# Taylor expansion of composite function

• Feb 20th 2010, 03:17 PM
Jones
Taylor expansion of composite function
Hi,

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

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

Do i have to substitute the expansion of sin x, in to the expansion of ln x?
• Feb 20th 2010, 05:22 PM
Prove It
Quote:

Originally Posted by Jones
Hi,

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

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

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

Yes, that would be a good idea. Then divide everything by the expansion for $\cos{x}$.
• Feb 20th 2010, 05:28 PM
Drexel28
Quote:

Originally Posted by Jones
Hi,

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

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

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

For certain interval restrictions $\frac{\ln(\sin(x))}{\cos(x)}=\frac{\ln\left(\sqrt{ 1-\cos^2(x)}\right)}{\cos(x)}=\frac{\ln\left(1-\cos^2(x)\right)}{2\cos(x)}$ and since $\left|\cos^2(x)\right|<1$ (with the proper interval restrictions) we see that $\frac{\ln\left(1-\cos^2(x)\right)}{\cos(x)}=\frac{1}{\cos(x)}\sum_{ n=1}^{\infty}\frac{\cos^{2n}(x)}{n}=\sum_{n=1}^{\i nfty}\frac{\cos^{2n-1}(x)}{n}$