Power and Taylor series

Question

1) Can someone explain how to use differentiation to find the power series representation for functions such as:

A) $1/(4+x)^2$

B) Moreover, would this have a pattern if I had $1/(4+x)^3$ or $x^2/(4+x)^3$. I only understand how to find the radius of convergence.

2) How can we convert indefinite integrals to power series such as $(x^5)\ln(1+x)$.

I understand that this may require differentiation but I am not too sure how to go ahead with this problem. Also, I understand how to find the radius of convergence, which I need to solve but I am stuck on the initial question itself.

3) Similarly, computing definite integrals such as: the integral from $0$ to $0.3$ of $x^2(1+x^6)$ ? How would this use power series?

4) $y = 1/3(x^{1.5})$ and $0 \le x \le 12$ , is rotated about y-axis . Find the area. I know this would be the integral for surface area . Any idea how to do this ?

Answer 1

For part A, the Taylor series expansion for a function around $0$ is $$\sum_{n=0}^{\infty}a_nx^n$$ where $$a_n = \frac{f^{(n)}(0)}{n!}$$ The term $f^{(n)}$ denotes the $nth$ derivative of $f$.

Answer 2

(1A) The trick: $1/(4+x)^2$ is "almost" the derivative of $1/(4+x)$ and $1/(4+x)$ is the sum of a geometric series for |x|<4 (a similar formula is valid for $|x|>4$): $$\frac1{4+x} = \frac14\frac1{1+x/4} = \frac14\sum_{n=0}^\infty\frac{(-1)^n}{4^n}x^n$$ $$\frac1{(4+x)^2} = -\frac{d}{dx}\frac1{4+x} = -\frac14\sum_{n=1}^\infty\frac{(-1)^nn}{4^n}x^{n-1} = \cdots$$ (1B) $1/(4+x)^3$ is "almost" the second derivative of $1/(4+x)$ and $x^2/(4+x)^3$ is simply $x^2\cdot($"almost" the second derivative of $1/(4+x))$. The generalization is obvious.

(2) Write the integrand as a series and integrate term to term.

(3) Idem, but your example is a polynomial

(4) Totally unrelated. Open another question for this.

Answer 3

A power series for $f(x)$ can be differentiated term-by-term to get $f'(x)$ for any $x$ in the interior of the circle of convergence. With $f(x)=-1/(4+x)$ and $|x|<4$ we have $$f(x)=-\frac {1}{4}(1-(x/4)+(x/4)^2-(x/4)^3+...),$$ $$ \text {Therefore }\quad 1/(4+x)^2=f'(x)=\frac {1}{16} (1-2(x/4)+3(x/4)^2-4(x/4)^3+...).$$

Justification: Let $R>0$ and let $g_n(x)=\sum_{j=0}^nA_j(x-x_0)^j$ for $|x-x_0|<R.$

If $g_n$ converges to $g(x)$ whenever $|x-x_0|<R,$ with $R>0$,then $g_n$ converges uniformly to $g(x)$ on any disc centered at $x_0$ of any radius $S<R.$ Therefore $g$ is continuous, and $$f_n(x)=\int_{t=0}^1g_n(x_0(1-t)+xt)\;dt$$ converges to $$f(x)=\int_{t=0}^1(g(x_0(1-t)+xt)\;dt $$ whenever $|x-x_0|<R.$ And $f$ is an anti-derivative of $g$.

So, given $f,$ let $g=f'.$ Applying the above argument, we see that we can differentiate the series for $f$ term-by-term to get the series for $g.$

Remark: I did not write $\;f_n(x)=\int_{x_0}^x g_n(y)\;dy\;$ nor $\;f(x)=\int_{x_0}^x g(y)\;dy\; $ so that the above argument holds in the complex plane, not just in the reals.