Taylor series formula

Aug 2008
15
0
I need some help with figuring out the Taylor series formula.

I have two equations:
f(x)=sin(x/2) at c=0
f(x)=ln(2x) at c=2

I need to find the equation for the Taylor series formula for each of them. So for the first one, I worked out the Taylor polynomial to be x/2-x^3/48.... I know n=1 from that. But how do I write the rest of the formula? Since it's Maclaurin, I know the formula is [f^(n)(0)(x)^n]/n!, but what numbers do I plug in?

The second is ln(4)+((x-2)/2)-(((x-2)^2)/8)+(((x-2)^3)/24)... So n=0, and the other formula applies: [f^(n)(c)(x-c)^n]/n!, but I have the problem with the numbers again.. I need this to get the interval of convergence via the Ratio test.

I would appreciate your help
 
Jul 2009
555
298
Zürich
I need some help with figuring out the Taylor series formula.

I have two equations:
f(x)=sin(x/2) at c=0
f(x)=ln(2x) at c=2

I need to find the equation for the Taylor series formula for each of them. So for the first one, I worked out the Taylor polynomial to be x/2-x^3/48.... I know n=1 from that. But how do I write the rest of the formula? Since it's Maclaurin, I know the formula is [f^(n)(0)(x)^n]/n!, but what numbers do I plug in?

The second is ln(4)+((x-2)/2)-(((x-2)^2)/8)+(((x-2)^3)/24)... So n=0, and the other formula applies: [f^(n)(c)(x-c)^n]/n!, but I have the problem with the numbers again.. I need this to get the interval of convergence via the Ratio test.

I would appreciate your help
If you know certain Taylor series already, for example \(\displaystyle \sin(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}\) and \(\displaystyle \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n\), then you can get the Taylor series of other functions very easily, like this:

\(\displaystyle \sin\left(\frac{x}{2}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\left(\frac{x}{2}\right)^{2n+1}=\sum_{n=0}^\infty\frac{(-1)^n}{2^{2n+1}(2n+1)!}x^n\)

and

\(\displaystyle \ln(2x)=\ln\big(2(x-2+2)\big)=\ln\left(4\left(1+\frac{x-2}{2}\right)\right)=\ln(4)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac{x-2}{2}\right)^n\)
\(\displaystyle =\ln(4)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n 2^n}(x-2)^n\)