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The Proof is Trivial!

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Reply 480
Avatar for Mladenov
Mladenov
1
Original post by shamika
...


Suppose x03−4x0+7=0x_{0}^{3}-4x_{0}+7=0, x0=pqx_{0}=\frac{p}{q}, gcd⁡(p,q)=1\gcd(p,q)=1. So, (pq)3−4pq+7=0⇒p3+7q3=4pq2\displaystyle \left(\frac{p}{q}\right)^{3}-4\frac{p}{q}+7=0 \Rightarrow p^{3}+7q^{3}=4pq^{2}. The last equation gives 7q3≡0(modp)7q^{3} \equiv 0 \pmod p, hence p=±1,±7 p= \pm1 ,\pm7. Next, p3≡0(modq)p^{3} \equiv 0 \pmod q, which implies q=±1q= \pm 1. We have to check that none of ±1,±7 \pm 1, \pm 7 is a solution.
There is a more general result, which is easy to prove. If I am not mistaken, it is called rational root theorem.
Original post by Felix Felicis
I think Boy Wonder's solution should get credited in the OP too :teehee:


But that solution is so ugly... :colonhash:
(edited 11 years ago)
Reply 482
Avatar for shamika
shamika
16
Original post by Mladenov
Suppose x03−4x0+7=0x_{0}^{3}-4x_{0}+7=0, x0=pqx_{0}=\frac{p}{q}, gcd⁡(p,q)=1\gcd(p,q)=1. So, (pq)3−4pq+7=0⇒p3+7q3=4pq2\displaystyle \left(\frac{p}{q}\right)^{3}-4\frac{p}{q}+7=0 \Rightarrow p^{3}+7q^{3}=4pq^{2}. The last equation gives 7q3≡0(modp)7q^{3} \equiv 0 \pmod p, hence p=±1,±7 p= \pm1 ,\pm7. Next, p3≡0(modq)p^{3} \equiv 0 \pmod q, which implies q=±1q= \pm 1. We have to check that none of ±1,±7 \pm 1, \pm 7 is a solution.
There is a more general result, which is easy to prove. If I am not mistaken, it is called rational root theorem.


Yep! I think this is one of those instances where the question would've been easier if they'd ask you to prove the general result, because its easier to see what do to there.

Also thanks, because I was too lazy to latex myself :tongue:
Reply 483
Avatar for und
und
OP
16
Problem 66*

For any fixed number qq, let An=qnn!∫0π[x(π−x)]nsinxdx\displaystyle A_n=\frac{q^n}{n!}\int^{\pi}_{0} \left[ x\left( \pi-x \right) \right]^{n}sin{x}dx.

Show that An=(4n−2)qAn−1−(qπ)2An−2\displaystyle A_n=\left(4n-2\right)qA_{n-1}-\left(q\pi\right)^{2}A_{n-2} and use this result to deduce that AnA_n is an integer for all nn if qq and qπq\pi are integers.

Show also that An→0A_n \to 0 as n→∞n \to \infty. Deduce that π\pi is irrational.
Original post by und
...

Interestingly you seem to have changed problem 60 from * to **/*** :eyeball: I was just looking at that and thinking how am I going to do this with A-level knowledge :colondollar:
Reply 485
Avatar for und
und
OP
16
Original post by Felix Felicis
Interestingly you seem to have changed problem 60 from * to **/*** :eyeball: I was just looking at that and thinking how am I going to do this with A-level knowledge :colondollar:

On further examination I decided that I should be kind to people by not giving them false hope. :tongue:
Original post by und
Problem 66*

For any fixed number qq, let An=qnn!∫0π[x(π−x)]nsinxdx\displaystyle A_n=\frac{q^n}{n!}\int^{\pi}_{0} \left[ x\left( \pi-x \right) \right]^{n}sin{x}dx.

Show that An=(4n−2)qAn−1−(qπ)2An−2\displaystyle A_n=\left(4n-2\right)qA_{n-1}-\left(q\pi\right)^{2}A_{n-2} and use this result to deduce that AnA_n is an integer for all nn if qq and qπq\pi are integers.

Show also that An→0A_n \to 0 as n→∞n \to \infty. Deduce that π\pi is irrational.


Done some introductory example sheets have we?
Original post by Indeterminate
See above - I have a problem with the 'and'.

If q∈Zq \in \mathbb{Z} then qπq\pi is not an integer.


you're using prior knowledge about pi. how do you know pi can't be expressed in the form p/q ?
Original post by ben-smith
you're using prior knowledge about pi. how do you know pi can't be expressed in the form p/q ?


Yes, it was an oversight on my part :smile:
Reply 489
Avatar for joostan
joostan
13
Original post by Star-girl
I haven't seen very many inequality questions around, so I'll throw some in.

Problem 62*

Show that if p>m>0,p>m>0, then
p−mp+m≤x2−2mx+p2x2+2mx+p2≤p+mp−m, ∀x∈R\dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m} ,\ \forall x \in \mathbb{R}


Solution 62
x2−2mx+p2x2+2mx+p2=(x−m)2−m2+p2(x+m)2−m2+p2\dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} = \dfrac{(x-m)^2-m^2+p^2}{(x+m)^2-m^2 + p^2}
This implies that the denominator and numerator are greater than 0 where p>m>0

Assume True.
Expansion and simplification. Of left hand inequality:
 2mx2+2mp2−4pmx≥0[br]⇒ x2−2px+p2≥0[br]⇒ (x−p)2≥0\ 2mx^2 + 2mp^2 - 4pmx \geq 0[br]\Rightarrow \ x^2 - 2px + p^2 \geq 0[br]\Rightarrow \ (x-p)^2 \geq 0
No Contradiction. True for all real x.

Expansion and simplification. Of Right hand inequality:
 2mx2+2mp2+4pmx≥0[br]⇒ x2+2px+p2≥0[br]⇒ (x+p)2≥0\ 2mx^2 + 2mp^2 + 4pmx \geq 0[br]\Rightarrow \ x^2 + 2px + p^2 \geq 0[br]\Rightarrow \ (x+p)^2 \geq 0
No contradiction. True for all real x.
Steps are reversible. Inelegant but functional :smile:
(edited 11 years ago)
Problem 67*

A lense maker wishes to measure the radius of a hemispherical lense by placing a 1 meter ruler on top of the lense and causing it oscillate with a small amplitude. Assuming the ruler is much larger than the radius of the lense, show, using appropriate approximations that:

r=13g(πT)2 \displaystyle r= \frac{1}{3g} \left(\frac{\pi}{T} \right) ^2

T T is the time period.

r r radius.
Original post by joostan
Solution 62
x2−2mx+p2x2+2mx+p2=(x−m)2−m2+p2(x+m)2−m2+p2\dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} = \dfrac{(x-m)^2-m^2+p^2}{(x+m)^2-m^2 + p^2}
This implies that the denominator and numerator are greater than 0 where p>m>0

Assume True.
Expansion and simplification. Of left hand inequality:
 2mx2+2mp2−4pmx≥0[br]⇒ x2−2px+p2≥0[br]⇒ (x−p)2≥0\ 2mx^2 + 2mp^2 - 4pmx \geq 0[br]\Rightarrow \ x^2 - 2px + p^2 \geq 0[br]\Rightarrow \ (x-p)^2 \geq 0
No Conradiction. True for all real x.

Expansion and simplification. Of Right hand inequality:
 2mx2+2mp2+4pmx≥0[br]⇒ x2+2px+p2≥0[br]⇒ (x+p)2≥0\ 2mx^2 + 2mp^2 + 4pmx \geq 0[br]\Rightarrow \ x^2 + 2px + p^2 \geq 0[br]\Rightarrow \ (x+p)^2 \geq 0
No contradiction. True for all real x.
Inelegant but functional :smile:


I see where you're coming from, but what you have actually shown is that if the result is true, then it implies another true result... but this is not actually a proof that the result is true itself because the implication here is the wrong way round.

Hint:

Spoiler

(edited 11 years ago)
Solution 60

∫−∞∞sin⁡(π4x2+1x2) dx\displaystyle\int_{-\infty}^{\infty} \sin \left(\pi^4x^2+\frac{1}{x^2} \right)\,dx

=cos⁡2π2∫−∞∞sin⁡(π2x−1x)2 dx⏟α+sin⁡2π2∫−∞∞cos⁡(π2x−1x)2 dx⏟β\displaystyle = \cos 2\pi^2\underbrace{\int_{-\infty}^{\infty} \sin \left(\pi^2x-\frac{1}{x}\right)^2\,dx}_{ \alpha}+\sin 2\pi^2\underbrace{\int_{-\infty}^{\infty} \cos \left(\pi^2x-\frac{1}{x}\right)^2\,dx}_{\beta}

In general, γ>0:\gamma >0:

∫0∞f(γ2x−1x)2 dx=1γ∫0∞f(γx−γx)2 dx(x→xγ)=1γ[∫01f(γx−γx)2 dx+∫1∞f(γx−γx)2 dx]=1γ[∫1∞1x2f(γx−γx)2 dx+∫1∞f(γx−γx)2 dx](x→1x)=1γ2∫0∞f(t2) dt(t=γx−γx)\begin{aligned}\displaystyle \int_{0}^{\infty}f \left(\gamma^2 x-\frac{1}{x}\right)^2\,dx &= \frac{1}{\gamma}\int_{0}^{\infty}f \left(\gamma x- \frac{\gamma}{x}\right)^2\,dx \quad(\textstyle x\to \frac{x}{\gamma} )\\ &=\frac{1}{\gamma} \left[ \int_{0}^{1} f \left(\gamma x-\frac{\gamma}{x} \right)^2\,dx+ \int_{1}^{\infty} f \left(\gamma x-\frac{\gamma}{x}\right)^2\,dx \right]\\&= \frac{1}{\gamma}\left[\int_{1}^{\infty} \frac{1}{x^2}f \left(\gamma x-\frac{\gamma}{x} \right)^2\,dx+\int_{1}^{\infty} f\left(\gamma x-\frac{\gamma}{x}\right)^2\,dx \right]\quad( x\to\textstyle \frac{1}{x})\\&=\frac{1}{\gamma^2} \int_{0}^{\infty} f (t^2 )\,dt \quad \big(t=\gamma x-\textstyle\frac{\gamma}{x}\big) \end{aligned}

Hence we have:

α=1π2∫−∞∞sin⁡t2dt\displaystyle\alpha =\frac{1}{\pi^2} \int_{-\infty}^{\infty} \sin t^2 dt and β=1π2∫−∞∞cos⁡t2dt\displaystyle\beta =\frac{1}{\pi^2} \int_{-\infty}^{\infty} \cos t^2 dt

Consider:

∫−∞∞cos⁡t2dt−i∫−∞∞sin⁡t2dt=∫−∞∞cos⁡(−t2)dt+i∫−∞∞sin⁡(−t2)dt=∫−∞∞e−it2dt=∫−∞∞e−[(22+i22)t]2dt=π(22+i22)−1=2π2−i2π2\begin{aligned}\displaystyle\int_{-\infty}^{\infty} \cos t^2 dt-i\int_{-\infty}^{\infty} \sin t^2 dt &=\int_{-\infty}^{\infty} \cos(- t^2) dt+i\int_{-\infty}^{\infty} \sin(- t^2) dt\\&=\int_{-\infty}^{\infty} e^{-it^2}dt\\&=\int_{-\infty}^{\infty} e^{-\left[\left(\frac{\sqrt{2}}{2}+i\frac{ \sqrt{2}}{2}\right)t\right]^2}dt\\&=\sqrt{\pi}\left( \frac{ \sqrt{2}}{2}+i\frac{ \sqrt{2}}{2}\right)^{-1}\\&=\frac{\sqrt{2\pi}}{2}-i\frac{\sqrt{2\pi}}{2}\end{aligned}

It follows that α=β=1π2⋅2π2=1π2π\alpha =\beta = \dfrac{1}{\pi^2}\cdot \dfrac{\sqrt{2\pi}}{2}= \dfrac{1}{\pi \sqrt{2\pi}} and hence:

∫−∞∞sin⁡(π4x2+1x2) dx=cos⁡2π2+sin⁡2π2π2π\displaystyle\int_{-\infty}^{\infty} \sin \left(\pi^4x^2+\frac{1}{x^2} \right)\,dx=\frac{\cos 2\pi^2+\sin 2\pi^2}{\pi\sqrt{2 \pi}}


⋆\star

In fact, using similar tricks one can show that p,q>0:p,q>0:

∫−∞∞sin⁡(px2+qx2) dx=π(cos⁡2pq+sin⁡2pq2p)\displaystyle\int_{-\infty}^{\infty} \sin \left(px^2+\frac{q}{x^2} \right)\,dx=\sqrt{\pi}\left( \frac{\cos 2\sqrt{pq}+\sin 2\sqrt{pq}}{\sqrt{2 p}}\right)

So for instance, we have the surprisingly clean result of:

∫−∞∞sin⁡(πx2+πx2) dx=12\displaystyle\int_{-\infty}^{\infty} \sin \left(\pi x^2+\frac{\pi}{x^2} \right)\,dx=\frac{1}{\sqrt{2}}
(edited 11 years ago)
Reply 493
Avatar for Mladenov
Mladenov
1
Solution 23

Let us denote Sn=∑i+j+k=nijk\displaystyle S_{n}= \sum_{i+j+k=n} ijk. Consider, F(x)=∑n=0∞Snxn=(∑n=0∞nxn)3\displaystyle F(x)= \sum_{n=0}^{\infty} S_{n}x^{n}= \left(\sum_{n=0}^{\infty} nx^{n}\right)^{3}.
From ∑n=0∞xn=11−x\displaystyle \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}, we have ∑n=0∞nxn−1=1(1−x)2\displaystyle \sum_{n=0}^{\infty} nx^{n-1}=\frac{1}{(1-x)^{2}}.
Therefore F(x)=(x(1−x)2)3=x3(1−x)6=x3(∑n=0∞(n+55)xn)=∑n=0∞(n+25)xn\displaystyle F(x)= \left(\frac{x}{(1-x)^{2}}\right)^{3}= \frac{x^{3}}{(1-x)^{6}}= x^{3}\left(\sum_{n=0}^{\infty} \dbinom{n+5}{5}x^{n}\right)= \sum_{n=0}^{\infty} \dbinom{n+2}{5}x^{n}.
Hence Sn=(n+25)S_{n}= \dbinom{n+2}{5}.
Double counting, I believe, also works; notwithstanding, I am in a quandary as to what to count to obtain the sum.

Solution 62

P(x)=x2−2mx+p2x2+2mx+p2=1−4mxx2+2mx+p2\displaystyle P(x)= \frac{x^{2}-2mx+p^{2}}{x^{2}+2mx+p^{2}}= 1-\frac{4mx}{x^{2}+2mx+p^{2}}. Further, P′(x)=−4m(p2−x2)(x2+2mx+p2)2\displaystyle P'(x)= -\frac{4m(p^2-x^2)}{(x^{2}+2mx+p^{2})^{2}}, P′′(x)=−8mx3−24mp2x−16m2p2(x2+2mx+p2)3\displaystyle P''(x)= -\frac{8mx^{3}-24mp^{2}x-16m^{2}p^{2}}{(x^{2}+2mx+p^{2})^{3}}. Hence, max⁡x∈RP(x)=P(−p)=p+mp−m≥P(x)≥min⁡x∈RP(x)=P(p)=p−mp+m\displaystyle \max_{x\in \mathbb{R}}P(x)=P(-p)= \frac{p+m}{p-m} \ge P(x) \ge \displaystyle \min_{x\in \mathbb{R}}P(x)=P(p)= \frac{p-m}{p+m}.
Reply 494
Avatar for shamika
shamika
16
Original post by Mladenov
Solution 23

Let us denote Sn=∑i+j+k=nijk\displaystyle S_{n}= \sum_{i+j+k=n} ijk. Consider, F(x)=∑n=0∞Snxn=(∑n=0∞nxn)3\displaystyle F(x)= \sum_{n=0}^{\infty} S_{n}x^{n}= \left(\sum_{n=0}^{\infty} nx^{n}\right)^{3}.
From ∑n=0∞xn=11−x\displaystyle \sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}, we have ∑n=0∞nxn−1=1(1−x)2\displaystyle \sum_{n=0}^{\infty} nx^{n-1}=\frac{1}{(1-x)^{2}}.
Therefore F(x)=(x(1−x)2)3=x3(1−x)6=x3(∑n=0∞(n+55)xn)=∑n=0∞(n+25)xn\displaystyle F(x)= \left(\frac{x}{(1-x)^{2}}\right)^{3}= \frac{x^{3}}{(1-x)^{6}}= x^{3}\left(\sum_{n=0}^{\infty} \dbinom{n+5}{5}x^{n}\right)= \sum_{n=0}^{\infty} \dbinom{n+2}{5}x^{n}.
Hence Sn=(n+25)S_{n}= \dbinom{n+2}{5}.
Double counting, I believe, also works; notwithstanding, I am in a quandary as to what to count to obtain the sum.

Solution 62

P(x)=x2−2mx+p2x2+2mx+p2=1−4mxx2+2mx+p2\displaystyle P(x)= \frac{x^{2}-2mx+p^{2}}{x^{2}+2mx+p^{2}}= 1-\frac{4mx}{x^{2}+2mx+p^{2}}. Further, P′(x)=−4m(p2−x2)(x2+2mx+p2)2\displaystyle P'(x)= -\frac{4m(p^2-x^2)}{(x^{2}+2mx+p^{2})^{2}}, P′′(x)=−8mx3−24mp2x−16m2p2(x2+2mx+p2)3\displaystyle P''(x)= -\frac{8mx^{3}-24mp^{2}x-16m^{2}p^{2}}{(x^{2}+2mx+p^{2})^{3}}. Hence, max⁡x∈RP(x)=P(−p)=p+mp−m≥P(x)≥min⁡x∈RP(x)=P(p)=p−mp+m\displaystyle \max_{x\in \mathbb{R}}P(x)=P(-p)= \frac{p+m}{p-m} \ge P(x) \ge \displaystyle \min_{x\in \mathbb{R}}P(x)=P(p)= \frac{p-m}{p+m}.


Finally someone does it!:biggrin:

This is the way I would do it - there is a simple (but ingenious) way of doing it too,

Spoiler

.

Note that once you think of using a generating function, the problem becomes rather straightforward :smile:
Reply 495
Avatar for Mladenov
Mladenov
1
Original post by shamika
...


Recently, I have started studying analytic number theory, and it occurred to me that I can use generating functions to solve this problem.

Spoiler



Another interesting application of generating functions is the following problem:
Determine whether there exist an infinite set SS, S⊂Z+S\subset \mathbb{Z^{+}} and a positive integer N∈Z+N \in \mathbb{Z^{+}} such that all positive integers which are greater than NN have the same number of representations in the form a+ba+b where a,b∈Sa,b \in S, a≥ba \ge b.
(edited 11 years ago)
Reply 496
Avatar for Mladenov
Mladenov
1
Solution 66

Why, in the world, is [.][.] used here? I spent some time to convince myself that the statement is not true if [.][.] is the well-known arithmetic function.

∫0π(x(π−x))nsin⁡xdx=∫0πn(x(π−x))n−1(π−2x)cos⁡xdx=−∫0πn(n−1)(π−2x)2(x(π−x))n−2sin⁡xdx+∫0π2n(x(π−x))n−1sin⁡xdx=−n(n−1)π2∫0π(x(π−x))n−2sin⁡xdx+(4n2−2n)∫0π(x(π−x))n−1sin⁡xdx\begin{aligned}\displaystyle \int_{0}^{\pi} \left(x(\pi-x)\right)^{n} \sin x dx&= \int_{0}^{\pi} n\left(x(\pi-x)\right)^{n-1}(\pi-2x)\cos x dx\\&=-\int _{0}^{\pi} n(n-1)(\pi-2x)^{2}\left(x(\pi-x)\right)^{n-2}\sin x dx + \int_{0}^{\pi} 2n\left(x(\pi-x)\right)^{n-1}\sin x dx\\& =-n(n-1)\pi^{2}\int_{0}^{\pi} \left(x(\pi-x)\right)^{n-2}\sin x dx + (4n^{2}-2n)\int_{0}^{\pi} \left(x(\pi-x)\right)^{n-1}\sin x dx \end{aligned}.

Hence, An=(4n−2)qAn−1−q2π2An−2\displaystyle A_{n}= (4n-2)qA_{n-1}-q^{2}\pi^{2}A_{n-2} .

Next, suppose that qq and qπq\pi are positive integers(WLOG). I assume n≥0n \ge 0. We have A0=2A_{0}=2 and A1=4qA_{1}=4q. Therefore, it follows, by induction, that AnA_{n} is an integer. Notice that An>0A_{n} >0.

∣(x(π−x))nsin⁡x∣≤π2n4n|(x(\pi-x))^{n} \sin x| \le \frac{\pi^{2n}}{4^{n}}.
So, An≤π(qπ24)nn!=BnA_{n} \le \pi \frac{\left(\frac{q\pi^{2}}{4} \right)^{n}}{n!}= B_{n}.
For nn sufficiently large, Bn>Bn+1B_{n} > B_{n+1}, and lim⁡n→∞Bn=0\displaystyle \lim_{n\to\infty} B_{n}= 0. Hence lim⁡n→∞An=0\displaystyle \lim_{n \to \infty} A_{n} =0.

Consequently, for any ϵ\epsilon there exists mm such that for n>mn >m, ∣An∣<ϵ|A_{n}| <\epsilon. This, together with the inequality An>0A_{n} >0 and the fact that AnA_{n} is an integer, leads to a contradiction.

We conclude that there in no integer qq such that qπ∈Zq\pi \in \mathbb{Z}. Therefore π\pi is irrational.
(edited 11 years ago)
Reply 497
Avatar for james22
james22
16
Original post by Mladenov

For nn sufficiently large, An>An+1A_{n} > A_{n+1}. Hence lim⁡n→∞An=0\displaystyle \lim_{n \to \infty} A_{n} =0.


I don't like this bit, the same could be said about 1+(1/n) but that converges to 1 not 0.
Reply 498
Avatar for Mladenov
Mladenov
1
Original post by james22
....


It should have been Bn>Bn+1B_{n}>B_{n+1}.

As we know, a bounded, monotonically decreasing sequence has limit. Call it BB. In our case, B=lim⁡n→∞Bn−1qπ24n=Blim⁡n→∞qπ24n=0B=\displaystyle \lim_{n \to \infty} B_{n-1} \frac{\frac{q\pi^{2}}{4}}{n}= B \lim_{n \to \infty} \frac{\frac{q\pi^{2}}{4}}{n}= 0, where
Unparseable latex formula:

B_{n}= \pi \frac{\left(\frac{q\pi^{2}}{4} \right)^{n}}{n!}\right)

.
Now, 0<An≤Bn0< A_{n} \le B_{n}. Since lim⁡n→∞Bn=0\displaystyle \lim_{n\to \infty} B_{n} = 0, we have lim⁡n→∞An=0\displaystyle \lim_{n\to \infty} A_{n}= 0.
(edited 11 years ago)
Original post by Mladenov
Solution 66

Why, in the world, is [.][.] used here? I spent some time to convince myself that the statement is not true if [.][.] is the well-known arithmetic function.



Sorry, I don't understand. Are you complaining about the square brackets?

I might post my solution at some point I did it slightly differently.

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