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 An alternate power series representation for ln(x) JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/07/2011, 08:41 PM (This post was last modified: 05/07/2011, 08:55 PM by 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. « Next Oldest | Next Newest »

 Messages In This Thread An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 08:41 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 09:43 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/07/2011, 10:45 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/07/2011, 11:20 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 01:38 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/08/2011, 07:54 PM RE: An alternate power series representation for ln(x) - by bo198214 - 05/08/2011, 08:28 PM RE: An alternate power series representation for ln(x) - by JmsNxn - 05/09/2011, 01:02 AM

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