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The Proof is Trivial!

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Original post by Felix Felicis
Pardon me if I'm wrong, but should the second term of your second bracket be 122x2\frac{1}{2^2} x^2?

Problem 353*/**
For a positive integer nn, let SnS_n be the total sum of the intervals of xx such that sin⁡4nx≥sin⁡x\sin 4nx \geq \sin x in 0≤x≤π20\leq x\leq \frac{\pi}{2}.

Find lim⁡n→∞Sn\displaystyle \lim_{n\to\infty} S_n.


Eagle eyed. Edited.

Solution 353

(edited 10 years ago)
Original post by Pterodactyl


Problem 352 */**
∑n=0∞(−1)n3n+1\displaystyle \sum^\infty_{n=0} \dfrac{\left(-1\right)^n}{3n+1}


Solution 352

Using the same method as LOTF's very pretty solution.

∑n=0∞(−1)n3n+1=∑n=0∞(−1)n∫01x3n dx= ∫01∑n=0∞(−1)nx3n dx=∫0111+x3 dx= 13(ln⁡2+33π) \displaystyle \begin{aligned} \sum^\infty_{n=0} \dfrac{\left(-1\right)^n}{3n+1} = \sum^\infty_{n=0} (-1)^n \int_0^{1} x^{3n} \, dx = \ & \int_0^{1} \sum^\infty_{n=0} (-1)^n x^{3n} \,dx = & \int_0^{1} \frac{1}{1+x^3} \,dx = \ & \frac{1}{3} \left( \ln 2 + \frac{\sqrt{3}}{3} \pi \right) \end{aligned}

I did the final integration by noting that x3+1=(x+1)(x2−x+1)\displaystyle x^3+1=(x+1)(x^2-x+1) then splitting using partial fractions.
(edited 10 years ago)
Reply 2282
Avatar for Jkn
Jkn
11
Original post by Pterodactyl
Problem 351 **
∫0∞(x−x32+x52⋅4−x72⋅4⋅6+...)(1+x222+x422⋅42+x622⋅42⋅62+...) dx\displaystyle \int^\infty_0 \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot4}-\frac{x^7}{2\cdot4\cdot6}+... \right)\left(1+\frac{x^2}{2^2}+ \frac{x^4}{2^2\cdot4^2}+\frac{x^6}{2^2\cdot4^2\cdot6^2}+... \right)\ dx

Sorry if this is wrong and/or inelegant - it's been a while!

Solution 351

Using sigma notation, putting the nth terms in to more familiar forms and then letting x22↦x\frac{x^2}{2} \mapsto x, we get:

∫0∞x∑m=0∞(−x2/2)mm!∑n=0∞(x2/4)n(n!)2 dx=∫0∞xe−x22∑n=0∞(x2/4)n(n!)2 dx=∑n=0∞12n(n!)2∫0∞e−xxn dx\displaystyle \begin{aligned} \int_0^{\infty} x \sum_{m=0}^{\infty} \frac{(-x^2/2)^m}{m!} \sum_{n=0}^{\infty} \frac{(x^2/4)^n}{(n!)^2} \ dx = \int_0^{\infty} x e^{-\frac{x^2}{2}} \sum_{n=0}^{\infty} \frac{(x^2/4)^n}{(n!)^2} \ dx = \sum_{n=0}^{\infty} \frac{1}{2^n (n!)^2} \int_0^{\infty} e^{-x} x^n \ dx \end{aligned}

We may evaluate the integral easily using the method of reduction. Alternatively the integral can be obtained instantly through using Ramanujan's master theorem (where ϕ(k)=1\phi(k)=1). This gives:

∑n=0∞Γ(n+1)2n(n!)2=∑n=0∞(1/2)nn!=e □\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma (n+1)}{2^n (n!)^2} = \sum_{n=0}^{\infty} \frac{(1/2)^n}{n!} = \sqrt{e} \ \square

Edit: Just realised this problem is the same as Problem 105 (though my solution is, dare I say, simpler).
(edited 10 years ago)
Original post by james22
This is harder than I first thought. I have now failed with 3 methods, all failing because the reccurance relation has an odd number of terms (I think they would work with all ones with an even number of terms), the sums just don't all cancel as 1 is always left out.

I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention.

Did you try using a 3x3 vector?
Reply 2284
Avatar for Jkn
Jkn
11
Original post by james22

I will use the identity

∑k=1nFk2=FnFn+1\displaystyle\sum_{k=1}^n F_k^2=F_nF_{n+1}

There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.

Gonna have to disagree with you there! Not only does that require having the explicit formula in the first place, but it takes a while. Consider the area of the Fibonacci rectangle (it has sides of length FnF_n and Fn+1F_{n+1} and yet it made up of the squares of the series of Fibonacci numbers) - beautiful! Alternatively a proof by induction follows immediately from Cassini's identity.

I've started wrestling with that tribonacci question. I agree that the obvious yet ugly technique can yield a solution pretty quickly, though obviously that isn't very satisfying. I've been trying to find some sort of geometric method (i.e. of arranging the squares) but as of yet am having no luck. Have you been trying to do it geometrically or do you think that's a lost cause?
Here is one way to do problem 350.

Solution 350

I shall use the following identities which are easily proved by induction (if time permits, I can add their proofs).

We have

∑0nTi=12(Tn+2+Tn−1)\displaystyle \sum_{0}^{n} T_{i} = \frac{1}{2}(T_{n+2}+T_{n}-1);
T2n=Tn−12+Tn(Tn+1+Tn−1+Tn−2)\displaystyle T_{2n}=T_{n-1}^{2}+T_{n}(T_{n+1}+T_{n-1}+T_{n-2});
T2n−1=Tn2+Tn−12+2Tn−1Tn−2\displaystyle T_{2n-1}=T_{n}^{2}+T_{n-1}^{2}+2T_{n-1}T_{n-2};
T2n−T2n−1=T2n−2+T2n−3\displaystyle T_{2n}-T_{2n-1}=T_{2n-2}+T_{2n-3}.

Now it follows that 12(T2n+T2n−2−1)=−2S+2TnTn+1+2TnTn−1\displaystyle \frac{1}{2}(T_{2n}+T_{2n-2}-1)=-2S+2T_{n}T_{n+1}+2T_{n}T_{n-1} which is 4S=4TnTn+1+4TnTn−1+1−T2n−T2n−2=4TnTn+1−(Tn+1−Tn−1)2+14S=4T_{n}T_{n+1} + 4T_{n}T_{n-1}+1-T_{2n}-T_{2n-2}=4T_{n}T_{n+1}-(T_{n+1}-T_{n-1})^{2}+1, where SS is the sum in question.

Spoiler



Problem 355*

Let SS be a set of positive integers. We say that SS has the property PP if no element of SS is a multiple of another. Find the largest subset SS of {1,⋯ ,2n−1}\{1,\cdots,2n-1\} having the property PP.

Problem 356*

In a town there are 2n2n people and mm clubs - A1,⋯ ,AmA_{1},\cdots, A_{m}. If ∣Ai∣|A_{i}| and ∣Ai∩Aj∣|A_{i} \cap A_{j}| are even for all i,j∈{1,⋯ ,m}i,j \in \{1,\cdots,m\}, then m≤2nm \le 2^{n}.
(edited 10 years ago)
Original post by Felix Felicis
Pardon me if I'm wrong, but should the second term of your second bracket be 122x2\frac{1}{2^2} x^2?

Problem 353*/**
For a positive integer nn, let SnS_n be the total sum of the intervals of xx such that sin⁡4nx≥sin⁡x\sin 4nx \geq \sin x in 0≤x≤π20\leq x\leq \frac{\pi}{2}.

Find lim⁡n→∞Sn\displaystyle \lim_{n\to\infty} S_n.

Problem 354*/**

Evaluate lim⁡n→∞(∫0πsin⁡2nxsin⁡x dx−∑k=1n1k)\displaystyle\lim_{n\to\infty} \left( \int_0^{\pi}\frac{\sin^2 nx}{\sin x} \ dx - \sum_{k=1}^{n}\frac{1}{k}\right).


Solution 354:

[spoiler]

First of all, let:

Unparseable latex formula:

F(n) = \displaystyle\int_0^{\pi}\frac{ \sin^2 nx}{\sin x} \ dx - \displaystyle\sum_{k=1}^{n}\frac{1}{k}\right)



Consider F(n+2)−F(n) F(n+2) - F(n) :

=∫0πsin2(n+2)x−sin2(nx)sinx dx−1n+1−1n+2 = \displaystyle\int_0^{\pi} \dfrac{sin^2(n+2)x-sin^2(nx)}{sinx} \ dx - \frac{1}{n+1}-\frac{1}{n+2}

Now, just work with the integral:

=∫0πsin2(n+2)x−sin2(nx)sinx dx= \displaystyle\int_0^{\pi} \dfrac{sin^2(n+2)x-sin^2(nx)}{sinx} \ dx

=∫0π(sin(n+2)x−sin(nx))(sin(n+2)x+sin(nx))sinx dx = \displaystyle\int_0^{\pi} \dfrac{(sin(n+2)x-sin(nx))(sin(n+2)x+sin(nx))}{sinx} \ dx

=∫0π4sinxcosxcos(n+1)xsin(n+1)xsinx dx = \displaystyle\int_0^{\pi} \dfrac{4sinxcosxcos(n+1)xsin(n+1)x}{sinx} \ dx

=∫0π2sin(2(n+1)x)cosx dx = \displaystyle\int_0^{\pi} 2sin(2(n+1)x)cosx \ dx

=∫0πsin(2n+3)x+sin(2n+1)x dx = \displaystyle\int_0^{\pi} sin(2n+3)x+sin(2n+1)x \ dx

So:

F(n+2)−F(n)=2(12n+3+12n+1)−1n+1−1n+2 F(n+2)-F(n) = 2(\dfrac{1}{2n+3}+\dfrac{1}{2n+1}) - \frac{1}{n+1}-\frac{1}{n+2}

Now, we can take the limit as n tends to infinity by summing this to infinity:

2F−F(2)−F(1)=1+2∑n=1∞12n+1−2∑n=2∞1n[br][br][tex]F(1)=∫0πsinx dx−1=1[/tex][br][br][tex]F(2)=∫0πsin2(2x)sinx dx−32[/tex][br][br]Summingfromn=1toinfinity,weget:[br][br][tex]2F−F(2)−F(1)=2∑n=1∞12n+1+12n+3−∑n=1∞1n+1+1n+2[/tex][br][br]Frepresentsthelimit,andwehave2FasaftersummingwehaveF(n+2)andF(n+1)remaining,whichareequalinthelimit.Wecanre−arrangethissumintotheform:[br][br][tex]4∑n=1∞12n−1−12n−136[/tex][br][br]WorkingoutF(1)(1)andF(2)(7/6)andaddingtheseonanddividingbytwo,ourfinallimitisln4.[br][br] 2F - F(2) - F(1) = 1 + 2\displaystyle\sum_{n=1}^{\infty} \frac{1}{2n+1} -2\displaystyle\sum_{n=2}^{\infty} \frac{1}{n}[br][br][tex] F(1) = \int_0^{\pi} sinx \ dx - 1 = 1 [/tex][br][br][tex] F(2) = \int_0^{\pi} \dfrac{sin^2(2x)}{sinx} \ dx - \frac{3}{2} [/tex][br][br]Summing from n=1 to infinity, we get:[br][br][tex] 2F - F(2) - F(1) = 2 \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{2n+1}+\dfrac{1}{2n+3} - \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n+1} + \dfrac{1}{n+2} [/tex][br][br]F represents the limit, and we have 2F as after summing we have F(n+2) and F(n+1) remaining, which are equal in the limit. We can re-arrange this sum into the form:[br][br][tex] 4 \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{2n-1}-\dfrac{1}{2n} - \frac{13}{6} [/tex][br][br]Working out F(1) (1) and F(2) (7/6) and adding these on and dividing by two, our final limit is ln4.[br][br]
(edited 10 years ago)
Original post by DJMayes
Solution 354:

Spoiler


Sorry I have to go soon so I don't have time to pinpoint an error - your limit is incorrect but you're along the right lines. The limit should be:

Spoiler

Original post by Felix Felicis
Sorry I have to go soon so I don't have time to pinpoint an error - your limit is incorrect but you're along the right lines. The limit should be:

Spoiler



Are you absolutely certain of this? I did make a few slips in the evaluation but my final limit is log2.
Original post by DJMayes
Are you absolutely certain of this? I did make a few slips in the evaluation but my final limit is log2.

Yes.

Original post by DJMayes

Spoiler


Here's your error.
Original post by Felix Felicis
Yes.


Here's your error.


I have now redone that part of the summation and obtained ln4 as a final answer. However, I'm still not entirely convinced that I made an error; I think infinity (and the fact that the harmonic series is not absolutely convergent) is screwing with me.

(See here: http://en.wikipedia.org/wiki/Riemann_series_theorem )
Original post by DJMayes
I have now redone that part of the summation and obtained ln4 as a final answer. However, I'm still not entirely convinced that I made an error; I think infinity (and the fact that the harmonic series is not absolutely convergent) is screwing with me.

(See here: http://en.wikipedia.org/wiki/Riemann_series_theorem )

Doh, you're right. :holmes: When I looked at that last bit of your initial solution, I just rearranged it in a way to get ln⁡4\ln{4} but after looking at it more closely, I see how you got ln⁡2\ln{2}, it wasn't an error on your part, my apologies.
Original post by Felix Felicis
Doh, you're right. :holmes: When I looked at that last bit of your initial solution, I just rearranged it in a way to get ln⁡4\ln{4} but after looking at it more closely, I see how you got ln⁡2\ln{2}, it wasn't an error on your part, my apologies.


Don't worry about it; this was an interesting excursion into one particular paradox of infinity. However I hope there is a way to evaluate this limit without summing infinite series like this as otherwise there is no way to tell which result is the correct limit, or indeed if there is a limit at all. :lol:

(Or what would be even more interesting is if the limit being the sum of an infinite series that is not absolutely convergent implied that a limit did not exist. However this is entirely speculative and probably more than a bit above my current pay grade. :tongue: )
Original post by DJMayes
Don't worry about it; this was an interesting excursion into one particular paradox of infinity. However I hope there is a way to evaluate this limit without summing infinite series like this as otherwise there is no way to tell which result is the correct limit, or indeed if there is a limit at all. :lol:

(Or what would be even more interesting is if the limit being the sum of an infinite series that is not absolutely convergent implied that a limit did not exist. However this is entirely speculative and probably more than a bit above my current pay grade. :tongue: )

Indeed, well my method was more or less the same as your's (I did F(n+1) - F(n) instead) so I've got nada as an alternative to the infinite series. :biggrin: I'll try and hunt the question down again (I found it on AoPS so there'll likely be a solution floating around somewhere, maybe one that doesn't involve the infinite series). Or maybe wait until someone who knows more about analysis shows up. :biggrin:

Sorry, I don't get what you mean on your last point, do you mean that if a limit ℓ=lim⁡n→∞∑i=1nai\ell = \lim_{n\to\infty} \sum_{i=1}^n a_i and ∑i=1∞ai\sum_{i=1}^{\infty} a_i is not absolutely convergent then the limit ℓ\ell does not exist? Well ∑i=1∞(−1)i+11/i\sum_{i=1}^{\infty} (-1)^{i+1} 1/i isn't absolutely convergent but the limit still exists. :tongue:
(edited 10 years ago)
Original post by Felix Felicis
Indeed, well my method was more or less the same as your's (I did F(n+1) - F(n) instead) so I've got nada as an alternative to the infinite series. :biggrin: I'll try and hunt the question down again (I found it on AoPS so there'll likely be a solution floating around somewhere, maybe one that doesn't involve the infinite series). Or maybe wait until someone who knows more about analysis shows up. :biggrin:

Sorry, I don't get what you mean on your last point, do you mean that if a limit ℓ=lim⁡n→∞∑i=1nai\ell = \lim_{n\to\infty} \sum_{i=1}^n a_i and ∑i=1∞ai\sum_{i=1}^{\infty} a_i is not absolutely convergent then the limit ℓ\ell does not exist? Well ∑i=1∞(−1)i+11/i\sum_{i=1}^{\infty} (-1)^{i+1} 1/i isn't absolutely convergent but the limit still exists. :tongue:


The last point was completely baseless speculation - if your limit can take any value, do you really have a limit? I do not know and my knowledge of infinite series isn't enough to give any kind of an answer. Either way I am not convinced our method is fine until we've justified which way we sum the series.
Original post by DJMayes
The last point was completely baseless speculation - if your limit can take any value, do you really have a limit? I do not know and my knowledge of infinite series isn't enough to give any kind of an answer. Either way I am not convinced our method is fine until we've justified which way we sum the series.

Neither is mine tbh. :lol: I've found this which may be of interest. Also, if we rewrite the series at the end as ∑k=1∞1k(2k−1)\sum_{k=1}^{\infty} \frac{1}{k(2k-1)}, then this may also be of interest (they even touch on the issue of cancelling terms on Ihf's solution).
(edited 10 years ago)
Original post by Jkn
Gonna have to disagree with you there! Not only does that require having the explicit formula in the first place, but it takes a while. Consider the area of the Fibonacci rectangle (it has sides of length FnF_n and Fn+1F_{n+1} and yet it made up of the squares of the series of Fibonacci numbers) - beautiful! Alternatively a proof by induction follows immediately from Cassini's identity.

I've started wrestling with that tribonacci question. I agree that the obvious yet ugly technique can yield a solution pretty quickly, though obviously that isn't very satisfying. I've been trying to find some sort of geometric method (i.e. of arranging the squares) but as of yet am having no luck. Have you been trying to do it geometrically or do you think that's a lost cause?


When I said straight forawrd, I meant it was the method that would be garanteed to work and requires the least mathematical ability or knowledge. It's like saying that the most straight forward way of adding up the first million integers is to do each sum. It's a stupid way to do it but it is a sure way to get the answer.

I never even considered a geometric method for the tribonacci (I didn't even think about the geometric properties of teh fibonacci numbers either).
Problem 357:**

Evaluate ∫01xlnx1+x2 dx \displaystyle\int_{0}^{1} \dfrac{xlnx}{1+x^2} \ dx
Solution 357

∫01xln⁡x1+x2 dx=∫01∑r≥0(−1)rx2r+1ln⁡x dx=∑r≥0∫01(−1)rx2r+1ln⁡x dx=∑r≥0((−1)rx2r+2ln⁡x2r+2∣01−12r+2∫01(−1)rx2r+1 dx)=−∑r≥0(−1)rx2r+2(2r+2)2∣01=−14∑r≥0(−1)r(r+1)2=−14(∑r≥11r2⏟=π2/6−2∑r≥11(2r)2⏟=π2/24)=−14⋅π212=−π248\displaystyle \begin{aligned} \int_0^1 \frac{x\ln x}{1+x^2} \ dx = \int_0^1 \sum_{r\geq 0} (-1)^r x^{2r+1} \ln x \ dx & = \sum_{r\geq 0} \int_0^1 (-1)^r x^{2r+1} \ln x \ dx \\ & = \sum_{r\geq 0} \left(\frac{(-1)^r x^{2r+2} \ln x }{2r+2}\bigg|_0^1 - \frac{1}{2r+2} \int_0^1 (-1)^r x^{2r+1} \ dx \right) \\ & = - \sum_{r\geq 0} \frac{(-1)^r x^{2r+2}}{(2r+2)^2} \bigg|_0^1 \\ & = - \frac{1}{4}\sum_{r\geq 0} \frac{(-1)^r}{(r+1)^2} \\ & = -\frac{1}{4} \left(\underbrace{\sum_{r\geq 1} \frac{1}{r^2}}_{=\pi^2 /6} - 2 \underbrace{\sum_{r\geq 1} \frac{1}{(2r)^2}}_{=\pi^2 /24}\right) \\ & = -\frac{1}{4} \cdot \frac{\pi^2}{12} \\ & = -\frac{\pi^2}{48} \end{aligned}
(edited 10 years ago)
Reply 2299
Avatar for MW24595
MW24595
Original post by Mladenov
Problem 355*

Let SS be a set of positive integers. We say that SS has the property PP if no element of SS is a multiple of another. Find the largest subset SS of {1,⋯ ,2n−1}\{1,\cdots,2n-1\} having the property PP.x]


Solution 355

The subset (x∣n≤x≤2n−1) \left( x| n \le x \le 2n-1 \right) . By the pigeonhole principle, any larger subset will have the property PP.
(edited 10 years ago)

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