Find Maclaurin series of $$\frac{x^2+1}{x^2-2x+1}$$
How should I approach this? I know that for a trigonometric function we derive the function $n$ times and write it as $\sum \frac{f^{(n)}(x_{0}}{n!}(x-x_{0})^n$ but what can we do in this case?
[Math] Finding Maclaurin series for a fraction
taylor expansion
Best Answer
A Maclaurin series is the Taylor series expansion of a function about $x_0=0$.
Hint: prove that the Maclaurin series of $1/(x-1)^2$ is $\sum_{n\geq 1} nx^{n-1}$ (remember that the Maclaurin series of $1/(1-x)$ is $\sum_{n\geq 0} x^{n}.$ Then $$\frac{x^2+1}{x^2-2x+1}=\frac{x^2-2x+1+2x}{x^2-2x+1}=1+\frac{2x}{(x-1)^2}$$ and its Maclaurin series is $$1+2x\sum_{n\geq 1} nx^{n-1}=1+2\sum_{n\geq 1} nx^n.$$