- #1
Marin
- 193
- 0
Hi there!
I 've been thinking about finding the exact value series converge to. How does one do this? I know the convergence proof methods (minor, grater, Leibniz's, quotien, square-root and Cauchy's criteria) - but they just prove the convergence itself and do not give us the exact value.
Here's an example if what I'm talking about
[tex]\displaystyle{\int}e^{-x^2}dx=?[/tex]
[tex]e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}[/tex]
so
[tex]\displaystyle{\int}e^{-x^2} dx=\displaystyle{\int}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-x^2)^n}{n!}dx=\displaystyle{\sum_{n=0}^{\infty}\int}\frac{(-1)^nx^{2n}}{n!}dx[/tex]
[tex]\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}[/tex]
Now this should be the series definition of the Gaussian error function. Suppose C=0 and we're going to take the normalised function for simplicity. How can the following two limits be calculated:
[tex]\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)[/tex]
[tex]\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)[/tex]
The answers I have seen from Wikipedia: Error function - Wikipedia, the free encyclopedia (cf. the graph)
Thanks in advance, Marine!
I 've been thinking about finding the exact value series converge to. How does one do this? I know the convergence proof methods (minor, grater, Leibniz's, quotien, square-root and Cauchy's criteria) - but they just prove the convergence itself and do not give us the exact value.
Here's an example if what I'm talking about
[tex]\displaystyle{\int}e^{-x^2}dx=?[/tex]
[tex]e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}[/tex]
so
[tex]\displaystyle{\int}e^{-x^2} dx=\displaystyle{\int}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-x^2)^n}{n!}dx=\displaystyle{\sum_{n=0}^{\infty}\int}\frac{(-1)^nx^{2n}}{n!}dx[/tex]
[tex]\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}[/tex]
Now this should be the series definition of the Gaussian error function. Suppose C=0 and we're going to take the normalised function for simplicity. How can the following two limits be calculated:
[tex]\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)[/tex]
[tex]\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)[/tex]
The answers I have seen from Wikipedia: Error function - Wikipedia, the free encyclopedia (cf. the graph)
Thanks in advance, Marine!