0
$\begingroup$

Let $n \in \mathbb{N}$ ,$$I_n=\int_{0}^{1}\sqrt{1+n^2x^{2n-2}}dx$$ we want to prove $$\forall n\in\mathbb{N} :I_n\leq2$$ I am looking for new Idea(s) or alternative proof.Thanks in advance.

$\endgroup$

2 Answers 2

6
$\begingroup$

For every $a, b \ge 0$, we have $\sqrt{a+b} \le \sqrt{a} + \sqrt{b}$. Then $I_n \le 1+\int_0^1 nx^{n-1}dx = 2$

$\endgroup$
2
$\begingroup$

Take $$y=c_n=x^n\\x\in[0,1] ,n\in\mathbb{N}$$ all of the $c_s$'s pass two points(0,0),(1,1),for example for n=1,2,3,4 as below . enter image description here $$x\in[0,1],\forall n\in \mathbb{N} : c''_n\geq 0 \\and , x^{n+1}<x^{n}$$ so $$length_{c_1}<length_{c_2}<length_{c_3}<...<length_{c_n}<OH+HM=2$$ length of curve formula was $$\int_{0}^{1}\sqrt{1+(y')^2}dx$$ in this case we have $$\forall n \in \mathbb{N} :length_{c_n} < 2\\so\\I_n=\int_{0}^{1}\sqrt{1+(y')^2}dx=\int_{0}^{1}\sqrt{1+(nx^{n-1})^2}dx<2\\I_n<2$$

$\endgroup$
3
  • 2
    $\begingroup$ This is a nice answer to your question. If you want fun, you could also play with $$I_n=\, _2F_1\left(-\frac{1}{2},\frac{1}{2 n-2};1+\frac{1}{2 n-2};-n^2\right)$$ $\endgroup$ Commented Aug 12, 2017 at 11:54
  • $\begingroup$ @ClaudeLeibovici :I do not get what you mean . would you please describe it more ? $\endgroup$
    – Khosrotash
    Commented Aug 12, 2017 at 12:02
  • 1
    $\begingroup$ What I wrote is the result in terms of hypergeometric function. But it does not prove anything ! I just made my comment for the fun of it. Your solution is very nice ! Cheers. $\endgroup$ Commented Aug 12, 2017 at 12:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .