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.
\( x \bigtriangleup y = ln(e^x + e^y) \)
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.
