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

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\begin{align*} \\ \erf x &=\erfi x\\ &=\\ &\,\vdots\\ &=\\ &=\frac2{\sqrt\pi}\left\{x\frac{x^3}3+\frac{x^5}{2!5}+\frac{x^7}{3!7}+\dots\right\} \end{align*}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
I'm not sure how to turn the integral into the infinite series,
 one year ago

AravindGBest ResponseYou've already chosen the best response.0
as always this is above my level always wish i could help !
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Looks like a Maclaurin Series bro.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
lol\[f(x) = \sum_{n = 0}^{\infty}\dfrac{{f^n}(0)}{n!}x^n\]Definitely not a Maclaurin.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Not sure. I am not good at this :(
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
ah yes Maclaurin Series (13) http://mathworld.wolfram.com/MaclaurinSeries.html
 one year ago

wioBest ResponseYou've already chosen the best response.4
looks like the series is \[ \frac{x^{2n+1}}{n!(2n+1)} \]
 one year ago

wioBest ResponseYou've already chosen the best response.4
But that  sign in there is throwing me off
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\newcommand\dd[1]{\,\mathrm d#1}% infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}}% integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)}% Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{u^2}}{u}}% Error function integral (x) \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \erf x =\erfi x \\ \de{\erf x}x =\de{}x\erfi x\\ \qquad\quad\quad=\frac2{\sqrt\pi}e^{x^2} \]
 one year ago

wioBest ResponseYou've already chosen the best response.4
Why not use: \[ \Large e^{x} = \sum_{n=0}^{\infty } \frac{x^n}{n!} \implies e^{x^2} = \sum_{n=0}^{\infty } \frac{(x^2)^n}{n!} \]
 one year ago

wioBest ResponseYou've already chosen the best response.4
It's really easy to find the anti derivative
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i think i got it now ! thanks!
 one year ago

wioBest ResponseYou've already chosen the best response.4
the part that threw me off was that there wasn't an alternating  +
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
oh dear i made an error in the question , it should be alternating , like you say @wio
 one year ago

wioBest ResponseYou've already chosen the best response.4
@UnkleRhaukus Where are you getting the questions from anyway?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
What is all this `\newcommand` thing? Are we allowed to introduce new variables in MathJax?
 one year ago

wioBest ResponseYou've already chosen the best response.4
Is it a class you're getting it from?
 one year ago

wioBest ResponseYou've already chosen the best response.4
@ParthKohli you're allowed to make command shortcuts, but I think it only works for your own post.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
MATH202 Differential equations sophomore subject \[\emph{Exercise 4D}1(d)\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
Are new commands allowed Ambassador?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
don't expand that series ... just integrate inside summation sign.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
have i done something wrong?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no it's okay ... just advised for simplicity. dw:1358071243367:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1358071379099:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x\frac{x^3}3+\frac{x^5}{2!5}\frac{x^7}{3!7}+\dots\right\}\\ \end{align*} \]
 one year ago
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