Wikipedia:Reference desk/Archives/Mathematics/2009 August 11

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August 11

Integral

Let $f\in L^{1}(0,1)$ and $\int _{0}^{1}e^{2n\pi it}f(t)\,dt=0$ for $n=0,\pm 1,\pm 2,\ldots$. Prove that $f=0$ a.e.

I have someone's solution but it quotes some result on Fourier series saying "Since G is continuous and G(0) = G(1), then the sequence of the arithmetic means of the partial sums of the Fourier series of G converges uniformly to it." They say this is a result of some theorem in baby Rudin. I am guessing we are not supposed to just know such a result. So, do any of you have an idea how to do this using qualifying exam real analysis? Thanks. StatisticsMan (talk) 14:38, 11 August 2009 (UTC)

The statement of your problem is equivalent to the statement that the Fourier basis is complete. I don't think there are too many elementary proofs of that floating around. The shortest one I found was in Zygmund's Trigonometric Series, section 1.5.

Briefly, Zygmund's proof starts by proving the statement for continuous periodic functions. This is done in the following way: if one assumes, to the contrary, that there is a continuous function whose Fourier coefficients are all zero, but which is nonzero, then it must be greater than some $\epsilon$ in a $\delta$-neighborhood of some point $x_{0}$. It is then only necessary to construct a trigonometric polynomial p which is bounded below one in absolute value outside this neighborhood, and exceeds 1 uniformly inside this neighborhood; this can be done by appropriate translations of the cosine function. If one then considers the interaction of $\int (p(x))^{n}f(x)dx$, this is supposed to be zero, but can be shown to grow to infinity, which is the desired contradiction.

Zygmund then considers the general integrable case by letting $F(x)=\int _{-\pi }^{x}f(y)dy$ and estbalishing that if the Fourier coefficients of f are all zero, then so are the Fourier coefficients of F. However, F is a continuous periodic function, so by the previous special case F is uniformly zero. That means that its derivative, f, is zero almost everywhere. Ray^Talk 15:24, 11 August 2009 (UTC)

So, G is continuous and 1-periodic, and we need to show that the Cesàro sum of its Fourier series converges uniformly to G(x). Let

$G_{k}=\int _{0}^{1}G(t)e^{2\pi ikt}\,dt$

be the kth Fourier coefficient. The Nth partial Cesàro sum is

$G(x,N)={\frac {1}{N}}\sum _{n=0}^{N-1}{\Bigl (}\sum _{k=-n}^{n}G_{k}e^{-2\pi ikx}{\Bigr )}=\int _{0}^{1}G(t){\frac {1}{N}}\sum _{n=0}^{N-1}\sum _{k=-n}^{n}e^{2\pi ik(t-x)}\,dt,$

and we have

${\begin{aligned}\sum _{n=0}^{N-1}\sum _{k=-n}^{n}e^{2\pi iku}&=\sum _{n=0}^{N-1}{\frac {e^{2\pi i(n+1/2)u}-e^{-2\pi i(n+1/2)u}}{e^{\pi iu}-e^{-\pi iu}}}={\frac {e^{2\pi iNu}-1-1+e^{-2\pi iNu}}{(e^{\pi iu}-e^{-\pi iu})^{2}}}\\&=\left({\frac {e^{\pi iNu}-e^{-\pi iNu}}{e^{\pi iu}-e^{-\pi iu}}}\right)^{2}={\frac {\sin ^{2}\pi Nu}{\sin ^{2}\pi u}},\end{aligned}}$

thus

$G(x,N)=\int _{0}^{1}G(t){\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt.\qquad (*)$

First, if we temporarily change G to the constant 1 function, we can easily compute directly G(x,N) = 1, hence

$\int _{0}^{1}{\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt=1.\qquad (**)$

Now, we return to our original G. For any $\varepsilon >0$ there exists $\delta \in (0,1/2)$ such that $|G(x)-G(y)|\leq \varepsilon$ whenever $|x-y|\leq \delta$. We have

${G(x,N)-G(x)=\int _{0}^{1}(G(t)-G(x)){\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt=\int _{x-1/2}^{x+1/2}(G(t)-G(x)){\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt}$

by (*) and (**). We can bound

$\left|\int _{x-\delta }^{x+\delta }(G(t)-G(x)){\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt\right|\leq \varepsilon \int _{x-\delta }^{x+\delta }{\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt\leq \varepsilon \int _{0}^{1}{\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\,dt=\varepsilon .$

For $t\in [x-1/2,x+1/2]\smallsetminus [x-\delta ,x+\delta ]$, we have

$|G(t)-G(x)|{\frac {\sin ^{2}\pi N(t-x)}{N\sin ^{2}\pi (t-x)}}\leq {\frac {c}{N\sin ^{2}\pi \delta }}$

for some c depending only on G, hence |G(x,N) − G(x)| ≤ 2ε if N is large enough. — Emil J. 15:33, 12 August 2009 (UTC)

Lottery odds

Hi,

I'm trying to work out the odds of the following situation:

49 balls in a draw, 6 balls drawn, and I want to know how to work out the probability of 3 and only 3 matching balls being drawn (with the other 3 being "duds") as a "condensed formula" and not a "series formula".

For (not) example:

P = 1/N * 1/(N-1) * .... * 1/(N-(n-2)) * 1/(N-(n-1))

For example:

P = N! / ((N-n)! * n!)

Rixxin (talk) 16:02, 11 August 2009 (UTC)

What do you mean by matching balls and duds? Bo Jacoby (talk) 16:53, 11 August 2009 (UTC).

A match ball is one that matches your lottery ticket and a dud is one that doesn't. --Tango (talk) 17:03, 11 August 2009 (UTC)

(outdent rewsponse) Hint. Figure out how many ways there are to draw 49 balls. Then figure out how many ways there are to draw exactly three matching. The second can be broken into two parts: How many ways to pick exactly three of the six you have, then how many ways to pick the rest (the other three) of the numbers you don't have. Baccyak4H (Yak!) 17:14, 11 August 2009 (UTC)

Thanks for the reply. I'm pretty familiar with the method to derive this, it's the formula itself I'm having trouble with.--Rixxin (talk) 18:17, 11 August 2009 (UTC)

Assuming that the lottery ticket also contains 6 numbers, the probability is that of the hypergeometric distribution for N=49, m=n=6, k=3. Bo Jacoby (talk) 08:32, 12 August 2009 (UTC).

The hypergeometric distribution is a continuous distribution though, is there not a discrete version of this, similar to the OP's initial example formula for working out the odds of picking $n$ numbers from a set of $N$, and matching all of them? ReeKorl (talk) 10:30, 12 August 2009 (UTC)

I can't see where your n and N come from. So lets assume there are 49 balls. You chose 6, and then 6 are drawn from the 49. What are the odds of getting exactly 3 of your 6 to match the 6 drawn. The odds are given by (Number of losing draws)/(number of winning draws). Here losing draws are the number of 6 ball combinations that don't include exactly 3 of your chosen 6. Winning draws are the number of 6 ball combinations that include exactly 3 of your chosen 6. Well, there are

$^{6}C_{3}={\frac {6!}{3!(6-3)!}}=20$

ways of chosing 3 from your 6. If you want to match exactly 3 then 3 of the 6 drawn must be different from your chosen 6. There are

$^{43}C_{3}={\frac {43!}{3!(43-3)!}}=12,341$

ways of chosing 3 from the 49 - 6 = 43 unchosen balls. It follows that there are 20 × 12,341 = 246,820 possible ways of drawing 6 from 49 so that exactly 3 will match your chosen 6. There are a total of

$^{49}C_{6}={\frac {49!}{6!(49-6)!}}=13,983,816$

ways to draw 6 numbers from 49. So the number of losing draws is total - winning = 13,736,996. The odds are the

${\frac {\mbox{No. losing}}{\mbox{No. winning}}}={\frac {13,736,996}{246,820}}\ .$

The odds are then about 55.7-to-1. If you want to introduce variables then you'll get an explicit formula but with the products and differences of factorials. Notice that odds of say m-to-n mean that you expect something will happen n times out of m + n. In the case above, you can expect not to match a single number more often than not. I forget the exact figure, it's about 70% of the time you won't get a single number. But don't quote me on that, I haven't worked it out for a long time. Also, the 55.7-to-1 means that if the lottery prize was £56 for 3 numbers then the lottery would still make a slight profit over time. Interesting that the English lottery pays £10 for exactly 3 numbers. Quite a profit margin, eh? ~~ Dr Dec (Talk) ~~ 11:27, 12 August 2009 (UTC)

The hypergeometric distribution is discrete. My n and N come from the article hypergeometric distribution.

${{{m \choose k}{{N-m} \choose {n-k}}} \over {N \choose n}}={{{6 \choose 3}{{43} \choose {3}}} \over {49 \choose 6}}={20\cdot 12341 \over 13983816}=0.0176504$

Bo Jacoby (talk) 14:47, 12 August 2009 (UTC).

I was asking about Rixxin's n and N. There aren't any variables in the formulation of his problem, but then s/he starts writing expressions in terms of n and N. Bo, you seem to have given the probability (0.018) instead of the odds (55.7-to-1). ~~ Dr Dec (Talk) ~~ 22:04, 13 August 2009 (UTC)

"N" was the total number of balls, and "n" was the number of balls drawn, so in my example this would be 49 and 6 respectively.--Rixxin (talk) 15:58, 14 August 2009 (UTC)

OK Dr Dec. Odds are 490607 to 8815 , to be exact. Bo Jacoby (talk) 20:43, 14 August 2009 (UTC).

True, but 56-to-1 would be better since you can either play the lottery or not. A whole number answer is good enough, a one decimal place answer is getting as accurate as useful. An answer like 490607-to-8815 is not very practicle. Like any question in maths: we should put out answers in the context of the question. ~~ Dr Dec (Talk) ~~ 21:52, 14 August 2009 (UTC)

Parametrisation

Hi. I've just finished a piece of work and would appreciate it if someone hear could check through it.

'A curve is given by $x^{\frac {2}{n}}+y^{\frac {2}{n}}=a^{\frac {2}{n}}$ where n is an odd integer and a is a positive constant. Find a parametrisation that describs the curve anticlockwise as t ranges from 0 to $2\pi$. The area enclosed by such a curve is given by $\int _{0}^{2\pi }[x{\frac {dy}{dt}}-y{\frac {dx}{dt}}]dt$. Determine the area for the case n=3.'

For the parametrisation I get $x=a{\sin }^{n}t$ and $y=a{\cos }^{n}t$ (though I see no reason why it couldn't be the other way around) and for the area I get $-{\frac {3a^{2}}{8}}$ (I'm slightly concerned about the negative sign). Is this correct? 92.3.137.99 (talk) 20:32, 11 August 2009 (UTC)

Switching x and y switches the direction that the curve goes with increasing t, which reverses the sign of the integral. The version you chose goes clockwise which is why the integral comes out negative. Rckrone (talk) 01:50, 12 August 2009 (UTC)

The area enclosed by such a curve is given by ${\frac {1}{2}}\int _{0}^{2\pi }[x{\frac {dy}{dt}}-y{\frac {dx}{dt}}]dt$. Bo Jacoby (talk) 05:37, 12 August 2009 (UTC).

Ahhhh, I see. It never crossed my mind to more closely examine why it specifically said 'anticlockwise'. I, hopefully, won't make that mistake again. Thank you Rckrone! 92.3.137.99 (talk) 12:57, 12 August 2009 (UTC)

BTW, with n=3 you get a classic curve, the Astroid. Other values of the exponent give less more or less known curves, still enjoying some geometric properties. --pma (talk) 16:01, 12 August 2009 (UTC)

With n = 1, you also get a fairly well-known curve. I daresay it is much better known that the astroid. — Emil J. 14:22, 13 August 2009 (UTC)

of course but he was intersted in n=3; what he wants is in the linked article... pma —Preceding unsigned comment added by 79.38.22.37 (talk) 19:44, 14 August 2009 (UTC)

As a second question, why is the area given by ${\frac {1}{2}}\int _{0}^{2\pi }[x{\frac {dy}{dt}}-y{\frac {dx}{dt}}]dt$ rather than $\int _{b}^{a}[y{\frac {dx}{dt}}]dt$ which is how I was taught to find the area under a curve when it's given in parametric form? 92.3.137.99 (talk) 19:01, 13 August 2009 (UTC)

For a closed curve, integration by parts shows that the two are equivalent. -- Meni Rosenfeld (talk) 20:52, 13 August 2009 (UTC)

By the way, the area is ${\frac {3a^{2}\pi }{8}}$ (note the $\pi$). -- Meni Rosenfeld (talk) 21:10, 13 August 2009 (UTC)