Here's the question you clicked on:
UnkleRhaukus
\[\newcommand\dd[1]{\,\mathrm d#1}% infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)}% Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}}% Error function integral (x) {-------------------}\\ \text{From the definition prove the result:} \\ \erf x=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}\color{brown}-\frac{x^7}{3!7}+\dots\right\}\\ {-------------------}\]
\[\begin{align*} \\ \erf x &=\erfi x\\ &=\\ &\,\vdots\\ &=\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}+\frac{x^7}{3!7}+\dots\right\} \end{align*}\]
I'm not sure how to turn the integral into the infinite series,
as always this is above my level always wish i could help !
Looks like a Maclaurin Series bro.
lol\[f(x) = \sum_{n = 0}^{\infty}\dfrac{{f^n}(0)}{n!}x^n\]Definitely not a Maclaurin.
Not sure. I am not good at this :-(
ah yes Maclaurin Series (13) http://mathworld.wolfram.com/MaclaurinSeries.html
looks like the series is \[ \frac{x^{2n+1}}{n!(2n+1)} \]
But that - sign in there is throwing me off
\[\newcommand\dd[1]{\,\mathrm d#1}% infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)}% Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}}% Error function integral (x) \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \erf x =\erfi x \\ \de{\erf x}x =\de{}x\erfi x\\ \qquad\quad\quad=\frac2{\sqrt\pi}e^{-x^2} \]
Why not use: \[ \Large e^{x} = \sum_{n=0}^{\infty } \frac{x^n}{n!} \implies e^{-x^2} = \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!} \]
It's really easy to find the anti derivative
i think i got it now ! thanks!
the part that threw me off was that there wasn't an alternating - +
oh dear i made an error in the question , it should be alternating , like you say @wio
@UnkleRhaukus Where are you getting the questions from anyway?
What is all this `\newcommand` thing? Are we allowed to introduce new variables in MathJax?
Is it a class you're getting it from?
@ParthKohli you're allowed to make command shortcuts, but I think it only works for your own post.
MATH202 Differential equations sophomore subject \[\emph{Exercise 4D}1(d)\]
Are new commands allowed Ambassador?
don't expand that series ... just integrate inside summation sign.
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.
have i done something wrong?
no it's okay ... just advised for simplicity. |dw:1358071243367:dw|
|dw:1358071379099:dw|
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(-u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right|_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\dots\right\}\\ \end{align*} \]