What is the meaning of the phrase "Power series expansion about $x = 0$"?

Question

I recently came across a problem that contained a phrase stating

Power series expansion about $x = 0$ of $(1 + x)^a$

What do "about $x = 0$" and "of $(1 + x)^a$" mean ?

Is it same as "at $x = 0$"?

If anybody could help it would really be appreciated.

Answer 1

It means write $$(1+x)^a = \sum_{i} c_i(x-b)^i$$

It is referring to the case when $b=0$.

Answer 2

Yes, "Power series expansion about $x = 0$" means the same as "Power series expansion at $x = 0$". I think "about" is a bit more accurate, since you're using derivatives which by definition care about what the function looks like not just at $x = 0$ but also in the immediate vicinity.

"of $(1+x)^a$" means that that's the function whose power series expansion you are asked to find.

Answer 3

In general, "a power series expansion about $x=\lambda$" means an infinite series in the form of $\sum_{n=0}^\infty c_n(x-\lambda)^n$. So, a power series expansion about $x=0$ means an infinite series of the form $\sum_{n=0}^\infty c_nx^n$.

Answer 4

This is known as a Taylor Series, the idea behind it is to write the function $(1 + x)^a$ as an (possibly) infinite sum of terms of them $c_k (x - x_0)^k$, $k=0,1,\cdots$ and in your case $x_0 = 0$ (that's the meaning of "around $x_0 = 0$")

$$ (1 + x)^a = \sum_{k=0}^{\infty} c_k x^k $$

The coefficients $c_k$ are calculated with the derivatives of the function

$$ c_k = \frac{1}{k!}\left.\frac{{\rm d}^k}{{\rm d}x^k}(1 + x)^a\right|_{x=x_0=0} $$

The idea is that the more terms you include in this sum, the closer it gets to the actual function

You can see from this figure that the approximation becomes increasingly better around $x=0$ with the number of terms added to the sum