# Simple Poles

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How can I prove that

$$\displaystyle \frac{1}{1-2^{1-s}}$$

has a simple pole at s=1.

Any help is much appreciated.

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I still dont understand it.

Is there no simple method to approach it,

#### mr fantastic

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I still dont understand it.

Is there no simple method to approach it,
Go to your classnotes. In them you will find a theorem that says that the singularity $$\displaystyle z = z_0$$ is a simple pole of f(z) if $$\displaystyle \lim_{z \to z_0} ((z - z_0) f(z))$$ exists and is finite. This is the simple approach I have suggested that you use.

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#### TheFinalPush

Hi,

You could show the pole has order 1 directly:

Pole (complex analysis) - Wikipedia, the free encyclopedia

Perhaps another possible method is as follows, though I could be wrong:

I have just posted up a question regarding the domain of the Riemann zeta function and am aware that we can show it has a singularity at s = 1 by rewriting it as the product of an infinite sum by the term you have posted up. The analytic continuation of the Riemann zeta function, valid in Re(z) > 0 gives it as 1 + 1/(s-1) + {some integral well defined integral} and hence we see straight away that is has a simple pole at s = 1.

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#### chisigma

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Lets start considering that is...

$$\displaystyle 2^{1-s}= e^{(1-s) \ln 2} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n} 2}{n!} (s-1)^{n}$$ (1)

From (1) we derive...

$$\displaystyle 1-2^{1-s} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \ln^{n} 2}{n!} (s-1)^{n}=$$

$$\displaystyle = (s-1) \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n+1} 2}{(n+1)!} (s-1)^{n} = (s-1)\cdot \ln 2 \cdot \varphi(s)$$ (2)

... where...

$$\displaystyle \varphi(s) = 1 - \frac{\ln 2}{2} \cdot (s-1) + ...$$ (3)

... is analytic in $$\displaystyle s=1$$ and is $$\displaystyle \varphi(1)=1$$. Consequence of that is that $$\displaystyle \frac{1}{\varphi(s)}$$ is also analytic and is...

$$\displaystyle \frac{1}{\varphi(s)} = 1 + \frac{\ln 2}{2} \cdot (s-1) + ...$$ (4)

... and finaly that is...

$$\displaystyle \frac{1}{1-2^{1-s}}= \frac{1}{\ln 2 \cdot (s-1)\cdot \varphi(s)} = \frac{1}{\ln 2 \cdot (s-1)} \{ 1 + \frac{\ln 2}{2} \cdot (s-1) + ...\}$$ (5)

If we observe the (5) it is easy tom conclude that $$\displaystyle \frac{1}{1-2^{1-s}}$$ has a single pole is $$\displaystyle s=1$$ and here the residue is $$\displaystyle \frac{1}{\ln 2}$$...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

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