# Math Challenge - December 2021

• Challenge
Mentor
2022 Award
This month's challenges will be my last thread of this kind for a while. Call it a creative break. Therefore, we will have a different format this month. I will post one problem a day, like an advent calendar, only for the entire month. I will try to post the questions as close as possible to 0:00 GMT.

1. (solved by @Infrared ) Let ##G## be a group with ##3129## elements. Prove it is solvable.

2. $$I(a):=\int_0^1 \left(\dfrac{\log x}{a+1-x}-\dfrac{\log x}{a+x}\right)\,dx \; ; \; a\in \mathbb{C}\backslash[-1,0]$$
3. (solved by @Infrared ) Let ##\mathfrak{g}## be a Lie algebra over a field of characteristic not ##2.## Prove that
$$\mathfrak{A(g)}=\{\alpha \in \mathfrak{gl(g)}\,|\,[\alpha (X),Y]+[X,\alpha (Y)]=0\text{ for all }X,Y\in\mathfrak{g}\}$$
is a Lie algebra. Determine ##\mathfrak{A(B)}## for the two-dimensional non-abelian Lie algebra ##\mathfrak{B}##.

4. (solved by @Infrared ) Show that a path-connected set is connected but not vice versa and not necessarily simply connected.

5. (solved by @julian ) $$\int_0^{\frac{\pi}{4}} \log(1+\tan x)\,dx$$
6. (solved by @TeethWhitener and @PeroK ) There are currently about ##7,808,000,000## people on earth. If we would enumerate them all, how many of them would have a prime number?

7. Let ##M=\mathbb{R}^2## and ##G=\mathbb{R}## and consider the map
$$\psi(\varepsilon ,(x,y)):=\left(\dfrac{x}{1-\varepsilon x},\dfrac{y}{1-\varepsilon x}\right)$$
defined on
$$U=\left\{(\varepsilon ,(x,y))\,|\,\varepsilon <\dfrac{1}{x}\text{ for }x>0\text{, or }\varepsilon >\dfrac{1}{x}\text{ for }x<0\right\}\subseteq \mathbb{R}\times \mathbb{R}^2$$
Show that ##\psi## defines a local group action of ##G## on the manifold ##M.## Does it have a global counterpart on ##\mathbb{R}^2##?

8. Give an example of a ring and a maximal ideal that isn't a prime ideal.

9. Let ##U,V\subseteq \mathbb{C}## open sets, ##\varphi \, : \,U\longrightarrow V## a holomorphic function, and ##\gamma \, : \,[0,1]\longrightarrow U## a closed, smooth path. Show that if ##\gamma ## is ##0##-homologue in ##U,## then ##\varphi \circ \gamma ## is ##0##-homologue in ##V.##

10. (solved by @julian ) Examine convergence:
$$\prod_{n=2}^\infty \left(1-\dfrac{1}{n}\right)\; , \;\prod_{n=3}^\infty \left(1-\dfrac{4}{n^2}\right)$$
11. (solved by @TeethWhitener ) The Heisenberg algebra can be viewed as
$$\mathfrak{H}=\left\{\begin{bmatrix} 0&x_1&x_3\\0&0&x_2\\0&0&0 \end{bmatrix}\, : \,x_1,x_2,x_3\in \mathbb{R}\right\}.$$
Calculate ##\exp(H)## for a matrix ##H\in \mathfrak{H}##.

12. (solved by @julian ) $$\int_{-\infty }^{\infty }\dfrac{|\sin(\alpha x)|}{1+x^2}\,dx\; , \; \alpha >0$$
13. (solved by @fishturtle1 ) Show that ##(n-1)!\equiv -1 \mod n## holds if and only if ##n## is prime. Determine the first two primes for which even ##(p-1)!\equiv -1 \mod p^2## holds.

14. (solved by @QuantumSpace ) Determine all possible topologies on ##X:=\{a,b\}##, and which of them are homeomorphic. Give an example of a topological space with more than one element such that all sequences converge.

15. (solved by @fishturtle1 ) Explain the difference between ##\mathbb{Z}_2\times \mathbb{Z}_3## and ##\mathbb{Z}_2 \ltimes \mathbb{Z}_3\,.## Is there also a group ##\mathbb{Z}_2 \rtimes \mathbb{Z}_3\,?##

16. Show that ##16## and ##33## are Størmer numbers, but no number ##N:=2n^2>2## can be one, e.g. ##32.##

17. (solved by @fishturtle1 ) Consider a number ##n## which is not a prime and
$$p\,|\,n\Longrightarrow p\,|\,\left(\dfrac{n}{p}-1\right)$$
E.g. ##30=2\cdot 3\cdot 5## is such a number, since ##2\,|\,14\, , \,3\,|\,9\, , \,5\,|\,5.##
Show that ##n## is square-free (all prime factors have exponent ##1##), and no semiprime (product of exactly two primes).

18. Prove that path integrals in ##\mathbb{R}^n## over gradient vector fields depend only on starting and endpoint, and not on the path itself.

19. (solved by @julian ) Let ##P_0=0,P_1=1,P_n=2P_{n-1}+P_{n-2}## for all ##n\in \mathbb{N},n\geq2##. Determine a closed form for ##P_n.##

20. Find the irreducible minimal polynomial for
$$\mathbb{Q} \subseteq \mathbb{Q}\left(\sqrt[3]{\dfrac{9+\sqrt{69}}{18}}+\sqrt[3]{\dfrac{9-\sqrt{69}}{18}}\right).$$
21. Show that the embedding ##\mathbb{S}^1\longrightarrow \mathbb{R}^2-\{0\}## is a homotopy equivalence, and that ##\mathbb{R}\longrightarrow \mathbb{R}^2-\{0\}## defined by ##x\mapsto (x,1)## is none.

22. (solved by @fishturtle1 ) Let ##\emptyset\neq X## be a set, ##\mathcal{P}(X)## its power set. Consider the following mappings
\begin{align*}
f\, : \,X &\longrightarrow \mathcal{P}(X)\\
x&\longmapsto \{x\}\\[6pt]
g\, : \,\mathcal{P}(X)\times \mathcal{P}(X)&\longrightarrow \mathcal{P}(X)\\
(A,B)&\longmapsto A\cup B
\end{align*}
and decide whether they are injective, surjective, and calculate the fiber (pre-image) of the empty set.

23. (solved by @TeethWhitener ) Find the smallest positive integer ##x## that solves
\begin{align*}
x &\equiv 2 \mod 3\\
x &\equiv 3 \mod 4\\
x &\equiv 2 \mod 5
\end{align*}

24. Let ##\vec{u},\vec{v},\vec{w}## be three different coplanar vectors of equal length, originating at a point ##O.## Their endpoints define a triangle ##\triangle UVW##. How can the barycenter ##S## be found?

25. Is a partially differentiable function ##f\, : \,\mathbb{R}^2\rightarrow \mathbb{R}## at some point ##x_0## also continuous at ##x_0##?

26. Let ##\mathfrak{g}## be the real Lie algebra generated by
$$A_1=\begin{bmatrix} 0&0&0\\0&1&0\\0&0&-1 \end{bmatrix}\, , \,A_2=\begin{bmatrix} 0&0&1\\-1&0&0\\0&0&0 \end{bmatrix}\, , \,A_3=\begin{bmatrix} 0&1&0\\0&0&0\\-1&0&0 \end{bmatrix}$$
Calculate its center ##\mathfrak{Z(g)}=\{X\,|\,[A_i,X]=0\,(i=1,2,3)\},## its commutator subalgebra ##[\mathfrak{g},\mathfrak{g}],## and a Cartan subalgebra ##\mathfrak{h}##.

27. (solved by @julian ) Let ##f\, : \,[a,b]\longrightarrow \mathbb{R}## be a continuous function and ##g\, : \,[a,b]\longrightarrow \mathbb{R}## integrable with ##g(x)\geq 0## for all ##x\in [a,b]##. Then there is a ##\xi\in [a,b]## such that
$$\int_a^b f(x)g(x)\,dx = f(\xi)\int_a^b g(x)\,dx$$
28. (solved by @Not anonymous ) Consider the circle segment above ##A=(-1,0)## and ##B=(1,0)## of
$$x^2+\left(y+\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{4}{3}\,.$$
The point ##P:=\left(\dfrac{1}{\sqrt{3}},1-\dfrac{1}{\sqrt{3}}\right)## lies on this segment. Calculate the height ##h## of the circle segment, and ##|AP|+|PB|.##

29. (solved by @ergospherical and @PeroK ) Let ##\varphi :V\longrightarrow V## a linear mapping. Prove
$$\operatorname{ker} (\varphi) \cap \operatorname{im}(\varphi) =\{0\} \Longleftrightarrow \operatorname{ker}(\varphi \circ \varphi )=\operatorname{ker}(\varphi )$$
30. (solved by @ergospherical ) Let ##A## by a cylindric surface (without base or cover) that rotates around the ##z##-axis and stands on the plane ##\{z=0\}##, with radius ##R>0## and height ##h>0##. Give a parameterization and calculate the surface integral
$$\int_A \langle F,n \rangle\,d^2r$$
for the vector field ##F\, : \,\mathbb{R}^3\rightarrow \mathbb{R}^3## defined by ##F(x,y,z)=(xz,yz,123).##

31. Let ##\mathcal{P}## be a finite set of points in a plane, that are not all collinear. Then there is a straight, that contains exactly two points.

.

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Mentor
On the first day of Christmas, my math teacher gave to me…

Wrichik Basu, SolarisOne, Hamiltonian and 2 others
jbergman
This month's challenges will be my last thread of this kind for a while. Call it a creative break. Therefore, we will have a different format this month. I will post one problem a day, like an advent calendar, only for the entire month. I will try to post the questions as close as possible to 0:00 GMT.

1. Let ##G## be a group with ##3129## elements. Prove it is solvable.
Love the threads. Hope they come back when you are a refreshed. Not sure how you come up with the problems.

pinball1970, MevsEinstein, Not anonymous and 1 other person
Mentor
2022 Award
On the first day of Christmas, my math teacher gave to me…
Christmas gave me at least most of the month by an advent calendar. Otherwise, I would have had to celebrate Hanukkah four times in a row!

pinball1970
valenumr
I have two questions about 1). It might be a little naive, but I'm slowly working on group theory and still in the basic stuff (I had to skip ahead to see what "solvable" even means). So first, when you say a group of n elements, is that the order of the group, or is that the size of the set used to compose the group? Second, do you mean that this should be provable for any group of the given order (or set elements), or just that it is provable that some solvable such group exists?

Thanks.

Mentor
2022 Award
I have two questions about 1). It might be a little naive, but I'm slowly working on group theory and still in the basic stuff (I had to skip ahead to see what "solvable" even means).
You will need the Sylow theorems, or find a reference of a theorem to use.
Hint: You can find one among the challenges, including a proof.
So first, when you say a group of n elements, is that the order of the group, ...
Yes.
... or is that the size of the set used to compose the group? ...
What is the difference? It is the number of all its elements, the order. It is not the number of generators if you thought of that.
... Second, do you mean that this should be provable for any group of the given order (or set elements), or just that it is provable that some solvable such group exists?

Thanks.
Every group of order ##3129## is solvable.

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valenumr
valenumr
I see. My confusion was, for example, Sn is order n!, But Cn is just order n, for example. Just wanted to clarify what was meant.

valenumr
Hmm... Based on other math background, I suspect any group of prime order is cyclical, and 3129 is square free. But Sylows theorem is several modules ahead of where I am. Anyhow, I was just investigating the question and found it interesting. Thanks for the insights.

Mentor
2022 Award
Hmm... Based on other math background, I suspect any group of prime order is cyclical, and 3129 is square free. But Sylows theorem is several modules ahead of where I am. Anyhow, I was just investigating the question and found it interesting. Thanks for the insights.
One of my goals is that the problems are useful and not only calculations. I didn't always succeed, but many results (inequalities, theorems, techniques) are worth keeping in mind. Or at least worth having a look at from time to time, which is why I published the solutions.

hutchphd
valenumr
One of my goals is that the problems are useful and not only calculations. I didn't always succeed, but many results (inequalities, theorems, techniques) are worth keeping in mind. Or at least worth having a look at from time to time, which is why I published the solutions.
Thanks so much for doing it. It also introduced topics that perhaps might be new to some!

hutchphd and fresh_42
Not anonymous
Love the threads. Hope they come back when you are a refreshed. Not sure how you come up with the problems.
Tons of thanks to fresh_42 for not only posting so many interesting questions but also patiently reviewing the answers. Though I am still unable to solve many of the problems and many of the topics are seemingly out of my reach, merely attempting to solve the problems was a great learning experience, and a refreshing one too! Will miss these math challenges and I hope to see them come back.

SolarisOne, hutchphd, Keith_McClary and 2 others
Gold Member
I'll try a few. It's been a while for me.

1. We have the prime factorization ##3129=3\times 7\times 149##. Let ##H## be a subgroup of order ##7.## By one of the Sylow theorems, the number of such subgroups is a divisor of ##3\times 149## and is ##1## modulo ##7.## Checking all possibilities, we see that there can only be one subgroup of order ##7,## and so it must be normal (as otherwise a conjugate subgroup would be another subgroup of order ##7##).

Consider the chain of normal subgroups ##\{1\}\leq H\leq G##. The quotient ##H/\{1\}\cong H## is abelian since it has prime order ##7## and is hence cyclic. The quotient ##G/H## has order ##3\times 149## and is therefore abelian as ##149## is not ##1## mod 3 by the following fact:

If ##p<q## are primes with ##q## not 1 modulo ##p,## then the only group ##K## of order ##pq## is the cyclic one. This is because by similar counting as above, there is a normal subgroup of order ##q##, so ##K## is a semidirect product of subgroups of order ##p## and ##q.## Since ##q## is not ##1## modulo ##p,## there is only the trivial homomorphism from a ##p##-subgroup to the automorphism group of the ##q## subgroup, so the semidirect product is trivial.

3. We have to check that this subspace of ##\mathfrak{gl}(\mathfrak{g})## is closed under bracket. If ##\alpha,\beta\in\mathfrak{A}(\mathfrak{g})##, then

##[[\alpha,\beta](X),Y]+[X,[\alpha,\beta](Y)]=[\alpha(\beta(X)),Y]-[\beta(\alpha(X)),Y]+[X,\alpha(\beta(Y))]-[X,\beta(\alpha(Y))].##

Note that ##[\alpha(\beta(X)),Y]=-[\beta(X),\alpha(Y)]## since ##\alpha\in\mathfrak{A}(\mathfrak{g})## and similarly for the other terms, so they cancel.

The non-abelian 2-dimensional Lie algebra has generators ##x,y## with the relation ##[x,y]=x.## If a matrix ##\begin{pmatrix}a & b\\c & d\end{pmatrix}## represents an element ##\sigma\in\mathfrak{gl}(\mathfrak{B})## with respect to this basis, in order for ##\sigma## to be in ##\mathfrak{A}(\mathfrak{B})## it is necessary and sufficient for ##[\sigma(x),y]+[\sigma(y),x]## to be zero. We calculate this to be ##a+d##, so ##\mathfrak{A}(\mathfrak{B})## is isomorphic to ##\mathfrak{sl}(2).##

4. Suppose that ##X## is path-connected. Suppose for contradiction that ##X=U\cup V## where ##U## and ##V## are disjoint, open, non-empty sets. Let ##u## and ##v## be points in these sets, respectively. Let ##\gamma:[0,1]\to X## be a path from ##u## to ##v.## Then ##\gamma^{-1}(U)\cup\gamma^{-1}(V)## is a disconnection of ##[0,1]##. Contradiction.

The space ##X=\{(x,\sin(1/x)):x\in\mathbb{R}^{>0}\}\cup\{0\}\times [-1,1]\subset\mathbb{R}^2## is connected but not path-connected. It is connected because if ##X=U\cup V## is as above, then since ##\{0\}\times [-1,1]## is connected, it must be entirely contained within one of these sets, say ##U.## The same must be true for the piece ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}.## Since every neighborhood of ##\{0\}\times [-1,1]## contains points in this second set, this part must also be contained in ##U.##

It is not path-connected because there is not path from any point in ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}## to any point in ##\{0\}\times [-1,1].## This is because if ##\gamma:[0,1]\to X## is such a path, then the second component of ##\gamma(t)## attains every value in ##[-1,1]## as ##t\to 1## and hence there is no point that can be assigned to ##\gamma(1)## in order for ##\gamma## to be continuous.

The circle ##S^1## is obviously path-connected but it is not simply connected. The (based) loop ##\gamma:[0,1]\to S^1, \gamma(t)=e^{2\pi it}## is not nulhomotopic since it lifts (with respect to the usual covering ##\mathbb{R}\to S^1##) to a map ##\tilde{\gamma}:[0,1]\to\mathbb{R}## in the obvious way, and now the endpoints ##0## and ##1## map to distinct elements of ##\mathbb{R}.## A based homotopy of ##\gamma## lifts to one of ##\tilde{\gamma}## and since the endpoints of the lifted map to ##\mathbb{R}## stay distinct throughout the lifted homotopy, the homotopy cannot end in a trivial map.

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fresh_42 and mathwonk
Mentor
2022 Award
I'll try a few. It's been a while for me.

1. We have the prime factorization ##3129=3\times 7\times 149##. Let ##H## be a subgroup of order ##7.## By one of the Sylow theorems, the number of such subgroups is a divisor of ##3\times 149## and is ##1## modulo ##7.## Checking all possibilities, we see that there can only be one subgroup of order ##7,## and so it must be normal (as otherwise a conjugate subgroup would be another subgroup of order ##7##).

Consider the chain of normal subgroups ##\{1\}\leq H\leq G##. The quotient ##H/\{1\}\cong H## is abelian since it has prime order ##7## and is hence cyclic. The quotient ##G/H## has order ##3\times 149## and is therefore abelian as ##149## is not ##1## mod 3 by the following fact:

If ##p<q## are primes with ##q## not 1 modulo ##p,## then the only group of order ##pq## is the cyclic one. This is because by similar counting as above, there is a normal subgroup of order ##q##, so ##G## is a semidirect product of subgroups of order ##p## and ##q.## Since ##q## is not ##1## modulo ##p,## there is only the trivial homomorphism from a ##p##-subgroup to the automorphism group of the ##q## subgroup, so the semidirect product is trivial.

Yes. I was testing my audience with this question. Sorry.

To my excuse: The questions this month are very colorful concerning their difficulty. There are many easy ones because there will be twice as much as usual.

3. We have to check that this subspace of ##\mathfrak{gl}(\mathfrak{g})## is closed under bracket. If ##\alpha,\beta\in\mathfrak{A}(\mathfrak{g})##, then

##[[\alpha,\beta](X),Y]+[X,[\alpha,\beta](Y)]=[\alpha(\beta(X)),Y]-[\beta(\alpha(X)),Y]+[X,\alpha(\beta(Y))]-[X,\beta(\alpha(Y))].##

Note that ##[\alpha(\beta(X)),Y]=-[\beta(X),\alpha(Y)]## since ##\alpha\in\mathfrak{A}(\mathfrak{g})## and similarly for the other terms, so they cancel.

The non-abelian 2-dimensional Lie algebra has generators ##x,y## with the relation ##[x,y]=x.## If a matrix ##\begin{pmatrix}a & b\\c & d\end{pmatrix}## represents an element ##\sigma\in\mathfrak{gl}(\mathfrak{B})## with respect to this basis, in order for ##\sigma## to be in ##\mathfrak{A}(\mathfrak{B})## it is necessary and sufficient for ##[\sigma(x),y]+[\sigma(y),x]## to be zero. We calculate this to be ##a+d##, so ##\mathfrak{A}(\mathfrak{B})## is isomorphic to ##\mathfrak{sl}(2).##

Note: Closure is sufficient since we know that the commutator ##[\alpha ,\beta ]=\alpha \beta -\beta \alpha ## satisfies the Jacobi-identity and anti-commutation from linear Lie algebras.

4. Suppose that ##X## is path-connected. Suppose for contradiction that ##X=U\cup V## where ##U## and ##V## are disjoint, open, non-empty sets. Let ##u## and ##v## be points in these sets, respectively. Let ##\gamma:[0,1]\to X## be a path from ##u## to ##v.## Then ##\gamma^{-1}(U)\cup\gamma^{-1}(V)## is a disconnection of ##[0,1]##. Contradiction.

The space ##X=\{(x,\sin(1/x)):x\in\mathbb{R}^{>0}\}\cup\{0\}\times [-1,1]\subset\mathbb{R}^2## is connected but not path-connected. It is connected because if ##X=U\cup V## is as above, then since ##\{0\}\times [-1,1]## is connected, it must be entirely contained within one of these sets, say ##U.## The same must be true for the piece ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}.## Since every neighborhood of ##\{0\}\times [-1,1]## contains points in this second set, this part must also be contained in ##U.##

It is not path-connected because there is not path from any point in ##\{(x,\sin(1/x):x\in\mathbb{R}^{>0}\}## to any point in ##\{0\}\times [-1,1].## This is because if ##\gamma:[0,1]\to X## is such a path, then the second component of ##\gamma(t)## attains every value in ##[-1,1]## as ##t\to 1## and hence there is no point that can be assigned to ##\gamma(1)## in order for ##\gamma## to be continuous.

The circle ##S^1## is obviously path-connected but it is not simply connected. The (based) loop ##\gamma:[0,1]\to S^1, \gamma(t)=e^{2\pi it}## is not nulhomotopic since it lifts (with respect to the usual covering ##\mathbb{R}\to S^1##) to a map ##\tilde{\gamma}:[0,1]\to\mathbb{R}## in the obvious way, and now the endpoints ##0## and ##1## map to distinct elements of ##\mathbb{R}.## A based homotopy of ##\gamma## lifts to one of ##\tilde{\gamma}## and since the endpoints of the lifted map to ##\mathbb{R}## stay distinct throughout the lifted homotopy, the homotopy cannot end in a trivial map.
I wonder if there is another famous counterexample beside ##(x,\sin(1/x)).## By another, I mean another construction principle.

Gold Member
5.

\begin{align*}
\int_0^{\frac{\pi}{4}} \log (1 + \tan x) dx & = \int_0^{\frac{\pi}{4}} \log \left( \dfrac{\cos x + \sin x}{\cos x} \right) d x
\nonumber \\
& = \int_0^{\frac{\pi}{4}} [ \log ( \cos x + \sin x) - \log (\cos x) ] d x
\nonumber \\
& = \int_0^{\frac{\pi}{4}} [ \log \left( \sqrt{2} \cos (- \frac{\pi}{4}) \cos x - \sqrt{2} \sin (- \frac{\pi}{4}) \sin x \right) - \log (\cos x) ] d x
\nonumber \\
& = \int_0^{\frac{\pi}{4}} [ \log \left( \sqrt{2} \cos (x - \frac{\pi}{4}) \right) - \log (\cos x) ] d x
\nonumber \\
& = \int_0^{\frac{\pi}{4}} [ \log ( \sqrt{2}) + \log (\cos (x - \frac{\pi}{4})) - \log (\cos x) ] d x
\nonumber \\
& = \frac{\pi}{4} \log (\sqrt{2}) + \int_0^{\frac{\pi}{4}} \log ( \cos (\frac{\pi}{4} - x)) d x - \int_0^{\frac{\pi}{4}} \log (\cos x) d x
\nonumber \\
& = \frac{\pi}{8} \log (2) + \int_0^{\frac{\pi}{4}} \log (\cos y) d y - \int_0^{\frac{\pi}{4}} \log (\cos x) d x \quad (y = \frac{\pi}{4} - x)
\nonumber \\
& = \frac{\pi}{8} \log (2)
\end{align*}

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Infrared and fresh_42
Gold Member
Do I get a chocolate from the advent calendar? Send over.

Last edited:
benorin and fresh_42
Gold Member
I wonder if there is another famous counterexample beside ##(x,\sin(1/x)).## By another, I mean another construction principle.

Of the examples I know/can think of, I find the dictionary order on the unit square to be the to be the one most different from the topologists' since curve in style. I think checking that this is not path-connected was an exam question in my undergrad.

Yes. I was testing my audience with this question. Sorry.
I'm having a hard time figuring out what the apology is for! I'm sure everyone here appreciates you submitting these problems.

fresh_42
Gold Member

Write

\begin{align*}
| \sin x | = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos (2kx)
\end{align*}

where

\begin{align*}
a_k = \frac{2}{\pi} \int_0^\pi \sin x \cos (2kx) dx
\end{align*}

as ##| \sin x |## is even and ##\pi##-periodic. We obtain

\begin{align*}
a_k & = \frac{1}{\pi} \int_0^\pi \left\{ \sin \big( (2k+1) x \big) - \sin \big( (2k-1) x \big) \right\}dx
\nonumber \\
& = \frac{1}{\pi} \left[ - \dfrac{\cos \big( (2k+1) x \big)}{2k+1} + \dfrac{\cos \big( (2k+1) x \big)}{2k-1} \right]_0^\pi
\nonumber \\
& = \frac{2}{\pi} \left( \frac{1}{2k+1} - \frac{1}{2k-1} \right)
\nonumber \\
& = - \frac{4}{\pi} \dfrac{1}{k^2 - 1}
\end{align*}

Therefore

\begin{align*}
| \sin x | = \frac{2}{\pi} - \frac{4}{\pi} \sum_{k=1}^\infty \dfrac{\cos (2kx)}{4k^2-1}
\end{align*}

Then

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = \frac{2}{\pi} \int_{-\infty}^\infty \frac{dx}{1+x^2} - \frac{4}{\pi} \sum_{k=1}^\infty \int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1 + x^2} \frac{1}{4k^2-1} dx
\nonumber \\
& = 2 - \frac{8 \alpha}{\pi} \sum_{k=1}^\infty \frac{k}{4k^2-1} \int_{-\infty}^\infty \dfrac{\cos y}{(2k \alpha)^2 + y^2} dy \qquad (y = 2k \alpha x)
\end{align*}

We need the integral

\begin{align*}
I = \int_{-\infty}^\infty \dfrac{\cos x}{a^2 + x^2} dx = \int_{-\infty}^\infty \dfrac{e^{ix}}{a^2 + x^2} dx
\end{align*}

Let

\begin{align*}
f(z) = \dfrac{e^{iz}}{a^2 + z^2}
\end{align*}

For ##z = x + i y## with ##y > 0## we have

\begin{align*}
|f(z)| = \dfrac{e^{-y}}{|a^2 + z^2|}
\end{align*}

So we consider a contour integral with a semi-circle in the UHP as the contribution from the semi-circle to the contour integral vanishes as ##R \rightarrow \infty##:

\begin{align*}
I & = \oint_C \dfrac{e^{iz}}{a^2 + z^2}
\nonumber \\
& = 2 \pi i \lim_{z \rightarrow ia} (z - i a) \dfrac{e^{iz}}{a^2 + z^2}
\nonumber \\
& = 2 \pi i \lim_{z \rightarrow ia} \dfrac{e^{iz}}{ia + z}
\nonumber \\
& = \frac{\pi}{a} e^{-a}
\end{align*}

So that

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 - \frac{8 \alpha}{\pi} \sum_{k=1}^\infty \frac{k}{4k^2-1} \frac{\pi}{2k \alpha} e^{-2 k \alpha}
\nonumber \\
& = 2 - 4 \sum_{k=1}^\infty \frac{e^{-2 k \alpha}}{4k^2-1}
\end{align*}

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Mentor
2022 Award

Write

\begin{align*}
| \sin x | = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos (2kx)
\end{align*}

where

\begin{align*}
a_k = \frac{2}{\pi} \int_0^\pi \sin x \cos (2kx) dx
\end{align*}

as ##| \sin x |## is even and ##\pi##-periodic. We obtain

\begin{align*}
a_k & = \frac{1}{\pi} \int_0^\pi \left\{ \sin \big( (2k+1) x \big) - \sin \big( (2k-1) x \big) \right\}dx
\nonumber \\
& = \frac{1}{\pi} \left[ - \dfrac{\cos \big( (2k+1) x \big)}{2k+1} + \dfrac{\cos \big( (2k+1) x \big)}{2k-1} \right]_0^\pi
\nonumber \\
& = \frac{2}{\pi} \left( \frac{1}{2k+1} - \frac{1}{2k-1} \right)
\nonumber \\
& = - \frac{4}{\pi} \dfrac{1}{k^2 - 1}
\end{align*}

Therefore

\begin{align*}
| \sin x | = \frac{2}{\pi} - \frac{4}{\pi} \sum_{k=1}^\infty \dfrac{\cos (2kx)}{4k^2-1}
\end{align*}

Then

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = \frac{2}{\pi} \int_{-\infty}^\infty \frac{dx}{1+x^2} - \frac{4}{\pi} \sum_{k=1}^\infty \int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1 + x^2} \frac{1}{4k^2-1} dx
\nonumber \\
& = 2 - \frac{8 \alpha}{\pi} \sum_{k=1}^\infty \frac{k}{4k^2-1} \int_{-\infty}^\infty \dfrac{\cos y}{(2k \alpha)^2 + y^2} dy \qquad (y = 2k \alpha x)
\end{align*}

We need the integral

\begin{align*}
I = \int_{-\infty}^\infty \dfrac{\cos x}{a^2 + x^2} dx = \int_{-\infty}^\infty \dfrac{e^{ix}}{a^2 + x^2} dx
\end{align*}

Let

\begin{align*}
f(z) = \dfrac{e^{iz}}{a^2 + z^2}
\end{align*}

For ##z = x + i y## with ##y > 0## we have

\begin{align*}
|f(z)| = \dfrac{e^{-y}}{|a^2 + z^2|}
\end{align*}

So we consider a contour integral with a semi-circle in the UHP as the contribution from the semi-circle to the contour integral vanishes as ##R \rightarrow \infty##:

\begin{align*}
I & = \oint_C \dfrac{e^{iz}}{a^2 + z^2}
\nonumber \\
& = 2 \pi i \lim_{z \rightarrow ia} (z - i a) \dfrac{e^{iz}}{a^2 + z^2}
\nonumber \\
& = 2 \pi i \lim_{z \rightarrow ia} \dfrac{e^{iz}}{ia + z}
\nonumber \\
& = \frac{\pi}{a} e^{-a}
\end{align*}

So that

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 - \frac{8 \alpha}{\pi} \sum_{k=1}^\infty \frac{k}{4k^2-1} \frac{\pi}{2k \alpha} e^{-2 k \alpha}
\nonumber \\
& = 2 - 4 \sum_{k=1}^\infty \frac{e^{-2 k \alpha}}{4k^2-1}
\end{align*}
The beginning looks fine, but I have a different answer at the end. Maybe it is just another notation of the same value, but I don't think so. I integrated a line earlier than you:
$$|\sin(\alpha x)|=\dfrac{2}{\pi}+\dfrac{2}{\pi}\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right)\cos(2n\alpha x)$$
and finally summed over ##(e^{-\alpha })^{2n+1}## plus some factors.

Gold Member

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin ( \alpha x) |}{1+x^2} dx = \frac{2}{\pi} \int_{-\infty}^\infty \frac{dx}{1+x^2} + \frac{2}{\pi} \sum_{k=1}^\infty \left( \frac{1}{2k+1} - \frac{1}{2k-1} \right) \int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx
\end{align*}

correct. Did you not then make the substitution ##y = 2k \alpha x##? You can give me a slight clue if you want.

But is this

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx = 2 k \alpha \int_{-\infty}^\infty \dfrac{\cos y}{(2 k \alpha)^2+y^2} dy
\end{align*}

is correct? Is this

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos x}{a^2+x^2} dx = \frac{\pi}{a} e^{-a}
\end{align*}

correct? I think it is.

About to go to bed in an hour or so.

Last edited:
Mentor
2022 Award
You can give me a slight clue if you want.
$$\int_{-\infty }^{\infty }\dfrac{|\sin(\alpha x)|}{1+x^2}\,dx= 2+2\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right) \underbrace{\dfrac{1}{\pi}\int_{-\infty }^\infty \dfrac{\cos(2n\alpha x)}{1+x^2}\,dx}_{=e^{-2n\alpha }}$$

Gold Member
$$\int_{-\infty }^{\infty }\dfrac{|\sin(\alpha x)|}{1+x^2}\,dx= 2+2\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right) \underbrace{\dfrac{1}{\pi}\int_{-\infty }^\infty \dfrac{\cos(2n\alpha x)}{1+x^2}\,dx}_{=e^{-2n\alpha }}$$
But this is what I got when I did my calculation!...I wrote

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx = 2 k \alpha \int_{-\infty}^\infty \dfrac{\cos y}{(2 k \alpha)^2+y^2} dy \qquad (*)
\end{align*}

and then used

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos x}{a^2+x^2} dx = \frac{\pi}{a} e^{-a}
\end{align*}

in ##(*)## to obtain:

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx & = 2 k \alpha \int_{-\infty}^\infty \dfrac{\cos y}{(2 k \alpha)^2+y^2} dy
\nonumber \\
& = 2 k \alpha \frac{\pi}{2 k \alpha} e^{-2 k \alpha}
\nonumber \\
& = \pi e^{-2 k \alpha}
\end{align*}

Mentor
2022 Award
But this is what I got when I did my calculation!...I wrote

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx = 2 k \alpha \int_{-\infty}^\infty \dfrac{\cos y}{(2 k \alpha)^2+y^2} dy \qquad (*)
\end{align*}

and then used

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos x}{a^2+x^2} dx = \frac{\pi}{a} e^{-a}
\end{align*}

in ##(*)## to obtain:

\begin{align*}
\int_{-\infty}^\infty \dfrac{\cos (2k \alpha x)}{1+x^2} dx & = 2 k \alpha \int_{-\infty}^\infty \dfrac{\cos y}{(2 k \alpha)^2+y^2} dy
\nonumber \\
& = 2 k \alpha \frac{\pi}{2 k \alpha} e^{-2 k \alpha}
\nonumber \\
& = \pi e^{-2 k \alpha}
\end{align*}
Then collect all and build a closed form. The only other trick I applied from there on was to sum over a single sum with ##2n+1## in the denominator. This gives two nice "elementary" functions.

Gold Member
Okey dokey, but it will have to wait until tomorrow.

Gold Member
In post original post, #17, I got the partial solution:

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 - 4 \sum_{k=1}^\infty \frac{e^{-2 k \alpha}}{4k^2-1}
\nonumber \\
& \equiv 2 + 2 \sum_{k=1}^\infty \left( \frac{1}{2k+1} - \frac{1}{2k-1} \right) e^{-2 k \alpha}
\end{align*}

I now write it is closed form.

\begin{align*}
\int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 + 2 e^\alpha \sum_{k=1}^\infty \frac{ (e^{-\alpha})^{2k+1} }{2k+1} - 2 e^{-\alpha} \sum_{k=0}^\infty \frac{(e^{-\alpha})^{2 k + 1}}{2k+1}
\nonumber \\
&= 2 + 2 ( e^\alpha \tanh^{-1} (e^{-\alpha}) - 1) - 2 e^{-\alpha} \tanh^{-1} (e^{-\alpha})
\nonumber \\
&= 2 e^\alpha \tanh^{-1} (e^{-\alpha}) - 2 e^{-\alpha} \tanh^{-1} (e^{-\alpha})
\nonumber \\
& = 4 \sinh \alpha \tanh^{-1} (e^{-\alpha})
\end{align*}

EDIT: The series expansion for ##\tanh^{-1} x## only converges if ##|x| \leq 1## and so the above formula applies when ##e^\alpha \geq 1##, i.e. when ##\alpha \geq \ln (1) = 0##.

I really do need to go to bed now!

Last edited:
benorin and fresh_42
Mentor
2022 Award
I really do need to go to bed now!
Yep. Tell me when I shall show the solution. And ##\alpha >0## is all that we need.

Hint: You show use the ##2## to spend the sums the index ##0## and distribute ##\tanh^{-1}## out of the sum.

Gold Member
Was about to go to sleep when I realized ##\alpha > 0## is all we need. I've edited my previous post.

Gold Member
10.

We have

\begin{align*}
\prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right) & = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( 1 - \frac{1}{n} \right)
\nonumber \\
& = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( \frac{n-1}{n} \right)
\nonumber \\
& = \lim_{m \rightarrow \infty} \frac{1}{m} = 0
\end{align*}

Hence the product

\begin{align*}
\prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right)
\end{align*}

converges.

Now we examine the convergence of

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

We need a lemma: If a sequence ##(\alpha_n)_{n=1}^\infty## satisfies ##\sum_{n=1}^\infty |1 - \alpha_n| \leq \frac{1}{2}##, then for all ##m \geq 1## we have

\begin{align*}
\left| 1 - \prod_{n=1}^m \alpha_n \right| \leq 2 \sum_{n=1}^m |1 - \alpha_n|
\end{align*}

Which is easily proved by induction: The base case ##m = 1## is obvious, so suppose the inequality holds for some ##m \geq 2##. By the inductive hypothesis we have

\begin{align*}
\left| \prod_{n=1}^m \alpha_n \right| & \leq 1 + \left| 1 - \prod_{n=1}^m \alpha_n \right|
\nonumber \\
& \leq 1 + 2 \sum_{n=1}^m |1 - \alpha_n| \leq 2
\end{align*}

We use this in:

\begin{align*}
\left| 1 - \prod_{n=1}^{m+1} \alpha_n \right| & = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n - \prod_{n=1}^{m+1} \alpha_n \right|
\nonumber \\
& = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n (1 - \alpha_{m+1}) \right|
\nonumber \\
& \leq \left| 1 - \prod_{n=1}^m \alpha_n \right| + \left| \prod_{n=1}^m \alpha_n \right| \cdot |1 - \alpha_{m+1}|
\nonumber \\
& \leq 2 \sum_{n=1}^m |1 - \alpha_n| + 2 |1 - \alpha_{m+1}|
\nonumber \\
& = 2 \sum_{n=1}^{m+1} |1 - \alpha_n|
\end{align*}

where we have used the inductive hypothesis again. Note that in particular that ##|1 - \prod_{n=1}^m \alpha_n| \leq 1## holds for all ##m \geq 1##.

We now show that there exists an ##N \geq 3## such that ##\sum_{n=N}^\infty \left| \frac{4}{n^2} \right| \leq \frac{1}{2}##. We do this by proving the series ##\sum_{n=3}^\infty \frac{4}{n^2}## convergences. Note

\begin{align*}
0 & \leq \frac{4}{n^2} \leq \frac{4}{n^2 - n} \quad \text{for } n \geq 2
\end{align*}

and

\begin{align*}
\sum_{n=3}^m \frac{4}{n^2 - n} & \leq 4 \sum_{n=3}^m \left( \frac{1}{n-1} - \frac{1}{n} \right)
\nonumber \\
& = 4 \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{m-1} - \frac{1}{m} \right) \right]
\nonumber \\
& = 2 - 4 \frac{1}{m}
\end{align*}

implying ##\lim_{m \rightarrow \infty} \sum_{n=3}^m \frac{4}{n^2 - n} = 2##, which in turn implies that ##\sum_{n=3}^\infty \frac{4}{n^2}## converges. As ##\sum_{n=3}^\infty \frac{4}{n^2}## converges there exists an ##N \geq 3## such that

\begin{align*}
\sum_{n=N}^\infty \frac{4}{n^2} & \leq \frac{1}{2}
\end{align*}

(In fact ##\sum_{n=9}^\infty \frac{4}{n^2} < \frac{1}{2}##). We can now use the above lemma:

As the sequence ##\left( 1 - \frac{4}{n^2} \right)_{n=N}^\infty## satisfies ##\sum_{n=N}^\infty \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq \frac{1}{2}##, then for all ##m \geq N## we have

\begin{align*}
\left| 1 - \prod_{n=N}^m \left( 1 - \frac{4}{n^2} \right) \right| \leq 2 \sum_{n=N}^m \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq 1
\end{align*}

So that

\begin{align*}
\left| 1 - \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right) \right| \leq 1
\end{align*}

Hence

\begin{align*}
\prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

convergences, and given that ##\prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right)## is finite for finite ##N##, we finally have that the product

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right) \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

converges.

Last edited:
Gold Member
Is 11. for high-schoolers?

Last edited:
Mentor
2022 Award
10.

We have

\begin{align*}
\prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right) & = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( 1 - \frac{1}{n} \right)
\nonumber \\
& = \lim_{m \rightarrow \infty} \prod_{n=2}^m \left( \frac{n-1}{n} \right)
\nonumber \\
& = \lim_{m \rightarrow \infty} \frac{1}{m} = 0
\end{align*}

Hence the product

\begin{align*}
\prod_{n=2}^\infty \left( 1 - \frac{1}{n} \right)
\end{align*}

converges.
Well, yes and no. My source says it does not converge, since we do not have ##0## in the range of multiplication, i.e. we say that the logarithm (minus infinity) does not converge, hence the product does not converge. I am not really convinced by this interpretation, but I wanted to mention it.

Now we examine the convergence of

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

We need a lemma: If a sequence ##(\alpha_n)_{n=1}^\infty## satisfies ##\sum_{n=1}^\infty |1 - \alpha_n| \leq \frac{1}{2}##, then for all ##m \geq 1## we have

\begin{align*}
\left| 1 - \prod_{n=1}^m \alpha_n \right| \leq 2 \sum_{n=1}^m |1 - \alpha_n|
\end{align*}

Which is easily proved by induction: The base case ##m = 1## is obvious, so suppose the inequality holds for some ##m \geq 2##. By the inductive hypothesis we have

\begin{align*}
\left| \prod_{n=1}^m \alpha_n \right| & \leq 1 + \left| 1 - \prod_{n=1}^m \alpha_n \right|
\nonumber \\
& \leq 1 + 2 \sum_{n=1}^m |1 - \alpha_n| \leq 2
\end{align*}

We use this in:

\begin{align*}
\left| 1 - \prod_{n=1}^{m+1} \alpha_n \right| & = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n - \prod_{n=1}^{m+1} \alpha_n \right|
\nonumber \\
& = \left| 1 - \prod_{n=1}^m \alpha_n + \prod_{n=1}^m \alpha_n (1 - \alpha_{m+1}) \right|
\nonumber \\
& \leq \left| 1 - \prod_{n=1}^m \alpha_n \right| + \left| \prod_{n=1}^m \alpha_n \right| \cdot |1 - \alpha_{m+1}|
\nonumber \\
& \leq 2 \sum_{n=1}^m |1 - \alpha_n| + 2 |1 - \alpha_{m+1}|
\nonumber \\
& = 2 \sum_{n=1}^{m+1} |1 - \alpha_n|
\end{align*}

where we have used the inductive hypothesis again. Note that in particular that ##|1 - \prod_{n=1}^m \alpha_n| \leq 1## holds for all ##m \geq 1##.

We now show that there exists an ##N \geq 3## such that ##\sum_{n=N}^\infty \left| \frac{4}{n^2} \right| \leq \frac{1}{2}##. We do this by proving the series ##\sum_{n=3}^\infty \frac{4}{n^2}## convergences. Note

\begin{align*}
0 & \leq \frac{4}{n^2} \leq \frac{4}{n^2 - n} \quad \text{for } n \geq 2
\end{align*}

and

\begin{align*}
\sum_{n=3}^m \frac{4}{n^2 - n} & \leq 4 \sum_{n=3}^m \left( \frac{1}{n-1} - \frac{1}{n} \right)
\nonumber \\
& = 4 \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{m-1} - \frac{1}{m} \right) \right]
\nonumber \\
& = 2 - 4 \frac{1}{m}
\end{align*}

implying ##\lim_{m \rightarrow \infty} \sum_{n=3}^m \frac{4}{n^2 - n} = 2##, which in turn implies that ##\sum_{n=3}^\infty \frac{4}{n^2}## converges. As ##\sum_{n=3}^\infty \frac{4}{n^2}## converges there exists an ##N \geq 3## such that

\begin{align*}
\sum_{n=N}^\infty \frac{4}{n^2} & \leq \frac{1}{2}
\end{align*}

(In fact ##\sum_{n=9}^\infty \frac{4}{n^2} < \frac{1}{2}##). We can now use the above lemma:

As the sequence ##\left( 1 - \frac{4}{n^2} \right)_{n=N}^\infty## satisfies ##\sum_{n=N}^\infty \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq \frac{1}{2}##, then for all ##m \geq N## we have

\begin{align*}
\left| 1 - \prod_{n=N}^m \left( 1 - \frac{4}{n^2} \right) \right| \leq 2 \sum_{n=N}^m \left|1 - \left( 1 - \frac{4}{n^2} \right) \right| \leq 1
\end{align*}

So that

\begin{align*}
\left| 1 - \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right) \right| \leq 1
\end{align*}

Hence

\begin{align*}
\prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

convergences, and given that ##\prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right)## is finite for finite ##N##, we finally have that the product

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{N-1} \left( 1 - \frac{4}{n^2} \right) \prod_{n=N}^\infty \left( 1 - \frac{4}{n^2} \right)
\end{align*}

converges.
Do you know the limit?

Mentor
2022 Award
Is 11. for high-schoolers?
More for physicists. The Heisenberg group comes up in physics and I wanted to show how the group and tangent space at ##1,## the algebra, look like. The way from a manifold to a tangent bundle and back via differentiation and exponentiation is important in physics. It is good to have at least one easy example in mind. It is also the easiest to see the difference between BCH and the commuting case of taking the exponential of a sum.

Why is it always December when I get a good idea for an insight article:
Why are integration and exponentiation the same thing?

Keith_McClary
Gold Member

Do you know the limit?
Using

\begin{align*}
\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left( 1 - \frac{z^2}{n^2} \right)
\end{align*}

So that

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) & = \lim_{z \rightarrow 2} \frac{\sin \pi z}{\pi z} \dfrac{1}{(1-z^2) \left( 1 - \dfrac{z^2}{2^2} \right)}
\nonumber \\
& = \frac{1}{1-2^2} \frac{1}{1 + \dfrac{2}{2}} \frac{1}{2} \lim_{z \rightarrow 2} \frac{\sin \pi z}{\pi} \dfrac{1}{1 - \dfrac{z}{2}}
\nonumber \\
& = - \frac{1}{12} \lim_{z \rightarrow 2} (-2) \dfrac{\dfrac{d}{dz} \sin \pi z}{\pi}
\nonumber \\
& = \frac{1}{6}
\end{align*}

fresh_42
Mentor
2022 Award
Using

\begin{align*}
\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left( 1 - \frac{z^2}{n^2} \right)
\end{align*}

So that

\begin{align*}
\prod_{n=3}^\infty \left( 1 - \frac{4}{n^2} \right) & = \lim_{z \rightarrow 2} \frac{\sin \pi z}{\pi z} \dfrac{1}{(1-z^2) \left( 1 - \dfrac{z^2}{2^2} \right)}
\nonumber \\
& = \frac{1}{1-2^2} \frac{1}{1 + \dfrac{2}{2}} \frac{1}{2} \lim_{z \rightarrow 2} \frac{\sin \pi z}{\pi} \dfrac{1}{1 - \dfrac{z}{2}}
\nonumber \\
& = - \frac{1}{12} \lim_{z \rightarrow 2} (-2) \dfrac{\dfrac{d}{dz} \sin \pi z}{\pi}
\nonumber \\
& = \frac{1}{6}
\end{align*}
And this is the version without the sine function:

\begin{align*}
\prod_{k=3}^n\dfrac{k^2-4}{k^2}&=\prod_{k=3}^n\dfrac{k+2}{k}\cdot\dfrac{k-2}{k}=\prod_{k=3}^n\dfrac{k+2}{k+1}\cdot\dfrac{k+1}{k}\cdot\dfrac{k-1}{k}\cdot\dfrac{k-2}{k-1}\\
&=\prod_{k=3}^n\dfrac{k+2}{k+1}\cdot\prod_{k=3}^n\dfrac{k+1}{k}\cdot\prod_{k=3}^n\dfrac{k-1}{k}\cdot\prod_{k=3}^n\dfrac{k-2}{k}\\
&\stackrel{\text{telescope}}{=}\dfrac{n+2}{4}\cdot\dfrac{n+1}{3}\cdot\dfrac{2}{n}\cdot\dfrac{1}{n-1}=\dfrac{n^2+3n+2}{6n^2-6n}
\end{align*}

julian
fishturtle1
Lemma: If ##p## is prime then the only solutions to ##x^2 = 1 \text{ (mod p)}## are ##x = -1, 1##.

Proof: Suppose ##x^2 = 1 \text{ (mod p)}##. Then ##p \vert (x^2 - 1)## i.e. ##p \vert (x-1)(x+1)##. Since ##p## is prime, we have ##p \vert x + 1## or ##p \vert x - 1##. This implies ##x = -1 \text{ (mod p)}## or ##x = 1 \text{ (mod p)}##. And we see ##1## and ##-1## really are solutions to ##x^2 = 1 \text{ (mod p)}##. []

Proof of problem in OP: ##(\Longleftarrow): ## Suppose ##p## is prime. Then

$$(p-1)! = (p-1)(p-2)\cdot \dots \cdot 2 \cdot 1 = -1\cdot (p-2) \cdot \dots \cdot 2 \cdot 1 = -1 \cdot 1 = -1 \text{ (mod p)}$$

The second equality is from ##p-1 = -1 \text{ (mod p)}## and the third inequality is using the fact the only squares are ##-1## and ##1##, therefore every other element is canceled by its inverse.

##(\Longrightarrow): ## Suppose ##(p-1)! = -1 \text{ (mod p)}##. We will show ##p## must be prime. Suppose ##1 \le d < p## is a divisor of ##p##. Then ##d## is also a divisor of ##(p-1)!##. We can rewrite the congruence as

$$-1 = (p-1)! + p\cdot k$$ for some integer ##k##. Since ##d## divides the right hand side, it must divide the left hand side. Hence, ##d \vert -1## which implies ##d = 1##. This implies the only divisors of ##p## are ##1## and ##p##. We may conclude ##p## is prime. []

For the next part, we claim the first two primes that satisfy ##(p-1)! = -1 \text{ (mod p^2)}## are ##p = 5## and ##p = 13##. Observe,

\begin{align}
(2-1)! &= 1 \text{ (mod 4)} \\
(3-1)! &= 2 \text{ (mod 9)} \\
(5-1)! &= 24 = -1 \text{ (mod 25)} \\
(7-1)! &= 720 = 34 \text{ (mod 49)} \\
(11-1)! &= 3628800 = 10 \text{ (mod 121)} \\
(13-1)! &= 479,001,600 = 168 = -1 \text{ (mod 169)}
\end{align}