In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$

which is

\begin{align*} \frac{1}{\sqrt{1+x^{x}}} =& \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}} -\frac{x^{2}\log^{2}(x)}{32\sqrt{2}}+ \frac{5x^{3}\log^{3}(x)}{384\sqrt{2}}\\\ \\\ &+ \frac{17x^{4}\log^{4}(x)}{6144\sqrt{2}} - \frac{121x^{5}\log^{5}(x)}{122880\sqrt{2}} - \frac{721x^{6}\log^{6}(x)}{2949120\sqrt{2}} \ldots \end{align*}

Before I realized that I couldn't really use this to help me with the sum, I found that the denominators (ignoring the $\sqrt{2}$, because all of them have it in common) correspond to $4^{n}n!$, what is baffling is that the numerators appear to correspond to the coefficients in the exponential generating function for $f(x)=e^{\tanh^{-1}(\tan(x))}$ (I believe that the 7th entry should be 1369 and not 6845), and I'm curious what the explanation is, because $f(x)=e^{\tanh^{-1}(\tan(x))}$ is a mighty weird looking function.

  • 1
    $\begingroup$ I don't get the question. $\endgroup$
    – Asaf Karagila
    Oct 16, 2010 at 8:56
  • 1
    $\begingroup$ @Asaf: The question is: Why does 1/√(1+x^x) have the power series it does? $\endgroup$ Oct 16, 2010 at 9:31
  • $\begingroup$ In SWP the 7th term is $+\dfrac{1369}{33\,030\,144}x^{7}\left( \ln^{7}x\right) \sqrt{2}=+\dfrac{1369}{33\,030\,144}x^{7}\left( \ln^{7}x\right) \dfrac{2}{\sqrt{2}}$ $\endgroup$ Oct 16, 2010 at 11:41
  • 4
    $\begingroup$ What does "w|α" mean? $\endgroup$ Oct 16, 2010 at 12:38
  • 5
    $\begingroup$ I just want to point out that this series expansion of the function $f$ is not a power series. In fact, $f$ is not differentiable at 0, so it doesn't have a power series expansion. $\endgroup$ Oct 17, 2010 at 9:21

2 Answers 2


Well, $$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$ where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$. Then $$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$ where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients are rationals over $\sqrt2$.

Now consider $f(x)=\exp(\tanh^{-1}(\tan x))$. Now $$\tanh z=\frac{e^{2z}-1}{e^{2z}+1}$$ so that $$e^z=\sqrt{\frac{1+\tanh z}{1-\tanh z}}.$$ Hence, $$\exp(\tanh^{-1} t)=\sqrt{\frac{1+t}{1-t}}.$$ Putting $t=\tan x$ gives $$f(x)=\sqrt{\frac{1+\tan x}{1-\tan x}}=\sum_{n=0}^\infty a_n x^n.$$ Then \begin{eqnarray*} &&(1+i)f(ix)+(1-i)f(-ix)\\ &=& 2\sum_{m=0}^\infty (a_{4m} x^{4m}-a_{4m+1} x^{4m+1}-a_{4m+2} x^{4m+2}+a_{4m+3} x^{4m+3}). \end{eqnarray*} But $$f(ix)=\sqrt{\frac{1+i\tanh x}{1-i\tanh x}} =\frac{1+i\tanh x}{\sqrt{1+\tanh^2x}}$$ and so \begin{eqnarray*} &&(1+i)f(ix)+(1-i)f(-ix)\\ &=& \frac{2-2\tanh x}{\sqrt{1+\tanh^2 x}} =\frac{2\cosh x-2\sinh x}{\sqrt{\cosh^2x+\sinh^2x}}\\ &=&\frac{2e^{-x}}{\sqrt{(e^{2x}+e^{-2x})/2}}=\frac{2\sqrt2}{\sqrt{1+e^{4x}}} \end{eqnarray*} which explains why the coefficients in the two series are the same up to signs and powers of 4.

  • 1
    $\begingroup$ thank you! this is what I was looking for. $\endgroup$ Oct 16, 2010 at 13:26

It's simple: $\rm\quad\quad f(x)\ =\ \cos x - \sin x \ =\ \frac{1+i}2 e^{ix} + \frac{1-i}2 e^{-ix}$

$\rm\quad\displaystyle \Rightarrow\quad\quad\quad\quad\ f(\tan^{-1} x)\ =\ \frac{1-x}{\sqrt{x^2+1}}\quad $ via $\rm\displaystyle\quad e^{\:i\:tan^{-1} x}\ =\ \frac{1+ x\: i}{\sqrt{x^2 + 1}}$

$\rm\quad\displaystyle \Rightarrow\quad f(\tan^{-1}\tanh x)\ =\ \ \frac{\sqrt 2}{e^{4x}+1}\quad\ $ via $\rm\displaystyle\quad\ \tanh{x}\ =\ 1 - \frac{2}{e^{2x}+1}$

  • $\begingroup$ Why is f(x) = cos x - sin x? $\endgroup$ Oct 19, 2010 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.