I am currently having difficulty in computing the MacLaurin series for the function $f(x) = (1+\frac{1}{x})^x$.

Please note: I am aware that this can be solved by using the binomial series expansion, and that by allowing $n$ to tend to infinity will produce the series expansion for $e$.

My primary concern/query is as to why I cannot produce this series using the MacLaurin method?

UPDATE: the first derivative to this function: ((x^-1+1)^x)*(ln(x^-1+1)−1/(x^-1+1)x) tends to infinity when x is set to zero (undefined). Would it then be appropriate to assume this function cannot be solved the Maclaurin way?

Note: I am aware that a Maclaurin series is simply a Taylor series approximated from 0

  • $\begingroup$ Typically $n$ is a natural number. Maclaurin series are for functions of real numbers. $\endgroup$
    – user223391
    Aug 10 '17 at 12:59
  • $\begingroup$ MacLaurin around $x=?$ $\endgroup$
    – G Cab
    Aug 10 '17 at 13:22
  • $\begingroup$ @GCab Maclaurin = Taylor around zero. $\endgroup$
    – GEdgar
    Aug 10 '17 at 13:27

You need to define the function in the complex plane, e.g. through: $$ f(x) = \exp \left( x \log\left( 1 + \frac{1}{x}\right) \right) .$$ Here, the tricky part is the log for which you need to specify a branch cut. For example, you may say that $\log(z)$ is defined for $z\in {\Bbb C}\setminus {\Bbb R}_-$, so that $f$ becomes defined for $x\notin [-1,0]$. You may choose other branch cuts but for $f$ you will always get some branch cut between $0$ and $-1$. Now $f(x)$ tends to $e$ as $|x|\rightarrow \infty$ so indeed $f$ does have a Laurent expansion which you may find by expansion of $\log$: $$ f(x) = \exp ( x \sum_{k\geq 1} \frac{(-1)^{k-1}x^{-k}}{k} ) = e \cdot \exp \sum_{p\geq 1} \frac{(-x)^{-p}}{p+1} = e \cdot \exp \left(-\frac{1}{2x} + \frac{1}{3x^2} \pm ...\right) $$ which you may then expand further... It converges for $|x|>1$.


A Maclaurin series is defined for a function only if it (and all its derivatives) are defined at zero... So what about $(1+\frac{1}{x})^x$ ??? for negative irrational $x$, it must be complex, right? So your function is not differentiable at zero.

  • $\begingroup$ It might however have a Laurent series. Does it? $\endgroup$
    – md2perpe
    Aug 10 '17 at 14:47
  • $\begingroup$ the first derivative term for this function: ((1+1/x)^x)*(ln(1+1/x) - (1/x)/(1/x +1)) When x is set to zero, this term tends to infinity so it is not defined as you have stated. I've yet to to try a Laurent series expansion, although I'm afraid I've never heard of this type of series until now. $\endgroup$ Aug 10 '17 at 18:39

I don't think that we can expand $(1+\frac1x)^x$ around $x=0$. We can however expand it around $x=\infty$. Setting $t = 1/x$ we have the expression $(1+t)^{1/t}$ that can be expanded using the generalized binomial formula: $$(x+y)^r = x^r + rx^{r-1}y + \frac{r(r-1)}{2}x^{r-2}y^2 + \frac{r(r-1)(r-2)}{6}x^{r-3}y^3+\cdots$$ Taking $x=1$, $y=t$ and $r=1/t$ we get $$ (1+t)^{1/t} = 1 + \frac1t t + \frac12 \frac1t \left(\frac1t-1\right) t^2 + \frac16 \frac1t\left(\frac1t-1\right)\left(\frac1t-2\right) t^3 + \cdots \\ = 1 + 1 + \frac12(1-t) + \frac16(1-t)(1-2t) + \cdots $$ The constant term is $1+1+\frac12+\frac16+\cdots=e$ as we could expect from $\lim_{t \to 0} (1+t)^{1/t} = e$.

Perhaps someone else will calculate the other terms.


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