Math Challenge - February 2021

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  • #26
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There's really nothing to prove other than the well known theorems :rolleyes:
By Modulus maximum principle [itex]f[/itex] must be constant on [itex]U_0\subseteq U[/itex].
By the Identity theorem, since [itex]U[/itex] is open and connected, [itex]f[/itex] must be constant on [itex]U[/itex].
That was the idea. Both proofs are easy enough such that they can be figured out, and the statements are important enough to be learnt.

Guess you all recognized that I begin to run out of good ideas. Algebra is usually not touched even if the problems are really easy. Geometry is a horror to check or correct without a common picture (plus that I do not know the correct English names of the objects, e.g. angle bisector). Functional analysis is either well-known, used, or a nightmare about the many categories (e.g. different convergences, Sobolev). Most of the interesting problems require an even more advanced framework than algebra does. Calculations like integrals, ODE, or inequalities are either used up or simply boring. And now that you just quote the theorems or corollaries I want to be proven, this source is exhausted, too. Any ideas?
 
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  • #27
martinbn
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That was the idea. Both proofs are easy enough such that they can be figured out, and the statements are important enough to be learnt.

Guess you all recognized that I begin to run out of good ideas. Algebra is usually not touched even if the problems are really easy. Geometry is a horror to check or correct without a common picture (plus that I do not know the correct English names of the objects, e.g. angle bisector). Functional analysis is either well-known, used, or a nightmare about the many categories (e.g. different convergences, Sobolev). Most of the interesting problems require an even more advanced framework than algebra does. Calculations like integrals, ODE, or inequalities are either used up or simply boring. And now that you just quote the theorems or corollaries I want to be proven, this source is exhausted, too. Any ideas?
How about a way for people to suggest problems? Of course it has to be in a way that doesn't give them away before the list is posted? Say we send you problems for you to choose from?
 
  • #28
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Say we send you problems for you to choose from?
I have nobody on my ignore list ...

I just need to know one answer to each suggestion:
Will you moderate the answers given, and if not please attach a solution.
 
  • #29
martinbn
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I have nobody on my ignore list ...

I just need to know one answer to each suggestion:
Will you moderate the answers given, and if not please attach a solution.
Does that mean that suggestions are welcomed? Once i did send you a problem, but heard nothing about it. May be it wasn't suitable, but you didn't say anything, so i assumed that you didn't want any.
 
  • #30
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Does that mean that suggestions are welcomed?
Sure. I think it was even mentioned in the early days.
Once i did send you a problem, but heard nothing about it. May be it wasn't suitable, but you didn't say anything, so i assumed that you didn't want any.
Sorry, that I don't remember this, so I can't say what happened. I only need to know who will moderate it: send me a PM if solved by whom and in which post, or give me the solution (for dummies) so that I can moderate it.
 
  • #31
martinbn
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Sure. I think it was even mentioned in the early days.

Sorry, that I don't remember this, so I can't say what happened. I only need to know who will moderate it: send me a PM if solved by whom and in which post, or give me the solution (for dummies) so that I can moderate it.
Yes, the problem should come with a detailed solution, so that you an deside if it fits the chalange.
 
  • #32
Office_Shredder
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There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.
 
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  • #33
wrobel
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big gun hint for 3b

Screenshot from 2021-02-06 13-05-32.png
 
  • #34
julian
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There are a bunch of unsolved questions from previous threads, maybe have a challenge month where you just repost them and we work together to try to crack them.

I went back and solved a couple of the previous problems I just didn't post the answers. I'd like another crack at some of the other previous problems myself. I think some of the questions require me learning a bit of new maths, and for that reason I'd like to have a go at them myself.

I thought problems originally were being posted by various members. Has @fresh_42 been setting all the problems of late? Impressive.

Now and then I've been putting together a few problems when I can, which I was going to give @fresh_42 at some point.

I posted one problem last month because the problems got done early. You can have a go at that if you want :smile:. It doesn't require that you learn new maths, you solve the problem by taking some quantity to be an integer and expressing it another way that allows you to make estimations (hint: integrals) and making postulates.
 
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  • #37
julian
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To elaborate on problem 4.

It was obvious that the AM-GM inequality would be involved, but like I said, a simple application leads to the Harmonic series:

\begin{align*}
\sum_{n=1}^\infty \prod_{k=1}^n (a_k)^{1/n} & \leq \sum_{n=1}^\infty \sum_{k=1}^n \frac{a_k}{n} \\
& = a_1 \sum_{n=1}^\infty \frac{1}{n} + a_2 \sum_{n=2}^\infty \frac{1}{n} + \cdots
\end{align*}

but the Harmonic series diverges, so I decided to consider a "weighted" version of the AM-GM inequality. It is not uncommon to see "weighted" versions of inequalities.

Replacing ##\sum_{n=1}^\infty \sum_{k=1}^n \rightarrow \sum_{k=1}^\infty \sum_{n=k}^\infty## is playing around with double sums, and has a simple interpretation in terms of dots on a 2d grid. You need the sum over ##k## to go from 1 to ##\infty##.

Putting ##b_k = \frac{c_k}{c_{k-1}}## and getting ##c_1 c_2 \cdots c_n = b_n## is simply telescoping products and obviously simplifies things.

And ##\left( 1 + \frac{1}{k} \right)^k \leq e## is a well known inequality.

This

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n(n+1)} = \sum_{n=k}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{k}
\end{align*}

is a very well known example of telescoping for series.
 
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  • #38
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To elaborate on problem 4.

It was obvious that the AM-GM inequality would be involved, but like I said, a simple application leads to the Harmonic series:

\begin{align*}
\sum_{n=1}^\infty \prod_{k=1}^n a_k & \leq \sum_{n=1}^\infty \sum_{k=1}^n \frac{a_k}{n} \\
& = a_1 \sum_{n=1}^\infty \frac{1}{n} + a_2 \sum_{n=2}^\infty \frac{1}{n} + \cdots
\end{align*}

but the Harmonic series diverges, so I decided to consider a "weighted" version of the AM-GM inequality. It is not uncommon to see "weighted" versions of inequalities.

Replacing ##\sum_{n=1}^\infty \sum_{k=1}^n \rightarrow \sum_{k=1}^\infty \sum_{n=k}^\infty## is playing around with double sums, and has a simple interpretation in terms of dots on a 2d grid. You need the sum over ##k## to go from 1 to ##\infty##.

Putting ##b_k = \frac{c_k}{c_{k-1}}## and getting ##c_1 c_2 \cdots c_n = b_n## is simply telescoping products and obviously simplifies things.

And ##\left( 1 + \frac{1}{k} \right)^k \leq e## is a well known inequality.

This

\begin{align*}
\sum_{n=k}^\infty \frac{1}{n(n+1)} = \sum_{n=k}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{k}
\end{align*}

is a very well known example of telescoping for series.
I'm having trouble sweeping this up to a proof, starting with a forgotten root in the very first term. It looks somehow right (even though I used Stirling), but can you line up your inequalities`?
 
  • #39
julian
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Sorry, I already gave the proof of everything on page 1. Here I'm just commenting on my methodology. (I've put the root in now).
 
  • #40
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Sorry, I already gave the proof on page 1. Here I'm just commenting on my methodology.
Oops! Your proof on page 1 was fine, no need to explain it. It is indeed Carleman's inequality. It is even a strict one.

The problems which have names can certainly be found somewhere, but that does not mean that they are famous enough to be automatically known, plus it means they are worth tackling them! I think it is still better to have a "I already have proven this" moment when encountering those statements in a book, than having solved a complicated but useless integral. I do not assume cheating in such cases. Why should I? Anyone who writes down an answer that they found elsewhere only cheats themselves, but at least has made the effort to learn a new theorem. I always try to put some value to the problems even if it means that they can be found somewhere.
 
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  • #41
julian
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I thought it was strict inequality. I used that ##1 + x \leq e^x## to derive ##\left( 1 + \frac{1}{k} \right)^k \leq e## with equality holding only when ##k \rightarrow \infty##.
 
  • #42
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I thought it was strict inequality. I used that ##1 + x \leq e^x## to derive ##\left( 1 + \frac{1}{k} \right)^k \leq e## with equality holding only when ##k \rightarrow \infty##.
You can simply consider GM=AM to get a contradiction. That's easier than to deal with the exponential function.
 
  • #43
julian
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In May 2008 I found that I could prove the inequality: ##(\theta^2 +x_1^2) \dots (\theta^2 +x_n^2) \geq [\theta^2 + (x_1^2 \dots x_n^2)^{1/k}]^k## with Cauchy's forward-backward induction proof, which I was pleased about:

https://www.physicsforums.com/threa...challenge-may-2018.946386/page-3#post-6006214

@StoneTemplePython pointed out a simpler proof: that the inequality is basically the superadditivity of the Geometric Mean. Something I wasn't aware of but I reckon might possibly be the mainstream proof. I have just now done a search for "superadditivity of the Geometric Mean" and the Cauchy's forward-backward induction proof and couldn't find anything (well, aside from the physicsforums "Math Challenge - May 2018" webpage). However, there is most be somebody else who has also had this idea.

Also, I have realised that the inequality can be derived by solving a constrained extremal problem. Which by the way is one of the challenge problems that I have come up with.

I have also come up with a combinatorial proof which was fairly easy to understand but takes ages to explain. It is also a good moment when you come up with proofs that might not be the mainstream proof.

p.s. Sorry if got defensive. It is just that there is a bunch of idiots following me around, but you dont want to know about that.
 
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  • #44
julian
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Anyway.
 
  • #45
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I have just now done a search for "superadditivity of the Geometric Mean" and the Cauchy's forward-backward induction proof and couldn't find anything (well, aside from the physicsforums "Math Challenge - May 2018" webpage). However, there is most be somebody else who has also had this idea.

Also, I have realised that the inequality can be derived by solving a constrained extremal problem. Which by the way is one of the challenge problems that I have come up with.
If you look at the previous solutions (currently 4 pdf files), you will find a lot of useful inequalities and equations around symmetries. Whenever possible I added their name.
 
  • #46
mathwonk
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In reference e.g. to problems generally, I suggest that the goal of this exercise might be to illuminate some key point, rather than to quote a powerful result that implies the problem. Problem #9 e.g., which is indeed essentially the maximum modulus principle, has several proofs, with different morals, each with something to teach us. The mean value property, which follows from integral representations, and I think shows that the value of an appropriate function at the center of a circle equals the average of its values on the boundary, is one that will appeal to analytically minded persons. The approach I hinted at is more geometric, and shows that every analytic function which is non constant near 0, (and with value zero there), looks locally, in some coordinates, like the function z^n, for some n ≥ 1. This makes the local geometric behavior of analytic complex functions very transparent, and implies also the useful "open mapping principle". To each his/her own. The mean value property also has the virtue of remaining true also for harmonic functions, and even in characterizing them, as I recall.

Remrk on #10: most group theory problems I solve follow from the Sylow theorems. But there are some more powerful results out there due to Burnside and others which some of us are not aware of. This problem #10 seems to yield to his "transfer theorem" or "N/C theorem", one that I only learned today. Is there a more clever, less advanced solution, does anyone know, using only Sylow? E.g. as an example of this sort, just try to show that every group of order 30 = 2.3.5 is solvable.
 
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  • #47
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3. Prove or find a counterexample to:
(a) ##L^2## convergence implies pointwise convergence.
(b) ##\displaystyle{\lim_{n \to \infty}\int_{0}^{\infty}\dfrac{\sin x^n}{x^n}\,dx}=1##
(c) Let ##(f_n)## be a sequence of measurable functions which converge uniformly to zero on ##[0,\infty).## Then
$$
\lim_{n \to \infty}\int_{[0,\infty)} f_n(x)\,dx=\int_{[0,\infty)} \lim_{n \to \infty} f_n(x)\,dx\,.
$$
(a) False; consider a trivial sequence of null sets ##X_n=Y## for some null set ##Y##, whence ##\int |\mathbf{1}_{X_n}|^2\,d\mu=0##. Thus ##\mathbf{1}_{X_n}## converges to zero in ##L^2(\mathbb{R})## (in fact ##X_n=0## wrt the ##L^2(\mathbb{R})## seminorm), but the pointwise limit is ##\mathbf{1}_Y##.

(b) First consider the integral ##\int_0^1\frac{\sin x^n}{x^n}dx##. From the alternating Maclaurin series of ##\sin x## one has the bound ##0\leq 1-\frac{x^{2n}}{6}\leq\frac{\sin x^n}{x^n}\leq 1## for ##x\in(0,1)##. This gives us bounds $$1-\frac{1}{12n}\leq\int_0^1\frac{\sin x^n}{x^n}dx\leq 1$$ so by the squeeze theorem, the limit as ##n\rightarrow\infty## is ##1##.

Now consider ##\int_1^\infty\frac{\sin x^n}{x^n}dx##. We have the simple chain of inequalities $$\Bigg|\int_1^\infty\frac{\sin x^n}{x^n}dx\Bigg|\leq \int_1^\infty\Bigg|\frac{\sin x^n}{x^n}\Bigg|dx\leq\int_1^\infty\frac{1}{x^n}dx=\frac{1}{n+1}$$ so our integral is bounded between ##0## and ##\frac{1}{n+1}##, and shrinks to ##0## as ##n\rightarrow\infty##.

Since $$\int_0^\infty\frac{\sin x^n}{x^n}dx=\int_0^1\frac{\sin x^n}{x^n}dx+\int_1^\infty\frac{\sin x^n}{x^n}dx,$$ the integral is equal to ##0+1## or ##1##.

(c) False: consider the fn. $$f_n(x)=\frac{(-1)^n\mathbf{1}_{[0,n)}}{n}.$$ ##f_n## converges uniformly to ##0##, but ##\int_0^\infty f_n(x)\,dx=(-1)^n##.
 
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  • #48
PeroK
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View attachment 277251

High Schoolers only

11.
Let ##z## be a natural number with ##1995## decimal digits and ##1\leq n\leq 1994.## Then we note the number, which we get by cutting off the first ##n## digits and append them in the same order at the end of ##z## by ##z^{[n]}.## Show that if ##z## is divisible by ##27,## then all ##z^{[n]}## are divisible by ##27,## too.

12. Let ##a,b,c,d## be positive real numbers. Prove (in the logically correct order)
$$
\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}}\leq \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}
$$

13. Let ##m\geq 2## be a given natural number. We define a sequence ##(x_0,x_1,x_2,\ldots)## of numbers by ##x_0=0,x_1=1,## and for ##n\geq 0## we set ##x_{n+2}## to be the remainder of ##x_{n+1}+x_n## by division by ##m,## chosen such that ##0\leq x_{n+2} < m.## Decide whether for every ##m\geq 2## there exists a natural number ##k\geq 1,## such that ##x_{k+2}=1\, , \,x_{k+1}=1\, , \,x_k=0.##

14. We define real functions
$$
f_n(x):=x^3+(n+3)\cdot x^2+2n\cdot x -\dfrac{n}{n+1}
$$
for every non-negative integer ##n\geq 0.## Determine all values of ##n,## such that all zeros of ##f_n(x)## are contained in an interval of length ##3.##

15. (a) Determine the number of all pairs of integers ##(x,y) \in \mathbb{N}_0^2## with ##\sqrt{x}+\sqrt{y}=1993.##
15. (b) Determine for every ##n\in \mathbb{N}## the greatest power of ##2## which divides ##[( 4+\sqrt{18} )^n ].##

I guess it's clear that the high-school questions have become too hard. Given that none of them has been solved!

These look like a sufficient challenge for anyone without the pure maths knowledge to tackle the main questions.
 
  • #49
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I guess it's clear that the high-school questions have become too hard. Given that none of them has been solved!

These look like a sufficient challenge for anyone without the pure maths knowledge to tackle the main questions.
I won't mind if anyone else tries to solve them in this case, i.e. at the end of month. But I am not sure whether your hypothesis (too hard) holds. There could be other reasons. I admit that some are tricky, but #11 and #12 are quite easy. Especially #12 is straight forward: start with the inequality, end up with something like ##f(a,b,c,d)^2\geq 0## and write it down backwards. It's more a question about proof logic than about arithmetic.

The questions in March will be easier, or at least significantly shorter to answer.
 
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  • #50
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Lemma: If ##a, a+k## are integers with ##a \ge 2## and ##k \ge 1##, we have ##a + (a+k) < a(a+k)##

Proof: We have ##a(a+k) \ge 2(a+k) = (a+k) + (a+k) \ge (a+1) + (a+k) > a + (a+k)##. []


Suppose ##G## is a group of order ##pqr## where ##p < q < r## are distinct primes. Then ##G## is solvable.

Proof: We note that a group of order ##ab## where ##a## and ##b## are distinct primes is solvable. So, it's enough to show ##G## has a normal subgroup of prime order. Let ##n_p## denote the number of Sylow p subgroups of ##G##. Define ##n_q, n_r## similarly. Sylow's theorems tells us
\begin{align*}
n_p &= 1 (\text{ mod p })\\
n_q &= 1 (\text{ mod q })\\
n_r &= 1 (\text{ mod r }) \\
\end{align*}
Sylow's theorems also tells us ##n_p \vert qr##, so ##n_p = 1, q, r,## or ##qr##. Similarly, ##n_q = 1, p, r## or ##pr## and ##n_r = 1, p, q## or ##pq##. If ##n_r = 1##, we're done. If ##n_r = p##, then ##p = 1 (\text{ mod r })##. But that means ##p - 1 = rk## for some positive integer ##k##. But ##p-1 < p < r \le rk##. We may conclude ##n_r \neq p## and ##n_r \neq q##. So we must have ##n_r = pq##. This shows ##G## has ##pq(r-1) = pqr - pq## elements of order ##r##. Subtracting that from ##pqr##, we have ##pqr - (pqr - pq) = pq##. Subtracting ##1## for the identity element, we have ##pq - 1## elements unaccounted for.


Assume by contradiction ##n_p, n_q \neq 1##. We'll show that no matter how small ##n_p## and ##n_q## are, there are still too many elements. If ##n_q = p##, then ##p - 1## is a positive multiple of ##q## which is a contradiction. So, the smallest ##n_q## can be is ##n_q = r##.


The smallest ##n_p## can be is ##n_p = q##. So, if ##n_q = r## and ##n_p = q##, we have ##r(q-1)## elements of order ##q## and ##q(p-1)## elements of order ##p##. But we only have ##pq - 1## elements left. We see
$$(pq - 1) - q(p-1) - r(q-1) = pq - 1 - pq + q - qr + r = q + r - qr - 1 < q + r - qr < 0$$
We've reached a contradiction and can conclude ##n_p = 1## or ##n_p = 1##. This gives a normal subgroup ##H## of prime order, say ##p##. Then ##G/H## is a group of order ##qr##. So, ##G/H## and ##H## are solvable. We may conclude ##G## is solvable.
 

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