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Consider the equation $\mathcal{F}(\lambda)=0\ \ \ \forall\ \lambda = \lambda_{n},\ n \in \mathbb{N}$.

I understand that the expression $\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}$ has poles of order 1 exactly at $\lambda_{n}$ because $\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'(\lambda)}}{(\lambda- \lambda_{1})...(\lambda-\lambda_{n})}$.

I wonder how I might expand the expression $\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}$ about $\lambda_{n}$ to find out that the residue of the poles is 1.

Any ideas?

Edit to the post with an alternative answer:

I've found out that, expanding about $\lambda = \lambda_{n}$, with $\mathcal{F'}(\lambda_{n})\neq0$, we obtain

$\frac{\mathcal{F'}(\lambda)}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})}{\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})} = \frac{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...}{(\lambda-\lambda_{n})\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})^{2}\mathcal{F''}(\lambda_{n})+...} = \frac{1}{\lambda - \lambda_{n}}+...$,

so that the residue at all eigenvalues is 1.

I took this evaluation out from a paper, so I am not really sure about a couple of things I have written - have we used the Taylor series to expand each of $\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})$ and $\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})$ about $\lambda=\lambda_{n}$? I know this is a silly question, as you've already used the Weierstrass factorisation theorem, but I still want to clarify things up.

nightmarish
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  • If we did use Taylor's series, I wonder why the factorials are missing from each coefficient. :( Also, the ellipsis after $\frac{1}{\lambda-\lambda_{n}}$ suggests that there are other terms afterwards. What might those terms be? – nightmarish Aug 18 '15 at 12:14
  • That is a valid point. There should be $f''/2!$ in the denominator, but that does not affect the solution since this and higher order terms vanish at $z=\lambda$ where you are calculating the residue. I'll add some lines on this type of proof to the answer. – Winther Aug 18 '15 at 12:23
  • Thanks for the edit to your answer! :) I've got one other query. To me, it seems that $\frac{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...}{(\lambda-\lambda_{n})\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})^{2} \mathcal{F''}(\lambda_{n})+...} = \frac{1}{\lambda-\lambda_{n}} \frac{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...}{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...} = \frac{1}{\lambda-\lambda_{n}}$. Does that mean $\frac{g'(z)}{g(z)}$ in your post equals 0? – nightmarish Aug 18 '15 at 13:13
  • No. In this language $g(z) = f'(\lambda) + \frac{f''(\lambda)}{2!}(z-\lambda) + \frac{f'''(\lambda)}{3!}(z-\lambda)^2 + \ldots$. Now $g(\lambda)\not=0$ so the last term $\frac{g'(z)}{g(z)}$ is finite at $z=\lambda$ thus the residue comes only from the first term $\frac{1}{z-\lambda}$. – Winther Aug 18 '15 at 13:28
  • I would like to ask one final question. Say, we now have $\frac{1}{\lambda^{s}}\frac{\mathcal{F'(z)}}{\mathcal{F(z)}} = \frac{1}{\lambda^{s}}\frac{1}{z-\lambda}+ \frac{1}{\lambda^{s}}\frac{g'(z)}{g(z)}$. In this case, are there n+1 poles in total, such that there are n poles $z = \lambda_{n}$ of order $1$ and one pole $z=0$ of order $s$? – nightmarish Aug 18 '15 at 13:49
  • This function is just a constant times what we had so the poles are not modified. Did you mean the function $\frac{1}{(z-\lambda)^s}\frac{f'(z)}{f(z)}$? This function has a pole of order $s+1$ at $z=\lambda$ and the residue can be read off from the Taylor expansion as $\frac{f^{(s+1)}(\lambda)}{s! g(\lambda)} = \frac{f^{(s+1)}(\lambda)}{(s+1)! f'(\lambda)}$ – Winther Aug 18 '15 at 14:11
  • Sorry, I meant $\frac{1}{z^{s}}\frac{\mathcal{F'(z)}}{\mathcal{F(z)}}$. But, I guess your answer applies with $\lambda = 0$. Thanks for the help! :) – nightmarish Aug 18 '15 at 14:17

1 Answers1

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Your last approach is correct. Here is a slight variation of the argument that avoids Taylor expanding. If $\mathcal{F}$ is analytic at $\lambda$ and $\mathcal{F}(\lambda) = 0$ with $\mathcal{F}'(\lambda) \not= 0$ then we can write $\mathcal{F}(z) = (z-\lambda)g(z)$ where $g(z) = \frac{\mathcal{F}(z)}{z-\lambda}$. Now since $\lim\limits_{z\to\lambda}g(z) = \mathcal{F}'(\lambda)$ we have that $g$ is analytic at $z=\lambda$ (it has a removable singularity) and

$$\frac{\mathcal{F}'(z)}{\mathcal{F}(z)} = \frac{1}{z-\lambda} + \frac{g'(z)}{g(z)}$$

Since $\mathcal{F}'(\lambda) \not= 0$ we have $g(\lambda)\not=0$ so the last term is analytic at $z=\lambda$ and the residue can be read off from the first term.

As a sidenote to your first attempt: when $\mathcal{F}$ is entire and has zeros for all $z\in\{\lambda_n\}_{n=1}^\infty$ you cannot simply write $\mathcal{F}(z) = A(z-\lambda_1)(z-\lambda_2)\cdots$ as we can do when $\mathcal{F}$ is a polynomial. However there is a very nice theorem called Weierstrass factorization theorem that gives us the functional form of $\mathcal{F}$. It is slightly more complicated:

$$\mathcal{F}(z)=z^m e^{h(z)} \prod_{n=1}^\infty \left(1 - \frac{z}{\lambda_n}\right)E_{p_n}\!\!\left(\frac{z}{\lambda_n}\right)$$

where $h$ is some analytic function, $m$ is the order of the zero at $z=0$, $p_n$ is some set of integers and

$$E_{n} = \left\{\matrix{1 & n=0\\\exp(z + \frac{z^2}{2} + \ldots + \frac{z^n}{n}) & n> 0 }\right.$$

If you want you can also use this formula to solve your problem. By taking the logarithmic-derivative we get

$$\frac{\mathcal{F}'(z)}{\mathcal{F}(z)} = \frac{m}{z} + \sum_{n=1}^\infty \frac{1}{z-\lambda_n} + \text{(analytic function)}$$

so the residue $\text{Res}\left[\frac{\mathcal{F}'(z)}{\mathcal{F}(z)};\lambda\right]$ is equal to the order of the zero of $\mathcal{F}$ at $z=\lambda$ which in your case is just $1$ since all zeros are simple.

Winther
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  • Hi Winther, I edited my post with an alternative answer. It would be very helpful to me if you might comment on my queries in the edit. – nightmarish Aug 18 '15 at 07:17