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

This Question is Closed

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure how to turn the integral into the infinite series,

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.0as always this is above my level always wish i could help !

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Looks like a Maclaurin Series bro.

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Not sure. I am not good at this :(

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0ah yes Maclaurin Series (13) http://mathworld.wolfram.com/MaclaurinSeries.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0looks like the series is \[ \frac{x^{2n+1}}{n!(2n+1)} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But that  sign in there is throwing me off

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why 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
 3 years ago
Best ResponseYou've already chosen the best response.0It's really easy to find the anti derivative

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i think i got it now ! thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the part that threw me off was that there wasn't an alternating  +

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus Where are you getting the questions from anyway?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is it a class you're getting it from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli you're allowed to make command shortcuts, but I think it only works for your own post.

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0Are new commands allowed Ambassador?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1don't expand that series ... just integrate inside summation sign.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1there 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
 3 years ago
Best ResponseYou've already chosen the best response.0have i done something wrong?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no it's okay ... just advised for simplicity. dw:1358071243367:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358071379099:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x\frac{x^3}3+\frac{x^5}{2!5}\frac{x^7}{3!7}+\dots\right\}\\ \end{align*} \]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.