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The Proof is Trivial!

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Reply 1020
Just came up with a really nice result (can I blag it as a theorem? :lol:)

Problem 152**

Given that i=1nxi=S\displaystyle\sum_{i=1}^n x_i=S and max(i=1nxi)=P\displaystyle max(\prod_{i=1}^n x_i) =P for real numbers xix_i such that xi>0 ix_i >0 \ \forall i, find a condition on S for which P is irrational.

Spoiler

(edited 10 years ago)
Reply 1021
For anyone taking Edexcel Statistics 4...

Problem 153
*/**

Prove that a t-distribution with degree of freedom ν\nu is asymptotically equivalent a the Normal distribution for sample size nn such that nn \to \infty, stating the parameters of the Normal distribution to which it is equivalent.

Hint:

Spoiler

(edited 10 years ago)
Problem 154 / *

Let a1<a2<a3<...a_1 < a_2 < a_3 < ... be positive, non-consecutive integers. Let

sm=a1+a2+a3+...+ams_m = a_1 + a_2 + a_3 + ... + a_m for mNm \in \mathbb{N}.

Prove that, for all nNn \in \mathbb{N}, the interval [sn,sn+1)[s_n, s_{n+1}) contains a perfect square.
Reply 1023
I did some really nice problems today and thought I would share them in an attempt to get people interested in this thread again:

Problem 155*

Find (without calculus) a fifth degree polynomial p(x)p(x) such that p(x)+1p(x)+1 is divisible by (x1)3(x-1)^3 and p(x)1p(x)-1 is divisible by (x+1)3(x+1)^3.

Spoiler



Problem 156*/**/***

Evaluate 0πln(12acos(x)+a2)dx\displaystyle\int_0^{\pi} \ln(1-2a\cos(x)+a^2) dx, where a is a real number, in the case such that a>1|a|>1.

Problem 157*/**/***

Evaluate 01x1ln(x)dx\displaystyle\int_0^1 \frac{x-1}{\ln(x)} dx

Problem 158*/**/***

Evaluate 0ln(1+x2)1+x2dx\displaystyle\int_0^{\infty} \frac{\ln(1+x^2)}{1+x^2} dx

and finally, because there's no point in me simply typing my solution up (especially since no-one seems too interested anyway)...

Problem 159**

Evaluate (sin(x)x)ndx\displaystyle\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x} \right)^n dx, where n is a positive integer, in terms of a combination of a finite number of elementary functions.
(edited 10 years ago)
Reply 1024
Original post by jack.hadamard
Problem 154 / *

Let a1<a2<a3<...a_1 < a_2 < a_3 < ... be positive, non-consecutive integers. Let

sm=a1+a2+a3+...+ams_m = a_1 + a_2 + a_3 + ... + a_m for mNm \in \mathbb{N}.

Prove that, for all nNn \in \mathbb{N}, the interval [sn,sn+1)[s_n, s_{n+1}) contains a perfect square.

Solution 154

As they are non-consecutive integers, the difference between to consecutive terms in at least 2.

For the size of the interval, sn+1sn=an+12(n+1)1=2n+1s_{n+1}-s_{n}=a_{n+1} \ge 2(n+1)-1=2n+1

Similarly, sn(2(n)1)+(2(n1)1)+(2(n2)1)+...+(2(1)1)=2(1+2+3+...+n)n=n(n+1)n=n2s_{n} \ge (2(n)-1)+(2(n-1)-1)+(2(n-2)-1)+...+(2(1)-1)=2(1+2+3+...+n)-n=n(n+1)-n=n^2.

From this we can deduce that sn+1=an+1+snn2+2n+1=(n+1)2s_{n+1}=a_{n+1}+s_{n} \ge n^2+2n+1=(n+1)^2.

Therefore, as the interval is always greater than or equal to (n+1)2n21(n+1)^2-n^2-1 and the smallest term in the interval is always greater than or equal to n2n^2, 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)[s_n,s_{n+1}) must contain a perfect square. \square

Edit: Where's this problem from btw? It was rather neat :smile:
(edited 10 years ago)
Solution 156

We denote I(a)=0πln(12acosx+a2)dx\displaystyle I(a) = \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx. Clearly, I(a)=I(a)I(a)=I(-a). Furthermore, for a0 a \to 0, I(a)0I(a) \to 0. Now, 2I(a)=I(a)+I(a)=1202πln(12a2cos2x+a4)dx=I(a2)\displaystyle 2I(a)= I(a)+I(-a) = \frac{1}{2} \int_{0}^{2\pi} \ln(1-2a^{2} \cos 2x+a^{4})dx = I(a^2). Hence, we obtain I(a)=I(a2n)2n\displaystyle I(a)= \frac{I(a^{2^{n}})}{2^{n}}, implying I(a)=0I(a)=0 for a<1|a| <1.
Let a>1|a|>1. So, ln(12acosx+a2)=ln(121acosx+1a2)+2lna\displaystyle \ln(1-2a \cos x + a^{2}) = \ln(1 - 2\frac{1}{a} \cos x + \frac{1}{a^{2}}) + 2\ln|a|. Therefore, for a>1|a|>1, we have I(a)=2πlnaI(a) = 2\pi\ln|a|.

Solution 157

We let lntt\ln t \mapsto t, and then ttt \mapsto -t. Thus obtaining 01x1lnxdx=0exe2xxdx=ln2\displaystyle \int_{0}^{1} \frac{x-1}{\ln x}dx= \int_{0}^{\infty} \frac{e^{-x}-e^{-2x}}{x}dx = \ln2. More generally, 01xa1xb1lnxdx=lnba\displaystyle \int_{0}^{1} \frac{x^{a-1}-x^{b-1}}{\ln x}dx = \ln\frac{b}{a}.

Solution 158

For a,b>0a,b >0, we define I(a,b)=0ln(1+a2x2)b2+x2dx\displaystyle I(a,b) = \int_{0}^{\infty} \frac{\ln(1+a^{2}x^{2})}{b^{2}+x^{2}}dx. I(a,b)I(a,b) is continuous, with respect to aa, for a>0a >0. Thus I(a,b)a=02ax2(b2+x2)(1+a2x2)dx=πab+1\displaystyle \frac{\partial I(a,b)}{\partial a} = \int_{0}^{\infty} \frac{2ax^{2}}{(b^{2}+x^{2})(1+a^{2}x^{2})}dx = \frac{\pi}{ab+1}. Thence, we have I(a,b)=πbln(ab+1)\displaystyle I(a,b) = \frac{\pi}{b}\ln(ab+1).

Solution 159

Notice 0sinnxxmdx=1(m1)!0dm1dxm1sinnxdxx\displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx= \frac{1}{(m-1)!}\int_{0}^{\infty} \frac{d^{m-1}}{dx^{m-1}}\sin^{n}x\frac{dx}{x}.
We need to treat the cases mn(mod2)m \equiv n \pmod 2 only.
If mn1(mod2)m \equiv n \equiv 1 \pmod 2, differentiating sinnx\sin^{n}x (m1m-1 times), we get 0sinnxxmdx=(1)(n1)(m1)4π2n(m1)!×(nm1n(n2)m1+)\displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx = \frac{(-1)^{\frac{(n-1)(m-1)}{4}}\pi}{2^{n}(m-1)!} \times \left(n^{m-1}-n(n-2)^{m-1}+ \cdots \right). The case mn0(mod2)m \equiv n \equiv 0 \pmod 2 is done analogically.
Hence, 0sinnxxndx=π2n(n1)!×(nn1n(n2)n1+)\displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{n}}dx = \frac{\pi}{2^{n}(n-1)!} \times \left(n^{n-1}-n(n-2)^{n-1}+\cdots \right).


By the way, Borwein have proved far more general result.
Fix n+1n+1 complex numbers z0,,znz_{0},\cdots,z_{n}.
Define a=(a1,,an)a= (a_{1},\cdots,a_{n}), ai{1,1}a_{i} \in \{-1,1\}; ba=z0+iziai\displaystyle b_{a} = z_{0}+ \sum_{i} z_{i}a_{i}, and ca=iai\displaystyle c_{a} = \prod_{i} a_{i}.
It follows that 0n=0nsinzixxdx=π2n+1n!acabansignba\displaystyle \int_{0}^{\infty} \prod_{n=0}^{n} \frac{\sin z_{i}x}{x}dx = \frac{\pi}{2^{n+1}n!}\sum_{a} c_{a}b_{a}^{n}{\rm sign b_{a}}.
Hence in our case, when z0=z1==zn1=1z_{0}=z_{1}=\cdots=z_{n-1}=1, we have 0(sinxx)ndx=π2n(n1)!0in2(1)i(ni)(n2i)n1\displaystyle \int_{0}^{\infty} \left(\frac{\sin x}{x} \right)^{n}dx= \frac{\pi}{2^{n}(n-1)!}\sum_{0 \le i \le \frac{n}{2}} (-1)^{i} \dbinom{n}{i} (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?

Spoiler

(edited 10 years ago)
Original post by Mladenov
May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?

Spoiler



Because the way which uses calculus was part of/similar to a previous STEP question that has obviously been solved many times on this forum. :laugh:
Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
Reply 1028
Original post by Mladenov

Spoiler


He returns :colone:

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 :biggrin: (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 :tongue: 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!) :lol: (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 :smile:

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 :lol:)

This one is really nice:

Problem 160*/**/***

Evaluate 01ln(x+1)x2+1dx\displaystyle\int_0^1 \frac{\ln(x+1)}{x^2+1} dx
(edited 10 years ago)
Reply 1029
Original post by Lord of the Flies
Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.

Post some! Post loads! I hereby declare the 29th May TSR Integration Day :pierre:
Very nice Mlad! I'm too slow again :tongue:
solution 156
0πln(12acosx+a2)dx=0πln((aeix)(aeix))dx=ππln(aeix)dx=ππlnadx+ππln(1eix/a)dx=2πlna\displaystyle \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx=\displaystyle \int_{0}^{\pi}ln((a-e^{ix})(a-e^{-ix}))dx=\displaystyle \int_{-\pi}^{\pi}ln(a-e^{ix})dx=\displaystyle \int_{-\pi}^{\pi}lnadx+\displaystyle \int_{-\pi}^{\pi}ln(1-e^{ix}/a)dx=2\pi lna
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 :colondollar:
Solution 160

f(t)=01ln(1+tx)1+x2dx\displaystyle f(t)=\int_0^1 \frac{\ln (1+tx)}{1+x^2}\,dx

f(t)=01x(1+tx)(1+x2)dx=1t2+1[01tx2+1dx+01xx2+1dx01tdx1+tx]=1t2+1[πt4+12ln2ln(1+t)]\displaystyle \begin{aligned}f'(t)=\int_0^1 \frac{x}{(1+tx)(1+x^2)}\,dx &= \frac{1}{t^2+1}\left[\int_0^1 \frac{t}{x^2+1}\,dx+\int_0^1 \frac{x}{x^2+1}\,dx-\int_0^1 \frac{t\,dx}{1+tx}\right]\\&=\frac{1}{t^2+1}\left[\frac{\pi t}{4}+\frac{1}{2}\ln 2 -\ln (1+t)\right]\end{aligned}

f(1)=01ln(1+t)1+t2dt=1201πt4(t2+1)+ln22(t2+1)dt=πln28\displaystyle f(1)= \int_0^1 \frac{\ln (1+t)}{1+t^2}\,dt=\frac{1}{2} \int_0^1 \frac{\pi t}{4(t^2+1)}+\frac{\ln 2}{2(t^2+1)}\,dt=\frac{\pi \ln 2}{8}

Alternate solution to previous integrals:

Solution 158

x=tant:    0ln(1+x2)1+x2dx=20π2lncostdt=πln2x=\tan t:\;\;\displaystyle \int_0^{\infty} \frac{\ln (1+x^2)}{1+x^2}\,dx= -2\int_0^{\frac{\pi}{2}}\ln \cos t\,dt =\pi \ln 2

Solution 157

f(s)=01xs1lnxdxf(s)=01xsdx=1s+1f(1)=01dss+1=ln2f(s)=\displaystyle \int_0^1 \frac{x^s-1}{\ln x}\,dx\Rightarrow f'(s)= \int_0^1 x^s\, dx=\frac{1}{s+1}\Rightarrow f(1)=\int_0^1 \frac{ds}{s+1}=\ln 2
(edited 10 years ago)
Reply 1032
Original post by ben-smith
Very nice Mlad! I'm too slow again :tongue:
solution 156
0πln(12acosx+a2)dx=0πln((aeix)(aeix))dx=ππln(aeix)dx=ππlnadx+ππln(1eix/a)dx=2πlna\displaystyle \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx=\displaystyle \int_{0}^{\pi}ln((a-e^{ix})(a-e^{-ix}))dx=\displaystyle \int_{-\pi}^{\pi}ln(a-e^{ix})dx=\displaystyle \int_{-\pi}^{\pi}lnadx+\displaystyle \int_{-\pi}^{\pi}ln(1-e^{ix}/a)dx=2\pi lna
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 :colondollar:

Looks decent to me!

I might as well add my solution into the soup :lol: ...

Solution 156 (3)

Let
Unparseable latex formula:

\displaystylef(a)=\int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx



af(a)=0π2a2cos(x)12acosx+a2dx\displaystyle\Rightarrow \frac{\partial}{\partial a} f(a) = \int_{0}^{\pi} \frac{2a-2\cos(x)}{1-2a \cos x+a^{2}}dx

Using the substitution u=tan(12x)u=\tan(\frac{1}{2}x):
af(a)=1a0πdx+2(a1)a(a+1)01u2+(a1a+1)2dx=2πa\displaystyle\Rightarrow \frac{\partial}{\partial a} f(a) =\frac{1}{a}\int_0^{\pi} dx + \frac{2(a-1)}{a(a+1)} \int_{0}^{\infty}\frac{1}{u^2+ \left(\frac{a-1}{a+1} \right)^2} dx=\frac{2\pi}{a},
for |a|>1.

...also, af(a)=0\frac{\partial}{\partial a} f(a) = 0, for |a|<1 (as the arctangent would take a different value).

Spoiler



So as f(1)=0, f(a)=1a2πtdt=2πln(a)\displaystyle f(a)=\int_1^a \frac{2\pi}{t} dt =2\pi \ln(a) for |a|>1. \square
(edited 10 years ago)
Reply 1033
Original post by Lord of the Flies
Solution 160

Solution 157

Exactly what I did :smile:

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! :eek: No idea why this method isn't widely taught/used!

POST SOME INTEGRALS!!!!!!!!!! :eek: :biggrin:
Original post by Jkn
As f(0)=0, f(a)=0a2πtdt=2πln(a)\displaystyle f(a)=\int_0^a \frac{2\pi}{t} dt =2\pi \ln(a) for |a|>1. \square


This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.
Reply 1035
Original post by Lord of the Flies
This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.

Typo :tongue: fixed!
Reply 1036
Problem 161**/***

Let k be an integer greater than 1. Suppose a0>0a_0>0 and an+1=an+1ank\displaystyle a_{n+1}=a_{n}+\frac{1}{\sqrt[k]{a_{n}}} for n>0n>0

Evaluate limnank+1nk\displaystyle\lim_{n \to {\infty}} \frac{a^{k+1}_n}{n^k}
(edited 10 years ago)
Problem 162 *

If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
Reply 1038
Original post by bananarama2
Problem 162 *

If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
On the ground below. :tongue: (?!?!?)
(edited 10 years ago)
Original post by Jkn
Solution 162

On the ground below. :tongue: (?!?!?)


Directly below?

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