Question

Use differentiation to find a power series representation for

f(x) = 1/(2 + x)^2

sum n=0 to infinity (-1)^n (n+2)x^n is that correct ??
.

Answer 1

Using
$$
f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.
$$
where \(a=0\):
$$
f(x)=1/4-x/4+(3 x^2)/16-x^3/8+(5 x^4)/64-(3 x^5)/64+O(x^6)
$$

Answer 2

what is the power series representation for sum n=0 to infinity________________ ?????

Answer 3

\[f(x)=\frac{1}{(2+x)^2}~~\Rightarrow~~\int f(x)~dx=-\frac{1}{2+x}+C\]
The power series of the antiderivative function is a geometric series:
\[\begin{align*}-\frac{1}{2+x}+C&=-\frac{\frac{1}{2}}{1-\left(-\frac{1}{2}x\right)}+C\\
&=-\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x}{2}\right)^n+C\end{align*}\]
Now differentiate:
\[\begin{align*}\frac{d}{dx}\left[\int f(x)~dx\right]&=\frac{d}{dx}\left[-\frac{1}{2}\sum_{n=0}^\infty \left(-\frac{x}{2}\right)^n+C\right]\\
f(x)&=-\frac{1}{2}\cdot\frac{d}{dx}\left[\sum_{n=0}^\infty \left(-\frac{x}{2}\right)^n\right]
\end{align*}\]
etc.