Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
natural_logarithm_of_complex_numbers [2015/06/26 22:07] nikolaj |
natural_logarithm_of_complex_numbers [2016/07/22 15:08] nikolaj |
||
|---|---|---|---|
| Line 6: | Line 6: | ||
| ----- | ----- | ||
| >todo: [[Complex argument]] | >todo: [[Complex argument]] | ||
| + | |||
| + | == Limits == | ||
| + | $\lim_{x\to 0}x\ln(x)=0$ | ||
| == Differentiation and integrals == | == Differentiation and integrals == | ||
| - | $\int \left(x^n\right)'\ln(x^n)\,{\mathrm d}x=x^n\left(\ln(x^n)-1\right)$ | + | $\int \ln(x^n)\,{\mathrm d}x^n=\int \left(x^n\right)'\ln(x^n)\,{\mathrm d}x=x^n\left(\ln(x^n)-1\right)$ |
| == Series == | == Series == | ||
| + | At least around $z=0$ (I think for $|z|<1$) | ||
| + | |||
| + | $\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$ | ||
| + | |||
| + | or | ||
| - | $\ln{\left(1+z\right)}=-\sum_{n=1}^\infty \frac{(-z)^n}{n}$ | + | $\ln{\left(1+z\right)} = -\sum_{n=1}^\infty \frac{(-z)^n}{n}$ |
| - | and so, from this | + | From this series in $(-z)^n$, if we always constitutive positive and negative germs, we get |
| - | $\ln{\left(\frac{1}{1-z}\right)}=\sum_{n=1}^\infty \frac{z^n}{n}$ | + | $\ln{\left(1+z\right)} = z - \sum_{n=1}^\infty \left(\dfrac{1}{2n}-\dfrac{z}{2n+1}\right) z^{2n}$ |
| Note that this implies that $\exp$ maps the series on the right to $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$. | Note that this implies that $\exp$ maps the series on the right to $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$. | ||
