1. ## Maclaurin series

Given $\displaystyle (sin^{-1}x)^2=x^2+\frac{1}{3}x^4+...$, use the standard series expansions for ln(1+x) and $\displaystyle e^x$ to show that when $\displaystyle x$ is small enough for powers of x above fourth to be neglected, $\displaystyle (sin^{-1}x)^2=\frac{1}{6}[ln(1+x^2)+5e^{x^2}-5]$

2. ## Re: Maclaurin series

Originally Posted by Punch
Given $\displaystyle (sin^{-1}x)^2=x^2+\frac{1}{3}x^4+...$, use the standard series expansions for ln(1+x) and $\displaystyle e^x$ to show that when $\displaystyle x$ is small enough for powers of x above fourth to be neglected, $\displaystyle (sin^{-1}x)^2=\frac{1}{6}[ln(1+x^2)+5e^{x^2}-5]$
following the given directions, I was able to approximate ...

$\displaystyle \ln(1+x^2) \approx x^2$

$\displaystyle e^{x^2} \approx 1 + x^2$

$\displaystyle \frac{1}{6}[\ln(1+x^2)+5e^{x^2}-5] \approx \frac{1}{6}[x^2 + 5(1+x^2) - 5] = x^2$

finish it ...

3. ## Re: Maclaurin series

Hence find an approximatevalue for $\displaystyle \pi^2-30e^{\frac{1}4}}$ giving your answer in the form cln\frac{5}{4}+$\displaystyle d$

4. ## Re: Maclaurin series

Originally Posted by Punch
Hence find an approximatevalue for $\displaystyle \pi^2-30e^{\frac{1}4}}$ giving your answer in the form cln\frac{5}{4}+$\displaystyle d$
why did you leave out the "hence" part in the original post?

5. ## Re: Maclaurin series

Originally Posted by skeeter
why did you leave out the "hence" part in the original post?
Sorry, I didn't thought I would have had a problem with this.