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.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
For every $a, b \ge 0$, we have $\sqrt{a+b} \le \sqrt{a} + \sqrt{b}$. Then $I_n \le 1+\int_0^1 nx^{n-1}dx = 2$
$\begingroup$
$\endgroup$
3
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 .
$$x\in[0,1],\forall n\in \mathbb{N} : c''_n\geq 0 \\and , x^{n+1}<x^{n}$$ 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 $$\forall n \in \mathbb{N} :length_{c_n} < 2\\so\\I_n=\int_{0}^{1}\sqrt{1+(y')^2}dx=\int_{0}^{1}\sqrt{1+(nx^{n-1})^2}dx<2\\I_n<2$$
answered Aug 12, 2017 at 10:20
-
2$\begingroup$ This is a nice answer to your question. If you want fun, you could also play with $$I_n=\, _2F_1\left(-\frac{1}{2},\frac{1}{2 n-2};1+\frac{1}{2 n-2};-n^2\right)$$ $\endgroup$ Aug 12, 2017 at 11:54
-
$\begingroup$ @ClaudeLeibovici :I do not get what you mean . would you please describe it more ? $\endgroup$ 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. $\endgroup$ Aug 12, 2017 at 12:07