How Does Log(z) Expand at Infinity?

[Belgium 12]

Hi,

see attached PdF file for my question concerning serie expansion of log(z) at infinity.

Thank you
Belgium 12

 

[stevendaryl]

So the usual expansion for $log(1+x)$ when $x$ is small is given by:

$log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$

We also know that $log(1+z) = log(z (1 + \frac{1}{z})) = log(z) + log(1 + \frac{1}{z})$

So you can put the two together (replacing $x$ by $\frac{1}{z}$):

$log(1+z) = log(z) + \frac{1}{z} - \frac{1}{2 z^2} + \frac{1}{3 z^3} ...$​

 

[Mark44]

Hi,

see attached PdF file for my question concerning serie expansion of log(z) at infinity.

Thank you
Belgium 12

Instead of posting an image of barely legible writing, please learn to use LaTeX. Everything you wrote can be entered directly into the input pane here. See our tutorial on LaTeX here: https://www.physicsforums.com/help/latexhelp/

Here are some examples of expressions you wrote and how they appear in LaTeX:
$\log(\frac{z - 1}{z})$
Script for the above: $\log(\frac{z - 1}{z})$

$$\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q}$$
Script for the above: $$\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q}$$

(Inline version of the above would be $\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q}$.)

BTW, in the summation that appears twice, it's very difficult to tell that the denominator is $qz^q$. What you wrote--twice--looks like $qzq$. The only clue that the exponent is q is that this letter appears slightly raised.