this is for holding javascript data
Benedict Irwin edited section_Another_Form_begin_equation__.tex
almost 8 years ago
Commit id: 1c1ba18e0f2f6e754fa6644ccfc8cf685340113e
deletions | additions
diff --git a/section_Another_Form_begin_equation__.tex b/section_Another_Form_begin_equation__.tex
index 9440a72..73240ce 100644
--- a/section_Another_Form_begin_equation__.tex
+++ b/section_Another_Form_begin_equation__.tex
...
and so on till we have an infinite token $Z$, or just a constant, and \begin{equation}
\log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) = Z\sum_{k=1}^\infty \frac{x^k}{f(k)} -\frac{Z^2}{2}\sum_{k=1}^\infty \frac{x^{2k}}{f(k)^2} +\frac{Z^3}{3}\sum_{k=1}^\infty \frac{x^{3k}}{f(k)^3}-\frac{Z^4}{4}\sum_{k=1}^\infty \frac{x^{4k}}{f(k)^4} +\cdots\\
\log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) = \sum_{j=1}^\infty \frac{(-1)^{j+1}Z^j}{j}\sum_{k=1}^\infty \frac{x^{jk}}{f(k)^j}\\
\log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) =
\log\left(1+Z\sum_{k=1}^\infty \frac{x^{jk}}{f(k)^j}\right) \log\left(1+Z\tau\right)\\
\end{equation}
On the condition that \begin{equation}
\sum_{k=1}^\infty \frac{x^{jk}}{f(k)^j} = \tau^j
\end{equation}
which is probably impossible...