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The Proof is Trivial!

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Solution 223

Clearly, f(2x)+1=∑k≥12pkcos⁡2xβk=2∑k≥1pk2cos⁡2βkxpk≥2(∑k≥1pkcos⁡βkx)2(∑k≥1pk)−1=2f2(x)\begin{aligned} \displaystyle f(2x)+1 = \sum_{k\ge 1} 2p_{k}\cos^{2} x\beta_{k} = 2 \sum_{k \ge 1} \frac{p^{2}_{k}\cos^{2} \beta_{k} x}{p_{k}} \ge 2\left(\sum_{k \ge 1} p_{k}\cos \beta_{k} x \right)^{2}\left(\sum_{k \ge 1} p_{k} \right)^{-1} = 2f^{2}(x) \end{aligned}.
(edited 10 years ago)
Reply 1501
Avatar for Jkn
Jkn
11
Original post by Mladenov
Solution 222

Clearly, f(2x)+1=∑k≥12pkcos⁡2xβk=2∑k≥1pk2cos⁡2βkxpk≥2(∑k≥1pkcos⁡βkx)2(∑k≥1pk)−1=2f2(x)\begin{aligned} \displaystyle f(2x)+1 = \sum_{k\ge 1} 2p_{k}\cos^{2} x\beta_{k} = 2 \sum_{k \ge 1} \frac{p^{2}_{k}\cos^{2} \beta_{k} x}{p_{k}} \ge 2\left(\sum_{k \ge 1} p_{k}\cos \beta_{k} x \right)^{2}\left(\sum_{k \ge 1} p_{k} \right)^{-1} = 2f^{2}(x) \end{aligned}.

****s sake, you're too fast! :lol: I'm wondering if there is a way to do it without Cauchy Schwartz...

Oh and, btw, do you know how the Engel form is derived?

Edit: Btw henpen, I didn't think this was too boring or easy, the result is rather nice! :smile:
(edited 10 years ago)
Original post by Jkn

Spoiler



From C-S, (b1+⋯+bk)(∑i≥1ai2bi)≥(b1a1b1+⋯+bkakbk)2\displaystyle \left(b_{1} + \cdots + b_{k} \right)\left(\sum_{i \ge 1} \frac{a^{2}_{i}}{b_{i}} \right) \ge \left(\sqrt{b_{1}}\frac{a_{1}}{ \sqrt{b_{1}}} + \cdots + \sqrt{b_{k}}\frac{a_{k}}{\sqrt{b_{k}}} \right)^{2}.
Reply 1503
Avatar for Jkn
Jkn
11
Original post by Mladenov
From C-S, (b1+⋯+bk)(∑i≥1ai2bi)≥(b1a1b1+⋯+bkakbk)2\displaystyle \left(b_{1} + \cdots + b_{k} \right)\left(\sum_{i \ge 1} \frac{a^{2}_{i}}{b_{i}} \right) \ge \left(\sqrt{b_{1}}\frac{a_{1}}{ \sqrt{b_{1}}} + \cdots + \sqrt{b_{k}}\frac{a_{k}}{\sqrt{b_{k}}} \right)^{2}.

You have a nice way of making things seem so easy :lol: :biggrin:

I'm such a fan of it now! It trivialises so many things! :awesome:
Reply 1504
Avatar for henpen
henpen
How about the form in Theorem 4.3? Engel form's easy enough to derive, but that form seems very far removed from the vanilla CS.
(edited 10 years ago)
Original post by henpen
How about the form in Theorem 4.3? Engel form's easy enough to derive, but that form seems very far removed from the vanilla CS.


Induction.
Reply 1506
Avatar for Jkn
Jkn
11
Original post by Mladenov

Problem 221**

Find all f:R→Rf: \mathbb{R} \to \mathbb{R} which satisfy f(x2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)f(x^{2})+f(xy)= f(x)f(y)+yf(x)+xf(x+y).

Solution 221

When x=0, f(0)(y+f(y)−2)=0\displaystyle f(0)(y+f(y)-2)=0.

Hence, for f(0)≠0f(0) \not= 0, f(x)=2−xf(x)=2-x

For f(0)=0f(0)=0, when y=0, f(x2)=xf(x)\displaystyle f(x^2)=xf(x). Letting x→−xx \to -x and equating the two equations gives
Unparseable latex formula:

\displystyle xf(x)=-xf(-x) \Rightarrow f(x)=-f(-x)

which we know also works when x=0, by assumption.

Letting y=-x gives f(x2)+f(−x2)=f(x)f(−x)−xf(x)+xf(0)\displaystyle f(x^2)+f(-x^2)=f(x)f(-x)-xf(x)+xf(0). In the case where f(0)=0, and using the equations just derived, this reduces to f(x)f(−x)=xf(x)⇒f(−x)=x⇒f(x)=−x\displaystyle f(x)f(-x)=xf(x) \Rightarrow f(-x)=x \Rightarrow f(x)=-x which also works for f(x)=0.

∴f(x)=2−x\displaystyle \therefore f(x)=2-x, f(x)=0f(x)=0 and f(x)=−x\displaystyle f(x)=-x are the only functions that satisfy the given functional equation. □\square

(maybe I should start doing functional equations questions. I've always avoided them for some reason...)
(edited 10 years ago)
I'm waiting for the mechanics :tongue:
Reply 1508
Avatar for Jkn
Jkn
11
Original post by bananarama2
I'm waiting for the mechanics :tongue:

Mechanics? Can't give you something that easy... here's a problem that takes place in a non-mathematical world:

Problem 223*

A spherical ball is dropped from the Eiffel Tower from rest. Give a convincing argument as to why Olber's Paradox gives strong evidence to support the idea that the observable universe has existed for a finite period of time. Note that, for such an argument to be convincing, it must rely solely on assumptions that are supported by strong evidence. In this question you are required to quote primary data sources to support all assumptions made.
(edited 10 years ago)
Solution 221

a,b,c>0:  x=a+b,  y=b+c,  z=a+ca,b,c>0:\;x=a+b,\;y=b+c,\; z=a+c

The inequality reduces to:

∑a,b,c8a4b(a+b)≥∑a,b,ca2+3ab\displaystyle \sum_{a,b,c}\frac{8a^4}{b(a+b)} \geq \sum_{a,b,c} a^2+3ab

This is easily shown using CS:

8a4b(a+b)+8b4c(b+c)+8c4a(c+a)≥8(a2+b2+c2)2a2+b2+c2+ab+ac+bc≥8(a2+b2+c2)22(a2+b2+c2)=4(a2+b2+c2)≥a2+b2+c2+3(ab+bc+ac)\displaystyle\begin{aligned} \frac{8a^4}{b(a+b)}+\frac{8b^4}{c(b+c)}+\frac{8c^4}{a(c+a)} &\geq \frac{8(a^2+b^2+c^2)^2}{a^2+b^2+c^2+ab+ac+bc}\\& \geq\frac{8(a^2+b^2+c^2)^2}{2(a^2+b^2+c^2)}\\ &= 4(a^2+b^2+c^2)\\ & \geq a^2+b^2+c^2+3(ab+bc+ac)\end{aligned}
(edited 10 years ago)
Original post by Jkn


Letting y=-x gives f(x2)+f(−x2)=f(x)f(−x)−xf(x)+xf(0)\displaystyle f(x^2)+f(-x^2)=f(x)f(-x)-xf(x)+xf(0). In the case where f(0)=0, and using the equations just derived, this reduces to f(x)f(−x)=xf(−x)⇒f(x)=−x\displaystyle f(x)f(-x)=xf(-x) \Rightarrow f(x)=-x which also works when f(x)=0, by assumption.


I do not understand this. From f(x)(f(−x)−x)=0f(x)(f(-x)-x)=0 it does not follow that f(−x)=xf(-x)=x for all xx or f(x)=0f(x)=0 for all xx. Can I not take ff to be zero over the rationals and −x-x over the irrationals?
Original post by Jkn
...


I've decided to give you a taste of what I do :tongue:

Problem 224*/**:

Whenever we say "function" we mean total function.

Denote for every n∈N0n \in \mathbb N _0 the set n‾={m∈N0∣0<m≤n}\underline n = \{ m \in \mathbb N _0 | 0 < m \leq n\}.

The mystery concept of a strange product m‾⊗n‾\underline m \otimes \underline n for two such sets m‾\underline m and n‾\underline n satisfies the following property:


There exist functions π1:m‾⊗n‾→m‾\pi_1 : \underline m \otimes \underline n \to \underline m and π2:m‾⊗n‾→n‾\pi_2 : \underline m \otimes \underline n \to \underline n such that:

For each k∈N0k \in \mathbb N_0 and each pair of functions f:k‾→m‾f: \underline k \to \underline m and g:k‾→n‾g: \underline k \to \underline n there exists a unique function h:k‾→m‾⊗n‾h : \underline k \to \underline m \otimes \underline n such that for all x∈k‾x\in \underline k we have f(x)=π1(h(x))f(x) = \pi_1(h(x)) and g(x)=π2(h(x))g(x) = \pi_2(h(x))


Show that regardless of what the exact definition for ⊗\otimes is, if it satisfies the above property then for all sets XX and (non-zero) natural nn, ∣X⊗n‾∣=n|X \otimes \underline n| = n iff ∣X∣=1|X| = 1 (in other words X=1‾X = \underline 1).

EDIT1: made it clearer, but question is still the same.

EDIT2: Hmmm I just realised that the elegant solution is far beyond *-level, and the inelegant one may be quite long. :colonhash:

EDIT3: Actually for the above reason I'm going to withdraw it as a formal problem for this thread, but it's open to anyone who wants to try it. :tongue:
(edited 10 years ago)
Original post by ukdragon37
]

EDIT1: made it clearer, but question is still the same.


Yeh. You just elucidated the whole problem :colone:
Original post by ukdragon37
I've decided to give you a taste of what I do :tongue:


EDIT1: made it clearer, but question is still the same.

EDIT2: Hmmm I just realised that the elegant solution is far beyond *-level, and the inelegant one may be quite long. :colonhash:

EDIT3: Actually for the above reason I'm going to withdraw it as a formal problem for this thread, but it's open to anyone who wants to try it. :tongue:

I must say I was shamelessly looking up most of the notation, and was wondering how I was meant to solve it :lol:
Original post by bananarama2
Yeh. You just elucidated the whole problem :colone:


It's a royal pain writing out the question with a concrete example and in words. Normally you would express the property as a diagram:

Original post by joostan
I must say I was shamelessly looking up most of the notation, and was wondering how I was meant to solve it :lol:


Nah don't worry about it. Once again I posed a problem that requires too much definitions to understand. :colonhash:
Original post by ukdragon37
It's a royal pain writing out the question with a concrete example and in words. Normally you would express the property as a diagram:



My mind just can't plough through that question :tongue: I'm sure someone else will have a go. I think I'll choose Matsci. over Compsci. :tongue:
(edited 10 years ago)
Original post by ukdragon37
Nah don't worry about it. Once again I posed a problem that requires too much definitions to understand. :colonhash:


Haha, it looks pretty :tongue:
Original post by joostan
Haha, it looks pretty :tongue:


My thoughts exactly for most of the solutions...
Original post by ukdragon37
...


So how many kids do you have? :tongue:

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