In trying to answer a recent MSE-question I came on the partial problem to determine the power series for the function $$ f(x) = \sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+x}}}} $$

I was not successful with developing this series myself but when I asked wolframalpha I got this $$ \small -(-1/2)^{3/4} x^{1/4} -{1 (-1/2)^{1/4} x^{3/4} \over 16} +{23 (-1/2)^{3/4} x^{5/4}\over 256} +{81 (-1/2)^{1/4} x^{7/4} \over 4096} -{5163 (-1/2)^{3/4} x^{9/4})\over 131072} -{23111 (-1/2)^{1/4} x^{11/4})\over 2097152} +{794707 (-1/2)^{3/4} x^{13/4}\over 33554432} +{3986865 (-1/2)^{1/4} x^{15/4} \over 536870912} -{563081651 (-1/2)^{3/4} x^{17/4}\over 34359738368} -{3029331403 (-1/2)^{1/4} x^{19/4} \over 549755813888} +{108017005065 (-1/2)^{3/4} x^{21/4} \over 8796093022208} +{610056649623 (-1/2)^{1/4} x^{23/4} \over 140737488355328} +O(x^{25/4}) $$

This expansion even has no constant term, so the fixpoint $x_0=0$ becomes immediately understandable.
(Remark: There are more fixpoints :$\small t\approx 0.222986+0.413364 i$ and $\small u\approx 0.222986-0.413364 i$ , see also the answer of G Edgar in that mentioned question. Using that complex-positive fixpoint $t$ we can get a power series without constant term - but with complex coefficients: $ \small f(x+t)-t = g(x) \approx 0.21947 x + (-0.10260 + 0.17758 i) x^2 + $ $\small (-0.13374 - 0.22046 i) x^3 + (0.37558 - 0.02179 i) x^4 + O(x^5)$ end remark)

How can such an expansion be found? When I used Carlemanmatrices to resolve the function-composition I ran into singularities, and as well when I tried to get the first derivative at zero I ran into the same singularities - so how is this been done?

  • $\begingroup$ @Mahdi: no, no duplicate: the question how one would determine the coefficients of a power series in such a difficult function was not discussed and not even mentioned in that question and its answers. It was only my idea, that the power series expansion which I wanted to find here might help to answer the other question - but I did not know how to do that until the answer of Matt E. $\endgroup$ – Gottfried Helms Aug 13 '13 at 7:52

How about just the binomial theorem?

This gives $\sqrt{1+x} = 1 + \dfrac{1}{2} x - \dfrac{1}{8} x^2 + \dfrac{1}{16}x^3+ \cdots.$ Thus $$\sqrt{1-\sqrt{1+x}} = \sqrt{ -\dfrac{1}{2} x + \dfrac{1}{8}x^2 - \dfrac{1}{16} x^3+ \cdots} $$ $$= \sqrt{-\dfrac{x}{2}}\sqrt{1 - \dfrac{1}{4}{x} + \dfrac{1}{8} x^2+ \cdots} =(-\frac{x}{2})^{1/2}(1 - \dfrac{1}{8} x + \dfrac{7}{128} x^2 + \cdots),$$ the last equality being another application of the binomial theorem.

Assuming the want a branch of $f(x)$ that is real-valued for at least some real values of $x$, we note that the above expression is real-valued when $-1 < x \leq 0$, and produces a positive answer. Since we are going to substitute this expression back into $\sqrt{1 - \sqrt{ 1 + x}}$, we should actually take the negative branch of the second square root.

Thus we substitute the negative of this expression into itself in place of $x$, and so compute that $$\sqrt{1 - \sqrt{1 + \sqrt{1 - \sqrt{1+x}}}} = \dfrac{(-x/2)^{1/4}(1 - x/8 + \cdots)^{1/2}}{2^{1/2}} ( 1 + \frac{1}{8}(-x/2)^{1/2} - \frac{7x}{256} + \cdots)$$ $$ = \dfrac{(-x)^{1/4}}{2^{3/4}}(1 -\frac{x}{16} + \cdots)(1 + \frac{(-x)^{1/2}}{2^{7/2}} -\frac{7x}{256} + \cdots) $$ $$ = \dfrac{(-x)^{1/4}}{2^{3/4}}(1 + \frac{(-x)^{1/2}}{2^{7/2}} - \frac{23}{256} x + \cdots).$$ (I have dropped some terms, since with the approximation I made at the beginning, the product of the parenthetical expressions is only valid up to second order in $(-x)^{1/2}$. If I had expanded more terms in the binomial theorem at the beginning, I would have had more terms in the final expression.)

This agrees with the first three terms of the posted answer, up to some signs having to do with the choice of branch. (If I had just substituted the first expression into itself, without changing the sign so as to get a real-valued function for $x \leq 0$, I think I would have gotten an expression that differs from WA's up to an overall sign.)

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  • $\begingroup$ Very nice, very well! Just recently I came across that method to extract the $x$-term as factor from a power series without constant term (I think it was in the fiddling with the "fractional/half-derivative"), but it seems now it didn't really arrive at my (mental) tools-box... I think I'll have to practice with this method a bit more. $\endgroup$ – Gottfried Helms Aug 13 '13 at 8:20

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