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Trying to solve for the area enclosed by $x^4+y^4=1$. A friend posed this question to me today, but I have no clue what to do to solve this. Keep in mind, we don't even know if there is a straightforward solution. I think he just likes thinking up problems out of thin air.

Anyway, the question becomes more general, since we think that

$lim_{n\to\infty}\int_0^1{(1-x^n)^{1/n}} = {1\over4}$    (it approaches a square / becomes linear)

can anyone confirm that this is true or not?

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4  
I should add: substitute $y=x^n$ first. Anyhow, I don't think this is quite an MO level question, so I am voting to close. –  Harald Hanche-Olsen Jun 2 '10 at 14:23
    
See the picture on the RHS: en.wikipedia.org/wiki/Lp_space#Motivation –  Steve Huntsman Jun 2 '10 at 14:43
    
@Harald: judging by the answers received, I would tend to agree with you about the level. I didn't know any better place to ask. Sorry! –  fuzzylintman Jun 2 '10 at 15:05
    
Closed. Please see the FAQ for a list of sites where you can ask questions of a similar level. –  S. Carnahan Jun 2 '10 at 15:14
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closed as too localized by Harald Hanche-Olsen, Steve Huntsman, Andrea Ferretti, Gjergji Zaimi, S. Carnahan Jun 2 '10 at 15:12

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2 Answers

up vote 6 down vote accepted

I always prefer not to skip $dx$: $$ I_n=\int_0^1(1-x^n)^{1/n}dx. $$ After the change of variable $t=x^n$, the integral becomes the beta integral, $$ I_n=\frac1n\int_0^1(1-t)^{1/n}t^{1/n-1}dt =\frac1n\frac{\Gamma(1+1/n)\Gamma(1/n)}{\Gamma(1+2/n)} =\frac1n\frac{\Gamma(1/n)^2\cdot 1/n}{\Gamma(2/n)\cdot 2/n} \to1 \quad\text{as $n\to\infty$}, $$ as $1/\Gamma(z)\sim z$ as $z\to 0$.

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Or easier: dominated convergence. –  Nate Eldredge Jun 2 '10 at 14:31
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Dominated convergence is overkill. Using monotonicity and positivity of the integrand, $I_n\ge\int_0^{a}(1-x^n)^{1/n}\,dx\ge a(1-a^n)^{1/n}\to a$ as $n\to\infty$ for any $a\in(0,1)$. Now let $a\to1$. (Oh, and putting $dx$ right next to the integral sign is standard physicists' notation. I dislike it too, but we have to live with it if we are to talk to physicists.) –  Harald Hanche-Olsen Jun 2 '10 at 15:03
    
Forgetting any limit, the area/volume of sets with $ x_1, x_2, \ldots, x_n \geq 0 $ and $ x_1^{a_1} + x_2^{a_2} + \ldots + x_n^{a_n} \leq 1 $ is a result of Dirichlet, for example with $n=3$ the volume is $ \frac{ \Gamma \left( 1 + \frac{1}{a_1} \right) \Gamma \left( 1 + \frac{1}{a_2} \right) \Gamma \left( 1 + \frac{1}{a_3} \right) }{ \Gamma \left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) } $ –  Will Jagy Jun 2 '10 at 17:59
    
@Nate & Harald: This was my pre-bed problem, so I immediately realised the uniform convergence of the integrand (it's even simpler to take $x$ from the interval $(a,b)\subset(0,1)$, so to exclude both $0$ and $1$). A big surprise to see 7 upvotes and 1 downvote... I am really puzzled of so huge interest to such an elementary question, while much more involved answers, like mathoverflow.net/questions/21643/…, remain not noticed. –  Wadim Zudilin Jun 2 '10 at 23:18
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@Wadim: This may not be a good place to discuss why one post attracts more notice than another, so I'll be brief: This one is simple to have an opinion on, so more people look in on it. That other one is complicated and requires real effort to understand, and most people won't bother unless the topic interests them. –  Harald Hanche-Olsen Jun 3 '10 at 2:08
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I think this question smells of homework, but another answer, which to me totally obscures the geometric nature of the question has been posted, and I feel that this justifies the following answer (even if the question is closed):

The $l^p$ norms $\lvert(x,y)\rvert_p = (\lvert x\rvert^p+\lvert y \rvert^p)^{1/p}$ are norms and satisfies that if $\lvert(x,y)\rvert_p=1$ and $q>p$ then $\lvert(x,y)\rvert_q\leq 1$. So the unit "circles" of which you want to find the area grows.

It is also a fact that $\lvert (x,y) \rvert_p \to \max (\lvert x\rvert,\lvert y\rvert)$ as $p\to \infty$. So the unit circles converges to the square which is the boundary of $[-1,1]\times [-1,1]$. This implies by monotone convergence theorem that your integral converges to 1. Because the entire square has area 4.

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A comment on LaTeX: \mid is a relation in TeX. It produces wrong spacing when you use it to write absolute values. Just type a vertical bar instead. –  Harald Hanche-Olsen Jun 2 '10 at 15:06
    
Or \lvert and \rvert... –  Steve Huntsman Jun 2 '10 at 15:21
    
@Harald: I did know this, but @Steve: \\lvert and \\rvert is better for me as I seem to have lost my vertical bar somewhere on the keyboard - so thanks. –  Thomas Kragh Jun 2 '10 at 16:24
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