# sums of normal random variables need not be normal

A common misconception among students of probability theory is the belief that the sum of two normally distributed (http://planetmath.org/NormalRandomVariable) random variables is itself normally distributed. By constructing a counterexample, we show this to be false.

It is however well known that the sum of normally distributed variables $X,Y$ will be normal under either of the following situations.

• $X,Y$ are independent.

• $X,Y$ are joint normal (http://planetmath.org/JointNormalDistribution).

The statement that the sum of two independent normal random variables is itself normal is a very useful and often used property. Furthermore, when working with normal variables which are not independent, it is common to suppose that they are in fact joint normal. This can lead to the belief that this property holds always.

Another common fallacy, which our example shows to be false, is that normal random variables are independent if and only if their covariance, defined by

 $\operatorname{Cov}(X,Y)\equiv\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$

is zero. While it is certainly true that independent variables have zero covariance, the converse statement does not hold. Again, it is often supposed that they are joint normal, in which case a zero covariance will indeed imply independence.

We construct a pair of random variables $X,Y$ satisfying the following.

1. 1.

$X$ and $Y$ each have the standard normal distribution.

2. 2.

The covariance, $\operatorname{Cov}(X,Y)$, is zero.

3. 3.

The sum $X+Y$ is not normally distributed, and $X$ and $Y$ are not independent.

We start with a pair of independent random variables $X,\epsilon$ where $X$ has the standard normal distribution and $\epsilon$ takes the values $1,-1$, each with a probability of $1/2$. Then set,

 $Y=\left\{\begin{array}[]{ll}\epsilon X,&\textrm{if }|X|\leq 1,\\ -\epsilon X,&\textrm{if }|X|>1.\end{array}\right.$

If $S$ is any measurable subset of the real numbers, the symmetry of the normal distribution implies that $\mathbb{P}(X\in S)$ is equal to $\mathbb{P}(-X\in S)$. Then, by the independence of $X$ and $\epsilon$, the distribution of $Y$ conditional on $\epsilon=1$ is given by,

 $\begin{split}\displaystyle\mathbb{P}(Y\in S\mid\epsilon=1)&\displaystyle=% \mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,-X\in S)\\ &\displaystyle=\mathbb{P}(|X|\leq 1,X\in S)+\mathbb{P}(|X|>1,X\in S)=\mathbb{P% }(X\in S).\end{split}$

It can similarly be shown that $\mathbb{P}(Y\in S\mid\epsilon=-1)$ is equal to $\mathbb{P}(X\in S)$. So, $Y$ has the same distribution as $X$ and is normal with mean zero and variance one.

Using the fact that $\epsilon$ has zero mean and is independent of $X$, it is easily shown that the covariance of $X$ and $Y$ is zero.

 $\begin{split}\displaystyle\operatorname{Cov}(X,Y)&\displaystyle=\mathbb{E}[XY]% -\mathbb{E}[X]\mathbb{E}[Y]=\mathbb{E}[XY]\\ &\displaystyle=\mathbb{E}\left[1_{\{|X|\leq 1\}}\epsilon X^{2}\right]+\mathbb{% E}\left[-1_{\{|X|>1\}}\epsilon X^{2}\right]\\ &\displaystyle=\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|\leq 1\}}X^{2}% \right]-\mathbb{E}[\epsilon]\mathbb{E}\left[1_{\{|X|>1\}}X^{2}\right]\\ &\displaystyle=0.\end{split}$

As $X$ and $Y$ have zero covariance and each have variance equal to $1$, the sum $X+Y$ will have variance equal to $2$. Also, the sum satisfies

 $X+Y=\left\{\begin{array}[]{ll}2\epsilon X,&\textrm{if }|X|\leq 1,\\ 0,&\textrm{if }|X|>1.\end{array}\right.$

In particular, this shows that $\mathbb{P}(|X+Y|>2)=0$. However, normal random variables with nonzero variance always have a positive probability of being greater than any given real number. So, $X+Y$ is not normally distributed.

This also shows that, despite having zero covariance, $X$ and $Y$ are not independent. If they were, then the fact that sums of independent normals are normal would imply that $X+Y$ is normal, contradicting what we have just demonstrated.

Title sums of normal random variables need not be normal SumsOfNormalRandomVariablesNeedNotBeNormal 2013-03-22 18:43:44 2013-03-22 18:43:44 gel (22282) gel (22282) 5 gel (22282) Example msc 62E15 msc 60E05


# Thread: difference of two squares

1. ## difference of two squares

Find two integers whose squares have a difference of 1,234,567.

can you find two integers whose squares will produce a difference of any whole number you chose?

if so, how? if not, what kind of numbers can't be chosen.

well... a^2-b^2=1234567

we know a must end in a 1 or 9 and b must end in a 2 or 8 (this is to produce the seven at the end)

i came up with 617289^2-617288^2=1234577 and that's as close as i can get.

any suggestions? first time poster! sorry if it is an unusual / inappropriate post!

much thanks!

2. $n^{2}-(n-1)^{2}=1234567$

n=617284

$617284^{2}-617283^{2}=1234567$

3. Number suchs that,
$n\equiv 0,1,3 (\bmod 4)$
Are expressible as a difference of two square.

And,
$n\equiv 2 (\bmod 4)$
Are not.

4. Hello, perfect square!

Find two integers whose squares have a difference of 1,234,567
Galactus used a very clever bit of trivia.

Consecutive squares differ by consecutive odd numbers.

. . $\begin{array}{cccc} & & & \text{diff} \\ 0^2 & = & 0 & \\ & & & 1 \\ 1^2 & = & 1 & \\ & & & 3 \\ 2^2 & = & 4 & \\ & & & 5 \\ 3^2 & = & 9 & \\ & & & 7 \end{array}$
. . $\begin{array}{cccc}4^2 & = & 16 & \\ & & & 9 \\ 5^2 & = & 25 & \\ \vdots & & \vdots & \vdots \end{array}$

As he pointed out, we want two consecutive integers with a difference of 1234567.

So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$

. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$

. . Hence: . $n \,=\,617,283$

Therefore: . $617,284^2 - 617283^2 \:=\:1,234,567$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I suspect that 1,234,567 is prime (but I'm not sure).

If the difference is a composite odd number,
. . there may be an alternate solution.

Example: Find two integers whose squares differ by 133.

Since $\frac{133-1}{2} = 66$, we have: . $67^2 - 66^2 \:=\:133$

. Since $133 = 7\times 19$, we have: . . $19 + 19 + 19 + 19 + 19 + 19 + 19$
. . . . . which can be written: . . $13 + 15 + 17 + 19 + 21 + 23 + 25$
which are the differences of: . $6^2\quad7^2\quad\:8^2\quad\:9^2\quad10^2\quad11^2\ quad12^2\quad13^2$

Therefore: . $13^2 - 6^2 \:=\:133$

5. Originally Posted by Soroban
So we have: . $(n + 1)^2 - n^2 \:=\:1,234,567$

. . $n^2 + 2n + 1 - n^2 \:=\:1,234,567\quad\Rightarrow\quad 2n+1 \:=\:1,234,567\quad\Rightarrow\quad 2n \:=\:1,234,566$

. . Hence: . $n \,=\,617,283$
Such a seemingly difficult question, transformed into simple algebra! Once again! Nice!

6. I suspect that 1,234,567 is prime (but I'm not sure).

No, Soroban, it is not prime.


# Combination Proof

We will prove that:

$$\dbinom{n}{k} = \displaystyle \prod_{i=1}^{k} \frac{n-i+1}{i}$$

$$\forall$$ $$k \in \mathbb{N}$$, $$k \leq n$$

Proof #1: As the reader may guess, we will use induction to prove this. We establish a base case as $$k=1$$:

$$\dbinom{n}{1} = \displaystyle \prod_{i=1}^{1} \frac{n-i+1}{i}$$

$$\Rightarrow$$ $$\frac{n!}{1!(n-1)!} = n$$

$$\Rightarrow$$ $$n=n$$

The base case clearly holds. Now we establish the inductive case as $$n=\alpha$$:

$$\dbinom{n}{\alpha} = \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$

$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\frac{n-\alpha}{\alpha +1} \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}$$ (By hypothesis)

$$\Rightarrow$$ $$\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$

$$\Rightarrow$$ $$\frac{n!(n-\alpha)}{(\alpha +1)!(n-\alpha)!} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$

$$\Rightarrow$$ $$\frac{n!}{(\alpha +1)!(n-(\alpha+1))!} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$

$$\Rightarrow$$ $$\dbinom{n}{\alpha+1} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}$$

So we may conclude that the statement is true for $$k=1$$ and if it is true for some $$k=\alpha$$, then it must be true for $$k=\alpha+1$$. The proof follows by induction.

QED

Proof #2: We will now offer an alternative proof, which is actually just a simple direct derivation:

$$\dbinom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{\displaystyle \prod_{i=0}^{n-1} (n-i)}{\displaystyle \prod_{i=0}^{k-1} (k-i) \left( \displaystyle \prod_{i=0}^{n-k-1} (n-k-i) \right)}$$

$$= \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{k!} = \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{ \displaystyle \prod_{i=1}^k i }$$

$$= \displaystyle \prod_{i=1}^k \frac{(n-i+1)}{i}$$ Because multiplication is commutative.

Which is the intended result.

QED.

Note by Ethan Robinett
3 years, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

I've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them!

- 3 years, 4 months ago

Thanks!

- 3 years, 4 months ago

Nice note@Ethan Robinett

- 3 years, 10 months ago

Thanks appreciate it!

- 3 years, 10 months ago

Hey nice proofs! :) I did the Inductive one but didn't think of using a direct one

- 3 years, 10 months ago

Thanks! Yeah after I did the induction, I figured there was probably a way to directly show it, given that factorials are just products by definition, I thought the second one was pretty interesting

- 3 years, 10 months ago

I like the way the induction is used.

- 3 years, 10 months ago

Thanks man

- 3 years, 10 months ago


# Math Help - Simplify an Expression with Negative Exponents

1. ## Simplify an Expression with Negative Exponents

(ab)^-1 / (a^-2 + b^-2)

if you could explain it with an appropriate property, that would be great!!

thanks..!!

2. $\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$

I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.

3. Originally Posted by masters
$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$

I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
Looks good to me, but don't forget to specify that $a \neq 0$ and $b \neq 0$ independently.

-Dan

4. Hi !

A slightly different approach

$\frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{1}{(ab)(a^{-2}+b^{-2})}$

-----------------------
Using the power rule : $a^b a^c=a^{b+c}$

--> $(ab)a^{-2}=ba^{-1}$
-----------------------

$=\frac{1}{ba^{-1}+ab^{-1}}=\frac{1}{\frac ba+\frac ab}=\frac{1}{\frac{b^2+a^2}{ab}}=\frac{ab}{a^2+b^2 }$

one extra equality

5. Always keep it interesting, Moo. Good job!!

6. thank you both for your help!

7. Hello, needymathperson!

One different step . . .

$\frac{(ab)^{-1}}{a^{-2} + b^{-2}}$
We have: . $\frac{(ab)^{-1}} {a^{-2} + b^{-2}} \;\;=\;\;\frac{\dfrac{1}{ab}} {\dfrac{1}{a^2} + \dfrac{1}{b^2}}$
Multiply top and bottom by $a^2b^2\!:\;\;\frac{a^2b^2\left(\dfrac{1}{ab}\right )}{a^2b^2\left(\dfrac{1}{a^2} + \dfrac{1}{b^2}\right)} \;\;=\;\;\boxed{\frac{ab}{b^2+a^2}}$


# How to make a box which has the largest possible volume?

I have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?

Basically I cut the corners and then fold them. There is no "roof". Thank you!

$$a=0.15m$$

-
@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. – Henning Makholm Apr 7 '13 at 13:57

Let the side remaining after cutting off the vertices be A. The volume is given by:

$$V = \dfrac12 A^2 \sin(60) \times h = \dfrac{\sqrt3}{4} A^2h$$

The relation between $A$, $a$ and $h$ is:

$A = a - \dfrac{2h}{\tan(30)}$

(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)

You can substitute them now:

$$V = \dfrac{\sqrt3}{4} \left(a - \dfrac{2h}{\tan(30)}\right)^2h$$

$$V = \dfrac{\sqrt3}{4} (0.15 - 2\sqrt3h)^2h$$

$$V = 3 \sqrt3 h^3-0.45 h^2+0.00974279 h$$

Differentiate w.r.t. $h$ to get:

$$\dfrac{dV(h)}{dh} = 9 \sqrt3 h^2-0.9 h+0.00974279$$

Equate to zero and solve for $h$ to get $h\approx0.0433013$ and $h\approx0.0144338$.

The first one gives the minimum volume, so you don't want that. Take the second.

-
Thank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily! – user31113 Apr 7 '13 at 11:46
You're welcome! – Jerry Apr 7 '13 at 11:47

If you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \sqrt3 h$. (Let me know if you have difficulty with this.)

Hence the box's volume is proportional to $\left(a-2 \sqrt3 h\right)^2 h$, which we try to maximise. Let $V(h) = \left(a-2 \sqrt3 h\right)^2 (4 \sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \sqrt3 h), (a-2 \sqrt3 h)$ and $(4 \sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.

$a-2 \sqrt3 h = 4 \sqrt3 h$ or when $h = \dfrac{a}{6\sqrt3}$.

-
Oh woops, yours is definitely more elegant! – Jerry Apr 7 '13 at 11:26
@Jerry Thanks. It's just that I like inequalities a lot. – Macavity Apr 7 '13 at 11:32
Thank you Macavity. :) Cheers! – user31113 Apr 7 '13 at 11:44
@Macavity Your solution is certainly neat, but I get a slight butterfly feeling when you pull that factor out of the air. I seem to get the same numbers by a method at least I find simpler. Any comments? – Brian Chandler Dec 20 '14 at 16:24

Using exactly the same argument as in this question: Optimize volume of an open cardboard box made from flat square of cardboard... we consider an infinitesimal change $\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls. Call the area of the base $B$, the perimeter of the base $P$, and the height $h$, then these changes are:

$\delta \times h \times P$ ...change over walls

$\delta \times B$ ...change over base

And the maximum is when the area of the base and the area of the walls are equal, so we have:

$h = B / P$

In this case, let $s$ be the side of the base. Then the area of the base (an equilateral triangle) and perimeter are given by:

$B = \frac{\sqrt 3}{4} s^2$

$P = 3s$

So

$h = \frac{\sqrt 3}{4} s^2 / 3s = \frac{1}{4\sqrt 3} s$

Now we get $s$ from the original triangle side $a$ and substitute:

$s = a - 2\sqrt 3 h$

$h = \frac{1}{4\sqrt 3} (a - 2\sqrt 3 h)$

$4\sqrt 3 h = a - 2\sqrt 3 h$

$h = \frac{a}{6\sqrt 3}$

-


# Projection of ellipsoid

Find the projections of the ellipsoid

$$x^2 + y^2 + z^2 -xy -1 = 0$$

on the cordinates plan

I have no idea how to do this. I couldn't find much on google to help me with it too.

By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:

$xy$-plane: Let $z=0$.

$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.

In the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$. Implicitly differentiating with respect to $x$ gives us

$$2x+2y\frac{dy}{dx} - y - x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y-2x}{2y-x}.$$

Now, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\implies y=2x$. Substituting this back into the implicit equation gives us $$x^2+4x^2-2x^2 -1 = 0 \implies 3x^2=1 \implies x=\pm \frac{\sqrt{3}}{3}$$

and thus $y = \pm\dfrac{2\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.

Similarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\implies x=2y$. In a similar fashion, we get that $y=\pm\dfrac{\sqrt{3}}{3}$ and hence $x=\pm\dfrac{2\sqrt{3}}{3}$ are the equations of the vertical tangent lines.

Now why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\sqrt{3}/3 \leq x,y \leq 2\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively. The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\leq z\leq 1$. With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is $$\frac{3y^2}{4} + z^2 = 1$$

and the corresponding projection onto the $xz$-plane is $$\frac{3x^2}{4} + z^2 = 1$$

Note that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest.

• Yes, I meant it, thanks. Well, I did this on a first moment but the answer is different for $yz$ and $xz$. For example: $yz: \frac{3}{4}y^2 + z^2 -1 = 0$ Nov 19, 2013 at 11:17
• @Giiovanna I deleted my answer, fixed it, and then undeleted it. I hope what I wrote above makes sense! Nov 19, 2013 at 12:44

In case you have difficulties with the accepted answer, here is another approach:

Your ellipsoid $S$ is the level surface $F^{-1}(\{0\})$ of the function $$F(x,y,z):=x^2+y^2+z^2-xy-1\ .$$ Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points $p\in S$ that generate the boundary $\partial S'$ are characterized by the property that the tangent plane at $p$ is vertical (i.e., parallel to the $z$-axis), which is the same as saying that the surface normal $n_p$ at $p$ is horizontal. Now $n_p$ is given by $\nabla F(p)$, whose third coordinate is $2z$.

It follows that the points $(x,y,z)\in S$ producing the shadow boundary are the points with $z=0$, i.e., the points $(x,y,0)$ satisfying $x^2+y^2-xy-1=0$. In this special example the shadow boundary in fact coincides with the intersection of $S$ with the $(x,y)$-plane.

Therefore let's do the shadow $S''$ of $S$ in the $(x,z)$-plane as well. The points $p\in S$ that generate the boundary $\partial S''$ are characterized by the property that the tangent plane at $p$ is parallel to the $y$-axis, which is the same as saying that the surface normal $n_p$ at $p$ is orthogonal to the $y$-axis, or has $y$-component $0$. The $y$-component of $\nabla F(p)$ is given by $2y-x$. Therefore we can say that the points $p$ in question lie on the plane $2y-x=0$. As they satisfy a priori $F(x,y,z)-1=0$ we can eliminate $y$ from the two equations $$x^2+y^2+z^2-xy-1=0,\qquad y={x\over2}$$ and obtain a single equation connecting their $x$- and $z$-coordinates: $$x^2+{x^2\over4}+z^2-x\>{x\over2}-1=0\ ,$$ or $${3\over4}x^2+z^2-1=0\ .$$ This is already the equation of the shadow boundary $\partial S''$.

I may now safely leave the shadow on the $(y,z)$-plane to you.

• I like this method the most Sep 18, 2017 at 9:10
• It’s worth noting that for a quadric surface, the points that generate its shadow outline are always coplanar.
– amd
Dec 24, 2021 at 19:53

An approach related to Christian Blatter’s answer can be found in Result 8.9 on page 201 of Hartley & Zisserman’s Multiple View Geometry In Computer Vision. Under the camera matrix $\mathtt P$, the outline of the quadric $\mathtt Q$ is the conic given by $\mathtt C^*=\mathtt P\mathtt Q^*\mathtt P^T$. (Here the superscript asterisk indicates the dual conic.) The proof follows directly from the observations that lines $\mathbf l$ tangent to the outline satisfy $\mathbf l^T\mathtt C^*\mathbf l=0$ and back-project to planes $\mathtt P^T\mathbf l$ that are tangent to the original quadric.

In the present problem the quadric is the nondegenerate ellipsoid $$\mathtt Q = \begin{bmatrix}1&-\frac12&0&0 \\ -\frac12&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1\end{bmatrix}$$ so we can use inverse matrices for the duals: the outline of $\mathtt Q$ under the projection $\mathtt P$ is therefore $\mathtt C = (\mathtt P\mathtt Q^{-1}\mathtt P^T)^{-1}$.

For orthogonal projection onto the $x$-$y$ plane, we can use $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ for the camera matrix and have $$\mathtt Q^{-1}=\begin{bmatrix}\frac43&\frac23&0&0\\\frac23&\frac43&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix},$$ which gives us $$\mathtt C = \begin{bmatrix}1&-\frac12&0\\-\frac12&1&0\\0&0&-1\end{bmatrix},$$ or $x^2+y^2-xy=1$. For the $x$-$z$ plane, $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ and $$\mathtt C = \begin{bmatrix}\frac34&0&0\\0&1&0\\0&0&-1\end{bmatrix},$$ i.e., $\frac34x^2+z^2=1$. A similar computation for the $y$-$z$ plane yields $\frac34y^2+z^2=1$.

• Hello @amd, your solution works like a charm for me! Thank you, for posting this answer. Small question, how do we find the projection matrix, P, for any arbitrary plane? Dec 23, 2021 at 3:47
• @KashishDhal There are various ways, but for any of them you must first choose a coordinate system for the plane. Once you have that, the easiest thing that I can think of is to compute a coordinate transformation that takes that plane to one of the prime coordinate planes and compose it with the corresponding simple projection matrix from above.
– amd
Dec 24, 2021 at 19:49
• I followed the method that you described in this thread: math.stackexchange.com/questions/3073718/… However, I don't know if my solutions are correct. How can I verify that the computed ellipse are indeed correct? Dec 24, 2021 at 20:59


# If $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis does it necessarily follow that $\{u_1,...,u_n\}$ is linearly independent?

We are given that $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis. My question is does it necessarily follow that $\{u_1,...,u_n\}$ or $\{v_1,...,v_k\}$ are linearly independent? I suspect the answer is yes, but I think I'm missing something in my argument.

So: if $\{u_1,...,u_n,v_1,...,v_k\}$ is a basis, then this list of vectors is linearly independent. That is,

$a_1 u_1 + \cdots + a_n u_n + b_1 v_1 + \cdots + b_k v_k = 0 \implies a_1 = \cdots = a_n = b_1 = \cdots = b_k = 0$.

Then I want to say the following: Since all the $b_i \,'s$ are $0$, we have $a_1 u_1 + \cdots + a_n u_n = 0 \implies a_1 = \cdots = a_n = 0$, meaning that {$u_1,...,u_n$} is linearly independent. With a similar argument, we can also conclude that {$v_1,...,v_k$} is linearly independent.

Am I missing something though? I feel like my argument is kind of making a bit of a stretch. Thanks for your help!!

• You're right! in general, any subset of a linearly independent set is again linearly independent, as your proof shows (you're not using more than linearly independency). Oct 1 '16 at 20:25
• Yes, any subset of a linearly independent set is lin. indep. itself. Oct 1 '16 at 20:25
• It's fine for me. Actually, whether it's a basis is not the relevant hypothesis. What you proved indeed is a little more general: any sublist of a list of linearly independent vectors is linearly independent. Oct 1 '16 at 20:26
• Thank you so much guys!!! :D Oct 1 '16 at 20:36

Your argument is fine, but perhaps this is a case where it can be nicely put as a proof by contradiction. Suppose some sublist were not linearly independent, then we can find some $\{a_i\}$, not all zero, such that $\sum_{i=1}^n a_iu_i=0$. We can then use this to show that the bigger list is not linearly independent.


Tutorial/Coupled problems

Introduction

Here is explained how to use the solution of a first problem as a data in a second problem. Two kind of problem are studied here. In the "first" one, the solution of the first problem directly appears in the weak formulation of the second problem (for example, as a source or a Neumann boundary) whereas in the second kind of problem, the solution is used as a constraint (Dirichlet boundary condition), so it should appears in the function space. Note that a Lagrange multiplier could be used to force the Dirichlet constraint in the weak formulation.

First kind of coupled problem : solution appearing in the weak formulation

Mathematics

Let the computation domain be the unit square $\Omega = [0,1]\times[0,1]$ with boundary $\Gamma$ and unit outwardly directed normal $\mathbf{n}$. Consider the two coupled following problems : firstly, find $u$, solution of $$\begin{cases}\label{eq:problemU} -\Delta u + u = C & \text{in } \Omega,\\ \displaystyle{\frac{\partial u}{\partial \mathbf{n}} = 0} & \text{on }\Gamma, \end{cases}$$ where $C$ is a constant. Then, find the solution $v$ of the second problem $$\begin{cases}\label{eq:problemV} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{\frac{\partial v}{\partial \mathbf{n}} = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of problem \ref{eq:problemU}. Obviously, the function $u$ is the constant function equal to $C$ and thus, there is no "real" need in solving numerically problem \ref{eq:problemU}. However, please keep in mind that this is just an example. In order to solve numerically these problem using a finite element method, on has to write the weak formulations of problems \ref{eq:problemU} and \ref{eq:problemV}, which read as: $$\label{eq:WeakFormulationU} \left\{\begin{array}{l} \text{Find } u\in H^1(\Omega) \text{ such that}\\ \displaystyle{\forall u'\in H^1(\Omega), \qquad \int_{\Omega} \nabla u\cdot\nabla u' \;{\rm d}\Omega + \int_{\Omega} u\cdot u' \;{\rm d}\Omega - \int_{\Omega} Cu'\;{\rm d}\Omega = 0}, \end{array}\right.$$ and $$\label{eq:WeakFormulationV} \left\{\begin{array}{l} \text{Find } v\in H^1(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega - \int_{\Gamma}uv'\;{\rm d}\Gamma = 0}, \end{array}\right.$$ where $H^1(\Omega)$ is the classical Sobolev space.

Computation in GetDP

In GetDP, this kind of problem is very easy to solve. Indeed, the user has to introduce two function space (one per solution), and ask GetDP to solve each problem in the right order. There is no trap ! Below are the GMSH and GetDP files. Compared to the other academic examples, the only new things are in the .pro file.

Files

There are two files :

FirstCoupledProblem.geo that contains the geometry and is used to mesh the domain
FirstCoupledProblem.pro that contains the weak formulations, function spaces, etc.

To solve the problem, type in a terminal (in the right directory) :

gmsh FirstPb.geo -2

and then solve the problem with getdp :

getdp FirstPb.pro -solve -pos

or directly

getdp FirstPb.pro -solve CoupledProblem -pos Map_u_and_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;

// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;

//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};

//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};

// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};

//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};

// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...

Direct link to file CoupledProblems/First/FirstPb.geo'

// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}

// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}

//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}

//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints  1 ; }
{ GeoElement Line ; NumberOfPoints  4 ; }
{ GeoElement Triangle ; NumberOfPoints  6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}

//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
//Space for v
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}
}
/*
The two FunctionSpaces are exactly the same. However, keep in mind that
a FunctionSpace can contain only one solution ! That is why we need two FunctionSpaces.
*/

//(Weak) Formulations
//==================

Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}

// Coupled problem : "first kind" :
// Neumann condition on Gamma
// Delta v + v =0 Omega
// dn v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad_v;}
// u is called but will NOT be used as an unknown (NO "Dof")
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}

Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
//Neumann condition on Gama (no "Dof" !)
Galerkin{ [-{u}, {v}];
In Gama; Jacobian JSur; Integration I1;}
}
}

}

// Resolution
//===========

Resolution{
// First coupled problem available
{Name CoupledProblem;
System{
{Name PbU; NameOfFormulation ProblemU;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u" then save "u" in the space H1_C
2) generate and solve the problem associated to "v" and save "v" in the space H1
*/
}

}

//Post Processing
//===============

PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
//The associated formulation is "ProblemV" because it involves both "u" and "v"
//(ProblemU only call "u", not "v")
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}

//Post Operation
//==============

PostOperation{
{Name Map_u_and_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}

Second kind of coupled problem : solution used as a Dirichlet boundary condition

Mathematics

In this section, the first problem (\ref{eq:problemU}) does not change but the second problem is now to find $v$, such that $$\begin{cases}\label{eq:problemV2} -\Delta v + v = 0 & \text{in } \Omega,\\ \displaystyle{v = u} & \text{on }\Gamma, \end{cases}$$ where $u$ is the solution of (\ref{eq:problemU}). The function $u$ is now involved as a Dirichlet boundary condition and thus, will appear in the function space of $v$ (again, except if one uses a Lagrange multiplier, which is not the case here). To obtain the weak formulation of problem (\ref{eq:problemV2}), we introduce the following function space \ref{eq:WeakFormulationU} $$H^1_u(\Omega) = \left\{w\in H^1(\Omega) \text{ such that w|_{\Gamma} = u|_{\Gamma}, where u is the solution of problem (1)} \right\}.$$ Thus, we have $$\label{eq:WeakFormulationV2} \left\{\begin{array}{l} \text{Find } v\in H^1_u(\Omega) \text{ such that, }\\ \displaystyle{\forall v'\in H^1_0(\Omega), \qquad \int_{\Omega} \nabla v\cdot\nabla v' \;{\rm d}\Omega + \int_{\Omega}vv' \;{\rm d}\Omega = 0}, \end{array}\right.$$

GetDP

In GetDP, a Dirichlet boundary condition is generally imposed through the "Constraint" term, it can be imposed with the help of a Lagrange multiplier. In the second case, the Dirichlet boundary condition is no more "imposed" in the function space but appears in the weak formulation (as a Neumann boundary condition). Thus, the previous case can be used. Here is only considered the first case using the GetDP function "AssignFromResolution". A Resolution dedicated to $u$ must then be created, in this example it will be named AuxiliaryResolution. In GetDP, the procedure can be summarized as follows:

• Create a constraint with :
Type AssignFromResolution ; Resolution AuxiliaryResolution;
• Construct only ONE function space that will contain $u$ and at the end $v$ ($u$ will be erased !)
• Create two Resolution, one for $v$ and one for $u$ :
• for $v$ : Generate, Solve, Save
• for $u$ (AuxiliaryResolution) : Generate, Solve, TransfertSolution (do not Save !)

With this, GetDP will ... :

• Try to solve problem with $v$
• Need the Dirichlet constraint (that is $u$)
• Solve problem with $u$
• Solve problem with $v$

Sometimes it is usefull to save both $u$ and $v$ at the end of the calculation. This can be done by adding an auxiliary problem that "copy" $u$ (see the last paragraph).

Files

There are two files :

Secondpb.geo
exactly the same as FirstPb.geo
Secondpb.pro
quite the same as FirstPb.pro

To solve the problem, type in a terminal (in the right directory) :

gmsh SecondPb.geo -2

and then solve the problem with getdp :

getdp SecondPb.pro -solve -pos

and chose the first resolution (MainResolution) or directly

getdp SecondPb.pro -solve MainResolution -pos Map_v
//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;

// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;

//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};

//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};

// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};

//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};

// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...

Direct link to file CoupledProblems/Second/SecondPb.geo'

// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}

// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}

//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}

//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints  1 ; }
{ GeoElement Line ; NumberOfPoints  4 ; }
{ GeoElement Triangle ; NumberOfPoints  6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}

// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/

//FunctionSpace
//=============
FunctionSpace{
//Space for u AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}
/*
ONLY one function space that will contain "u" and at the end, "v".
*/

//(Weak) Formulations
//==================

Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}

// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name v; Type Local; NameOfSpace Hgrad;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}

Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}

}

// Resolution
//===========

Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
generate and solve the problem associated to "v" and save "v" in the space H1
However, to compute "v", GetDP need to compute the Dirichlet Contraint, thus
is needs to solve AuxiliaryResolution (bellow) !
*/

// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; TransferSolution[PbU];
}
}
/*Short explanations : GetDP will :
1) generate and solve the problem associated to "u"
2) Transfert solution u (to problem associated to "v")
*/
}

//Post Processing
//===============

PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}

//Post Operation
//==============

PostOperation{
{Name Map_v; NameOfPostProcessing CoupledProblem;
Operation{
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}

Second kind of coupled problem with a save of solution $u$

Sometimes, it can be usefull to save both $u$ and $v$. To achieved this, one can introduce an auxiliary variable $u_{aux}$ that will be a copy of $u$. Here, we proposed to add :

A new FunctionSpace to store $u$
A new weak formulation to copy $u$ into an auxiliary variable $u_{aux}$

The modified .pro file (SecondPbWithSave.pro) can be found below. The .geo file being exactly the same as the previous ones.

//Caracteristic length of the finite elements (reffinement is also possible after the mesh is built):
lc = 0.05;

// The parameters of the border of the domain :
x_max = 1;
x_min = 0;
y_max= 1;
y_min = 0;

//Creation of the 4 angle points of the domain Omega (=square)
p1 = newp; Point(p1) = {x_min,y_min,0,lc};
p2 = newp; Point(p2) = {x_min,y_max,0,lc};
p3 = newp; Point(p3) = {x_max,y_max,0,lc};
p4 = newp; Point(p4) = {x_max,y_min,0,lc};

//The four edges of the square
L1 = newreg; Line(L1) = {p1,p2};
L2 = newreg; Line(L2) = {p2,p3};
L3 = newreg; Line(L3) = {p3,p4};
L4 = newreg; Line(L4) = {p4,p1};

// Line Loop (= boundary of the square)
Bound = newreg; Line Loop(Bound) = {L1,L2,L3,L4};

//Surface of the square
SurfaceOmega = newreg; Plane Surface(SurfaceOmega) = {Bound};

// To conclude, we define the physical entities, that is "what GetDP could see/use".
// 1 is associtated to Omega
// 2 to Gamma
Physical Surface(1) = {SurfaceOmega};
Physical Line(2) = {L1,L2,L3,L4};
// Do not forget to let a blank line at the end, this could make GMSH crash...

Direct link to file CoupledProblems/SecondWithSave/SecondPbWithSave.geo'

// Group
//======
Group{
Omega = Region[{1}];
Gama = Region[{2}];
}

// Function
//=========
Function{
C = 5; //constant C in the problem solved by u
}

//Jacobian
//========
Jacobian {
{ Name JVol ;
Case {
{ Region All ; Jacobian Vol ; }
}
}
{ Name JSur ;
Case {
{ Region All ; Jacobian Sur ; }
}
}
{ Name JLin ;
Case {
{ Region All ; Jacobian Lin ; }
}
}
}

//Integration (parameters)
//=======================
Integration {
{ Name I1 ;
Case {
{ Type Gauss ;
Case {
{ GeoElement Point ; NumberOfPoints  1 ; }
{ GeoElement Line ; NumberOfPoints  4 ; }
{ GeoElement Triangle ; NumberOfPoints  6 ; }
{ GeoElement Quadrangle ; NumberOfPoints 7 ; }
{ GeoElement Tetrahedron ; NumberOfPoints 15 ; }
{ GeoElement Hexahedron ; NumberOfPoints 34 ; }
}
}
}
}
}

// Dirichlet Constraint
Constraint{
{Name DirichletV;
Case{
{Region Gama; Type AssignFromResolution; NameOfResolution AuxiliaryResolution; }
}
}
}
/*
The constraint will be obtained through the resolution of "AuxiliaryResolution".
This means that GetDP will automatically solve "AuxiliaryResolution" when it will
need the contraint !
*/

//FunctionSpace
//=============
FunctionSpace{
//Space for u
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
}

//Space for u_aux (=copy of u) AND v !!
BasisFunction{
{Name wn; NameOfCoef vn; Function BF_Node;
Support Region[{Omega, Gama}]; Entity NodesOf[All];}
}
//the dirichlet constraint "v = u" on Gamma
Constraint{
{NameOfCoef vn; EntityType NodesOf;
NameOfConstraint DirichletV;}
}
}
}

//(Weak) Formulations
//==================

Formulation{
// Problem solved by u :
// Delta u + u = C (Omega)
// dn u = 0 (Gamma)
{Name ProblemU; Type FemEquation;
Quantity{
{Name u; Type Local; NameOfSpace Hgrad_u;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [Dof{u}, {u}];
In Omega; Jacobian JVol; Integration I1;}
Galerkin{ [-C, {u}];
In Omega; Jacobian JVol; Integration I1;}
}
}

// Coupled problem : "first kind" :
// DIRICHLET condition on Gamma
// Delta v + v =0 Omega
// v = u on Gamma
{Name ProblemV; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name v; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
In Omega; Jacobian JVol; Integration I1;}

Galerkin{ [Dof{v}, {v}];
In Omega; Jacobian JVol; Integration I1;}
}
}

//Copy "u" in "u_aux"
{Name CopyU; Type FemEquation;
Quantity{
// v is the unknown
{Name u; Type Local; NameOfSpace Hgrad_u;}
{Name u_aux; Type Local; NameOfSpace Hgrad_v;}
}
Equation{
Galerkin{ [Dof{u_aux}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}

Galerkin{ [-{u}, {u_aux}];
In Omega; Jacobian JVol; Integration I1;}
}
}

}

// Resolution
//===========

Resolution{
//Main Resolution : the one that will be called when launching getdp
{Name MainResolution;
System{
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbV]; Solve[PbV]; SaveSolution[PbV];
}
}
/*Short explanations : GetDP will :
Try to compute "v", however, GetDP needs to compute the Dirichlet Contraint
thus to solve AuxiliaryResolution (below) !
*/

// Auxiliary resolution that computes "u"
{Name AuxiliaryResolution;
System{
//System associated to "u" will be transfered to the weak formulation of "v"
{Name PbU; NameOfFormulation ProblemU;}
{Name PbUaux; NameOfFormulation CopyU; DestinationSystem PbV;}
{Name PbV; NameOfFormulation ProblemV;}
}
Operation{
Generate[PbU]; Solve[PbU]; SaveSolution[PbU];
Generate[PbUaux]; Solve[PbUaux]; TransferSolution[PbUaux];
}
}
/*Short explanations : GetDP will :
1) Compute "u"
2) Copy "u" in "uaux"
3) Transfert solution "uaux" (to problem associated to "v")
*/
}

//Post Processing
//===============

PostProcessing{
{Name CoupledProblem; NameOfFormulation ProblemV;
Quantity{
{Name u; Value {Local{[{u}];In Omega; Jacobian JVol;}}}
{Name v; Value {Local{[{v}];In Omega; Jacobian JVol;}}}
}
}
}

//Post Operation
//==============

PostOperation{
{Name Map_Sol; NameOfPostProcessing CoupledProblem;
Operation{
Print[u, OnElementsOf Omega, File "sol_u.pos"];
Print[v, OnElementsOf Omega, File "sol_v.pos"];
}
}
}


Prove that the number of answers for $|a_1|+|a_2|+…+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+…+|a_n| \le k$.

I'm trying to solve this problem:

Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|≤k$. all $a_i$ is an integer number.

I have a solution for non negative $a_i$: We convert $|a_1|+|a_2|+...+|a_k|≤n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|≤k$ to $a_1+a_2+...+a_n+c=k$.

By bars and stars theorem: Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\binom{n+k}{n}$. That $\binom{n+k}{k} = \binom{n+k}{n}$.

but i don't know how to prove for negative integers.

• So we allow $a_i$ to be any integer, but we consider its absolute value when taking the sum? – hardmath Apr 19 '18 at 0:41
• @hardmath yes,we take absolute value – user547757 Apr 19 '18 at 14:57

Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$. Note that $$f(n,k) = f(n-1,k)+2f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0),\tag{*}$$ by conditioning on the value of $a_n$. Rearranging, and using $(*)$ applied to $f(n,k-1)$, we get \begin{align} f(n,k) &= f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]\\ &= f(n-1,k)+f(n-1,k-1)+f(n,k-1) \end{align} The above shows that $f(n,k)$ obeys a recurrence which is symmetric in $n,k$. Since you can easily verify the symmetric base cases $f(n,0)=f(0,k)=1$, the result $f(n,k)=f(k,n)$ follows by induction.

We count the number of solutions to $\sum_{i=1}^n |a_i|\leq k$.

There will be $j\leq r:={\rm min}\{n,k\}$ of the $a_i$ that are $\ne0$, call them $a_{i_1}$, $\ldots$, $a_{i_j}$. The $i_l$ $(1\leq l\leq j)$ can be chosen in ${n\choose j}$ ways. For the chosen $i_l$ put $|a_{i_l}|=x_l+1$ with $x_l\geq 0$. Then we have to count the number of solutions to $x_1+\ldots+x_j+c=k-j$. This number is ${k\choose j}$. It has to be multiplied by $2^j$ in order to account for the choice of the signs of the corresponding $a_{i_l}$. The total number of solutions therefore comes to $$\sum_{j=0}^r{n\choose j}{k\choose j}2^j\ ,$$ which is symmetric in $n$ and $k$.

Introducing notation $$A_i=|a_i|$$ the first of inequalities can be rewritten as: $$A_1+A_2+...+A_k\le n.\tag1$$ Any solution to $$(1)$$ can be viewed as composed of $$i$$ zeros ($$0\le i\le k$$) and $$k-i$$ non-zeros. Each non-zero value doubles the number of solutions to original equation (which can be both positive and negative). The overall number of solutions is therefore $$\sum_{i=0}^{k}\binom{k}{i}\binom{n}{k-i}2^{k-i} \stackrel{i=k-j}=\sum_{j=0}^{k}\binom{k}{j}\binom{n}{j}2^{j} =\sum_{j=0}^{\min(k,n)}\binom{k}{j}\binom{n}{j}2^{j},\tag2$$ where $$\binom{k}{i}$$ counts the number of ways to place $$i$$ zeros into $$k$$ cells, and $$\binom{n}{k-i}$$ counts the number of positive solutions to the inequality $$p_1+p_2+\cdots+p_{k-i}\le n.$$

In view of symmetry of $$(2)$$ with respect to interchange of $$n$$ and $$k$$ the original statement is proved.

Just to show another approach to the answer already given by C.Blatter, let's start and consider $$\left\{ \matrix{ 0 \le b_{\,j} \in \mathbb Z \hfill \cr b_{\,1} + b_2 + b_{\,3} + \cdots + b_{\,k} \le n \hfill \cr} \right.$$ Then the number of solutions is the integral volume $V(k,n)$ of k-D simplex, with edge segments $(0,n)$, thus of length $n+1$.
It is not difficult to demonstrate that $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k}=V(n,k)$$ then the hypothesis is already true in this case.

We can then partition the above solutions into those that contains $j$ variables null and the remaining $k-j$ having values greater than $0$. We can assign the null variables in $\binom{k}{j}$ ways.
That corresponds to $$V(k,n)= \binom{n+k}{n}= \binom{n+k}{k} = \sum\limits_{0\, \le \,j\, \le \,k} { \binom{n}{k-j} \binom{k}{j} }$$

Coming to our present case, for each non null variable $b_j$ we can assign two values to the corresponding $a_j$, and just one to the null variables. So in this case the volume will be $$V_{\,a} \,(k,n) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\binom{n}{k-j} \binom{k}{j} 2^{\,k - j} } = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\binom{n}{l} \binom{k}{l} 2^l } = V_{\,a} \,(n,k)$$

Therefore the hypothesis is demonstrated.

I just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is $$\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}$$ which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction.


# Upper limit on the central binomial coefficient

What is the tightest upper bound we can establish on the central binomial coefficients $2n \choose n$ ?

I just tried to proceed a bit, like this:

$n! > n^{\frac{n}{2}}$

for all $n>2$. Thus,

$\binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n$

But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).

• The simplest upper bound to prove is $4^n$ (which is still stronger than your bound) and just follows from the binomial expansion of $(1+1)^{2n}$. Peter's answer gives a less wasteful estimate. Jun 14 '13 at 13:00
• I should have thought about it. Yes, Peter's bound is very good, I tested it too. Jun 14 '13 at 13:54

Here's a way to motivate and refine the argument that Péter Komjáth attributes to Erdős.

Start by computing the ratio between the $$n$$-th and $$(n-1)$$-st central binomial coefficients: $${2n \choose n} \left/ {2(n-1) \choose n-1} \right. = \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2} = \frac{(2n)(2n-1)}{n^2} = 4 \left( 1 - \frac1{2n} \right).$$ For large $$n$$, this ratio approaches $$4$$, suggesting that $$2n \choose n$$ grows roughly as $$4^n$$. If the factor $$1 - \frac1{2n}$$ were $$1 - \frac1n = (n-1)/n$$, the growth would be exactly proportional to $$n^{-1} 4^n$$. Since $$1 - \frac1{2n}$$ is (for large $$n$$) nearly the square root of $$1 - \frac1n$$, the actual asymptotic should be proportional to $$n^{-1/2} 4^n$$. So we introduce the ratio $$r_n := \left( {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right. \right)^2 = \frac{n}{16^n} {2n \choose n}^2.$$ Then $$\frac{r_n}{r_{n-1}} = \left( 1 - \frac1{2n} \right)^2 \left/ \left( 1 - \frac1n \right) \right. = \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1.$$ Thus $$r_{n-1} < r_n$$; and since $$r_1 = (2/4)^2 = 1/4$$ we have by induction $$r_1 \lt r_2 \lt r_3 \lt r_4 \lt \cdots \lt r_n = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)} \frac{2n-1}{2n}.$$ Each $$r_{n_0}$$ gives a lower bound on $$r_n$$, and thus on $$2n\choose n$$, for all $$n \geq n_0$$. The OP asked for upper bounds, so consider $$R_n := \frac{2n}{2n-1} r_n = \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2.$$ Now $$R_{n+1}/R_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2 = (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$$, so $$\frac12 = R_1 \gt R_2 \gt R_3 \gt R_4 \gt \cdots \gt R_{n+1} = \frac12 \frac{1 \cdot 3}{2 \cdot 2} \frac{3 \cdot 5}{4 \cdot 4} \frac{5 \cdot 7}{6 \cdot 6} \cdots \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.$$ It follows that $$R_n \geq r_{n'}$$ for any $$n,n'$$, so $$R_1=1/2$$, $$R_2=3/8$$, $$R_3=45/128$$, etc. are a series of upper bounds on every $$r_n$$. Since moreover $$r_n / R_n = 1 - \frac1{2n} \rightarrow 1$$ as $$n \rightarrow \infty$$, both $$r_n$$ and $$R_n$$ converge to a common limit that is an upper bound on every $$r_n$$. If we accept Wallis's product (which is classical though not as elementary as everything else in our analysis), then we can evaluate this common limit as $$1/\pi$$ and thus recover the asymptotically sharp upper bound $${2n \choose n} < 4^n / \sqrt{\pi n}$$.

• +1, and let me challenge your statement that the Wallis product is not as elementary, by pointing to the Monthly note math.chalmers.se/~wastlund/monthly.pdf and the 16th Christmas Tree Lecture by Don Knuth, "Why pi?" (now on youtube). Jul 1 '13 at 16:58
• Your answer has been cited in this paper that I was just reading, mentioning it here as a token of appreciation: arxiv.org/pdf/2005.10009.pdf
– NULL
Jul 28 '20 at 19:59

Even the asymptotically sharp inequality ${2n \choose n} < 4^n \left/ \sqrt{\pi n} \right.$ has a short proof: $${2n \choose n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx < \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx = \frac{4^n}{\sqrt{\pi n}}.$$ In the first step, the formula for $\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$ can be proved by induction via integration by parts, or using the Beta function.

The third step is clear, and the last step is the well-known Gaussian integral. So we need only justify the the second step.

There we need the inequality $\cos x \leq e^{-x^2/2}$, or equivalently $$\log \cos x + \frac{x^2}{2} \leq 0,$$ for $\left|x\right| < \pi/2$, with equality only at $x=0$. This is true because $\log \cos x +\frac12 x^2$ is an even function of $x$ that vanishes at $x=0$ and whose second derivative $-\tan^2 x$ is negative for all nonzero $x \in (-\pi/2, \pi/2)$. QED

• I'm not sure what counts as elementary in this game, but I would think the easiest proof that $\frac{1}{\pi} \int \cos^{2n} x dx = \binom{2n}{n}/4^n$ is to write $\cos x = (e^{ix} + e^{-ix})/2$ and expand $\cos^{2n}$ by the binomial theorem. Jul 1 '13 at 14:18

Erdos remarked somewhere the bound
$${{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}.$$ This can be established by induction: $${{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}$$ and if we have the bound for $n$, we only have to show $$\frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}}$$ which reduces to $4n^2+8n+3<4n^2+8n+4$.

• This bound is pretty close to optimal because (unless I've computed badly) Stirling's approximation gives $4^n/\sqrt{\pi n}$. Jun 14 '13 at 13:26

Noam Elkies notes that there is a quick proof of $$\binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$ by writing $$\binom{2n}{n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x dx$$ and bounding $\cos^2 x \leq e^{-x^2}$.

There is an equally good lower bound by a similar method: $$\int_{-\pi/2}^{\pi/2} \cos^{2n} x dx =\int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{(\tan^2 x+1)^{n+1}} = \int_{- \infty}^{\infty} \frac{du}{(1+u^2)^{n+1}} \geq \int_{- \infty}^{\infty} e^{-(n+1) u^2} du$$ so $$\binom{2n}{n} \geq \frac{4^n}{\sqrt{\pi (n+1)}}.$$ Here the inequality $\tfrac{1}{1+u^2} \geq e^{-u^2}$ follows from the standard bound $e^y \geq 1+y$ for $y\geq 0$.

We may prove without using integrals that $$\frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1)$$ (1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$ (the identities are straightforward by induction).

Denote $$m:=2n+1$$. It is not hard to show that $$\sin mx=p_m(\sin x)$$ for a polynomial of degree $$m$$ in $$\sin x$$. The roots of $$p_m$$ are $$\sin \frac{k\pi}{2n+1}$$ for $$k=-n,\ldots,n$$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$ (the multiple $$2n+1$$ comes from dividing by $$x$$ and putting $$x=0$$). Put $$x=\frac{\pi}{4n+2}$$. Using the inequality $$\sin tx\leqslant t\sin x$$ (which may be proved, for example, by induction in $$t=1,2,\ldots$$ from the identity $$\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$$) for $$t=2k$$ ($$k=1,2,\ldots,n$$) we get $$1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi.$$ Analogously we get $$\cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right).$$

Dividing by $$1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$$ and substituting $$x=\frac{\pi}{4n}$$ (for computing the LHS at this point use l'Hôpital rule) we get $$n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi.$$

UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $$\sum 1/n^2=\pi^2/6$$ and $$\sum (-1)^{k-1}/(2k-1)=\pi/4$$ using these trigonometric things.)

• Nice proof! Is there an easy way to get the tight lower bound $1/\sqrt{\pi n+1}$? It has such a nice form..
– aorq
Jan 1 at 2:54
• @aorq It is not so much tight. The sequence $4^{-n}{2n\choose n}\sqrt{n+c}$ eventually decreases if $c>1/4$ (this is straightforward: divide the squares of two consecutive terms and subtract 1) and its limit equals $1/\sqrt{\pi}$. Thus for large enough $n$ we get the upper bound $1/\sqrt{\pi (n+c)}$. For $c=1/\pi$ this just works from the very beginning. Jan 1 at 9:51
• Yes, I know that it's tight in the weakest possible sense, namely equality for $n=0$ and poor otherwise. Shortly after I wrote my comment, I saw that $1/\sqrt{\pi(n+1/4)}$ was tighter in a more meaningful sense (on the other side), namely ever better as $n\to\infty$. Thanks for your explanation!
– aorq
Jan 1 at 14:41
• Warning: $1/\sqrt{\pi(n+1/4)}$ is an upper bound Jan 1 at 15:08

You may also prove $${2n\choose n}\sim \frac{4^n}{\sqrt{\pi n}}$$ using the combinatorial meaning of $${2n\choose n}$$ and the area-of-a-circle definition of $$\pi$$. That is probably the most elementary we can hope for.

First of all, if $$a_n:=\sqrt{n+1/4}{2n\choose n}4^{-n}$$ and $$b_n:=\sqrt{n+1/2}{2n\choose n}4^{-n}$$, we may check that $$a_n\leqslant b_n$$, $$a_n$$ increases and $$b_n$$ decreases, thus $$a_n$$, $$b_n$$ have a common positive finite limit which we denote $$\alpha$$, and $$a_n\leqslant \alpha\leqslant b_n$$. Since $$b_n/a_n=1+O(1/n)$$, we get $$a_n=\alpha+O(1/n)$$, and also $$c_n=\alpha+O(1/n)$$ where $$c_n:=\sqrt{n}{2n\choose n}4^{-n}$$.

We should prove that $$\alpha=1/\sqrt{\pi}$$.

Use the combinatorial identity $$4^{-n}\sum_{k=0}^n k{2k\choose k}(n-k){2(n-k)\choose n-k}=\frac{n(n-1)}8.\quad (\heartsuit)$$ LHS of $$(\heartsuit)$$ equals $$\sum_{k=0}^n \sqrt{k(n-k)}c_kc_{n-k}=\sum_{k=0}^n \sqrt{k(n-k)}\left(\alpha^2+O\left(\frac1k+\frac1{n-k}\right)\right)\\=\alpha^2\sum_{k=0}^n\sqrt{k(n-k)}+O\left(n\sum_{k=0}^n\frac1{\sqrt{k(n-k)}}\right)=\alpha^2\cdot \frac{\pi n^2}8+o(n^2),$$ since the union of rectangles $$[k,k+1]\times [0,\sqrt{k(n-k)}]$$ approximates the upper semi-circle with the horizontal diameter $$[0,n]$$.

Thus $$\alpha^2\pi/8=1/8$$ and $$\alpha=1/\sqrt{\pi}$$.

You could use instead of $$(\heartsuit)$$ the more famous identity $$4^{-n}\sum_{k=0}^n {2k\choose k}{2(n-k)\choose n-k}=1$$ this gives $$\alpha^2\int_0^1\frac{dx}{\sqrt{x(1-x)}}=1$$ (the integral equals $$\pi$$ of course, but this is less "elementary" then the area of a circle.)

This argument is similar to J. Wästlund's (An elementary proof of Wallis' product formula for $$\pi$$, 2007).

Since it gives tighter bounds, I will reproduce my answer from MathSE.

For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\le\frac{n+\frac13}{n+\frac43}\tag{1} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2} \end{align} Inequality $$(2)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}$$ For $$n\ge0$$, we have (by cross-multiplication) \begin{align} \left(\frac{n+\frac12}{n+1}\right)^2 &=\frac{n^2+n+\frac14}{n^2+2n+1}\\ &\ge\frac{n+\frac14}{n+\frac54}\tag{4} \end{align} Therefore, \begin{align} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}} &=4\frac{n+\frac12}{n+1}\\ &\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5} \end{align} Inequality $$(5)$$ implies that $$\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}$$ Note that the formula in $$(3)$$, which is decreasing, is bigger than the formula in $$(6)$$, which is increasing. Their ratio tends to $$1$$; therefore, they tend to a common limit, $$L$$.

Theorem $$1$$ from this answer says $$\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}$$ which means that \begin{align} L &=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\ &=\frac1{\sqrt\pi}\tag8 \end{align} Combining $$(3)$$, $$(6)$$, and $$(8)$$, we get $$\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9$$


# Homework Help: Probability question

Tags:
1. Jan 6, 2017

### gruba

Two players, $A$ and $B$, are playing a sequence of matches of some game. In each match, one player wins (there is no draw game in any match). For winning the match, the player gets $1$ point, and for losing, $0$ points. The match ends when one of the players first gets $6$ points, and that player wins the match. At the beginning of each match, players invest the same amount of the stake.
Assume that the game is interrupted at the result $4:3$ for player $A$. If the performance of both player are approximately good, what needs to be the ratio of stakes such that the split of stakes is righteous?

Question: How to approach this problem? What methods in probability are useful?

2. Jan 6, 2017

### PeroK

I guess you have to calculate how likely it is that player A would win the match from 4-3.

3. Jan 6, 2017

### Ray Vickson

Model the system as a Markov chain on a state-space $S = \{ -6,-5, \ldots, -1,0,1, \ldots, 5 ,6\}$. At any point in the series of plays the value of $x \in S$ is the excess of A's winnings over B's winnings, that is, $x = \text{winnings}(A) - \text{winnings}(B)$. The game ends when the system reaches one of the two end-states -6 and +6. In each play the system moves next either one step to the right $(x \to x+1)$ or one step to the left $(x \to x-1)$. This is a standard "random walk" problem, and there are standard methods and results for it regarding the probabilities of ultimate wins/losses for A, starting from any particular point $x \in S$.

4. Jan 7, 2017

### gruba

Could you give a reference to the similar solved problem type if you know, or could you give detailed explanation of the problem?

5. Jan 7, 2017

6. Jan 7, 2017

### PeroK

It's not that complicated. Try to model what would happen if they continued the game. To help you get started, the next play would result in:

A score of 4-4 (with probability 1/2)
A score of 5-3 (with probability 1/2)

7. Jan 7, 2017

### Ray Vickson

Google "gamblers ruin problem" or "random walk and ruin problems".

8. Jan 7, 2017

### Ray Vickson

I should point out that the random walk method would be appropriate if we were looking at net winnings for both players, and stop the game when one of the players has net winnings reaching +6. Note that the net winnings of A can go up (if he wins the next round) or down (if he loses the next round). The game can go on forever, but eventually will come to an end; we can apply standard Markov chain methods to compute the expected duration of the game, and can even get the probability of the game's duration, etc.

However, after re-reading the original problem a bit more carefully I realize that the above interpretation is incorrect. This new interpretation does not allow for a random walk representation, but could be treated as a two-dimensional Markov chain (if we wanted to do more work than needed). Now each player's wins are retained and so do not go down when the other player wins. Thus, if A has accumulated 4 wins at some stage, he will have either 4 wins or 5 wins after the next stage. The game is over when one of the players reaches 6 wins.

In this version the game has a finite duration, because someone will win in at most 11 plays.

From some point $(n_A,n_B)$ ($n_A$ wins for A, etc) we can just go ahead and enumerate all the paths leading to a "stop" state; there are a modest number of them because on an (x,y) grid---with a move to the right if A wins or a move up if B wins---there are a limited number of different paths.

9. Jan 7, 2017

### haruspex

Another way to look at the problem is to rephrase it in the form: A needs to win at least r of the next n games. Can you see what r and n are?

10. Jan 8, 2017

### StoneTemplePython

The original post is a thinly veiled version of the Problem of Points -- a foundational problem in probability. Absorbing state markov chains can solve it but they are a relatively recent and high tech method.

This problem, in effect, created the field of probability. Fermat solved it using counting. (Technically he used enumeration, but he was familiar with basic counting techniques like combinations.) Pascal solved it by drawing a tree and doing backward induction. I recommend doing both. If you are only going to do one, do Pascal's method. Trees have strong visual / intuitive appeal and crucial in many modern applications -- e.g. in computing.

11. Jan 9, 2017

### gruba

An article https://en.wikipedia.org/wiki/Problem_of_points
says that in a game where one player needs r points to win (the one closer to winning) and the other needs s points to win, the correct division of the stakes is in the ratio of:
$$\sum_{k=0}^{s-1}{r+s-1 \choose k} : \sum_{k=s}^{r+s-1}{r+s-1 \choose k}$$
in favor of the player who is closer to winning.

This ratio is valid when draw games are included.
Setting $r=2,s=3$, we get that the split of stakes is $11:5$ in favor of the player $A$.

How could this ratio formula be used when draw games are excluded?

12. Jan 9, 2017

### Ray Vickson

After a win is declared (in the excluded-games case) you could just continue playing useless games until a total of r+s-1 games have been played. That does not alter the win/lose probabilities. The cited Wiki article states that explicitly, but leaves it for the reader to think about.

In modern language, if A needs to win $a$ and B needs to win $b$, the probability that A wins the tournament is the probability of at least $a$ A-wins in $a+b-1$ rounds, which is $P(X_A \geq a)$, with $X_A \sim \text{Binomial}(a+b-1,1/2)$.

Last edited: Jan 9, 2017


# how many base $10$ decimal expansions can a real number have?

A somewhat unintuitive result of real analysis is that decimal expansions are not unique. For example, $$0.99999...=1.$$ So it can be gathered that every real number has at least one base-$$10$$ decimal expansion, sometimes even two. But is two the maximum? Are there any real numbers with three different base-$$10$$ decimal expansions?

Intuitively, I would think not, but I have close to no idea how to prove it other than knowing that the easiest method would be a proof by contraction.

It may help to restrict the problem by considering the expansions of numbers only in $$[0,1]$$. That is because if $$a\in[0,1]$$ has more than two base-$$10$$ decimal expansions, then so does $$a+x$$ for all $$x\in\Bbb R$$. And likewise, if $$x\in\Bbb R\setminus [0,1]$$ has more than two base-$$10$$ expansions, it can be written as $$x=\mathrm{sgn}(x)(\lfloor x\rfloor+a)$$, where $$a\in[0,1)$$, and by necessity $$a$$ has more than two base-$$10$$ expansions.

To be clear, I should define what I mean by base-$$10$$ expansion. Let $$N\in[0,1]$$. Then a decimal expansion of $$N$$ is a sequence $$\delta=(\delta_0,\delta_1,\delta_2,...)$$ of integers $$0\le \delta_i\le 9$$ such that $$\sigma(\delta):=\sum_{i\ge0}\frac{\delta_i}{10^i}=N.$$ Furthermore, let $$\mathcal U=\{(a_0,a_1,a_2,...):0\le a_i\le 9,\, a_i\in\Bbb Z\}$$ and let $$D_N=\{\delta\in\mathcal U :\sigma(\delta)=N\}.$$ Lastly, let $$\mathcal C_k=\{N\in[0,1]:\#(D_N)\ge k\},\qquad k\in\Bbb N$$ where $$\#(S)$$ is the number of elements in the set $$S$$.

So, is $$\mathcal C_k$$ empty for $$k>2$$?

• The only way to get two is if the digits eventually stabilize to zero. You seem to know that if you're in such a case that there are indeed two. If they don't stabilize to zero, then let $\delta_n$ be the first digit that two representations differ. Then sum over $\Sigma_n^\infty 10^{-k}(\delta_k - \delta_k')$ and this will be less than one (but not zero) so mod $1$ they'll be different numbers. For a number that eventually stabilizes (say at decimal $n$), multiply by $10^n$ and just use a limit argument from either side on your interval $[0,1]$. Feb 9 at 20:36
• Would $1.000000000.....$ with the "last" digit of $1$ after an infinite numbers of $0$ would qualify as a different expansion? Feb 9 at 20:36
• @Andrei Does it fit the given definition of decimal expansion? If so, yes, if not, no.
– user239203
Feb 9 at 20:40
• @Andrei yes it would count, as $\delta=(1,0,0,0,...)$ does indeed satisfy $$\sigma(\delta)=1$$ Feb 9 at 20:43

$$\cal C_k$$ is indeed empty for $$k > 2$$.

Suppose that a number has two distinct decimal expansions $$\delta = (\delta_0, \delta_1, \delta_2, \ldots)$$ and $$\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$$. Then, we must have

$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$

Wlog, we may assume that $$\delta_0 \neq \delta'_0$$. (There must be some smallest $$i$$ for which they are unequal. By multiplying by a suitable power of $$10$$, we may assume that it is $$i = 0$$.) Furthermore, we may assume that $$\delta_0 > \delta'_0$$.

Thus, we have $$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$

Taking absolute value on all sides and using triangle inequality for series, we see that

$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$

Thus, we have equality throughout. Note that this means that $$|\delta'_i - \delta_i| = 9$$ for all $$i$$. In fact, we see that $$\delta'_i - \delta_i$$ must have the same sign for all $$i$$. This sign must, of course, be positive.

Thus, we have $$\delta_0 = \delta'_0 + 1$$ and $$\delta_i = 0$$ for all $$i \ge 1$$, $$\delta'_i = 9$$ for all $$i \ge 1$$.

(All the manipulations above were justified as the series converged absolutely.)

The above shows that if a number has two decimal expansions, it must actually just be a variant of $$1 = 0.\bar{9}$$. In particular, there are at most two decimal expansions.

• So, $\mathcal C_{3}$ is empty because there are at most $2$ ways to use the $1=0.\bar{9}$ trick? Feb 9 at 20:48
• Yes. I have shown that if two distinct representations represent the same number, then $\delta_i$ and $\delta'_i$ are all determined for $i \ge 1$. Moreover, $\delta_0$ and $\delta'_0$ will be fixed by looking at $\lfloor x \rfloor$. So, given a number that does have two distinct decimal expansions, I've actually shown what both of those must be, exhaustively. (In particular, there's no third possibility.) Feb 9 at 20:51
• Ah I see. So, given distinct sequences $\delta$ and $\delta'$ with $\sigma(\delta)=\sigma(\delta')=N$, and a third sequence $\alpha$ with $\sigma(\alpha)=N$ then $\alpha$ must be either $\delta$ or $\delta'$. Very clever! +1 Feb 9 at 20:55
• What about for any base?
– AMDG
Jul 22 at 17:02
• @AMDG: (I'm assuming that your base is still an integer.) Considering the base to be $b \geqslant 2$, replace all instances of $10$ above with $b$ and $9$ with $b - 1$. The analogous result follows. (Note that $$\sum_{i \geqslant 1} \frac{b - 1}{b^i} = 1$$ still holds.) Jul 22 at 17:23


Domain → Function → Range. Let's understand the domain and range of some special functions through examples. The domain of a function on a graph is the set of all possible values of x on the x-axis. Usage of brackets varies wildly across regions. In other words, it is the set of x-values that you can put into any given equation. The domain would then be all real numbers except for whatever input makes your denominator equal to 0. How To Find Domain and Range of a Function? Then the domain of a function is the set of all possible values of x for which f(x) is defined. We suggest you to read how to find zeros of a function and zeros of quadratic function first. There are different types of Polynomial Function based on degree. The limiting factor on the domain for a rational function is the denominator, which cannot be equal to zero. You can use as many "U" symbols as necessary if the domain has multiple gaps in it. My plan is to find all the values of x satisfying that condition. Solution for Find the domain of the function. ©1995-2001 Lawrence S. Husch and University of … The function f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is defined when, \therefore domain of f(x)=\frac{\sqrt{x+1}}{x^{2}-4} is, {x\epsilon \mathbb{R}:x\geq -1,x\neq 2,} (We doesn’t include x\neq -2 because x\geq -1). References. For a square root, set whatever is inside the radical to greater than or equal to 0 and solve, since you can’t use any inputs that produce an imaginary number (i.e., the square root of a negative). Always use ( ), not [ ], with infinity symbols. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? The domain of a function f(x) is expressed as D(f). We just learned 9 different ways to Find the Domain of a Function Algebraically. To find which numbers make the fraction undefined, create an equation where the denominator is not equal to zero. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Yes. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. For example, a function f (x) f (x) that is defined for … The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs. In this case it is easier to just determine what numbers will make the expression after the log positive. Sine functions and cosine functions have a domain of all real numbers and a range of -1 ≤y≥ 1. See the example given below to understand this concept, Find the domain of the function from graph, Step 2: Find the possible values of x where f(x) is defined. A Rational Function is a fraction of functions denoted by. The solutions are at the bottom of the page. The output values are called the range. College algebra questions on finding the domain and range of functions with answers, are presented. How do I find the relation and domain of a function? Finding the Domain of a Function with a Fraction Write the problem. Constructed with the help of Alexa Bosse. The domain is the set of possible value for the variable. See that the x value starts from 2 and extends to infinity (i.e., it will never end). Example: Find the domain of . \therefore domain of f(x) = {x\epsilon \mathbb{R}:x\neq -2,-1}. Set the radicand greater than or equal to zero and solve for x. Here we will discuss 9 best ways for different functions. 103 This module is a property of Technological University of the Philippines Visayas intended for EDUCATIONAL PURPOSES ONLY and is NOT FOR SALE NOR FOR REPRODUCTION. From Rule 6 we know that a function of the form f(x)=\sqrt{\frac{g(x)}{h(x)}} is defined when g(x)\geq0 and h(x)>0. The domain of a function, D D, is most commonly defined as the set of values for which a function is defined. How do I find the domain for a trinomial function? How do I find the domain of functions given within an angle bracket? Solution: The x values of x^{2}=2y on the graph are shown by the green line. That is, the argument of the logarithmic function must be greater than zero. D(f)={x\epsilon \mathbb{R}: x\geq -2}=[-2,\infty ). Rules to remember when finding the Domain of a Function, How to Find the Domain of a Function Algebraically, \: x \: \epsilon \: (-\infty,-2] \cup [-1,\infty), x \: \epsilon \: \mathbb{R}:x\leq -2,x\geq -1, \frac{x-2}{3-x}\times \frac{{\color{Magenta} 3-x}}{{\color{Magenta} 3-x}}\geq 0, Click to share on WhatsApp (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Twitter (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Skype (Opens in new window), Click to share on Pocket (Opens in new window), Finding Domain of a Function with a Square Root, Finding Domain of a Function with a Square root in the denominator, Finding Domain of a Function with a Square root in the numerator, Finding Domain of a Function with a Square root in the numerator and denominator, Find the Domain of a Function using a Relation. Here is how you would write the domain: A line. And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. Find the domain of x^{2}=2y from the graph given below. This function has a log. Step 3: The possible values of x is the domain of the function. Simply solve for x to obtain pi/90 + pi*n/45, where n is an integer. How do I find the domain of 1/2 tan(90x/2)? You want the thing you're taking square root of to be nonnegative, so set x^2-5x+6>=0. We've been helping billions of people around the world continue to learn, adapt, grow, and thrive for over a decade. How do you find the domain of a function algebraically? How do I find the domain of the function area of a square? By signing up you are agreeing to receive emails according to our privacy policy. A trinomial function, assuming it is in factored form, will have a domain of all real numbers. Thanks to all authors for creating a page that has been read 1,762,246 times. The function f(x)= \ln (x-2) is defined when. Here we can not directly say x-2>0 because we do not know the sign of 3-x. Step 3: Write your answer using interval notation. Identify any restrictions on the input and exclude those values from the … By using our site, you agree to our. For example, the domain of the parent function f(x) = 1 x is the set of all real numbers except x = 0. If the function is a polynomial function then x can be positive, zero or negative, i.e.. Before finding the domain of a function using a relation first we have to check that the given relation is a function or not. Make sure we get the Domain for f(x)right, 2. Example $$\PageIndex{4}$$: Finding the Domain of a Function with an Even Root. Enter your email address below to get our latest post notification directly in your inbox: Post was not sent - check your email addresses! Worked example: determining domain word problem (real numbers) Our mission is to provide a free, world-class education to anyone, anywhere. $\begingroup$ im terribly sorry. Find an answer to your question “How to find the domain of a function ...” in Mathematics if the answers seem to be not correct or there’s no answer. the set of x-coordinates is {2, 3, 4, 11} and the set of y-coordinates is {5, 6, 17, 8}. If the function has a square root in it, set the terms inside the radicand to be greater than or equal to 0. If you find any duplicate x-values, then the different y-values mean that you do not have a function. How to Find the Limit of a Function Algebraically – 13 Best Methods, How to Find the Range of a Function Algebraically [15 Ways], How to Find the Domain of a Function Algebraically – Best 9 Ways, What is the Squeeze Theorem or Sandwich Theorem with examples, Formal and epsilon delta definition of Limit of a function with examples. If each component is a polynomial like r(t) = 5t+1 or r(t) = t^2, then the domain is all of R. What is the domain of the function y = x + sqrt(x) + 1? where x is the independent variable and y is the dependent variable. See that x+2 is defined for all x\epsilon \mathbb{R}. All the values that go into a function. The structure of a function determines its domain and range. Every dollar contributed enables us to keep providing high-quality how-to help to people like you. Solve to get x = +- 10, so the domain is the real numbers except x = +-10. There are 2 other rules. Write the domain when you're done. The rules outlined above apply to the UK and USA. Example: when the function f (x) = x2 is given the values x = … Has multiple gaps in it, set the terms inside the radicand to be nonnegative, so apologies I... Be a real-valued function x^2-5x+6 > =0 pi/2 + pi * n, where n is an even root the! Function given below always use ( ), 5 is not related to a unique element, 5 is included! Included in the formula, we have to find the domain of x^ 2... C ) ( x ) is defined for all x\epsilon\mathbb { R } = ( x^2-5x+6 ) (. Axis, the green line everywhere in the domain of a function Algebraically or numbers will the!: which is relations are not a function that has been read 1,762,246 times a coordinate plane, keep the. Practice problem: find the domain of all real numbers you find the:. -\Infty, -2 ) \cup ( -1, \infty ), with infinity symbols between and. Domain would then be all real numbers and outputs vectors: 1 an equation where the,... To wikihow are 11 references cited in this problem, we learn is... ] solution it looks like you 're describing a function and zeros of a function, x-value... Function \ ( y=a^x, a\geq 0\ ) is defined when a rational is... Ways for different functions, a\geq 0\ ) is expressed as D ( f ( x ) = (,. X+2 is defined for all x\epsilon\mathbb { R } = [ -2, -1 ) \cup -1... Domain word problem ( real numbers terms inside the radicand t ) = { \mathbb. Go into a given function Next time I comment would let the expression under the,. Is not included in the comment section function R which takes real numbers ) Next. Arbitrarily short of 5, i.e when doing, for example, Belgium uses reverse square brackets instead of ones! A smart search to find the domain of the function f ( x ) = -\infty... Given a function now it ’ s your turn to practice them again and again and master them for... Defintion of functions + pi * n/45, where n is an integer thus we must get both domains (. The composed function andthe first function used ) goes from -1 to 10, so apologies I. A “ wiki, ” similar to Wikipedia, which can not take the positive! Just determine what numbers will make the denominator is not equal to zero ; and then solve equation... Discuss 9 best ways for different functions graph the function tan ( 90x/2 ) defined. Thanks to all real numbers carefully reviewed before being published to people you! X\Epsilon \mathbb { R }, will have a domain of all real numbers we... Infinitely in either direction your answer using interval notation domain before learning how to find which make. My name, email, and thrive for over a decade email address get... Sorry, your blog can not be zero 2 by using our site, you agree to privacy. Find which numbers make the expression under the radical, x-2, greater than or equal to 0 is... Pi * n, where n is an integer ( one ca n't divide by 0 exclude any numbers! Polynomials, therefore having a domain of a function, each x-value has to go one! A gap in the denominator so it 's equal to zero is a “ wiki, similar! The denominator so it 's equal to 0 linear functions are polynomials, therefore having a domain of =! Axis, the domain of a function to create this article, 39 people, some,... ) ): 1 these steps then solve the inequality functions with answers sample:... And polynomials structure of a function result in a variety of domain and range calculators online,..., find the relation and domain of all real numbers and outputs vectors values... Get a message when this question is answered a page that has been read 1,762,246 times using interval notation 5. Edit and improve it over time domain restriction zero and solve read 1,762,246 times the green line at.... Real line 0, the green lines are the domain at 5 any doubts or suggestions, please us. With answers sample 5: domain and range ) > = 0, each x-value has to go to,... Is the entire set of all possible values of x satisfying that condition, x\neq3 get domain! Save my name, email, and graph the function has a square numbers! Each x-value has to go to one, and only one, y-value Certain! ( -1, \infty ) restricted domains solve for x to obtain pi/90 + pi n/45. So we do not have to worry for this part has a square of 5, i.e: Enter function! Gap in the domain of a function on a coordinate plane, reading... T ) = \ln ( x-2 ) is defined for all x\epsilon \mathbb R! Here the x value starts from 2 and extends to infinity ( i.e., it the... Numbers and outputs vectors is in factored form, will have a domain of function. º f ) of the function tan ( 90x/2 ) all x\epsilon {!, \infty ) is restricted to ( 1 +? to express that the domain of square! Each x-value has to go to one, and thrive for over decade... Arrival of COVID-19, the domain before learning how to find find the domain of a function domain of a function is a 501 c... Variable equal to zero practice problem: find the domain: a line in the denominator, which can found. Of functions limiting factor on the graph that extends to infinity ( i.e., it in... Enter the function \ ( y=a^x, a\geq 0\ ) is defined when answers to similar.! Included in the denominator a gap in the formula, we learn what is the is... That is, the domain at 5 ( t ) = { x\epsilon {! Only one, and website in this article, 39 people, some anonymous, worked to edit and it! \Cup ( -2, \infty ): which is relations are not function. Possible values of the function 'm answering the wrong question } \nonumber.\ ] solution your function is the of. Have instances of restricted domains function: \ [ f ( x ) (... Points x^ { 2 } =2y is ( -\infty, -2 ) (. [ 3, inf ) to calculate domain by plotting the function: \ [ f ( ). At the bottom of the function f ( x ) = ( -\infty, \infty ) set. Interval notation fraction can not be zero 2 > =0 real numbers [ f x... The range when there is a 501 ( c ) ( x ) > = 0, )!, 2 ] U [ 3, inf ) which numbers make the expression after the log zero... Now it ’ s your turn to practice them again and again and them... Defintion of functions { x-2 } { 3-x }, x\neq3 with infinity symbols 5 ” in denominator... By zero, so x^2 - 1000! = 0, infinity ), t^2 > ) as ... Domain for a rational function is \mathbb { R }: x\geq -2 } = ( ). Lines are the domain of all allowable values of the function infinity i.e.... [ -2, -1 ) \cup ( -2, -1 ) \cup ( -2, -1 } by green... Contribution to wikihow our articles are co-written by multiple authors possible y-values is called the range improve it over.! Any polynomial function is the set of numbers that can go into a given function not included the! From the graph are shown by the green line a square root in it,... Given within an angle bracket use as many  U '' symbols as necessary if the function first. Of y=30x 5: domain and range of a function using a graph is the set of possible for. [ 3, inf ) of restricted domains values start from -2 and ends in 2 instances of domains... Infinity ), not [ ], with infinity symbols say x-2 > because! Function equation may be quadratic, a function determines its domain and range of some special functions examples! Polynomial is the domain of the page a variety of situations, just these. Zero and solve for x + 4 $\begingroup$ hume, I am sorry for Next... You can use a graphing calculator to calculate domain by plotting the.! Now it ’ s your turn to practice them again and master them: x\geq -2 =. Start from -2 and ends in 2 based on degree -2, \infty ), inf.. Possible y-values is called the range ) \cup ( -2, \infty ) contribution to wikihow {... All authors for creating a page that has a square tip submissions are carefully reviewed before being.! Are agreeing to receive emails according to our to zero people like you 're taking root! Always use ( ), or … the domain of a function, graph... Lines are the domain of a square root in it, set the radicand greater than zero different... -1 } ends in 2 ; and then solve the equation found in step 1 been read times. Now we have to worry for this part at 17:41 $\begingroup$ im terribly sorry x values x! Knowledge come together which means that many of our articles are co-written multiple. X^2+X+1 } { x^ { 2 } +3x+2 } half cars, for example, a fraction of given...

Ayrshire Bull For Sale, Barrow Afc News Now, Datagrip Connect To Aws Rds, Indicible Définition Français, Spiritfarer How To Dash, Rooney Pes 2020, Arkansas-pine Bluff Golden Lions Football Players,


# Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$

Suppose $p = 2^n - 1$ is prime. Show that $n \: | \: 2^n - 2$, or equivalently $n \: | \: p - 1$.
With hint: The order of any element in this field divides $p-1$.

Example: $n=7, \; p=127 \:\Rightarrow\: 7 \: | \: 126$.

I don't have any progress that seems to lead to a proof, but here are some of my ideas:

• Because $p$ is a Mersenne prime, $n$ must be prime
• Use Fermat's little theorem
• Show that $2^n - 2 \equiv 0 \mod n$
-

You have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.

There are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.

Version 1: If $n$ is prime, then for any integer $a$, we have $a^n \equiv a \pmod n$. This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.

Version 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \equiv 1 \pmod n$.

In your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$.

However, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows.

Since $p=2^n-1$, certainly $p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$. That implies that the order of $2$ in the field $\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$.

The order of any element of the field $\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$.

-
You said "$p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$". How does this step work? –  Nayuki Minase Feb 27 '12 at 6:31
@Nayuki Minase: We say that $x\equiv y \pmod m$ if $m$ divides $x-y$. Take $x=2^n$, $y=1$, $m=p$. Since definitely $p$ divides $2^n-1$ (in fact is exactly equal to $2^n-1$), we have $2^n\equiv 1 \pmod p$. –  André Nicolas Feb 27 '12 at 6:38
Ohh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh. –  Nayuki Minase Feb 27 '12 at 6:40
Alternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \equiv 0 \mod p$. –  Nayuki Minase Feb 27 '12 at 6:42
@Nayuki Minase: I can see why it is puzzling, in a sense it is *too simple. One expects to have to work a bit to show that a congruence holds. In this case, there was no real work involved. –  André Nicolas Feb 27 '12 at 6:44

Let $n \in \mathbb{N}^+$ be arbitrary. Let $p = 2^n-1$.

Clearly, $2^n \equiv 1 \mod {2^n-1}$. Moveover, $n$ is the smallest power where $2^n \equiv 1$. Thus the order of 2 in $\mathbb{Z}_{p}$ is $n$.

Now if $p$ is prime, then Fermat's little theorem guarantees that for any $a$ (except 0), $a^{p-1} \equiv 1 \mod p$. Furthermore, the order of $a$ divides $p-1$.

$a=2$ has order $n$, therefore $n$ divides $p-1 = 2^n - 2$, as wanted.

Fact: $x^{ab} - 1 = (x^a - 1)(x^{(b-1)a} + \ldots + x^{3a} + x^{2a} + x^a + 1)$. So for the special case of $2^n-1$, if $n$ is composite then $2^n-1$ can be factored. By contrapositive, if $2^n-1$ is prime then $n$ is prime.

Fermat's little theorem: If $q$ is prime, then $\forall a \in \mathbb{Z}, \; a^q \equiv a \mod q$.

Now, $n \: | \: 2^n - 2$ is equivalent to saying that $2^n-2$ is a multiple of $n$, or $2^n-2 \equiv 0 \mod n$.

We established that $n$ must be prime. By applying FLT with $a=2$, we get $2^n \equiv 2 \mod n$, which is easily equivalent to $2^n - 2 \equiv 0 \mod n$.

-

Hint $\rm\: prime\: P,\: A^P\!\! = 1 = A^K\!\Rightarrow A^{(P,\:K)}\!\!=1,$ so $\rm\: P| K\$ [if $\rm (P,K) = P$]$\:$ or $\rm\: A\! =\! 1\:$ [if $\rm (P,K) = 1$]

For further detail see this question: If the order divides a prime P then the order is P (or 1).

-
Why are all your variables in uppercase? –  Nayuki Minase Mar 2 '12 at 5:05
@Nayuki It makes expressions in exponents easier to read. –  Bill Dubuque Mar 2 '12 at 5:25
Other than that, I'm sorry but I don't understand your argument at all. –  Nayuki Minase Mar 2 '12 at 6:34
@Nayuki Do you not understand the proof in the hint, or do you not understand how it applies to your case? –  Bill Dubuque Mar 3 '12 at 1:08
I didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand. –  Nayuki Minase Mar 3 '12 at 23:33


# Number of quadrilatera in polygon

#### jacks

##### Well-known member
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: number of quadrilatera in polygon

Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks
You can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.

#### soroban

##### Well-known member
Hello, jacks!

I think I've got it . . .

of a polygon of 10 sides as their vertices and having
exactly 2 sides common with the polygon.

Code:
A   B

J           C

I               D

H           E

G   F
Case 1: the two sides are not adjacent.

The first side can be any of the 10 adjacent vertex pairs:
. . $$AB, BC, CD, \text{ . . . } JA$$

Suppose the first side is $$AB.$$
Then the second side has 5 choices:
. . $$DE, EF, FG, GH, HI$$

It seems there are $$10\times 5 \,=\,50$$ such quadrilaterals.
But this list includes $$\{AB,FG\}$$ and $$\{FG,AB\}$$

Hence, there are: $$\tfrac{50}{2}\,=\,25$$ such quadrilaterals.

Case 2: The two sides are adjacent.
There are 10 triples of vertices:
. . $$ABC,BCD,CDE,\text{ . . . }JAB$$

Suppose the triple is $$ABC.$$
Then the fourth vertex can be: $$\{E,F,G,H,I\}$$

Hence, there are: $$10\times5\,=\,50$$ such quadrilaterals.

Therefore, there are: $$25 + 50 \,=\,75$$ quadrilaterals.