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

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

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Other answers:

Looks like a Maclaurin Series bro.
are you series ?
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)
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!
No problem,
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?
Or a book or what?
@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.
1 Attachment
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|
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(-u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right|_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\dots\right\}\\ \end{align*} \]

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