# Series expansion of ##(1-cx)^{1/x}##

• A
• Arman777
In summary, the wolfram alpha seems to solve the problem by using taylor series for ##x\rightarrow 0## and Puiseux series for ##x\rightarrow \infty##.

#### Arman777

Gold Member
I am trying to understand the series expansion of $$(1-cx)^{1/x}$$ The wolframalpha seems to solve the problem by using taylor series for ## x\rightarrow 0## and Puiseux series for ##x\rightarrow \infty##. Any ideas how can I calculate them ?

https://www.wolframalpha.com/input?i=(1-cx)^(1/x)+series

Here is the link of the problem. Or anyone can provide detailed explanation about how Puiseux series works ( I am not looking for Wiki page info but more of an example type info).

You can start rewriting the function ##f(x)=(1-cx)^{\frac{1}{x}}## as ##f(x)=e^{\frac{\log{(1-cx)}}{x}}## that is equivalent to your function because exponential is the inverse of logarithm (and using a property of log). Now you can calculate the Taylor expansion by the formula:

## f(x)= f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^2+\frac{f'''(x_{0})}{3!}(x-x_{0})^3 + ... ##

where ##f',f'',f''', ...## are the derivatives in ##x_{0}## (in your case ##x_{0}=0##). You will find the Taylor expansion (or better Mc Laurin expansion)...

The Puiseux series here is obtained from the expression of ##f(x)=e^{\frac{\log{(1-cx)}}{x}}## considering only the term ##-cx## as ##x\rightarrow +\infty## and not the number ##1## (doesn't contribute to infinity) and after, taylor expanding (formally) the exponential function ##e^{\frac{\log{(-cx)}}{x}}## so:

##e^{\frac{\log{(1-cx)}}{x}}\sim e^{\frac{\log{(-cx)}}{x}} \sim 1+ \frac{\log{(-cx)}}{x} + \frac{1}{2!}\left(\frac{\log{(-cx)}}{x}\right)^2 + ...##

for ##x\rightarrow +\infty##.
Ssnow

Arman777
$f(x) = (1 - cx)^{1/x}$ is of the form $e^{g(x)}$ so $$f'(x) = g'(x) f(x)$$ and higher derivatives can be calculated recursively using the product rule: $$f^{(n)}(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} g^{(k+1)}(x) f^{(n-1-k)}(x).$$

Arman777
Given that you wrote ##x## and not ##z##, I'm not sure if you even think the function is defined at infinity, since you are computing arbitrary roots of a negative number.

The easiest way is to use series you already know, and write the function in terms of something that is close to 0 when it's close to your target.

For example when x goes to zero,
##(e^{\log(1-cx)/x} \approx e^{(-cx-c^2x^2/2-c^3x^3/3)/x}## using the first couple terms of the taylor series of ##log(1-x)##.

##=e^{-c-c^2x/2-c^3x^2/3}=e^{-c} e^{-(c^2x/2+c^3x^2/3}##
Using the Taylor series for ##e^{-x}##
##\approx e^{-c} (1-(c^2x/2+c^3x^2/3) + (
c^2x/2+c^3x^2/3)^2/2##

Now at this point you wish you could expand everything out and have the first bunch of terms. But remember, we truncated our Taylor series early at the start, so that ##c^2x/2+c^3x^2/3## should continue next with cubic term. Because it's missing, I just can't know the cubic term of the approximation. I did write enough down to get the quadratic term, since I know the next term in ##c^2x/2+c^3x^2/3)## is cubic so doesn't contribute, and the final ##e^{-x}## expansion is only missing powers of 3 or higher applied to terms that all have a ##x## at least in them. So I should square that last summary, but only keep the quadratic part of it.

So the final approximation i get is
##\approx e^{-c}(1-c^2x/2-c^3x^2/3+c^4x^2/8)##

Which with some Algebra does match wolfram alpha.

To be a little more rigorous, youshould write ##+o(x^3)## in lots of places in my solution. You can try to include the ##x^3## term as an exercise if you want.

Sssnow did the same thing for infinity, but since you keep the log, it is a lot simpler.

Arman777