Represent $ f(x) = 1/x $ as a power series around $ x = 1 $

Question

As stated on the title, my question is: (a) represent the function $ f(x) = 1/x $ as a power series around $ x = 1 $. (b) represent the function $ f(x) = \ln (x) $ as a power series around $ x = 1 $.

Here's what I tried:

(a) We can rewrite $ 1/x $ as $ \frac{1}{1 - (1-x)} $ and thus using the series $ \frac{1}{1-k} = \sum_{n=0}^\infty k^n, |k| < 1 $, we can write that:

$ \frac{1}{x} = \frac{1}{1 - (1-x)} = \sum_{n=0}^\infty (1-x)^n, |1-x| < 1 $

I have a doubt because when I type "power series of 1/x when x = 1" on WolframAlpha the result is $ \sum_{n=1}^\infty (-1)^n \cdot (-1+x)^n $.

Am I wrong?

(b) Since $ (\ln (x))' = \frac{1}{x} $, all I have to do is integrate both sides of (a)' answer:

$ \int \frac{1}{x} dx = \int \sum_{n=0}^\infty (1-x)^n dx \therefore \ln(x) = \sum_{n=0}^\infty \frac{(1-x)^{n+1}}{n+1} + C $ and by putting $ x = 1 $ we get $ C =0 $ and thus $ \ln(x) = \sum_{n=0}^\infty \frac{(1-x)^{n+1}}{n+1} $.

Are my answers correct?

Really appreciate the help.

Have a good night, mates.

Answer 1

I don't have enough reputation to comment, so here I am.

a) The your answer and the one you get from Wolfram|Alpha are identical.

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

b) I think you have forgotten some minus signs.

$\int (1-x)^n dx = \frac{-(1-x)^{n+1}}{n+1} + C$.

Answer 2

$$ \begin{align} \frac1x &=\frac1{1+(x-1)}\\ &=1-(x-1)+(x-1)^2-(x-1)^3+\dots\\ &=\sum_{k=0}^\infty(-1)^k(x-1)^k \end{align} $$ (a) You are correct; your series is the same as mine, however, usually we expand in powers of $(x-a)^n$.

(b) integrating $\frac1t$ between $t=1$ and $t=x$ gives $$ \begin{align} \log(x) &=\sum_{k=0}^\infty\frac{(-1)^k}{k+1}(x-1)^{k+1}\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k(x-1)^k \end{align} $$

Answer 3

For the function $f:(0,\infty)\to(0,\infty)$ defined as:

$$f(x)=1/x$$

and given the general Taylor's expansion formula:

$$\sum_{k=0}^{\infty}\left(D^kf(a)\right){(x-a)^k\over k!}$$

First, calculate some of the derivaties:

$$D^0f(x)=1/x$$ $$D^1f(x)=-1/x^2$$ $$D^2f(x)=1\cdot2/x^3$$ $$D^3f(x)=-1\cdot2\cdot3/x^4$$ $$\vdots$$ $$D^kf(x)=(-1)^k\ k!/x^{k+1}$$

So the Taylor's expansion of $1/x$ around $a$ is:

$$\sum_{k=0}^{\infty}\left(D^kf(a)\right){(x-a)^k\over k!}=\sum_{k=0}^{\infty}\left((-1)^k\ k!\over a^{k+1}\right){(x-a)^k\over k!}$$

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

so for $a=1$, your answer is correc. And for the expansion of $\ln x$ should be:

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

but note that for the first term $k=0$, it's false that

$$\ln a=0 \ \forall a\neq 1 $$

while the other terms are true... so you need to shift the index $k$ by $1$:

$$\ln x=\ln a+\sum_{k=1}^{\infty}{(-1)^{k\pm 1}(x-a)^{k}\over a^{k}\ k}$$