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

Question

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

unklerhaukus · Answer

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

unklerhaukus · Answer

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

aravindg · Answer

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

parthkohli · Answer

Looks like a Maclaurin Series bro.

unklerhaukus · Answer

are you series ?

parthkohli · Answer

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

parthkohli · Answer

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

unklerhaukus · Answer

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

anonymous · Answer

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

anonymous · Answer

But that - sign in there is throwing me off

unklerhaukus · Answer

$$\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}

$$

anonymous · Answer

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

anonymous · Answer

It's really easy to find the anti derivative

unklerhaukus · Answer

i think i got it now ! thanks!

anonymous · Answer

No problem,

anonymous · Answer

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

unklerhaukus · Answer

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

anonymous · Answer

@UnkleRhaukus Where are you getting the questions from anyway?

unklerhaukus · Answer

attachment

parthkohli · Answer

What is all this `
ewcommand` thing? Are we allowed to introduce new variables in MathJax?

anonymous · Answer

Is it a class you're getting it from?

anonymous · Answer

Or a book or what?

anonymous · Answer

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

unklerhaukus · Answer

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

unklerhaukus · Answer

Are new commands allowed Ambassador?

experimentx · Answer

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

unklerhaukus · Answer

attachment

experimentx · Answer

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.

unklerhaukus · Answer

have i done something wrong?

experimentx · Answer

no it's okay ... just advised for simplicity.
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