# Math Challenge - February 2020

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## Summary:

A lot of calculus questions this month, but a bit (linear and abstract) algebra, too. The functional analysis question is a bit tricky.

## Main Question or Discussion Point

Questions

1.
(solved by @archaic ) Determine $\lim_{n\to \infty}\cos\left(t/\sqrt{n}\right)^n$ for $t\in \mathbb{R}$.

2. (solved by @Antarres ) Let $a_0,\ldots,a_n$ be distinct real numbers. Show that for any $b_0,\ldots,b_n\in\mathbb{R}$, there exists a unique polynomial $p$ of degree at most $n$ such that $p(a_i)=b_i$ for each $i$.

3. (solved by @Antarres ) Let $r(t)=(x(t),y(t))$ be an arc-length parameterization of a smooth closed curve $C$ with length $2\pi$. Let $A$ be the area enclosed by $C$
a.) Using Green's formula, show $A^2\leq\left(\int_0^{2\pi}x(t)^2dt \right)\ \left(\int_0^{2\pi}y'(t)^2 dt\right)$
b.) By translating $C$, we may assume without loss of generality that $\int_0^{2\pi}x(t) dt=0$. Use this assumption together with the Wirtinger inequality (question 1 of the January challenge) to show that $A\leq\pi.$
c.) By examining the equality condition for the Wirtinger inequality (Problem No. 1 in 01/2020), show that $A=\pi$ can only happen if $C$ is a circle.

4. (solved by @Antarres ) Let $f:[0,\infty)\to\mathbb{R}$ be a continuous function such that the sequence of functions $f_n(x)=f(x+n)$ converges uniformly to a function $g:[0,\infty)\to\mathbb{R}$. Show that $f$ and $g$ are uniformly continuous.

5.
a.)
(solved by @etotheipi ) Let $A$ = $\begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}$ Find $A^{99}$, $A^{2n}$ and $A^{2n + 1}$, $n \in \mathbb{N}$
b.) (open) How would we conclude in case of any commutative ring using ideals? ($\operatorname{char} \neq 2$)

6. (solved by @Antarres )
a.) If $a$,$b$,$k$ are integers with $b \neq 0$ show that $(a + kb, b) = (a,b)$
b.) We take the Fibonacci sequence $(F_n)$ (starting from the third term): $1, 2, 3, 5, 8, 13, \dots$, where each term after the second one, is the sum of the two previous terms. If $F_n$ is the $n-$th term of the sequence, show that every two consecutive Fibonacci numbers are coprime.

7.
a.)
(solved by @archaic ) If $f: \mathbb{R} \to \mathbb{R}$ is an integrable function in every closed interval of $\mathbb{R}$ show using calculus that
$$\int_{a}^{b} f(x)\,dx = \int_{a + d}^{b + d} f(x - d)\,dx$$
with $a \leq b$ and $d \geq 0$.
b.) (open) The function $\varphi\, : \,\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ with $\varphi(t,r)=(t(r+2),t^2-r)$ is injective on $U:=(0,1)\times (-1,1)\,.$ Show that $\varphi\, : \,U\longrightarrow V:= \varphi(U)$ is a diffeomorphism.
Next let $f\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}$ be integrable over $V$. Write
$$\int_V f\,d\lambda = \int_{\ldots}^{\ldots} \int_{\ldots}^{\ldots} \ldots f(\ldots\, , \,\ldots)\,dr\,dt$$
and calculate the area of $V$.

8. (solved by @julian ) Calculate $\displaystyle{\sum_{k=1}^\infty} \dfrac{1}{\binom{2k}{k}}$

9. (solved by @Antarres ) Let $a$ be an integer and $p$ an odd prime which does not divide $a$. The left multiplication
$$\lambda_{a,p}\, : \,\mathbb{Z}_p^\times \longrightarrow \mathbb{Z}_p^\times \, ; \,x \longmapsto ax \operatorname{mod} p$$
is then a permutation on $\{\,1,\ldots,p-1\,\}\,.$ Prove
$$\left(\dfrac{a}{p}\right) = \operatorname{sgn}\left(\lambda_{a,p}\right)$$

10. (solved by @suremarc and @Antarres ) Let $\mathcal{H}$ be a real Hilbert space and $\beta$ a continuous bilinear form, $\mathcal{H}^*$ its dual space of continuous functionals on $\mathcal{H}$, and $\beta(f,f)\geq C\|f\|^2>0.$
Prove that for any given continuous functional $F \in \mathcal{H}^*$ there is a unique vector $f^\dagger \in \mathcal{H}$ such that
$$F(g)=\beta(f^\dagger,g)\quad \forall\,\,g\in \mathcal{H}$$ High Schoolers only

11.
a.) (solved by @mmaismma ) How many knights can you place on a $n\times m$ chessboard such that no two attack each other?
b.) (solved by @Not anonymous ) In how many different ways can eight queens be placed on a chessboard, such that no queen threatens another? Two solutions are not different, if they can be achieved by a rotation or by mirroring of the board. (For this part there is no proof required.)

12. (solved by @Not anonymous ) Determine (with justification, but without explicit calculation) which of
a.) $1000^{1001}$ and ${1002}^{1000}$
b.) (solved by @etotheipi ) $e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001}$ and $e^{0.000008}-e^{0.000005}$
is larger.

13. (solved by @etotheipi , @Not anonymous)
a.) Let $A=(-2,0)\, , \,B=(0,4)$ and $M=(1,3)\,.$ What is $\alpha = \sphericalangle (AMB)$?
b.) Let $C=(-1,2+\sqrt{5})\, , \,D=(-1,2-\sqrt{5})$ and $M=(1,3)\,.$ What is $\beta = \sphericalangle (CMD)$?

14. (solved by @Not anonymous ) Find the exact formula of the function $f(x) =\sup_{t \in \mathbb{R}}(2tx - t^2)$, $x \in \mathbb{R}$.

15. (solved by @etotheipi ) In what way does the oscillation period of a pendulum change, if the suspension pivot moves
a.) vertically up with acceleration $a$,
b.) vertically down with acceleration $a < g$,
c.) horizontally with acceleration $a$?

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• etotheipi, JD_PM and berkeman

etotheipi
Gold Member
2019 Award
5. Let $A$ = $\begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}$. Find $A^{99}$, $A^{2n}$ and $A^{2n + 1}$, $n \in \mathbb{N}$
$A^{2} = \begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} = I$

$A^{99} = ({A^{2}})^{49}A = I^{49}A = IA = A$

$A^{2n} = ({A^{2}})^{n} = I^{n} = I$

$A^{2n+1} = A^{2n}A = IA = A$

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• Erico Romaric and QuantumQuest
1. Determine $\lim_{n\to \infty}\cos\left(t/\sqrt{n}\right)^n$ for $t\in \mathbb{R}$.
\begin{align*} \lim_{n\to\infty}\cos\left(\frac{t}{\sqrt{n}}\right)^n&=\lim_{n\to\infty}e^{n\ln\left(\cos\left(\frac{t}{\sqrt{n}}\right)\right)}\\ &=\lim_{n\to\infty}e^{n\ln\left(1-\frac{t^2}{2n}+o(\frac{1}{n})\right)}\\ &=\lim_{n\to\infty}e^{n\left(-\frac{t^2}{2n}+o(\frac{1}{n})\right)}\\ &=e^{-\frac{t^2}{2}} \end{align*}

• Abhishek11235 and Erico Romaric
7. If $f: \mathbb{R} \to \mathbb{R}$ is an integrable function in every closed interval of $\mathbb{R}$, show using calculus that $$\int_{a}^{b} f(x)\,dx = \int_{a + d}^{b + d} f(x - d)\,dx$$with $a \leq b$ and $d \geq 0$.
\begin{align*} I&=\int_{a}^{b} f(x)\,dx,\text{ let } u = x+d \Leftrightarrow x=u-d\\ &=\int_{a+d}^{b+d} f(u-d)\,du \end{align*}
Just call $u$ as $x$. This seemed to be too straightforward. Did I miss something?

QuantumQuest
Gold Member
$A^{2} = \begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} = I$

$A^{99} = ({A^{2}})^{49}A = I^{49}A = IA = A$

$A^{2n} = ({A^{2}})^{n} = I^{n} = I$

$A^{2n+1} = A^{2n}A = IA = A$
Well done @etotheipi. It's a very nice shortcut way for this particular case. However, there is a more general way we can solve problems like this. I won't tell what is it but if anyone else wants to try the problem and come up with a correct solution, he / she'll also get credit.

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• etotheipi
etotheipi
Gold Member
2019 Award
Well done @etotheipi. It's a very nice shortcut way for this particular case. However there is a more general way we can solve problems like this. I won't say what it is but if anyone else wants to try the problem and come up with a correct solution, he / she'll also get credit.
Naïve me should have known there be more to it! Of course, I have no idea what...

QuantumQuest
Gold Member
Naïve me should have known there be more to it! Of course, I have no idea what...
I wouldn't call you naive. According to the educational background you have in your info, you're doing really well. It's just that, what is needed for the solution I said in post #5, is usually taught in the first year(s) of College / University for mathematicians and of course, as far as I know. Things may be different in some countries for both High School and College /University.

• etotheipi
QuantumQuest
Gold Member
\begin{align*} I&=\int_{a}^{b} f(x)\,dx,\text{ let } u = x+d \Leftrightarrow x=u-d\\ &=\int_{a+d}^{b+d} f(u-d)\,du \end{align*}
Just call $u$ as $x$. This seemed to be too straightforward. Did I miss something?
What I ask for, is an approach / solution using some sort of limits. I say "using calculus" in the wording of the question, which is not clear enough but if I make it totally clear, I'll give the way of the solution ;)

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13. a.) Let $A=(-2,0), B=(0,4)$ and $M=(1,3)$. What is $\alpha = \sphericalangle (AMB)$?
b.) Let $C=(-1,2+\sqrt{5})\, , \,D=(-1,2-\sqrt{5})$ and $M=(1,3)$. What is $\beta = \sphericalangle (CMD)$?
A silly doubt but I have to ask because I had never heard of "spherical angle" till I saw the symbol $\sphericalangle$ in the quoted question. I could find some articles on the web describing spherical angle but my understanding is still vague. To the doubt - given 3 points, there would be infinite number of spheres on whose surface those lie. Had the 3 points in the question been 3D points specified w.r.t. some center of sphere, the problem would be fully specified. With 2D points and no mention of the reference sphere, I do not understand how to decide on the sphere to choose. Any additional hints that you can provide?

Mentor
A silly doubt but I have to ask because I had never heard of "spherical angle" till I saw the symbol $\sphericalangle$ in the quoted question. I could find some articles on the web describing spherical angle but my understanding is still vague. To the doubt - given 3 points, there would be infinite number of spheres on whose surface those lie. Had the 3 points in the question been 3D points specified w.r.t. some center of sphere, the problem would be fully specified. With 2D points and no mention of the reference sphere, I do not understand how to decide on the sphere to choose. Any additional hints that you can provide?
Don't blame me for the word LaTeX uses to show an angle! Read it as 2D $\angle (AMB)$. I just liked the symmetry of $\sphericalangle (AMB)$ more. Handwritten, I would have used Anyway, it is a plane angle meant in the problem.

14. Find the exact formula of the function $f(x) =\sup_{t \in \mathbb{R}}(2tx - t^2), x \in \mathbb{R}$.
The simple nature of the function that I get as the solution gives me the suspicion that I might have misunderstood the question . Posting that solution anyway

For a fixed value of $x$, we find the maxima for $g(t) = (2tx - t^2), t \in \mathbb{R}$. At the maximum points, $g'(t) = 0 \implies 2x - 2t = 0 \implies x = t$. This is the maximum and not the minimum because $g''(t) = -2$ is negative (alternative explanation - there is no finite minimum; as $g(t) \rightarrow \infty$ as $t$ tends to $\pm \infty$ for any finite value of $x$). Substituting this value of $t$ for which maximum is attained, we get $\sup_{t \in \mathbb{R}}(2tx - t^2) = x^2$. Therefore, $f(x) = x^2$

• Abhishek11235 and QuantumQuest
etotheipi
Gold Member
2019 Award
Don't blame me for the word LaTeX uses to show an angle! Read it as 2D $\angle (AMB)$. I just liked the symmetry of $\sphericalangle (AMB)$ more.
In that case,
a.) $\vec{MA} \cdot \vec{MB} = (-1)(-3) + (1)(-3) = 0 \implies \alpha = \frac{\pi}{2}$
b.) $\vec{MC} \cdot \vec{MD} = (-2)(-2) + (\sqrt{5} -1)(-\sqrt{5} -1) = 0 \implies \beta = \frac{\pi}{2}$

Mentor
In that case,
a.) $\vec{MA} \cdot \vec{MB} = (-1)(-3) + (1)(-3) = 0 \implies \alpha = \frac{\pi}{2}$
b.) $\vec{MC} \cdot \vec{MD} = (-2)(-2) + (\sqrt{5} -1)(-\sqrt{5} -1) = 0 \implies \beta = \frac{\pi}{2}$
Right. Can you also give a geometric argument?

etotheipi
Gold Member
2019 Award
Right. Can you also give a geometric argument?
Does cosine rule suffice?

$\cos{\theta} = \frac{a^{2} + b^{2} - c^{2}}{2ab}$

where for the part a), $a = 3\sqrt{2}, b= \sqrt{2}, c= 2\sqrt{5}$

and for part b), $a = \sqrt{10 + 2\sqrt{5}}, b =\sqrt{10 - 2\sqrt{5}}, c = 2\sqrt{5}$

These both give $\cos{\theta} = 0 \implies$ both angles are $\frac{\pi}{2}$.

• fresh_42
Mentor
Does cosine rule suffice?

$\cos{\theta} = \frac{a^{2} + b^{2} - c^{2}}{2ab}$

where for the part a), $a = 3\sqrt{2}, b= \sqrt{2}, c= 2\sqrt{5}$

and for part b), $a = \sqrt{10 + 2\sqrt{5}}, b =\sqrt{10 - 2\sqrt{5}}, c = 2\sqrt{5}$

These both give $\cos{\theta} = 0 \implies$ both angles are $\frac{\pi}{2}$.
I thought of classical geometry - another one - but it seems there are many ways to see it.
Let's see whether someone comes up with a third method.

hilbert2
Gold Member
For problem 2:

For instance, if there are 3 points $a_i$ and $b_i$ each, the question is whether you can always find a polynomial $P(x) = c_2 x^2 + c_1 x + c_0$ such that

$c_2 a_{1}^2 + c_1 a_1 + c_0 = b_1 ,$
$c_2 a_{2}^2 +c_1 a_2 + c_0 = b_2 ,$
$c_2 a_{3}^2 + c_1 a_3 + c_0 = b_3$

This is equivalent to the linear system

$\begin{bmatrix}a_{1}^2 & a_1 & 1 \\ a_{2}^2 & a_2 & 1 \\ a_{3}^2 & a_3 & 1\end{bmatrix}\begin{bmatrix}c_2 \\ c_1 \\ c_0\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix}$

and it has a solution if the coefficient matrix is invertible, i.e.

$\left|\begin{array}{ccc}a_{1}^2 & a_1 & 1 \\ a_{2}^2 & a_2 & 1 \\ a_{3}^2 & a_3 & 1\end{array}\right|\neq 0$

The determinant of that type is called a Vandermonde determinant, and it is always nonzero if numbers $a_i$ are distinct, no matter what's the $n\in\mathbb{N}$ in the size of the square matrix $n\times n$. This proves the claim.

etotheipi
Gold Member
2019 Award
I thought of classical geometry - another one - but it seems there are many ways to see it.
Let's see whether someone comes up with a third method.
The line passing through points $A$ and $M$ has equation $y=x+2$, and intercepts the y axis at $y=2$. Let's call this point $Q$. Now let $Q$ and $M$ be the two ends of a diameter of a circle of radius $1$, which we construct about a centre of $(0,3)$. The point $M$, $(1,3)$ then lies on the circumference on this circle and using circle theorems we deduce that $\angle BMQ$ is a right angle. By extension, since $A$, $Q$ and $M$ are on the same straight line, the necessary angle is also right.

Mentor
The line passing through points $A$ and $M$ has equation $y=x+2$, and intercepts the y axis at $y=2$. Let's call this point $Q$. Now let $Q$ and $M$ be the two ends of a diameter of a circle of radius $1$, which we construct about a centre of $(0,3)$. The point $M$, $(1,3)$ then lies on the circumference on this circle and using circle theorems we deduce that $\angle BMQ$ is a right angle. By extension, since $A$, $Q$ and $M$ are on the same straight line, the necessary angle is also right.
Close, and on the right track, but it's a little bit easier. What are $AB$ and $CD$?

• etotheipi
etotheipi
Gold Member
2019 Award
Close, and on the right track, but it's a little bit easier. What are $AB$ and $CD$?
Appears they are also the endpoints of the diameter of a circle, e.g. in part a) a circle of radius $\sqrt{5}$ and centre $(-1,2)$. And as the distance from $(-1,2)$ to $(1,3)$ is also $\sqrt{5}$, the result follows much more easily!

• fresh_42
Infrared
Gold Member
@hilbert2 This is right, but the identity for the Vandermonde determinant is a more difficult result than the original question!

There is a very simple solution.

Don't blame me for the word LaTeX uses to show an angle! Read it as 2D $\angle (AMB)$. I just liked the symmetry of $\sphericalangle (AMB)$ more. Handwritten, I would have used
View attachment 256496
Anyway, it is a plane angle meant in the problem.
Thanks for clarifying. I had initially found the solution assuming plane angle but on noticing the LaTeX code of the symbol when posting the solution, I stopped and went to find what spherical angle means but stopped again after getting confused about the choice of sphere.

For both the questions, the answer is 90°.

13a. Lengths of line segments $AB$, $BM$ and $AM$ are $\sqrt{20}, \sqrt{2}, \sqrt{18}$ respectively and it may be noticed it $AM^2 + BM^2 = 18 + 2 = 20 = AB^2$ (in the equation, I used $AB$ to denote the length of line segment $AB$ and so on). By Pythagoras theorem, this implies that the triangle is right angled with $AB$ as the hypotenuse. $∠(AMB)$ being its opposite angle must be 90°.

13b. $C=(−1,2+√5),D=(−1,2−√5)$ and $M=(1,3)$. Lengths of the line segments are $CD= 2\sqrt{5} = \sqrt{20}, DM = \sqrt{2^2 + {(1+√5)}^2}, CM = \sqrt{2^2 + {(1-√5)}^2}$. $DM^2 + CM^2 = (4 + (1 + 5 + 2√5)) + (4 + (1 + 5 - 2√5)) = 20 = CD^2$. Again by Pythagorean theorem, this implies that $\Delta CDM$ is a right-angled triangle with $CD$ as the hypotenuse. $∠(CMD)$ is the hypotenuse's opposite angle and must be 90°.

Mentor
Appears they are also the endpoints of the diameter of a circle, e.g. in part a) a circle of radius $\sqrt{5}$ and centre $(-1,2)$. And as the distance from $(-1,2)$ to $(1,3)$ is also $\sqrt{5}$, the result follows much more easily!
They are both diameters of $(x+1)^2+(y-2)^2=5$ and $M$ is part of the circle, too. The statement follows with the theorem of Thales.

• etotheipi
etotheipi
Gold Member
2019 Award
If we transform into the accelerating frame of the pivot, and consider the extra fictitious force contributing to an effective weight, we obtain the following.
a.) $g_{eff} = a + g \implies T = 2\pi \sqrt{\frac{l}{a+g}}$
b.) $g_{eff} = g - a \implies T = 2\pi \sqrt{\frac{l}{g - a}}$
c.) $g_{eff} = \sqrt{a^{2} + g^{2}} \implies T = 2\pi \sqrt{\frac{l}{\sqrt{a^{2} + g^{2}}}}$ about $\theta = \arctan{\frac{a}{g}}$ from the vertical
I don't know whether this argument is sufficient, though, or whether you were after a derivation from first principles especially for part $c$!

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• QuantumQuest
hilbert2
Gold Member
@hilbert2 This is right, but the identity for the Vandermonde determinant is a more difficult result than the original question!

There is a very simple solution.
Ok, I guess it's something similar to finding a polynomial with zeros at $a_i$ by forming a product of binomials $(x-a_i)$.

StoneTemplePython
Gold Member
2019 Award
@hilbert2 This is right, but the identity for the Vandermonde determinant is a more difficult result than the original question!
I happen to like Vandermonde matrices, so I'll point out there is a very easy, computation free, and sneaky finish here.
note: the LaTeX below renders fine locally where I wrote it, but a lot of it is not rendering on PF for reasons I don't understand

for a Vandermonde matrix
$$V = \begin{bmatrix} 1 & 1 & \dots & 1 & 1 \\ v_1 & v_2 & \dots & v_{n-1} & v_{n}\\ v_1^2 & v_2^2 & \dots & v_{n-1}^2 & v_{n}^2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ v_{1}^{n-1} & v_{2}^{n-1} & \dots & v_{n-1}^{n-1} & v_{n}^{n-1}\end{bmatrix}$$

its immediate that the matrix is singular if $v_i = v_j$ for $j\neq i$ because one column is a copy of the other. To prove that this is the only way the matrix is singular, suppose first n-1 columns are distinct, insert a free variable $x$ in the final column and compute the determinant

$$\det\left( \begin{bmatrix}1 & 1 & \dots & 1 & 1 \\v_1 & v_2 & \dots & v_{n-1} & x\\v_1^2 & v_2^2 & \dots & v_{n-1}^2 & x^2 \\\vdots & \vdots & \ddots & \vdots & \vdots \\v_{1}^{n-1} & v_{2}^{n-1} & \dots & v_{n-1}^{n-1} & x^{n-1}\end{bmatrix}\right) = p(x) = k\cdot\prod_{i=1}^{n-1}(x-v_i)$$

which is a degree n-1 polynomial and by inspection has exactly n-1 distinct roots and no more. so $p(x) \neq 0$ for $x\not \in \{v_1,v_2,...,v_{n-1}\}$ because the product of finitely many non-zero numbers is a non-zero number. This proves $\det\big(V\big)\neq 0$ if all columns are distinct.

(For avoidance of doubt, we know $k\neq 0$ say because, up to a sign, it is given by the leading n-1 x n-1 principal minor which is non-zero by induction hypothesis -- which is obviously $= 1\neq 0$ in the implicit 2x2 base case.)

I found the outlines of this in Feller vol 2 a while back, so maybe the argument is sophisticated in some sense but it really is quite simple and sneaky.

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• hilbert2 and Infrared