An alternate power series representation for ln(x)

[bo198214]

This proof involves the use of a new operator:
\( x \bigtriangleup y = ln(e^x + e^y) \)

and it's inverse:
\( x \bigtriangledown y = ln(e^x - e^y) \)

and the little differential operator:
\( \bigtriangleup \frac{d}{dx} f(x) = \lim_{h\to\ -\infty} [f(x \bigtriangleup h) \bigtriangledown f(x)] - h \)

(The notation is more unambiguous than in your previous thread )

But your operator can be expressed with the classical differentiation, see:

\( \begin{eqnarray}
\bigtriangleup \frac{d}{dx} f(x) &=& \lim_{h\to\ -\infty} [f(x \bigtriangleup h) \bigtriangledown f(x)] - h\\
&=& \lim_{h\to\ -\infty} \quad\ln[\exp(f(x \bigtriangleup h)) - \exp(f(x))] - h\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(x \bigtriangleup \log(d))) - \exp(f(x))}{d}\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(\ln(e^x + d)) - \exp(f(x))}{d}\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(\ln(e^x + d)) - \exp(f(\ln(\exp(x)))))}{d}\\
& =& \ln((\exp\circ f\circ \ln)'(\exp(x)))
\end{eqnarray} \)

Or purely functional with the composition operation \( \circ \): \( \bigtriangleup \frac{d}{dx} f = \ln\circ(\exp\circ f\circ \ln)'\circ\exp \)

PS: when you write ln with backslash in front:

Code:

[tex]\ln(x)[/tex]

you get a better ln-typesetting.