What Taylor series has the sum of $-ln\left ( \frac{8}{5} \right )$

76 Views Asked by At

How to compute the Taylor series which has the sum $-ln\left ( \frac{8}{5} \right )$. Is this done by finding the series for $-ln\left (x \right )$ centered around the point $a=\frac{8}{5}$?

1

There are 1 best solutions below

0
On BEST ANSWER

$$\ln (1+x)= \int \frac {1}{1+x}= \int 1-x+x^2-x^3+....=$$

$$x-x^2/2+x^3/3-x^4/4+...$$

$$\ln (8/5) = \ln (1+3/5)= (3/5)-(3/5)^2/2 +(3/5)^3/3-...$$

$$-\ln (8/5) = - (3/5)+(3/5)^2/2-(3/5)^3/3+...$$