# Thread: Extracurricular Elementary Mathematics Marathon

1. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
$LHS^2=\prod_{k=1}^n \left(\frac{2k-1}{2k}\right)^2 \\ \\= \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\frac{7}{8}\cdot\frac{9}{10}\cdot\frac{9}{10}\cdot\frac{11}{12}\cdot \left(\frac{11}{12}\cdot\frac{13}{14}\right)\cdot\left(\frac{13}{14}\cdot\frac{15}{17}\right)\ldots \left(\frac{2n-3}{2n-2}\cdot \frac{2n-1}{2n}\right)\cdot \frac{2n-1}{2n}\\ \\ < \left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\frac{7}{8}\cdot\frac{9}{10}\cdot\frac{9}{10}\cdot\frac{11}{12}\right)\cdot \left(\frac{12}{13}\cdot\frac{13}{14}\right)\cdot\left(\frac{14}{15}\cdot\frac{15}{17}\right)\ldots \left(\frac{2n-2}{2n-1}\cdot \frac{2n-1}{2n}\right)\\ \\ < \left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdot\frac{7}{8}\cdot\frac{9}{10}\cdot\frac{9}{10}\cdot\frac{11}{12}\right)\cdot \frac{12}{2n}$

$=\frac{43659}{131072n}<\frac{1}{3n}.$

This establishes the claim for all n larger than 6.

Verifying the inequality for the five values of n between 2 and 6 with a calculator or by hand completes the proof.
Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.

2. ## Re: Extracurricular Elementary Mathematics Marathon

Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.
Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.

3. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.
The intermediate inequality you induct on is placing 1/sqrt(3n+1) between the two. It may seem arbitrary but the problem is that the right hand side doesn't shrink fast enough for an inductive proof to be able to compare the two sides.

4. ## Re: Extracurricular Elementary Mathematics Marathon

$5. It is clear that x=-y is a solution for any integer y. We now assume x+y\neq 0 and so we can divide by this to reduce our equation to x^2+y^2-xy=x+y. \\ \\ The reason this is promising is that for x and y large, the LHS should grow a lot faster than the RHS and equality should be impossible.\\ \\ Indeed: LHS\geq x^2+y^2-\frac{x^2+y^2}{2}=\frac{x^2+y^2}{2}.\\ \\ Now if |t|>2, then t^2/2>t, so we cannot have both |x| and |y| larger than 2. \\ \\ As the equation is symmetric, we might as well assume without loss of generality that |y|\leq |x| (keeping in mind that we need to symmetrise the solution set we end up with after this assumption).\\ \\ We now need to simply find the solution set for the 5 possible values of y.\\ \\ y=0 \Rightarrow x^3=x^2 \Rightarrow x=1.\\ \\ y=1 \Rightarrow x^2=x+2 \Rightarrow x=-2,2\\ \\y=-1\Rightarrow (x^2+2)(x-1)=x^3-x^2+2x-2=0 \Rightarrow x=1\\ \\ y=2 \Rightarrow (x-2)(x+2)(x-1)=x^3-x^2-4x+4=0 \Rightarrow x=\pm 2$

$y=-2 \Rightarrow (x^2+4)(x-1)=0 \quad \textrm{(no solutions)}.\\ \\ Symmetrising this solution set completes the problem.$

5. ## Re: Extracurricular Elementary Mathematics Marathon

$7. \\ \\(2n)!^2=4n^2 \prod_{k=1}^{2k-1} k(2n-k)\\ \\ \leq 4n^2 \prod_{k=1}^{2n-1} n^2=4n^{4n}.\\ \\ where the inequality came from applying AM-GM to each factor in the product.\\ \\ Taking square roots of both sides completes the proof.$

6. ## Re: Extracurricular Elementary Mathematics Marathon

You should probably specify the domain of the functional equation you posted Paradoxica. It it the reals?

Assuming this is the case (and in fact this working is also valid if the domain is the integers, rationals or complex numbers), then:

6/

Let x=0 and let y=-f(0)+z, so

f(z)=f(f(0)+y)=y=-f(0)+z.

This implies that all solutions are of the form f(z)=z+c for some constant c.

However, if we apply this formula to our original functional equation, we get:

x+y+2c=x+y.

Hence c must be zero, and the only solution to the functional equation is the identity function f(x)=x.

7. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent 1. \forall n \in \mathbb{N} , prove \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1} \geq n (\sqrt[n]{2}-1) (done by seanieg98)\\ 2. Prove (2n-1)!\geq n^n (done by Integrand) \\ 3. Prove that for every integer x, \frac{x^5}{5}+\frac{x^3}{3}+\frac{7x}{15} is also an integer. (done by KingOfActing) \\ 4. Prove for all n >1, \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n-1}{2n}<\frac{1}{\sqrt{3n}} (case bashed by seanieg98)\\ 5. Find all integer pairs (x,y) that satisfy x^3+y^3 = (x+y)^2 \\ 6. Solve for all f(x) the following functional equation: f(f(x)+y) = x+y \\ 7. 2(n^{2n}) > (2n)!$
$\noindent 6. Bit rushed since I need to go for a bit but I hope it's right. We have f:\mathbb{R}\to \mathbb{R}, with f(f(x)+y) = x+y, for all x,y\in \mathbb{R}.$

$\noindent Let y=0, then f(f(x)) =x for all real x. Hence f is an involution from \mathbb{R}\to \mathbb{R}. In other words, it is a bijection with f=f^{-1}. Now, applying f=f^{-1} to both sides of f(f(x)+y) = x+y gives f(x) + y = f(x+y). Now, for x,t\in \mathbb{R}, let y\equiv t-x, so x+y = t. Then f(x) + (t-x) = f(t) for all x,t\in \mathbb{R}. So f(x)-f(t)=x-t for all t,x\in \mathbb{R}. Let t=0, then f(x)-f(0) = x \Rightarrow f(x)=x+f(0). Since we have an involution, we must have f(0) = 0. So f(x) = x for all x\in \mathbb{R}. (It is easy to check this does also satisfy the equation.)$

Edit: done above

8. ## Re: Extracurricular Elementary Mathematics Marathon

Lols, mega procrastination on my behalf. Am supposed to be brushing up on something for a supervisor meeting. These questions are fun though.

9. ## Re: Extracurricular Elementary Mathematics Marathon

$\noindent 1. \forall n \in \mathbb{N} , prove \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1} \geq n (\sqrt[n]{2}-1) (AM-GMed by seanieg98)\\ 2. Prove (2n-1)!\geq n^n (inducted by Integrand) \\ 3. Prove that for every integer x, \frac{x^5}{5}+\frac{x^3}{3}+\frac{7x}{15} is also an integer. (modulo-ed by KingOfActing) \\ 4. Prove for all n >1, \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2n-1}{2n}<\frac{1}{\sqrt{3n}} (case bashed by seanieg98)\\ 5. Find all integer pairs (x,y) that satisfy x^3+y^3 = (x+y)^2 (WLOGGED by seanieg98)\\ 6. Solve for all f(x) the following functional equation: f(f(x)+y) = x+y (Solved simultaneously by Integrand and seanieg98)\\ 7. 2(n^{2n}) > (2n)! (AM-GMed by seanieg98)$
And we are done!

someone else post next, I've had my due run for the time being.

10. ## Re: Extracurricular Elementary Mathematics Marathon

1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k?

11. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k?

I'm not sure if this works but:

Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)

12. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by KingOfActing
(1 is not a prime number, right?)
Correct, 1 is not prime.

13. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by KingOfActing
I'm not sure if this works but:

Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)
Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?

14. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?
I think I'm grasping at straws here, but:

P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.

15. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by KingOfActing
I think I'm grasping at straws here, but:

P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.
Bingo , back yourself more!

Will let someone else post one now.

16. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
This isn't English.

Here, we say "Prove"

Anyway, I digress.

$\noindent Do there exist infinitely many f : \mathbb{R} \mapsto \mathbb{R} such that \\f(x+y) = f(x)f(y)f(xy) \, ?$

17. ## Re: Extracurricular Elementary Mathematics Marathon

This isn't English.

Here, we say "Prove"
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.

18. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
You could also say "prove your result" or "prove or disprove"

I see the second one a lot in Putnam questions.

19. ## Re: Extracurricular Elementary Mathematics Marathon

You could also say "prove your result" or "prove or disprove"
I could. I could also say: "Justify your answer by way of proof", or countless variations of this.

I made a choice, and the meaning was unambiguous. You will be frequently frustrated in future studies if such variations in wording bother you.

20. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.

21. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by InteGrand
I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.
Nah he was defs srs lol.

22. ## Re: Extracurricular Elementary Mathematics Marathon

Also, there are only finitely many solutions to that functional equation I believe.

It is clear that if a solution vanishes somewhere it vanishes everywhere (let y be a root to make the RHS vanish and then vary x so the LHS vanishing tells you that f is identically zero).

Let us assume now that f is a nonvanishing solution.

Applying the functional equation tells us:

f(x+y+z)=f(x+(y+z))=f(x)f(y+z)f(xy+xz)=f(x)f(y)f(z )f(yz)f(xy)f(xz)f(x^2yz).

On the other hand, if we group the terms differently, we obtain f(x+y+z)=f(y+(x+z))=f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x y^2z).

Setting z=1 and dividing out the factors (which are nonzero!) tells us f(x^2y)=f(xy^2) for all real x and y.

But for nonzero a,b we can always solve x^2y=a, xy^2=b. (x^3=a^2/b, y^3=b^2/a).

This implies that f must be constant on R \ {0}. Moreover, by setting x = y =/= 0 in the functional equation, we obtain that this constant must satisfy x^3=x, which has only three solutions.

Similarly, setting x=y=0 tells us that the value at the origin must also solve this same cubic.

So we have narrowed down possible solutions to the functional equation to a finite set of functions and we are done.

(It would not take much additional effort to find ALL solutions to the functional equation if one so desired.)

23. ## Re: Extracurricular Elementary Mathematics Marathon

Here's a cute one:

Find all solutions f: R\{2/3}->R to:

504x-f(x)=f(2x/(3x-2))/2.

24. ## Re: Extracurricular Elementary Mathematics Marathon

Originally Posted by seanieg89
Here's a cute one:

Find all solutions f: R->R to:

$504x-f(x)=\frac{f(\frac{2x}{3x-2})}{2}$
$\noindent Observe g(x) = \frac{2x}{3x-2} is an involution. So if f(x) is a solution, then f(\frac{2x}{3x-2}) is a solution. Substituting in x \mapsto \frac{2x}{3x-2}, we have\\ \frac{1008x}{3x-2}-f(\frac{2x}{3x-2}) = \frac{f(x)}{2} \\ Solving simultaneously, we get 1008x(1 - \frac{1}{3x-2}) = \frac{3}{2}f(x) \\ 1008x (\frac{3(x-1)}{3x-2}) = \frac{3}{2}f(x) \\ f(x) = \frac{2016x (x-1)}{3x-2} \\ Substituting in x \mapsto \frac{2x}{3x-2} gives us f(x) = \frac{1008x (2-x)}{3x-2} \\ Substituting again returns us back to where we started. Cycling through the functional input doesn't give us anything different. This verifies all the possible solutions to the functional equation. By substitution, we see both functions satisfy the functional equation. \\ So the only two solutions are:\\f(x) = \frac{1008x (2-x)}{3x-2} \\ f(x) = \frac{2016x (x-1)}{3x-2}$

25. ## Re: Extracurricular Elementary Mathematics Marathon

I do not believe your first solution is correct, neither is the statement that if f(x) is a solution then f(2x/(3x-2)) is one too.

This would be the case if the functional equation was symmetric in these two unknowns, but the factor of 1/2 stops that from happening.

A quick way to verify that this solution doesn't work without plugging into the functional equation directly is by considering the limit L of f(x)/x as x->0. The functional equation implies that L must be 1008 if it exists, whereas the limit in your first proposed solution is -1008.

The second solution is the unique solution and you are done after solving the simultaneous equations. (Because all of the previous lines are quantified over all x, you can actually be sure that this is the unique solution after reaching the line f(x)=blah).

(I obviously cooked up the numbers so that 2016 featured in the unique solution.)

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