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Practicing the manipulation of recurrence relations, I'm stuck on this :

Defining $I(n)=\int_{0}^{\pi/2}sin^n(x)dx$, I got the recurrence relation $nI(n)=(n-1)I(n-2)$ for $n\ge2$.

Now I'm also trying to prove that $I(n) \le I(n-1)$ for all $n\ge 1$.

[I've tried by induction on the odd and even possibilities but it doesn't give anything concluent.]

and that $2n/2n+1 \le I(2n+1)/I(2n) \le 1$

The second part of the equality can be obtained easily from above but I have no idea for the first one.

Can you hint me? I've never been good with these...

Timmy
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2 Answers2

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With your induction relation, you can write $I(2n)=\frac{(2n-1)(2n-3)...}{2n(2n-2)...}I(0)$. You can symplify this in $I(2n)=\frac{(2n)!}{4^n(n!)^2}I(0)$. The same kind of relation exists for I(2n+1).

Adrien Laurent
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  • I've just seen that the edit that was made is wrong... – Timmy May 14 '14 at 20:08
  • That explains everything... – Adrien Laurent May 14 '14 at 20:09
  • It is corrected now... – Timmy May 14 '14 at 20:11
  • mmmm the last inequality is not this easy... But with the stirling approximation it works for huge n. – Adrien Laurent May 14 '14 at 20:25
  • Oh I found. You can use the inequality $\frac{2}{\pi}x \le sin(x) \le x$. It gives $\frac{I(2n+1)}{I(2n)} \ge \frac{2n+1}{2n+2}$. Then this gives the result because $\frac{2n+1}{2n+2} \ge \frac{2n}{2n+1}$. – Adrien Laurent May 14 '14 at 20:45
  • oh right! I couldn't find it either from what you had done above ^^ I'm still on the algebra of the first inequality... – Timmy May 14 '14 at 20:50
  • sorry but I don't see how you get what you say it gives, can you give more details? – Timmy May 15 '14 at 07:01
  • Well you're right, I forgot a $(\frac{2}{\pi})^n$... Anyway this approaches the result : $\int_0^{\frac{\pi}{2}} x^n dx \ge I(n)=\int_0^{\frac{\pi}{2}} sin(x)^n dx \ge (\frac{2}{\pi})^n \int_0^{\frac{\pi}{2}} x^n dx$.

    So $\frac{\pi^{n+1}}{2^{n+1}(n+1)} \ge I(n) \ge \frac{\pi}{2(n+1)}$.

    Then $\frac{I(2n+1)}{I(2n)} \ge \frac{\pi}{2(2n+2)} \frac{2^{2n+1}(2n+1)}{\pi^{2n+1}} = \frac{2n+1}{2n+2} (\frac{2}{\pi})^{2n}$. But this is not what I was expecting...

     – Adrien Laurent May 15 '14 at 07:54
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I finally found something =)

I(n) is a decreasing sequence so $\frac{I(n+1)}{I(n)} \ge \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}$.

Then $\frac{I(2n+1)}{I(2n)} \ge \frac{2n+1}{2n+2}$!!!

and $\frac{2n+1}{2n+2} \ge \frac{2n}{2n+1}$.

Adrien Laurent
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