For anyone taking Edexcel Statistics 4... Problem 153*/**
Prove that a t-distribution with degree of freedom ν is asymptotically equivalent a the Normal distribution for sample size n such that n→∞, stating the parameters of the Normal distribution to which it is equivalent.
From this we can deduce that sn+1=an+1+sn≥n2+2n+1=(n+1)2.
Therefore, as the interval is always greater than or equal to (n+1)2−n2−1 and the smallest term in the interval is always greater than or equal to n2, it follows trivially that the interval is always larger than or equal to the difference between any two square numbers.
Therefore, the interval [sn,sn+1) must contain a perfect square. □
Edit: Where's this problem from btw? It was rather neat
We denote I(a)=∫0πln(1−2acosx+a2)dx. Clearly, I(a)=I(−a). Furthermore, for a→0, I(a)→0. Now, 2I(a)=I(a)+I(−a)=21∫02πln(1−2a2cos2x+a4)dx=I(a2). Hence, we obtain I(a)=2nI(a2n), implying I(a)=0 for ∣a∣<1. Let ∣a∣>1. So, ln(1−2acosx+a2)=ln(1−2a1cosx+a21)+2ln∣a∣. Therefore, for ∣a∣>1, we have I(a)=2πln∣a∣.
Solution 157
We let lnt↦t, and then t↦−t. Thus obtaining ∫01lnxx−1dx=∫0∞xe−x−e−2xdx=ln2. More generally, ∫01lnxxa−1−xb−1dx=lnab.
Solution 158
For a,b>0, we define I(a,b)=∫0∞b2+x2ln(1+a2x2)dx. I(a,b) is continuous, with respect to a, for a>0. Thus ∂a∂I(a,b)=∫0∞(b2+x2)(1+a2x2)2ax2dx=ab+1π. Thence, we have I(a,b)=bπln(ab+1).
Solution 159
Notice ∫0∞xmsinnxdx=(m−1)!1∫0∞dxm−1dm−1sinnxxdx. We need to treat the cases m≡n(mod2) only. If m≡n≡1(mod2), differentiating sinnx (m−1 times), we get ∫0∞xmsinnxdx=2n(m−1)!(−1)4(n−1)(m−1)π×(nm−1−n(n−2)m−1+⋯). The case m≡n≡0(mod2) is done analogically. Hence, ∫0∞xnsinnxdx=2n(n−1)!π×(nn−1−n(n−2)n−1+⋯).
By the way, Borwein have proved far more general result. Fix n+1 complex numbers z0,⋯,zn. Define a=(a1,⋯,an), ai∈{−1,1}; ba=z0+i∑ziai, and ca=i∏ai. It follows that ∫0∞n=0∏nxsinzixdx=2n+1n!πa∑cabansignba. Hence in our case, when z0=z1=⋯=zn−1=1, we have ∫0∞(xsinx)ndx=2n(n−1)!π0≤i≤2n∑(−1)i(in)(n−2i)n−1.
May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
You used the same method I did for Problem 158 but the rest I did differently. I used differentiation under the integral sign for 156 which pops the solution out almost instantly (yours looks a little complicated!) Really like the elegance in 157 (I simply used the same approach for that one - was fairly simple but surely took a little longer than yours!)
And I don't see what's going on in 159. Where does the first line come from? And how does it link to the second and third? The differentiation thing doesn't seem to make sense (though I'm sure it does), could you clarify it for me My solution looks so different to yours, it's in you form of two symmetrical series expression involving all the square numbers (and no factorials!) (I checked it though and it does work!)
And yeah, as said above, it's far too easy and boring to do it that way
Please post a load of really nice integrals! (None of that crap involving 2013s and stuff where you're never going to come across a similar thing )
Very nice Mlad! I'm too slow again solution 156 ∫0πln(1−2acosx+a2)dx=∫0πln((a−eix)(a−e−ix))dx=∫−ππln(a−eix)dx=∫−ππlnadx+∫−ππln(1−eix/a)dx=2πlna the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic. <insert rigour here> Not sure if this works but I thought I'd give it a go
Very nice Mlad! I'm too slow again solution 156 ∫0πln(1−2acosx+a2)dx=∫0πln((a−eix)(a−e−ix))dx=∫−ππln(a−eix)dx=∫−ππlnadx+∫−ππln(1−eix/a)dx=2πlna the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic. <insert rigour here> Not sure if this works but I thought I'd give it a go
160 is from the Putnam exam (undergrad comp for the USA) and only 20 of the top 200 scorers got more than 2/10 marks on that question! No idea why this method isn't widely taught/used!