Difference between revisions of "Derivative of prime zeta"
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==Proof== | ==Proof== | ||
| + | Recall the definition of the prime zeta function: | ||
| + | $$P(z)=\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{1}{p^z}.$$ | ||
| + | Rewrite | ||
| + | $$\dfrac{1}{p^z} = p^{-z} = e^{-z \log(p)}.$$ | ||
| + | Differentiate term-by-term using the [[derivative of the exponential function]] and the [[chain rule]]: | ||
| + | $$\begin{array}{ll} | ||
| + | \dfrac{\mathrm{d}}{\mathrm{d}z} P(z) &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{p^z} \\ | ||
| + | &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} e^{-z \log(p)} \\ | ||
| + | &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} e^{-z\log(p)} (-\log(p)) \\ | ||
| + | &= -\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\log(p)}{p^z}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. | ||
==References== | ==References== | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
| + | [[Category:Justify]] | ||
Latest revision as of 17:46, 19 October 2016
Theorem
The following formula holds: $$P'(z)=-\displaystyle\sum_{p \mathrm{\hspace{2pt} prime}} \dfrac{\log(p)}{p^z},$$ where $P$ denotes the prime zeta function and $\log$ denotes the logarithm.
Proof
Recall the definition of the prime zeta function: $$P(z)=\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{1}{p^z}.$$ Rewrite $$\dfrac{1}{p^z} = p^{-z} = e^{-z \log(p)}.$$ Differentiate term-by-term using the derivative of the exponential function and the chain rule: $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} P(z) &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{p^z} \\ &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} e^{-z \log(p)} \\ &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} e^{-z\log(p)} (-\log(p)) \\ &= -\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\log(p)}{p^z}, \end{array}$$ as was to be shown.
