# Power series representation of $f(x) = \frac{1}{x+2}$

In attempting to find the power series representation of $$f(x)$$, using the fact:

$$\frac{1}{1 - t} = \sum_{n=0}^{\infty}{t^n}$$

I simply set $$t = -x - 1$$, which when substituting into the above formula gives $$f(x)$$. Therefore, I presumed that the power series representation of $$f(x)$$ is $$\sum_{n=0}^{\infty}{(-x - 1)^n} = \sum_{n=0}^{\infty}{(-1)^n(x + 1)^n}$$.

But apparently this is wrong, and have seen a more complicated derivation, that also seems correct, that gives $$\sum_{n = 0}^{\infty}{\frac{(-1)^n}{2^{n+1}}x^n}$$

Can anyone provide any insight into why my more simplistic derivation is wrong?

EDIT: To clarify, I would like to point that I understand that the second derivation is correct and I understand how it is derived. What I want to know is why my derivation is wrong.

• As stated, your problem doesn't make sense. You're after the power series representation of $f$ centered at which point? – José Carlos Santos Jan 23 at 8:39
• @JoséCarlosSantos I'm sorry, I don't understand what you mean? – esotechnica Jan 23 at 8:42
• A power series is a series of the type $\sum_{n=0}^\infty a_n(x-a)^n$. What is your $a$? – José Carlos Santos Jan 23 at 8:44
• What makes you think that your result is wrong? Your result yields a power series representation, but not centered a 0, but centered at $-1$ $(x+1=x-(-1))$. – Severin Schraven Jan 23 at 8:46
• I'm completely confused, the question is simply 'Find the power series representation for $\frac{1}{x+2}$'. Maybe I'm not understanding the idea properly – esotechnica Jan 23 at 8:54

It simply gives the power series developed about a different point. The series

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}} x^n$$

is developed about the point $$x = 0$$, while

$$\sum_{n=0}^{\infty} (-1)^n (x + 1)^n$$

is developed about the point $$x = -1$$. Typically, one "likes" to have a series expansion of the first type since simple powers are easier to deal with than powers involving an addition, but there is no need for the first kind of expansion and moreover, they may not exist in many cases due to singular behavior at the origin, e.g. the function $$x \mapsto \frac{1}{x}$$ directly, or $$\log$$. If you were specifically asked for the series to be developed at $$x = 0$$ then the answer would be wrong. If no development point is advanced, and yet this is considered wrong, then the problem wasn't stated well to begin with.

ADD: from the comments, the meaning of "development point" is because the general power series has the form

$$\sum_{n=0}^{\infty} a_n (x - c)^n$$

The point "$$c$$" is called the "center". The reason for calling it such is two-fold: one reason is because such series, when they converge, do so within a circle (most literally in the complex plane, but on the real numbers, an interval constitutes a 1-d "circle") about this point. Another reason is that this point is the one of maximum rate of convergence (namely, infinitely fast, since at $$x = 0$$ it "instantly" converges to the value of $$a_0$$).

• It seems this book has let me down in terms of giving a proper understanding. I didn't know that functions had multiple power series representations. So what is the meaning behind 'developed' or 'centred'? In other words, what is the importance or usefulness in deciding/choosing this value? – esotechnica Jan 23 at 9:11
• @esotechnica May I ask which book it is? – The_Sympathizer Jan 23 at 9:16
• Stewart's Calculus 8th Ed - I realise it's not the most rigorous of books, but it's what I've got. It's Example 11.9.2 – esotechnica Jan 23 at 9:19

The function doesn't have a unique series representation; you found the one centered at $$-1$$ and the formula is correct.

If you were asked for the representation around $$0$$, then you gave the wrong answer. The phrase “power series” without qualifications usually means a series of the form $$\sum_{n}a_nx^n$$ which yours isn't.

• OK, so a power series without qualification means 'power series centred at zero'? – esotechnica Jan 23 at 9:03
• @esotechnica I'd say so, but it's a convention that you should check in your textbook. – egreg Jan 23 at 9:39

There could be three cases and one can use infinite geometric series: $$(1+z)^{-1}=1-z+z^2-z^3+\dots+(-z)^k+\cdots,$$ if $$|z|<1.$$

1. When $$-2, $$f(x) =\frac{1}{2} \left(1+\frac{x}{2}\right)^{-1}= =\sum_{k=0}^{\infty} \frac{(-x)^k}{2^{k+1}};$$
2. When $$x<-2$$ or $$x>2$$: $$f(x)=\frac{1}{x} \left(1+\frac{2}{x}\right)^{-1}= \sum_{0}^{\infty} \frac{(-2)^k}{x^{n+1}};$$
3. When $$a (e.g., when $$a=2$$, and $$b=3$$) $$f(x)=\frac{1}{4+y}=\frac{1}{4} \left(1+\frac{y}{4} \right)^{-1} = \sum_{k=0}^{\infty} \frac{(-y)^k}{4^{k+1}},~ y=(x-2).$$