Marin
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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
\displaystyle{\int}e^{-x^2}dx=?
e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}
so
\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
\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}
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:
\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)
\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)
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
\displaystyle{\int}e^{-x^2}dx=?
e^x=\displaystyle{\sum_{n=0}^{\infty}}\frac{x^n}{n!}
so
\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
\displaystyle{\int}e^{-x^2} dx=C+\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}
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:
\displaystyle{\lim_{x\rightarrow +\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: 1)
\displaystyle{\lim_{x\rightarrow -\infty}}\frac{2}{\sqrt{\pi}}\displaystyle{\sum_{n=0}^{\infty}}\frac{(-1)^nx^{2n+1}}{(2n+1)n!}=?(answer: -1)
The answers I have seen from Wikipedia: Error function - Wikipedia, the free encyclopedia (cf. the graph)
Thanks in advance, Marine!