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.
$\endgroup$ 22 Answers
$\begingroup$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.
$\int (1-x)^n dx = \frac{-(1-x)^{n+1}}{n+1} + C$.
$\endgroup$ 1 $\begingroup$$$ \begin{align} \frac1x &=\frac1{1+(x-1)}\\ &=1-(x-1)+(x-1)^2-(x-1)^3+\dots\\ &=\sum_{k=0}^\infty(-1)^k(x-1)^k \end{align} $$ (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 $$ \begin{align} \log(x) &=\sum_{k=0}^\infty\frac{(-1)^k}{k+1}(x-1)^{k+1}\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k(x-1)^k \end{align} $$
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