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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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  • $\begingroup$ en.wikipedia.org/wiki/Beta_function $\endgroup$ Jun 2, 2010 at 14:20
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    $\begingroup$ 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. $\endgroup$ Jun 2, 2010 at 14:23
  • $\begingroup$ See the picture on the RHS: en.wikipedia.org/wiki/Lp_space#Motivation $\endgroup$ Jun 2, 2010 at 14:43
  • $\begingroup$ @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! $\endgroup$ Jun 2, 2010 at 15:05
  • $\begingroup$ Closed. Please see the FAQ for a list of sites where you can ask questions of a similar level. $\endgroup$
    – S. Carnahan
    Jun 2, 2010 at 15:14

2 Answers 2

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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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    $\begingroup$ Or easier: dominated convergence. $\endgroup$ Jun 2, 2010 at 14:31
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    $\begingroup$ 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.) $\endgroup$ Jun 2, 2010 at 15:03
  • $\begingroup$ 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) } $ $\endgroup$
    – Will Jagy
    Jun 2, 2010 at 17:59
  • $\begingroup$ @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. $\endgroup$ Jun 2, 2010 at 23:18
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    $\begingroup$ @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. $\endgroup$ Jun 3, 2010 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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  • $\begingroup$ 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. $\endgroup$ Jun 2, 2010 at 15:06
  • $\begingroup$ Or \lvert and \rvert... $\endgroup$ Jun 2, 2010 at 15:21
  • $\begingroup$ @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. $\endgroup$ Jun 2, 2010 at 16:24

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