Hi,

If i want to use Taylor expansion on a function like:

$\displaystyle \frac{ln(sin(x))}{cosx}$

Do i have to substitute the expansion of sin x, in to the expansion of ln x?

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- Feb 20th 2010, 02:17 PMJonesTaylor expansion of composite function
Hi,

If i want to use Taylor expansion on a function like:

$\displaystyle \frac{ln(sin(x))}{cosx}$

Do i have to substitute the expansion of sin x, in to the expansion of ln x? - Feb 20th 2010, 04:22 PMProve It
- Feb 20th 2010, 04:28 PMDrexel28
For certain interval restrictions $\displaystyle \frac{\ln(\sin(x))}{\cos(x)}=\frac{\ln\left(\sqrt{ 1-\cos^2(x)}\right)}{\cos(x)}=\frac{\ln\left(1-\cos^2(x)\right)}{2\cos(x)}$ and since $\displaystyle \left|\cos^2(x)\right|<1$ (with the proper interval restrictions) we see that $\displaystyle \frac{\ln\left(1-\cos^2(x)\right)}{\cos(x)}=\frac{1}{\cos(x)}\sum_{ n=1}^{\infty}\frac{\cos^{2n}(x)}{n}=\sum_{n=1}^{\i nfty}\frac{\cos^{2n-1}(x)}{n}$