An alternate power series representation for ln(x)

[JmsNxn]

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

see here for more: http://math.eretrandre.org/tetrationforu...40#pid5740

I'll use these specifically:
\( \bigtriangleup \frac{d}{dx} [f(x) \bigtriangleup g(x)] =(\bigtriangleup \frac{d}{dx} f(x)) \bigtriangleup (\bigtriangleup \frac{d}{dx} g(x)) \)
and
\( \bigtriangleup \frac{d}{dx} xn = x(n-1) +ln(n) \)

The proof starts out by first proving:
\( \bigtriangleup \frac{d}{dx} e^x = e^x \)

first give the power series representation of e^x
\( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)

And given: \( ln(x + y) = ln(x) \bigtriangleup ln(y) \)

We take the ln of e^x to get an infinite series of deltations, if:
\( \bigtriangleup \sum_{n=0}^{R} f(n) = f(0) \bigtriangleup f(1) \bigtriangleup ... f® \) represents a series of deltations

then
\( x = \bigtriangleup \sum_{n=0}^{\infty} nln(x) - ln(n!) \)

and therefore if we let x = e^x

\( e^x = \bigtriangleup \sum_{n=0}^{\infty} nx - ln(n!) \)

and now we have an infinite lowered polynomial which when little differentiated equals itself. It's the little polynomial equivalent of e^x's perfect taylor series.

therefore:
\( \bigtriangleup \frac{d}{dx} e^x = e^x \)

and using the chain rule:
\( \bigtriangleup \frac{d}{dx} f(g(x)) = f'(g(x)) + g'(x) \) where f'(x) is taken to mean the little derivative of f(x).

we get the result:
\( \bigtriangleup \frac{d}{dx} ln(x) = -x \)

and now we can solve for the k'th little derivative of ln(x) using the little power rule, k E N

\( \bigtriangleup \frac{d^k}{dx^k} ln(x) = -kx + ln((-1)^{k-1} (k-1)!) \)

and so if the little derivative Taylor series is given by:
\( f(x) = \bigtriangleup \sum_{n=0}^{\infty} f^{(n)}(a) + n(x \bigtriangledown a) - ln(n!) \)
where \( f^{(n)}(x) \) is the n'th little derivative of \( f \).

we can take the little derivative taylor series of ln(x) centered about 1.
The first term is equal to 0, so I'll start the series from n = 1.

Here it is in two steps:
\( ln(x) = 0 \bigtriangleup (\bigtriangleup \sum_{n=1}^{\infty} n(x \bigtriangledown 1) + ln((-1)^{n-1}(n-1)!)-ln(n!)-n) \)

\( ln(x) = 0 \bigtriangleup (\bigtriangleup \sum_{n=1}^{\infty} ln(\frac{(-1)^{n-1}}{n}) + n(x \bigtriangledown 1) - n) \)

now let's plug this in our formula for \( x \bigtriangleup y \); it works for an infinite sum because \( \bigtriangleup \) is commutative and associative.

\( ln(x) = ln(1 + \sum_{n=1}^{\infty} e^{ln(\frac{(-1)^{n-1}}{n}) + n(x \bigtriangledown 1)-n}) \)

now since:
\( x \bigtriangledown 1 = ln(e^x - e) \)


\( ln(x) = ln(1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(e^x - e)^n) \)

now take the lns away and

\( x = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(e^x - e)^n \)

and now if we let x = ln(x) we get:
\( ln(x) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(x - e)^n \)

And there, that's it. I haven't been able to check if this series converges, I don't have many convergence tests. I'm just sure that the algebra is right. I haven't tried computing it. In my gut it doesn't look like it will converge, but I have faith.

[JmsNxn]

Doing some tests it converges for 1, 2, 4, 5 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works.

Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of \( \bigtriangleup \)) I find the normal taylor series of ln(x): \( ln(x) = \sum_{n=1}^{\infty} (x-1)^n \frac{(-1)^{n+1}}{n} \)

so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0.

[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.

[JmsNxn]

Yeah, I was aware of a direct relation to differentiation:

\( \frac{d}{dx} f(x) = e^{(\bigtriangleup \frac{d}{dx} f(x)) + x - f(x)} \)

And I've extended the definition of the series to:

\( \ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n \)

And I've found a beautiful return:

\( \ln(x) = e + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{ne}}(x-e^e)^n \)

Converges for values as high as 30. My computer overflows before it stops converging.

I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x)

edit:
sadly, doesn't converge for values less than 1

[bo198214]

\( \ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n \)

But this follows directly from the definition of the logarithm, by the following equivalent transformations:
\( \begin{eqnarray}
\ln(x) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n\\
&=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x/e^a-1)^n\\
\ln(e^a y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
a + \ln(y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
\ln(y) &=& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n
\end{eqnarray} \)

sadly, doesn't converge for values less than 1

The \( \ln \) series converges for \( |y-1|<1 \), hence your series converges for \( |x/e^a-1|<1 \), this should include all values \( x\in (0,2e^a) \)?

[JmsNxn]

I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the \( e^{an} \) went to in the denominator. Otherwise, though, that's a nice proof.

[bo198214]

I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the \( e^{an} \) went to in the denominator. Otherwise, though, that's a nice proof.

Oh thats just:
\( \frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n \)

[JmsNxn]

Oh thats just:
\( \frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n \)

Oh that's so simple! Thanks for the help.