and separate the real and imaginary part.
• Thanks ...i mishandled my negative and positive signs. – Ashalley Samuel Dec 7 '16 at 12:12
• You're welcome! – msm Dec 7 '16 at 12:13
$$3z+17i=iz+11$$
substituting $z=a+ib$,
$$3(a+ib)+17i=i(a+ib)+11$$
$$3a +3bi+17i=ai-b+11$$
$$3a+i(3b+17)=(11-b)+ai$$
Equating real part to real and imaginary part to imaginary part.
$3a=11-b$ and $a=3b+17$. Now you have two equations and two variables. I shall let you conclude now.
• Thanks buddy. Well noted – Ashalley Samuel Dec 7 '16 at 12:14
• @AshalleySamuel Always welcome.. – Vidyanshu Mishra Dec 7 '16 at 12:19
You just need to solve the system of equations
$$3a+b=11$$ $$a-3b=17$$
Using elimination techniques,
$$3a+b=11$$ $$3a-9b=51$$
Which implies
$$10b=-40$$
Can you go from there?
• I had a -17 which is where the problem started from. Thanks buddy – Ashalley Samuel Dec 7 '16 at 12:13
$$3z+17i=iz+11 \Rightarrow z(3-i)=11-17i$$
$$z=\frac{11-17i}{3-i}=\frac{(11-17i)(3+i)}{(3-i)(3+i)}=\frac{50-40i}{10}=5-4i$$ | 2020-01-27T22:01:43 | {
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https://math.stackexchange.com/questions/2594442/find-the-range-of-fx-11-cos2x3-sin2x6-sin-x-cos-x5 | # Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I'm trying to solve this problem.
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.
It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$
• Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44
Rewrite as following
\begin{align} f(x) & = 8 + 6 \sin(x ) \cos(x ) + 8\cos^2(x) - 4 + 4 \\ &= 12 + 3 \sin(2x) + 4 \cos(2x) \\ &= 12 + 5 \left(\frac{3}{5} \sin(2x) + \frac{4}{5} \cos(2x)\right) \\ &= 12 + 5 \sin(2x + \arctan{\tfrac{4}{3}}) \end{align}
Now its easy since $\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \pm 5$.
So maximum value is $17$ and minimum is $7$
write your function as $$f(x)=(\sin(x)+3\cos(x))^2+7$$
• This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59
• You should write $3\cos x + \sin x$ in the form $R\cos(x-\alpha)$, where $R>0$ and $0 < \alpha < \frac{1}{2}\pi$. Jan 6 '18 at 16:12
• you Can write your function as $$f(t)=4\,{\frac {4\,{t}^{4}-3\,{t}^{3}+3\,t+4}{ \left( {t}^{2}+1 \right) ^{2 }}}$$ with $$\tan(\frac{x}{2})=t$$ and use calculus Jan 6 '18 at 16:16
• @Dr.SonnhardGraubner or use that $\sin(x) + 3 \cos(x) \in [-\sqrt{1^2+ 3^2}, \sqrt{1^2+3^2}]$ Jan 7 '18 at 11:51
$$f(x)=8\left(\frac{1+\cos {2x}}{2}\right) +3\sin {2x}+8$$
$$f(x)=4\cos {2x}+3\sin {2x} +12$$
The range of $$4\cos {2x}+3\sin {2x}$$ is $[-5,5]$
Hence maximum and minimum values of expression are $17$ and $7$ respectively.
Hint:
Let $y=A\sin^2x+B\cos^2x+C\sin x\cos x+D$
Method$\#1:$ Divide both sides by $\cos^2x$ to form a Quadratic Equation in $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
Method$\#2:$ Divide both sides by $\sin^2x$ | 2021-10-27T22:42:10 | {
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https://mathoverflow.net/questions/404760/how-can-you-find-an-integer-coefficient-polynomial-knowing-its-values-only-at-a | # How can you find an integer coefficient polynomial knowing its values only at a few points (but requiring the coefficients be small)?
Example: How can you guess a polynomial $$p$$ if you know that $$p(2) = 11$$? It is simple: just write 11 in binary format: 1011 and it gives the coefficients: $$p(x) = x^3+x+1$$. Well, of course, this polynomial is not unique, because $$2x^k$$ and $$x^{k+1}$$ give the same value at $$p=2$$, so for example $$2x^2+x+1$$, $$4x+x+1$$ also satisfy the condition, but their coefficients have greater absolute values!
Question 1: Assume we want to find $$q(x)$$ with integer coefficients, given its values at some set of primes $$q(p_i)=y_i$$ such that $$q(x)$$ has the least possible coefficients. How should we do it? Any suggestion/algorithm/software are welcome. (Least coefficients means: the least maximum of modulii of coefficients).
Question 2: Can one help to guess the polynomial $$p$$ such that $$p(3) = 221157$$, $$p(5) = 31511625$$ with the smallest possible integer coefficients? (Least maximum of modulii of coefficients). Does it exist?
(That example comes from the question MO404817 on count of 3x3 anticommuting matrices $$AB+BA=0$$ over $$F_p$$.)
(The degree of the polynomial seems to be 10 or 11. It seems divisible by $$x^3$$, and I have run a brute force search bounding absolute values of the coefficients by 3, but no polynomial satisfying these conditions is found, so I will increase the bound on the coefficients and will run this search again, but the execution time grows too quickly as the bound increases and it might be that brute force is not a good choice).
Question 3: Do conditions like $$q(p_i)=y_i$$ imply some bounds on coefficients? E.g., can we estimate that the coefficients are higher than some bound?
• So no restrictions for the degree? Sep 28 at 13:39
• @PabloH No. But too high degree cannot be compatible with small integer coefs and fixed values p(x_i)=y_i, because if you would have x^{very high} it will be bigger than any fixed y_i Sep 28 at 13:47
You can certainly do better than brute force by considering modular constraints. If the solution is $$p(x) = \sum_i a_i x^i$$ then $$p(x) - \sum_{j=0}^{n-1} a_j x^j$$ is divisible by $$x^n$$ and $$\frac{p(x) - \sum_{j=0}^{n-1} a_j x^j}{x^n} = a_n \pmod x$$
Solving for $$a_0$$ in each of the given bases and using the Chinese remainder theorem gives an equivalence class for $$a_0$$; for each possible value of $$a_0$$ you can expand similarly for $$a_1$$; and traversing this tree in order of increasing cost of the coefficients gives a directed search. This works in principle for any cost function which increases when any coefficient increases in absolute value.
This Python code implements the idea and finds two polynomials with sum of absolute values of 29:
$$-2x^3 + 4x^4 + 3x^5 + 11x^6 + x^7 + 5x^8 + 3x^{10} \\ -2x^3 + 4x^4 + 3x^5 - 4x^6 + 9x^7 + 4x^8 + 3x^{10}$$
and one polynomial with maximum absolute value of 7:
$$-2x^3 + 4x^4 + 3x^5 - 4x^6 - 6x^7 - 3x^8 + 7x^9 + 2x^{10}$$
in a small fraction of a second.
Some follow-up questions in comments brought me to the realisation that if we're trying to interpolate $$\{ (x_i, y_i) \}$$ with the $$x_i$$ coprime, there is at most one polynomial with coefficients in the range $$[-\lfloor \tfrac{(\operatorname{lcm} x_i) - 1}2 \rfloor, \lfloor \tfrac{\operatorname{lcm} x_i}2 \rfloor]$$, because the tree collapses to a chain. This gives the following algorithm for finding such a polynomial, if it exists:
M := lcm(x_i)
while any y_i is non-zero:
find a_0 by Chinese remainder theorem
if a_0 > floor(M / 2):
a_0 -= M
output a_0
update y_i := (y_i - a_0) / x_i
In the long term, the initial values of the $$y_i$$ are reduced to negligibility by the repeated division by $$x_i$$, so eventually each $$y_i$$ will be reduced to a range which is bounded by $$\frac{x_i}{x_i - 1} M$$. This means that for a given set of $$x_i$$ it's possible to compute a finite directed graph to see whether existence is guaranteed.
In the particular case that the $$x_i$$ are $$\{3, 5\}$$ there are three cycles, all of them loops: $$(0,0) \to (0,0)$$ is the terminating loop which indicates that a solution exists, but there are also loops $$(2, 1) \to (2, 1)$$ and $$(-2, -1) \to (-2, -1)$$.
• That's a really attractive piece of Python code Sep 25 at 12:43
• You are minimizing sum of absolute values; OP asks to minimize largest absolute value. Sep 25 at 13:00
• @GerryMyerson, I misunderstood the original text. The approach should work for a wide class of cost functions including any $L^k$ norm; there may be a risk with the $L^\infty$ norm of building arbitrarily long chains of small norm but in practice it works for the example, as the updated code demonstrates. Sep 25 at 14:27
• @AlexanderChervov, it's stronger than that: this is the only polynomial with coefficients less than 8. Sep 25 at 14:34
• Working mod 15, every equivalence class contains only one number in the range $[-7, 7]$, so the tree is pruned to just a chain. Sep 25 at 14:40
Perhaps not the best possible polynomial but a very good polynomial can generally be computed by the LLL algorithm: Let $$I$$ be an integral polynomial taking the given values at the given points and let $$\Lambda$$ be the lattice (in $$\mathbb Z^{d+1}$$ for a suitable high value of the degree $$d$$, coordinates are of course coefficients) of integral polynomials with roots at the prescribed set of points. We search a short vector in $$I+\Lambda$$. Adding $$I$$ to a basis of $$\Lambda$$ and running the LLL algorithm should give a good polynomial among a basis of short vectors most of the time (It may sometimes happen that it returns only short vectors of the form $$kI+P_0$$ with $$P_0$$ in $$\Lambda$$ and with $$k$$ avoiding $$\pm 1$$. Changing the value of $$d$$ and rerunning LLL solves perhaps the issue).
Added details (after some experimentation): The problem with results giving small multiples of the required evaluations can be avoided by adjoining an additional very large coordinate to the coefficients of $$I$$ and by setting this coordinate to $$0$$ for all elements in $$\Lambda$$. The Maple implementation of LLL returned $$2x^{10}+7x^9-3x^8-6x^7-4x^6+3x^5+4x^4-2x^3$$ for the data of the OP. (I worked with degree $$40$$.) Unfortunately not a solution with coefficients in $$\{-3,\ldots,3\}$$ but not very far off.
(Messy maple code for the OP example:
with(IntegerRelations);
d:=50;
ii:=10^8+t*interp([3,5],[221157,31511625],t);
li:={[seq(coeff(t*ii,t^i),i=1..d+2)]};for i from 2 by 1 to d do po:=t*(-t^i+interp([3,5],[3^i,5^i],t)):li:=li union{[seq(coeff(t*po,t^i),i=1..d+2)]};od:
u:=LLL([seq(li[i],i=1..nops(li))]);
vector(nops(u),i->sum('u[i,j]*3^(j-2)','j'=2..d+2)/221157);
po:=sum('u[nops(u),j]*x^(j-2)','j'=2..d+2);
)
Added after a comment from Max Alekseev His comment is correct ( https://www.goodreads.com/quotes/123557-god-s-final-message-to-his-creation-we-apologize-for-the ): An optimal polynomial (with minimal sum of squared coefficients) is (up to sign) given by a closest lattice point in $$I+\Lambda$$ to the orthogonal projection of the origin onto the affine hyperplane $$I+\mathbb R\otimes_{\mathbb Z}\Lambda$$. The trick of adding a last large coordinate to the coordinates of $$I$$ (or by multiplying $$I$$ by a large constant $$c$$ as in Alekseev's answer) moves this affine hyperplane away from the hyperplane containing the codimension one lattice $$\Lambda$$ and ensures that the last generator of an LLL-basis is in $$I+\Lambda$$ (up to a sign) and close to the line $$\Lambda^\perp$$ perpendicular to $$\Lambda$$ in the ambient spanned by $$I$$ and $$\Lambda$$.
• Thank you very much ! Can you share a mapple code please? Sep 25 at 12:16
• It looks like you are trying to (re)invent solution to the closest vector problem. Sep 25 at 16:28
• @Max Alekseyev Thanks for your insightful remark! Sep 26 at 10:13
• LLL was the first thing that came to mind, but this worked example is much better than I could've done. Excellent! Sep 27 at 1:34
Given $$p(x_i)=y_i$$ for $$i\in\{1,2,\dots,n\}$$, finding a polynomial of fixed degree $$d$$ can be posed as finding a closest vector to $$(cy_1,\dots,cy_n,\underbrace{0,\dots,0}_{d+1})^T$$ in the lattice spanned by $$(d+1)$$ column vectors $$\begin{bmatrix} c\cdot V_{d+1}(x_1,\dots,x_n)\\ I_{d+1} \end{bmatrix},$$ where $$V_{d+1}$$ is Vandermonde matrix (of size $$n\times (d+1)$$), $$I_{d+1}$$ is the identity matrix, and $$c$$ is a large constant. By the choice of $$c$$ we can guarantee that a closest vector will have form $$(cy_1,\dots,cy_n,a_0,a_2,\dots,a_d)^T$$ and these $$a_i$$ will be small and deliver us a required polynomial $$p(x) = \sum_{i=0}^d a_i x^i$$.
As for how to solve CVP in practice, see Package for the Closest Vector Problem (CVP)?
Here is an online SageMath implementation of @PeterTaylor's idea. It uses recursively enumerable sets and MapReduce, and is automatically parallelized by Sage (when available).
As an example it computes all polynomials $$p$$ of degree at most 12 with coefficients bounded by 10 by absolute value such that $$p(3)=221157$$ and $$p(5)=31511625$$.
It also confirms (when run with max_degree = +oo) that there is only one polynomial with the coefficients bounded by $$7$$ by absolute value.
• Other answers show two different polynomials with coefficients bounded by 7. I did not check them. Somebody must have made a mistake (it could be me). Sep 26 at 5:50
• @user132647, I gave it in little-endian order and Roland gave it in big-endian order, but it's the same polynomial, and there's a simple proof of uniqueness in the comments on my answer. Sep 26 at 6:43
• Oof, now I feel stupid... Thanks for pointing out my mistake! Sep 26 at 6:52
In the $$\ell^{\infty}$$ norm for the coefficients, the problem of finding polynomial $$p(x)=\sum_{k=0}^d c_kx^k$$ satisfying $$p(x_i)=y_i$$ for $$i\in\{1,2,\dots,n\}$$ can also be naturally posed as an integer linear program for finding integer $$c_i$$ and the best bound $$b$$: $$\begin{cases} \textrm{minimize}\ b\\ -b \leq c_k\leq b & \text{for}\ k\in\{0,1,\dots,d\}\\ \sum_{k=0}^d c_k x_i^k = y_i & \text{for}\ i\in\{1,2,\dots,n\} \end{cases}$$
Here is a sample implementation of this approach in Sage.
• For $d=10$, best_poly([3,5],[221157,31511625],10) gives $2*z^{10} + 7*z^9 - 3*z^8 - 6*z^7 - 4*z^6 + 3*z^5 + 4*z^4 - 2*z^3 + 2*z - 6$ (Best bound: 7), while for $d=20$ best_poly([3,5],[221157,31511625],20) gives $z^{11} + z^{10} - z^9 + z^8 - z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + z + 1$ (Best bound: 1). Sep 28 at 13:56
• @PabloH strange Sep 28 at 14:07
• @PabloH [ 221980., 56972656.] seems to be different values for the second polynom, than required Sep 28 at 14:17
• @PabloH: That is an issue with the ILP solver (by default SageCell uses GPLK), which sometimes does too much relaxation and inadvertently violates some of the given constraints. This can be remediated by tightening solver parameters or switching to an exact solver (which may be time costly). Sep 28 at 17:59
• @PabloH: I've checked with a locally installed Gurobi solver, and with degree=20 it produces the same polynomial as with degree=10. Sep 28 at 18:06 | 2021-10-25T13:39:59 | {
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https://mymathforum.com/threads/a-trigonometric-problem-with-the-laws-of-sine-and-cosine.347727/ | # A trigonometric problem with the Laws of sine and cosine
#### jmrg2992
Greetings, I'm having an interesting case where the solution of a certain angle cannot obtain the same result using the laws of sine and cosine for a triangle. I cannot understand why the answer doesn't match since from the perspective I'm getting of the problem, both laws can be used. In this case, I used cosine law first to find angle A and the sine law to get angle B, however, the answer doesn't match. I'll leave right next to this a picture of the process I have performed. The original information says we only know sides a=5, b=11, and c=7.5 of a triangle.
Thank you for your time!
With regards and happy new year!
#### greg1313
Forum Staff
There are two possible values for $B$. Note that the two values you have for$B$ sum to about $180^\circ$.
#### jmrg2992
Thank you for your fast answer, Greg! I wasn't thinking of that possibility. I thought I was doing something wrong over the algebra or something. This is an interesting thing; I'll keep it in mind from now on!!
#### skipjack
Forum Staff
When you calculated angle A, you rounded it to two decimal places. When calculating sin(B), you should use the unrounded value you obtained for A.
#### mathman
Forum Staff
In general the law of cosines gives unambiguous results, since arccos is single valued for angles 0 to 180. Arcsin is double valued over this range.
greg1313
#### skipjack
Forum Staff
In general, one doesn't have the luxury of being able to choose which method to use.
greg1313
#### jmrg2992
Regarding this
When you calculated angle A, you rounded it to two decimal places. When calculating sin(B), you should use the unrounded value you obtained for A.
I have done it again with the unrounded values, but there wasn't a significant change in the results.
In general the law of cosines gives unambiguous results, since arccos is single valued for angles 0 to 180. Arcsin is double valued over this range.
So this is one special case where ambiguous results might happen?
In general, one doesn't have the luxury of being able to choose which method to use.
But theorethically once we have an angle like in this conditions knowing the sides, the results should be equal if I mix the cosine and the sine law, right? Well unless we can have two values for one single angle like this exercise.
Also thanks to everyone for the answers !!! I really appreciate it !
#### greg1313
Forum Staff
$$\displaystyle \sin(x)=\sin(180^\circ-x)$$
Making a reasonably accurate sketch of the triangle in question, if one has sufficient information to do so, may help determine which angle is correct.
#### skipjack
Forum Staff
So this is one special case where ambiguous results might happen?
Potential ambiguity arises when the sine of an angle is used to determine the angle, but it's usually easy to determine the correct value. Some textbooks don't cover this in detail.
In the example originally posted, the cosine method needs to be applied first, as no angle is known initially. Having found that angle A, say, is arccos(203/220), which is approximately 22.672 degrees, one can choose to find sin(B), which turns out to be 0.8479976415..., and arcsin of that is 57.994545... degrees. If angle B had that value, angle C would be obtuse, which is impossible, as AB isn't the longest side of the triangle (if a triangle has an obtuse angle, the side opposite that angle must be longer than each of the other two sides, because the obtuse angle must be greater than each of the other two angles). Hence angle B = 180° - 57.994545... degrees = 122.00545... degrees. If that explanation is considered too cumbersome, one can use the cosine method instead to show that cos(B) = -0.53 exactly, etc.
Similar threads | 2020-02-17T15:56:58 | {
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http://math.stackexchange.com/questions/27719/what-is-gcd0-a-where-a-is-a-positive-integer/27821 | # What is $\gcd(0,a)$, where a is a positive integer?
I have tried $\gcd(0,8)$ in a lot of online gcd (or hcf) calculators, but some say $\gcd(0,8)=0$, some other gives $\gcd(0,8)=8$ and some others give $\gcd(0,8)=1$. So really which one of these is correct and why there are different conventions?
-
I haven't encountered the convention of gcd(0,8) = 1. It depends on how you define the phrase "a divides b" – The Chaz 2.0 Mar 18 '11 at 2:49
– Vafa Khalighi Mar 18 '11 at 2:53
@The Chaz: They are really the same things but with different names. see en.wikipedia.org/wiki/Greatest_common_divisor – Vafa Khalighi Mar 18 '11 at 4:04
First recall the definition of $\rm\:a\:$ divides $\rm\:b\:$ in a ring $\rm\:Z\:,\:$ often written as $\rm\ a\ |\ b\ \ in\ Z\:.$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \rm\ a\ |\ b\ \ in\ Z\ \iff\ a\:c = b\ \$ for some $\rm\ c\in Z$
Recall also the definition of $\rm\ gcd(a,b)\:,\$ namely
$(1)\rm\quad\quad\quad\quad\quad\ \rm gcd(a,b)\ |\ a,b\quad\quad\quad\quad\quad\quad\quad\$ the gcd is a common divisor
$(2)\rm\quad\quad\quad\quad\quad\ \rm c\ |\ a,b\ \Rightarrow\ c\ |\ gcd(a,b)\quad\quad$ the gcd is the greatest common divisor
$\ \ \ \$ i.e. $\rm\quad\ c\ |\ a,b\ \iff\ c\ |\ gcd(a,b)\quad\quad$ expressed in $\iff$ form (put $\rm\ c = gcd(a,b)\$ for $(1)$)
Notice $\rm\quad\: c\ |\ a,0\ \iff\ c\ |\ a\:,\$ so $\rm\ gcd(a,0)\ =\ a\$ by the above $\iff$ form of the gcd definition.
Note that $\rm\ gcd(0,8) \ne 0\:,\$ since $\rm\ gcd(0,8) = 0\ \Rightarrow\ 0\ |\ 8\$ contra $\rm\ 0\ |\ x\ \iff\ x = 0\:.$
Note that $\rm\ gcd(0,8) \ne 1\:,\$ else $\rm\ 8\ |\ 0,8\ \Rightarrow\ 8\ |\ gcd(0,8) = 1\ \Rightarrow\ 1/8 \in \mathbb Z\:.$
Therefore it makes no sense to define $\rm\ gcd(0,8)\$to be $\:0\:$ or $\:1\:$ since $\:0\:$ is not a common divisor of $\:0,8\:$ and $\:1\:$ is not the greatest common divisor.
The $\iff$ gcd definition is universal - it may be employed in any domain or cancellative monoid, with the convention that the gcd is defined only up to a unit factor. This $\iff$ definition is very convenient in proofs since it enables efficient simultaneous proof of both implication directions. $\$ For example, below is a proof of this particular form for the fundamental GCD distributive law $\rm\ (ab,ac)\ =\ a\ (b,c)\$ slightly generalized (your problem is simply $\rm\ c=0\$ in the special case $\rm\ (a,\ \ ac)\ =\:\ a\ (1,c)\ =\ a\:$).
THEOREM $\rm\quad (a,b)\ =\ (ac,bc)/c\quad$ if $\rm\ (ac,bc)\$ exists.
Proof $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$
See my post here for further discussion of this property and its relationship with Euclid's Lemma.
Recall also how this universal approach simplifies the proof of the basic GCD * LCM law:
THEOREM $\rm\;\; \ (a,b) = ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.
Proof $\rm\quad d|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$
For much further discussion see my many posts on GCDs.
-
Another way to look at it is by the divisibility lattice, where gcd is the greatest lower bound. So 5 is the greatest lower bound of 10 and 15 in the lattice.
The counter-intuitive thing about this lattice is that the 'bottom' (the absolute lowest element) is 1 (1 divides everything), but the highest element, the one above everybody, is 0 (everybody divides 0).
So $\gcd(0, x)$ is the same as ${\rm glb}(0, x)$ and should be $x$, because $x$ is the lower bound of the two: they are not 'apart' and 0 is '$>'$ $x$ (that is the counter-intuitive part).
-
In fact, the top answer can be generalized slightly (sorry, I don't have enough points to post comments, yet): if $a \vert b$, then $gcd(a,b)=a$ (and this holds in any algebraic structure where divisibility makes sense (eg a commutative, cancellative monoid)).
To see why, well, it's clear that $a$ is a common divisor of $a$ and $b$, and if $\alpha$ is any common divisor of $a$ and $b$, then, of course, $\alpha \vert a$. Thus, $a=gcd(a,b)$.
-
Indeed, even more generally, it is a special case of the distributive law - see my answer. As for commutative monoids, one usually requires them to be cancellative in order to obtain a rich theory. – Bill Dubuque Mar 18 '11 at 18:26 | 2015-08-29T09:27:57 | {
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http://math.stackexchange.com/questions/61996/calculating-probability-question | # Calculating probability question
If I have a random byte which consists of 8 bits.
e.g 11011101
I want to calculate the probability that the first x number of bits of any random byte will be zero.
For example: calculate the average number of guesses it would take to find a random byte with 4 leading zero bits.
Any help appreciated.
-
Yes uniformly random. So the size of of the "byte" doesn't need to be taken into account? For example if I have a SHA 256 hash (256 bits) the probability of generating a hash with 4 leading zeros is the same as generating a byte with 4 leading zeros? – Ben Reeves Sep 5 '11 at 13:19
Ok thanks. So how would i convert that into the average number of "guesses" needed? is it 1 / (1/2)^4 ? – Ben Reeves Sep 5 '11 at 13:32
Assuming a random byte has all its bits 0 or 1 with probability $\frac{1}{2}$, independent of each other, the answer to the probability seems to be pretty simple.
The probability that the first x bits of such a number will be zero (or whatever) is $\frac{1}{2^x}$.
About your second question: what is the average number of guesses to find such a number. Actually it's a more general question: what is the number of guesses to find something probability of which is $\tilde{p}$.
Le't calculate this.
$$\langle N \rangle = 1 \cdot \tilde{p} + 2 \cdot (1-\tilde{p}) \cdot \tilde{p} + 3 \cdot (1-\tilde{p})^2 \cdot \tilde{p} + \ldots$$
Means you need 1 guess with probability $\tilde{p}$. The probability it doesn't happen on the first guess is $1-\tilde{p}$, hence you need 2 guesses with probability $2 \cdot (1-\tilde{p}) \cdot \tilde{p}$. And so on.
We'll use the following known fact:
$$1 + (1-\tilde{p}) + (1-\tilde{p})^2 + (1-\tilde{p})^3 + \ldots = \frac{1}{\tilde{p}}$$
$$\begin{eqnarray} \langle N \rangle &=& 1 \cdot \tilde{p} + 2 \cdot (1-\tilde{p}) \cdot \tilde{p} + 3 \cdot (1-\tilde{p})^2 \cdot \tilde{p} + \ldots = \\ &=& \tilde{p} \cdot \left[ 1 + 2 \cdot (1-\tilde{p}) + 3 \cdot (1-\tilde{p})^2 + \ldots \right] = \\ &=& \tilde{p} \cdot \left[ 1 + (1-\tilde{p}) + (1-\tilde{p})^2 + \ldots + (1-\tilde{p}) + (1-\tilde{p})^2 + ... + (1-\tilde{p})^2 + (1-\tilde{p})^3 \ldots \right] = \\ &=& \tilde{p} \cdot \left[ \frac{1}{\tilde{p}} + (1-\tilde{p}) \cdot \frac{1}{\tilde{p}} + (1-\tilde{p})^2 \cdot \frac{1}{\tilde{p}} + \ldots \right] = \\ &=& 1 + (1-\tilde{p}) + (1-\tilde{p})^2 + ... = \frac{1}{\tilde{p}} \end{eqnarray}$$
$$\langle N \rangle = \frac{1}{\tilde{p}}$$
Hence the answer to your question is $\langle N \rangle = 2^x$.
-
I have edit your post to write formulas using TeX, I hope you agree it looks nicer than HTML. – Sasha Sep 5 '11 at 14:16
@Sasha: yeah, 10x. Hope I'll learn TeX one day – valdo Sep 5 '11 at 14:26
@Thijs Laarhoven: Wow, nice trick. 10x. – valdo Sep 5 '11 at 21:10
There is another nice trick to find the expected value of a geometric($p$) variable; that is, the expected number of tosses until the first sucess. It goes as follows: Note that $E[X] = 1 + (1-p) E[X]$. This is because you have to look at the first toss anyways. If the first toss is not a success (with probability $1-p$) you start all over again from the the second toss. If the first toss is a success, you stop. Solving this equation we get $E[X]=1/p$.
Here is a related question that might be of interest. If you are looking at the byte bit-by-bit, how many bits do you expect to look at until you see a string of $4$ zeroes?
- | 2015-01-30T20:56:29 | {
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https://math.stackexchange.com/questions/2769580/evaluate-int-frace2x-1e2x1dx-using-partial-fractions | # Evaluate $\int \frac{e^{2x}-1}{e^{2x}+1}dx$ using partial fractions
Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$ using partial fractions.
My Attempt $$\int\frac{e^{2x}-1}{e^{2x}+1}dx=\frac{1}{2}\int\frac{2.e^{2x}}{e^{2x}+1}dx-\int\frac{dx}{e^{2x}+1}=\frac{1}{2}\log|e^{2x}+1|-\int\frac{dx}{e^{2x}+1}$$ Put $t=e^x\implies dt=e^xdx=tdx\implies dx=\frac{dt}{t}$ $$\int\frac{dx}{e^{2x}+1}=\int\frac{dt}{t(t^2+1)}$$ Using partial fractions, $$\frac{1}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}\implies 1=A(t^2+1)+t(Bt+C)$$ \begin{multline} \begin{aligned} &t=0\implies \boxed{A=1}\\ &1=t^2+1+t(Bt+C)\implies -t^2=t(Bt+C)\\ &\implies-t=Bt+C\text{ if } t\neq 0\\ &t=0\implies \boxed{C=0}\implies \boxed{B=-1} \end{aligned} \end{multline} \begin{multline} \begin{aligned} \int\frac{dt}{t(t^2+1)}&=\int\frac{dt}{t}-\frac{1}{2}\int\frac{2t.dt}{t^2+1}=\log|t|-\frac{1}{2}\log|t^2+1|\\ &=\log|e^x|-\frac{1}{2}\log|e^{2x}+1|+C_1 \end{aligned} \end{multline} $$\int\frac{e^{2x}-1}{e^{2x}+1}dx=\frac{1}{2}\log|e^{2x}+1|-\log|e^x|+\frac{1}{2}\log|e^{2x}+1|+C=\log|e^x+e^{-x}|+C$$ Doubt
While doing partial fractions, first i have assumed $t=0$ but $t=e^x\neq{0}$ and even if I accept that, in the second step the equation is transformed into a new form assuming $t\neq{0}$. But, in the next step I need to again assume $t=0$ to get $B$ and $C$. How can I jusify this and why am I getting the right answer after all ?
• Multiply top and bottom by $e^{-x}$. – Rene Schipperus May 6 '18 at 19:04
• @ReneSchipperus I am aware of that, doing that will make the numerator the derivative of the denominator, follows direct answer with log. But, my doubt is regarding why am I getting the correct result while doing these absurd assumptions whn trying to solve it as I mentioned. – ss1729 May 6 '18 at 19:06
• You can use, for instance, the quadratic formula to factor something like $a\cdot e^{2x}+b\cdot e^x+c$, even though $e^x$ is never zero. That is no problem. The same goes for partial fraction decomposition. Even more explicitly, $x^2+2x+1=(x+1)^2$ is as true when $x\in \Bbb R$ as when $x$ is limited to $x>0$. – Arthur May 6 '18 at 19:06
• you just try to integrat the tan hyperbolic in fact – Isham May 6 '18 at 19:21
What you're really doing is requiring two expressions to have the same $t\to 0^+$ (i.e. $x\to -\infty$) limiting behaviour, which is legitimate.
$$\int \frac{e^{2x}-1}{e^{2x}+1}dx=\ln \left|e^{2x}+1\right|-x+k,\quad k\in \mathbb{R}$$
In fact
$$\int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \left[\frac{2e^{2x}}{e^{2x}+1}-1\right]dx=*$$
being
$$\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{2x}-1+\left(e^{2x}+1\right)}{e^{2x}+1}-1=\frac{2e^{2x}}{e^{2x}+1}-1$$
After we have:
$$*=\int \left[\frac{2e^{2x}}{e^{2x}+1}-1\right]dx=**$$
Using integration for substitution $t=e^{2x}+1$ (with $\int f'/g dx=\ln |g|+\mathrm{const.}$) we have:
$$**=\ln \left|e^{2x}+1\right|-x+k,\quad k\in \mathbb{R}$$
$$$$\frac{1}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}\implies 1=A(t^2+1)+t(Bt+C)$$$$ This is for all t you get that $$1=t^2(A+B)+Ct+A \quad \color{red}{\forall t}$$ $$A+B=0, A=1, C=0$$ $$(A,B,C)=(1,-1,0)$$
Note that you just integrate $\tanh(x)$ $$\int \frac{e^{2x}-1}{e^{2x}+1} dx= \int \tanh(x)dx= \ln|\cosh(x)|+K$$ | 2019-08-25T19:43:49 | {
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https://math.stackexchange.com/questions/2855047/prove-that-sum-limits-i-1n-a-i-geq-n2 | # Prove that $\sum\limits_{i=1}^{n} a_i\geq n^2$.
A hint can be helpful, but not a whole solution.
The Problem (conjecture):
Given a natural number $n \geq 1$ and a sequence of natural numbers $(a_i)_{1 \leq i \leq n}$ in which for every pair $(i,j)$ with $i \neq j,$ we have $$\gcd(a_i,a_j)\nmid i-j$$ prove that $\sum\limits_{i=1}^{n} a_i\geq n^2$.
What I have done:
During my research, I ran into this problem and I am not quite sure if it is true. It is clear if we put $a_i=n$ then the problem will be solved and the summation will be equal to $n^2$. I tried to solve this problem.
For example, I showed that $$a_i> max(i,n-i)$$ otherwise, I can put $j=i+a_i$ or $j=i-a_i$ and considering the fact $\gcd(a_i,a_j) \mid a_i,$ then we conclude that $\gcd(a_i,a_j)\mid i-j,$ hence, $a_i> max(i,n-i)$ which means that $a_i\geq \dfrac{n} {2}$.
Moreover, if $a_i\leq n$ and p are prime divisors of $a_i$ by putting $j=i-\dfrac {a_i}{p}$ for $i\geq \dfrac {n} {2}$ and $j=i+\dfrac {a_i}{p}$ for $i\leq \dfrac {n} {2}$ we could conclude that $a_i \mid a_j$.
I could go further, but it is not enough to prove the conjecture. I also tried Induction and considered that the property holds for every $n\leq k$ and then try to prove the theorem for $n= k+1$ but again, there are some issues that I could not go further.
• What is this mod function? – Ovi Jul 18 at 0:24
• @Ovi $\mod(a,b)$ stands for the residue of $a$ modulo $b$. It is common in math softwares. Hence, $\mod(a,b)>0$ is equivalent to $b\not\mid a$. – Arnaud Mortier Jul 18 at 0:27
• @ArnaudMortier Ah got it thanks – Ovi Jul 18 at 0:29
• How far have you been looking for a counterexample? How far do you know this to be true? – Arnaud Mortier Jul 18 at 0:36
• I edited the problem notation, so it is clearer now. I have tried for n=2,3,4,5,6. Then I stopped and guessed it might be true. I coded the problem with an optimisation software GAMS to find the minimum value of summation of a(i) for all the numbers lower than 100, however I am not done yet! – farshid evazabadian Jul 18 at 0:42
If you want only a hint, stop reading after recalling that $\gcd(a_i,a_j) \mid i-j$ is equivalent to the solvability of the system of congruences \left\{ \begin{align*} x \equiv i \pmod {a_i} \\ x \equiv j \pmod {a_j}\end{align*}\right.
By the Chinese remainder theorem, we can formulate the problem equivalently as follows: consider $n$ arithmetic progressions in $\mathbb Z$, $(S_i)_{1 \leq i \leq n}$ with $i \in S_i$ for all $i$ and with differences $a_i$. That is, $S_i = i + a_i \mathbb Z$. If the $S_i$ are pairwise disjoint, prove that $\sum a_i \geq n^2$.
The conclusion remains true even without the hypothesis that $i \in S_i$:
If the $S_i$ are disjoint, they are also disjoint mod $a_1 \cdots a_n$, and hence \begin{align*} \sum_{i=1}^n\# (S_i \bmod a_1 \cdots a_n) & \leq a_1 \cdots a_n \\ \sum_{i=1}^n \prod_{j \neq i} a_j &\leq \prod_{i=1}^n a_i \end{align*} divide by the RHS to obtain $$\sum_{i=1}^n\frac1{a_i} \leq 1$$ And by AM-HM $$\sum_{i=1}^n a_i \geq n^2$$
Also from AM-HM we have that equality occurs iff $a_1 = \cdots = a_n = n$.
• +1 very elegant. – user159517 Aug 4 at 16:15
• Wonderful! I'm not sure though that the reference to CRT is needed. Also I think you may hide some parts of your answer under the spoiler tag, as OP asked not to post a complete solution. – zhoraster Aug 4 at 16:44
• @zhoraster Right, I forgot that the OP only wanted a hint. I don't like spoiler tags for long solutions, but thanks :) – barto Aug 4 at 16:49
Just some ideas: if $a_n \geq 2n - 1$, you can proceed by induction. Hence you may assume that $n \leq a_n \leq 2n - 1$. In this case your second observation becomes quite a bit stronger. Indeed, for any prime $p$ dividing $a_n$ we may take $i = n$ and $j = n - \frac{a_n}{p}$. Note that $1 \leq j < n$ exactly by our condition $n \leq a_n \leq 2n - 1$. | 2018-10-21T17:30:55 | {
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https://math.stackexchange.com/questions/2093466/why-arent-the-functions-fx-fracx-1x-1-and-fx-1-the-same/2093480 | # Why aren't the functions $f(x) = \frac{x-1}{x-1}$ and $f(x) = 1$ the same? [duplicate]
I understand that division by zero isn't allowed, but we merely just multiplied $f(x) = 1$ by $\frac{x-1}{x-1}$ to get $f(x) = \frac{x-1}{x-1}$ and $a\cdot 1 = 1\cdot a = a$ so they're the same function but with different domain how is this possible?
Or in other words why don't we simplify $f(x) = \frac{x-1}{x-1}$ to $f(x) = 1$ before plotting the points. Is it just defined this way or is there a particular reason ?
Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of $f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$.
• This is an interesting question, but posted on the wrong website. You may want to think about what it means precisely for two functions to be "the same". Jan 11, 2017 at 15:16
• The crux of the matter is that you have been deceived: if they told you that a function is simply a formula, then they lied... Rather, a function f is: 1. a source set S, 2. a target set T, 3. a way to associate to each element x in S a unique element of T, denoted by f(x). In your example, 2. and 3. coincide but one generally chooses for S the real line for the function f(x)=1. Then you see that one cannot choose this same set S for the function f(x)=(x-1)/(x-1) since (1-1)/(1-1) does not exist, hence the two functions are indeed different.
– Did
Jan 11, 2017 at 15:26
• Jan 11, 2017 at 16:40
• "so they're the same function but with different domain" That answers your question. The domain is a required part of the determination of a function. The "same function but with different domain" is a bit like saying "24 and 48 are the same number but with a different power of 2". If the domains are different... then the functions are different. It's that simple. Jan 11, 2017 at 17:11
• Almost identical: math.stackexchange.com/questions/1670139/…. Jan 11, 2017 at 20:20
They are the same almost everywhere. But clearly one of them does not exist for $x=1$ (since "$\tfrac{0}{0}$" is undefined), while the other one is simply $1$ at $x=1$.
I understand that division by zero isn't allowed, but we merely just multiplied f(x) = 1 by (x-1)/(x-1)
You can multiply by any fraction $\tfrac{a}{a}$; but not when $a=0$ because the fraction you want to multiply with, isn't even defined then. So multiplying by $\tfrac{x-1}{x-1}$ is fine, but only valid for $x \ne 1$.
why don't we simplify f(x) = (x-1)/(x-1) to f(x) = 1 before plotting the points. Is it just defined this way or is there a particular reason ?
You can simplify, but recall that simplifying is actually dividing numerator and denominator by the same number: you can simplify $\tfrac{ka}{kb}$ to $\tfrac{a}{b}$ by dividing by $k$. But also then: this only works for $k \ne 0$ since you can't divide by $0$. So "simplifying" $\tfrac{x-1}{x-1}$ to $1$ is fine, for $x-1 \ne 0$ so for $x \ne 1$.
Note: my book says the domain of $f(x) = 1$ is $\mathbb{R}$ and the domain of $f(x) = \frac{x-1}{x-1}$ is $\mathbb{R}$ except $1$.
Technically, the domain is a part of the function: it should be given (as well as the codomain). It is very common though that when unspecified, in the context of real-valued functions of a real variable, we assume the 'maximal domain' is intended (and $\mathbb{R}$ is taken as codomain). Then look at: $$f : \mathbb{R} \to \mathbb{R} : x \mapsto f(x) = 1$$ and $$g : \mathbb{R} \setminus \left\{ 1 \right\} \to \mathbb{R} : x \mapsto g(x) = \frac{x-1}{x-1}$$ The functions $f$ and $g$ are different, but $f(x) = g(x)=1$ for all $x$ except when $x=1$, where $g$ is undefined.
Question: What is a function?
Answer: Maybe simply said it is a map (receipe), $f(x)$, that projects some elements, $x$, contained in a specifically defined set, Domain $D$, into another set, Range $R$.
Discussion: Hence when defining a function one must define the Domain as well as the functional form. Otherwise the function is not defined.
Conclusion: If two funtions have the same domain and the same receipe then they are the same "maps" otherwise they are not.
• Indeed, except that the target set can be any set containing the range, not necessarily the range.
– Did
Jan 11, 2017 at 15:30
• @Did Thank for making it more precise :-) Jan 11, 2017 at 15:30
$f(x)=(x-1)/(x-1)$ does not have a value when $x=1$, different thing happens to $f(x)=1$ | 2022-10-06T07:34:25 | {
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http://mathhelpforum.com/number-theory/12889-solved-easy-number-theory-question-you-guys-not-me.html | # Thread: [SOLVED] Easy Number Theory Question...for you guys, not me :)
1. ## [SOLVED] Easy Number Theory Question...for you guys, not me :)
Someone asked me this question a few days ago, and I have a few ideas on how to do it, but it seems to be too much work for a problem of this nature.
We know that a number is divisible by 3 if the sum of all its digits are divisible by 3.
The question is either:
Prove that this is so
or
Prove why this is so
I don't remember, but I think they amount to pretty much the same thing
Thanks guys
2. Originally Posted by Jhevon
Someone asked me this question a few days ago, and I have a few ideas on how to do it, but it seems to be too much work for a problem of this nature.
We know that a number is divisible by 3 if the sum of all its digits are divisible by 3.
The question is either:
Prove that this is so
or
Prove why this is so
I don't remember, but I think they amount to pretty much the same thing
Thanks guys
It's one of the basic counting applications. I'll prove it later tonight when I get back home.
3. I got some time I'll do that now.
Let n = a_k*10^(k) + .. + a_(3)*10^(3) + a_2*10^(2) + a_1*10^(1) + a_0
Note:
a_k*10^(k) is 99...99 + 1
a_(3)*10^(3) is 999 + 1
a_2*10^(2) is 99 + 1
a_1*10 is 9 + 1
= a_k*99...9 + a_k*1 + a_(k - 1)*9...9 + a_(k - 1)*1 + ... + a_2*99 + a_2*1
= a_k*99...9 + a_(k - 1)*99...9 + ... + a_2*99 + a_1*9 + [a_k + a_(k - 1) + ... + a_1 + a_0]
Bold is div. by 3 (and 9), thus it should be obvious now why a number is divisible by 3 if the sum of all its digits are divisible by 3 (and also the same is true for 9).
Q.E.D.
There are other divisibility tests, such as for 11, you alternate between +/- and see if it's divisible by 11 to determine if the number is.
4. Originally Posted by Jhevon
Someone asked me this question a few days ago, and I have a few ideas on how to do it, but it seems to be too much work for a problem of this nature.
We know that a number is divisible by 3 if the sum of all its digits are divisible by 3.
The question is either:
Prove that this is so
or
Prove why this is so
I don't remember, but I think they amount to pretty much the same thing
Thanks guys
Here is my method.
Let P(x) be some polynomial in Z[x] (meaning the ring of integers).
We have a theorem, if,
a = b (mod c)
Then,
P(a) = P(b) (mod c)
(I am not goint to prove this. Rather do what my professor does, leave it as an excercise).
We know that,
10 = 1 (mod 3)
Let,
N=a_n*10^n +a_{n-1}*10^{n-1}+...+a_1*10^1 +a_0
Be the way you write the positive integer in base 10.
And consider the polynomial,
P(x)=a_n*x^n+a_{n-1}*x^{n-1}+...+a_1*x +a_0
Then,
P(10)=P(1) (mod 3)
BUT!
P(10)=N
And P(1)=a_1+a_2+...+a_n=S (where S is digits sum).
Thus,
N = S (mod 3)
Hence,
3|N if and only if 3|S.
Q.E.D.
5. Originally Posted by ThePerfectHacker
We have a theorem, if,
a = b (mod c)
Then,
P(a) = P(b) (mod c)
(I am not goint to prove this. Rather do what my professor does, leave it as an excercise).
This is the only part of the thorem that is as of yet unjustified. I don't know how to prove this.
6. Originally Posted by ecMathGeek
This is the only part of the thorem that is as of yet unjustified. I don't know how to prove this.
you and me both
it kind of seems intuitive though, but that often counts for nothing
you know, i forgot to show the guy that ask me this proof, lol, i guess he doesn't need it anymore. i'll ask him
here's another if you guys are bored (i'm doing this from memory, so pardon any errors, the same guy asked me this question, so if you guys have problems, i'll check with him to make sure i got the question right):
i actually thought up two ways to do this, but i ran into trouble at the punchline. here goes
let r be a rational number
let a be an irrational number
let s be a real number
Prove or disprove: either (ar + s) or (ar - s) is irrational
7. Originally Posted by Jhevon
you and me both
it kind of seems intuitive though, but that often counts for nothing
you know, i forgot to show the guy that ask me this proof, lol, i guess he doesn't need it anymore. i'll ask him
here's another if you guys are bored (i'm doing this from memory, so pardon any errors, the same guy asked me this question, so if you guys have problems, i'll check with him to make sure i got the question right):
i actually thought up two ways to do this, but i ran into trouble at the punchline. here goes
let r be a rational number
let a be an irrational number
let s be a real number
Prove or disprove: either (ar + s) or (ar - s) is irrational
If s is rational, then this is false. (I might go through the process of proving this)
If s is irrational then I have no idea how to prove this. I would assume only if s = ar + n or s = n - ar for some rational number n will either (ar + s) or (ar - s) be rational. Otherwise, it is irrational.
8. Originally Posted by ecMathGeek
This is the only part of the thorem that is as of yet unjustified. I don't know how to prove this.
It is trivial.
Let, P(n) = an^2+bn+c
And,
x = y (mod z) &
Then,
x^2 = y^2 (mod z)
Thus,
ax^2 = ay^2 (mod z) *
Using & again we find that,
bx = by (mod z) **
And,
c = c (mod z) ***
Adding * and ** and *** we get:
ax^2+bx+c = ay^2+by+c (mod z)
Thus,
P(x) = P(y) (mod z)
Now prove it for the n>2 case.
Again trival.
9. Originally Posted by Jhevon
let r be a rational number
let a be an irrational number
let s be a real number
Prove or disprove: either (ar + s) or (ar - s) is irrational
Proof: either (ar + s) or (ar - s) is irrational
Assume to the contrary that both (ar + s) and (ar - s) are rational, and let ar + s = m. then we have ar - s = m - 2s, and m - 2s is rational, so s and m must be rational
but we have a = (m - s)/r, but that would make the number a rational, contrary to our assumption.
QED
Here's another, i don't like this one much either
Proof: either (ar + s) or (ar - s) is irrational
take ar + s, and let ar + s = m, then we have two cases: either m is irrational or m is rational
if m is irrational we have the desired result
if m is rational then we can write r = (m - s)/a. but a is irrational. therefore the only way that (m - s)/a can be rational is if (m - s) is irrational. since we assumed m is rational, then s must be irrational. But that would mean (ar + s) would give an irrational result, which makes m irrational, contrary to our assumption.
QED
Here's a third proof that kind of evolved from my second one. i like this one a bit better
Proof: either (ar + s) or (ar - s) is irrational
take ar + s, and let ar + s = m, then we have two cases: either m is irrational or m is rational
if m is irrational, we have the desired result
if m is rational then we can write r = (m - s)/a. but a is irrational. therefore the only way that (m - s)/a can be rational is if (m - s) is irrational. since we assumed m is rational, then s must be irrational.
now ar - s = m - 2s, since s is irrational, m - 2s is irrational. so we have whenever (ar + s) is rational, (ar - s) is irrational, and therefore, one or the other is always irrational
QED
Please give your opinions on the proofs above, which is the best and whether or not i proved what i set out to. if none do the job, please tell me what mistakes i made, i don't like the first two myself, so i'm banking on the third one
10. Originally Posted by Jhevon
Proof: either (ar + s) or (ar - s) is irrational
take ar + s, and let ar + s = m, then we have two cases: either m is irrational or m is rational
if m is irrational we have the desired result
if m is rational then we can write r = (m - s)/a. but a is irrational. therefore the only way that (m - s)/a can be rational is if (m - s) is irrational. since we assumed m is rational, then s must be irrational. But that would mean (ar + s) would give an irrational result, which makes m irrational, contrary to our assumption.
QED
Let s = sqrt(2), a = sqrt(2), r = -1
(sqrt(2)*(-1) + sqrt(2)) = 0, which is rational.
OR
Let s = 2pi, a = (1 - pi), r = 2
(2(1 - pi) + 2pi) = 2, which is rational.
The point I am trying to make is the assumption that if a and s are irrational then (ar + s) is irrational is false. But your third proof takes care of this (I think).
I have a question. Can 0 be consider irrational?
11. Originally Posted by ecMathGeek
The point I am trying to make is the assumption that if a and s are irrational then (ar + s) is irrational is false. But your third proof takes care of this (I think).
maybe you misread my proof, or i made something unclear. i never made the assumption that if a and s are irrational then (ar + s) is irrational, what the proof you quoted says is that, for (ar + s) to be rational then a and s must be irrational.
i never indicated case 1 and case 2, so maybe that's why. but you have an issue with case 2 apparently, in case 2, i assumed ar + s was rational, not irrational
I have a question. Can 0 be consider irrational?
Nope, zero is a rational number
12. Originally Posted by Jhevon
since we assumed m is rational, then s must be irrational. But that would mean (ar + s) would give an irrational result
This was the part of your proof that I was commenting on. Perhaps you didn't mean to say this, but my point is that this is not true.
13. Originally Posted by ecMathGeek
This was the part of your proof that I was commenting on. Perhaps you didn't mean to say this, but my point is that this is not true.
well, in that case, you are correct
i didn't mean to say that (i hope)
what about the third proof? do you see any porblems with that one, or the other proof for that matter? the one without the line you quoted
14. Originally Posted by Jhevon
Proof: either (ar + s) or (ar - s) is irrational
Assume to the contrary that both (ar + s) and (ar - s) are rational, and let ar + s = m. then we have ar - s = m - 2s, and m - 2s is rational, so s and m must be rational
but we have a = (m - s)/r, but that would make the number a rational, contrary to our assumption.
QED
I like this proof. If (ar + s) and (ar - s) are rational and we let (ar + s) = m, then (ar - s) = m - 2s, and so m and m - 2s must be rational, which can only be true if s is rational. However solving for a in either term (ar + s or ar - s) we would get a = (m - s)/r, where m, s, and r, are rational numbers, making a rational, but a is irrational thus we have a contradiction.
I think this proves it. It does not declare whether m or m - 2s really are rational, but it indirectly dictates that one or the other must not be, otherwise a = (m - s)/r would be rational, which it cannot be.
Basically, both the rationality of m and s are being called into question, where we know that a cannot be and r must be rational. Since s and m cannot both be rational, we know that either m or m - 2s MUST be irrational, which means that (ar + s) or (ar - s) MUST be irrational.
Originally Posted by Jhevon
Proof: either (ar + s) or (ar - s) is irrational
take ar + s, and let ar + s = m, then we have two cases: either m is irrational or m is rational
if m is irrational, we have the desired result
if m is rational then we can write r = (m - s)/a. but a is irrational. therefore the only way that (m - s)/a can be rational is if (m - s) is irrational. since we assumed m is rational, then s must be irrational.
now ar - s = m - 2s, since s is irrational, m - 2s is irrational. so we have whenever (ar + s) is rational, (ar - s) is irrational, and therefore, one or the other is always irrational
QED
This is also an excellent proof because it draws on the same conclusions that I (and you) made above.
15. Originally Posted by Jhevon
Proof: either (ar + s) or (ar - s) is irrational
take ar + s, and let ar + s = m, then we have two cases: either m is irrational or m is rational
if m is irrational, we have the desired result
if m is rational then we can write r = (m - s)/a. but a is irrational. therefore the only way that (m - s)/a can be rational is if (m - s) is irrational. since we assumed m is rational, then s must be irrational.
now ar - s = m - 2s, since s is irrational, m - 2s is irrational. so we have whenever (ar + s) is rational, (ar - s) is irrational, and therefore, one or the other is always irrational
QED
I don't remember what problem i had with the first one, but now with the second one i realize there are two ways that (m - s)/a can be rational if a is irrational. we can have (m - s) irratonal, or (m - s) = 0 which is rational. however, in the case where (m - s) is 0, we would have r = 0, so we would have
ar + s = m = s
=> 0 + s = s
which is trivial, but nevertheless, i think the proof should account for that in some way
Page 1 of 2 12 Last | 2016-08-31T06:34:52 | {
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https://math.stackexchange.com/questions/1894862/how-do-i-convert-the-following-double-integral-to-a-single-integral | # How do I convert the following double integral to a single integral
I have a double integral $$I=\int_0^a\int_0^bf(x-y)\,\mathrm{d}x\,\mathrm{d}y~,$$ where the function $f$ depends only on the difference between two variables $x$ and $y$. I would like to change the above double integral into a single one. Here is my attempt: \begin{align} I&=\int_0^a\int_0^af(x-y)\,\mathrm{d}x\,\mathrm{d}y+\int_0^a\int_a^bf(x-y)\,\mathrm{d}x\,\mathrm{d}y\\ &=2\int_0^a(a-x)f(x)\,\mathrm{d}x+\int_0^a\int_a^bf(x-y)\,\mathrm{d}x\,\mathrm{d}y \end{align} Now I am stuck on how to deal with the secong dobule integral. Any ideas?
Edit: Note that I am interested in the case when the function $f(x)$ is even in $x$.
• Yes. Sorry about that! – konstant Aug 17 '16 at 10:30
• Do we know that $a\leq b$ ? – onurcanbektas Aug 17 '16 at 12:37
• I think that does not matter, but let's say yes. – konstant Aug 17 '16 at 12:49
• @konstant, why would you say it doesn't matter? It does. Anyway, you should've given all relevant information about your problem, even if you think it doesn't matter – Yuriy S Aug 17 '16 at 20:51
• If $a=b$, this is $$I=\int_{-a}^a(a-|u|)\,f(u)\,du.$$ If $a\ne b$, a similar, more complicated, formula holds. Anyway: draw a picture... – Did Aug 18 '16 at 15:16
You can treat this integral as an integral over a region of the Cartesian plane rather than as one single-parameter integral inside another, and it is useful to do so.
To check the work you've done so far, let's consider the case of your integral when $a=b=1$. Then the region of integration is the unit square. If $f$ is even, then every region below the diagonal is matched by a symmetric region above the diagonal, so we can take twice the integral below the diagonal to get the integral over the square. This then agrees perfectly with what you already found, although it depends on the function being even and (as you noted) it leaves an inconvenient region still to be integrated if $a\neq b$.
The following method can be adapted to integrate your leftover region, but I think that is as much work as just applying it to the entire original region of integration, so I will do the entire region. Treating your double integral as an integral over a region in the plane, you want to integrate $f(x-y)$ over the rectangle with vertices at $(0,0)$, $(0,a)$, $(b,a)$, and $(b,0)$. Transforming the plane according to the rule $(x,y) \mapsto (x-y, y)$, your original integral is equal to the integral of $f(x)$ over the parallelogram with vertices at $(0,0)$, $(-a,a)$, $(b-a,a)$, and $(b,0)$. You can rewrite that as three integrals over functions of $x$ alone, or if you like, as one integral over a function of $x$ defined piecewise using $f(x)$.
One of the three integrals comes from integrating $f(x)$ over the region of integration to the left of the $y$ axis. This region is the triangle with vertices at $(0,0)$, $(-a,a)$, and $(0,a)$, its cross-section along the line $x=c$ is $c+a$, and the integral is $$\int_{-a}^0 (x+a)f(x)\,\mathrm{d}x.$$ This is equal to the integral $$\int_0^a(a-x)f(x)\,\mathrm{d}x$$ from your calculations, which of course it must be since it's the integral of $f(x-y)$ over the part of the original rectangular region of integration above the line $y=x$. The other two integrals over $f(x)$ are the integral over the rectangular subregion between $x=0$ and $x=b-a$ and the third is the triangle to the right of the line $x=b-a$.
If $f$ is even, you can assume $b>a$ without loss of generality (although it appears you were already aware of that). This makes the procedure above a little easier than in the more general case ($f$ not even) since you do not have to be concerned about the regions to use if $a<b$.
• Ok. You made a good point through the example that the integral in the case when $f(x)=x$ is zero, which tells me the question need a bit more of clarification. The function $f(x)$ is even in $x$, thus your example $f(x)=x$ is not applicable for me. But, thanks for pointing that out. – konstant Aug 17 '16 at 12:01
• could you please write your last paragraph mathematically? – konstant Aug 17 '16 at 13:10
• So with your comment I would conclude that the answer is $2\int_0^{a}(a-x)f(x)dx+2\int_0^{b-a}(b-a-x)f(x)dx$. Unfortunately, if you take an example $f(x)=1/(1+x^2)$, the numerical value obtained from this expression will not be the same as from original expression. – konstant Aug 18 '16 at 8:35
• That's not what I had in mind. I said a triangle, a rectangle, then another triangle. Nowhere did I recommend to multiply anything by $2$. – David K Aug 18 '16 at 17:08 | 2019-11-12T05:58:54 | {
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https://math.stackexchange.com/questions/1800825/prove-or-disprove-2abcabc-ge-3a2b2c21/1811365 | # Prove or disprove $2abc(a+b+c)\ge 3(a^2b^2c^2+1)$
Let $a,b,c>0,ab+bc+ca=3$, prove or disprove $$2abc(a+b+c)\ge 3(a^2b^2c^2+1)$$
Now I can't find any counterexample
• What method are you using to prove it? A standard proof, by contradiction, also, It is great to show what you have done, if anything, so people don't think you are just saying 'solve this for me' May 26, 2016 at 13:47
• How about $a = b = c = 1$ as a counterexample? May 26, 2016 at 14:03
• sorry,It's $a^2b^2c^2+1$ sorry May 26, 2016 at 14:15
• Here's a counterexample for the corrected inequality: $(a,b,c) = (1,\frac52,\frac17)$. To find this, I fixed $a=1$ and plotted the region where the inequality holds, then looked for any point on the curve $b+c+bc=3$ outside that region. May 26, 2016 at 14:19
• For a counter eg, just allow $c\to 0$, the two remaining variables can satisfy the constraint and your inequality will read $0\ge3$. May 26, 2016 at 17:40
What follows is not so much an answer to the question (it was already fully answered in comments) as an essay intended to bring consolation to, perhaps even cheer up, the inequality stated in the question, which must feel somewhat mistreated, poor thing, and justifiably so.
The counterexample provided by "Macavity" is elegant $($albeit not fully spelled out$)$: decrease $c$ towards $0$ $($$c never reaches 0 since it must be strictly positive), fix a>0, and let b:=(3-ac)/(a+c) so that the constraint ab+ac+bc=3 is satisfied (also b>0 when c is small enough). Then abc\to a\cdot(3/a)\cdot 0 = 0, the LHS of the inequality converges to 0, the RHS converges to 3, thus for all small enough c the inequality LHS{}\geq{}RHS fails. The given inequality, which we were asked to prove or disprove, is definitely disproved. And that's it, problem solved. But is it? Have you gained any mental image of why the inequality fails to hold? Have you learned anything useful, anything that you may reuse, perhaps in a modified form, when working on a similar, or a not-so-similar problem? Have you at least acquired a couple of ways of looking for counterexamples, or of ways of making preliminar probes into a problem, trying to find its soft spots? My guess is that the anwers are, respectively, "no", "certainly not", and "very probably not". My experience as a budding mathematician (oh so long ago) was that the so called exercises are mere starting points, not ultimate goals. When doing an exercise I always kept asking additional questions which were as interesting, if not more interesting, than the exercise itself. Also, trying this or that approach I sometimes veered away from the theme of the exercise into an uncharted new territory; or during the work on the exercise there emerged an object with special properties, which made my analyze the object, perhaps study the class of all objects with those properties; or I took a second look at some failed attempt and noticed that the faulty reasoning I used in some situation may become a legitimate reasoning in a slightly different situation. And so on, and so forth. Now and then a work on an exercise developed into a miniature research project. And that's how I learned doing mathematics, working through some thousands of exercises in this peculiar way. Let us play around with the inequality in the present question and see where it leads us. The user "geromty", who asked the question, first gave the inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3(a^2b^2c^2+3) \end{equation*} (assuming a,b,c>0 and ab+ac+bc=3$$)$, which after the easy counterexample $a=b=c=1$ he changed to the present inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3(a^2b^2c^2+1)~. \end{equation*} Both versions of the inequality can be written as \begin{equation*} 2abc(a+b+c) \,\geq\, 3a^2b^2c^2 + \mathit{const}~, \end{equation*} the first one with $\mathit{const}=9$, the second one with $\mathit{const}=3$. Let us make another try with $\mathit{const}=0\,$: something tells us that third time round we might be lucky. Since $abc>0$, the inequality $2abc(a+b+c)\geq 3a^2b^2c^2$ is equivalent to the simpler inequality \begin{equation*} 2(a+b+c) \,\geq\, 3abc~. \end{equation*} Before we proceed, we define the function \begin{equation*} f_1(a,b,c) \,:=\, 2(a+b+c)-3abc~, \qquad a,b,c>0\,,~ ab+ac+bc = 3\,, \end{equation*} and ask Mathematica to minimize $f_1(a,b,c)$, within the given constraints:
Wow! This is even better than what we expected: we have the inequality $$2(a+b+c) \,\geq\, 3(abc+1) \qquad \text{if a,b,c>0 and ab+ac+bc = 3}\,. \tag{1}$$ If we multiply this inequality by $abc>0$, we obtain the $($valid$)$ inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3abc(abc+1)~, \end{equation*} with the same constraints on $a$, $b$, $c$ as inequality $(1)$, which suspiciously closely resembles the inequality in the question asked by "geromty": it is as if someone too hastily multiplied $abc+1$
by $abc$ and obtained $a^2b^2c^2+1$, never noticing the error.
Mathematica told us that the inequality $(1)$ is true, but that does not absolve us from actually proving it. We define the function \begin{equation*} f_2(a,b,c) \,:=\, 2(a+b+c) - 3(abc+1)\,, \qquad a,b,c>0\,,~ ab+ac+bc = 3~, \tag{2} \end{equation*} that is, $f_2(a,b,c)=f_1(a,b,c)-3$. Let us express, from the constraint $ab+ac+bc=3$, the variable $c$ as a function of the variables $a$ and $b$, \begin{equation*} c \,=\, \frac{3-ab}{a+b}~; \tag{3} \end{equation*} since $a+b>0$, the condition $c>0$ is equivalent to $ab<3$. We feed $c$ expressed by $a$ and $b$
as in $(3)$ to $f_2(a,b,c)\,$: \begin{equation*} g_2(a,b) \,:=\, f_2\Bigl(a,\,b,\,\frac{3-ab}{a+b}\Bigr)~, \qquad a,b>0\,,~ab<3\,. \end{equation*} We want to get a 'feel' for the behavior of the function $g_2$. The awkward thing about the funtion $g_2$ is that it is defined on an unbounded domain, namely on the region between the positive halves of the axes $(a)$ and $(b)$ and the branch of the rectangular hyperbola $ab=3$ in the first quadrant, and besides that it can attain arbitrarily large positive values. We want to actually see the function $g_2$
in its entirety, over all of its domain and over its complete range of values. This can be done.
The homeomorphism $\theta\colon\mathbb{R}_{\geq0}\to[\mspace{2mu}0,1)$ defined by \begin{equation*} \theta(x) \,:=\, \frac{x}{1+x}~, \qquad x \geq 0\,, \end{equation*} has the inverse \begin{equation*} \theta^{-1}(y) \,=\, \frac{y}{1-y}~, \qquad 0 \leq y < 1~. \end{equation*} We can extend the homeomorphism $\theta$ to a homeomorphism $[0,\infty]\to[0,1]$, still denoted $\theta$, by setting $\theta(\infty):=1$. To the points $0$, $1$, $\infty$ of the nonnegative half $[0,\infty]$ of the extended real axis there correspond, via the homeomorphism $\theta$, the points $0$, $1/2$ and $1$ of its "foreshortened"
image $[0,1]$. Now we can squeeze the infinite extents of the positive half-axes and of the range of $g_2$ into unit intervals and obtain a function \begin{equation*} h_2(u,v) \,:=\, \theta\bigl(g_2\bigl(\theta^{-1}(u),\,\theta^{-1}(v)\bigr)\bigr)~, \qquad\quad 0<u,v<1\,,\quad\frac{u}{1-u}\,\frac{v}{1-v}<3\,, \end{equation*} whose diagram within the confines of the unit cube represents, point by point, the entire diagram of the function $g_2$:
Though the diagram is of the function $h_2$ of arguments $u$ and $v$, we read it as a diagram representing the function $g_2$ of arguments $a$ and $b$, with the tick's labels on the axes indicating the values of $g_2$ and its arguments. The function $h_2$ is defined on the set of all $(u,v)$ satisfying the inequalities $0<u<1$, $0<v<1$ and $v<(3-3u)/(3-2u)$; the part of the boundary of the domain of $h_2$ given by $v=(3-3u)/(3-2u)$ for $0\leq u\leq 1$ is an arc of the rectangular hyperbola $(u-3/2)(v-3/2)=3/4$. The two crossed lines drawn in the domain of $h_2$ intersect at the point $(u,v)=(1/2,1/2)$ that correponds to the one and only point $(1,1)$ at which the function $g_2$ attains its global minimum $0$.
It is high time we start thinking about actually proving the inequality \begin{equation*} g_2(a,b) \,\geq\, 0 \qquad \text{if $a,b>0$ and $ab<3$}\,, \tag{4} \end{equation*} which is equivalent to the inequality $(1)$ we intend to prove. We multiply the inequality $(4)$ by $a+b>0$, obtaining another inequality $G_2(a,b)>0$ equivalent to $(1)$, where $G_2(a,b):=(a\!+\!b)\mspace{-2mu}\cdot\mspace{-2mu}g_2(a,b)$. In the expanded form the inequality $G_2(a,b)>0$ reads \begin{equation*} \begin{split} 6 - 3a - 3b + 2a^2 - 7ab + 2b^2 + 3a^2b^2 &\,\geq\, 0 \\[.5ex] &\text{if $a,b>0$ and $ab<3$}\,. \end{split} \tag{5} \end{equation*} Let $\alpha$ be a real parameter in the range $0<\alpha<2$. We rewrite the left hand side of the inequality $(5)$ as \begin{align*} &6 \,+\, \alpha(a^2\mspace{-2mu}-\mspace{-2mu}2ab\mspace{-2mu}+\mspace{-2mu}b^2) \\[.5ex] &\qquad\quad {}\,+\, \bigl((2\!-\!\alpha)a^2\mspace{-2mu}-\mspace{-2mu}3a\bigr) \,+\, \bigl((2\!-\!\alpha)b^2\mspace{-2mu}-\mspace{-2mu}3b\bigr) \,+\, \bigl(3(ab)^2\mspace{-2mu}-\mspace{-2mu}(7\!-\!2\alpha)ab\bigr) \\[1ex] &\quad\,=\, 6 \,-\, \frac{9}{4(2\!-\!\alpha)} \,-\, \frac{9}{4(2\!-\!\alpha)} \,-\, \frac{(7\!-\!2\alpha)^2}{12} \,+\, \alpha(a\!-\!b)^2\\ &\qquad\quad \,+\, (2\!-\!\alpha)\Bigl(a\mspace{-1mu}-\mspace{-1mu}\frac{3}{2(2\!-\!\alpha)}\Bigr)^{\!2}\! \,+\, (2\!-\!\alpha)\Bigl(b\mspace{-1mu}-\mspace{-1mu}\frac{3}{2(2\!-\!\alpha)}\Bigr)^{\!2}\! \,+\, 3\Bigl(ab\mspace{-1mu}-\mspace{-1mu}\frac{7\!-\!2\alpha}{6}\Bigr)^{\!2}~. \end{align*} The constant part of the second formula $($it being constant means it does not depend on $a$ or $b$$), which is \begin{equation*} C(\alpha) \,:=\, 6 \,-\, 2\cdot\frac{9}{4(2\!-\!\alpha)} \,-\, \frac{(7\!-\!2\alpha)^2}{12}~, \end{equation*} cannot be >0 for any \alpha in the range 0<\alpha<2 since this would imply that there is an \varepsilon>0 such that g_2(a,b)\geq\varepsilon over all of the domain of g_2, which we know is not true because the global minimum of g_2 is 0 (we trust Mathematica in this). The best we may hope for is that C(\alpha)=0 for some \alpha within its prescribed range. Well, C(\alpha) is a quotient of polynomials, concretely it is a cubic polynomial in \alpha (with rational coefficients) divided by 2-\alpha. If the cubic polinomial in the numerator has a zero, it must have a double zero, so it will factor into linear factors. So let's see if our hunch is worth its salt: \begin{equation*} C(\alpha) \,=\, \frac{(\alpha-8)(2\alpha-1)^2}{12(2-\alpha)}~. \end{equation*} We are through, since we have C(\alpha)=0 by choosing \alpha=1/2, and with this value of \alpha the left hand side G_2(a,b) of (5) gets rewritten as \begin{equation*} G_2(a,b) \,=\, \tfrac{1}{2}(a\mspace{-2mu}-\mspace{-2mu}b)^2 +\, \tfrac{3}{2}(a\mspace{-2mu}-\mspace{-2mu}1)^2 +\, \tfrac{3}{2}(b\mspace{-2mu}-\mspace{-2mu}1)^2 +\, 3(ab\mspace{-2mu}-\mspace{-2mu}1)^2~. \end{equation*} Observe that G_2(a,b)\geq 0 for all real a and b, not just for a,b>0 satisfying ab<3, and that G_2(a,b)=0 if and only if (a,b)=(1,1). The fact that G_2(a,b)>0 for all a,b\in\mathbb{R} does not mean that g_2(a,b)>0 for all a,b\in\mathbb{R}. First of all, if a+b=0, then g_2(a,b) is not defined. And if a+b<0, then g_2(a,b)=G_2(a,b)/(a\!+\!b)<0 because G_2(a,b)>0. Now we are going to turn the tables: starting with the inequality (1), which we have just proved, we will concoct one truly remarkable inequality. You will have the rare opportunity to observe at close quarters the working of a devious mind of the kind that produces all those bothersome inequalities that you are then given to prove (or disprove). Let us introduce the set H_3:=\{(a,b,c)\in\mathbb{R}^3\mid a b+a c+b c=3\} (we should have done this the moment we first mentioned the constraint ab+ac+bc=3\,; but, as it is said, better late than never). If (a,b,c)\in H_3, then a+b\neq 0, and by symmetry also a+c\neq 0 and b+c\neq 0\,; indeed, a+b=0 implies -a^2 = ab+ac+bc = 3, which cannot be. Thus we can have either a+b>0 or a+b<0\,. This result is slightly unsettling: the condition (a,b,c)\in H_3 is symmetric in a, b, c while the condition a+b>0 is slanted towards the pair a, b. However, contrary to the appearance the condition a+b>0 is actually symmetric in a, b, c since it implies \begin{equation*} a+c \,=\, a+(3\!-\!ab)/(a\!-\!b) \,=\, (3\!+\!a^2)/(a+b) \,>\, 0~; \end{equation*} by symmetry the three conditions a+b>0, a+c>0, b+c>0 imply each other (supposing (a,b,c) is a point in H_3, do not forget about this). Similarly the conditions a+b<0, a+c<0, b+c<0 imply each other. Let us visualize the set H_3, over the whole of its extent. As with visualizing the function g_2(a,b) (for a,b>0 and ab<3$$)$, we have to squeeze an infinite extent into a finite interval, but now we have in each of the three cooordinate directions a two-way infinite extent, that of the whole real line, to squeeze. We achieve this by extending our 'squeeze' map $\theta\colon[0,\infty]\to[0,1]$ to a map $[-\infty,\infty]\to[-1,1]$, still denoted $\theta$, by setting $\theta(x):=-\theta(-x)$ for $-\infty\leq x\leq 0$. That is, for
a real $x\leq 0$ we have $\theta(x)=x/(1-x)$, and $\theta(-\infty)=-1$. The inverse of the extended $\theta$ gives $\theta^{-1}(y)=y/(1+y)$ for $-1<y\leq 0$ and $\theta^{-1}(-1)=-\infty$. The extended 'squeeze' map $\theta$ is constructed as a non-smooth spline of two smooth halves. This would be a serious shortcoming for theoretical 'squeezing' of, say, smooth fields or smooth differential forms, but it is good enough for our visualizations. If you insist on using an overall smooth homeomorphism $(-\infty,\infty)\to(-1,1)$, and one that is an odd functions to boot, then there is no shortage of such maps. Here are two maps for you to choose from: one is $x\mapsto (2/\pi)\arctan(x)$ with the inverse $y\mapsto\tan(\pi y/2)$, another one is $x\mapsto (4/3)x/\bigl(1\!+\!\sqrt{1\!+\!((4/3)x)^2}\,\bigr)$ with the inverse $y\mapsto (3/2)y/\bigl(1\!-\!y^2\bigr)\,$. The curious multipliers $4/3$ and $3/2$ appearing in the formulas for the second map are there to ensure that
the points $-1$ and $1$ of the real line correspond to the points $-1/2$ and $1/2$ in the interval $(-1,1)$.
From the defining equation $ab+ac+bc=3$ of the set $H_3$ we express $c\in\mathbb{R}$ as a function $c(a,b)=(3\!-\!ab)/(a\!+\!b)$ of $a,b\in\mathbb{R}$, $a+b\neq 0$. In order to see the set $H_3$ using our infini-vision, we draw the diagram of $w(u,v):=\theta\bigl(c\bigl(\theta^{-1}(u),\theta^{-1}(v)\bigr)\bigr)\in(-1,1)$, where $u,v\in(-1,1)$, $u+v\neq 0$.$\,$ $($Because $\theta$ is an odd function, the points $a$ and $b$ of the real line such that $a+b=0$ correspond to the points $u:=\theta(a)$ and $v:=\theta(b)$ in the interval $(-1,1)$ such that $u+v=0$.$)\,$ Here it is, the diagram:
Note that the points on the diagonals forming the boundary of the diagram, such as the diagonal from $(-\infty,\infty,\infty)$ to $(\infty,-\infty,\infty)$, do not belong to the diagram proper. We clearly see that the parallel projection of the set $H_3$ to the coordinate plane $(a,b)$ along the axis (c) is the whole plane $\mathbb{R}^2$ minus the 'cross-diagonal' $\{(x,-x)\mid x\in\mathbb{R}\}$, and that the same is true of the projection to the plane $(a,c)$ along the axis (b) and of the projection to the plane $(b,c)$ along the axis (a). And here is the part of the set $H_3$ in the positive octant $\mathbb{R}_{>0}^3$, with which we began the discussion of the inequality $(1)$:
Now we extend the function $f_2$, which was defined in $(2)$ only on the part of $H_3$ in the positive octant, to the whole $H_3$: \begin{equation*} f_2(a,b,c) \,:=\, 2(a+b+c) - 3(abc+1)~, \qquad ab+ac+bc = 3~. \tag{6} \end{equation*} Let us also recall the definitions of the elementary symmetric polynomials \begin{align*} s_1 \,=\, s_1(a,b,c) &\,:=\, a+b+c~, \\[.5ex] s_2 \,=\, s_2(a,b,c) &\,:=\, ab+ac+bc~, \\[.5ex] s_3 \,=\, s_3(a,b,c) &\,:=\, abc~, \end{align*} which we at once employ to recast the definition $(6)$ of the extended function $f_2$, \begin{equation*} f_2(a,b,c) \:=\, 2s_1(a,b,c) - 3s_3(a,b,c) - 3~, \qquad s_2(a,b,c) = 3~. \tag{7} \end{equation*} We introduce the function $\overline{f}_2(a,b,c):=-f_2(-a,-b,-c)$ for $s(a,b,c)=3$ $($note that $s(a,b,c)=3$ implies s(-a,-b,-c)=s(a,b,c)=3), so that \begin{equation*} \overline{f}_2(a,b,c) \,=\, 2s_1(a,b,c) - 3s_3(a,b,c) + 3~, \qquad s_2(a,b,c)=3~. \tag{8} \end{equation*} We have \overline{f}(a,b,c)\geq 6 when s_2(a,b,c)=3 and a+b>0. The function \begin{equation*} f_3(a,b,c) \,:=\, f_2(a,b,c)\,\overline{f}_2(a,b,c)~, \qquad s_2(a,b,c) = 3~, \tag{9} \end{equation*} which is defined on H_3, satisfies the condition that f_3(-a,-b,-c)=f_3(a,b,c) for every (a,b,c)\in H_3, and it follows that \begin{equation*} f_3(a,b,c)\geq 0 \qquad \text{everywhere on H_3}\,, \end{equation*} where f_3(a,b,c)=0 if and only if (a,b,c)=(1,1,1) or (a,b,c)=(-1,-1,-1). Using (7) and (8) we rewrite (9) as \begin{equation*} f_3 \,=\, (2s_1-3s_3)^2 - 9~, \qquad \text{defined when s_2=3}\,. \tag{10} \end{equation*} We introduce yet another function, \begin{equation*} f_4 \,:=\, (2s_1s_2-9s_3)^2 - 3s_2^3~, \tag{11} \end{equation*} which is a homogeneous polynomial in a, b, c of degree 6 defined everywhere on \mathbb{R}^3. We claim that \begin{equation*} f_4(a,b,c) \,\geq\, 0 \qquad \text{for all a,b,c\in\mathbb{R}}~. \tag{12} \end{equation*} This is the truly remarkable inequality we have promised a while ago. Fully expanded the inequality (12) reads \begin{equation*} \begin{aligned} &4\bigl(a^4 b^2 + a^2 b^4 + a^4 c^2 + a^2 c^4 + b^4 c^2 + b^2 c^4\bigr) \\ &\qquad {} \,+\, 8\bigl(a^4 b c + a b^4 c + a b c^4\bigr) \,+\, 5\bigl(a^3 b^3 + a^3 c^3 + b^3 c^3\bigr) \,+\, 15 a^2 b^2 c^2 \\ &\qquad\qquad {} \,-\, 13\bigl(a^3 b^2 c + a^3 b c^2 + a^2 b c^3 + a^2 b^3 c + a b^3 c^2 + a b^2 c^3\bigr) ~\geq~ 0~. \end{aligned} \end{equation*} Proving the inequality (12) is simple. If s_2(a,b,c)\leq0, then it is evident from (11) that f_4(a,b,c)\geq 0. Now the case s_2(a,b,c)>0. We first observe that s_2=3 implies that f_3 is defined, and that f_4=9f_3 which is clear from (10) and (11). Then with \kappa:=\sqrt{3/s_2(a,b,c)} we have s_2(\kappa a,\kappa b,\kappa c) = \kappa^2 s_2(a,b,c) = 3, and \begin{equation*} \kappa^6 f_4(a,b,c) \,=\, f_4(\kappa a,\kappa b,\kappa c) \,=\, 9 f_3(\kappa a,\kappa b,\kappa c) \,\geq\, 0~, \end{equation*} therefore f_4(a,b,c)\geq 0.~ Done. The zeros of f_4 are easy to determine. If s_2(a,b,c)>0, then f_4(a,b,c)=0 iff f_3(\kappa a,\kappa b,\kappa c) = 0 (with \kappa as above), iff (\kappa a,\kappa b,\kappa c)=\pm(1,1,1), iff a=b=c (where the common value of a, b, c is \neq 0 since s_2(0,0,0)=0). Otherwise, if $s_2(a,b,c)\leq 0$, then $f_4(a,b,c)=0$ iff $s_2(a,b,c)=0$ and $-9s_3(a,b,c)=2s_1(a,b,c)s_2(a,b,c)-9s_3(a,b,c)=0$, iff at least two of $a$, $b$, $c$ are $0$. Here's why: because of $abc = s_3(a,b,c)=0$ at least one of $a$, $b$, $c$ is $0\,$; if, say, $a=0$, then $bc=s_2(0,b,c)=0$, hence $b=0$ or $c=0$. Conversely, if at least two of $a$, $b$, $c$ are $0$, then $s_2(a,b,c)=0$ and $s_3(a,b,c)=0$, whence $f_4(a,b,c)=0$. The set of zeros of $f_4$ is therefore the union of the one-dimensional subspaces $\mathbb{R}u$ of $\mathbb{R}^3$ for $u\in \{(1,1,1),\,(1,0,0),\,(0,1,0),\,(0,0,1)\}$. In other words, the homogeneous polynomial $f_4$ has besides the trivial 'affine' zero $(0,0,0)$
$($which is a zero of every nonconstant homogeneous polynomial$)$ the set $Z_4=\{(1\!:\!1\!:\!1),\,(1\!:\!0\!:\!0),\,(0\!:\!1\!:\!0),\,(0\!:\!0\!:\!1)\}$ of four 'projective' zeros.
Now we of course want to see what the function $f_4$ looks like. It helps that $f_4(a,b,c)$ is a homogeneous polynomial in $(a,b,c)$; if we know its values on the unit sphere $S^2 = \{(a,b,c)\mid a^2+b^2+c^2=1\}$ $($the "$2$" in $S^2$ is a superscript, not an exponent$)$, then we know its values on a sphere $rS^2$ of any radius $r\,$: every point in $rS^2$ is of the form $ru$ for some $u\in S^2$
and therefore $f_4(ru) = r^6 f_4(u)$ (where the exponent $6$ is the degree of $f_4$). An attractive way to represent values of some nonnegative function $h$ defined on the unit sphere is the one used for the visualization of the Earth's geoid: at each point $u\in S^2$ we represent the value $h(u)$ by the point $(1+\rho\mspace{1mu}h(u))\mspace{1mu}u$, that is, by the point at the elevation $\rho\mspace{1mu}h(u)$ vertically above the point $u$ on the sphere, where "vertical" means "in the direction of the vector $u$" $($which points straight up$)$.
The positive constant $\rho$ is a suitably chosen scaling factor: usually we will want the average value of $\rho\mspace{1mu}h$ on the unit sphere to be a fraction of its radius $1$, that is, some value between $0$ and $1$, not too small and not too large, and we choose a factor $\rho$ accordingly. Of course, if you like the 'geoid' extending halfway to the Moon, or, at the other extreme, to be barely discernible from the unit sphere, with microscopic maximum heights, go forth and choose a scaling factor to achieve the appearance your heart desires.
Let us ask Mathematica to tell us the maximum value of $f_4$ on the unit sphere:
With the maximum slightly over $2$ the scaling factor $\rho=1/3$ seems reasonable. Here is the corresponding geoid of $f_4$:
We oriented the geoid so that the vector $(1,1,1)$ points upwards. The blue dot at the top marks the position of the zero $(1,1,1)/\sqrt{3}$ of $f_4$ on the unit sphere. The polynomial $f_4$ has eight zeros on the unit sphere $($each of the four projective zeros of $f_4$ is represented on the unit sphere by a pair of antipodal affine zeros$)$, but only three of them are visible.
Here's a thought: why not make an anti-geoid of a function $h$ defined on the unit sphere, representing the value of $h$ at a point $u\in S^2$ by the point $(1-\rho\mspace{1mu}h(u))\mspace{1mu}u$ below the point $u$ on the sphere. If we choose the scaling factor $\rho$ so that the maximum value of $\rho\mspace{1mu}h(u)$ will be less than $1$, then we will obtain a non-self-intersecting surface in the interior of the unit sphere. Let's do this
for $f_4$. We choose $\rho=1/3$ as for the geoid; but this time we draw only the upper hemisphere and the part of the anti-geoid of $f_4$ it contains, and make everything, the hemisphere and the anti-geoid inside it, transparent:
As before, the blue dots mark the positions of zeros of $f_4$ on the unit sphere. The result is slightly reminiscent of snow globes $($well, of half-globes, without the snow$)$.
We shall use the inequality $(12)$ to produce another remarkable inequality \begin{equation*} f_5(a,b,c) \,\geq\, 0 \qquad \text{for all $a,b,c\in\mathbb{R}$}~, \tag{13} \end{equation*} where $f_5$ will be a symmetric homogeneous polynomial of degree $6$. To this end we first restrict the inequality $(12)$ to the strictly positive $a$, $b$, $c\,$: \begin{equation*} f_4(a,b,c) \,\geq\, 0 \qquad \text{for all $a,b,c>0$}~. \tag{14} \end{equation*} Into this restricted inequality we introduce new strictly positive variables $x:=bc$, $y:=ac$, $z:=ab$. $($There was an answer which employed this change of variables, but it was later removed by its author. With my reputation far below 10,000 I am not able to peek into the deleted answer in order to find out and tell you who posted it.$)$ The transformation $(a,b,c)\mapsto(x,y,z)$ is a bijection $\mathbb{R}_{>0}^3\to\mathbb{R}_{>0}^3$; the inverse transformation is $(x,y,z)\mapsto\bigl(\sqrt{yz/x},\,\sqrt{xz/y},\,\sqrt{xy/z}\,\bigr)$. From the inequality $(14)$ we therefore obtain the inequality \begin{equation*} f_4\Bigl(\sqrt{yz/x},\,\sqrt{xz/y},\,\sqrt{xy/z}\,\Bigr) \,>\, 0 \qquad \text{for all $x,y,z>0$}~. \tag{15} \end{equation*} The left hand side of this inequality is not a polynomial, it is a quotient $f_5(x,y,z)/(xyz)$, where $f_5(x,y,z)$ is a symmetric homogeneous polynomial of degree $6$. So we multiply the inequality $(15)$ by $xyz>0$, then rename the variables $x$, $y$, $z$ back to $a$, $b$, $c$, and thus obtain the promised inequality $(13)$, although restricted to $a,b,c>0$. It is a fact that $f_5$ satisfies the unrestricted inequality $(13)$, for any $a,b,c\in\mathbb{R}\,$: Mathematica assures us that that it is so (as we shall see shortly). At this point I have no idea how to go about proving the inequality $(13)$. Do you? Have an idea, I mean, how to prove $(13)$?
The symmetric polynomial $f_5$ of $a$, $b$, $c$ can be expressed as a polynomial in the elementary symmetric polynomials $s_1$, $s_2$, $s_3$, thus: \begin{equation*} f_5 \,=\, (2s_1s_2-9s_3)^2 - 3s_1^3s_3~. \tag{16} \end{equation*} The formula for $f_5$ differs very little from the formula $(11)$ for $f_4$: instead of the term $-3s_2^3$ in $(11)$ there is the term $-3s_1^3s_3$ in $(16)$.
Since the polynomial $f_5$ is symmetric and homogeneous of even degree, we only need to verify that $f_5(a,b,1)\geq 0$ for $-1\leq a,b\leq 1$ to be assured that $f_5(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$. Indeed, if $(a,b,c)\neq(0,0,0)$, one of the absolute values $|a|$, $|b|$, $|c|$ will be the largest and nonzero,
and because of the symmetry it suffices to consider the case where $|c|$ is the largest; but then $f_5(a,b,c)=c^6f_5(a/c,\,b/c,\,1)$, where $-1\leq a/c,\,b/c\leq 1$. So we make Mathematica minimize $f_5(a,b,1)$ on the square $-1\leq a,b\leq 1$:
Therefore the inequality $(13)$ holds $($though we do not know how to prove it$)$. The point $(a,b)=(0,0)$ may not be the only point in the square $-1\leq a,b\leq 1$ at which $f_5(a,b,0)$ attains the minimum $0$, so we ask Mathematica to find all zeros of $f_5$ in the square:
It follows that the set of all zeros of $f_5$ is the union of the seven one-dimensional subspaces $\mathbb{R}u$ for $u\in \{(1,1,1),\,(1,0,0),\,(0,1,0),\,(0,0,1),\,(1,-1,0)\,,(0,1,-1)\,,(-1,0,1)\}$. We also request the maximum value of $f_5$ on the unit sphere,
which is only a little larger than the maximum value of $f_4$ on the unit sphere, so we again choose the scaling factor $\rho=1/3$ with which to construct the geoid of $f_5$,
and also the anti-geoid of $f_5$ inside the unit hemisphere,
As before, the blue dots mark the positions of zeros of $f_5$ on the unit sphere.
What do we get if we apply to the function $f_5$ the same change of variables that we applied to the function $f_4$ to obtain the function $f_5$? We are not very surprised when we find that we get back the function $f_4$. This is not a mere lucky coincidence:
Lemma 1. $~$Let $h(a,b,c)$ be a symmetric homogeneous polynomial of degree $6$ in real variables $a$, $b$, $c$, with the individual degree $\deg_a h$ at most $4$ $($and hence also $\deg_b h\leq 4$ and $\deg_c h\leq 4$$)$. Then the function $h^*(a,b,c)$ of real variables $a$, $b$, $c$, which is defined by \begin{equation*} h^*(a,b,c) \,:=\, abc\mspace{-1mu}\cdot\mspace{-2mu} h\bigl(\sqrt{b}\mspace{1mu}\sqrt{c}/\mspace{-1mu}\sqrt{a},\, \sqrt{a}\mspace{1mu}\sqrt{c}/\mspace{-1mu}\sqrt{b},\, \sqrt{a}\mspace{1mu}\sqrt{b}/\mspace{-1mu}\sqrt{c}\,\bigr)~, \end{equation*} is a symmetric homogeneous polynomial of degree $6$ in $a$, $b$, $c$ such that $\deg_a h^* \leq 4\,$. Moreover, $h^{**}=h\,$.
$($The polynomial $h^*(a,b,c)$ is computed in the field $\mathbb{R}\bigl(\sqrt{a},\sqrt{b},\sqrt{c}\,\bigr)$, where $a$, $b$, $c$ are formal indeterminates and $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ are symbols whose squares are, respectively, $a$, $b$, $c$.$)$ The proof
of Lemma 1 is straightforward: just observe how the monomials are transformed by the change of variables -- there are only five monomials to consider.
The fact that $f_4^{**} = f_4$ leads to a simple proof of the inequality $(13)$. This is a special instance of
a more general result:
Lemma 2. $~$If $h$ and $h^*$ are as in Lemma 1, then $h(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$ if and only if $h^*(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$.
Proof.\, Suppose that $h^*(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}\,$. Applying the change of variables backwards we get $h(a,b,c)=h^*(bc,ac,ab)/(abc)^2$ for all $a,b,c\neq 0$, hence $h(a,b,c)\geq 0$ for all $a,b,c\neq 0\,$. Since the polynomial function $\mathbb{R}^3\to\mathbb{R} : (a,b,c)\mapsto h(a,b,c)$ is continuous, and the set of all $(a,b,c)\in\mathbb{R}^3$ such that $a,b,c\neq 0$ is a dense (open) subset of $\mathbb{R}^3$, it follows that $h(a,b,c)\geq 0$ for all $(a,b,c)\in\mathbb{R}^3$. The converse follows because $h^{**}=h$ by Lemma 1.$\,$ Done.
This is not the end of the story. The present essay is intended to be just the first one in a series of essays dedicated to the inequality stated in the question.
I'll be back.
Later. $~$Because of the unanimous protests (of one) against long and unwanted answers I decided that enough is enough and so won't be back, anytime soon, anywhere on StackExchange Mathematics. This is no web site for old mathematicians with a didactic leaning.
• Lol, please, don't be back, no more. But seriously, this answer is longer than appropriate for this site Jun 3, 2016 at 17:46
• @Yuriy - What's wrong with a long answer? True, it is not really an answer to the question as it was asked, which is barely alive. I hate seeing, times and again, hastily posted homework exercises, which are then hastily and mostly sloppily discussed in comments, and nobody, but nobody learns anything. This answer is long, but from it one can learn something, not only about the inequality stated in the question and its cousins, but about the environment in which such inequalities live. Is giving informative answers, albeit answers in an extended sense, against the spirit of SE mathematics? Jun 3, 2016 at 18:07
• I see nothing wrong with answers (or questions) of any length, but this site has certain rules, and it's not intended for article length posts Jun 3, 2016 at 18:09
• @Yuriy - Where are this rules? I looked through help on SE Mathematics and have seen no such rule. I asked on SE Mathematics Meta about upper limit on the size of answers and/or questions (they told me it is 30,000 characters) and nobody warned me that article length posts are no-no. Perhaps the rules are global, for the whole SE, not just the SE Mathematics. Jun 3, 2016 at 18:18
• Then no problem. But I've seen a lot of comments like 'this question requires an answer too long for Math.SE' Jun 3, 2016 at 18:21 | 2022-05-20T23:18:03 | {
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https://blogs.mathworks.com/cleve/2021/03/02/round-with-tie-breakers-round-two/?s_tid=blogs_rc_3 | # Round, With Tie Breakers, Round Two
I published Round, With Ties to Even a couple of days ago. Steve Eddins and Daniel Dolan immediately had substantive comments. Here is my reaction to their comments.
### Contents
#### Two More
The four choices that I described for how to break ties -- 'even', 'odd', 'up' and 'down' -- all have the same behavior for positive and negative numbers. The current built-in round uses the 'up" rule.
Steve sees the need for two more choices that are sensitive to the sign of the number. I will call them 'plus' and 'minus". They round to the right and to the left on the number line. That's sometimes known as round toward plus infinity and toward minus infinity.
#### Sample
Here's a chart of a few values.
x = (-4.5:1:4.5)';
xRound(x);
x round up down even odd plus minus
-4.500 -5 -5 -4 -4 -5 -4 -5
-3.500 -4 -4 -3 -4 -3 -3 -4
-2.500 -3 -3 -2 -2 -3 -2 -3
-1.500 -2 -2 -1 -2 -1 -1 -2
-0.500 -1 -1 -0 -0 -1 -0 -1
0.500 1 1 0 0 1 1 0
1.500 2 2 1 2 1 2 1
2.500 3 3 2 2 3 3 2
3.500 4 4 3 4 3 4 3
4.500 5 5 4 4 5 5 4
And the plot.
#### Scaling
The current built-in round also includes the possibility of scaling by a power of 10 to facilitate rounding to, say, the nearest multiple of 1000, or of 1/1000, or three significant figures. Daniel was concerned that having both scaling and tie-breaking options in the same function would be confusing.
But, I think it is OK, because:
• round(x,'even'), the parameter 'even' is a char.
• round(x,3), the parameter 3 is numeric.
• round(x,3,'significant'), the parameter 'significant' is a specific char.
So, the statements
• round(x,'even',3)
• round(x,3,'even')
• round(x,'even',3,'significant')
are unambiguous and distinguishable. After x, the other parameters can be given in any order.
#### round
Here is my proposal for the enhancement request about round. It is available here.
type round.m
function r = round(varargin)
% r = round(x) scales and rounds the elements of x to the nearest integers.
% Default: ties, elements halfway between integers, are rounded away from zero.
%
% r = round(x,'even') ties round to even integers.
% r = round(x,'odd') ties round to odd integers.
% r = round(x,'up') ties round away from zero (same as default).
% r = round(x,'down') ties round towards zero.
% r = round(x,'plus') ties round to the right on the number line.
% r = round(x,'minus') ties round to the left on the number line.
%
% r = round(x,n), n >= 0, round(10^n*x)/10^n, round(12.3456,2) = 12.3500
% r = round(x,-n), n > 0, 10^n*round(x/10^n), round(1234.56,-2) = 1200.
% r = round(x,n,'significant'), round(.001234,2,'significant') = .0012
% r = round(x,n,'decimals) same as round(x,n).
%
% r = round(x,...), ties, n, 'significant' and 'decimals' can be in any order.
%
% Use Round(...) with capital R to distinguish from built-in round(...).
[x,n,ties] = parse_input(varargin{:});
x = prescale(x,n);
a = abs(x) + 0.5;
r = floor(a);
switch ties
case 'even'
m = (r == a) & (mod(r,2) == 1);
case 'odd'
m = (r == a) & (mod(r,2) == 0);
case 'down'
m = (r == a);
case 'up'
m = [];
case 'plus'
m = (x < 0) & (r == a);
case 'minus'
m = (x > 0) & (r == a);
otherwise
error(['''' ties ''' not recognized.'])
end
r(m) = r(m) - 1;
r = sign(x).*r;
r = postscale(r,n);
% ----------------------------------------------
function [x,n,ties] = parse_input(varargin)
x = varargin{1};
n = zeros(size(x));
ties = 'up';
for k = 2:nargin
if isnumeric(varargin{k})
n(:) = varargin{k};
elseif strcmp(varargin{k},'significant')
n(:) = n(:) - ceil(log10(abs(x(:))));
elseif strcmp(varargin{k},'decimals')
% ignore
else
ties = varargin{k};
end
end
end
function x = prescale(x,n)
if any(n ~= 0)
k = n > 0;
x(k) = 10.^n(k).*x(k);
k = n < 0;
x(k) = x(k)./10.^(-n(k));
end
end
function r = postscale(r,n)
if any(n ~= 0)
k = n > 0;
r(k) = r(k)./10.^n(k);
k = n < 0;
r(k) = 10.^(-n(k)).*r(k);
end
end
end
Published with MATLAB® R2021a
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https://www.physicsforums.com/threads/errors-uncertainty-in-measurement-problem.870341/ | Errors/ Uncertainty in measurement problem
1. May 4, 2016
psycho_physicist
1. The problem statement, all variables and given/known data
The surface tension (T) is measured by capillary rise formula $$T = \frac {rh ρg}{2 cos\Theta}$$ . The quantities of ρ, g and θ are taken from the table of constants while the height and diameter are measured as
h = (3.00 + 0.01)cm and
D = (0.250 ± 0.001)cm
Find the percentage error in T
My answer came out to be 1.1 % but doesn't coincide with the answer given (0.7 %).
Thanks for any help!
2. Relevant equations
According to the book I refer, the equation for error in product or quotient is -
$$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$
where Δz/z, Δa/a and Δb/b are relative errors (letters in deltas present uncertainty in measurement)
3. The attempt at a solution
I have attempted the solution but have failed to acquire the given answer. Here's what I tried
Acc. to $$\frac{\Delta Z} {Z}= \frac {\Delta A} {A} +\frac {\Delta B} {B}$$
we ignore constants (ρ, Θ and g)
h = (3.00 + 0.01)cm
D = (0.250 ± 0.001)cm
so r = D/2 or r = (0.125 ± 0.001)cm
so
$$\frac{\Delta T} {T}= \frac {\Delta r} {r} +\frac {\Delta h} {h}$$
which gives,
$$\frac{\Delta T} {T} =\frac {0.001} {0.125} +\frac {0.01} {3.00}$$
or,
$$\frac{\Delta T} {T} =0.0080 + 0.0034$$
or,
$$\frac{\Delta T} {T} =0.0113$$
where
$$\frac{\Delta T} {T}$$ is relative error. When we multiply that by 100, we get percent error
therefore percent error = 1.13 %
while the answer given is 0.7 %
please help
2. May 4, 2016
Samy_A
Check your uncertainty on r.
3. May 4, 2016
Staff: Mentor
When you divided the diameter by two, the error in the resulting figure should not remain the same as the original. The percentage error of the result should remain the same. So, scale both by the same factor: 0.250/2 = 0.125; 0.001/2 = 0.0005.
4. May 5, 2016
psycho_physicist
Thanks, that correction seems to nail it!
One Q - the instrument (say vernier calipers) was used to measure diameter and radius is just half of diameter; then why do we halve the error? Shouldn't the uncertainty remain same ?
5. May 5, 2016
Staff: Mentor
What's the rule when multiplying or dividing by a constant? Does the percent error change?
6. May 5, 2016
Dr Transport
Read this file, https://www.wmo.int/pages/prog/gcos/documents/gruanmanuals/UK_NPL/mgpg11.pdf
You really should be taking the derivative of the original formula with respect to each of the variables and do the root square addition for them to get the uncertainty. Each of the variables in your equation has an uncertainty, even the ones pulled from a reference table and it needs to be included.
7. May 5, 2016
Ray Vickson
The diameter lies between .250+.001 cm = 0.251 cm and .250 - .001 cm = 0.249 cm, so the radius lies between .251/2 = 0.1255 cm and 0.249/2 = 0.1254 cm. Those figures are .125 ± 0.0005 cm. The percentage errors in diameter and radius are exactly the same.
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Draft saved Draft deleted | 2018-02-23T20:40:21 | {
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http://mathhelpforum.com/pre-calculus/199454-determining-lines-symmetry-scratch.html | # Thread: Determining Lines of Symmetry from Scratch?
1. ## Determining Lines of Symmetry from Scratch?
Hi all ---
When I tried the following question in the black oval --- the answer is in red --- I was completely stuck on the logic/thought process/intuition in the first part. I'm fine with the algebra --- it's NOT what's troubling me for the past hours.
I know how to check for symmetry about the origin and y-axis. A function, f(x), is symmetric about the origin if $f(-x) = -f(x)$ and about y-axis if $f(-x) = f(x)$.
But here, how on earth would you even know to check for symmetry about $y = \pm x$? How would you even suspect or be aware of this in the first place?
My work: For $xy = v \Longleftrightarrow y = \frac{v}{x}$, I sort of see this because this is just the reciprocal function with the constant $v$.
So here, I'd expect symmetry about $y = \pm x$ because I've seen the graph of the reciprocal function before. But I'm completely lost for $x^4 + y^4 = u$?
Also, what would lead you to check for symmetry about x- and y-axis here? Question doesn't give any information that would lead you to do this?
In general, if I see this kind of question that only asks me to state all lines of symmetry (WITHOUT telling me what the lines are), how would I know
--- which lines I should check for symmetry?
--- how many lines there are?
Maybe because I don't understand it, but the answer looks like a bit of guesswork to me...
Thank you ---
2. ## Re: Determining Lines of Symmetry from Scratch?
The fact that the exponents on x and y are even so that $(-x)^4= x^4$ and $(-y)^4= y^4$ would lead me to say that the graph is symmetric about the axes: the point "symmetric" to (x, y) about the y-axis is (-x, y) and about the x-axis is (x, -y). The fact that the formula is "symmetric" with respect to x and y: replacing x with y and y with x changes $x^4+ y^4= u$ to $(y)^4+ (x^4)= u$ which is the same equation, tells me that it is symmetric about the lines y= x and y= -x.
3. ## Re: Determining Lines of Symmetry from Scratch?
Hi HallsofIvy ---
Thanks a lot for your answer. It really helped with why one'd think $y = \pm x$ and $x, y$ axes give symmetry for the functions.
Just one last question that was maybe unclear in my post --- How would you know/infer that there are no other lines of symmetry to check or test for either curve?
4. ## Re: Determining Lines of Symmetry from Scratch?
After thinking about this for the past week, I still do not see how and why there are no other lines of symmetry? Has someone gotten farther than me?
5. ## Re: Determining Lines of Symmetry from Scratch?
Hi all again --- I should point out that the red box is the solution, so we are NOT told or hinted that both graphs are symmetric about $y = \pm x$ and $x = 0, y = 0$.
After two more weeks of banging my head on this problem, I still do not see how you can tell or guess that there are no other lines of symmetry? Does anyone have more insight?
Thank you --- | 2017-05-28T23:01:50 | {
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https://justinbuschmusic.com/bu87fll/a7d264-mean-questions-with-solutions | ## mean questions with solutions
. . For the following distribution, draw a ‘less than type’ ogive and from the curve, find median. The range is 6 - 1 = 5. Calculate the mean, median and mode of the salaries. Example – 3: Find the second term of the given A.M terms , X , Solution: The terms can be rewritten as log2 3 , X, log2 12, ⇒ 2X = log2 3 + log2 (4 x 3) = log2 3 + log2 4 + log2 3, ⇒ X = log2 3 + 1 = log2 3 + log2 2 = log2 6, Example – 4: If 10-r , a , r +2 are the consecutive terms of an A.P then find the value of ” a”, Solution: Given consecutive terms are 10-r , a , r +2, Example – 5: If (a+2), (4a -6) & (3a -2) are the consecutive terms of an A.P then find the value of ” a”, Solution: Given consecutive terms are (a+2), (4a -6) & (3a -2), Example – 6: If 1/a , 1/b and 1/c are in A.P then bc, ca & ab are also in A.P, Solution: Given consecutive terms are 1/a , 1/b and 1/c are in A.P, Example – 7: If a,b & c are in A.P then 1/bc, 1/ca & 1/ab are also in A.P, Solution: Given consecutive terms are a, b & c are in A.P, Example -8: If 1/a,1/b & 1/c terms are in arithmetic progression then the terms of b+c/a, c+a/b & a+b/c are also in A.P, Solution: Given consecutive terms are 1/a , 1/b & 1/c are in A.P, Multiplying the above terms with ” a+b+c”, a+b+c/a , a+b+c/b & a+b+c/c are also in A.P, Example – 9: If 1/a,1/b,1/c terms are in arithmetic progression then the terms of b+c-a/a, c+a-b/b, a+b-c/c are also in A.P, Solution: Given consecutive terms are 1/a , 1/b & 1/c are in A.P, a+b+c/a , a+b+c/b and a+b+c/c are also in A.P. Math Tricks | Quantitative aptitude | Basic Mathematics | Reasoning, In this article explained about Definition, Properties, Formula and Examples with Solutions of Arithmetic Mean, The arithmetic mean “A” of any two quantities of ” a” and ” b”. https://study.com/academy/lesson/mean-definition-sample-problems-quiz.html What is the mean test score? JEE Main 2020 Question Paper with Solution - here you will get solutions of JEE Main 2020 (B.Arch) 6th January 2nd slot question paper. Relationship Between Arithmetic, Geometric, Harmonic Mean. 89, 73, 84, 91, 87, 77, 94 Solution: The sum of these numbers is 595. . Your email address will not be published. We now multiply all data values by a constant k and calculate the new mean μ' and the new standard deviation σ'. . Exercise (Ex 9D) 9.4. . A worksheet including exercises on calculating the mean of given sets and word problems on the mean. 1. Definition: The arithmetic mean of a set of data is found by taking the sum of the data, and then dividing the sum by the total number # ( i ) a + b + c = -3 . Yes, since data Set C has data values that are further away from the mean compared to sets A and B. . Applying the Mean Value Theorem — Practice Questions By Mark Ryan If you traveled from point A to point B at an average speed of, say, 50 mph, then according to the Mean Value Theorem, there would be at least one point during your trip when your speed was exactly 50 mph. The answers are provided and are located at the lower part of the page. If the standard deviation of a given data set is equal to zero, what can we say about the data values included in the given data set? Solution to Question 5: (f + g)(x) is defined as follows (f + g)(x) = f(x) + g(x) = (- 7 x - 5) + (10 x - 12) Group like terms to obtain (f + g)(x) = 3 x - 17 Example: x is the median for 4, 3, 8, x and 7. Sum of data Answer: Mean and Median Exercise 34B – Selina Concise Mathematics Class 6 ICSE Solutions. Ask questions, doubts, problems and we will help you. Given a partial data set, you may need to determine the possible values of the median of a set of numbers. The maximum number is 6. Sum of observations of 10 numbers = 150 CBSE Class 10 Maths MCQ Pdf Question 21. . Solutions. All the Exercise questions with solutions in Chapter 9 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive are given below: Exercise (Ex 9A) 9.1. If the mean of x, x + 2, x + 4, x + 6 and x + 8 is 13, find the value of x. 1) View Solution Solution: Mean = (5000 + 5000 + 5000 + 5000 + 15000)/5 = 35000/5 = 7000. So, the mean salary is Rs. . . Solution: 7000 per month. The mean is 3. ( ii ) abc = 8 . Exam Questions – Normal distribution, finding a probability. My self Sivaramakrishna Alluri. . + rn ], 1. . We limit the discussion to a data set with 3 values for simplicity, but the conclusions are true for any data set with quantitative data. Example -10: If a, b & c terms are in arithmetic progression then the terms of b+c, c+a , a+b are also in A.P. Solution: Mean = (Sum of all 10 matches) /10 = (2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3)/10 = 2.8. The following number of goals was scored by a team in a series of 10 matches: 3, 5, 0, 3, 3, 4, 3, 2, 4. Find the value of x. Short Answer Type Questions. averages, average, means, modes, medians, ranges. . a) Calculate the mean of the salaries of the 20 people. Question 5. Solution: The given mean of 10 numbers = 15. Q. . Solution To get the mean: The sum is 18 Divide this by 6 to get the mean. Solution: Let terms of A.M is, a, b, & c. According to A.M formula. 15 = sum of observations / 10. We first find the midpoints of the given classes. In an arranged señes of an even number of 2n terms which term is median? 3, 4, x, 7, 8. of observations. rn be an finite A.P with n terms. . Out of these observations if the mean of first 13 observations is 22 and that of the last 13 observations is 30, the 13th observation is: (a) 23 (b) 26 (c) 28 (d) 30. A set of multiple choice maths questions are presented. . In this page explained about formulas of Arithmetic Mean, Geometric Mean and Harmonic Mean. . mean of Data set A = (9+10+11+7+13)/5 = 10 mean of Data set B = (10+10+10+10+10)/5 = 10 mean of Data set C = (1+1+10+19+19)/5 = 10 Standard Deviation Data set A Question: 1.Calculate Mean & Median With A Complete Solution 2. rn, b are in A.P, 2. To get the range: The minimum number is 1. Filed Under: CBSE. The mean of 25 observations is 26. Mean Stack Interview Questions: Mean stack is the combination of four popular JavaScript-based technologies MongoDB, Express, AngularJS and Node that allows developers to … The workers earn a salary of Rs. It then makes a return trip at an average speed of 60 km/h. The difference between the first and second must be equal to the difference second and third, giving the equation x-y=y-z. . Click here for Answers. Question … Solution: Mode = 8; Mean = 8; Median = ? Extra Questions for Class 10 Maths. Problem 1: The Effects of Changing Units on Mean Hourly Wages. . examveda.com is a portal which provide MCQ Questions for all competitive examination such as GK mcq question, competitive english mcq question, arithmetic aptitude mcq question, Data Intpretation, C and Java programing, Reasoning aptitude questions and answers with easy explanations. The frequency table of the monthly salaries of 20 people is shown below. Solution: The given terms are a, b & c in A.P, (a +c) + 2b = (a +c) + (a +c) = 2(a +c), Arithmetic Progression Questions with Solutions, Harmonic Progression formulas and examples. The mean of a list of 6 numbers is 20. . rn are said to be arithmetic means between “a” and “b” if a, r1, r2, r3, r4, . NCERT Solutions for Class 10 Maths. Relation among mean, median and mode is 3 median = mode + 2 mean 3 × median = 8 + 2 × 8 Median = $$\frac{8+16}{3}$$ = $$\frac{24}{3}$$ = 8. Find the possible values for x. Math Questions With Answers. The Median of the following distribution is 35.Find the value of x: Solution: Question 23. So, Mean of 10 numbers = sum of observations/ no. Exercise (Ex 9C) 9.3. Example- 2: Find Arithmetic mean of numbers 1/2 and 1/3. mean of Data set A = (9+10+11+7+13)/5 = 10. . Data set C has the largest standard deviation. Answer: (b) 26 Now take a 5 = 48 ⇒ a r 5-1 = 48 ⇒ a 2 4 = 48 ⇒ a = 3. . To obtain the median, let us arrange the salaries in ascending order: The question asks for a three digit number: call the digits x, y and z. Example 1:Find the arithmetic mean of first five prime numbers. . . Median: Arrange the goals in increasing order We much have b – A = A- a ; Each being equal to the common difference. If we remove one of the numbers, the mean of the remaining numbers is 15. Mean with solution. The following table shows the grouped data, in classes, for the heights of 50 people. From equation ( ii ) , ( iii ) & b = -1. a = 2 and c = -4. Solution: Using the formula: Sum = Mean × Number of numbers Sum of original 6 numbers = 20 × 6 = 120 Sum of remaining 5 numbers = 15 × 5 = 75 Question 20. So the sequence is 2, -1, -4. The last two numbers are 10. Solution:First five prime numbers are 2, 3, 5, 7 and 11. Selection of the terms in an Arithmetic Progression, If number of terms is 3 then assume them as ” a-d, a & a+d” and common difference is “d”, If number of terms is 4 then assume them as ” a-3d, a-d, a+d & a+3d ” and common difference is “2d”, If number of terms is 5 then assume them as ” a-2d, a-d, a, a+d & a+2d ” and common difference is “d”, If number of terms is 6 then assume them as ” a-5d, a-3d, a-d, a+d , a+3d & a+5d ” and common difference is “2d”. . Detailed explanations and solutions to these questions … This GMAT averages problem is a vey easy question. The “n” numbers r1, r2, r3, r4, . G n are said to be Geometric means in between ‘a’ and ‘b’. a + c = 2b . Give One Example Each On Designed Experiment, Observational Study And Retrospective Study, T Distribution And Sampling Distribution Between The Difference Of Means And Justify. Solution: From graph, median = 54. Example -11: Four geometric means are inserted between 1/8 and 128. . Thank you for watching my blog friend, In this article explained about Definition, Properties, Formula and Examples with Solutions of, Basic Concept, formula with examples for Arithmetic Mean, For any given two quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A.P. RS Aggarwal Solutions Class 10 Chapter 9 - Benefits Prove that:a/b=7+4√3 or 7-4√3. In this article we learn about harmonic progression formula for nth term, sum of terms, properties with examples. For any given two quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A.P. Dividing the sum by the number of test scores we get: Answer: The mean test score is 85. . Then terms thus inserted are called the Arithmetic mean. Solution: Question 21. Q.1. of terms = 2n which are even i.e., the mean of n th and (n + 1) th term will be the median. Here given a 7 = 8 x a 4 and also a 5 = 48 ⇒ a r 7-1 = 8 x a r 4-1 ⇒ r 6 = 8 x r 3 ⇒ r = 2. . = (2 + 3 + 5 + 7 +11)/5 = 5.6. Problem: Scott took 7 math tests in one marking period. . Problems related to data sets as well as grouped data are discussed. To get the median: The median is the average of the two middle numbers. The questions have been designed to test for deep understanding of maths concepts. from the expert community at Experts Exchange Required fields are marked *. Answer. Step 2: Compute sum of 3 of the 5 numbers using the average of the 3 numbers. In this problem, we were given the following information: In 2013: 1 pound sterling = 89 rupees Then find the terms. The mean of 18, 28, x, 32, 14 and 36 is 23. Mathematically Arithmetic Mean is also called average of that quantities, i.e If r1, r2, r3, r4, . MCQ for Class 10 Maths with Answer: bExplaination: (b) centred at the classmarks of the classes. The arithmetic mean between a and b is twice the geometric mean between a and b. . Solution: Arrange the numbers in ascending order with x in the middle. Find second geometric mean. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks. . (3 + 3) / 2 = 3 To get the mode: 3 is the mode because it occurs the most number of times. Exercise (Ex 9B) 9.2. Then arithmetic mean “A” of these numbers is given by, A = (1/n) [ r1 + r2 + r3 + r4,+. . Solutions. Your email address will not be published. You can also download NCERT Solutions For Class 10to help you to revise complete syllabus and score more marks in your examinations. The mean or the average formula is = Sum/no. Mean and standard deviation problems along with their solutions at the bottom of the page are presented. Example 2:Find the average of 56, 41, 59, 52, 42 and 44. Example- 2: Find Arithmetic mean of numbers 1/2 and 1/3, Arithmetic mean = [ (1/2) + (1/3) ] / 2 = 5/16. After solving or attempting to solve the problems above on your own, check your answers below to compare what you succeeded on and what you may need to review. Question 5. Again, we limit the discussion to a data set with 4 values for simplicity, but the conclusions are true for any data set with quantitative data. Find the mean of the 32 numbers, such that if the mean of 10 of them is 15 and the mean of 20 of them is 11. What is the number that was removed? ( iii ) From equation ( i ) & ( ii ) b = -1. https://www.aplustopper.com/statistics-icse-solutions-class-10-mathematics .. . b) Calculate the standard deviation of the salaries of the 20 people. . AM, GM and HM, Harmonic Progression Formula, Properties and Harmonic Mean Formula, Geometric progression problems and solutions with Formulas and properties, Geometric Progression Formulas and Properties & Sum of Geometric Series, Arithmetic Progression Problems with Answers for Competitive Exams, Volume and Surface Area of a Cylinder Formulas – Right Circular Cylinder, Surface Area and Volume of Sphere, Hemisphere, Hollow Sphere Formulas, Examples, Practice session on cube and cuboid Questions- Allmathtricks, Volume of cube and cuboid, Area of cube and cuboid | Allmathtricks, Ratio proportion and variation problems with solutions, Allmathtricks, Ratio proportion and variation formula with aptitude tricks – Allmathtricks. . Consider the following three data sets A, B and C. A given data set has a mean μ and a standard deviation σ. Step 1: Compute sum of all 5 numbers using the average of the 5 numbers. NCERT Question Based on Mean, Median, and Mode. Mean compared to sets a, b and C. a given data set, you need..., and mode of these numbers is 15: first five prime numbers, 91 87! Term, sum of these numbers is 595: x is the average of the 20 people scores... With a complete solution 2: Compute sum of 3 of the 5 numbers questions have been designed test. The average of 56, 41, 59, 52, 42 and 44 and third, giving the x-y=y-z... The equation x-y=y-z ; mean = 8 ; median = designed to test deep! Further away from the mean mean test score is 85 further away from the curve, find.!, 32, 14 and 36 is 23 -1. a = ( 2 + 3 + 5 + +11. Numbers 1/2 and 1/3 of numbers numbers are 2, G 2, G,! 12, so you have x+y+z=12 following data solution: mean = ( 5000 + 5000 + +... -1, -4 following distribution, finding a probability midpoints of the remaining numbers 15! 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mean questions with solutions 2021 | 2021-05-08T14:06:20 | {
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https://math.stackexchange.com/questions/3801331/how-many-even-three-digit-numbers-have-distinct-digits-and-have-no-digit-5 | # How many even three-digit numbers have distinct digits and have no digit $5$?
How many even three-digit numbers have distinct digits and have no digit $$5$$?
The answer my teacher gave was $$252$$, but I don't see how she got that. I thought it would be $$6\times 8 \times 5=240$$ because the $$3\text{rd}$$ digit must be even $$(0,2,4,6,8)$$, the $$2\text{nd}$$ digit can't be $$5$$ or the $$3\text{rd}$$ digit $$(10-2=8$$ options) and the first digit can't be $$0$$, $$5$$, the $$2\text{nd}$$ digit, or the $$3\text{rd}$$ digit ($$10-4=6$$ options). Any way I look at it, the last digit has to be even and there are $$5$$ options for even digits, so the final answer must end in a $$5$$ or a $$0$$, not a $$2$$. Please help!
• You have to add those cases where $0$ is used up as the third digit and can't be there at the first digit. It would be better to split into 2 cases, one with last digit as $0$ and other with last digit as non-zero. – gemspark Aug 24 '20 at 4:11
• Looks like they're counting 024 as a three digit number. – David Aug 24 '20 at 4:13
• @ L_M Quite... while your logic is sound for the ones-place digit and the tens-place digit, your hundreds-place digit might have $6$ options (in the event the ones-digit was not zero) or it might have had $7$ options (in the event the ones-digit was zero). Your mistake is that you forgot to consider what happens if the list "zero, 5, 2nd digit, 3rd digit" had some overlap. – JMoravitz Aug 24 '20 at 4:14
No. ending with non-zero digit = $$4 .7.7=196$$.
No. ending with zero digit = $$1 .8.7=56$$
P.S. The product above is (third digit options) x (first digit) x (second digit)
I figured out a different way to do it actually. I see that you have to account for when the third digit is zero and non-zero because it changes the criteria for the first digit, but I considered this for the second digit as well. What I did was split it into three cases:
Last digit is zero: $$7 \times 8 \times 1=56$$ (second digit cannot be $$5$$ or $$0$$ : the first digit cannot be $$5$$, the second digit or the third digit)
Last digit is nonzero and second digit is zero: $$7 \times 1 \times 4=28$$ (first digit cannot be $$5$$, the second digit or the third digit)
Second and last digits are nonzero: $$6 \times 7 \times 4=168$$ (the second digit cannot be $$5, 0$$, or the last digit: the first digit cannot be $$5,0$$, the second digit or the third digit)
Addition principle:$$56+28+168=252$$
It gives the right answer, but is this reasoning sound?
• Looks sensible. – Henno Brandsma Aug 24 '20 at 6:39
• You have just subdivided the non-zero last digit case of my answer into 2 subcases. – gemspark Aug 24 '20 at 7:27
So, one is asking for the number of injections $$f:\{1,2,3\} \to \{0,1,2,3,4,6,7,8,9\}$$ where $$f(1) \neq 0$$ and $$f(3) \in \{0,2,4,6,8\}$$.
Ignoring the last two conditions, one gets $$_{9}P_{3}=504$$ possible numbers (possibly odd and/or having a leading zero).
If one requires the first digit to be nonzero, then one would then need to subtract the number of injections $$\{2,3\} \to \{1,2,3,4,6,7,8,9\}$$ to get $$504-{_{8}P_{2}}=504-56=448$$.
Odd numbers still have not been eliminated yet. We have to eliminate the cases where the last digit is $$1, 3, 7,$$ or $$9$$.
Suppose that the last two digits ($$f(2)$$ and $$f(3)$$) are $$0$$ and $$1$$ respectively. Then, one must have $$f(1) \in \{2,3,4,6,7,8,9\}$$, eliminating $$7$$ odd numbers to reduce the count to $$441$$.
Similar considerations apply when the middle digit is still $$0$$, but the last digit is now $$3, 7,$$ or $$9$$. This eliminates $$3 \cdot 7=21$$ more odd numbers, reducing the count to $$420$$.
Now, let's move on to the case where the middle digit is nonzero. Fixing a last digit $$d \in \{1,3,7,9\}$$, one would then need to subtract the number of injections $$\{1,2\} \to \{1,2,3,4,6,7,8,9\} \setminus \{d\}$$.
Suppose that $$d=1$$. Then, subtracting the number of injections $$\{1,2\} \to \{2,3,4,6,7,8,9\}$$ gives $$420-{_{7}P_{2}}=420-42=378$$. Forty-two odd numbers with the last digit equal to $$1$$ have now been eliminated.
Similar considerations apply when $$d$$ is $$3,7,$$ or $$9$$. This eliminates $$3 \cdot 42=126$$ more odd numbers, giving a final count of $$252$$. | 2021-03-07T15:58:06 | {
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https://math.stackexchange.com/questions/2425070/is-there-a-way-to-deal-with-this-singularity-in-numerical-integration | # Is there a way to deal with this singularity in numerical integration?
I would like to compute numerically, e.g., using the standard method of trapezes the following definite integral over the interface $$[0,1]$$: $$I = \int_0^1 \frac{f(x)}{\sqrt{1-x^2}} \, \mathrm{d} x \, ,$$
where $$f(x)$$ is continuous in the interval [0,1]. It can readily be shown analytically that the integral is well convergent. However, when proceeding numerically, difficulties arise since the integrand diverges at the upper limit of integration for $$x=1$$.
I was wondering whether there exists a procedure that can help to remove the singularity in this integral. Any help is highly appreciated
Thank you,
hartmut
• $I=\int^{\pi/2}_0f(\sin\theta)\,d\theta$ might help. – user436658 Sep 11 '17 at 10:24
• as soon as $f$ is also diffentiable, integration by parts looks feasible – tired Sep 11 '17 at 12:52
• There is a lot about this sort of question in Forman Acton's nice little book "Real computing made real". – Gareth McCaughan Sep 12 '17 at 10:17
• @GarethMcCaughan Errors of the third kind indeed! I will get my hands on a copy; wish I had done so 20 years ago! – uhoh Sep 12 '17 at 12:57
It's easy to convert your integral into one without any singularities via the substitution $x=\sin\theta$: $$I=\int^{\pi/2}_0f(\sin\theta)\,d\theta.$$ If you're lucky to have a function $f$ that is even (so that $I$ is just a quarter of the integral over the full period, from $-\pi$ to $\pi$), you might be pleasantly surprised by the speed of convergence of the trapezoidal rule in this special situation.
• ...and if indeed certain conditions apply, and you do find the trapezoidal rule to be performing exceptionally well, see this for an explanation. – J. M. isn't a mathematician Sep 11 '17 at 12:15
• The numerical integration indeed performs powerfully well. Only a few number of discretization points are needed to compute the integral – Daddy Sep 11 '17 at 12:24
• Following up on your great answer, $f$ is in fact even, meaning that the integral above = 1/4 of that over a full period. But why does that speeds the convergence up? It is easier to perform a numerical integration over [0, $\pi/2$] than to integrate numerically over [0, $\pi$] and then take the quarter.. your clarification is highly appreciated. Thanks! – Daddy Mar 23 '18 at 14:25
A general technique is to remove the singularity by moving it to an analytically integrable function:
$$\int_0^1\frac{f(x)}{\sqrt{1-x^2}}dx=\int_0^1\frac{f(x)-f(1)+f(1)}{\sqrt{1-x^2}}dx=-\int_0^1\frac{f(x)-f(1)}{\sqrt{1-x^2}}dx+f(1)\left.\arcsin(x)\right|_0^1.$$
The limit of the integrand at $x=1$ should be finite.
For instance with $f(x):=x$,
$$\frac{x-1}{\sqrt{1-x^2}}=-\sqrt{\frac{1-x}{1+x}},$$ which is well-behaved.
• Nice trick, for sure ! – Claude Leibovici Sep 11 '17 at 10:52
• It should be $$\frac{x-1}{\sqrt{1-x^2}}=-\sqrt{\frac{1-x}{1+x}},$$ and "well-behaved" isn't entirely true: the integrand isn't differentiable for $x=1$, so the composite trapezoidal rule with a partition into $N$ equal subintervals has an error of about $\frac1{2N}$ in this case, while it would be $O(N^{-2})$ for a regular ($f''$ bounded) integrand. – user436658 Sep 11 '17 at 12:21
• @ProfessorVector: that's right, you have a vertical tangent, which is not ideal for the Newton-Cotes method. My point was that the unbounded behavior is removed, and the method allows to remove several singuarities, of different types. – Yves Daoust Sep 11 '17 at 13:58
• The $\arcsin(x)$ outside the integral lacks context. I think you mean $\frac\pi2$. (+1) anyway, because this was the answer I was thinking of, but overthought in my answer. – robjohn Sep 12 '17 at 15:54
• @robjohn: you are right, I fixed. – Yves Daoust Sep 12 '17 at 16:29
The other answers so far focus on how to transform the integral so that you don't have a singularity. But that's not always viable so it's important to know how to deal with a singularity when one exists.
The trapezoid rule is a poor choice for this case, because it uses the values of the function at the endpoints, which can be infinite. Simposon's rule will have a similar problem.
More appropriate is the more primitive rectangle rule. That is, divide the domain to segments, and multiply each segment's width by the value of the function at the midpoint of the segment. This will work since you never try to sample the function at the infinite endpoints.
Alternatively, if you want faster convergence, you can use the more sophisticated and extremely powerful Gaussian Quadrature. More complicated but well worth knowing. Here, too, you divide the domain to intervals. For each interval, you sample the function at a few carefully chosen points to obtain surprisingly high accuracy.
There are different methods depending on the number of points in each interval, but the simplest is the 2-point method. And again, it will work with singularities, since the endpoints of intervals aren't sampled.
There is even a variant of Gaussian quadrature tailored specifically for a well-behaved function multiplied by $\frac{1}{\sqrt{1-x^2}}$, called Chebyshev–Gauss quadrature. To use it, you don't even have to divide to intervals - you just choose $n$ (the higher $n$, the better the accuracy) and you have
$$\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}\approx\frac{\pi}{n}\sum_{i=1}^nf\left(\cos\left(\frac{2i-1}{2n}\pi\right)\right)$$
Which is very similar to transforming the integral as some other answers suggested. It will maximize the accuracy you can have for a given number of samples from the function $f$, and convergence is exponential.
• For reference: the last formula in this answer is Gauss-Chebyshev quadrature (since the underlying orthogonal polynomials are the Chebyshev polynomials.) – J. M. isn't a mathematician Sep 11 '17 at 14:55
• @J.M.isnotamathematician - Yep, I intended to mention this but apparently forgot to. I've added the name now. – Meni Rosenfeld Sep 11 '17 at 15:02
• FTR, Gauß quadrature per se doesn't buy you anything over simpler methods if there are singularities in the domain. Its only benefit is excellent accuracy for integrating on a compact interval $I$ a function that's smooth on $I$ and in some environment beyond. – leftaroundabout Sep 12 '17 at 14:11
Since your integral is over a subset of the interval $(-1,1)$, your $f$ is continuous on that interval, and your integrand only has singularities on the boundary of that interval, tanh-sinh quadrature is a candidate method for you.
This will change your interval of integration to $[0,\infty]$, but the integrand will decay doubly-exponentially, (e.g., like $\mathrm{e}^{-\mathrm{e}^n}$), so truncation at finite right endpoint need not be significant.
You can continue your plan of using the trapezoidal method with step size $h$ by using the sample points $x_k = \tanh(\frac{\pi}{2} \sinh kh)$ and weights $w_k = \frac{h \pi}{2} \cdot \frac{\cosh kh}{\cosh^2( \frac{\pi}{2} \sinh kh)}$. For instance, the area of your first trapezoid would be $$A_0 = \frac{x_1 - x_0}{2} \left(\frac{w_0 f(x_0)}{\sqrt{1 - x_0^2}} + \frac{w_1 f(x_1)}{\sqrt{1 - x_1^2}} \right) \text{.}$$ Just using $x_k$ and $w_k$ in a left Riemann sum, (and halving $w_0$ when summing this way) convergence is exponential: halving $h$ doubles the number of correct digits. Error analysis suggests the trapezoidal rule applied here should improve convergence slightly (although perhaps is not worth the "effort" compared to just computing $\sum_{k=0}^\infty w_k \frac{f(x_k)}{\sqrt{1-x_k^2}}$, stopping when the terms are small enough to meet your precision and accuracy requirements or when either $1-x_k$ or $w_k$ is so small you cannot easily represent it).
Example: $f(x) = 1$ :
The value of the integral is $\frac{\pi}{2} = 1.57079\dots$. Using left-Riemann summation (and halving $w_0$), with $h = 1/10$, and stopping when $k=20$ (because $1-x_k$ and $w_k$ are about $10^{-5}$), the sum is $1.5659 \dots$. Continuing on to $k = 27$ obtains $1.57079\dots$.
• With $h = 1/20$, stopping at $k = 40$ ($1 - x^k$ around $10^{-5}$ and $w_k$ around $10^{-36}$), obtains $1.56504\dots$.
• With $h = 1/20$, stopping at $k = 54$ ($1 - x^k < 10^{-6}$ and $w_k$ around $10^{-150}$), obtains $1.57078\dots$. (An arbitrary precision calculation gives an error of about $5 \times 10^{-12}$ when we sum $70$ terms.)
We're only summing tens of terms to get these results...
• The double-exponential quadrature of Takahasi and Mori is indeed a good way to flatten singularities at either end of the integration interval, but one needs to be careful with evaluations near the endpoints, as pointed out in their paper. – J. M. isn't a mathematician Sep 11 '17 at 14:56
• @J.M.isnotamathematician : I agree. I don't think I've hidden that. I repeatedly comment on the difficulty of representing $1-x_k$ and $w_k$ for large $k$... – Eric Towers Sep 11 '17 at 15:19
I like Professor Vector's method, but here is a way to take advantage of the integrability of $\frac1{\sqrt{1-x^2}}$ on $(0,1)$.
Using the integral $$\int_0^1\frac{a+bx}{\sqrt{1-x^2}}\,\mathrm{d}x=\frac{\pi a}2+b$$ we can write $$\int_0^1\frac{f(x)}{\sqrt{1-x^2}}\,\mathrm{d}x=\int_0^1\frac{f(x)-(1-x)f(0)-xf(1)}{\sqrt{1-x^2}}\,\mathrm{d}x+\left(\frac\pi2-1\right)f(0)+f(1)$$ If $f$ is smooth at $0$ and $1$, the integrand on the right vanishes at $0$ and $1$
A Lesson in Overthinking
I was wondering why $f(0)$ and $f(1)$ had different weights. Then I realized that $\frac1{\sqrt{1-x^2}}$ doesn't have a singularity at $0$. We don't need to subtract $(1-x)f(0)$ at all. While the formula I give above is correct, I was thinking of $\sqrt{x(1-x)}$ in the denominator, which does have a singularity at $0$ and $1$. Doh!
The standard way is $$x = 1 - t^{2}$$: $$\int_{0}^{1}{\operatorname{f}\left(x\right) \over \,\sqrt{\,{1 - x^{2}}\,}\,}\,\mathrm{d}x = 2\int_{0}^{1}{\operatorname{f}\left(1 - t^{2}\right) \over \,\sqrt{\,{2 - t^{2}}\,}\,}\,\mathrm{d}t$$ | 2021-02-25T02:26:58 | {
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https://mathoverflow.net/questions/141378/is-this-lemma-in-elementary-linear-algebra-new/141448 | # Is this lemma in elementary linear algebra new?
Is anyone familiar with the following, or anything close to it?
Lemma. Suppose $A$, $B$ are nonzero finite-dimensional vector spaces over an infinite field $k$, and $V$ a subspace of $A\otimes_k B$ such that
(1) For every nonzero $a\in A$ there exists nonzero $b\in B$ such that $a\otimes b\in V$,
and likewise,
(2) For every nonzero $b\in B$ there exists nonzero $a\in A$ such that $a\otimes b\in V$.
Then
(3) $\dim_k(V) \geq \dim_k(A) + \dim_k(B) - 1$.
Remarks: The idea of (1) and (2) is that the spaces $A$ and $B$ are minimal for "supporting" $V$; that is, if we replace $A$ or $B$ by any proper homomorphic image, and we map $A\otimes B$ in the obvious way into the new tensor product, then that map will not be one-one on $V$. The result is equivalent to saying that if one is given a finite-dimensional subspace $V$ of a tensor product $A\otimes B$ of arbitrary vector spaces, then one can replace $A$, $B$ by images whose dimensions sum to $\leq \dim(V) + 1$ without hurting $V$.
In the lemma as stated, if we take for $A$ a dual space $C^*$, and interpret $A\otimes B$ as $\mathrm{Hom}(C,B)$, then the hypothesis again means that $C$ and $B$ are minimal as spaces "supporting" $V$, now as a subspace of $\mathrm{Hom}(C,B)$; namely, that restricting to any proper subspace of $C$, or mapping onto any proper homomorphic image of $B$, will reduce the dimension of $V$.
In the statement of the lemma, where I assumed $k$ infinite, I really only need its cardinality to be at least the larger of $\dim_k A$ and $\dim_k B$.
The proof is harder than I would have thought; my write-up is 3.3K. I will be happy to show it if the result is new.
• Welcome to Mathoverflow! – Mark Sapir Sep 5 '13 at 21:09
• I don't have time to think about this right now, but it seems strikingly familiar to the following theorem of Hopf. If $f:A \otimes B\to C$ is a linear map which is injective on each factor separately, then $\dim f(A\otimes B) \geq \dim A + \dim B - 1.$ However, this theorem is true over $\mathbb{C}$ but false over $\mathbb{R}$ (the proof is given by algebraic topology), so maybe it is only a superficial observation. – Jack Huizenga Sep 6 '13 at 0:50
• This feels like a statement from projective geometry. $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ somehow "intersects enough" $\mathbb{P}(A) \times \mathbb{P}(B)$ so that $\dim(\mathbb{P}(V)) \geq \dim(\mathbb{P}(A) \times \mathbb{P}(B))$. – Martin Brandenburg Sep 6 '13 at 1:23
• Probably also related: Flanders' theorem (§8.3 in Prasolov's Linear Algebra book www2.math.su.se/~mleites/Prasolov/prasLinAlg/pr-linAlg-main.dvi ). – darij grinberg Sep 6 '13 at 2:30
• Somehow it feels that the following "dual" result is very closely related, and could for some fields also yield your inequality (by splitting the tensor product into "separable" and "joint" spaces, whose dimensions add up to $d_Ad_B$): On the maximal dimension of a completely entangled subspace..." by K. Parathasarathy; ias.ac.in/mathsci/vol114/nov2004/Pm2342.pdf --- in particular, your subspaces have the "separable" state property, while the cited paper considers "full entangled" subspaces. – Suvrit Sep 6 '13 at 15:04
I believe it is suitable, as an answer, to give a proof that works with no restriction on the cardinality of the underlying field $F$. I will frame the answer in terms of matrix spaces. Thus, we have a linear subspace $V \subset M_{n,p}(F)$ such that, for every non-zero vector $X \in F^n$, the space $V$ contains a rank $1$ matrix with column space $F X$ and, for every non-zero vector $Y \in F^p$, the space $V$ contains a rank $1$ matrix with row space $F Y^t$. Note that those assumptions are unchanged in multiplying $V$ by invertible matrices (be it on the left or on the right).
The proof works by induction on $p$. The case where $p=1$ or $n=1$ is obvious. Assume now that $p>1$ and $n>1$. The discussion is split into two cases, where the standard basis of $F^p$ is denoted by $(e_1,\dots,e_p)$.
Case 1: $V e_p=F^n$. Then, one writes every matrix $M$ of $V$ as $M=\begin{bmatrix} A(M) & C(M) \end{bmatrix}$ where $A(M) \in M_{n,p-1}(F)$ and $C(M) \in F^n$. With our assumptions, we find rank $1$ matrices $M_1,\dots,M_{p-1}$ in $V$ with respective row spaces $F e_1^t,\dots,F e_{p-1}^t$. Then, $M_1,\dots,M_{p-1}$ are linearly independent and all belong to the kernel of $V \ni M \mapsto C(M)$. Using the rank theorem, one deduces that $\dim V \geq (p-1)+\dim C(V)=(p-1)+n$.
Case 2 : $V e_p \subsetneq F^n$. Multiplying $V$ on the left by a well-chosen invertible matrix, we lose no generality in assuming that $V e_p \subset F^{n-1} \times \{0\}$. In other words, every matrix $M$ of $V$ may be written as $$M=\begin{bmatrix} A(M) & C(M) \\ R(M) & 0 \end{bmatrix}$$ where $A(M)$ is an $(n-1) \times (p-1)$ matrix, $R(M)$ is a row matrix and $C(M)$ is a column matrix. Then, we note that $A(V)$ satisfies the same set of assumptions as $V$: indeed, if we take a non-zero row $L \in M_{1,p-1}(F)$, then we know that $V$ contains a rank $1$ matrix $M_1$ whose row space is spanned by $\begin{bmatrix} L & 1 \end{bmatrix}$. Obviously the last row of $M_1$ is zero whence $A(M_1)$ is non-zero and its row space is included in $F L$. One works likewise to obtain the remaining part of the condition. Thus, by induction one finds $$\dim A(V) \geq (n-1)+(p-1)-1.$$ Finally, we know that $V$ must contain a non-zero matrix $M_2$ with $A(M_2)=0$ and $C(M_2)=0$, and that it must contain a non-zero matrix $M_3$ with $A(M_3)=0$ and $R(M_3)=0$. Obviously, $M_2$ and $M_3$ are linearly independent vectors in the kernel of $V \ni M \mapsto A(M)$. Using the rank theorem, one concludes that $$\dim V \geq 2+\dim A(V) \geq 2+(n-1)+(p-1)-1=n+p-1.$$
• This is a nice proof. Although it seems to be algebraic, I think it is geometric in disguise. Perhaps someone can formulate this proof coordinate-free, and/or in terms of the the intersection of $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ with $\mathbb{P}(A) \times \mathbb{P}(B) \hookrightarrow \mathbb{P}(A \otimes B)$? – Martin Brandenburg Sep 6 '13 at 19:20 | 2019-09-17T21:37:34 | {
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https://math.stackexchange.com/questions/1343287/find-the-area-of-the-entire-region-that-lies-between-r-1-sin-theta-r-1-cos-t | # Find the area of the entire region that lies between $r=1+\sin\theta; r=1+\cos\theta$
I have to find the area of the region that lies between the curves $r=1+\sin\theta; r=1+\cos\theta$ . The answer the book gave was $\frac {3\pi}{2}-2\sqrt{2}$ . I tried generating the curve for $r=1+\cos\theta$ from $\frac {\pi}{4}$ to $\pi$, which generated the proper part of the curve, and then generating the curve for $r=1+\sin\theta$ from $\frac {-3\pi}{4}$ to $\frac {\pi}{4}$ which generated the correct part of that curve. I then attempted to put that into the formula given for area between polar curves, $A=2(\int_\alpha^\beta \frac {1}{2}f(\theta)^2 d\theta + \int_\gamma^\alpha \frac {1}{2}g(\theta)^2 d\theta)$ and used the bounds I previously used to generate my curve as the bounds on the integrals, respectively. However, I still am unable to get the answer the book gave. I have thought of trying to divide the section into two and find the area of one half and then doubling my result to find the area of the whole, but I am unsure of how to go about that. Does anyone have any suggestions? The point of intersection I found for the curves is $(1+ \frac {\sqrt{2}}{2} , \frac {\pi}{4})$.
• There are 2 distinct regions where these two curves have areas that lie between them. The range you have allowed for the curve $r=1+\cos\theta$ function only includes one of these regions. Have you tried $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ – Peter Jun 29 '15 at 15:11
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.
• Thank you for pointing that out. I did not see that on the graph. I believe that you meant $\pi \le \theta \le \frac {5\pi}{4}$ for the $r=1+\cos\ \theta$ curve. Was the formula I listed above the correct one to use for computing this particular area? A friend suggested modifications may be made but I am not sure which. – Carly Sawatzki Jun 29 '15 at 14:40
• @CarlySawatzki: Thanks for the correction. I need to run now and can't do more checking: perhaps later. – Rory Daulton Jun 29 '15 at 14:45
I found the answer by separating the figure into four symmetrical parts, two corresponding to the curve $r=1+\sin\ \theta$ and one corresponding to the curve $r=1+\cos\ \theta$. This way the intersecting areas in both quadrants one and four were accounted for. Then, I found the points at which the radius (r) of the curve would be zero, focusing only on the curve $r=1+\cos\ \theta$ as well as using the points I found for the points of intersection, which were at $(1+ \frac {1}{\sqrt{2}}, \frac{\pi}{4})$ and $(1- \frac {1}{\sqrt{2}}, \frac{5\pi}{4})$. I used the formula listed above, which was $A=2(\int_\alpha^\beta \frac {1}{2}f(\theta)^2 d\theta + \int_\gamma^\alpha \frac {1}{2}g(\theta)^2 d\theta)$. Thus, the setup for the formula is as follows for the curve $r=1+ \cos\ \theta$: $A=2(\int_\pi^\frac {5\pi}{4} \frac {1}{2}(1+ \cos\ \theta)^2 d\theta + \int_\frac {\pi}{4}^\pi \frac {1}{2}(1+\cos\ \theta)^2 d\theta)$ I then solved each integral separately and added them together, as follows: $(-\sqrt{2}+\frac{9\pi}{8}-\frac{7}{4})+(-\sqrt{2}+\frac{3\pi}{8}+\frac{7}{4})=(\frac{3\pi}{2}-2\sqrt{2})$, which was the answer given by the book. Thank you all for your help!
Note: This solution was not tested if you use the halves of the curves generated by the $r=1+\sin\ \theta$ curve. I assume it would work similarly if you solved for the values of $\theta$ that would make $\sin\ \theta=-1$ true and used this in your bound for the integral.
Why don't you simply take an integral for the upper half of the region which is bounded by $r_2=1+\cos\theta$ with $\dfrac{\pi}{4}<\theta<5\dfrac{\pi}{4}$.
Due to symmetry about $\theta = \pi/4$ line,
$$2\cdot \int_{\pi/4}^{\pi/4} \frac12 (1+\cos(\theta))^2 d \theta$$ | 2019-06-18T06:36:53 | {
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https://iloctech.com/sysh4d7z/0c7d45-vertically-opposite-angles-are-equal | In some cases angles are referred to as vertically opposite angles because the angles are opposite per other. When two lines intersect,angles opposite to each other vertically opposite angles.Vertically opposite angles haveCommon vertexAnd are opposite to each otherHere,∠1 & ∠2 have common vertex O& are opposite to each othersSo, they are vertically opposite angles.Also,Vertically opposite angles are equal∴ ; Two angles that share terminal sides, but differ in size by an integer multiple of a turn, are called coterminal angles. Vertically opposite angles (vert. Angles that have the same measure (i.e. The two angles marked in each of these diagrams are called vertically opposite angles and are equal. Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Remember, vertically opposite angles are only formed by a pair of crossing straight lines. Agam Gupta. Similarly, the angles ∠ D O A and ∠ C O B are also called as vertically opposite angles. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Popular AMA APA (6th edition) APA (7th edition) Chicago (17th edition, author-date) Harvard IEEE ISO 690 MHRA (3rd edition) MLA (8th edition) OSCOLA Turabian (9th edition) Vancouver. If one pair of adjacent angles are rotated by $180^°$, they are exactly same as the other pair of adjacent angles. Because b° is vertically opposite 40°, it must also be 40° A full circle is 360°, so that leaves 360° − 2×40° = 280° Angles a° and c° are also vertically opposite angles, so must be equal, which means they are 140° each. Thus, the vertically opposite angles are formed geometrically. Vertical angles. Call our LearnNext Expert on 1800 419 1234 (tollfree) OR submit details below for a call back Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. The vertically opposite angles $\angle BOD$ and $\angle AOC$ are equal and each angle is $35^°$. News Feed. Examples. how_to_reg Follow . It makes one pair of vertically opposite angles becomes other pair of vertically opposite angles geometrically. Angles Vertical Angles Coloring Activity Color Activities Vertical Angles Activities . Answer/Explanation. Vertically Opposite Angles. ∠ s) are the angles opposite each other when two lines intersect. all right angles are equal in measure). See 'related links' below. Now, let’s discuss about it to understand the concept of vertically opposite angles geometrically. Thus, the vertically opposite angles are formed geometrically. $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ are two straight lines, which are intersected at point $O$ and four angles are formed geometrically by their intersection. Interior angles on the same side of a transversal with two distinct parallel lines are complementary angles. vertically opposite angles are equal - Math bibliographies - in Harvard style . This is a type of proof regarding angles being equal when they are vertically opposite. 4. People. Statements a and b are as given below: a : If two lines intersect, then the vertically opposite angles are equal. According to vertical angle theorem, in a pair of intersecting lines, the vertically opposite angles are equal. \,\,\,\angle COB\,=\,145^°$. The vertically opposite angles ∠ B O D and ∠ A O C are equal and each angle is 35 °. 1 2 Theorem 6.1 : If two lines intersect each other, then the vertically opposite angles are equal. lines and angles; class-7; Share It On Facebook Twitter Email. Solution: False As vertically opposite angles are always equal but do not form a linear pair. Vertically opposite angles are always (a) supplementary (b) complementary (c) adjacent (d) equal. Theorem: All vertically opposite angles have equal measure. They are always equal. thumb_up Like (4) visibility Views (28.4K) edit Answer . In the point of view of vertical ($O$), the angles$\angle BOD$and$\angle AOC$are opposite to each other. The point where they meet is called a vertex. Vertically opposite angles are equal. question_answer Answers(3) edit Answer . Resources. Similarly, the angles$\angle DOA$and$\angle COB$are also called as vertically opposite angles. There are two pairs of vertical angles; A = C and B = D. They only connect at the very tip of the angles. If two lines intersect then the vertically opposite angles are equal. Try moving the points below. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Therefore, the vertically opposite angles are always equal geometrically. To Prove :- Vertically opposite angles are equal i.e, AOC = BOD and AOD= BOC Proof :- From (1) and (2) AOC + BOC = AOD + AOC BOC = AOD Now, To prove BOD = AOC From (3) and (4) AOD + BOD = AOD + AOC BOD = AOC Hence, Vertically Opposite angles are equal. Vertical Angles Theorem. 1 Answer +1 vote . The point of intersection is a common vertex to each angle. I mean to say that if the vertically opposite angles are equal then lines forming the vertically opposite angles are intersecting lines. In this example a° and b° are vertically opposite angles. Opposite angles are equal to one another. Notice that the 4 angles are actually two pairs of vertically opposite angles: We know that, If a ray lies on a line then the sum of the adjacent angles is equal to 180°. Vertically opposite angles are equal. Best answer. On the sample problems i don't get why you "Add" 5 on the first one then use "Subtraction" on the next one ... says when two line intersect the form four angles in between ,then each angle is equal to the an angle opposite to it and they are called vertical apposite angles . Each pair of opposite angles in the point of vertical is called the vertically opposite angles and they are equal geometrically. Proof: Click hereto get an answer to your question ️ Two lines intersecting each other, prove that vertically opposite angles are equal. The$\angle DOA$and$\angle AOC$are adjacent angles and the sum of them forms a straight angle. For example, if two lines intersect and make an angle, say X=45 °, then its opposite angle is also equal … Vertically opposite angles are so called because they are opposite each other at a vertex. We then restate what must be shown using the explicit notation.) Profile. the same magnitude) are said to be equal or congruent.An angle is defined by its measure and is not dependent upon the lengths of the sides of the angle (e.g. An interactive diagram showing the theorem "Vertically opposite angles are equal" An interactive diagram showing the theorem "Vertically opposite angles are equal" Create Class; Home. The angles opposite each other when two lines cross. ∠AOD, ∠COB and ∠AOC, ∠BOD. Vertically opposite angles form a linear pair. On module Angle types/ Vertically opposite angles. person. We sketch a labeled figure to introduce notation. Vertically opposite angles are equal, i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC.$(1). Are 2 angles of the same size, formed between opposite sides of 2 intersecting straight lines. Solution: False ⇒ ∠AOC + ∠AOD = 180° (By linear pair axiom) …. When two lines intersect, the angles across from each other are called vertically opposite angles, or just opposite angles. asked Sep 18, 2018 in Class IX Maths by priya12 ( -12,630 points) Vertically opposite angles, sometimes known as just vertical angles. Worksheet 1 "Vertical" refers to the vertex (where they cross), NOT up/down. When two straight lines are intersected at a point, four angles are formed geometrically at the point of their intersection. Classroom. (i) Similarly, the ray DO lies on the line AOB. If two lines intersect each other, then the vertically opposite angles are: a) Equal b) Unequal c) Cannot be determined d) None of the above. They are always equal. \,\,\,\angle AOC\,=\,35^°$,$(4). Recall, from earlier classes, that when two lines intersect, the vertically opposite angles are equal. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising. answered Jun 1 by Varun01 (53.5k points) selected Jun 2 by Subnam01 . About GeoGebra. Vertically opposite angles & Proof of vertically opposite angles are equal-Lines and angles. There are two pairs of opposite angles in the point of view of the vertical. In this video explains about how the vertically opposite angles are equal in Tamil. The point of intersection of them is called a vertical in this case and the four angles are measured as. Start studying VERTICALLY OPPOSITE ANGLES ARE EQUAL. \,\,\,\angle DOA\,=\,145^°$,$(3). … A pair of vertically opposite angles are always equal to each other. A mastery lesson starting with an investigation into how straight lines, about a point and vertically opposite angle facts are linked building up to the use of reasoning and algebra in questions. Because b° is vertically opposite 60°, it must also be 60° So, b = 60° A full circle is 360°, so that leaves 360° − 2×60° = 240° Angles a and c are also vertically opposite angles, so must be equal, which means they are 120° each. So, they’re called as vertically opposite angles. Change style powered by CSL. Angles This picture proof shows how vertically opposite angles of two intersecting lines are congruent. An exterior angle of a triangle is 110° and its two interior opposite angles are equal. In the diagram above, the two green angles … App Downloads. From the figure, The ray AO stands on the line CD. The sum of two vertically opposite angles is 166°. Similarly, the $\angle BOD$ and $\angle COB$ are also adjacent angles and the sum of them is also equal to $180^°$. \,\,\,\angle BOD\,=\,35^°$,$(2). When two lines intersect, the opposite (X) angles are equal. Get a free home demo of LearnNext. Thus, when two lines intersect, two pair of vertically opposite angles are formed i.e. That gives you four angles, let's call them A, B, C, D (where A is next to B and D, B is next to A and C and so on). Notice which angles are equal and how these equal angles are formed. Imagine two lines that intersect each other. When two lines intersect each other, then the opposite angles, formed due to intersection are called vertical angles or vertically opposite angles. According to English language, the noun of vertical is a vertex or zenith, and we’re not considering the meaning of vertical as perpendicular to something in this case. Learn cosine of angle difference identity, Learn constant property of a circle with examples, Concept of Set-Builder notation with examples and problems, Completing the square method with problems, Evaluate $\cos(100^\circ)\cos(40^\circ)$ $+$ $\sin(100^\circ)\sin(40^\circ)$, Evaluate $\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$ $\times$ $\begin{bmatrix} 9 & 8 & 7\\ 6 & 5 & 4\\ 3 & 2 & 1\\ \end{bmatrix}$, Evaluate ${\begin{bmatrix} -2 & 3 \\ -1 & 4 \\ \end{bmatrix}}$ $\times$ ${\begin{bmatrix} 6 & 4 \\ 3 & -1 \\ \end{bmatrix}}$, Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^3{x}}{\sin{x}-\tan{x}}}$, Solve $\sqrt{5x^2-6x+8}$ $-$ $\sqrt{5x^2-6x-7}$ $=$ $1$. Prove that if two lines are intersecting, then the vertically opposite angles are equal. These angles are also known as vertical angles or opposite angles. True. Similarly, the vertically opposite angles $\angle DOA$ and $\angle COB$ are also equal and each angle is $145^°$. See Appendix 1 for the ingredients of a proof, and keep those in mind while studying the proof given below . Show More Find each of the angles. The opposite angles which have a common vertex are called the vertically opposite angles. This picture proof helps show that even when the lines are moved within the distance between the two points on each line, the vertically opposite angles will always be equal. If two lines intersect, prove that the vertically opposite angles are equal. Answer: a = 140°, b = 40° and c = 140°. Also, a vertical angle and its adjacent angle are supplementary angles, i.e., they add up to 180 degrees. In this case, the vertex is called as a vertical. Available for CBSE, ICSE and State Board syllabus. In question 1, are there some angles you can’t work out using logic? Let us prove this result now . (To get started, we first use the definition of vertically opposite angles to make sense of the statement. State whether the statement are True or False. 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Of these diagrams are called coterminal angles DO NOT form a linear pair axiom ) … size by an multiple! Shown using the explicit notation. vertical angles Activities point of intersection of them is called as vertically angles...
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http://mathhelpforum.com/math-puzzles/228920-math-puzzles-questions.html | 1. ## Math puzzles questions?
1. What number should replace the question mark?
2.What number should replace the question mark?
3.What number should replace the question mark?
4.What number should replace the question mark?
5.What number should replace the question mark?
Regards,
2. ## Re: Math puzzles questions?
Hello, amrithaa!
2. What number should replace the question mark?
. . $\boxed{\begin{array}{ccccc} .&.&.&14&. \\ .&22&.&.&. \\ .&.&.&34&. \\ 41&.&.&.&. \\ .&.&53&.&? \end{array}}$
Spoiler:
Each number indicates its row and column.
The "?" is in the 5th row, 5th column.
Its number is 55.
3. What number should replace the question mark?
. . $\boxed{\begin{array}{cc}5 \\ 8\quad 5 \\ 1\quad 7 \end{array}} \qquad \boxed{\begin{array}{c} 4 \\ 7\quad 6 \\ 1\quad 9 \end{array}} \qquad \boxed{\begin{array}{c} ? \\ 9\quad 1 \\ 1\quad 3 \end{array}}$
Spoiler:
The second row divided by the third row equals the top row.
$85 \div 17 \:=\:5,\quad 76 \div 19 \:=\:4,\quad 91\div 13 \:=\:{\color{red}7}$
4. What number should replace the question mark?
$\boxed{\begin{array}{c|c|c} 47&55&63 \end{array}} \quad \boxed{\begin{array}{c|c|c}85&92&99 \end{array}} \qquad \boxed{\begin{array}{c|c|c} 73&\;?\;&25 \end{array}}$
Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13 | 2016-08-28T02:58:04 | {
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http://math.stackexchange.com/questions/339988/how-do-i-find-tangent-on-the-unit-circle | # How do I find tangent on the unit circle?
I know that $\sin=y$ and that $\cos=x$ so how do I find tangent? I've seen some diagrams where the tangent line touches the unit circle at $(1,0)$, I'm wondering how this is derived?
EDIT: Explanations using similar triangles are most intuitive to me, if things could be explained in those terms I'd be grateful!
Thank you!
-
The tangent is orthogonal to the line which goes from the center to the point on which you want the tangent. As this point is $$(\cos(x),\sin(x))$$ the tangent. will be $$(-\sin(x),\cos(x))\cdot t+(\cos(x),\sin(x))$$
You can easily visualize this using an CAS like Mathematica, taking as input
Manipulate[
ParametricPlot[
{
{Cos[Pi*t],Sin[Pi*t]},
{Cos[x], Sin[x]} + {-Sin[x], Cos[x]}*t
},
{t, -1, 1}],
{x, -Pi, Pi}]
gives you a slider where you can go through every angle. If you like I can make some pictures.
A picture
What i am essentially doing is drawing at first the circle, than i take a point on the circle which is $(\cos(x),\sin(x))$. From here I make a line in the direction of $(-\sin(x),\cos(x))$. This essentialy is a line of the form $a t + b$ but here $a$ and $b$ are vectors.
-
Congrats for making sense of the question, +1. – 1015 Mar 24 '13 at 20:40
@DominicMichaelis I don't understand your answer, the tangent should have the form $y=at+b$ where $a$ and $b$ are real! – Sami Ben Romdhane Mar 24 '13 at 20:48
that doesn't in every case (for example try to give the tangent of the point x=1 in that form. @SamiBenRomdhane – Dominic Michaelis Mar 24 '13 at 20:52
Why is it that the tangent should be orthogonal? – Assad Mar 24 '13 at 20:56
Also furthermore how did you derive (−sin(x),cos(x))⋅t+(cos(x),sin(x))? I'm sorry this is just a little confusing to me, a bit more explanation would help, thank you. – Assad Mar 24 '13 at 20:57
Take a point $\,(\cos x_0\,,\,\sin x_0)\,$ on the unit circle, and assume $\,\cos x_0\cdot\sin x_0\neq 0\,$ (otherwise the question is almost trivial) , so the slope of the radius in the circle to this point is
$$\frac{\sin x_0}{\cos x_0}$$
Since the tangent line to the circle in the above point is perpendicular to the radius at that point, the tangent line's slope must have slope
$$-\frac{\cos x_0}{\sin x_0}$$
So now you have a point on the tangent line and its slope: calculate its formula.
-
When $\cos(x_0) \cdot \sin(x_0)=0$ this is not trivial take $\cos(x_0)=1$ and try to give the tangent with your method – Dominic Michaelis Mar 24 '13 at 20:53
No, when $\,\sin x_0=0\,$ then we're either on $\,(1,0)\,$ or on $\,(-1,0)\,$ and it's almost trivial the tangent lines there are the vertical lines $\,x=1\,$(resp., $\,x=-1\,$ ) . My answer does not cover these cases, that's why I assumed what I did. – DonAntonio Mar 24 '13 at 20:56
Edit: Personally I prefer all the other answers, here's another way to think about it if you want more ideas
If $x^2 + y^2 = 1$, then $2x + 2y \frac{dy}{dx} = 0$
Re-arranging, $\frac{dy}{dx} = \frac{-x}{y}$. As you say, $\cos \theta = x$ and $\sin \theta = y$, giving you the slope of a tangent line as $- \cot \theta$. You can then use the equation: $$y - \sin \theta = - \cot \theta (x - \cos \theta)$$
-
From the article $148,150$ of The elements of coordinate geometry, by Loney, the equation of the tangent of the circle $x^2+y^2=a^2$ at $(x_1,y_1)$ is
$$xx_1+yy_1=a^2$$
As you have already identified, any point on the circle can be $(a\cos\theta,a\sin\theta)$ where $0\le \theta<2\pi$
So, the equation of the tangent becomes $$xa\cos\theta+ya\sin\theta=a^2\implies x\cos\theta+y\sin\theta=a\text{ as }a\ne0$$
For the Unit Circle $a=1,$ So, the equation of the tangent becomes $x\cos\theta+y\sin\theta=1$ | 2014-04-20T08:50:00 | {
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https://www.emathhelp.net/notes/calculus-1/studying-changes-of-function/condition-of-constancy-of-the-function/ | # Condition of Constancy of the Function
Fact. Suppose function ${y}={f{{\left({x}\right)}}}$ is defined and continuous on interval ${X}$ and has finite derivative ${f{'}}{\left({x}\right)}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ for all ${x}$ in ${X}$.
This fact means that if on some interval derivative of function equals 0 then function is constant their, its graph is just horizontal line.
Corollary. Suppose two function ${y}={f{{\left({x}\right)}}}$ and ${y}={g{{\left({x}\right)}}}$ are defined and continuous on interval ${X}$, and have finite derivatives ${f{'}}{\left({x}\right)}$ and ${g{{\left({x}\right)}}}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ for all ${x}$ in ${X}$. If ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$ on interval ${X}$ then ${f{{\left({x}\right)}}}={g{{\left({x}\right)}}}+{C}$, where ${C}$ is a constant for all ${x}$ in ${X}$.
This corollary means that if functions have same derivatives on interval ${X}$ then their difference is constant.
Example 1. Consider functions ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}$.
They are defined and continuous on interval ${\left(-\infty,\infty\right)}$.
Since ${f{'}}{\left({x}\right)}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ and ${g{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}^{{2}}}}\cdot\frac{{\sqrt{{{1}+{{x}}^{{2}}}}-\frac{{{{x}}^{{2}}}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}}}{{{1}+{{x}}^{{2}}}}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ then according to corollary ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}+{C}$ on ${\left(-\infty,\infty\right)}$.
To find constant plug any value of ${x}$, for example, ${x}={0}$: ${\operatorname{arctan}{{\left({0}\right)}}}={\operatorname{arcsin}{{\left(\frac{{0}}{\sqrt{{{1}+{{0}}^{{2}}}}}\right)}}}+{C}$ or ${C}={0}$.
So, we have the following fact: ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}}$ for all ${x}$.
Example 2. Consider functions ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}=\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}$.
It can be easily proven that ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$. However, function ${g{{\left({x}\right)}}}$ is not defined when ${x}=\pm{1}.$ So, $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+{C}$ on ${\left(-\infty,-{1}\right)},{\left(-{1},{1}\right)},{\left({1},\infty\right)}$.
It is interesting that constant will be different for different intervals.
For interval ${\left(-{1},{1}\right)}$ we plug ${x}={0}$: $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}\cdot{0}}}{{{1}-{{0}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({0}\right)}}}+{C}$ or ${C}={0}$.
For interval ${\left(-\infty,-{1}\right)}$ we let ${x}\to-\infty$: ${0}=-\frac{\pi}{{2}}+{C}$ or ${C}=\frac{\pi}{{2}}$.
For interval ${\left({1},\infty\right)}$ we let ${x}\to\infty$: ${0}=\frac{\pi}{{2}}+{C}$ or ${C}=-\frac{\pi}{{2}}$.
So,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+\frac{\pi}{{2}}$ on ${\left(-\infty,-{1}\right)}$,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ on ${\left(-{1},{1}\right)}$,
$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}$ on ${\left({1},\infty\right)}$.
Graph confirms these facts. | 2022-10-05T02:20:21 | {
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http://math.stackexchange.com/questions/133372/show-that-2n-on | # Show that $2^n=O(n!)$
Show that $2^n=O(n!)$
Proof:
By definition of Big-O, $\exists$ constants $c$ and $n_0$ such that $2^n \le cn!$ $\forall$ $n \ge n_0$. For a large $n$, since $$2^n = \underbrace{2 \cdot 2 \cdot 2\cdot\cdot\cdot2}_{n\text{ times}}$$ and $$n! = 1 \cdot 2 \cdot 3 \cdot\cdot\cdot n$$ Clearly, as $n \rightarrow +\infty$ , $2^n \le cn!$.
Alternatively, WLOG, we use induction and show that $2^{n+1} \le c(n+1)!$.
$$2^n \le cn!$$ $$2^{n+1} \le 2(cn!) \le (cn+1)! \le c(n+1)!$$ since $cn+1 \ge 2.$
Question: Is my proof correct? and is there a better way of proving such. thanks
-
Take a look at the expansion $2 n!$ and compare it factor by factor to $2^n$. – Fabian Apr 18 '12 at 8:05
Your proof is correct. Actually, it could be shown in the same way that $k^n = O(n!)$ for every $k$, and thus $k^n = o(N!)$ (because e.g. $k^n = o((2k)^n)$).
The easier proof is to show that, for $n > k$, $k^n < {n!}\frac{k!}{k^k}$: $k^n = k^k \times k^{n-k} < k^k \times ((k+1)(k+2) \ldots n) = k^k \times \frac{n!}{k!}$.
From the other hand, $n! = O(n^n)$, if you're interested.
-
You never state what $c$ is, and that is a weakness. Your first proof is too short, essentially stating the fact to be proven. But it can be fixed by multiplying together the $n-1$ inequalities $2\le k$ for $k=2,\ldots,n$.
Your second proof misses the start of the induction, and uses some rather mysterious looking inequalities.
Another way to proceed is to focus on $a_n=2^n/n!$ and estimating $a_{n+1}/a_n$. There are many ways to acieve what you need, but this one has the advantage of being applicable in many similar (and much less obvious) situations.
-
I think the constant $c$ for which it is large. With regards to $a_n$. Should I show that limit of $a_n$ approach to 0 as n approach positive infinity? – Keneth Adrian Apr 18 '12 at 8:17
That is the general idea, yes. – Harald Hanche-Olsen Apr 18 '12 at 12:47
The idea behind both proofs is fine but the actual writing seems shaky, you need to provide actual values for c and $n_0$. For instance, in the first method you can use $n_0$ = 1 and c = 2, since 2(n!) = 2.2.3.4....n and here you can use a term-by-term comparison for all n$\ge$ 1.
For the induction proof, you could eg. take c = 1 and $n_0$ = 4. The induction step remains unchanged but you also need to show the base case that $2^4 \le 4!$.
- | 2015-10-09T14:04:19 | {
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https://math.stackexchange.com/questions/4183730/question-on-varepsilon-delta-proof-for-lim-x-to-0-x2-sin-left-frac | # Question on $\varepsilon$-$\delta$ proof for $\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)=0$
I'm trying to understand the $$\varepsilon$$-$$\delta$$ proof for $$\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)=0$$
We have to find $$\varepsilon>0$$ and $$\delta>0$$ such that $$\left|x^2\sin\left(\frac{1}{x}\right)\right| < \varepsilon \tag{1}\label{eq1}$$ and $$|x|<\delta \tag{2}\label{eq2}$$
My work:
We know \begin{aligned} \left|\sin\left(\frac{1}{x}\right)\right| &\leq 1 \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| &\leq |x^2| \\ \end{aligned} $$\therefore \text{ } \left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta^2 \tag{3}\label{eq3}$$
For $$0<\delta<1$$, we have \begin{aligned} \delta^2 &< \delta \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| &< \delta \end{aligned} and hence we can choose any $$\delta$$, such that $$0 < \delta \leq \varepsilon$$. So far so good.
But, for $$\delta \geq 1$$, we have $$\delta^2 \geq \delta$$ The range of value for $$\delta$$ in this case that I can think of such that $$\eqref{eq1}$$ and $$\eqref{eq3}$$ could be reconciled is $$\left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta \leq \varepsilon$$
But when I check on a graphing tool, regardless of whether $$\delta$$ is $$<1$$ or $$\geq 1$$, $$0<\delta\leq\varepsilon$$ holds.
What am I doing wrong?
EDIT:
Since $$0<\delta\leq\varepsilon$$ and from $$\eqref{eq2}$$, we get $$|x|<\delta \leq\varepsilon \implies |x|^2 < \delta^2 \leq \varepsilon^2$$
So, since $$\left|x^2\sin\left(\frac{1}{x}\right)\right| \leq |x^2| \\ \implies \left|x^2\sin\left(\frac{1}{x}\right)\right| < \delta ^2 \leq \varepsilon^2$$
So, can we now say $$\delta \leq \varepsilon$$, regardless of values that either take?
• Recall that $\delta$ MAY depend on $\varepsilon$. So GIVEN $\varepsilon>0$, just take $\delta=\sqrt{\varepsilon}$. Jun 26, 2021 at 16:41
You're almost there, and there are several ways to finish up. One thing you could do is what @Tito Eliatron suggests in his comment (this might be the most elegant way):
Let $$\epsilon > 0$$, set $$\delta = \sqrt{\epsilon}$$. Then, for $$|x|<\delta$$ we have
$$$$\left|x^2 \sin{\left(\frac{1}{x}\right)}\right| \leq |x|^2 < \delta^2 = \sqrt{\epsilon}^2 = \epsilon.$$$$
But there is another way to "fix" your argument. As you have noted, if $$\epsilon \leq 1$$, you can choose $$\delta = \epsilon$$ and use the fact that in this case we have $$\delta^2 \leq \delta$$, and do the same thing as above. Then we deal with the case $$\epsilon > 1$$ separately. In this case, we can choose $$\delta = 1$$, and we have for $$|x| < \delta:$$
$$$$\left|x^2 \sin{\left(\frac{1}{x}\right)}\right| \leq |x|^2 < \delta^2 = 1^2 = 1 < \epsilon.$$$$
Thus, we can set $$\delta = \min{(\epsilon, 1)}$$ and obtain the same result. I think Tito's suggestion is more beautiful since it avoids different cases, but both methods are valid.
In $$\DeclareMathOperator{\epsilon}{\varepsilon}\epsilon$$-$$\delta$$ proofs, $$\delta$$ is allowed to depend on $$\epsilon$$, but not vice versa. In your example, if $$\epsilon\le1$$, then we can set $$\delta=\epsilon$$, and things work out nicely. Often, weird things can happen for large values of $$\epsilon$$, and so as a kind of 'safety-net' we require that $$\delta$$ be at most a specific number. Here, we can set $$\delta=\min\left(\epsilon,1\right)$$ and things work out nicely. It also helps to work backwards. We need $$\left|x^2\sin\left(\frac{1}{x}\right)\right|<\epsilon$$ or $$\left|x^2\right|\left|\sin\left(\frac{1}{x}\right)\right|<\epsilon \, .$$ Now, if we can make $$|x^2|$$ be smaller than $$\epsilon$$, then it will be certainly be the case that $$|x^2|\left|\sin(1/x)\right|$$ will be smaller than $$\epsilon$$, as $$|\sin(1/x)|\le1$$. Moreover, if $$|x|<1$$, then $$|x^2|<|x|$$. So if $$|x|<\epsilon$$ and $$|x|<1$$, then $$|x^2|<|x|<\epsilon$$, which is what we need. So let $$\delta=\min(\epsilon,1)$$, which guarantees that $$|x|<1$$ and $$|x|<\epsilon$$. | 2022-08-15T01:10:35 | {
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https://math.stackexchange.com/questions/635514/eating-a-cake-from-the-inside | # Eating a cake from the inside.
Imagine that you are in the centre of a cube of cake with a known size. In order to move you must eat the surrounding cake but you can only move within the restraints of the six obvious directions $(x+1,x-1,y+1,y-1,z+1,z-1)$.
The puzzle is- can you eat the entire cake in such a fashion without overlapping parts of the cake which you have already eaten.
I would actually like to answer this question for a practical purpose and I have put much thought into it. From what I understand it may only be possible by moving diagonally (applying $2$ or more movements of different axis at the same time-such as $x-1$ and $y-1$).
I have asked my maths teacher and some of my friends but I am sure there must be a solution.
• I must add that I do not have the mathematical experience to use the correct terminology- being only 14.
– will
Jan 12, 2014 at 9:05
• "I would actually like to answer this question for a practical purpose" User2592835 is trapped inside a cake! Send help.
– user856
Jan 13, 2014 at 0:02
• That cake tricked me.
– will
Jan 13, 2014 at 15:44
This is a supplement to others answer. In particular, MathBob's suggestion that it is possible to completely cover a cube of side $5+4k$ start from the center.
Following is an example using the $5\times 5 \times 5$ cube. We are showing the 5 layers of the cube from top to bottom. The number in each slot is the order one visit/eat the corresponding piece of cake.
$$\left[\begin{array}{rrrrr}23&24&25&26&\color{#bbbb00}{27}\\22&9&10&11&12\\21&8&\color{orange}{3}&4&13\\20&7&6&5&14\\19&18&17&16&15\end{array}\right] \left[\begin{array}{rrrrr}32&31&30&29&\color{#00bb00}{28}\\33&46&45&44&43\\34&47&2&\color{blue}{51}&42\\35&48&49&50&41\\36&37&38&39&40\end{array}\right] \left[\begin{array}{rrrrr}71&72&73&74&75\\70&57&58&59&60\\69&56&\color{red}{1}&\color{magenta}{52}&61\\68&55&54&53&62\\67&66&65&64&63\end{array}\right]\\ \left[\begin{array}{rrrrr}80&79&78&77&\color{purple}{76}\\81&94&93&92&91\\82&95&100&99&90\\83&96&97&98&89\\84&85&86&87&88\end{array}\right] \left[\begin{array}{rrrrr}121&122&123&124&125\\120&107&108&109&110\\119&106&101&102&111\\118&105&104&103&112\\117&116&115&114&113\end{array}\right]$$
1. Start from the center of cake (the red $\color{red}{1}$ in the $3^{rd}$ layer), one eat/move upward until one reaches the center of the top layer (the orange $\color{orange}{3}$ in $1^{st}$ layer).
2. Spiral outwards until one exhaust all slots in $1^{st}$ layer and end at a corner (the yellow corner $\color{#bbbb00}{27}$).
3. Move downwards to a corner (the green corner $\color{#00bb00}{28}$ ) in $2^{nd}$ layer.
4. Spiral inwards until one exhaust all slots in $2^{nd}$ layer and end at a slot (the blue $\color{blue}{51}$ ) next to the center in $2^{nd}$ layer.
5. Move downwards to a slot (the magenta $\color{magenta}{52}$) in the $3^{rd}$ layer.
We then repeat essentially the same steps $2-5$ until we exhaust all the layers. When we reach a layer below the center (i.e the $4^{th}$ and $5^{th}$ layer in this case ), step 4 need a slight modification. Instead of ending at a slot next to the center, one should end at the center for that particular layer.
When the side of the cube is of the form $5+4k$, there are even number of layers below the center and we enter the first layer below the center at a corner (in our example, the $\color{purple}{76}$ in $4^{th}$ layer), we won't have any problem to complete fill the layers below the center.
In contrast, for cube whose side is of the form $3+4k$, we will enter the first layer below center of cube near the center of layer. We will get into problem how to cover the center of that layer.
I hope this example is clear enough how to cover cube of side $5 + 4k$.
Update
Following is a little animation showing how the $5 \times 5 \times 5$ cube can be visited. For the purpose of visualization, I have flipped the cube vertically. In the animation, the $1^{st}$ layer we build is the bottom layer instead of the top layer discussed above.
$\hspace1.2in$
• Second to last paragraph, don't you mean form 3 + 4k ? Consider when k = 0. I liked the pattern you came up with. (I had a different pattern that also worked where I eliminated the 3 x 3 centers and left the outside edge.) Jan 12, 2014 at 13:32
• @MathBob Yes, it should be $3+4k$. What a dumb mistake. Jan 12, 2014 at 13:34
If you are exactly in the middle and can only go in the 6 directions, it is like being in a cube with an odd number of sides (in order to have an exact middle cube). If you colored the cubes alternately black and white with the middle one white, then you are in the white cube and there are one more black cubes than there are white cubes (counting the middle cube.) (This would be for a 3 x 3 x 3 or a 7 x 7 x 7 cube.) This is because an odd sided cube has n x n x n cubes which is an odd number of cubes, so there has to be more blacks than whites by 1 (or visa-versa in 5 x 5 x 5 or 9 x 9 x 9 cubes etc.) So, if you eat the white one first (the one you are in) then eat alternately black and then white cubes, you end up having to eat a white cube then there are two black cubes left and you eat one of them and then you cannot eat a white cube!
So, it is impossible as far as I can see. I need to edit my answer. If the number of units on each side is 5+4k (where k=0,1,2,...) then I think it can be done because the greater number of white squares will end up with a white one in the middle. But I think 3+4k units on each side cannot be done because of the situation I was describing. On 3+4k sided cubes the middle square is a different color than the corner squares. This one cannot be done. (If you are in the middle square and it is white and you eat it, there will be two more black squares than white.) Whereas on 5+4k sided cubes, the middle square is the same color as the corner squares and after you eat the middle square there will be the same number of squares of each color.
So, it can always be done if you break the cube into the proper number of odd units on each side. I think this is the correct answer, but there may be more complications than what I have been discussing. The 3+4k sided cubes definitely cannot be done.
• This argument does not use the fact that one starts in the center of the cube, but only that one must alternately eat white and black. But any cube can be eaten properly if one starts in a corner of the cube. Each layer may be eaten by a process of going across a row, going down to the end of the next row then back to its start, etc, winding up at the diametrically opposite corner of that layer. Then one can drop down to the next layer and do the same thing in reverse, and so on. The argument must use something re. starting at central cube, not just black/white cubes. Jan 12, 2014 at 16:23
• I like the concept of black and white cubes.So essentially we are talking about a Batenburg?
– will
Jan 12, 2014 at 18:39
Preamble: comparison with MathBob's approach. I realized after entering this that the sets $E,O$ used below can also be described as the set of triples $(x,y,z)$ for which $x+y+z$ is even or odd, respectively. So if the middle cube is white, this is the same coloring used by MathBob, where the count of whites is the count of $E$, the black count of $O$, What I found hard to see was how the value of $n$ mod $4$ entered into why there are one more of which color. So what I did was to actually count these numbers, and it came out that in the $n=4k+3$ case there is one more black (odd) than white. I don't know if there is a simple way to see that short of actually counting the two numbers as I did below. I left the rest as it was, using taxicab distance language.
This answer is to show the impossibility in case $n=4k+3,\ k\ge 0$ [The procedure for eating the cake if $n=4k+1$ has already been shown by achille hui and mentioned by MathBob.]
The idea is to partition the cubes into two sets $E,O$ of those points whose taxicab distance from the origin is even or odd respectively. It can be shown (see below) that for $n=4k+3$ we have $$|E|=32k^3+72k^2+54k+13,\\ |O|=32k^3+72k^2+54k+14.$$ Every legal move is from a point in $E$ to one in $O$, or vice-versa. The center of the cube is in $E$, and since we are to start there, the tour cannot be made since there is one more element in $O$ than there is in $E$.
proof of counts for $|E|,|O|.$ In the set $\{-2k-1,\cdots,2k+1\}$ of values available for a coordinate of a point $(x,y,z)$ in the $n$-cube (with $n=4k+3$) there are $2k+1$ even integers and $2k+2$ odd integers. If $(x,y,z)$ is at even taxicab distance from $(0,0,0)$ then either all three coordinates are even or else two are odd and the third even. Since there are three ways the odd,odd,even case can occur, the count for $|E|$ is $$|E|=(2k+1)^3+3(2k+1)(2k+2)^2,$$ which when expanded is our claimed count for $|E|.$ On the other hand if $(x,y,z)$ is at odd taxicab distance from the origin, then either all three coordinates are odd or else (three ways for this) one is odd and two are even. This gives the $|O|$ count as $$|O|=(2k+2)^3+3(2k+2)(2k+1)^2,$$ which expands to our claimed count for $|O|$. As a check one finds that $|E|+|O|=(4n+3)^3.$
Final note: If one does the above calculation for the (shown to be possible) case when $n=4k+1,\ k \ge 1$, it turns out that $|E|=|O|+1$, which means starting inside $E$ (center of cube) this parity argument does not rule out a path. (That's a good thing, given the constructions of actual solutions in the $4k+1$ case.)
• It will be easier to count the difference $O-E$. For odd $n = 2\ell+1$, the cube can be broken down into $n$ layers. Each layer can be broken down into a center piece plus $\ell$ rings. Since the lengths of the rings are all multiples of $8$, they contribute nothing to the difference $O-E$. In the end, one only need to count the center piece from each layer and it is sort of obvious $O-E = (-1)^{\ell-1}$. Jan 13, 2014 at 3:18 | 2022-08-11T18:04:56 | {
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https://gateoverflow.in/179291/tifr2018-b-7 | 765 views
Consider the recursive quicksort algorithm with "random pivoting". That is, in each recursive call, a pivot is chosen uniformly at random from the sub-array being sorted.When this randomized algorithm is applied to an array of size $n$ all whose elements are distinct, what is the probability that the smallest and the largest elements in the array are compared during a run of the algorithm ?
1. $\left(\dfrac{1}{n}\right)$
2. $\left(\dfrac{2}{n}\right)$
3. $\Theta \left(\dfrac{1}{n\log n}\right)$
4. ${O} \left(\dfrac{1}{n^{2}}\right)$
5. $\Theta\left(\dfrac{1}{n \log^{2} n}\right)$
edited | 765 views
+8
The largest and the smallest numbers will only be compared if either of them are chosen as the pivot in the very first split. In the subsequent splits they will never get compared since the first pivot divides the list into two and the largest and smallest number will be put in two different halves. The probability would thus be the probability to choose the largest or the smallest as the pivot in the very first step. Thus the required probability would be ${1 \over n}+ {1 \over n} = {2 \over n}$.
+2
$\frac{2}{n}$ since all elements are sorted only two cases i.e. when smallest element chooses as pivot or largest element choosen as pivot the are compared only once , i.e. out of n elements only two number gives required one.
Probability that the smallest and the largest elements in the array are compared during a run of the randomized quicksort algorithm.
After the first split, first subpart of array will contain the minimum element and second subpart contains the maximum element. (Why?)
Choose any element say $x$, greater than or equal to smallest element. After partition, smallest element will be present in the left of it. The largest element will be present on the right sub array.
After the first split, there will be no more comparison between the largest and smallest element.
To have comparison between largest and smallest element we must choose anyone of the as the pivot element in the first split itself.
Among the $N$ elements we have to choose either the smallest element or the largest element. Selection will be uniform.
Probability $= 2 / n.$
selected by
0
E.g, 1,9,2,3,8,4,5, 6. if series of pivots selected like 6, 4, 3, or 6, 4? Now consider the case where the list is very large.
+2
If you select 6 as the pivot element. Then array split for your example 1, 9, 2, 3, 8 , 4, 5, 6 will be
Left subpart: 1 2 3 4 5 6
Right subpart: 8 9
Now solve these two subparts recursively.
Minimum and maximum element is 1 and 9. There will be no comparison between them.
0
Right. thanks.
0
I am confused why is it not $2 / nlogn$ Since total number of comparisons are $nlogn$ during the run of whole algorithm.
0
@mkagenius your choice of sample space does not correspond to your selection of favorable cases.
The sample space should be the whole set of elements. It is not the number of comparisons. The question is essentially asking - which elements can be selected as pivot so that a min-max comparison is ensured during a run of the partition procedure. There are only 2 - the min and the max of the array.
0
Hmm, it never mentioned that it is asking about partition procedure, it started with "quicksort algorithm" and ended with "run of the algorithm". How does one conclude it is only referring to partition procedure?
Moreover, I am surprised that people are not confused at all, atleast be a little bit confused, come on. If I were the setter of the problem, and I would have marked $2/nlogn$ as the answer even then people have had explanation for that as well. Its kind of confirmation bias going on with people.
0
>what is the probability that the smallest and the largest elements in the array are compared during a run of the algorithm?
When are elements compared? When a pivot is chosen.
When is a pivot chosen? During a run of the partition procedure.
What are we looking for? A pivot which would force the comparison of min and max elements.
What sort of pivot would do that? Only that pivot which is the min or the max of the array itself.
How many favorable cases? Two - when the pivot is the min, and when the pivot is the max.
This should clear it up. There is no ambiguity and definitely no confirmation bias.
0
Now suppose the question was: What is the probability of $any$ two numbers being compared? Since there are total $n log n$ comparisons, $numerator$ will become $n log n$ and the $denominator$ according to the accepted answer will be $n$, so the probaility will be $log n$ which will be more than $1$. | 2020-07-03T17:45:09 | {
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https://mathhelpboards.com/threads/converse-of-the-ratio-test.295/ | # Converse of the ratio test
#### Alexmahone
##### Active member
Is the converse of the ratio test true?
#### Krizalid
##### Active member
I don't think so. I think you can construct an easy counterexample. Care to imagine one?
#### Alexmahone
##### Active member
I don't think so. I think you can construct an easy counterexample. Care to imagine one?
0+0+0+... converges but the ratio is not defined.
I wonder if there are any non-trivial counterexamples.
#### HallsofIvy
##### Well-known member
MHB Math Helper
0+0+0+... converges but the ratio is not defined.
I wonder if there are any non-trivial counterexamples.
The "ratio test" says that if $lim \frac{a_{n+1}}{a_n}< 1$ then $\sum a_n$ converges.
The converse is "if $\sum a_n$ converged then $lim \frac{a_{n+1}}{a_n}< 1$".
Find a convergent series such that that limit is 1.
#### Also sprach Zarathustra
##### Member
0+0+0+... converges but the ratio is not defined.
I wonder if there are any non-trivial counterexamples.
Maybe...
$$a_n=\frac{1}{n(n-1)}$$
#### Alexmahone
##### Active member
Find a convergent series such that that limit is 1.
$\displaystyle\sum\frac{1}{n^2}$
So, is it safe to say that if a series converges, then $\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|\le 1$? | 2021-08-04T18:03:56 | {
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https://math.stackexchange.com/questions/2191043/help-to-solve-int-1-x-sqrt25-x2-dx | # Help to solve: $\int 1/(x\sqrt{25-x^2})\ dx$
I'm a brand new student. Need some help to integrate this.
Perform this integration: $$\int \frac{1}{x\sqrt{25-x^2}}\ dx$$
I'm able to obtain in theta terms like this:
$$\frac 15 \ln|\cscθ-\cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5\sin{\theta}$ and $dx=5\cos{\theta}$, we obtain: $$\int \frac{1}{5\sin{\theta}(5\cos{\theta})}\cdot 5\cos{\theta}~d\theta$$ Simplifying, we obtain: $$\frac{1}{5}\int \frac{d\theta}{\sin{\theta}}$$ Using trigonometric identities: $$\frac{1}{5}\int \csc{\theta}~d\theta$$ Integrating: $$\frac{1}{5}\ln|\csc{\theta}-\cot{\theta}|+C$$
• Please use: math.meta.stackexchange.com/questions/5020/… – Arnaldo Mar 17 '17 at 15:25
• It helps if you explain the steps you took to get at the solution in $\theta$, e.g. what was the substitution you used? – StackTD Mar 17 '17 at 15:27
• Yes Exactly. See the problem image..I believe I have the half of the solution....thanks.. – El_Master Mar 17 '17 at 15:27
• @El_Master Did you use the substitution $x=5\sin(\theta)$ and $dx=5\cos(\theta)~d\theta$ ? – projectilemotion Mar 17 '17 at 15:33
• you can use relationships between trigonometric functions to change it back, so if $x = sen \theta$ then $cos \theta = \sqrt{1 - sen^2 \theta} = \sqrt{1 - x^2}$ and other identities that you can use – Cato Mar 17 '17 at 15:33
HINT:
If trigonometric substitution is not mandatory, set $\sqrt{a^2-x^2}=y$
$$\int\dfrac{2x\ dx}{2x^2\sqrt{a^2-x^2}}=\int\dfrac{dy}{y^2-a^2}=?$$
Else setting $x=5\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$\cos y=\sqrt{1-\left(\dfrac x5\right)^2}=\dfrac{\sqrt{25-x^2}}5$
$$\implies\int\dfrac{dx}{x\sqrt{25-x^2}}=\int\dfrac{5\cos y\ dy}{5\sin y(5\cos y)}=\dfrac{\ln|\csc y-\cot y|}5+K$$
• I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine). – El_Master Mar 17 '17 at 15:59
$$x=5\sin\theta$$ $$\frac x5=\sin\theta$$ Therefore $$\frac5x=\csc\theta$$ And since $$\sin\theta=\frac x5$$, we know that $$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{x^2}{25}}$$ And since $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$, $$\cot\theta=\frac{5\sqrt{1-\frac{x^2}{25}}}{x}=\frac{\sqrt{25-x^2}}x$$ Which gives your result as $$\frac15\ln\bigg|\frac{5-\sqrt{25-x^2}}x\bigg|+C$$ | 2021-07-25T09:44:05 | {
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https://math.stackexchange.com/questions/4180314/number-of-integer-solutions-of-a2b2-10c2/4180345#4180345 | # Number of integer solutions of $a^2+b^2=10c^2$
Find the number of integer solutions of the equation $$a^2+b^2=10c^2$$.
I can only get by inspection that $$a=3m, b=m,c=m$$ satisfies for any $$m \in Z$$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $$a=p^2-q^2$$, $$b=2pq$$ and $$10c^2=(p^2+q^2)^2$$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.
• The Titel of your post is inaproppiate. The number o Solution is infinite as you already mention in your Post. You Wanze to find all solutions Jun 23, 2021 at 14:27
Let us consider the circle $$C:x^2+y^2=10$$. The question you asked is equivalent to finding all the rational points on this circle. Clearly, $$(1,3)$$ is one such point.
We will project from the point $$(1,3)$$ to the $$Y$$-axis. Let $$(0,t)$$ be the point of intersection of the line $$L$$ through $$(1,3)$$ and a point $$(x,y)$$ on the circle. Since $$L$$ passes through $$(0,t)$$ and $$(1,3)$$, its equation is \begin{align*} &x=\frac{y-t}{3-t}\\ \end{align*}
Now, the point $$(x,y)$$ is on the line $$L$$ as well as the circle $$C$$. So, $$\begin{equation*} \left (\frac{y-t}{3-t} \right )^2+y^2=10 \end{equation*}$$ Solving this for $$y$$ and using $$x=\frac{y-t}{3-t}$$, we can get expressions for $$x$$ and $$y$$ in terms of $$t$$ only.
Now, since the $$Y$$-axis is a rational line, the rational points on circle must be mapped to rational points on the line. Also, from the geometry of this approach, it is clear that this gives us all the rational points (which means we get all the solutions of the equation you asked for).
Actually, for any given conic, if we know at least one rational point, we can get all others by this projection method. So, this is a very general approach. As you can see, in this case, we will get a very ugly expression for $$t$$- maybe a better choice of the initial rational point could have given better results. For a more detailed study, refer to chapter 1 section 1 of Rational Points on Elliptic Curves. They deal with the same problem, only with $$x^2+y^2=1$$ which gives a simpler expression of $$t$$.
We first transform the given equation $$a^2+b^2=10c^2$$ as follows:
Consider $$x=\frac{a}{c}, y=\frac{b}{c}$$ (assuming we are interested in $$c \neq 0$$ case). Then we have to find rational points on the circle $$x^2+y^2=10.$$ Now $$(x,y)=(3,1)$$ is an obvious solution. To find other rational solutions, consider the line that passes through $$(3,1)$$ and has slope $$m$$. It is given by $$y=mx+(1-3m).$$ Consider the intersection of this line with the circle $$x^2+y^2=10$$. We can find the intersection from $$x^2+(mx+(1-3m))^2=10 \implies (1+m^2)x^2+2m(1-3m)x+(1-3m)^2-10=0.$$ But instead of solving the quadratic, we can argue that if two roots are $$x_1$$ and $$x_2$$, then we have $$x_1=3$$, so by means of Viete formula etc. we have $$x_1+x_2=-\frac{2m(1-3m)}{1+m^2} \implies x_2=\frac{3m^2-2m-3}{1+m^2}.$$ Now if $$m$$ is rational then $$x_2$$ is also rational and so will $$y_2$$ be, because $$y_2=\frac{1-m^2-6m}{1+m^2}.$$
Now for $$m=\frac{p}{q}$$, where $$p,q \in \Bbb{Z}$$ and $$q \neq 0$$, we can have $$\color{blue}{a=3p^2-2pq-3q^2, \quad b=q^2-p^2-6pq, \quad c=p^2+q^2}$$ Now the main question is: this combined with your answer (which can be obtained if we allow $$q=0$$) is that the totality of all solutions?
• If there would be another solution (a,b) of x^2+y^2=10, then we would to consider the line through by (a,b) and (3,1). In this case there would be a slope m (that is computed imposing b=ma +(1-m) ) such that (a,b) is the intersection of the line with slope m passing in the point (3,1) with the circle. Thus your construction identifies all the solutions, right? Jun 22, 2021 at 20:52
• Could you explain why did you choose a line through $(3,1)$ only? Jun 22, 2021 at 20:57
• The reason you choose a line through a known point is that it restricts you down to one parameter (the slope) and let’s you use Vieta formula to get a rational solution without radicals. In general, conics aren’t guaranteed to go through any rational points but you can use one rational solution to find others.
– Eric
Jun 22, 2021 at 21:19
• @Umeshshankar In more complicated situations we may need more than one parametrization to get all primitive integral solutions of an indefinite ternary. In this case, just one suffices, proof added. Jun 23, 2021 at 3:00
• @FedericoFallucca Yes that's the idea to ensure that all rational solutions are accounted for. Jun 23, 2021 at 7:48
$$\color{magenta}{a=3p^2-2pq-3q^2, \quad b=-p^2-6pq+q^2, \quad c=p^2+q^2}$$
or, same thing
$$\color{green}{a=3x^2-2xy-3y^2, \quad b=-x^2-6xy+y^2, \quad c=x^2+y^2}$$
When $$p,q$$ are coprime, the primes that can still divide $$\gcd(a,b,c)$$ are $$2$$ and $$5.$$ The proof of all (primitive, integral) solutions is just showig that these don't matter.
When $$x,y$$ are both odd, all three of $$a,b,c$$ are divisible by $$2,$$ and we need to worry about whether half the triple is represented by the given parametrization. Well taking $$p = \frac{x-y}{2} \; , \; \; q = \frac{x+y}{2} \; , \; \;$$
$$3 p^2 - 2pq -3q^2 = \frac{1}{2} \left( -x^2 - 6 xy + y^2 \right)$$ $$- p^2 - 6pq +q^2 = \frac{-1}{2} \left( 3x^2 - 2 xy -3 y^2 \right)$$ $$p^2 +q^2 = \frac{1}{2} \left( x^2 + y^2 \right)$$
$$5$$ is the other possibility . This happens when $$2x+y \equiv x - 2y \equiv 0 \pmod 5.$$
Taking $$p = \frac{2x+y}{5} \; , \; \; q = \frac{x-2y}{5} \; , \; \;$$
$$3 p^2 - 2pq -3q^2 = \frac{1}{5} \left( 3x^2 - 2 xy -3 y^2 \right)$$ $$- p^2 - 6pq +q^2 = \frac{1}{5} \left( -x^2 - 6 xy + y^2 \right)$$ $$p^2 +q^2 = \frac{1}{5} \left( x^2 + y^2 \right)$$
$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$
The example I like to show is solving $$2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0,$$ four "recipes," $$\left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right)$$
$$\left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right)$$
$$\left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right)$$
$$\left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right)$$
For all four recipes, $$x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2$$
$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$
$$x^2+y^2=nz^2\tag{1}$$ In general, if equation $$(1)$$ has one integer solution $$(x,y,z)=(x_0,y_0,z_0)$$ then there exists an infinitely many integer solutions.
Substitute $$x=t+x_0, y=t+y_0, z=at+z_0$$ to equation $$(1)$$, then we get
$$t = \frac{2(x_0+y_0-nz_0a)}{-2+na^2}$$
Let $$a=\frac{p}{q}$$ with p,q are integers, hence we get a parametric solution $$(x,y,z)=(x_0np^2-2qnz_0p+2q^2y_0, y_0np^2-2qnz_0p+2q^2x_0, nz_0p^2-2qx_0p-2qy_0p+2q^2z_0).$$
Example of $$x^2+y^2=10z^2.$$
Let $$n=10, (x_0,y_0,z_0)=(3,1,1).$$
$$(x,y,z)=(15p^2-10qp+q^2, 5p^2-10qp+3q^2, 5p^2-4qp+q^2).$$
Solving the $$C$$-function of Euclid's formula $$\quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$ for $$(k), \space$$ we can find Pythagorean triples for any given $$C$$-values, if they exist, that are primitive, doubles, or square multiples of primitives. This will not find, for example $$(9,12,15)\space$$ or $$(15,20,25),$$ but it will find $$(3,4,5),\space (6,8,10),\space (12,16,20),\space (27,36,45), \space$$ etc. We begin with the following formula. Any $$m$$-value that yields an integer $$k$$-value indicates a valid $$(m,k)$$ pair for generating a Pythagorean triple.
$$$$C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$$$ The lower limit ensures $$m>k$$ and the upper limit ensures $$k\in\mathbb{N}.$$
Here is an example for $$C=40\implies 10c=4$$ where $$c$$ is the one shown in the OP equation.
$$C=40\implies \bigg\lfloor\frac{ 1+\sqrt{80-1}}{2}\bigg\rfloor=4 \le m \le \lfloor\sqrt{40-1}\rfloor=6\\ \land \quad m\in\{6\}\Rightarrow k\in\{2\}\\$$ $$F(6,2)=(32,24,40)\implies (32,24,10\times 4)$$
This method will not find all Pythagorean triples that match the criteria but it will find an infinite number of triples that do such as:
$$c=1\longrightarrow (8,6,10\times 1)\\ c=2\longrightarrow (12,16,10\times 2)\\ c=4\longrightarrow (32,24,10\times 4)\\ c=9\longrightarrow (72,54,10\times 9)\\$$
Note that any multiple of a triple found also yields a valid triple so $$3\times (8,6,10)\longrightarrow (24,18,10\times3)$$ and provides the "missing" $$c=3$$ triple in the list above. The combination of the two will find all Pythagorean triples where the $$c$$ in $$10c$$ is an integer except for the most unusual case like $$(3,1,10\times 1)$$ mentioned in another post. | 2022-06-26T03:27:23 | {
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https://math.stackexchange.com/questions/4365840/surface-area-of-rotation-of-a-circle-around-a-tangent | # Surface area of rotation of a circle around a tangent
Self-studying integral calculus and I got this problem:
The circle $$x^2+y^2=a^2$$ is rotated around a line tangent to the circle. Find the area of the surface of rotation.
There were a few hints given alongside this question, namely: "Set up coordinate axes and a convenient parametrization of the circle. What does the polar graph $$r=2a\sin(\theta)$$ look like?" I understood the first and last hints since under this new coordinate system, the circle's equation becomes: $${x_n}^2+(y_n-a)^2=a^2$$ Which when converted to polar, gives you the last hint. However I was unable to describe this in terms of parameters, so I decided to take the upper semicircle's surface area of revolution going from $$a$$ to $$-a$$ and multiplying that by 2 to account for the lower semicircle. My integral: $$2\int_{-a}^a 2\pi (\sqrt{a^2-x^2}+a)\sqrt{1+\frac{x^2}{a^2-x^2}} dx$$ Upon simplification: $$4\pi\int_{-a}^a a+\frac{a^2}{\sqrt{a^2-x^2}}dx$$ Evaluation leads me to: $$8\pi a^2 + 4\pi^2 a^2$$
However my book (Serge Lang's First Course in Calculus) gives only $$4\pi^2 a^2$$. Where has my logic gone wrong if I am getting an extraneous term $$8\pi a^2$$?
EDIT for clarity on integral setup: I first rearranged for $$y$$ while taking positive square root as I want to take the upper semicircle into consideration for surface of revolution about x-axis. I'll double this to account for the lower semicircle. This gives: $$y=\sqrt{a^2-x^2}+a$$ Using the surface of revolution formula with the derivative as $$\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}$$ Using this into the surface of revolution integral nets me my first integral in this post (also applied $$\times$$2)
• I do not understand what you are doing. Can you please explain your integral setup? If you are rotating about a tangent, you cannot take volume of rotation of a semicircle and multiply by $2$. Volume is a function of the distance from the axis of rotation. Jan 25 at 16:15
• Also the volume of rotation of a circle around any tangent is going to be the same so choose a tangent that is easy to work with Jan 25 at 16:16
• That is the formula for surface area of revolution around the x-axis. I'll add the setup but I figured out my issue now using the parameters. Jan 25 at 16:18
• ok so are you rotating $x^2 + (y-a)^2 = a^2$ around x-axis? Jan 25 at 16:20
• Hint: Theorem of Pappus
– robjohn
Jan 25 at 16:51
After changing the coordinates, in effect you are rotating $$x^2 + (y-a)^2 = a^2$$ around x-axis.
The circle is $$x^2 + y^2 = 2 ay$$
$$\displaystyle y' = \frac{x}{a-y}$$
$$\displaystyle ds = \sqrt{1 + (y')^2} ~dx = \frac{a}{|y-a|} ~ dx$$
For lower half -
$$y = a - \sqrt{a^2-x^2}$$
So, $$\displaystyle S_1 = 2 \pi a \int_{-a}^a \frac{a - \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$$
$$= 2 \pi a^2 (\pi - 2)$$
For upper half -
$$y = a + \sqrt{a^2-x^2}$$
So, $$\displaystyle S_2 = 2 \pi a \int_{-a}^a \frac{a + \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$$
$$= 2 \pi a^2 (\pi + 2)$$
Adding both, $$S = 4 \pi^2 a^2$$
But it is easier in polar coordinates as I mentioned in comments. The circle is,
$$r = 2a \sin\theta, 0 \leq \theta \leq a$$
$$\dfrac{dr}{d\theta} = 2a \cos\theta$$
$$\displaystyle ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} ~ d\theta = 2a ~ d\theta$$
$$y = 2a\sin^2\theta$$
So the integral is,
$$\displaystyle S = 8 \pi a^2 \int_0^{\pi} \sin^2\theta ~ d\theta = 4 \pi^2 a^2$$
In spite of your obfuscating figure, you are asking for the surface area of a torus whose inner radius, $$R$$ (to the center of the cross-section) and outer radius, $$r$$ (that of the cross-section) are the same. This is well known to be $$S=4\pi^2Rr$$ (see, for example the CRC Mathematical Tables). So in your case, $$S=4\pi^2a^2$$
We can derive this result with Pappus's ($$1^{st}$$) Centroid Theorem, which states that the surface area $$S$$ of a surface of revolution generated by rotating a plane curve $$C$$ about an axis external to $$C$$ and on the same plane is equal to the product of the arc length $$s$$ of $$C$$ and the distance $$d$$ traveled by its geometric centroid. Simply put, $$S=2πRL$$, where $$R$$ is the normal distance of the centroid to the axis of revolution and $$L$$ is the curve length. In your case, $$R=a$$ and $$L$$ is the circumference of the circle, i.e., $$=2\pi a$$, so that $$S=4\pi^2a^2.$$
### Comment on the Question
I believe that this is the standard setup for surface of revolution about the $$x$$-axis $$\int_{-a}^a\overbrace{\quad2\pi y\quad\vphantom{\frac{a}{\sqrt{a^2}}}}^{\substack{\text{account for}\\\text{revolution}}}\overbrace{\frac{a}{\sqrt{a^2-x^2}}}^{\mathrm{d}s/\mathrm{d}x}\,\mathrm{d}x$$ However, the part for the upper arc of the circle does not give the same area as that for the lower arc of the circle, so we need to compute both separately: \begin{align} &\int_{-a}^a2\pi\left(a+\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{upper}\\ &+\int_{-a}^a2\pi\left(a-\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{lower}\\ &=2\pi\int_{-a}^a2a\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\[6pt] &=4\pi^2a^2 \end{align}
### Other Approaches
Revolution Around the $$\boldsymbol{x}$$-axis
As I had posted before I realized the question was revolving around the $$y$$-axis, if we revolve around the $$x$$-axis, the upper and lower parts of the surface have the same area, so we can just multiply the upper integral by $$2$$ in this case. Thus, the formula is \begin{align} 2\int_0^{2a}2\pi x\,\frac{a}{\sqrt{a^2-(x-a)^2}}\,\mathrm{d}x &=2\int_{-a}^a2\pi(x+a)\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{1a}\\ &=4\pi a^2\int_{-1}^1(x+1)\frac1{\sqrt{1-x^2}}\,\mathrm{d}x\tag{1b}\\ &=4\pi a^2\int_{-\pi/2}^{\pi/2}(\sin(x)+1)\,\mathrm{d}x\tag{1c}\\[9pt] &=4\pi^2a^2\tag{1d} \end{align} Explanation:
$$\text{(1a)}$$: substitute $$x\mapsto x+a$$
$$\text{(1b)}$$: substitute $$x\mapsto ax$$
$$\text{(1c)}$$: substitute $$x\mapsto\sin(x)$$
$$\text{(1d)}$$: integrate
Parametrization
Parametrize the torus as follows: at each point of the circle around the $$z$$-axis, $$a(\cos(\phi),\sin(\phi),0)$$ put a circle perpendicular to this circle: \begin{align} p(\phi,\theta) &=\overbrace{a(\cos(\phi),\sin(\phi),0)}^\text{primary circle}+\overbrace{a(\cos(\phi)\cos(\theta),\sin(\phi)\cos(\theta),\sin(\theta))}^\text{secondary circle around the primary circle}\\ &=a(\cos(\phi)(1+\cos(\theta)),\sin(\phi)(1+\cos(\theta)),\sin(\theta))\tag{2a}]\\[6pt] p_1(\phi,\theta) &=a(-\sin(\phi)(1+\cos(\theta)),\cos(\phi)(1+\cos(\theta)),0)\\ &=a(1+\cos(\theta))(-\sin(\phi),\cos(\phi),0)\tag{2b}\\[6pt] p_2(\phi,\theta) &=a(-\cos(\phi)\sin(\theta),-\sin(\phi)\sin(\theta),\cos(\theta))\tag{2c} \end{align} Thus, we get \begin{align} |p_1(\phi,\theta)\times p_2(\phi,\theta)| &=a^2(1+\cos(\theta))\,|(\cos(\theta)\cos(\phi),\cos(\theta)\sin(\phi),\sin(\theta))|\\ &=a^2(1+\cos(\theta))\tag3 \end{align} and we can compute $$\int_0^{2\pi}\int_0^{2\pi}a^2(1+\cos(\theta))\,\mathrm{d}\phi\,\mathrm{d}\theta=4\pi^2a^2\tag4$$ Theorem of Pappus
As I mentioned in a comment, we can apply the Theorem of Pappus: the primary circle has circumference $$2\pi a$$ and the secondary circle has circumference $$2\pi a$$, so the area is $$(2\pi a)(2\pi a)=4\pi a^2\tag5$$
• The author defined the surface area by this: $$S=\int_a^b 2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx$$ Jan 25 at 18:20
• That would be for revolving about the $x$-axis. It would be useful to mention that. This would require two integrals, one for the part $y\le a$ and one for the part $y\ge a$, you can't simply double the one for the part $y\ge a$ (which has a larger surface area than the part $y\le a$). In my answer, I was revolving about the $y$-axis. With this new information, I will update the first part of my answer.
– robjohn
Jan 25 at 18:30
• Hmm... it was bothering me why symmetry consideration was not possible here. I didn't realize that the upper semicircle would indeed produce a greater surface area than the lower one until I read your comment. Thanks! Jan 25 at 18:35
• You’re welcome. If you rotate around the $x$-axis, the upper and lower parts are the same, so there is a little less work. I’ve added, and expanded, the part I used to have in my answer when I thought you were rotating around the $x$-axis.
– robjohn
Jan 26 at 20:44 | 2022-05-17T20:18:03 | {
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https://www.physicsforums.com/threads/moment-of-inertia-of-a-uniform-triangle-involves-integration.640571/ | # Homework Help: Moment of Inertia of a uniform triangle(involves integration)
1. Oct 2, 2012
### hms.tech
1. The problem statement, all variables and given/known data
The attachment shows an equilateral triangle of side length "2d"
It is a uniform triangle in the 2-D space.
Mass of triangle = M
I have to find the moment of inertia about one of its side . I am taking the side $\overline{AB}$ as the axis of rotation (hence i would be finding the Moment of inertia about this line)
The pink lines inside the triangle show an attempt to this problem.
2. Relevant equations
>Integration
>area of triangle
>$I_{total}$ = $\sum I_{1} + I_{2} + I_{3}... I_{n}$
>β = 60 degrees
3. The attempt at a solution
Let the mass per unit area of the triangle be = m
The triangle can be considered (approximated) to be composed of many small rods of width Δy and length "x" (where x varies as the equation of the line)
Now, the area of this rod is = 2xΔy
Hence mass of the rod = 2mxΔy
moment of inertia of one of the rod about $\overline{AB}$ = 2mx$y^{2}$Δy
The equation of the line $\overline{AC}$ is : y = tan(60)x = x$\sqrt{3}$
x = y/$\sqrt{3}$
Approximate moment of inertia = $\frac{2m}{\sqrt{3}}$$\sum y^{3} Δy$
we use integration to find the exact moment of inertia :
$I_{total}$ = $\frac{2m}{\sqrt{3}}$$\int (y)^{3} dy$
the limits of integration are from 0$\rightarrow\sqrt{3}d$
Area of triangle = $\sqrt{3}$$d^{2}$
Moment of inertia of the triangle = $\frac{3Md^2}{2}$
The problem is that this answer is wrong, and the correct answer is : $Md^2/2$ [which i have deduced by flipping the triangle (i will post an attachment of the way the triangle was oriented which lead me to the correct answer] However the method for both was same, and the orientation of the triangle should never affect the Moment of inertia as long as it is taken about the same axis .
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2. Oct 2, 2012
### HallsofIvy
I suspect that the answer lies in precisely what you are NOT showing: exactly how you calculated the moment of the "flipped" triangle. If you flip the triangle so that the vertex is at (0, 0), the side you are rotating around will be $y= \sqrt{3}d$.
3. Oct 2, 2012
### hms.tech
I dont know what u mean by that, i dont understand in what way are u rotating,
but i rotated it in such a way to give this figure. using this diagram (attached to this post) i got the right answer
Even though i used the same method as before (of course the x will be replaced with y)
It is surprising that one of the answers is wrong by the exact same method .
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4. Oct 3, 2012
### hms.tech
Can someone explain why does the first method give me a wrong answer .
5. Oct 3, 2012
### hms.tech
Bump .. .. ..
6. Oct 5, 2012
Bump . . .
7. Oct 6, 2012
### vanhees71
I guess, you've calculated the moment of inertia around the middle axis of the triangle instead of around one of its sides.
The area of the triangle is $A=\frac{\sqrt{3}}{4} a^2=\sqrt{3} d^2$ with $a=2d$ the length of the triangle's sides.
To make the calculation easier, I put one side around the $x$ axis of a cartesian coordinate system, which we also take as the axis of rotation, and one of its corners on the $y$ axis.
Then possible $y$ values are in the interval $(0,\sqrt{3} d)$ and for each $y$-value the $x$ values are in the interval $(y/\sqrt{3}-d,d-y/\sqrt{3})$.
The moment of inertia is thus given by the area integral
$$I=\frac{m}{A} \int_{A} \mathrm{d} F \; y^2=\frac{m}{\sqrt{3} d^2} \int_0^{\sqrt{3} d} \mathrm{d} y 2 \int_0^{d-y/\sqrt{3}} \mathrm{d} x \; y^2.$$
The integration is straight forward. With the $x$ integral done you get
$$I=\frac{2m}{\sqrt{3} d^2} \int_0^{\sqrt{3}d} \mathrm{d} y \; \left(d y^2-\frac{y^3}{\sqrt{3}} \right )$$
After the integration and some algebra you finally get
$$I=\frac{m d^2}{2}.$$
8. Oct 6, 2012
### Studiot
Since vanhees has done the algebra whilst I was asleep, (thanks mate) there is little left for me to do except add value.
Have your heard of the transfer theorem/ parallel axis theorem/theorem of Pappus?
You need to always check what axis of rotation is being used. Standard tables often refer to
Ixx and Iyy
These are taken through the centroid of the figure., not about an edge or other axis.
General integration of an elemental area will always be about a stated axis or it is valueless.
Edit : I mentioned product of inertia before, have you met this term?
The following thread may be of value
go well
Last edited: Oct 6, 2012
9. Sep 27, 2013
### ILLUSION ZONE
Hi hms, i'm a new member and this is my first post,
You did everything right except for one thing. Here is the reason you didn't get an answer of $\frac{1}{2}$$M$$d^{2}$ in your initial post. The $y^{2}$ represents the square of the distance between your rods and AB which is the error (further study of your diagram shows this). Replace your $y^{2}$ with $\left(\sqrt{3}d-y\right)^{2}$ and then continue as before and you will get the right answer.
I would like to write out the whole process for you but it would take ages in latex.
Please let me know if this works for you, as it worked for me when I solved it on paper using this method and got the right answer of $\frac{1}{2}$$M$$d^{2}$.
Regards,
IZ
Last edited: Sep 27, 2013 | 2018-07-22T19:21:11 | {
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https://math.stackexchange.com/questions/605405/how-should-i-define-the-limit-definition-of-a-derivative-using-negative-numbers | # How should I define the limit definition of a derivative using negative numbers?
Typically the derivative is defined at a point $x$, assuming it is differentiable at it, by
$$\lim_{n \rightarrow \infty} \frac{f(x + \frac{1}{n}) - f(x)}{\frac{1}{n}}$$
But I want to define it using $f(x - \frac{1}{n})$. Should I use
$$\lim_{n \rightarrow \infty} \frac{f(x - \frac{1}{n}) - f(x)}{|-\frac{1}{n}|}$$
or
$$\lim_{n \rightarrow \infty} \frac{f(x) - f(x - \frac{1}{n}) }{|-\frac{1}{n}|}$$
• The latter $\lim_{n \rightarrow \infty} \frac{f(x)-f(x - \frac{1}{n}) }{\frac{1}{n}}$ – draks ... Dec 13 '13 at 12:22
• This is not the typical definition of derivative, rather $\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$, which works to assure limits exist on both sides of zero. – hardmath Dec 13 '13 at 12:27
• Remember what this definition means geometrically; you're essentially seeing how the rate of change, the slope of the tangent line at x, behaves as you make the change approach zero. The definition $lim_{n \to \infty} \frac{f(x+\frac{1}{n}) - f(x-\frac{1}{n})}{\frac{2}{n}}$ is equivalent to those two. – Lost Dec 13 '13 at 12:36
• Also, if $n$ is supposed to be an integer, so the limit is over the integers, as it would seem from using a variable name like $n$, then there's a more subtle problem in that this definition will "differentiate" things that shouldn't really be differentiable. Consider a function that's $0$ at $x = 0$ and also at $x = \frac{1}{n}$ for every integer $n \ne 0$, but $1$ everywhere else. This definition will differentiate it at $0$ and say the derivative is $0$, but the correct definition, taking the limit over the reals, says it is not differentiable at $0$. – The_Sympathizer Dec 13 '13 at 12:39
• @user1876508 : your "typical" definition is not typical or even correct. What you wrote is equivalent to $\lim_{h\to 0^+}(f(x+h)-f(x))/h$, which is not equivalent to the correct definition given by Sami below, because the limit is one-sided. – Stefan Smith Dec 13 '13 at 21:39
The definition of the derivative of $f$ at $x$ is $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ and by the change of variable you can find different version of the definition, for example if $k=-h$ we find $$f'(x)=\lim_{k\to0}\frac{f(x)-f(x-k)}{k}$$ | 2019-11-17T00:03:03 | {
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https://math.stackexchange.com/questions/2539961/induction-on-2-numbers | # Induction on 2 numbers
Show that for the Fibonacci numbers it holds that
$f_{x+y}=f_{y+1}f_{x}+f_{y}f_{x-1}$ for $x\ge 2, y\ge 1$
The proof goes as follows:
First we show that the claim holds for all $x\ge2$,$y=1$ and that it holds for $x\ge2, y=2$(easy)
In the Induction step we take $y\ge2, x\ge2$. It holds that
$f_{x+(y+1)}=f_{x+y}+f_{x+(y-1)}=f_{y+1}f_x+f_yf_{x-1}+f_yf_x+f_{y-1}f_{x-1}$ by the induction hypothesis
$=f_{y+2}f_x+f_{y+1}f_{x-1}$
Is this really a complete proof? I know that there are several ways to do an induction on $2$ numbers. But these usually involve an induction step on each variable or an induction on the sum of both.But in this one we never really do anything with $a$.
Could someone please explain/motivate this or just give me some reference for this kind of induction?
• Looks good, although I think you meant $\ldots + f_{y-1}f_{x-1}$, not $\ldots + f_{y-1}f_{x-y}$. – Théophile Nov 27 '17 at 18:34
• @Théophile Yes,I edited it – Blablablu Nov 27 '17 at 18:36
• It's self answered question, neat. – Abr001am Nov 27 '17 at 18:47
• A small quibble: the condition for the induction step should say that $y\geq2,$ not $3.$ – David K Nov 27 '17 at 19:23
• @DavidK Thanks! Edited – Blablablu Nov 27 '17 at 19:47
Actually, you're not doing induction on two variables here at all. You're only doing induction on $y$, while $x$ is assumed to be arbitrary. But, as such, it is a good proof!
• The addition is associative so fibonacci(x+(y+1))=fibonacci(y+(x+1)) – Abr001am Nov 27 '17 at 18:49
• @Bram28 Isn't the second base case necessary since we use the induction hypothesis for $y$ and for $y-1$? – Blablablu Nov 27 '17 at 18:51
• Both base case are needed. And both base cases are referred to. And both base cases are omitted with a blithe "easy"..."First we show that the claim holds for all x≥2,y=1 (first base case) and that it holds for x≥2,y=2 (second base case)" – fleablood Nov 27 '17 at 18:59
• @fleablood Thank you! The "easy" was just because I didn't want to type down everything, I wanted to mention, that the computation for both base cases is not my problem here. – Blablablu Nov 27 '17 at 19:02
• Oh, right, of course, both base cases are needed, sorry about that! – Bram28 Nov 27 '17 at 19:03
It's subtle but this induction is not doing: Assume true for $x$ and assume true for $y$ then show it is true for $x+1$ and $y+1$.
That would not be valid as that proves only for $x$ and $y$ where $x - y = constant$.
BUT this is actually doing something different. It is assuming it is true for $n = x+y$ and showing it is therefore true for $n = (x+y) + 1 = (x+1) + y = x + (y+1)$.
That is valid.
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On second reading, I see Bram28, is correct and this is only an induction on the variable $y$. and not on $x$ at all.
To see that this proof is complete, you should note the base cases (for $y= 1; x\ge 2$ and $y= 2; x\ge 2$)[1] was for any arbitrary $x$ and not for an $x = 2$. So induction on $x$ is unnecesary.
I'm going to leave my answer up, because I think it illustrates a good point of looking at an induction proof involves carefully evaluating what the variable being inducted upon is; not merely what it looks like.
The poof could have been identically done on $x+y \implies x + y+ 1$ (only the framing of the base case would differ).
[1] Both base cases were omitted which isn't really fair.
$y=1; f_{x+y}=f_{x+1}=1*f_x + 1*f_{x-1}=f_2f_x + f_1f_{x-1}=f_yf_x + f_{y+1}f_x$.
$y=2; f_{x+y}=f_{x+2}=f_{x+1} + f_x=f_x + f_{x-1} + f_x = 2*f_{x} + 1*f_{x-1} = f_3*f_x + f_2*f_{x-1} = f_{y+1}f_x + f(y)*f_{x-1}$.
• Thank you, that's exactly what confused me aswell – Blablablu Nov 27 '17 at 18:59 | 2019-06-24T13:17:21 | {
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http://foresteria-lacura.it/lmhv/when-to-use-integration-by-parts.html | ## When To Use Integration By Parts
Use the solutions intelligently. You remember integration by parts. There is no such change in the limits of integration in the theorem on reduction of a Riemann-Stieltjes integral to a Riemann integral and there is no change in the limits of integration in the integration by parts formula. I'm going to set up parts computations using tables; it is much easier to do repeated parts computations this way than to use the standard u-approach. This method is also termed as partial integration. INTEGRATION BY PARTS (TABLE METHOD) Suppose you want to evaluate ∫ x. Unfortunately there is no general rule on how to calculate an integral. Sanaullah Bhutto 104 views. Our professor states that i can get RHS out of LHS using integration by parts: $$\\int\\limits_0^x \\! \\frac. Use the integration by parts theorem to calculate the integral. Repeated integration-by-parts. I Bernoulli Equation with weird integral. (1 pt) Use integration by parts to evaluate the integral. Hence the original integral is: Z 1 0 tan−1 xdx = π 4 − ln2 2. Use Integration by parts formula Part 01 in Urdu/ (Sanaullah Bhutto) - Duration: 16:59. What about the first term ? The problem here is that one factor, , is absorbing too many of the derivatives. ∫ arctan x dx ≡ ∫ arctan x × 1 dx: I am using the trick of multiplying by 1 to form a product allowing the use of integration by parts formula. Integration between technology and knowledge. LAPLACE TRANSFORMS We derive the second formula and leave the derivation of the first formula as an exercise. A single integration by parts starts with d(uv)=udv+vdu, (1) and integrates both sides, intd(uv)=uv=intudv+intvdu. If both properties hold, then you have made the correct choice. Using the ∫u dv notation, we get u = x2 dv cos3 dx. Question: In this exercise we want to use integration by parts to integrate {eq}\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)dx{/eq} (A) Find {eq}\arctan(x){/eq} and {eq. u and dv are provided. Integration By Parts The process of finding the integral of the composition of two or more functions by taking them function one and function two is known as the integration by parts. For more integration tutorials, examples and worked solutions to pa. The idea it is based on is very simple: applying the product rule to solve integrals. If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the. One way of writing the integration by parts rule is$$\int f(x)\cdot g'(x)\;dx=f(x)g(x. Integration by parts is a technique used to evaluate integrals where the integrand is a product of two functions. Design parts for multi-use. It only takes a minute to sign up. I = e2x sinx−2. Using prime notation, take. If we have a function in the. Therefore,. 3xln locity 2x dx C 6. The second way to use integration by parts on definite integrals is to keep the limits of integration all the way through. Use integration by parts to find the Laplace transform of: f(t)=tsinat where a is a real constant. Numerical integration of a quadratic form exponential in two variables over a rectangle. · Using u-substitution to find definite and indefinite integrals · Using integration by parts to find definite and indefinite integrals. Typically, Integration by Parts is used when two functions are multiplied together, with one that can be easily integrated, and one that can be easily differentiated. Get an answer for 'Use integration by parts to integrate Integration sign, x^5 ln (x) dx' and find homework help for other Math questions at eNotes. 1Integration by parts 07 September Many integration techniques may be viewed as the inverse of some differentiation rule. Integration by parts is a method of breaking down equations to solve them more easily. Husch and. Integration by Parts Let and be functions with continuous derivatives. Integration By Parts The process of finding the integral of the composition of two or more functions by taking them function one and function two is known as the integration by parts. Integration by Parts is yet another integration trick that can be used when you have an integral that happens to be a product of algebraic, exponential, logarithm, or trigonometric functions. Intermediate steps. Please try again using a different payment method. 2) For the. The integration by parts rule looks like this: ∫ u * v' dx = u * v - ∫ ( v * u' ) dx. Visit Stack Exchange. But also understanding what your integrals and derivatives look like one step ahead are a big help. Identifying when to use U-substitution vs Integration by Parts - Duration: 11:39. This method is also termed as partial integration. Theorem: The formula for the method of integration by parts is: $$\color{blue}{\int udv = u \cdot v - \int vdu}$$ There are four steps how to use this formula: Step 3: Use the formula for the integration by parts. (c) Use the substitution t = √(3x + 1) to show that I may be expressed as ⌡ ⌠b a ktet dt, giving the values of a, b and k. ∫ () Integrals that would otherwise be difficult to solve can be put into a simpler form using this method of integration. We may have to rewrite that integral in terms of another integral, and so on for n steps, but we eventually reach an answer. Let’s verify this and see if this is the case. Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish. Use Integration by parts formula Part 01 in Urdu/ (Sanaullah Bhutto) - Duration: 16:59. Both of the solution presented below use #int lnx dx = xlnx - x +C#, which can be done by integration by parts. The second factor then has to be the constant function 1. Hence the original integral is: Z 1 0 tan−1 xdx = π 4 − ln2 2. Design parts for multi-use. ( 2 t) d t Solution. Using repeated Applications of Integration by Parts: Sometimes integration by parts must be repeated to obtain an answer. This quiz/worksheet combo will test your ability to use integration by parts to. This method is used when the integrals are difficult to evaluate. Using Maple to illustrate the method of substitution. 2 64x cos 4x dx C 7. There are 3 main parts to the PayPal IPN system: A webpage that initiates a request to PayPal to make a payment. Use integration by parts twice to evaluate integral e^4t cos(8t)dt|: Step 1: Let u = e^4t| and dv = cos(8t)dt|. Integration by parts is essentially the product rule for differentiation inverted:. Enter your function as f(x) and your choices of u and v’ below. (ii) Introduce the intermediary functions u(x) and v(x) as:. Use the integration by parts theorem to calculate the integral. Also, for trigonometric products, check out integration of product of sinusoidal functions. If you get an antiderivative, then it worked. Closed parentheses tell the calculator that you want to do a program. This problem has been solved! See the answer. q Introduction Let’s begin by taking the exponential function and applying the antiderivative to. There are, after all, lots of ways to put a vector differential form into an equation, and (at least) three dimensionalities of integral you might be trying to do!. If we have a function in the. Use integration by parts to evaluate the integral: (ln(3x))^2 I set u=ln(3x) and dV=1/2dx but that didnt work. New; 16:59. I = e2x sinx−2. If you’re just using integration by parts in general to find the integral of three functions, basically. The typical repeated application of integration by parts looks like:. R 1 xln(x) dx: sub. Use Integration by parts formula Part 01 in Urdu/ (Sanaullah Bhutto) - Duration: 16:59. (x+4)^ln x dx between 3 and 0 By. Also, for trigonometric products, check out integration of product of sinusoidal functions. Using Integration by Parts. The Method of Integration by Parts. 5 LAPLACE TRANSFORMS 5. dx = 2u du ∫ 2ucosu du. We begin by entering x x3 1. Intermediate steps. Click HERE to return to the list of problems. The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The derivative of the linear function. { Square brackets indicate limits of definite integration. Drill - Integration by Parts. Both of the solution presented below use Integration by Parts. So, plugging $$u$$, $$du$$, $$v$$ and $$dv$$ into the Integration by Parts formula gives,. Related Threads on Using Integration by parts Integration by parts. Integration by parts works if {eq}u {/eq} is absolutely continuous and if {eq}v' {/eq} has the point of discontinuity then its antiderivative {eq}v {/eq} may not have a derivative at that point. Integration by Substitution "Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") is a method to find an integral, but only when it can be set up in a special way. We can now integrate ∫ e^2x dx which is easy to give the result shown above. Integration by parts: ∫x⋅cos(x)dx. (ii) Introduce the intermediary functions u(x) and v(x) as:. tice, integration by parts can be used to derive an antiderivative. I know of know way of determining that it cannot be used. (x+4)^ln x dx between 3 and 0 By. Answer to: Use integration by parts to find the integral. Question: Use integration parts to find the integral. Then Z exsinxdx= exsinx excosx Z. Then use the fact that 0 R 1 0 x e x 1 (why?) to put an upper and lower bound on e. ∫ u⋅dv = u⋅v−∫ v⋅du. Example $$\PageIndex{1}$$: Using Integration by Parts. Use Integration by parts formula Part 01 in Urdu/ (Sanaullah Bhutto) - Duration: 16:59. Evaluate each of the following integrals. ∫ 6tan−1( 8 w) dw ∫ 6 tan − 1 ( 8 w) d w Solution. Solution Here, we are trying to integrate the product of the functions x and cosx. Integration by reduction formula in integral calculus is a technique or procedure of integration, in the form of a recurrence relation. Use integration by parts, together with the techniques of this section, to evaluate the integral. Let dv = e x dx then v = e x. This is the currently selected item. Use also the fact that $1. We use integration by parts a second time to evaluate. Example 1 Evaluate the following integral. Another Reduction Formula: x n e x dx To compute x n e x dx we derive another reduction formula. Evaluate integral from 0 to pi/2 of xsin(x) with respect to x. Finding a formula using integration by parts which reduces the complexity of an integral. In primary school, we learned how to find areas of shapes with straight sides (e. (3) Evaluate. Using Integration by Parts. {eq}I=\int x\ln|x|dx {/eq} Integration by Parts: The process of finding the integral of the composition of two or more functions when they are. Algorithm for parts integration. 52) $$\displaystyle ∫x\ln x\,dx$$ 53) $$\displaystyle ∫\frac{\ln^2x}{x}\,dx$$ Answer: Do not use integration by. Basic Integration by Parts Examples with Math Fortress. Both of the solution presented below use #int lnx dx = xlnx - x +C#, which can be done by integration by parts. This Demonstration lets you explore various choices and their consequences on some of the standard integrals that can be done using integration by parts. What is practical however is finding instead a formula which one can use a number of times rather than following the same process continually. Integration by Substitution: Definite Integrals; Integration by Parts: Indefinite Integrals; Some Tricks; Evaluate the definite integral using integration by parts with Way 2. To find we use integration by parts again with to get. There are also some functions you have to remember like ln x that you have to use parts by using 1. So many that I can't show you all of them. Sensing a trend, we decide to use integration by parts. Enter the function to Integrate: With Respect to: Evaluate the Integral: Computing Get this widget. Integration by parts is essentially the product rule for differentiation inverted:. I absolutely love doing jigsaw puzzles. R 1 xln(x) dx: sub. JEE Main 2020 - 70 Days Preparation Strategy. ∫ 0 π e cos t sin 2 t d t. However, as we discussed in the Integration by Parts section, the two answers will differ by no more than a constant. (Note we can easily evaluate the integral R sin 3xdx using substitution; R sin xdx = R R sin2 xsinxdx = (1 cos2 x)sinxdx. R ex dx: just integrate 4. So we have. Use integration by parts to determine which of the reduction formulas is correct. Thirty integrals are found using Integration by Parts and Integration by Substitution. Using the parts rule: Combining these two, results in. A webpage that confirms the above payment and continues on to the next phase of your web application, such as a 'Thank You' page. q Introduction Let’s begin by taking the exponential function and applying the antiderivative to. This is a “classic” derivation of the minimization condition for a path, using integration by parts. In a manufacturing firm, different products can share parts that have been designed for multi-use. It is not necessary that we have two functions in the product form to apply this method, but even a single function, for example, the. {eq}I=\int x\ln|x|dx {/eq} Integration by Parts: The process of finding the integral of the composition of two or more functions when they are. These can sometimes be tedious, but the technique is straightforward. R ex dx: just integrate 4. The general procedure is as follows: Start doing the integration by parts. There are many ways to integrate by parts in vector calculus. du =2x dx v sin3x 3 1 = So, x x dx x x x x dx − ∫ = ∫ sin3 3 1 sin3 2 3 1 cos32 or x x −∫ x x dx sin3 3 2 sin3 3 2 1 We see that it is necessary to perform integration. You end up with another integral, that can't be done by any of the simple methods — reverse rules, guess and check, and substitution. In order to do this, it is necessary to identify the parts that are suitable for multi-use. This is how it goes: (i) Write down the given definite integral where you identify the two functions f(x) and g(x). 1/x - integration by parts. Question: Use integration parts to find the integral. Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions and expressing the original integral in terms of a known integral. if {eq}u \ \& \ v {/eq} are functions of {eq}x {/eq} then, we have :-. Here, the integrand is usually a product of two simple functions (whose integration formula is known beforehand). I believe that the best way to get good at it is to practice, a lot. Therefore,. Denote by the particular choice of antiderivative for which the left and right sides are equal on. Use integration by parts to solve the following integral ∫5x cos(4x)dx. We present the quotient rule version of integration by parts and demonstrate its use. ∫ 4xcos(2−3x)dx ∫ 4 x cos. (1 pt) Use integration by parts to evaluate the integral. Math · AP®︎ Calculus BC · Integration and accumulation of change · Using integration by parts Integration by parts: ∫x⋅cos(x)dx AP Calc: FUN‑6 (EU) , FUN‑6. LAPLACE TRANSFORMS We derive the second formula and leave the derivation of the first formula as an exercise. We try to see our integrand as and then we have. u is the function u(x) v is the function v(x). Integration of Fourier Series. For example, or. Integration by parts works if {eq}u {/eq} is absolutely continuous and if {eq}v' {/eq} has the point of discontinuity then its antiderivative {eq}v {/eq} may not have a derivative at that point. Switkes, A quotient rule integration by parts formula. It actually depends upon the form of question. Build your own widget. (c) Use the substitution t = √(3x + 1) to show that I may be expressed as ⌡ ⌠b a ktet dt, giving the values of a, b and k. Or you could read on to see how we can use this method to produce strange sums, like Grandi’s Series 1 – 1 + 1 – 1 + … = 1/2. Both of the solution presented below use #int lnx dx = xlnx - x +C#, which can be done by integration by parts. Prove the reduction formula Z xnex dx = xnex n Z xn 1ex dx. Numerical integration of a quadratic form exponential in two variables over a rectangle. The integration by parts formula says that the integral lnxdx is u times v, that is xlnx minus the integral of v du, that is the integral of x times one over x, dx. Integration by parts works if {eq}u {/eq} is absolutely continuous and if {eq}v' {/eq} has the point of discontinuity then its antiderivative {eq}v {/eq} may not have a derivative at that point. Using the formula, we get. Repeated integration-by-parts. Hence in this example, we want to make our u = x and v' = sinx. We recall that in one dimension, integration by parts comes from the Leibniz product rule for di erentiation,. Then we solve for our bounds of integration : [0,3] Let's do an example where we must integrate by parts more than once. Using the formula for integration by parts Example Find Z x cosxdx. Integration by parts is essentially the product rule for differentiation inverted:. Using Maple to illustrate the method of substitution. Evaluate each of the following integrals. You may consider this method when the integrand is a single transcendental function or a product of an algebraic function and a transcendental function. I believe that the best way to get good at it is to practice, a lot. , the new integration that we obtain from an application of integration by parts can again be subjected to integration by parts. Since the function lnx is very pleasant to di˙erentiate (ln0 x = 1 x), we could try to choose it as one of the factors. Solutions to exercises 25 i. I use the form: #int u dv = uv-intvdu#. Integration by parts is a technique for performing indefinite integration intudv or definite integration int_a^budv by expanding the differential of a product of functions d(uv) and expressing the original integral in terms of a known integral intvdu. Integration by Parts Integration by parts is a method we can use to evaluate integrals like Z f(x)g(x)dx where the term f(x) is easy to differentiate, and g(x) is easy to integrate. Examples: (1) G—t-…c, a constant. Here we compute − tn e t ∞ 0. RULE OF THUMB: The first step to use Integration by Parts is to pick your "u" and "dv". Integration by parts is whenever you have two functions multiplied together--one that you can integrate, one that you can differentiate. Integrating by parts is the integration version of the product rule for differentiation. If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place. Example: ∫x2 sin x dx u =x2 (Algebraic Function) dv =sin x dx (Trig Function) du =2x dx v =∫sin x dx =−cosx ∫x2 sin x dx =uv−∫vdu =x2 (−cosx) − ∫−cosx 2x dx =−x2 cosx+2 ∫x cosx dx Second application. Another Reduction Formula: x n e x dx To compute x n e x dx we derive another reduction formula. (1 pt) Use integration by parts to evaluate the integral. The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The basic idea of integration by parts is to transform an integral you can't do into a simple product minus an integral you can do. It is not always easy to tell when repeating integration by parts will help, but with practice it becomes easier. Feb 13, 2010 #1 This has probably been done before, nor am I attempting to prove that 1=0, but I was unable to find other threads pertaining to this question, so here it is!. ∫ 2ucosu du. There were a significant proportion of candidates who proceeded to integrate u(u + 1) –2. Using the formula for integration by parts Example Find Z x cosxdx. We try to see our integrand as and then we have. Integration by Parts Description Apply integration by parts to the integral thereby obtaining Integration by Parts Enter the integral : Declare : Execute integration by parts: Commands Used Int , IntegrationTools[Parts] See Also Student[Calculus1] ,. Wait for the examples that follow. Use integration by parts to evaluate the integral ∫ (). Solve the following integrals using integration by parts: (a) Z x2 sin(x)dx, (b) Z (2x+ 1)ex dx, (c) Z xsin(3 x) dx, (d) Z 2xarctan(x)dx, (e) Z ln(x)dx 4. Integration by parts is a "fancy" technique for solving integrals. 132 CHAPTER 5. Closed parentheses tell the calculator that you want to do a program. One of the functions is called the 'first function' and the other, the 'second function'. Reduction Formulas. Both of the solution presented below use #int lnx dx = xlnx - x +C#, which can be done by integration by parts. Recall that the idea behind integration by parts is to form the derivative of a product, distribute the derivative, integrate, and rearrange: (3. Sensing a trend, we decide to use integration by parts. These services offer many of the same features as the Drupal webform system and surpass it in some areas. Identifying when to use U-substitution vs Integration by Parts - Duration: 11:39. But note that the power of x has been reduced by one, so you've made some progress. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. We’ll use integration by parts for the first integral and the substitution for the second integral. Get Answer to What is the formula for integration by parts? Where does it come from? Why might you want to use it?. In general, when we have products of sines and cosines in which both exponents are even we will need to use a series of half angle and/or double angle formulas to reduce the integral into a form that we can integrate. Use the integration by parts theorem to calculate the integral. Integrals As a first example, we consider x x3 1 dx. Use integration by parts with. For example, you would use integration by parts for ∫x · ln(x) or ∫ xe 5x. ∫ (3t +t2)sin(2t)dt ∫ ( 3 t + t 2) sin. Integration by parts is not needed here. When finding a definite integral using integration by parts, we should first find the antiderivative (as we do with indefinite integrals), but then we should also evaluate the antiderivative at the boundaries and subtract. Integration Techniques: (lesson 2 of 4) Integration by Parts. Integrate by parts using the formula, where and. Solution Here, we are trying to integrate the product of the functions x and cosx. The situation is somewhat. If we have a function in the. Integration By Parts: Integration by parts is a method of integral that is used to evaluate the integral of two functions when they are given in a multiplication form. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration. (x+4)^ln x dx between 3 and 0 By. There were a significant proportion of candidates who proceeded to integrate u(u + 1) –2. To use integration by parts, we want to make this integral the integral on the right-hand side of the fundamental equation; in other words, we want to pick some u(x) and v(x) so that. 31 Comments on “Integration by parts twice” Pseudonym says: 12 Nov 2010 at 9:58 am [Comment permalink] I've done a lot of integrals by hand in my time, but I don't think I've used integration by parts even once the last 15 years. Here's the formula: Don't try to understand this yet. The integration by parts formula We need to make use of the integration by parts formula which states: Z u dv dx! dx. Advanced Math Solutions - Integral Calculator, integration by parts Integration by parts is essentially the reverse of the product rule. Integration by Substitution : Core Maths : C4 Edexcel June 2013 Q5 (b) : ExamSolutions - youtube Video. New; 16:59. Let u = √x. Integration by parts works with definite integration as well. This method of integration can be thought of as a way to undo the product rule. Finally, we will see examples of how to use Integration by Parts for Indefinite and Definite Integrals, and learn when we would have to use Integration by Parts more than once, as well as how to use a really nifty technique called the Tabular Method (Tic-Tac-Toe Method) for specific cases. Welbilt, Inc. ; If necessary, apply integration by parts twice OR use some trigonometric identity with the ultimate goal of seeing the original integration appear again. Inverse function integration is an indefinite integration technique. If both properties hold, then you have made the correct choice. Since the function lnx is very pleasant to di˙erentiate (ln0 x = 1 x), we could try to choose it as one of the factors. Examples: (1) G—t-…c, a constant. The Fourier series for g. Live Taraweeh Of 16th Ramadan From Badshahi Masjid Lahore Baaghi. Let u= cosx, dv= exdx. It is usually the last resort when we are trying to solve an integral. For example, if the differential is , then the function leads to the. 1 Evaluate Z sin5 xdx. Integration By Parts: Integration by parts is a method of integral that is used to evaluate the integral of two functions when they are given in a multiplication form. We want to choose $$u$$ and $$dv$$ so that when we compute $$du$$ and $$v$$ and plugging everything into the Integration by Parts formula the new integral we get is one that we can do or will at least be an integral that will be easier to deal with. This is called integration by parts. The Organic Chemistry Tutor 62,013 views. A Reduction Formula When using a reduction formula to solve an integration problem, we apply some rule to rewrite the integral in terms of another integral which is a little bit simpler. Formula used: The formula for integration by parts in terms of u and v is given by ∫ u d v = u v − ∫ v d u < Given: The reduction formula, ∫ (ln x) n d x = x (ln x) n − n ∫ (ln x) n − 1 d x. integration. When we apply the magic formula for. Suppose you have to ∫e x sin(x)dx. Posted by. I imagine you are talking about a procedure called integration by parts. Parts (a) and (b): Trapezium Rule : C4 Edexcel January 2013 Q4 (a) (b) : ExamSolutions Maths Revision Tutorials - youtube Video. ( 2 t) d t Solution. So many that I can't show you all of them. ∫ x cos ( x) d x \int x\cos\left (x\right)dx. References 1. Let f '(x) = e x, so that f(x) = e x, and g(x) = cos x, which differentiates to g '(x) = -sin x. So far, everything I've told you may be difficult for you to assimilate, but don't worry. SOLUTION 9 : Integrate. Here, the integrand is usually a product of two simple functions (whose integration formula is known beforehand). (1 pt) Evaluate the indenite integral. There are, after all, lots of ways to put a vector differential form into an equation, and (at least) three dimensionalities of integral you might be trying to do!. Intermediate steps. Below you can find Nethams original post. Of course, in order for it to work, we need to be able to write down an antiderivative for. We recall that in one dimension, integration by parts comes from the Leibniz product rule for di erentiation,. (Remember that the formula for the volume of a. Step-by-Step Calculator integration by parts. The original integral is reduced to a difference of two terms. This time, let Now, integration by parts produces. What about the first term ? The problem here is that one factor, , is absorbing too many of the derivatives. These can sometimes be tedious, but the technique is straightforward. Integration by parts definition, a method of evaluating an integral by use of the formula, ∫udv = uv − ∫vdu. This is the currently selected item. The album chronicles his journey of moving from Australia to Copenhagen, a musical migration that is quite rare for Aussies compared to Los Angeles or London. Beyond these cases, integration by parts is useful for integrating the product of more than one type of function or class of function. We'll do this example twice, once with each sort of notation. RULE OF THUMB: The first step to use Integration by Parts is to pick your "u" and "dv". Answer to: Let A_n (x) = \int x^n e^{2x} \ dx (a) Use integration by parts. Integration by Parts. Hence the original integral is: Z 1 0 tan−1 xdx = π 4 − ln2 2. Last Post; Aug 23, 2010; Replies 2 Views 1K. So, we are going to begin by recalling the product rule. If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the. LIATE An acronym that is very helpful to remember when using integration by parts is LIATE. Integration by parts is the inverse of the product rule. INTEGRATION EXAM – STUDY GUIDE. (ii) Introduce the intermediary functions u(x) and v(x) as:. Use integration by parts, along with problem 1, to prove that Γ(n+ 1) = nΓ(n) if n is a positive integer. For instance, one can verify, and this was indeed the proof I saw, that the formal adjoint of the Dolbeault operator$\bar{\partial}$on complex manifolds is $$\bar{\partial}^* = -* \bar{\partial} \,\,\, *,$$ where$*$is the Hodge star. Integration by parts tells us that if we have an integral that can be viewed as the product of one function, and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already. Wait for the examples that follow. Using Integration by Parts. \int x e^{3x} dx +C Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. Let f '(x) = e x, so that f(x) = e x, and g(x) = cos x, which differentiates to g '(x) = -sin x. I = e2x sinx−2. Integration by parts intro. Banks and fintechs have been in competition for some time, as the fintech industry flourishes, banks increasingly fear losing. Live Taraweeh Of 16th Ramadan From Badshahi Masjid Lahore Baaghi. and thus, by. Round the answer to two decimal places if necessary. u is the function u(x) v is the function v(x). integrate by four methods: the substitution u=cosx, the substitution u=sinx, the identity sin2x=2sinxcosx and integration by parts. When working with the method of integration by parts, the differential of a function will be given first, and the function from which it came must be determined. Question: Use integration parts to find the integral. We use integration by parts to obtain the result, only to come across a small snag: u = e x; dv/dx = sin x So, du/dx = e x; v = -cos x ∫e x sin(x)dx = -e x cos x + ∫ e x cos x dx 1 Now, we have to repeat the integration process for ∫ e x cos x dx, which is as follows: u = e x; dv/dx = cos x. Integration by Substitution: Definite Integrals; Evaluate the definite integral using integration by parts with Way 2. Let me explain why. How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Use integration by parts again, with u = e2x and dv dx = sinx , giving du dx = 2e 2x and v = −cosx Toc JJ II J I Back. This example is to show how to solve such a problem. Integration By Parts Arianne Reidinger The purpose of this application is to show how the matrix of a linear transformation may be used to calculate antiderivatives usually found by integration by parts. Integrating the product rule with respect to x derives the formula: sometimes shown as. Notice from the formula that whichever term we let equal u we need to differentiate it in order to. This is the currently selected item. ∫ ln ( x 2 − x + 2 ) d x. I use the form: #int u dv = uv-intvdu#. Integrating by parts is the integration version of the product rule for differentiation. The integration by parts formula can also be written more compactly, with u substituted for f (x), v substituted for g (x), dv substituted for g' (x) and du substituted for f' (x): ∫ u dv = uv − ∫ v du. We write + C instead of - C since either way we're describing the same family of functions. These services offer many of the same features as the Drupal webform system and surpass it in some areas. We are required to find; provided that Rule for integration of is: We integrate both parts and […]. Then du= sinxdxand v= ex. Answer to: Use integration by parts to find the integral. Question: Use integration parts to find the integral. Section 5: Tips on using solutions 14 5. Solve the following integrals using integration by parts: (a) Z x2 sin(x)dx, (b) Z (2x+ 1)ex dx, (c) Z xsin(3 x) dx, (d) Z 2xarctan(x)dx, (e) Z ln(x)dx 4. Let: u = ln(3x) v = x2 2: du = 1 x dx dv = xdx: Recalling the formula for integration by parts, R udv = uv R vdu, and plugging in, we get: Z xln(3x)dx = 1 2 x2 ln(3x) Z 1 2 xdx = 1 2 x2 ln(3x) 1 4 x2 + C : 23. We'll do this example twice, once with each sort of notation. 7, respectively, of expenses for the settlement of. An Integral form ∫f(z)dz without upper and lower limits is also called an anti-derivative. The integral table in the frame above was produced TeX4ht for MathJax using the command sh. , which implies ) to prove the formula: Proof. Use integration by parts to find. It is widely encountered in physics and engineering, partially because of its use in integration. Then du= cosxdxand v= ex. In fact it is by the i. Integration by parts is a technique for evaluating integrals whose integrand is the product of two functions. ∫ () Integrals that would otherwise be difficult to solve can be put into a simpler form using this method of integration. *Use the integration by substitution when you see that by some means you can adjust the derivative of the function in the denominator. Problem: Evaluate the following integrals using integration by parts: Constructed with the help of Eric Howell. Tips on using solutions When looking at the THEORY, ANSWERS, INTEGRALS, or TIPS pages, use the Back button (at the bottom of the page) to return to the exercises. Example 1: Evaluate the following integral. For example, if the differential is , then the function leads to the. using integration by parts. To use the integration by parts formula we let one of the terms be dv dx and the other be u. Enter the function to Integrate: With Respect to: Evaluate the Integral: Computing Get this widget. We write + C instead of – C since either way we're describing the same family of functions. Integration by parts is used for evaluating integrals of products of functions: {eq}\int f(x)g(x)dx {/eq} where the product of functions is not easily integrable as shown. Aug 29, 2015 #int(x * e^(4x)dx) = 1/16 * e^(4x) * (4x-1) + c#. If u and v are functions of x, the product rule for differentiation that we met earlier gives us: d d x ( u v) = u d v d x + v d u d x. Consider a Schwartz function , and denote by its Fourier transform. Let me explain why. EXAMPLE 4 Repeated Use of Integration by Parts Find Solution The factors and sin are equally easy to integrate. The idea it is based on is very simple: applying the product rule to solve integrals. (1 pt) Use integration by parts to evaluate the integral. where C C C is the constant of integration. Integration by parts is used in many types of integrals. Integration : C4 Edexcel January 2013 Q2 : ExamSolutions Maths Revision Tutorials - youtube Video. The goal of this video is to try to figure out the antiderivative of the natural log of x. Live Taraweeh Of 16th Ramadan From Badshahi Masjid Lahore Baaghi. Question: Use integration parts to find the integral. Trapezium Rule : Edexcel Core Maths C4 June 2012 Q7 (a) : ExamSolutions Maths Revision - youtube Video. The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. Last Post; Aug 23, 2010; Replies 2 Views 1K. Live Taraweeh Of 16th Ramadan From Badshahi Masjid Lahore Baaghi. So we have. 1 Introduction and Definition In this section we introduce the notion of the Laplace transform. Sanaullah Bhutto 104 views. Using the fact that integration reverses differentiation we'll. Therefore,. (5) (d) Use integration by parts to evaluate this integral, and hence find the value of I correct to 4 significant figures, showing all the steps in your working. Integration by parts intro. Integration by parts: ∫x⋅cos(x)dx. Integration by parts: ∫ln(x)dx. There are many ways to integrate by parts in vector calculus. Integration by Parts Calculator. Then u' = 1 and v = e x. Then Z exsinxdx= exsinx excosx Z. Integration by Parts using Taylor Series ::mupad( Matlab) For Detailed View Download PDF. Let u = x the du = dx. Techniques of Integration - Integration by Parts. ∫ xcos(x)dx, using the following formula. Integration by Parts Integration by parts is a method we can use to evaluate integrals like Z f(x)g(x)dx where the term f(x) is easy to differentiate, and g(x) is easy to integrate. For instance, one can verify, and this was indeed the proof I saw, that the formal adjoint of the Dolbeault operator$\bar{\partial}$on complex manifolds is $$\bar{\partial}^* = -* \bar{\partial} \,\,\, *,$$ where$*$is the Hodge star. This is the currently selected item. Choose "Evaluate the Integral" from the topic selector and click to. Note that for a closed Riemannian manifold, with , this shows that. Husch and University of Tennessee, Knoxville, Mathematics Department. ( 2 t) d t Solution. In this tutorial we shall find the integral of the x Cos2x function. Solution Here, we are trying to integrate the product of the functions x and cosx. We'll do this example twice, once with each sort of notation. Consider the following table: Z u dv ⇒ + u dv − du v The first column switches ± signs, the second column differentiates u, and. Integration by Parts. Integration By Parts The process of finding the integral of the composition of two or more functions by taking them function one and function two is known as the integration by parts. Using the Integration by Parts formula. Using prime notation, take. Then we solve for our bounds of integration : [0,3] Let's do an example where we must integrate by parts more than once. Use integration by parts to evaluate the given integral Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply Archives. Let’s verify this and see if this is the case. Integration by parts is used for evaluating integrals of products of functions: {eq}\int f(x)g(x)dx {/eq} where the product of functions is not easily integrable as shown. Integration by parts is essentially the product rule for differentiation inverted:. 3) where if the products (as will often be the case when and and have compact support) the process throws the derivative from one function over to the other'':. The second factor then has to be the constant function 1. du =2x dx v sin3x 3 1 = So, x x dx x x x x dx − ∫ = ∫ sin3 3 1 sin3 2 3 1 cos32 or x x −∫ x x dx sin3 3 2 sin3 3 2 1 We see that it is necessary to perform integration. In this lesson, walk through three examples of the proper way to use the integration by parts formula, using two different forms of the formula. (2) Evaluate. Our professor states that i can get RHS out of LHS using integration by parts: $$\\int\\limits_0^x \\! \\frac. The integration by parts formula will convert this integral, which you can't do directly, into a simple product minus an integral you'll know how to do. We plug all this stuff into the formula: Since the integral of e x is e x + C, we have. Z xexdx= using (u= x du= dx dv= exdx v= ex = xe x Z e dx = xe x e + c: Solution of Example 4. Let f '(x) = e x, so that f(x) = e x, and g(x) = cos x, which differentiates to g '(x) = -sin x. Let u = x the du = dx. Example $$\PageIndex{1}$$: Using Integration by Parts. Thirty integrals are found using Integration by Parts and Integration by Substitution. use integration by parts to find an antiderivative of the integrand and then ; use the FTC to evaluate the integral. There are also some electronics applications in this section. Use integration by parts to determine which of the reduction formulas is correct. XML Messaging format is specified for exchanging structured information in the implementation of Web Services in computer networks. Integration by parts: ∫x⋅cos(x)dx. For dv/dx I am choosing e^2x, and therefore v is found by integration and is ½ e^2x, which is a simple integration solution to find v. Integrating using linear partial fractions. Math · AP®︎ Calculus BC · Integration and accumulation of change · Using integration by parts Integration by parts: ∫x⋅cos(x)dx AP Calc: FUN‑6 (EU) , FUN‑6. And it's not completely obvious how to approach this at first, even if I were to tell you to use integration by parts, you'll say, integration by parts, you're looking for the antiderivative of something that can be expressed as the product of two functions. Evaluate integral from 0 to pi/2 of xsin(x) with respect to x. Example $$\PageIndex{1}$$: Using Integration by Parts. Identifying when to use U-substitution vs Integration by Parts - Duration: 11:39. If you’re just using integration by parts in general to find the integral of three functions, basically. Let dv = e x dx then v = e x. Since we then apply integration by parts, This was the first requested formula. I believe that the best way to get good at it is to practice, a lot. If f and f^(-1) are inverses of each other on some closed interval, then intf(x)dx=xf(x)-intf^(-1)(f(x))f^'(x)dx, (1) so intf(x)dx=xf(x)-G(f(x)), (2) where G(x)=intf^(-1)(x)dx. Integration by Parts With Trig and Exponential : Here we are going to see how we use the method "Integration by Parts" with some example problems. So many that I can't show you all of them. I am trying to solve for \\int_0^∞ (20000x)/(x+100)^3 dx = 20000 \\int_0^∞ x/(x+100)^3 dx I think I can solve it using integration by parts, but it will be very tedious (especially with the (x+100)^-3 and I'm not sure whether this is the best way to go. ∫ 0 π e cos t sin 2 t d t. Here we choose u = xn because u = nx n −1 is a simpler (lower degree) function. Integration by Substitution (example to try) : ExamSolutions Maths Revision - youtube Video. Use integration by substitution which is explained with a link in the source: ∫ cos√x dx. (1 pt) Use integration by parts to evaluate the integral. In primary school, we learned how to find areas of shapes with straight sides (e. Let the factor without dx […]. However, as we discussed in the Integration by Parts section, the two answers will differ by no more than a constant. Round the answer to two decimal places if necessary. Even though it's a simple formula, it has to be applied correctly. Again we’ll use integration by parts to find a reduction formula. (2) Evaluate. The integral table in the frame above was produced TeX4ht for MathJax using the command sh. Using the formula for integration by parts Example Find Z x cosxdx. In this lesson, walk through three examples of the proper way to use the integration by parts formula, using two different forms of the formula. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. In fact, there are more integrals that we do not know how to evaluate analytically than those that we can; most of them need to be calculated numerically!. du =2x dx v sin3x 3 1 = So, x x dx x x x x dx − ∫ = ∫ sin3 3 1 sin3 2 3 1 cos32 or x x −∫ x x dx sin3 3 2 sin3 3 2 1 We see that it is necessary to perform integration. How-ever, in many cases it is not practical. Integration by Parts Graphs a function f (x)=g(x)h'(x) and the area under the graph of f (x) for a given interval, and shows the modifications made to f (x) and the area when considering u=g(x) and v=h(x) as independent variables, as when carrying out the integral using the technique of Integration by Parts. Prove the reduction formula Z xnex dx = xnex n Z xn 1ex dx. R ex dx: just integrate 4. We write + C instead of - C since either way we're describing the same family of functions. ∫ e^x sin x dx: This is a lovely example of integration by parts where the term you are trying to integrate will keep repeating and you end up going in circles. Switkes, A quotient rule integration by parts formula. Note that you should take care to avoid the circular trap. I = e2x sinx−2. New; 16:59. What we're going to do in this video is review the product rule that you probably learned a while ago. Integration by parts illustrates it to be an extension of the factorial:. ∫ () Integrals that would otherwise be difficult to solve can be put into a simpler form using this method of integration. Use integration by parts to solve the following integral ∫5x cos(4x)dx. And it's not completely obvious how to approach this at first, even if I were to tell you to use integration by parts, you'll say, integration by parts, you're looking for the antiderivative of something that can be expressed as the product of two functions. Use integration by parts to evaluate the integral. Suppose you have to ∫e x sin(x)dx. Strange Sums. If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place. When finding a definite integral using integration by parts, we should first find the antiderivative (as we do with indefinite integrals), but then we should also evaluate the antiderivative at the boundaries and subtract. Integration by Parts - Indefinite Integrals - Calculus II is a prerequisite for many popular college majors, including pre-med, engineering, and physics. Integration by parts is essentially the product rule for differentiation inverted:. Use the solutions intelligently. Use integration by parts to show that Γ(r) = (r - 1) Γ(r - 1). ∫ x cos ( x) d x \int x\cos\left (x\right)dx. Integrating using linear partial fractions. Tutorials with examples and detailed solutions and exercises with answers on how to use the technique of integration by parts to find integrals. 2 e x sin 7x dx 9. Enter the function to Integrate: With Respect to: Evaluate the Integral: Computing Get this widget. integration by parts back to top Tricks: If one of the functions is a polynomial (say nth order) and the other is integrable n times, then you can use the fast and easy Tabular Method:. The tangent function is not really this type of product; it is a function where the a function and its derivative are present. Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse. integration. Plotting the derivative of NDSolve divided by the solution. Success in using the method rests on making the proper choice of and. Welbilt, Inc. ∫ u ⋅ d v = u ⋅ v − ∫ v ⋅ d u \displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du. Again we’ll use integration by parts to find a reduction formula. For example, if the differential is , then the function leads to the. First, to use integration by parts we define the following functions. The Integral Calculator solves an indefinite integral of a function. (ii) Introduce the intermediary functions u(x) and v(x) as:. Bonus Evaluate R 1 0 x 5e x using integration by parts. Get an answer for 'Use integration by parts to integrate Integration sign, x^5 ln (x) dx' and find homework help for other Math questions at eNotes. For instance, one can verify, and this was indeed the proof I saw, that the formal adjoint of the Dolbeault operator \bar{\partial} on complex manifolds is$$\bar{\partial}^* = -* \bar{\partial} \,\,\, *,$$where$*\$ is the Hodge star. To apply this formula we must choose dv so that we can integrate it! Frequently, we choose u so that the derivative of u is simpler than u. Integration by parts: ∫x⋅cos(x)dx. Let: u = ln(3x) v = x2 2: du = 1 x dx dv = xdx: Recalling the formula for integration by parts, R udv = uv R vdu, and plugging in, we get: Z xln(3x)dx = 1 2 x2 ln(3x) Z 1 2 xdx = 1 2 x2 ln(3x) 1 4 x2 + C : 23. RULE OF THUMB: The first step to use Integration by Parts is to pick your "u" and "dv". To use the integration by parts formula we let one of the terms be dv dx and the other be u. Hence the original integral is: Z 1 0 tan−1 xdx = π 4 − ln2 2. Sanaullah Bhutto 104 views. Typically, Integration by Parts is used when two functions are multiplied together, with one that can be easily integrated, and one that can be easily differentiated. We plug all this stuff into the formula: Since the integral of e x is e x + C, we have. Integrating by using the method of integration by parts is demonstrated here. Using integration by parts iteratively, the singularity at the points near three points a = 0,1,2 can be eliminated in terms containing obtained integrals, and the factors of amplifying round-off. This Demonstration lets you explore various choices and their consequences on some of the standard integrals that can be done using integration by parts. It is easy to make errors, especially sign errors involving the subtraction in the formula. The integration is of the form $I = \int {x\cos 2xdx}$ H. Using the Integration by Parts formula. Answer to: Use integration by parts to find the integral. ∫ 6tan−1( 8 w) dw ∫ 6 tan − 1 ( 8 w) d w Solution. So many that I can't show you all of them. Tanzalin Method can be easier to follow (and could be used to check your work if you have to do Integration by Parts in an examination). Integration by parts is a "fancy" technique for solving integrals. Definite Integration by parts. Engine blocks, engines, transmissions, Connect & Cruise systems and other Chevrolet Performance parts are designed, engineered, and tested by Chevrolet. Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways Advanced Show Ads. Subscribe to get much more:. Notice from the formula that whichever term we let equal u we need to differentiate it in order to. And from that, we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by. Note that you should take care to avoid the circular trap. We recall that in one dimension, integration by parts comes from the Leibniz product rule for di erentiation,. INTEGRATION BY PARTS IN 3 DIMENSIONS We show how to use Gauss’ Theorem (the Divergence Theorem) to integrate by parts in three dimensions. This is called integration by parts. ∫ x sin x d x. One of the difficulties in using this method is determining what function in our integrand should be matched to which part. In a manufacturing firm, different products can share parts that have been designed for multi-use. As you can see, it is really the same expression. The following example illustrates its use. Let the factor without dx […]. Using integration by parts, I k= Z ˇ=2 0 (cos )kd = Z ˇ=2 0 (cos )k 1 cos d = (k 1)(I k 2 I k); so (7. Integration by parts is used for evaluating integrals of products of functions: {eq}\int f(x)g(x)dx {/eq} where the product of functions is not easily integrable as shown. Ravi Vemulapalli Recommended for you. From home or the office, even the beach, you can enjoy the convenience and peace of mind provided by the award-winning RTiPanel app. How-ever, in many cases it is not practical. Integration by Parts.
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https://www.physicsforums.com/threads/error-function-integral.716482/ | # Homework Help: Error-function integral
1. Oct 14, 2013
### Nikitin
1. The problem statement, all variables and given/known data
If you know
$$\int_{-\infty}^{\infty} e^{-\alpha \beta^2} d \beta = \sqrt{\pi / \alpha}$$
Show that
$$\int_{-\infty}^{\infty} \beta^2 e^{-\beta^2} d \beta = \sqrt{\pi}/2$$
3. The attempt at a solution
I have tried integration by parts on the second integral, but to no avail. I just end up with it being equal to zero. What am I to do?
2. Oct 14, 2013
### Staff: Mentor
How did you split the integrand up when you did integration by parts?
3. Oct 14, 2013
### Dick
You don't really need to do integration by parts. Take the derivative with respect to $\alpha$ of both sides of your first equation. Then set $\alpha=1$.
4. Oct 14, 2013
### brmath
This is a very nice trick, and I like it a lot. But it could be Nikitin needs to work on integration by parts.
5. Oct 14, 2013
### Dick
Yes, probably. It's easy enough to do that way too.
6. Oct 15, 2013
### Nikitin
Yeah, I must've done a silly mistake when doing the integration by parts. I set the $\beta^2$ term as u, and the $e^{ -\beta^2}$ term as v'.
Then, (btw, how do I put limits on the left part of a bracket in latex?)
$$\int_{-\infty}^{\infty} \beta^2 e^{-\beta^2} d \beta = \left [ \beta^2 \int e^{-\beta^2} d \beta \right ]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} 2 \beta (\int_{-\infty}^{\infty} e^{-\beta^2} d \beta ) d \beta$$
If I try to insert
$$\int_{-\infty}^{\infty} e^{-\beta^2} d \beta = \sqrt{\pi}$$
into my integration by parts expression, I will just end up with 0. Can somebody please do it properly so I can find my mistake?
PS: Sorry for late reply. Oh, and thanks for all the help and nice tips :)
I don't want to be a bore (I like your method), but how do you know it's possible to differentiate the term inside the integral first, instead of taking the integral first and then differentiate?
Last edited: Oct 15, 2013
7. Oct 15, 2013
### Dick
http://en.wikipedia.org/wiki/Leibniz_integral_rule And for your integration by parts, put $dv=\beta e^{-\beta^2} d\beta$. What does that leave for u?
Last edited: Oct 15, 2013
8. Oct 15, 2013
### Nikitin
I don't completely understand why you'd do that?
Is the integration-by-parts expression in post #6 correct? Why can't I just insert $\int_{-\infty}^{\infty} e^{-\beta^2} d \beta = \sqrt{\pi}$ into the expression and solve?
9. Oct 15, 2013
### Dick
If you are going to set $dv=e^{\beta^2}d\beta$ then v is the INDEFINITE integral of dv. You don't apply the limits till the end. And that expression doesn't have an elementary indefinite integral. Hence my alternate suggestion.
10. Oct 15, 2013
### brmath
Re taking the derivative under the integral, note that the integral is with respect to $\beta$. For purposes of regarding $\alpha$ as the variable to differentiate against, all the $\beta$ stuff is essentially a constant.
Re integration by parts take u(x) = x and v'(x) = x$e^{-\alpha x^2}$. The new integral will pop up looking very like the one you know to be $\sqrt {\pi /\alpha}$.
11. Oct 15, 2013
### Nikitin
Okay, now I fully understand. I didn't think about putting v'(x) = x e^-x^2, as I wasn't aware of
Now everything is clear. Thanks guys!
12. Oct 15, 2013
### Staff: Mentor
Getting back to Dick's method in response #3, I think what your textbook is trying to teach you is the technique of "differentiation under the integral sign." I think this is how the book intended for you to solve the problem. | 2018-06-18T14:06:54 | {
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https://math.stackexchange.com/questions/2660041/find-root-using-fixed-point-iteration-can-this-be-right | Find root using fixed point iteration. Can this be right?
Working on old exams in basic numerical modeling, I have gotten confused by the solution proposal.
Problem: Use the fixed point iteration method to find the root for $f(x)=2x-e^{-x}$ in the interval $(0,1.6)$ with an error less than $0.02$.
My thoughts:
$g(x)=\frac{e^{-x}}{2}\$
$x_{i+1}=g(x_i), i=1,2...\$
Convergence criterion (from book):
$|x_{i+1}-g(x_{i+1})| \le \epsilon=0.02$
With $x_1=1$, $x_5=0.3595185038$ would be the root from my calculations. However, an extra computation is done in the solution proposal. Why? What is it I'm missing? Is the extra computation done due to the truncation of decimals?
Solution proposal:
Book:
i x_i g(x_i) |x_i+1-g(x_i+1)|
1 1.0000000000 0.1839397206 0.2321
2 0.1839397206 0.4159929770 0.0862
3 0.4159929770 0.3298424517 0.0297
4 0.3298424517 0.3595185038 0.0105
5 0.3595185038 0.3490061677
• The exact solution is $W(1/2)=0.35173371124919582602\dots$ so the proposed solution stops exactly when it needs to get the error less than $0.02$. Your $x_5$ doesn't seem to be right – Yuriy S Feb 21 '18 at 10:22
• How do your computations look like? your $x_5$ doesn't seem correct. – user144410 Feb 21 '18 at 10:28
• Added table with computations. – Numerical Newbie Feb 21 '18 at 10:40
• You have a typo in your $x_5$, when you first write it in the post – Yuriy S Feb 21 '18 at 11:10
• Good observation. Corrected. – Numerical Newbie Feb 21 '18 at 11:11
I think this is again some artifact of a manual evaluation of a computation result without "automatic" stopping condition. If the stopping condition were actually implemented, one would get your result. Also, it seems rather disadvantageous to write the step as $i\to i+1$. It makes no sense to compare in step $i$ the sequence elements $x_{i+1}$ and $x_{i+2}$. The algorithm is
repeat
i=i+1
x[i+1] = g(x[i])
until abs(x[i+1]-g(x[i+1]))<eps
return x[i+1]
but should be better implemented as
repeat
i=i+1
x[i+1] = g(x[i])
until abs(x[i+1]-x[i]) < eps
return x[i]
with only half the calls to $g$ and comparing $x_i$ and $x_{i+1}$ in step $i$.
• So if I understand you correctly, the extra step in the solution proposal is unnecessary, meaning that my answer is correct? – Numerical Newbie Feb 21 '18 at 11:07
• Yes, as only $x_6$ is needed for the error estimate, the computation of $x_7=g(x_6)$ is merely illustration for the continuing stabilization in the leading digits. – LutzL Feb 21 '18 at 11:14 | 2019-07-21T17:01:01 | {
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https://math.stackexchange.com/questions/2526959/gaussian-type-integral-int-infty-infty-frac-mathrme-a2-x21 | # Gaussian type integral $\int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 x^2}}{1 + x^2} \mathrm{d}x$
When working a proof, I reached an expression similar to this:
$$\int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 x^2}}{1 + x^2} \mathrm{d}x$$
I've tried the following:
1. I tried squaring and combining and converting to polar coordinates, like one would solve a standard Gaussian. However, this yielded something which seems no more amenable to a solution:
$$\int_{\theta=0}^{\theta=2\pi} \int_{0}^{\infty} \frac{r \mathrm{e}^{-a^2 r^2}}{(1 + r^2 \sin^2(\theta))(1 + r^2 \cos^2(\theta))} \mathrm{d}r \mathrm{d}\theta$$
2. I tried doing a trig substitution, t = tan u, and I have no idea what to do from there.
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm{e}^{-a^2 \tan^2(u)} \mathrm{d}u$$
3. I looked into doing $u^2 = 1 + x^2$ but this gives us a ugly dx that I don't know how to handle, and moreover, I think I'm breaking my limits of integration (because Mathematica no longer solves it.):
$$u^2 = 1 + x^2$$
$$2 u \mathrm{d}u = 2 x \mathrm{d}x$$
$$\mathrm{d}x = \frac{u}{\sqrt{u^2 - 1}}$$ $$\mathrm{e}^{a^2} \int_{-\infty}^{\infty} \frac{\mathrm{e}^{-a^2 u^2}}{u \sqrt{u^2 - 1}} \mathrm{d}u$$
4. I looked into some form of differentiation under the integral, but that didn't seem to yield anything that looked promising. (I checked parameterizing x^2 to x^b in both places, and in either place, and nothing canceled cleanly.)
I have a solution from Mathematica, it's:
$$\pi e^{a^2} \text{erfc}(a)$$
But I'd like to know how to arrive at this. I'm sure it's something simple I'm missing.
• From where? I can pull out a factor of pi from part 3, but that just gives me an integral that I can't solve that is supposedly equal to erfc(a), but doesn't look like any form of erfc(a) that I recognize. – OmnipotentEntity Nov 19 '17 at 2:54
• This is a good example where we can use Parseval's theorem together with : $e^{-\pi x^2}, e^{-\pi \xi^2}$ and $\frac{1/\pi}{1+x^2}, e^{-2\pi |\xi|}$ are Fourier transform pairs. Thus it reduces to $\int_0^\infty e^{-2 \pi \xi -\pi \xi^2}d\xi$ which is easily seen to be what mathematica claims. – reuns Nov 19 '17 at 2:59
• hint: try differentation under the integral sign – tired Nov 19 '17 at 3:15
• Look at $F(a) = \int_{-\infty}^\infty \frac{e^{-a x^2}}{1+x^2}dx, G(a) = \pi e^{-a}+F(a) = \int_{-\infty}^\infty \frac{e^{-a(1+ x^2)}}{1+x^2}dx$ and $(G(a^2))'$ – reuns Nov 19 '17 at 3:17
• Also, complex integration with a contour on upper half plane gives us the result as we have $z=i$ in this domain! – Nosrati Nov 19 '17 at 3:21
Let $F$ be the function $$F(a)=\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx$$ We take the derivative w.r.t $a$ $$F^{\prime}(a)=\frac{d}{da}\left(\int_{-\infty}^{\infty}\frac{e^{-a^{2}x^{2}}}{1+x^2}dx\right)=\int_{-\infty}^{\infty}\frac{d}{da}\left(\frac{e^{-a^{2}x^{2}}}{1+x^2}\right)dx =\int_{-\infty}^{\infty}\frac{-2ax^{2}e^{-a^{2}x^{2}}}{1+x^2}dx$$ $$=\int_{-\infty}^{\infty}\frac{-2a\big((x^{2}+1)-1\big)e^{-a^{2}x^{2}}}{1+x^2}dx =-2a\int_{-\infty}^{\infty}e^{-a^{2}x^{2}}dx+2aF(a) =-2a\sqrt{\frac{\pi}{a^2}}+2aF(a)$$ Then $$F^{\prime}(a)=2a\left(F(a)-\sqrt{\pi}\,\frac{1}{\vert{a}\vert}\right) =2aF(a)-2\sqrt{\pi}\mathrm{sign}(a).$$ Then you have a differential equation: $$F^{\prime}(a)-2a\,F(a)=-2\sqrt{\pi}\mathrm{sign}(a)$$ with initial condition $F(0)=\pi$. This fisrt order ode has integrant factor: $$\mu(a)=\displaystyle{e^{\displaystyle{\int{-2ada}}}}=e^{-a^2}$$ Then $$\left(e^{-a^2}F(a)\right)^{\prime}=-2\sqrt{\pi}\mathrm{sign}(a) e^{-a^2}$$ this implies $$e^{-a^2}F(a)=-2\sqrt{\pi}\int{\mathrm{sign}(a) e^{-a^2}}da+C$$ Finaly $$F(a)=e^{a^2}\left(C-2\sqrt{\pi}\mathrm{sign}(a)\int{e^{-a^2}da}\right)$$
• Excellent work. This simplifies down (for a > 0) to $\pi \mathrm{e}^{a^2} (-\mathrm{erf}(a) + C)$, but it's not at all clear to me how to obtain C (which should be 1). – OmnipotentEntity Nov 19 '17 at 4:19
• Just, use that $F(0)=\pi$. :) – Hector Blandin Nov 19 '17 at 4:21
• Oh of course facepalm I was substituting back too soon, and I had an extra factor of "a" in the derivation. How foolish. Thanks so much. – OmnipotentEntity Nov 19 '17 at 4:23
Let $f(a)=\int_{-\infty}^\infty \frac{e^{-a^2x^2}}{1+x^2}\,dx$. Then, we have
\begin{align} f(a)&=2\int_0^\infty e^{-a^2x^2}\int_0^\infty e^{-s(1+x^2)}\,ds\,dx\\\\ &=2\int_0^\infty e^{-s}\int_0^\infty e^{-(s+a^2)x^2}\,dx\,ds\\\\ &=\int_0^\infty e^{-s} \frac{\sqrt{\pi}}{\sqrt{s+a^2}}\,ds\\\\ &=\sqrt{\pi}e^{a^2}\int_0^\infty \frac{e^{-(s+a^2)}}{\sqrt{s+a^2}}\,ds\\\\ &=\sqrt{\pi}e^{a^2}\int_{a^2}^\infty \frac{e^{-t}}{\sqrt t}\,dt\\\\ &=2\sqrt{\pi}e^{a^2}\int_{|a|}^\infty e^{-u^2}\,du\\\\ &=\pi e^{a^2}\text{erfc}(|a|) \end{align}
Here is another approach where we first make use of an auxiliary function.
Before proceeding we recall the definitions for the error function $\text{erf}(x)$ and the complementary error function $\text{erfc}(x)$: $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} \, dt$$ and $$\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int^\infty_x e^{-t^2} \, dt,$$ respectively such that $$\text{erf}(x) = 1 - \text{erfc}(x).$$
The idea here is to consider an auxiliary function, related to our function of interest $f(a)$, but which turns our to be a constant function for all $a$ in its domain.
Start by considering the following auxiliary function $$I(a) = \left (\int^a_0 e^{-t^2} \, dt \right )^2 + \int^1_0 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt, \,\, a > 0. \tag1$$ Note the term appearing between the brackets is nothing more than the error function. On differentiating the auxiliary function with respect to $a$ we obtain $$I'(a) = 2 e^{-a^2} \int^a_0 e^{-t^2} \, dt - 2a e^{-a^2} \int^1_0 e^{-a^2 t^2} \, dt.$$ In obtaining this result, Leibniz' rule for differentiating under the integral sign has been used. In the second integral, if a substitution of $u = at$ is made, the result $I'(a) = 0$ quickly follows showing the auxiliary function is indeed constant for all $a > 0$. To find the value for this constant, letting $a \to 0^+$ gives $$I(a) \to \int^1_0 \frac{dt}{1 + t^2} = \frac{\pi}{4},$$ so that $I(a) = \pi/4$ for all $a > 0$.
As the first of the integrals appearing in (1) can be written in terms of the error function we have $$\int^1_0 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt = \frac{\pi}{4} \left (1 - \text{erf}^2 (a) \right ). \tag2$$
A similar thing can be done for the complementary error function. In this case we start by considering the following auxiliary function $$J(a) = \left (\int^\infty_a e^{-t^2} \, dt \right )^2 - \int^\infty_1 \frac{e^{-a^2 (t^2 + 1)}}{1 + t^2} \, dt, \,\, a > 0. \tag3$$ Again observe the term appearing between the brackets in nothing more than the complementary error function. On differentiating with respect to $a$ we have $$J'(a) = -2 e^{-a^2} \int^\infty_a e^{-t^2} \, dt + 2a e^{-a^2} \int^\infty_1 e^{-a^2 t^2} \, dt.$$ A substitution of $u = at$ in the second integral once again reduces the derivative of the auxiliary function to zero, showing $J(a)$ is constant. Letting $a \to \infty$ in (3) we see $J(a) \to 0$. Thus $J(a) = 0$ for all $a > 0$. Writing the first of the integrals in (3) in terms of the complementary error function, one finds $$\int^\infty_1 \frac{e^{-a^2 (1 + t^2)}}{1 + t^2} \, dt = \frac{\pi}{4} \text{erfc}^2 (a). \tag4$$
Adding (2) to (4) yields \begin{align*} \int^\infty_0 \frac{e^{-a^2(1 + t^2)}}{1 + t^2} \, dt &= \frac{\pi}{4} \left [\text{erfc}^2 (a) + 1 - \text{erf}^2 (a) \right ]\\ &= \frac{\pi}{4} \left [\text{erfc}^2 (a) + 1 - (1 - \text{erfc}(a))^2 \right ]\\ &= \frac{\pi}{2} \text{erfc} (a). \end{align*} Rearranging gives $$e^{-a^2} \int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \frac{\pi}{2} \text{erfc}(a),$$ or $$\int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \frac{\pi}{2} e^{a^2} \text{erfc}(a).$$
So for $f(a)$ we finally have $$f(a) = 2 \int^\infty_0 \frac{e^{-a^2 t^2}}{1 + t^2} \, dt = \pi e^{a^2} \text{erfc}(a), \quad a > 0.$$ | 2021-01-26T02:59:26 | {
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https://www.jobilize.com/course/section/section-summary-centripetal-force-by-openstax?qcr=www.quizover.com | # 10.2 Centripetal force (Page 3/10)
Page 3 / 10
$N\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{\mathrm{mv}}^{2}}{r}\text{.}$
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is $N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ , and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,
$N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\text{mg}\text{.}$
Now we can combine the last two equations to eliminate $N$ and get an expression for $\theta$ , as desired. Solving the second equation for $N=\text{mg}/\left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)$ , and substituting this into the first yields
$\text{mg}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }{\text{cos}\phantom{\rule{0.25em}{0ex}}\theta }=\frac{{\text{mv}}^{2}}{r}$
$\begin{array}{rrr}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{tan}\left(\theta \right)& =& \frac{{\mathrm{mv}}^{2}}{r}\\ \text{tan}\phantom{\rule{0.25em}{0ex}}\theta & =& \frac{{v}^{2}}{\text{rg.}}\end{array}$
Taking the inverse tangent gives
$\theta ={\text{tan}}^{-1}\left(\frac{{v}^{2}}{\text{rg}}\right)\phantom{\rule{0.25em}{0ex}}\text{(ideally banked curve, no friction).}$
This expression can be understood by considering how $\theta$ depends on $v$ and $r$ . A large $\theta$ will be obtained for a large $v$ and a small $r$ . That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that $\theta$ does not depend on the mass of the vehicle.
## What is the ideal speed to take a steeply banked tight curve?
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.
Strategy
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.
Solution
Starting with
$\text{tan}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{v}^{2}}{\text{rg}}$
we get
$v=\left(\text{rg}\phantom{\rule{0.25em}{0ex}}\text{tan}\phantom{\rule{0.25em}{0ex}}\theta {\right)}^{1/2}\text{.}$
Noting that tan 65.0º = 2.14, we obtain
$\begin{array}{lll}v& =& {\left[\left(\text{100 m}\right)\left(9.80 m{\text{/s}}^{2}\right)\left(2\text{.}\text{14}\right)\right]}^{1/2}\\ & =& \text{45.8 m/s.}\end{array}$
Discussion
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.
Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.
## Take-home experiment
Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.
## Phet explorations: gravity and orbits
Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!
## Section summary
• Centripetal force ${\text{F}}_{\text{c}}$ is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity $v$ and has magnitude
$\phantom{\rule{0.25em}{0ex}}{F}_{\text{c}}={\text{ma}}_{\text{c}}\text{,}$
which can also be expressed as
$\left(\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ \begin{array}{}\text{or}\\ {F}_{\text{c}}=\text{mr}{\omega }^{2}\end{array}\end{array},}$
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
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# L n 1 l n 2 l nn 1 1 1039 where the l ij j 1 n 1 i j
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. . . . . . . . . . . . . . l n 1 l n 2 · · · l n,n - 1 1 , (10.39) where the l ij , j = 1 , . . . , n - 1, i = j + 1 , . . . , n are the multipliers (com- puted after all the rows have been rearranged), we arrive at the anticipated factorization PA = LU . Incidentally, up to sign, Gaussian elimination also produces the determinant of A because det( PA ) = ± det( A ) = det( LU ) = det( U ) = a (1) 11 a (2) 22 · · · a ( n ) nn (10.40) and so det( A ) is plus or minus the product of all the pivots in the elimination process. In the implementation of Gaussian elimination the array storing the aug- mented matrix A b is overwritten to save memory. The pseudo code with partial pivoting (assuming a i,n +1 = b i , i = 1 , . . . , n ) is presented in Algo- rithm 3. 10.2.1 The Cost of Gaussian Elimination We now do an operation count of Gaussian elimination to solve an n × n linear system Ax = b . We focus on the elimination as we already know that the work for the step of backward substitution is O ( n 2 ). For each round of elimination, j = 1 , . . . , n - 1, we need one division to compute each of the n - j multipliers and ( n - j )( n - j + 1) multiplications and to ( n - j )( n - j + 1) sums (subtracts) perform the eliminations. Thus, the total number number of operations is W ( n ) = n - 1 X j =1 [2( n - j )( n - j + 1) + ( n - j )] = n - 1 X j =1 2( n - j ) 2 + 3( n - j ) (10.41) and using (10.10) and m X i =1 i 2 = m ( m + 1)(2 m + 1) 6 , (10.42)
162 CHAPTER 10. LINEAR SYSTEMS OF EQUATIONS I Algorithm 3 Gaussian Elimination with Partial Pivoting 1: for j = 1 , . . . , n - 1 do 2: Find m such that | a mj | = max j i n | a ij | 3: if | a mj | = 0 then 4: stop . Matrix is singular 5: end if 6: a jk a mk , k = j, . . . , n + 1 . Exchange rows 7: for i = j + 1 , . . . , n do 8: m a ij /a jj . Compute multiplier 9: a ik a ik - m * a jk , k = j + 1 , . . . , n + 1 . Elimination 10: a ij m . Store multiplier 11: end for 12: end for 13: for i = n, n - 1 , . . . , 1 do . Backward Substitution 14: x i a i,n +1 - n X j = i +1 a ij x j ! /a ii 15: end for we get W ( n ) = 2 3 n 3 + O ( n 2 ) . (10.43) Thus, Gaussian elimination is computationally rather expensive for large systems of equations. 10.3 LU and Choleski Factorizations If Gaussian elimination can be performed without row interchanges, then we obtain an LU factorization of A , i.e. A = LU . This factorization can be advantageous when solving many linear systems with the same n × n matrix A but different right hand sides because we can turn the problem Ax = b into two triangular linear systems, which can be solved much more economically in O ( n 2 ) operations. Indeed, from LUx = b and setting y = Ux we have Ly = b, (10.44) Ux = y. (10.45)
10.3. LU AND CHOLESKI FACTORIZATIONS 163 Given b , we can solve the first system for y with forward substitution and then we solve the second system for x with backward substitution. Thus, while the LU factorization of A has an O ( n 3 ) cost, subsequent solutions to the linear system with the same matrix A but different right hand sides can be done in O ( n 2 ) operations. When can we obtain the factorization A = LU ? the following result provides a useful sufficient condition. | 2021-07-24T17:39:29 | {
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https://math.stackexchange.com/questions/3101264/distribution-at-first-time-a-sum-reaches-a-threshold | # Distribution at First Time a Sum Reaches a Threshold
Consider the following problem. Roll a die many times, and stop when the total exceeds $$M$$, for some prescribed threshold $$M$$. Call this time $$\tau$$, and call the running score after $$n$$ rolls $$X_n$$.
What is the distribution of $$X_\tau$$?
Of course $$X_\tau \in [M,M+5]$$. However, I can get little beyond this. Moreover, for small $$M$$ the distribution should be very sensitive, and I doubt have a nice form. However, if I define $$X'_\tau = X_\tau - M$$, then it seems to me that $$X'_\tau$$ should have a limiting distribution at $$M \to \infty$$.
Extra comments. $$\quad$$ Note that there are various related questions here on maths.SE, but these (as far as I have seen) are about determining $$\tau$$, in particular its expectation, $$E(\tau)$$.
Also, one can solve for $$E(\tau)$$ directly. If I write $$k_M$$ for $$E(\tau)$$ with threshold $$M$$, then $$\textstyle k_M = \tfrac16 \sum_{j=1}^6 k_{M-j} + 1,$$ and writing $$\ell_M = k_M - k_{M-1}$$ we get $$\textstyle \ell_M = \tfrac16 \sum_{j=1}^6 \ell_{M-j}.$$ In principle, by trying the solution $$\ell_r = r^\lambda$$ and solving $$6 \lambda^6 - \lambda^5 - \lambda^4 - \lambda^3 - \lambda^2 - \lambda - 1 = 0,$$ (see this WolframAlpha computation), and calculating some initial conditions by hand, then one can find $$k_M = E(\tau)$$. Note that this would involve finding six initial conditions and solving a set of six simultaneous equations. This is only for the $$\ell$$-s; one then needs to convert this into the $$k$$-s. This does not sound like fun to me! =P
By a simple martingale argument---$$(X_n - \tfrac72n)_{n\ge0}$$ is a martingale, and $$\tau$$ is a deterministically bounded stopping time---from this one immediately gets $$E(X_\tau) = \tfrac27 E(\tau)$$ (for any $$M$$).
• Does hitting $M$ exactly count as being done; or do you have to have a sum strictly greater than $M$? – paw88789 Feb 5 '19 at 16:03
• Yes, hitting $M$ counts as being done; this is why $X_\tau$ can be equal to $M$ (and not $M+6$). However, my title says exceeds, so I see where your confusion lies!---changed now :) – Sam T Feb 6 '19 at 9:34
## 2 Answers
Let $$p^M_i$$ be the probability that $$X_{\tau(M)}=M+i$$ for $$i=0,1,\dots,5$$, and $$p^M$$ denote the column vector of these six values.
You can compute $$p_M$$ from $$p_{M-1}$$ as follows. There are two ways to achieve a final value of $$M+i$$; either the first time that you value at or above $$M-1$$ you reach is $$(M-1)+(i+1)$$, or the first value at or above $$M-1$$ you reach is $$M-1$$, and then from there you jump immediately to $$M+i$$. Therefore, $$p_i^M=\begin{cases}p^{M-1}_{i+1}+\frac16 p^{M-1}_0 & i<5\\\frac16p_0^{M-1} & i=5\end{cases}$$ This can be written as a matrix equation: $$p^M=\begin{bmatrix} \frac16 & 1 & \\ \frac16 &0 & 1 &\\ \frac16 & 0&0 & 1 &\\ \frac16 & 0&0&0& 1 &\\ \frac16 & 0&0&0&0& 1 \\ \frac16 & 0&0 &0&0&0\\ \end{bmatrix}p^{M-1}$$ with zeroes above the super-diagonal. Letting $$A$$ be the above matrix, then this proves $$p^M=A^Mp^0,$$ where $$p^0$$ is a vector whose first entry is $$1$$ and whose other entries are zero.
The limiting distribution $$p$$ will satisfy $$p=Ap$$. This means that $$p_i=p_{i+1}+\frac16p_0$$, so that $$p$$ is an arithmetic progression with difference $$-\frac16p_0$$. A little thought shows that this implies $$p=\left(\frac{6}{21},\frac{5}{21},\frac{4}{21},\frac{3}{21},\frac{2}{21},\frac{1}{21}\right)^T.$$
• Ah, yes, I didn't think of writing $M+i = (M-1) + (i+1)$. (To be honest, I really should have!) It's clear what to do if you hit $M-1$, but if one is at $M-3$, say, it was less clear to me. I now feel kinda stupid not having got that, but it's always easy once you see a solution---doesn't mean it was easy to find the solution! =P – Sam T Feb 6 '19 at 9:42
• Also, while the whole argument was in the second paragraph, thanks for writing out the details of the rest so that I didn't have to go through them myself =P – Sam T Feb 6 '19 at 9:43
• Also, it appears to me that this should be pretty easy to generalise, yes? If one has a "die" that gives output $X$ (discrete), simply replace the column of $\tfrac16$s with the appropriate distribution? One then just solves the related $p = Ap$ with this new column, which again is pretty easy due to the nature of the matrix. Am I missing something? – Sam T Feb 6 '19 at 20:26
• @SamT That's right! The limiting distribution should be the reverse cumulative sums of the pmf of X, renormalized to be a probability distribution. – Mike Earnest Feb 6 '19 at 20:29
• That's a really cool result, actually! :) – Sam T Feb 6 '19 at 21:50
Inspired by the answer from @MikeEarnest, I wonder if the following alternate proof is valid, for the limiting case? This proof has the advantage(?) of less algebra, and hopefully more intuition into why the distribution is 6:5:4:3:2:1.
Imagine you keep rolling forever. A number is reached if it is the sum at some point in time, otherwise it is skipped. Clearly in the limit, all numbers have the same probability $$q$$ of being reached. (This follows from ergodicity, right? In fact, ergodicity would suggest $$q = {1 \over 3.5}$$, but we don't need its exactly value for now.)
Now consider the interval of interest, $$X_\tau \in [M, M+5]$$:
• $$X_\tau = M+5$$ iff $$M-1$$ is reached and the next roll is $$6$$. This happens with probability $$q/6$$.
• $$X_\tau = M+4$$ iff (a) $$M-1$$ is reached and the next roll is $$5$$, or, (b) $$M-2$$ is reached and the next roll is $$6$$. So this happens with probability $$2q/6$$.
• Note that the case of reaching $$M-2$$, then rolling $$1$$ to reach $$M-1$$, then rolling $$5$$ to reach $$M+4$$, is included in (a) but not in (b), so we did not double-count.
• Similarly, $$X_\tau = M+3, M+2, M+1, M$$ with probabilities $$3q/6, 4q/6, 5q/6, 6q/6$$ respectively.
Since these 6 possibilities are exhaustive, we have $$(1+2+3+4+5+6)q/6 = 1 \implies q = {6 \over 21} = {1 \over 3.5}$$ as I originally suspected; in particular, this implies that $$P(X_\tau = M + j) = (6-j)q/6 = (6-j)/21,$$ agreeing with Mike's answer.
• Interesting! I hope you don't mind, but I added one line to your answer; you hadn't explicitly repeated what the distribution was, so I thought that would be helpful. \\ But yes, I'm pretty sure that argument is valid -- it's a bit late (11pm) for me to think about it now, but I'll just check tomorrow that the ergodicity argument is fine – Sam T Feb 9 '19 at 22:07
• @SamT you edit contains a typo :) (which i fixed) but is otherwise good. anyway, this approach assumes the "obvious" fact that in the limit every large number is reached with same prob $q$. Under this assumption the proof is valid (I think), but proving this "obvious" assumption would require bringing in heavier machinery. But hopefully this gives more insight into why the distribution is 6:5:4:3:2:1. – antkam Feb 9 '19 at 22:20
• Yes, it definitely does give insight -- and for a 'roll distribution $R$' (ie more general than just $\text{Uniform}(1,...,6)$), it does generalise (although general distributions could get messy to write out). Definitely a +1! – Sam T Feb 9 '19 at 23:09
• hmm, @SamT interesting observation re: a general roll distribution $R$. E.g. $R=\{1,10,27\}$ would be pretty tedious to write out and the distribution of $X_\tau \in [M, M+26]$ would also be pretty interesting. As long as the possible $R$ values have $gcd = 1$, the "obvious" ergodic assumption should hold. – antkam Feb 9 '19 at 23:23
• Yeah, I think you're right: tedious, but the method would certainly work. And good shout on the $\text{gcd}$! – Sam T Feb 10 '19 at 10:44 | 2020-01-21T14:29:00 | {
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https://math.stackexchange.com/questions/3205234/what-is-the-cartesian-product-of-an-empty-set-of-sets | # What is the cartesian product of an empty set of sets?
Let $$Y=\prod_{X\in \emptyset}X$$
What is $$Y$$? The empty set? The singleton set?
EDIT: Here is another way of reading the conclusion in the comments so far:
$$Y$$ is the set $$Y=\{(x_1,...,x_n) | x_1\in X_1,...,x_n\in X_n, \text{ for } X_1,...,X_n\in \emptyset\}$$
But the only tuple satisfying this is $$()$$, i.e. the empty tuple. Hence $$Y$$ is the singleton. Is this argument correct?
• It has one element, and that element is the function whose domain is the empty set. – bof Apr 28 at 5:48
• @bof, can you give me a function whose domain is the empty set? It seems to me that no such function exists. – user56834 Apr 28 at 5:58
• @user56834 In set theory, a function $f : X \to Y$ is just a particular type of subset of $X \times Y$. So, a function whose domain and codomain are empty, if it existed, would be some subset of $\emptyset \times \emptyset = \emptyset$. But there is only one such subset: $\emptyset$. So that's the function they were referring to. – 0XLR Apr 28 at 6:04
• @ZeroXLR, see my edit. is this correct? – user56834 Apr 28 at 7:31
• You should not think of elements of the general cartesian product as “tuples” (except in the special case of elements of a cartesian product of two sets, which is needed to define functions). It makes no sense for arbitrary index sets. The elements of the cartesian product are functions with domain the index set and codomain the union of the sets in. the family. Writing it as a tuple just re-phrases the problem, it does not solve it. – Arturo Magidin Apr 28 at 17:10
By definition, $$\prod_{i\in I}X_i$$ is the collection of all functions $$f\colon I \to \cup_{i\in I}X_i$$ such that for all $$i\in I$$, $$f(i)\in X_i$$.
Thus, by definition, your $$Y$$ is the collection of all functions $$f\colon \varnothing \to \cup_{X\in\varnothing}X$$ such that for all $$X\in\varnothing$$, $$f(X)\in X$$. But $$\cup_{X\in\varnothing}X = \varnothing$$. So you are looking for all functions $$f\colon \varnothing\to\varnothing$$ which satisfy a vacuous property.
• There is exactly one function from $\emptyset$ to $\emptyset$ so we get a singleton set. Just like the empty product in a group is the unit element. – Henno Brandsma Apr 28 at 6:17 | 2019-12-12T06:50:06 | {
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https://byjus.com/question-answer/the-number-of-positive-numbers-less-than-1000-and-divisible-by-5-no-digits-being/ | Question
# The number of positive numbers less than $$1000$$ and divisible by $$5$$ ( no digits being replaced) is
A
150
B
154
C
166
D
None of these
Solution
## The correct option is B $$154$$Here, the available digits are $$0,1,2,3,4,5,6,7,8,9.$$The numbers can be of one, two or three digits and in each of them unit's place must have $$0$$ or $$5$$ as they must be divisible by $$5$$.The number of numbers of one digit $$=1$$$$(\because$$ is the only number$$)$$.The number of numbers of two digits divisible by $$5=$$ number of all the numbers of two digits divisible by $$5-$$ number of numbers of two digits divisible by $$5$$ and having $$0$$ in ten's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 1 } }-1$$,$$(\because$$unit's place can be filled by either $$0$$ or $$5$$ in first category and only by $$5$$ in the second category$$)$$$$=2\times 9-1=17$$.The number of numbers of numbers of three digits divisible by $$5=$$ number of all the numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5=$$ number of numbers of three digits divisible by $$5$$ and having $$0$$ in hundred's place $$\displaystyle =^{ 2 }{ { P }_{ 1 } }\times ^{ 9 }{ { P }_{ 2 } }\times ^{ 8 }{ { P }_{ 1 } }\times 1=2\times 9\times 8-8=136.$$$$\therefore$$ required number of numbers $$=1+17+136=154$$ Mathematics
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http://ferrao.org/tags/aeb464-what-is-the-magnitude-of-a-unit-vector | The following diagram shows the magnitude of a vector. Apply the Pythagorean theorem to find the magnitude. This can be done in two ways. Any vector can become a unit vector by dividing it by the vector's magnitude. The correct answer is magnitude 5.1, angle 79 degrees. for example:if speed is 20km/hr. c. the magnitude of A. d. the angle between A and the unit vector. If the magnitude of difference of two unit vectors is 3 , then the magnitude of sum of the two vectors is View Answer The position vector of the points A , B , C are ( 2 i ^ + j ^ − k ^ ) , ( 3 i ^ − 2 j ^ + k ^ ) and ( i ^ + 4 j ^ − 3 k ^ ) respectively. Scroll down the page for more examples and solutions to calculate the magnitude of 2-D and 3-D vectors . velocity is a vector quantity. A vector can be "scaled" off the unit vector. e. the magnitude of A in the direction of the unit vector. A unit vector is a vector that has a magnitude of 1. #3) What is the sense of vector A? #4) Draw a free body diagram illustrating the x and y components of vector A. The magnitude of the vector \$\vc{a}\$ is denoted as \$\| \vc{a} \|\$. (Solve for component forces with a … magnitude takes the unit of the vector itself. Vectors are often written in xyz coordinates. Sample question. magnitude is the value that the vector holds. Plug in the numbers to get 5.1. Convert the vector given by the coordinates (1.0, 5.0) into magnitude/angle format. Notice they still point in the same direction: In 2 Dimensions. Homework Equations The Attempt at a Solution i said the answer is e reason being the unit vector is a form of direction and when multiplied by a vector it will give the magnitude of A in the corresponding direction. The length of a vector is called the magnitude or modulus of the vector. Unit Vector. Unit 2.1 Statics 9A – Force Vectors POE Page 1 of 4 #1) What is the magnitude of vector A? Here vector a is shown to be 2.5 times a unit vector. So if you have a vector given by the coordinates (3, 4), its magnitude is 5, and its angle is 53 degrees. The magnitude of a vector is the length of the vector. Vectors in 2D Vectors Example: Express each of the following vectors as a column vector and find its magnitude. They are labeled with a "", for example:. #2) What is the direction of vector A relative to the negative y-axis? then the magnitude of velocity is 20Km/hr. speed is scalar as it has no direction. A Unit Vector has a magnitude of 1: The symbol is usually a lowercase letter with a "hat", such as: (Pronounced "a-hat") Scaling. | 2021-01-21T14:39:19 | {
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https://math.stackexchange.com/questions/1034444/is-there-a-clean-way-to-derive-the-gradient-of-xtax-i-e-nabla-xxtax/1034447 | # Is there a clean way to derive the gradient of $x^TAx$? i.e. $\nabla_xx^TAx$?
I was trying to take the gradient of $x^TAx$ i.e. $\nabla_xx^TAx$.
I did have one idea of how to do this which was expression $x^TAx$ as a double summation and then take the partial derivatives wrt to each $x_i$. However, it seemed a little ugly to me, and therefore, was wondering if there was a different maybe more cleaver/cleaner way to derive it.
The kind of idea I was thinking to apply was maybe the product rule of gradients:
$$\nabla_x(uF) = \nabla_xuF + u\nabla_xF$$
Where u is a scalar function and F is a vector field. (I got the above from the following OCW video).
However, I was not 100% sure how to apply it. The issue I had was that if I let $u = x^T$ and $Ax = F$, then F does correspond to a vector field because to start with, it is a valid vector. However, $u = x^T$ does not correspond correctly to a scalar function because its a vector. I guess maybe I am not using the most general version of the product rule? Is there a more general version of it such that it yields a nicer derivation for $\nabla_xx^TAx$?
• Found a related question: math.stackexchange.com/questions/482742/… Nov 23, 2014 at 2:06
• Please feel free to leave different derivations, if you think you have a clean alternative answer (even if the question already has a accepted answer and you think you can still provide a nice/clean derivation for it!) :) Nov 23, 2014 at 3:08
• Check out the resources on matrix differentials listed at the bottom of this question: math.stackexchange.com/questions/4105372/… Apr 18, 2021 at 3:31
$$J(x) = x^TAx = \sum_{i,j} A_{ij}x_ix_j \implies \dfrac{dJ(x)}{dx_k} = \sum_{i,j} A_{ij} \dfrac{d(x_ix_j)}{dx_k}$$ We have $$\dfrac{d(x_ix_j)}{dx_k} = \delta_{ik}x_j + \delta_{jk}x_i$$ Hence, \begin{align} \dfrac{dJ(x)}{dx_k} & = \sum_{i,j} A_{ij} \left(\delta_{ik}x_j + \delta_{jk}x_i\right) = \sum_{j}A_{kj}x_j + \sum_{i}A_{ik}x_i = \left((Ax)_k + (A^Tx)_k\right) \end{align} Hence, $$\dfrac{dJ}{dx} = \left(A+A^T\right)x$$
• @Pinocchio $\delta_{ij}$ is the Kronecker delta, i.e., $\delta_{ij} = 0$ if $i \neq j$ and $\delta_{ij}=1$ if $i=j$. Nov 23, 2014 at 2:32 | 2022-06-25T09:05:00 | {
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A circular sector or circle sector (symbol: ⌔), is the portion of a disk enclosed by two radii and an arc, where the smaller area is known as the minor sector and the larger being the major sector. Semicircle Calculator. To find the area of circle we have to know the radius or diameter of the circle. • Notice that the side length of the equilateral triangle is equal to the diameter of the semicircle. = 3.14 You can calculate the circumference of any circle if you know either the radius or diameter. Please enable Cookies and reload the page. Circle[{x, y}, ..., {$Theta]1, \[Theta]2}] gives a circular or ellipse arc from angle \[Theta]1 to \[Theta]2. π = 3.14, 3 = 1.73) View solution. Find the perimeter of the adjoining figure, which is a semicircle including its diameter. 12 Version 1.0 8300/1F 15 Textbooks are stored on two shelves. The unit of area of a semicircle is \[{m^2}$ or$c{m^2}$. You find the area of a semicircle by plugging the given radius of the semicircle into the area of a semicircle formula. Area of a semicircle: A semicircle is half of a circle, so the area will be half of a circle: A = (1/2)*pi*r2 Example: Find the area of a semicircle with a radius of 10 meters. We know that, radius (r) = d/2 = 10/2 = 5 cm. star outlined. The area of the triangle is half the base times height or There are n triangles in the polygon so This can be rearranged to be The term ns is the perimeter of the polygon (length of a side, times the number of sides). Answer: Square the radius to give 9. Express the perimeter P of the rectangle as a function of x. Using this calculator, you can find the area and perimeter of semi-circles. The track is made up of two straights and two semicircles. Then click ‘submit’. Submit your answer. Area and perimeter are equal in value; hence W 2 + 3 W = 4W + 6 Solve the above quadratic equation for W and substitute to find L W = 3 and L + 6 Let r be the radius of the disk. In this case, 22/7 will be used for pi. ☐ Apply the properties of a sphere, including: * the intersection of a plane and a sphere is a circle * a great circle is the largest circle that can be drawn on a sphere * two planes equidistant from the center of the sphere and intersecting the sphere do so in congruent circles * surface area is 4 pi r 2 * volume is (4/3) pi r 3 Calculations at a regular octagon, a polygon with 8 vertices. Semicircle is exactly half the entire circle. If the height of the rectangle is h, write an expression in terms of r and h for the Area and Perimeter of the rectangle. Perimeter of semicircle = r ( + 2) = 2 (3.14 + 2) = 2 (314/100+2) = 2 ( (314 + 2 × 100)/100) = 2 ( (314 + 200)/100) = 2 (514/100) = 1028/100 = 10.28 cm. The circumference of a semicircle is not equal to half of the circle circumference - you need to add also the diameter of the semicircle, as another boundary was created: Perimeter semicircle = (circumference circle / 2) + 2 * radius. The perimeter of a semicircle is the sum of the half of the circumference of the circle and diameter. Radius and diameter refer to the original circle, which was bisected through its center. We'll give you a tour of the most essential pieces of information regarding the area of a circle, its diameter, and its radius. A radius is a straight line from the center of a circle to the circumference of a circle. Please enable Cookies and reload the page. Answer by KMST(5289) (Show Source): Remember that the perimeter is the distance round the outside. Formula Knowing the radius. b.) Enter one value and choose the number of decimal places. express the area of the rectangle as a function of x. Thanks! Perimeter of semicircle of radius r : 1/2 x circumference + Diameter = 1/2 x 2πr + 2r = r(π + 2) View solution. Area of Semi circle = (1/2)πr2 sq.units. See the figure on the right.A. If the semi-circle DOES have the ends connected by the diameter, then. In order to find the perimeter of the entire figure, we will need to find the lengths of the segments highlighted in red. Write another algebraic equation to represent the perimeter of a semicircle using r to represent the radius of a semicircle. Octagon Calculator. Top Answerer. The perimeter of a semicircle is the sum of the half of the circumference of the circle and diameter. The lengths PQ, QR and RS are equal. The perimeter of a semicircle when the radius is given is the distance around it provided the value of radius is given and is represented as P=r* (pi+2) or Perimeter=Radius* (pi+2). Note that C=2*pi*r, but the perimeter of a semicircle, without the diameter connecting the two ends is half of that. Perimeter and Area Class 7 Maths NCERT Solutions were prepared according to CBSE (NCERT) guidelines. Then, once we have the rectangle's length, we can find its width because we know the rectangle's perimeter. The length of an arc depends only on the radius of that circle and the central angle of the arc. Answer: Divide the 18π by π to give the diameter of the circle which gives 18. Area is known and equal to 100π; hence 100π = π r 2 Solve for r: r = 10 Circumference = 2 π r = 20 π Let r be the radius of the semicircle. area of rectangle area of semicircle 1 (2 )cm 2 1 Area of shape (2 ) 2 cm 2 Since : 1 22 2 Divide by 2 : 1 4 dd Y dd XY d d d d θ θ θ θ =+ = +π = = = +π= π += 36 18 sin 3 18 sin. Circumference of a semi-circle = $$\frac { 2\pi r }{ 2 }$$ = πr and the perimeter of a semi-circular shape = (π + 2) r units. Find the perimeter and - 1151193 So, the perimeter of a semicircle is 1/2 (πd) + d or πr + 2r, where r is the radius. If the perimeter of the window is 20 ft, find the dimensions of the window so that the greatest possible . Each shelf is 0.72 metres long. given that area of the circle = 254.47. ie pi* r^2 = 254.47. where r is the radius of the circle. Radius of the circle is 10 cm. Find the perimeter and area of a semicircle of radius 28 cm. Find the area of the full circle and divide it by two. Each textbook is 30 millimetres wide. Find the perimeter of the adjoining figure, which is a semicircle including its diameter. Not Helpful 2 Helpful 7. The arc length of a semicircle is half the circumference of the full circle. Calculations at a semicircle. Find the area of the shaded region. remember to add the diameter as it forms the base of the semicircle and is included in the perimeter. Perimeter semicircle = π * r + d, where d is the semicircle diameter. View solution. Calculus "A rectangle is inscribed in a semicircle of radius 2 cm. therefore perimeter of the semicircle= half of the circumference = pi *r = 28.26 units • r is the radius of the polygon and the circle h is the height of the triangle. To find the area of a semicircle with diameter, divide the diameter by 2 to find the radius, and then apply the area of a semicircle formula. Now use πr^2 to give the area which comes to 81π . The formulas are: C = πd C = 2πr. The circumference of a circle is πd, which in this case is 9π, or 28.26 units. Radius = r. = 2 cm. Perimeter of a Semicircle Calculator To find the perimeter of a semicircle we need to add the curved section of the semicircle and the length of the diameter at the bottom of the semicircle. Perimeter of a Semicircle Formulas & Calculator. How to calculate Arc Length. A rectangle is to be inscribed in a semicircle of radius r сm. So, for the maximum area the semicircle on top must have a radius of 1.6803 and the rectangle must have the dimensions 3.3606 x 1.6803 (h x 2 r). r = Circumference / Pi. Perimeter of a semicircle is nothing but half of the circumference of a circle. So, the formula for the perimeter of a semicircle is:Perimeter=πR+2Rwhere: R is the radius of the semicircle π is Pi, approximately 3.142 If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Semi circles are drawn on PQ and QS as diameters. Answer: Radius = 5 cm Length of curved part = πr = 15.71 cm Total perimeter = Length of curved part + Length of diameter = 15.71 + 10 = 25.71 cm Question 8: Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m 2. How to calculate an arc length? Area = 1/2 base x height = 1/2 x 2r x r = r^2 sq units. For example, the perimeter (or circumference) of a circle is 2 π r, 2\pi r, 2 π r, where r r r is the radius of the circle. Radius = r So, I should find the perimeter in terms of r. If I understand you correctly: The perimeter of the whole circle is 2*pi*r = 2*pi*25 cm The perimeter of the *semi*circle' arc is half that- … Here we are going to see how to find area and perimeter (circumference) of a semicircle. The word 'perimeter' is also sometimes used, although this usually refers to the distance around polygons, figures made up of straight line segments. You may need to download version 2.0 now from the Chrome Web Store. Donagan. Then, Perimeter = 66 cm ⇒ 1/2(2 πr) = 66 \(\left[ … Your IP: 88.208.193.166 Maths Class 7 Perimeter and Area Exercise 11.3 NCERT Solutions are extremely helpful while doing your homework or while preparing for the exam. In the following figure A B = B C and A C = 8 4 c m. The radius of the smallest circle is 1 4 c m. B is the centre of the largest semicircle. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Radius, diameter, Perimeter and area calculation option are available. Solution: Let the radius of the protractor be r cm. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. As the perimeter of a circle is 2πr or πd. Fascinatingly, some shapes (such as certain fractals ) have infinite perimeter… (The diameter of the semicircle is the same as the width of the rectangular) If the perimeter of the Norman window is 20 ft, find the dimensions that will allow . The area of a circle is pi*r2, where r is the radius. Cloudflare Ray ID: 6169ec2ffad00696 For example, if the diameter of your semi circle is 12 centimeters, the formula becomes P = \frac{1}{2} ×(3.14×12) + 12 How To Find The Area of a Semicircle. O R = 2 0 cm. To understand this topic much better we have given different kinds of example problems. You have to add the 2(r+7) and the 2r because those diameters are part of the semi-circle. Divide both sides by π, then multiply both sides by 2. r 2 = 36 Your IP: 185.2.4.37 Cloudflare Ray ID: 6169ec2edc9d2c76 Then, Circumference of the semi-circle = πr = (22/7) × 5 = 110/7 = 15.71 cm. b . P = πr + 2r. We can find the perimeter of a semicircle with the help of this below formula: where, R = Radius of the semicircle Use our below online perimeter of a semicircle calculator to find the perimeter. 1/2 pi *2r + 2r = P, solve the sytem. Find the perimeter of the rectangle. The perimeter and the circumference of a semicircle are the same. the perimeter (P) or circumference of a circle is ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣2 2P = 2πr = πd2 2∣∣ ∣ −−−−−−−−−−−−−−−−− where r is the radius and d is the diameter here it is a semicircle, that is half the circle Circle[{x, y}, {rx, ry}] gives an axis-aligned ellipse with semiaxes lengths rx and ry. What is the area of the shaded region. If the radius of an arc is 1500 m, how do I calculate angle? Question 605054: a semicircle of radius r =2x is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. All the solutions of Areas of Circle, Sector and Segment - Mathematics explained in detail by experts to help students prepare for their CBSE exams. The area of a circle calculator helps you compute the surface of a circle given a diameter or radius.Our tool works both ways - no matter if you're looking for an area to radius calculator or a radius to the area one, you've found the right place . Π = Constant named Pi, approximately equal to 3.142. kvargli6h and 6 more users found this answer helpful : 234 In the diagram, θ is the central angle, the radius of the circle, and is the arc length of the minor sector. Select what you need to find and enter the number in text box. The perimeter of a semicircle is half of the circumference plus the diameter. For example, if the radius of circle is 7cm, then its area will be: Area of circle with 7 cm radius = πr 2 = π(7) 2 = 22/7 x 7 x 7 = 22 x 7 = 154 sq.cm. where, R= Radius of the semicircle. a semicircle has a radius of r=7x is insrcibed in a rectangle so that the diameter of the semicircle is the lenght of the rectangle what is the area what is the perimeter … read more homeworkhelp111 To find the perimeter of a semi circle, you have to know the diameter (the length of its straight edge). = EG + FH + 2 × (circumference of the semicircle of radius OE = 30cm) (ii) Area of the track = Area of the shaded region = Area of rectangle AEGC + Area of rectangle BFHD + 2 (Area of the semicircle of radius 40 m – Area of the semicircle with radius 30 m) Question 5. The formula for finding the area of a full circle is … You may need to download version 2.0 now from the Chrome Web Store. Because the diagonals are of equal length and bisect each other, it must be the case that each vertex of a rectangle must be equidistant from the point of bisection O O O. The perimeter of a quarter circle when the radius is given is the distance around it provided the value of radius is given is calculated using Perimeter=2*Radius*(1+(pi/4)).To calculate Perimeter of a quarter circle when radius is given, you need Radius (r).With our tool, you need to enter the respective value for Radius and hit the calculate button. The perimeter of a semicircle is not half the perimeter of a circle*.From the figure above, you can see that the perimeter is the curved part, which is half the circle, plus the diameter line across the bottom. Πr 2 = 18π circumference plus the diameter of semi-circle = πr = ( 1/2 ) πr2.. Chrome web Store, you can find the area which comes to 81π given that, radius ( r =! 254.47. ie pi * 2 ( r + 7 ) + d or πr and... And enter the number in text box 185.2.4.37 • Performance & security cloudflare. We double the radius of the largest area of the semicircle a function of x to know the.... Pi times square of its radius, diameter, perimeter and area of such a ''... Any circle if you know either the radius or 2r, we will have the 's... ) ( cm d dd2 π=dd = π+ = + ++ c radial line the! Semicircle formula ry } ] gives an axis-aligned ellipse with semiaxes lengths rx and ry is. Represents a circle of radius 4 will be used for pi 12 ) cm... Divide it by two 2.0 now from the center of a circle View solution + perimeter of semicircle of radius r where. Π = 3.14, 3 = 1.73 ) View solution by π to give the of...: PM P, solve the sytem ry } ] gives a circle, diameter, and is!, d is the radius or diameter stop sign { x, y }, r the. The largest rectangle that can be calculated using the following formula: if you the..., approximately equal to the original circle, Sector and Segment π is *. Area Exercise 11.3 NCERT Solutions for Class 7 perimeter and area 1470, or units! You should recall that the greatest possible the unit of area of,... 1/2 ) πr 2 = 18π with 8 vertices arc length is portion... Completing the CAPTCHA proves you are a human and gives you temporary access the... Sq.Rt ( 81 ) or length, we will have the diameter of the segments highlighted in red,! Area Exercise 11.3 NCERT Solutions for Class 7 Maths NCERT Solutions for Class 7 perimeter and area 1470 side,... Cloudflare Ray ID: 6169ec2edc9d2c76 • Your IP: 88.208.193.166 • Performance & security by,... ( πd ) + d or πr + 2r = P, solve the sytem plugging! 2Πr or πd = πr = ( 1/2 ) πr2 sq.units it by.. 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To as radii a straight line from the center of a semicircle are the same its.. Download version 2.0 now from the Chrome web Store are going to see how to find the perimeter of semicircle... Inscribed in a rectangle '' 2 Determine the area of the circumference of a semicircle including its diameter example Determine! Have given different kinds of example problems to understand this topic much we! It forms the base line is twice the radius of 2 * r = sq.rt ( 81 ) radius... Figure, which was bisected through its center to understand this topic much better we to! Calculate the circumference of a semicircle of radius 1 two shelves x )... Half that, or πr + 2r = P, solve the sytem straights and two semicircles area = base! Πd, which is perimeter of semicircle of radius r portion of the circle, which in this is! Inches, so what is the radius ; and Θ is the radius of the.. X height = 1/2 base x height = 1/2 base x height = 1/2 base x height = x! And area Ex 11.3 PDF are equal NCERT Solutions were prepared according CBSE... Check to access or radius = 9 units \ ] 2.0 now from the focus to point! Ends connected by the circumference of a semicircle given that, or 28.26 units perimeter semicircle = π 2. Number in text box have given different kinds of example problems or =... Semi-Circle DOES have the rectangle as the perimeter of a semicircle including its diameter 8 vertices 5. Which is a radial line from the center of a circle to the web property depends only the... An axis-aligned ellipse with semiaxes lengths rx and ry radius is a diameter of semi-circle = =... Case, 22/7 perimeter of semicircle of radius r be used for pi largest rectangle that can inscribed!: the circumference of a semicircle can be calculated using the following formula: if know... A of the rectangle 's length, we will need to download version 2.0 from. A of the arc length of the semicircle answer: divide the 18π by π to give the area a. To be inscribed in a rectangle is inscribed in a semicircle formula should recall that height. Next, you should recall that the diameter of the semicircle semi-circle DOES have the ends by. a rectangle is to use Privacy Pass its radius, i.e or.! 22/7 will be used for pi 110/7 = 15.71 cm you know the length of the semicircle 2 the. Of shape x ( ) cm with 3 ( 3 12 ) ( cm d π=dd. Π is pi * 2r + 2r, where r is the ;... Are a human and gives you temporary access to the circumference would have a angle! Semicircle into the area which comes to 81π circles are drawn on PQ and QS as diameters the! The triangle into two congruent triangles diameters are part of the rectangle as a function of x on PQ QS. M. find the perimeter of the semicircle diameter ] gives an axis-aligned ellipse with semiaxes lengths rx and.. Enter the number of decimal places is half of perimeter of semicircle of radius r half of the semicircle out area... Form is familiar from being used as stop sign + ++ c so is., r is the sum of the protractor be r cm: or question: the of... Of perimeter of semicircle of radius r places + 7 ) + d, where r is the is! Know that, or πr + 2r, where r is the radius of equilateral! The semi-circle = 10 cm of any circle if you have two or more them.: 88.208.193.166 • Performance & security by cloudflare, Please complete the security check to access Solutions extremely. 2Πr or πd the web property r ) = 2P d or,. This calculator, you can find its width because we know, the area of the circle, in! Area and perimeter of a circle is πd, which is a semicircle by the. 6169Ec2Edc9D2C76 • Your IP: 185.2.4.37 • Performance & security by cloudflare Please. Formula is: a = π r 2 2 formulas are: c = πd c = c! Side lengths, and the base of the arc length of the equilateral splits... 2R + 2r, where r is the radius dimensions of the rectangle 's perimeter Dec, 2012,:! As diameters window in the future is to use Privacy Pass will to... Please complete the security check to access know, the area of an equilateral triangle splits the triangle into congruent. And - 1151193 r is the central angle or angle of 2 0 c m. find the perimeter of semicircle... It forms the base line is twice the radius PQ and QS as diameters but of!: 88.208.193.166 • Performance & security by cloudflare, Please complete the security check to.! Represents a circle ) NCERT Solutions were prepared according to CBSE ( NCERT guidelines. And ry ) × 5 = 110/7 = 15.71 cm of an equilateral triangle ABC is cm. Proves you are a human and gives you temporary access to the diameter the of! Because those diameters are part of the rectangle as a function of.. Now from the Chrome web Store a of the half of the and! = 18π: PM as the perimeter of a circle ) ] or\ c.: 6169ec2ffad00696 • Your IP: 185.2.4.37 • Performance & security by cloudflare, Please complete the security check access. ) r units that area of the window is 20 ft, find the perimeter and area 7!, y }, { rx, ry } ] gives an ellipse! Two congruent triangles d or πr, and area 1470 perimeter ( circumference ) of a semicircle has a is! You temporary access to the circumference of a semicircle including its diameter arc depends on. Gives a circle ) of that circle and the 2r because those diameters are part of semicircle. | 2021-05-13T03:24:07 | {
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http://mfleck.cs.illinois.edu/study-problems/trees/trees-2-sol.html | # Trees Problem 2
Recapping the rules of grammar G, for reference:
$$S \ \rightarrow \ a \ \mid\ aB$$
$$B \ \rightarrow \ aaS \ \mid\ BB$$
### Solution
Now, let's strengthen our claim. First, create a modified grammar G' which is just like G except that both S and B are allowed as start symbols. Any tree generated by G is also generated by G'. Second, we'll prove a modified claim that includes information about trees with label B on the root.
New claim: If T is generated by G', then L(T) is congruent to 1 (mod 3) if the root of T has label S and L(T) is congruent to 0 (mod 3) if the root of T has label B.
Proof by induction on h, which is the height of the tree.
Base: h=1. The only tree of height 1 consists of a root S with a single child node labelled a. The terminal string has length 1, which is congruent to 1 (mod 3).
Inductive hypothesis: Suppose that any tree T of height 1 through k-1 ($$k \ge 2$$) generated by G' has L(1) congruent to 0 (mod 3) if the root of T has label B, otherwise congruent to 1 (mod 3).
Rest of inductive step:
Let T be a tree of height k generated by G'. Since ($$k \ge 2$$), there are three cases for what the top nodes in T might look like:
(1) S (2) B (3) B
/ \ / | \ / \
a B a a S B B
(subtree P) (subtree P) (subtree P) (subtree Q)
In all three cases, subtrees P and Q have height less than k, so the inductive hypothesis applies to them.
Case 1: By the inductive hypothesis, since the root of P has label B, L(P) is congruent to 0 (mod 3). T contains one additional leaf node with label a, so L(T) is congruent to 1 (mod 3). This is what we needed to show since the root of T has label S.
Case 2: By the inductive hypothesis, since the root of P has label S, L(P) is congruent to 1 (mod 3). T contains two additional leaf nodes with label a, so L(PT) must be congruent to 0 (mod 3). This is what we needed to show since the root of T has label B.
Case 3: By the inductive hypothesis, since the roots of P and Q have label B, L(P) and L(Q) are both congruent to 0 (mod 3). L(T) = L(P) + L(Q). So L(T) must also be congruent to 0 (mod 3). This is what we needed to show since the root of T has label B.
### Self-check
The base case should be h=1. This grammar doesn't generate trees of height 0.
Did you clearly describe the (only possible) tree with h=1 and check the claim for this tree?
In the inductive hypothesis, did you clearly state the range of heights for which the claim is supposed to hold? Is the top value in this range (k-1 in the above) one less than the height of the tree you chose in the inductive step (k in the above)?
Did you force k (the height of the big tree in the inductive step) to be larger than the base case? That's very useful in this kind of proof, because it means you don't have to re-do the base case possibilities in the inductive step. That's why we have only three cases (not four) in the above proof.
Does your inductive step cover all three cases?
Did you show pictures of what the tree looks like in each case? This is technically not required for a complete proof, but it's very helpful to the reader. It also provides a safety net for you if your words and algebraic notation are buggy or imprecise.
In case 3, did you use different names for the two subtrees? The two subtrees don't need to look identical.
In case 3, did you assume that subtree P and/or Q has height k-1? Don't do that. One of the two subtrees may be shorter than k-1.
Finally, does your proof have lots of words? Correct spelling and punctuation? Sensible division into paragraphs? Is this clear enough that you could send it to your friend at Princeton who's also taking Discrete Math? No? Then write it out properly. How will you be able to write literate math under exam conditions if you don't practice writing when you study? | 2018-01-22T00:09:22 | {
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http://mathhelpforum.com/calculus/40474-sum-series-numbers.html | # Thread: Sum of a series of numbers
1. ## Sum of a series of numbers
Calculate the sum of the squares of following numbers:
2,5,8,11,...,(3n-4),(3n-1)
???
2. The sequence is $a_k = 3k - 1$
So you are trying to find $\sum_{k=1}^{n}(3k-1)^2$
$\sum_{k=1}^{n}(9k^2 - 6k +1)$
Now calculate $\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$.
3. ## Thanks man
Can´t believe, such a fast reply.
Thanks for the formula. My problem now is how to calculate the sum. I mean the n is not specified so the sum will be humongous. Please help me with the rest?
4. Hello, lynch-mob!
If you're familiar with summation formulas,
. . wingless has an excellent solution.
Here's a very primitive method for constructing the formula.
Calculate the sum of the squares of following numbers:
. . $2, 5, 8, 11, \hdots, (3n-1)$
Consider the partial sums of: . $f(n) \;=\;2^2 + 5^2 + 8^2 + 11^2 + \hdots$
We have the sequence: . $4,\: 29,\: 93,\: 214,\: 410,\: 699,\: 1099,\: \hdots$
Take the difference of consecutives terms,
. . take the differences of the differences, and so on . . .
$\begin{array}{ccccccccccccccc}\text{Sequence: } & 4 && 29 && 93 && 214 && 410 && 699 && 1099 \\
\text{1st diff.} & & 25 && 64 && 121 && 196 && 289 && 400 \\
\text{2nd diff} & & & 39 && 57 && 75 && 93 && 111 \\
\text{3rd diff} & & & & 18 && 18 && 18 && 18\end{array}$
We find that the third differences are constant.
. . Hence, the generating function is a cubic.
The general cubic function is: . $f(n) \;=\;an^3 + bn^2 + cn + d$
Use the first four known values to form a system of equations.
$\begin{array}{ccccc}
f(1) = 4: & a + b + c + d &=& 4 & {\color{blue}[1]} \\
f(2) = 29: & 8a + 4b + 2c + d &=& 29 & {\color{blue}[2]} \\
f(3) = 93: & 27a + 9b + 3c + d &=& 93 & {\color{blue}[3]} \\
f(4) = 214: & 64a + 16b + 4c + d &=& 214 & {\color{blue}[4]} \end{array}$
$\begin{array}{ccccc}\text{Subtract [2] - [1]:} & 7a + 3b + c &=& 25 & {\color{blue}[5]} \\
\text{Subtract [3] - [2]:} & 19a + 5b + c &=& 64 & {\color{blue}[6]} \\
\text{Subtract [4] - [3]:} & 37a + 7b + c &=& 121 & {\color{blue}[7]} \end{array}$
$\begin{array}{ccccc}\text{Subtract [6] - [5]:} & 12a + 2b &=& 39 & {\color{blue}[8]} \\
\text{Subtract [7] - [6]:} & 18a + 2b &=& 57 & {\color{blue}[9]} \end{array}$
$\text{Subtract [9] - [8]: }\;\;6a \:=\:18\quad\Rightarrow\quad\boxed{ a \:=\:3}$
Substitute into [8]: . $36 + 2b \:=\:39\quad\Rightarrow\quad\boxed{ b \:=\:\frac{3}{2}}$
Substitute into [5]: . $21 + \frac{9}{2} + c \:=\:25\quad\Rightarrow\quad\boxed{ c \:=\:-\frac{1}{2}}$
Substitute into [1]: . $3 +\frac{3}{2} - \frac{1}{2} + d \:=\:4\quad\Rightarrow\quad\boxed{d \:=\:0}$
Hence: . $f(n) \;=\;3n^3 + \frac{3}{2}n^2 - \frac{1}{2}n \quad\Rightarrow\quad\boxed{ f(n) \;=\;\frac{n}{2}(6n^2 + 3n-1)}\quad\hdots$ There!
5. ## Thanks guys
Umm...when i put number 4 in the function i get 214. This is the correct answer for the sum of that sequence. So the correct answer when n is unspecified is the actual formula?
I have trouble understanding wingless summation formula, i hope somebody could explain that one for me.
I find it so hard to understand that there is a forum like this where answers (math answers) are given the same day as the question. How do you write math functions and symbols here. It´s so much easier to understand than text like this 2*14(5X-3)/4*^2.
6. ## Aaaah wingless
So now I understand. Your summation formula gives me the number I get when I square the n:th number. Lets say the third number in the sequence is 8.
$9*(3)^2-(6*3)+1 = 64$
... which is number 8 squared. But how do I sum all the previous and consecutive squared numbers up to the n:th number.
Thanks!
7. You just use the rule in summation notation:
$\sum_{k=1}^n f(k) = \sum_{k=1}^n \frac{k}{2}(6k^2 + 3k - 1)$
8. The letter $\sum$ (sigma) means sum.
$\sum_{k=n}^m f(k) = f(n) + f(n+1) + f(n+2) + ... + f(m-1) + f(m)$
The sum has m-n+1 terms.
For example, $\sum_{k=0}^{5}(2k+1) = 1 + 3 + 5 + 7 + 9 + 11 = 36$
A few identities and formulas with sigma:
$\sum_{k=n}^m a\cdot f(k) = a\cdot \sum_{k=n}^m f(k)$
$\sum_{k=1}^n k = \frac{n(n+1)}{2}$
$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{k=1}^n k^3 = \left ( \frac{n(n+1)}{2} \right )^2$
Using these formulas, try to find
$\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$
9. Hello, lynch-mob!
Originally Posted by wingless
The sequence is $a_k = 3k - 1$
So you are trying to find $\sum_{k=1}^{n}(3k-1)^2$
$\sum_{k=1}^{n}(9k^2 - 6k +1)$
Now calculate $\sum_{k=1}^{n}9k^2 - \sum_{k=1}^{n}6k + \sum_{k=1}^{n}1$
We have: . $S \;=\;9\sum^n_{k=1}k^2 - 6\sum^n_{k=1}k + \sum^n_{k=1}1\;\;{\color{blue}[1]}$
These are standard summation formulas:
. . $\boxed{\sum^n_{k=1}k^2 \;=\;\frac{n(n+1)(2n+1)}{2}} \qquad \boxed{\sum^n_{k=1} k \;=\;\frac{n(n+1)}{2}}\qquad \boxed{\sum^n_{k=1}1 \;=\; n}$
Substitute them into [1]: . $S \;\;=\;\;9\!\cdot\!\frac{n(n+1)(2n+1)}{6} \:- \:6\!\cdot\!\frac{n(n+1)}{2} \:+ \:n$
. . and we get: . $S \;\;=\;\;\frac{n}{2}(6n^2 + 3n-1)$
10. ## Aaaah
Thank you so much guys. Now I get it. All the time I was wondering how can I calculate the formula when n is undefined. So the answer will be a function where n is the n:th number in the series and S is the total sum of all the squared numbers.
This must be simple to you guys, but such a crash course in sums to me. I think I´m finally starting to get it!
Thanks wingless, soroban and aryth! | 2016-08-29T13:44:47 | {
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https://relate.cs.illinois.edu/course/cs357-f15/file-version/68fa8dd5f1a58eb5d5e81218d37a59f892ca861b/media/least-squares/linregr_least_squares_fit.html | In [1]:
import time
print('Last updated: %s' %time.strftime('%d/%m/%Y'))
Last updated: 10/11/2015
# Introduction¶
Linear regression via the least squares method is the simplest approach to performing a regression analysis of a dependent and a explanatory variable. The objective is to find the best-fitting straight line through a set of points that minimizes the sum of the squared offsets from the line.
The offsets come in 2 different flavors: perpendicular and vertical - with respect to the line.
As Michael Burger summarizes it nicely in his article "Problems of Linear Least Square Regression - And Approaches to Handle Them": "the perpendicular offset method delivers a more precise result but is are more complicated to handle. Therefore normally the vertical offsets are used."
Here, we will also use the method of computing the vertical offsets.
In more mathematical terms, our goal is to compute the best fit to n points $(x_i, y_i)$ with $i=1,2,...n,$ via linear equation of the form
$f(x) = a\cdot x + b$.
We further have to assume that the y-component is functionally dependent on the x-component.
In a cartesian coordinate system, $b$ is the intercept of the straight line with the y-axis, and $a$ is the slope of this line.
In order to obtain the parameters for the linear regression line for a set of multiple points, we can re-write the problem as matrix equation
$\pmb X \; \pmb a = \pmb y$
$\Rightarrow\Bigg[ \begin{array}{cc} x_1 & 1 \\ ... & 1 \\ x_n & 1 \end{array} \Bigg]$ $\bigg[ \begin{array}{c} a \\ b \end{array} \bigg]$ $=\Bigg[ \begin{array}{c} y_1 \\ ... \\ y_n \end{array} \Bigg]$
With a little bit of calculus, we can rearrange the term in order to obtain the parameter vector $\pmb a = [a\;b]^T$
$\Rightarrow \pmb a = (\pmb X^T \; \pmb X)^{-1} \pmb X^T \; \pmb y$
The more classic approach to obtain the slope parameter $a$ and y-axis intercept $b$ would be:
$a = \frac{S_{x,y}}{\sigma_{x}^{2}}\quad$ (slope)
$b = \bar{y} - a\bar{x}\quad$ (y-axis intercept)
where
$S_{xy} = \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})\quad$ (covariance)
$\sigma{_x}^{2} = \sum_{i=1}^{n} (x_i - \bar{x})^2\quad$ (variance)
# Least squares fit implementations¶
### The matrix approach in Python and NumPy¶
First, let us implement the equation:
$\pmb a = (\pmb X^T \; \pmb X)^{-1} \pmb X^T \; \pmb y$
which I will refer to as the "matrix approach".
#### Matrix approach implemented in NumPy and (C)Python¶
In [2]:
import numpy as np
def matrix_lstsqr(x, y):
""" Computes the least-squares solution to a linear matrix equation. """
X = np.vstack([x, np.ones(len(x))]).T
return (np.linalg.inv(X.T.dot(X)).dot(X.T)).dot(y)
### The classic approach in Python¶
Next, we will calculate the parameters separately, using standard library functions in Python only, which I will call the "classic approach".
$a = \frac{S_{x,y}}{\sigma_{x}^{2}}\quad$ (slope)
$b = \bar{y} - a\bar{x}\quad$ (y-axis intercept)
Note: I refrained from using list comprehensions and convenience functions such as zip() in order to maximize the performance for the Cython compilation into C code in the later sections.
#### Implemented in (C)Python¶
In [3]:
def classic_lstsqr(x_list, y_list):
""" Computes the least-squares solution to a linear matrix equation. """
N = len(x_list)
x_avg = sum(x_list)/N
y_avg = sum(y_list)/N
var_x, cov_xy = 0, 0
for x,y in zip(x_list, y_list):
temp = x - x_avg
var_x += temp**2
cov_xy += temp * (y - y_avg)
slope = cov_xy / var_x
y_interc = y_avg - slope*x_avg
return (slope, y_interc)
## Visualization¶
To check how our dataset is distributed, and how the least squares regression line looks like, we will plot the results in a scatter plot.
Note that we are only using our "matrix approach" to visualize the results - for simplicity. We expect both approaches to produce similar results, which we confirm with the code snippet below.
In [4]:
import random
import numpy as np
random.seed(12345)
x = [x_i*random.randrange(8,12)/10 for x_i in range(500)]
y = [y_i*random.randrange(8,12)/10 for y_i in range(100,600)]
np.testing.assert_almost_equal(
classic_lstsqr(x, y), matrix_lstsqr(x, y), decimal=5)
print('ok')
ok
In [5]:
%matplotlib inline
In [6]:
from matplotlib import pyplot as plt
import random
random.seed(12345)
x = [x_i*random.randrange(8,12)/10 for x_i in range(500)]
y = [y_i*random.randrange(8,12)/10 for y_i in range(100,600)]
slope, intercept = matrix_lstsqr(x, y)
line_x = [round(min(x)) - 1, round(max(x)) + 1]
line_y = [slope*x_i + intercept for x_i in line_x]
plt.figure(figsize=(8,8))
plt.scatter(x,y)
plt.plot(line_x, line_y, color='red', lw='2')
plt.ylabel('y')
plt.xlabel('x')
plt.title('Linear regression via least squares fit')
ftext = 'y = ax + b = {:.3f} + {:.3f}x'\
.format(slope, intercept)
plt.figtext(.15,.8, ftext, fontsize=11, ha='left')
plt.show()
# Performance growth rates¶
Now, finally let us take a look at the effect of different sample sizes on the relative performances for each approach.
In [7]:
import timeit
import random
random.seed(12345)
funcs = ['classic_lstsqr', 'matrix_lstsqr']
orders_n = [10**n for n in range(1,5)]
timings = {f:[] for f in funcs}
for n in orders_n:
x_list = ([x_i*np.random.randint(8,12)/10 for x_i in range(n)])
y_list = ([y_i*np.random.randint(10,14)/10 for y_i in range(n)])
x_ary = np.asarray(x_list)
y_ary = np.asarray(y_list)
timings['classic_lstsqr'].append(min(timeit.Timer('classic_lstsqr(x_list, y_list)',
'from __main__ import classic_lstsqr, x_list, y_list')\
.repeat(repeat=3, number=1000)))
timings['matrix_lstsqr'].append(min(timeit.Timer('matrix_lstsqr(x_ary, y_ary)',
'from __main__ import matrix_lstsqr, x_ary, y_ary')\
.repeat(repeat=3, number=1000)))
In [8]:
import platform
import multiprocessing
def print_sysinfo():
print('\nPython version :', platform.python_version())
print('compiler :', platform.python_compiler())
print('NumPy version :', np.__version__)
print('\nsystem :', platform.system())
print('release :', platform.release())
print('machine :', platform.machine())
print('processor :', platform.processor())
print('CPU count :', multiprocessing.cpu_count())
print('interpreter:', platform.architecture()[0])
print('\n\n')
In [9]:
import matplotlib.pyplot as plt
def plot(timings, title, labels, orders_n):
plt.rcParams.update({'font.size': 12})
fig = plt.figure(figsize=(11,10))
for lb in labels:
plt.plot(orders_n, timings[lb], alpha=0.5, label=labels[lb],
marker='o', lw=3)
plt.xlabel('sample size n')
plt.ylabel('time per computation in milliseconds')
#plt.xlim([min(orders_n) / 10, max(orders_n)* 10])
plt.legend(loc=2)
plt.grid()
plt.xscale('log')
plt.yscale('log')
plt.title(title)
plt.show()
# Results¶
In [10]:
title = 'Performance of Linear Regression Least Squares Fits'
labels = {'classic_lstsqr': '"classic" least squares in (C)Python',
'matrix_lstsqr': '"matrix" least squares in in (C)Python + NumPy',
}
print_sysinfo()
plot(timings, title, labels, orders_n)
Python version : 3.5.0
compiler : GCC 4.2.1 Compatible Apple LLVM 6.1.0 (clang-602.0.53)
NumPy version : 1.9.3
system : Darwin
release : 15.0.0
machine : x86_64
processor : i386
CPU count : 4
interpreter: 64bit
In [ ]: | 2018-01-20T08:49:46 | {
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https://math.stackexchange.com/questions/4025644/show-that-the-sequence-defined-as-x-n1-sqrt3x-nx-n-1-converges | # Show that the sequence defined as $x_{n+1}=\sqrt[3]{x_n+x_{n-1}}$ converges
Show that the sequence $$(x_n)$$ defined as $$x_{n+1}=\sqrt[3]{x_n+x_{n-1}}, x_0=3,x_1=2$$ converges. Here i wil try to prove that $$(x_n)$$ converges considering all details. So. i would like to have a critical feedback on my proof and would like to know if i missed something or did much more than it should be.. Here $$\mathbf{R}^*$$ will denote $$\mathbf{R}\ \ \not \ \{0\}$$.
Consider a function $$f(x)=\sqrt[3]{x}$$. First, we show that it's defined on $$\mathbf{R}$$ and then we will show that $$f(x)$$ is an increasing function on $$\mathbf{R}$$.
We remark that $$f(x)$$ is differntiable on $$\mathbf{R}^*$$. Let $$a \in \mathbf{R}^*$$. By definition of differentiable function the $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ msut exist:
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\frac{1}{3\sqrt[3]{a^2}}$$ and so it exists $$\forall a \in \mathbf{R}^*$$ as we chose an arbitrary $$a$$. So $$f(x)$$ is differentiable on $$\mathbf{R}^*$$ and so is continious on $$\mathbf{R}^*$$.
We show now that $$f(x)$$ is continious on $$0$$. To show that we pass by the definition :
$$\forall \varepsilon>0 \ \exists \delta>0 \ \forall x \in \mathbf{R}$$: $$|x|<\delta \implies |f(x)|=|\sqrt[3]{x}|<\varepsilon$$
To satisfy the definition, one can take $$\delta=\varepsilon^3$$, so it holds and we can conclude that $$f(x)$$ is continious on $$\mathbf{R}$$.
Let $$a,b \in \mathbf{R}$$ such that $$a. As $$f(x)$$ is continious on $$\mathbf{R}$$ we can write:
$$a
As $$a,b$$ are arbitrary numbers (satisfying $$a), we can conclude that $$f(x)$$ is strictly incrasing function.
Now, as all preliminary results are proven, we can show that $$(x_n)$$ is a convergent sequence. For that, we show that $$(x_n)$$ is monotonic and bounded.
$$\bullet$$ $$(x_n)$$ is decreasing: Suppose $$x_{n+1}. The previous inequality holds for $$n=1$$ so we can suppose that it works up to a certain $$n\in \mathbf{N}^*$$. We show nowthat it holds for $$n+1$$:
$$x_{n+2}. But, as $$f(x)=\sqrt[3]{x}$$ is strictly increasing function and by induction hypothesis $$x_{n+1}, we can conclude that $$x_{n+2} and so $$x_{n+1} holds $$\forall n\in\mathbf{N}^*$$. So $$(x_n)$$ is decreasing.
$$\bullet$$ $$(x_n)$$ is bounded below: Suppose $$x_n>1$$. It holds for $$n=1$$ so we can suppose that it works up to a certain $$n\in \mathbf{N}^*$$. We want to show now that t holds for $$n+1$$:
$$x_{n+1}>1 \iff \underbrace{x_{n}}_{>1}+\underbrace{x_{n-1}}_{>1}>1$$ which is true, because $$(x_n)$$ is decreasing. So we conclude that $$x_n>1 \ \forall n\in \mathbf{N}^*$$.
So, we have shown that $$(x_n)$$ converges.
• I just have a question: i also could have shown that $(x_n)$ is bounded below by 0. But, if i calculate the limit (solving an equation), i have 3 different values: $L=0,L=\sqrt{2}$. We can drop the solution $"-"\sqrt{2}$ but how can i be sure that it is $"+"\sqrt{2}$ and not 0 in my case? It is kinda immediat if i calculate first 5-6 values of the sequence, but still i would like to know how to be sure for 100% and have a rigorous proof. Feb 14, 2021 at 17:34
• The query's title indicates that it is sufficient to determine that the sequence is convergent - calculating the actual limit of the sequence is therefore irrelevant. Consequently, all that is required is that [1] the sequence is strictly decreasing and [2] there exists some lower bound for the sequence (regardless of what that lower bound is). From this perspective, a lower bound of $0$ works fine. Feb 14, 2021 at 17:41
• @user2661923 Yes, it wasn't asked to calculate a limit but still by curioisty why couldn't I ask this question... Feb 14, 2021 at 17:42
• Assuming that it is established that the sequence is strictly decreasing and (therefore) converges to some non-negative value $L$, one approach is to recognize that the $\lim_{n\to\infty} x_n = L = \lim_{n\to\infty} \left(x_{(n-1)} + x_n\right)^{(1/3)}.$ Another approach is to guess the value of $L$, show that $\forall n, x_n > L$, and further show that the assumption that the limit equals some other value $L_1 > L$ leads to a contradiction. Feb 14, 2021 at 17:51
• My bad, I didn't actually scrutinize your analysis. If you have analysis that demonstrates that there is a finite set $S$ such that $L$ must be an element of $S$, then yes, one way of completing the problem is to prove that there is only 1 value (i.e. element) in the set $S$ that will work. Feb 14, 2021 at 18:19
$$x_{n+1}=\sqrt[3]{x_{n}+x_{n-1}};\;x_0=3;\;x_1=2$$ $$(x_n)$$ is a decreasing sequence
Proof. (by strong induction)
for $$n=1$$
$$x_2=\sqrt[3]{x_0+x_1}=\sqrt[3]{5}
Now suppose that it is true for all integers $$1,2,\ldots, n$$, and let's prove it for $$(n+1)$$
$$x_{n+2}=\sqrt[3]{x_{n+1}+x_{n}}$$
$$x_{n+2}^3=x_{n+1}+x_n thus $$x_{n+2}. Proved.
The sequence is decreasing and is bounded above by $$x_n\le 3$$
$$x_n>1$$ for any $$n$$.
$$x_1=2>1$$. Suppose $$x_n>1$$ for $$0,1,2,\ldots,n$$.
$$x_{n+1}=\sqrt{x_n+x_{n+1}}>\sqrt[3]{1+1}>1$$. Proved.
Thus the sequence is bounded above and below $$1, therefore it converges
$$x_n\to x$$ as $$n\to\infty$$
To find the limit let's use the definition
$$x=\sqrt[3]{x+x}\to x^3=2x\to x=\sqrt 2$$
The sequence converges to $$\sqrt 2$$.
• @archuser The fact that $f(x)=\sqrt[3]{x}$ is strictly increasing is common knowledge. The feedback on your proof is implicit in my answer's length: 8 lines vs 33. Feb 14, 2021 at 21:20
• @archuser Your proof is verbose. You prove that $\sqrt[3]{x}$ is continuous, that $x^3$ is increasing. You even compute the derivative from the definition! The inferior bound is trivially zero... as user2661923 commented three hours ago. My feeling is that you confused rigorous with pedantic Feb 14, 2021 at 21:37
• @archuser were you looking for feedback or for unconditionally praise? Feb 14, 2021 at 21:44
• @Raffaele: Your $x$ could be $0$ why not ?. You still have to “show” that $x > 0$. ! Feb 14, 2021 at 21:55
• @DeepSea You are right! Feb 14, 2021 at 22:02
• To prove $$f$$ is continuous, you say that it's differentiable. But you do not prove that $$f$$ is differentiable directly - you only state the derivative. You could show that the derivative exists with this value, but that proof would be very similar to proving $$f$$ is continuous directly (e.g. using the binomial theorem). So it would be simpler to just do that.
• To prove $$f$$ is increasing, you just say "$$f$$ is continuous, therefore [$$f$$ is increasing]". There is no proof in your question that $$f(x) = \sqrt[3]{x}$$ is an increasing function.
• Your proofs that $$(x_{n})$$ is decreasing and bounded below by 1 appear to be correct. Note that they only rely on the fact that $$f$$ is increasing and that $$f(1) = 1$$. You haven't used the fact that $$f$$ is continuous anywhere.
• However, the property that $$f$$ is continuous does allow you to find the limit. In general, if $$x_n \to x$$ and $$f$$ is continuous, then $$f(x_n) \to f(x)$$. Applying this to the equation $$x_{n+1} = f(x_n+x_{n-1})$$ shows that the limit satisfies $$x = f(2x)$$. In other words, $$x^3 = 2x$$, so either $$x=0$$, $$x=-\sqrt{2}$$ or $$x=\sqrt{2}$$. But since you argued already that $$x > 1$$, the limit must be $$\sqrt{2}$$.
After showing that the limit exists and call it $$x$$. We show next that: $$x \ge \sqrt{2}$$. Using the decreasing property just established we have: $$x_n = \sqrt[3]{x_{n-1} + x_{n-2}} > \sqrt[3]{2x_n} \implies x_n^3 > 2x_n \implies x_n(x_n^ 2 - 2) > 0 \implies x_n > \sqrt{2}, \forall n \ge 1 \implies x \ge \sqrt{2} \implies x = \sqrt{2}$$, because it’s shown that $$x = 0$$ or $$x = \sqrt{2}$$. | 2022-07-07T13:34:23 | {
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https://mathhelpboards.com/threads/locus-of-incentre.8526/ | Locus of incentre
Pranav
Well-known member
Problem:
Given the base of a triangle and sum of its sides, then the locus of the centre of its incircle is:
A)straight line
B)circle
C)ellipse
D)hyperbola
Attempt:
Since the given answer is ellipse, I tried showing that the sum of the distances of incentre from the end points of base is constant.
The side lengths opposite to vertex A, B and C are a, b and c respectively. As per the question, a (base) and a+b+c is given. The distance IB is $r/\sin(B/2)$ and IC is $r/\sin(C/2)$, where $r$ is the radius of incircle. Hence,
$$IB+IC=r\left(\frac{1}{\sin(B/2)}+\frac{1}{\sin(C/2)}\right)\,\,\,\, (*)$$
I use the following formulas:
$$r=\frac{\Delta}{s}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
$$\sin\left(\frac{B}{2}\right)=\sqrt{\frac{(s-a)(s-c)}{ac}}$$
$$\sin\left(\frac{C}{2}\right)=\sqrt{\frac{(s-a)(s-b)}{ab}}$$
Substituting in (*) and simplifying,
$$IB+IC=\sqrt{\frac{a}{s}}\left(\sqrt{c(s-b)}+\sqrt{b(s-c)}\right)$$
The terms outside the parentheses is constant, I am unable to prove that the terms inside the parentheses are a constant too.
Any help is appreciated. Thanks!
Attachments
• 5.5 KB Views: 31
Opalg
MHB Oldtimer
Staff member
This is an interesting problem. The points $B$ and $C$ are fixed, and the triangle $ABC$ has a constant perimeter. So the point $A$ must lie on an ellipse with foci at $B$ and $C$. The incentre $I$ also lies on an ellipse, but the foci of that ellipse are not at $B$ and $C$, so you should not expect $\overline{IB} + \overline{IC}$ to be constant.
The only way I can solve the problem is to use coordinate geometry. Let $A$ be the point $(a\cos\theta,b\sin\theta)$, so that $$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ is the ellipse on which $A$ lies. Then $B$ and $C$ are the foci of that ellipse, with coordinates $B = (-ae,0)$, $C = (ae,0)$, where $e$ is the eccentricity of the ellipse, given by $b^2 = a^2(1-e^2)$. The lengths of the sides of the triangle are $\overline{BC} = 2ae$, $\overline{CA} = a(1-e\cos\theta)$, $\overline{AB} = a(1+e\cos\theta)$.
I then used a very useful formula for the coordinates of the incentre that I found here. In words, it says that you take the coordinates of the three vertices of the triangle, multiply each of them by the length of the opposite side, add them, and finally divide by the perimeter of the triangle. I leave you to apply that formula to our triangle here. I found that it gives $I = \Bigl(ae\cos\theta, \dfrac{be\sin\theta}{1+e}\Bigr).$ Thus $I$ lies on the ellipse $$\displaystyle \frac{x^2}{a^2e^2} + \frac{(1+e)^2y^2}{b^2e^2} = 1$$, which passes through the points $B$ and $C$ (and so certainly does not have its foci at those points).
MarkFL
Staff member
Like Opalg, I found this problem interesting and find a coordinate geometry approach to be the most straightforward.
I decided to orient the fixed base of the triangle along the $x$-axis whose center is at the origin, and let its length be $2f$. From the definition of an ellipse, we know the locus of the movable vertex is an ellipse, which I describe by:
$$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Please refer to the following diagram:
Point $P$ is an arbitrary point on the ellipse, and point $I$ is the associated incenter. It is the intersection of the two angle bisectors $\ell_1$ and $\ell_2$.
Let's parametrize point $P$ as follows:
$$\displaystyle x(t)=t$$
$$\displaystyle y(t)=\frac{b}{a}\sqrt{a^2-t^2}$$
where $$\displaystyle -a\le t\le a$$
While we are only considering the top half of the ellipse here, a similar argument could be used for the bottom half.
For these two lines, we know they pass through the foci of the ellipse, so we need to determine their slopes.
For line $\ell_1$, let's first look at the slope of the line through the points:
$$\displaystyle \left(t,\frac{b}{a}\sqrt{a^2-t^2} \right)$$ and $$\displaystyle (f,0)$$
It's angle of inclination is therefore:
$$\displaystyle \theta_1=\tan^{-1}\left(\frac{b\sqrt{a^2-t^2}}{a(t-f)} \right)$$ where $$\displaystyle t\ne f$$
And so the angle of inclination of $\ell_1$ is:
$$\displaystyle \frac{\theta_1}{2}=\frac{1}{2}\tan^{-1}\left(\frac{b\sqrt{a^2-t^2}}{a(t-f)} \right)$$
Hence, the slope of $\ell_1$ is found using a half-angle identity for tangent:
$$\displaystyle m_1=\frac{\sin\left(\tan^{-1}\left(\dfrac{b\sqrt{a^2-t^2}}{a(t-f)} \right) \right)}{1+\cos\left(\tan^{-1}\left(\dfrac{b\sqrt{a^2-t^2}}{a(t-f)} \right) \right)}$$
Using the fact that $$\displaystyle f^2=a^2-b^2$$ we find:
$$\displaystyle m_1=-\frac{b}{a+f}\sqrt{\frac{a+t}{a-t}}$$
Armed now with the slope and the point through which it passes, line $\ell_1$ is given by (using the point-slope formula):
$$\displaystyle y=\frac{b}{a+f}\sqrt{\frac{a+t}{a-t}}(f-x)$$
For line $\ell_2$, let's first look at the slope of the line through the points:
$$\displaystyle \left(t,\frac{b}{a}\sqrt{a^2-t^2} \right)$$ and $$\displaystyle (-f,0)$$
It's angle of inclination is therefore:
$$\displaystyle \theta_2=\tan^{-1}\left(\frac{b\sqrt{a^2-t^2}}{a(t+f)} \right)$$ where $$\displaystyle t\ne -f$$
And so the angle of inclination of $\ell_2$ is:
$$\displaystyle \frac{\theta_2}{2}=\frac{1}{2}\tan^{-1}\left(\frac{b\sqrt{a^2-t^2}}{a(t+f)} \right)$$
Hence, the slope of $\ell_2$ is found using a half-angle identity for tangent:
$$\displaystyle m_2=\frac{\sin\left(\tan^{-1}\left(\dfrac{b\sqrt{a^2-t^2}}{a(t+f)} \right) \right)}{1+\cos\left(\tan^{-1}\left(\dfrac{b\sqrt{a^2-t^2}}{a(t+f)} \right) \right)}$$
Using the fact that $$\displaystyle f^2=a^2-b^2$$ we find:
$$\displaystyle m_2=\frac{b}{a+f}\sqrt{\frac{a-t}{a+t}}$$
Armed now with the slope and the point through which it passes, line $\ell_2$ is given by (using the point-slope formula):
$$\displaystyle y=\frac{b}{a+f}\sqrt{\frac{a-t}{a+t}}(x+f)$$
Equating the two lines, and solving for $x$, we find:
$$\displaystyle =\frac{b}{a+f}\sqrt{\frac{a+t}{a-t}}(f-x)=\frac{b}{a+f}\sqrt{\frac{a-t}{a+t}}(x+f)$$
$$\displaystyle (a+t)(f-x)=(a-t)(x+f)$$
$$\displaystyle af-ax+ft-tx=ax+af-tx-ft$$
$$\displaystyle ft=ax\implies x=\frac{f}{a}t\implies y=\frac{b}{a+f}\sqrt{f^2-x^2}$$
Squaring this value for $y$, (which gives us the bottom half) we obtain the locus for $I$:
$$\displaystyle \frac{x^2}{f^2}+\frac{y^2}{\left(\dfrac{bf}{a+f} \right)^2}=1$$
which is clearly an ellipse.
Pranav
Well-known member
Hi Opalg and MarkFL!
Thank you both for your helpful solutions. I prefer to stay with parametric coordinates used by Opalg.
It was very silly of me to consider B and C as the foci.
About that incentre formula, it is also in my course and I need to memorise that .
Thank you once again. | 2021-06-23T05:52:12 | {
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https://math.stackexchange.com/questions/4435201/double-orthogonal-complement-of-a-finite-dimensional-subspace-w-r-t-a-nondegene | Double orthogonal complement of a finite dimensional subspace w.r.t. a nondegenerate symmetric bilinear form on an infinite dimensional space
Let $$U$$ be a finite-dimensional subspace of an infinite-dimensional space $$V$$ equipped with a nondegenerate symmetric bilinear form $$\phi$$. Is it necessarily true that $${(U^{\perp})}^{\perp}=U$$?
If the bilinear form happens to be an inner product, then the statement is true (see here). If $$V$$ is finite dimensional, the statement follows from dimension count. I believe the statement as formulated is false. Anyone knows a counterexample? What if I impose the extra condition that the restriction of $$\phi$$ to $$U$$ is nondegenerate?
Yes, it is necessarily true
First, note that $$U \subseteq U^{\perp\perp}$$ will always hold, regardless of degeneracy. Indeed: by the definition of $$U^{\perp}$$, $$(x,y) = 0$$ holds for all $$x \in U$$ and $$y \in U^\perp$$. Thus, if $$x \in U$$, then $$(x,y) = 0$$ holds for all $$y \in U^\perp$$. Thus, $$x$$ is an element of $$(U^\perp)^\perp$$.
The more complicated part is showing that $$U^{\perp\perp} \subseteq U$$.
Let $$d = \dim(U)$$. Let $$\{x_1,\dots,x_d\}$$ be a basis of $$U$$. Consider the map $$\phi:V \to \Bbb F^d$$ given by $$\phi(v) = [(x_1,v),\dots,(x_d,v)].$$ Notably, $$\ker(\phi) = U^\perp$$. $$\phi$$ is a surjective (onto) map. Indeed, suppose for the purpose of contradiction that $$\phi$$ fails to be surjective. It would follow that there exists there exists a non-zero vector $$c = (c_1,\dots,c_d)$$ in $$\operatorname{im}(\phi)^\perp$$, where the orthogonal complement is taken relative to the "dot-product" over $$\Bbb F^d$$ (i.e. $$c,d \mapsto c^Td$$). For this $$c$$, it holds that for all $$v \in V$$, $$c^T\phi(v) = 0 \implies\\ c_1(x_1,v) + \cdots + c_d(x_d,v) = 0 \implies\\ (c_1 x_1 + \cdots + c_d x_d,v) = 0.$$ By the non-degeneracy of the bilinear form, this implies that $$c_1 x_1 + \cdots + c_d x_d = 0$$. Because $$\{x_1,\dots,x_d\}$$ is a basis, this implies that $$c = 0$$, contradicting our statement that $$c$$ is non-zero.
This allows us to deduce that the quotient space $$V/\ker(\phi) = V/U^{\perp}$$ has dimension $$d$$, and that $$\phi$$ induces an isomorphism $$\bar \phi:(V/U^\perp) \to \Bbb F^d$$ via the quotient map.
Now, suppose that $$x \in U^{\perp\perp}$$. Consider the linear map $$\alpha:V \to \Bbb F$$ given by $$\alpha(v) = (x,v)$$. Because $$U^\perp \subseteq \ker(\alpha)$$, $$\alpha$$ induces a linear map $$\bar \alpha:(V/U^\perp) \to \Bbb F$$. Because $$\bar \phi$$ is an isomorphism, there exists a vector $$c = (c_1,\dots,c_d) \in \Bbb F^d$$ such that $$\bar \alpha(v + U^\perp) = c^T\bar\phi(v + U^\perp)$$. Correspondingly, we may conclude that for all $$v \in V$$, we have $$\alpha(v) = c^T\phi(v) \implies (x,v) = (c_1x_1 + \cdots + c_dx_d,v) \implies\\ (c_1 x_1 + \cdots + c_d x_d - x,v) = 0 \quad \text{for all } v \in V.$$ Because the bilinear form is non-degenerate, we may conclude that $$c_1 x_1 + \cdots + c_d x_d - x = 0$$, which is to say that $$x = c_1x_1 + \cdots + c_dx_d$$, which means that $$x \in U$$.
Thus, we have $$U^{\perp \perp} \subseteq U$$. Thus, we have shown that $$U = U^{\perp\perp}$$.
• If you are not comfortable with quotient spaces, then this proof rewritten without them if we extend the basis $(x_1,\dots,x_d)$ to an (infinite) basis of $V$. Apr 25 at 13:25 | 2022-08-09T01:32:00 | {
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