My Math Forum An alternate power series representation for ln(x)

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 May 7th, 2011, 11:59 AM #1 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 An alternate power series representation for ln(x) 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/tetrationfor ... 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: $\bigtriangleup \sum_{n=0}^{R} f(n) = f(0) \bigtriangleup f(1) \bigtriangleup ... f(R)$ 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: $\bigtriangleup \frac{d}{dx} f(g(x))= f'(g(x)) + g#39;(x)$ 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 feels like it will converge, so I have faith. If anyone has any questions about the proof please feel free to ask them, I may have skimmed over some important details.
 May 7th, 2011, 12:38 PM #2 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 Re: An alternate power series representation for ln(x) Doing some tests it converges for 1, 2, 4 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works.
 May 7th, 2011, 12:57 PM #3 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: An alternate power series representation for ln(x) jamesuminator, The radius of convergence of the last series in your first post is R = e, so it is expected that it won't converge for any integer argument x ? 6, for instance. -Ormkärr-
 May 7th, 2011, 03:54 PM #4 Senior Member   Joined: Aug 2008 Posts: 133 Thanks: 0 Re: An alternate power series representation for ln(x) If you don't mind me asking, how did you solve the radius of convergence? I've successfully generalized it further, using the same proof but centering the little taylor series about a, I get: $\ln(x)= a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}} (x -e^a)^n$ which is the usual logarithmic power series when a = 0 The radius of convergence has to be at least 30 when a = e; I verified this by calculation.
 May 7th, 2011, 09:00 PM #5 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: An alternate power series representation for ln(x) jamesuminator, The standard way to compute the radius and interval of convergence is by means of the Ratio Test. There is also the Cauchy-Hadamard Formula. Both of these can be found in any good book on advanced calculus, or otherwise on Wikipedia, I imagine. I could give the calculation if you really like, but I get the feeling you might be content to try your hand yourself, based on your enterprising spirit evident in this presentation. -Ormkärr-

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