Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the following function of a variable $\theta\in[0,\frac{\pi}{2}]$ $$f(\theta)=\frac{A}{\sqrt{2\pi}\sigma\sin\theta\sin\theta_{\mu}}\exp(-\frac{(\sin\theta-\sin\theta_{\mu})^{2}}{2\sigma^{2}})$$ Numerically it seems that for small $\sigma$ this can be approximated by a Gaussian $$f(\theta)\approx\frac{A'}{\sqrt{2\pi}\sigma'}\exp(-\frac{(\theta-\theta_{\mu}')^{2}}{2\sigma'^{2}})$$ How do $\theta_{\mu}'$, $A'$ and $\sigma'$ relate to $\theta_{\mu}$, $A$ and $\sigma$? Trail and error suggests $$\theta_{\mu}'=\theta_{\mu}$$ $$\sigma'=\frac{\sigma}{\cos\theta_{\mu}}$$ $$A'=\frac{A}{\cos\theta_{\mu}\sin^{2}\theta_{\mu}}$$ Any idea on how to justify this approximation?

Edit: original question already assumed $\sin\theta\approx\sin\theta_{\mu}$ for small $\sigma$ in the factor before the exponent, thereby making $f(\theta)$ a Gaussian of $\sin\theta$

share|improve this question

1 Answer 1

This approximation uses the first-order Taylor series approximation of the sine around x=0 (you can see it here Wikipedia).

Since you are just using the approximation for the sine function, I think the constants would remain the same. Please take into account that, in order to approximate this function for higher values, you could use more terms in the Taylor series.

Hope this helps.

share|improve this answer
    
Thanks for your suggestion: it led me to reconsider an approximation I already made (changed it in the question). –  Wox Sep 20 '12 at 9:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.