09/25/2015, 08:26 AM
(This post was last modified: 02/18/2016, 06:17 PM by sheldonison.)
\( g(x) = \ln\( f( e^x) \) \;\;\; h_n = \left(g'\right)^{-1}(n) \)
\( a_n \; = \; \frac{1}{2\pi} \oint \frac{f(x)}{x^{n+1}}
\; = \; \frac{1}{2\pi i} \int_{x-\pi i}^{x+\pi i} {\exp \( g(x) - n\cdot x \)} \)
\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \;\;\; h_n = \left(g'\right)^{-1}(n)\;\;\; \) The Gaussian approximation for the Taylor series of f(x)
Now consider another better approximation where m is the minimum of \( \Re \(g(h_n + ix)\)\;\; \), which might be less than \( \pm \pi i \); or in the case of \( f(x)=\exp^{0.5}(x)\; \) m is actually greater than \( \pm \pi i \)
\( a_n \approx \frac{1}{2\pi i} \int_{h_n-mi}^{h_n+mi} {\exp \( g(h_n +x) - n\cdot x \)}\;\;\; a_0 = f(0)\;\; \) This is pretty much version V of the half iterate approximation first mentioned in post#31
\( f2(z) = \sum_{n=0}^{\infty} a_n z^n \)
Since this approximation actually involves an integral, it should be much more accurate than the Gaussian approximation. It works for generic analytic functions, not just entire functions. And, if f(z) is entire, but maybe f(z) is an even or an odd function, than f2(z) is an approximation with all positive derivatives. For example, take \( f(x)=e^x + e^{-x} \), than I think \( f2(x)=e^x+1\; \), or at least something very close to that.
Lets assume f is entire, with g'(x) an increasing function for x>0 at the real axis that gets arbitrarily large, then what is the ratio of \( \frac{f2(x)}{f(x)}\; \) as x gets arbitrarily large? How well does it converge to 1?
How about \( \|f2(x)-f(x)\| \) as \( \Re(x) \) gets arbitrarily large? The difference could be dominated by the error terms in the smaller derivatives.
\( a_n \; = \; \frac{1}{2\pi} \oint \frac{f(x)}{x^{n+1}}
\; = \; \frac{1}{2\pi i} \int_{x-\pi i}^{x+\pi i} {\exp \( g(x) - n\cdot x \)} \)
\( a_n \approx \frac{\exp(g(h_n) - n h_n)}{\sqrt{2 \pi g''(h_n) }} \;\;\; h_n = \left(g'\right)^{-1}(n)\;\;\; \) The Gaussian approximation for the Taylor series of f(x)
Now consider another better approximation where m is the minimum of \( \Re \(g(h_n + ix)\)\;\; \), which might be less than \( \pm \pi i \); or in the case of \( f(x)=\exp^{0.5}(x)\; \) m is actually greater than \( \pm \pi i \)
\( a_n \approx \frac{1}{2\pi i} \int_{h_n-mi}^{h_n+mi} {\exp \( g(h_n +x) - n\cdot x \)}\;\;\; a_0 = f(0)\;\; \) This is pretty much version V of the half iterate approximation first mentioned in post#31
\( f2(z) = \sum_{n=0}^{\infty} a_n z^n \)
Since this approximation actually involves an integral, it should be much more accurate than the Gaussian approximation. It works for generic analytic functions, not just entire functions. And, if f(z) is entire, but maybe f(z) is an even or an odd function, than f2(z) is an approximation with all positive derivatives. For example, take \( f(x)=e^x + e^{-x} \), than I think \( f2(x)=e^x+1\; \), or at least something very close to that.
Lets assume f is entire, with g'(x) an increasing function for x>0 at the real axis that gets arbitrarily large, then what is the ratio of \( \frac{f2(x)}{f(x)}\; \) as x gets arbitrarily large? How well does it converge to 1?
How about \( \|f2(x)-f(x)\| \) as \( \Re(x) \) gets arbitrarily large? The difference could be dominated by the error terms in the smaller derivatives.
- Sheldon
