Can someone help to prove this? $${\pi\over4}\leq\int_0^{\pi\over2}e^{-\sin^2{x}}dx\leq{11\over32}\pi$$
I totally have no idea how to approach.
Thanks.
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A related question. – Start wearing purple Jun 02 '14 at 13:34
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1Sorry. I didn't get you. How is that question related? – user149367 Jun 02 '14 at 13:42
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1You are asking to prove that something approximately equal to $1.01$ is between something approximately equal to $0.79$ and $1.08$. Why? – Start wearing purple Jun 02 '14 at 13:47
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1"approximately 1.01" is $\frac{\pi}{2 \sqrt{e}} \sum_{k=0}^{\infty} 2^{-4k}/{(k!)^2} $, for interest. – John Fernley Jun 02 '14 at 13:52
3 Answers
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Integrating the following inequality gives you that result. $$1 - \sin^2(x) \le e^{-\sin^2(x)} \le 1 - \sin^2(x) + \frac{\sin^4(x)}{2}$$
S L
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Use the fact that
$$ \exp (x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$
Jeb
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From Taylor's expansion, we have: $$ e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\mathcal{O}(x^8) $$ It means that: $$ 1-\sin(x)^2\leq e^{-x^2}\leq 1-\sin(x)^2+\frac{\sin(x)^4}{2} $$ Integral the inequality from $0$ to $\pi/2$, we have: $$ \frac{\pi}{4}=\int_0^{\pi/2}1-\sin(x)^2dx\leq\int_0^{\pi/2}e^{-\sin(x)^2}dx\leq\int_0^{\pi/2}1-\sin(x)^2+\frac{\sin(x)^4}{2}dx=\frac{11\pi}{32} $$ Then your proposition is follow.
Lion
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