calculate the roots of $z = 1 + z^{1/2}$ using Lagrange expansion

Question

I am trying to find the roots of equation $z = 1 + \sqrt{z}$ using Lagrange's expansion given on this book page no 15.

I expanded $z = 1 + \zeta z^p$ as $$z = 1 + \zeta + \frac{2p}{2!}\zeta^2 + \frac{3p(3p-1)}{3!}\zeta^3+\frac{4p(4p-1)(4p-2)}{4!}\zeta^4 + \dots$$ putting $\zeta = 1$ and $p = 1/2$ I obtained the series $$z = 1 + 1 + \frac{1}{2} + \frac18 + \frac{-1}{128} + 0 + \dots$$

The series seems to have formula $\displaystyle 2 + \sum_{n=2}^\infty \frac 1{n!}\prod_{k=0}^{n-2} (n \cdot \frac12 - k)$ which I found it to be $2 + \frac12(1 + \sqrt 5)$ from Mathematica while the root of $z = 1 + \sqrt z$ is $1 + \frac12(1 + \sqrt 5)$. Can anyone help me where it went wrong? Also can anyone help me to evaluate that series with hand?

Answer 1

You can solve the equation explicitly by squaring both sides then solving the resulting quadratic. Compare the root you find with sum of the first terms of the series you gave.

In more detail.

We have

$$ z - 1 = \sqrt{z} \quad \Longrightarrow \quad z^2 - 2z + 1 = z, \tag{1} $$

so that, by the quadratic formula, the two possible roots are

$$ z = \frac{3 \pm \sqrt{5}}{2}. $$

The smaller candidate root is not a solution to the original equation since

$$ \frac{3 - \sqrt{5}}{2} < \frac{3 - \sqrt{4}}{2} = \frac{1}{2}, $$

which would make the left-hand side negative and the right-hand side positive in the first equation in $(1)$. Thus we conclude that the only possible root is

$$ z = \frac{3 + \sqrt{5}}{2} \approx 2.61803. $$

The sum of the first terms of the series you found is approximately

$$ 1 + 1 + \frac{1}{2} + \frac18 - \frac{1}{128} \approx 2.61719, $$

so it looks like you've got everything right.

Evaluating the series directly.

If you'd like, you can get the closed form for your series the hard way by noticing that

$$ \prod_{k=0}^{n-2} (np - k) = \frac{(np)!}{(np-n+1)!}, $$

for $n \geq 2$, so that the general term in your sum is

$$ \frac{(np)!}{n!(np-n+1)!} \zeta^n = \frac{1}{n} \binom{np}{n-1} \zeta^n \tag{2} $$

Let's take $p=1/2$ now and consider only the odd terms, so that $n = 2m+1$. The expression in $(2)$ becomes

$$ \frac{1}{2m+1} \binom{m+1/2}{2m} \zeta^{2m+1}. $$

Thanks to Sasha's answer here this is equal to

$$ \frac{1}{4^m} \binom{1/2}{m} \zeta^{2m+1} = \zeta \binom{1/2}{m} \left(\frac{\zeta^2}{4}\right)^m, $$

so that the sum of the odd terms in your series is

$$ \zeta \sum_{m=0}^{\infty} \binom{1/2}{m} \left(\frac{\zeta^2}{4}\right)^m = \zeta \sqrt{1+\zeta^2/4} = \frac{\zeta}{2}\sqrt{4+\zeta^2}. \tag{3} $$

Next we consider the even terms. By taking $n = 2m$ in $(2)$ we see that the general even term is

$$ \frac{1}{2m} \binom{m}{2m-1} \zeta^{2m}. $$

For all $m > 1$ the quantity $2m-1$ is a negative integer, and in that case $\binom{m}{2m-1} = 0$. Thus the sum of the even terms is just

$$ 1 + \frac{1}{2} \binom{1}{1} \zeta^2 = 1 + \frac{1}{2} \zeta^2. \tag{4} $$

Summing $(3)$ and $(4)$ we find that

$$ \begin{align} z &= 1 + \zeta + \frac{2p}{2!}\zeta^2 + \frac{3p(3p-1)}{3!}\zeta^3+\frac{4p(4p-1)(4p-2)}{4!}\zeta^4 + \dots \\ &= 1 + \frac{1}{2} \zeta^2 + \frac{\zeta}{2}\sqrt{4+\zeta^2}. \end{align} $$

By taking $\zeta = 1$ we see that this agrees with the result we found earlier, that $z = (3+\sqrt{5})/2$.