Show for any $s \in \mathbb{R}$ that $C_1(1+x^2)^s \le (1+x)^{2s} \le C_2 (1+x^2)^s$.

Asked May 03 '14 at 10:46

Active May 03 '14 at 10:46

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Let $x \in \mathbb{R}$, then for any $s \in \mathbb{R}$ we must show that there exist $C_1,C_2 \in \mathbb{R}$ such that $$C_1(1+x^2)^s \le (1+x)^{2s} \le C_2 (1+x^2)^s.$$

It seems pretty obvious for the case where $s \in \mathbb{N}$, but after playing with it for a while, I don't seem to be getting anywhere for the general case.

asked May 03 '14 at 10:46

Zorngo

1,671

3

Try studying the quantity $$\frac{(1+x)^2}{1+x^2}.$$ – Siminore May 03 '14 at 10:48
This was asked very recently (and answered) with a mistake in the exponents (corrected in the answer). – Did May 03 '14 at 10:52
For some reason that never occurred to me, thanks! I think it's time to get some sleep. – Zorngo May 03 '14 at 10:52
Here's some discussion on this: http://math.stackexchange.com/q/341621/8157 (This question coincides with the other upon setting $x=\frac{a}{b}$) – Giuseppe Negro May 03 '14 at 10:59

Show for any $s \in \mathbb{R}$ that $C_1(1+x^2)^s \le (1+x)^{2s} \le C_2 (1+x^2)^s$.

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