## UnkleRhaukus 3 years ago $\newcommand\dd[1]{\,\mathrm d#1}% infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)}% Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}}% Error function integral (x) {-------------------}\\ \text{From the definition prove the result:} \\ \erf x=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}\color{brown}-\frac{x^7}{3!7}+\dots\right\}\\ {-------------------}$

1. UnkleRhaukus

\begin{align*} \\ \erf x &=\erfi x\\ &=\\ &\,\vdots\\ &=\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}+\frac{x^7}{3!7}+\dots\right\} \end{align*}

2. UnkleRhaukus

I'm not sure how to turn the integral into the infinite series,

3. AravindG

as always this is above my level always wish i could help !

4. ParthKohli

Looks like a Maclaurin Series bro.

5. UnkleRhaukus

are you series ?

6. ParthKohli

lol$f(x) = \sum_{n = 0}^{\infty}\dfrac{{f^n}(0)}{n!}x^n$Definitely not a Maclaurin.

7. ParthKohli

Not sure. I am not good at this :-(

8. UnkleRhaukus

ah yes Maclaurin Series (13) http://mathworld.wolfram.com/MaclaurinSeries.html

9. wio

looks like the series is $\frac{x^{2n+1}}{n!(2n+1)}$

10. wio

11. UnkleRhaukus

$\newcommand\dd[1]{\,\mathrm d#1}% infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)}% Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}}% Error function integral (x) \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \erf x =\erfi x \\ \de{\erf x}x =\de{}x\erfi x\\ \qquad\quad\quad=\frac2{\sqrt\pi}e^{-x^2}$

12. wio

Why not use: $\Large e^{x} = \sum_{n=0}^{\infty } \frac{x^n}{n!} \implies e^{-x^2} = \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!}$

13. wio

It's really easy to find the anti derivative

14. UnkleRhaukus

i think i got it now ! thanks!

15. wio

No problem,

16. wio

the part that threw me off was that there wasn't an alternating - +

17. UnkleRhaukus

oh dear i made an error in the question , it should be alternating , like you say @wio

18. wio

@UnkleRhaukus Where are you getting the questions from anyway?

19. UnkleRhaukus

20. ParthKohli

What is all this \newcommand thing? Are we allowed to introduce new variables in MathJax?

21. wio

Is it a class you're getting it from?

22. wio

Or a book or what?

23. wio

@ParthKohli you're allowed to make command shortcuts, but I think it only works for your own post.

24. UnkleRhaukus

MATH202 Differential equations sophomore subject $\emph{Exercise 4D}1(d)$

25. UnkleRhaukus

26. experimentX

don't expand that series ... just integrate inside summation sign.

27. UnkleRhaukus

28. experimentX

there are many cases where you have to change functions into infinite series to evaluate integral with real methods. I think ... expanding the series will just mess up.

29. UnkleRhaukus

have i done something wrong?

30. experimentX

no it's okay ... just advised for simplicity. |dw:1358071243367:dw|

31. experimentX

|dw:1358071379099:dw|

32. UnkleRhaukus

\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(-u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right|_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\dots\right\}\\ \end{align*}