An alternate power series representation for ln(x) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: An alternate power series representation for ln(x) (/showthread.php?tid=628) An alternate power series representation for ln(x) - JmsNxn - 05/07/2011 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/tetrationforum/showthread.php?tid=627&pid=5740#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. RE: An alternate power series representation for ln(x) - JmsNxn - 05/07/2011 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. RE: An alternate power series representation for ln(x) - bo198214 - 05/07/2011 (05/07/2011, 08:41 PM)JmsNxn Wrote: 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:$$\ln(x)$$ you get a better ln-typesetting. RE: An alternate power series representation for ln(x) - JmsNxn - 05/07/2011 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 RE: An alternate power series representation for ln(x) - bo198214 - 05/08/2011 (05/07/2011, 11:20 PM)JmsNxn Wrote: $\ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n$ But this follows directly from the definition of the logarithm, by the following equivalent transformations: $\begin{eqnarray} \ln(x) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n\\ &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x/e^a-1)^n\\ \ln(e^a y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\ a + \ln(y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\ \ln(y) &=& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n \end{eqnarray}$ Quote: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 $x\in (0,2e^a)$? RE: An alternate power series representation for ln(x) - JmsNxn - 05/08/2011 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. RE: An alternate power series representation for ln(x) - bo198214 - 05/08/2011 (05/08/2011, 07:54 PM)JmsNxn Wrote: 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: $\frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n$ RE: An alternate power series representation for ln(x) - JmsNxn - 05/09/2011 (05/08/2011, 08:28 PM)bo198214 Wrote: 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.