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As a specific example of this question and a follow up to this one does anyone know a nice way to calculate the germs at $z=1$ of

$$f(z)=\sqrt{1+\sqrt{z}}$$

My attempts have been messy at best, and I'd rather avoid trying to wade through Taylor series if I can! Any ideas would be most welcome!

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  • $\begingroup$ What exactly do you mean by "calculating the germ"? Do you want the power series of $f$ around $z= 1$ or are you asking for something else? $\endgroup$
    – mrf
    Jun 1, 2012 at 16:09
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    $\begingroup$ Of course, the point is that there's more than one power series, because this is a multi-valued "function"... $\endgroup$
    – Zhen Lin
    Jun 1, 2012 at 16:43

3 Answers 3

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I'm not sure what you're looking for since you asked two distinct questions in the other posts, so I'll answer both.

Power series of the principal branch

For all $|t|<1$: $$\left(\sqrt{1+t}+\sqrt{1-t}\right)^2=2\left(1+\sqrt{1-t^2}\right)$$ $$\frac{\sqrt{1+t}+\sqrt{1-t}}{2}=\frac{1}{\sqrt 2}\sqrt{1+\sqrt{1-t^2}}$$ The left hand side is simply the even part of $\sqrt{1-t}$, thus if $\sqrt{1-t}=\sum_{n=0}^\infty a_n t^n$ we have: $$\frac{1}{\sqrt 2}\sqrt{1+\sqrt{1-t^2}}=\sum_{n=0}^\infty a_{2n} t^{2n}$$ $$\sqrt{1+\sqrt z}=\sum_{n=0}^\infty \sqrt 2 a_{2n} (1-z)^n$$ (Note: $a_0=1$ and $a_n=-1/2 \frac{(2n-2)!}{n!(n-1)!4^{n-1}}$ for $n>0$.)

Call this germ $g_0$.

Other germs

If we interpret the square root as a multi-valued function, there are in principle 3 other branches obtained by flipping the sign of one or both radicals. But $\sqrt{1-\sqrt{z}}~\sim \sqrt{(1-z)/2}$ is not analytic around $z=1$, so around $z=1$ there are only two (distinct) germs $g_0$ and $-g_0$.

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  • $\begingroup$ Thank you - this is just what I was looking for! Am I right in thinking that when $z \neq 1$ we do get $4$ germs? If I'm right, what's the best way to rigorously show it? Once I understand that I shall certainly award the bounty! Many thanks again! $\endgroup$ Jun 2, 2012 at 15:52
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    $\begingroup$ Around $z=z_0\not\in\{0,1\}$, $1\pm\sqrt z$ is analytic and non-zero in a neighborhood of $z_0$, so $\pm\sqrt{1\pm\sqrt z}$ is analytic around $z_0$. You can trivially prove these 4 germs are distinct by looking at the value at $z=z_0$. $\endgroup$ Jun 3, 2012 at 9:10
  • $\begingroup$ Ah excellent - I understand now! Many thanks! $\endgroup$ Jun 3, 2012 at 10:03
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We have $$ \sqrt{1+\sqrt{z}} = \sqrt{2}\left(1+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1} (z-1)^n\right) $$

Here is a very general method, in the particular case of an algebraic function, to prove the identity.

I — Notations

Let $f$ be the function $z \mapsto \sqrt{1+\sqrt{1+z}}/\sqrt 2$, which is holomorphic in a neighbourhood of zero. We take a determination of the square root holomorphic around 1 and such that $\sqrt 1 = 1$. Note that in order to simply the computation, I shifted the variable and normalized the value at zero.

II — Algebraic equation

The function $f$ obviously satisfies the algebraic equation $$ (f(z)^2 - 1/2)^2 = \frac{1+z}{4}, $$ or, equivalently, $P(f(z), z) = 0$, where $$ P(Y, z) = 4Y^4 - 4Y^2 -z. $$

III — Differential equation

The function $f$ satisfies the following linear differential equation : $$ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 0 $$

This is a general fact that an algebraic function satisfies a linear differential equation with polynomial coefficient.

IV — Recurrence

Write $f(z) = \sum_{n\geqslant 0} u_n z^n$. Thus $$ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 16n(n-1)u_n + 16 (n+1)n u_{n+1} + 8 (n+1) u_{n+1} + 16 n u_n - u_n $$ which implies that $$ (4 n - 1)(4n+1)u_n + 8(n+1)(2n+1)u_{n+1} = 0. $$

V — Resolution

We check easily that the sequence defined by $$v_n = \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1}$$ if $n>0$ and $v_0 = 1$ satisfies the first order recurrence above. Since $u_0 = 1 = v_0$, we can conclude that $$u_n = v_n$$

VI — Automation

Here is a Maple session showing how to automate the proof of steps II to V.

> with(gfun):
> f := sqrt(1+sqrt(1+z))/sqrt(2);
                                                              1/2 1/2  1/2
                                                  (1 + (1 + z)   )    2
                                             f := ------------------------
                                                             2

> holexprtodiffeq(f, y(z));      
                                                                         / 2      \
                                              /d      \        2         |d       |
                          {-y(z) + (8 + 16 z) |-- y(z)| + (16 z  + 16 z) |--- y(z)|, y(0) = 1}
                                              \dz     /                  |  2     |
                                                                         \dz      /

> diffeqtorec(%, y(z), u(n));    
                                         2                         2
                              {(-1 + 16 n ) u(n) + (24 n + 8 + 16 n ) u(n + 1), u(0) = 1}
> rsolve(%, u(n));
                                                                        n
                                                   GAMMA(2 n - 1/2) (-1)
                                              -1/2 ----------------------
                                                      1/2
                                                    Pi    GAMMA(2 n + 1)

You have to use the duplication formula for $\Gamma$ to retrieve the binomial coefficient.

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  • $\begingroup$ Great answer! I think there's a small mistake in your final expression since you get $1/8^n$ instead of $1/16^n$. $\endgroup$ Jun 2, 2012 at 8:39
  • $\begingroup$ You're right, thanks ! $\endgroup$
    – Lierre
    Jun 2, 2012 at 8:45
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Let $w = \sqrt{1 + \sqrt{z}}$. Since taking square roots of complex numbers is unpleasant, let's get rid of them to get a nicer equation: $$(w^2 - 1)^2 = z$$ Since we want to expand around $z = 1$, we should change variables: set $u = z - 1$. Our problem now is to find $a_0, a_1, a_2, \ldots$ so that $w = a_0 + a_1 u + a_2 u^2 + \cdots$ satisfies $$(w^2 - 1)^2 = 1 + u$$

First, by neglecting terms of order $u$ or higher, we find that $({a_0}^2 - 1)^2 = 1$, so $$a_0 \in \{ \pm \sqrt{2}, 0 \}$$ (Officially, what we are doing is working in the power series ring $\mathbb{C} [[ u ]]$ modulo $u$.) Now, we divide into cases.

  1. We ignore the case $a_0 = 0$, because this corresponds to a ramification point in the associated Riemann surface and so $w$ has no power series here. (If you try to do the computation by brute force, you will find that all the coefficients are $0$!)

  2. Take $a_0 = \sqrt{2}$. Neglecting terms of order $u^2$ or higher, we get $$(w^2 - 1)^2 = ((\sqrt{2} + a_1 u)^2 - 1)^2 = (1 + 2 \sqrt{2} a_1 u)^2 = 1 + 4 \sqrt{2} a_1 u = 1 + u \pmod{u^2}$$ so $a_1 = \frac{1}{8} \sqrt{2}$. Now, neglecting terms of order $u^3$ or higher, we get $$(w^2 - 1)^2 = 1 + u + \left( \frac{5}{16} + 4 \sqrt{2} a_2 \right) u^2 = 1 + u \pmod{u^3}$$ so $a_2 = - \frac{5}{128} \sqrt{2}$. And so on.

  3. Take $a_0 = -\sqrt{2}$. Observe that nothing changes if we substitute $-w$ for $w$, so the coefficients for this expansion must be the same as the ones computed in (2), except with the opposite sign.

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