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UnkleRhaukus

  • 3 years ago

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

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  1. UnkleRhaukus
    • 3 years ago
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    \[\begin{align*} \\ \erf x &=\erfi x\\ &=\\ &\,\vdots\\ &=\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}+\frac{x^7}{3!7}+\dots\right\} \end{align*}\]

  2. UnkleRhaukus
    • 3 years ago
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    I'm not sure how to turn the integral into the infinite series,

  3. AravindG
    • 3 years ago
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    as always this is above my level always wish i could help !

  4. ParthKohli
    • 3 years ago
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    Looks like a Maclaurin Series bro.

  5. UnkleRhaukus
    • 3 years ago
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    are you series ?

  6. ParthKohli
    • 3 years ago
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    lol\[f(x) = \sum_{n = 0}^{\infty}\dfrac{{f^n}(0)}{n!}x^n\]Definitely not a Maclaurin.

  7. ParthKohli
    • 3 years ago
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    Not sure. I am not good at this :-(

  8. UnkleRhaukus
    • 3 years ago
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    ah yes Maclaurin Series (13) http://mathworld.wolfram.com/MaclaurinSeries.html

  9. wio
    • 3 years ago
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    looks like the series is \[ \frac{x^{2n+1}}{n!(2n+1)} \]

  10. wio
    • 3 years ago
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    But that - sign in there is throwing me off

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

  12. wio
    • 3 years ago
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    Why not use: \[ \Large e^{x} = \sum_{n=0}^{\infty } \frac{x^n}{n!} \implies e^{-x^2} = \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!} \]

  13. wio
    • 3 years ago
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    It's really easy to find the anti derivative

  14. UnkleRhaukus
    • 3 years ago
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    i think i got it now ! thanks!

  15. wio
    • 3 years ago
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    No problem,

  16. wio
    • 3 years ago
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    the part that threw me off was that there wasn't an alternating - +

  17. UnkleRhaukus
    • 3 years ago
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    oh dear i made an error in the question , it should be alternating , like you say @wio

  18. wio
    • 3 years ago
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    @UnkleRhaukus Where are you getting the questions from anyway?

  19. UnkleRhaukus
    • 3 years ago
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  20. ParthKohli
    • 3 years ago
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    What is all this `\newcommand` thing? Are we allowed to introduce new variables in MathJax?

  21. wio
    • 3 years ago
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    Is it a class you're getting it from?

  22. wio
    • 3 years ago
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    Or a book or what?

  23. wio
    • 3 years ago
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    @ParthKohli you're allowed to make command shortcuts, but I think it only works for your own post.

  24. UnkleRhaukus
    • 3 years ago
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    MATH202 Differential equations sophomore subject \[\emph{Exercise 4D}1(d)\]

  25. UnkleRhaukus
    • 3 years ago
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    Are new commands allowed Ambassador?

  26. experimentX
    • 3 years ago
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    don't expand that series ... just integrate inside summation sign.

  27. UnkleRhaukus
    • 3 years ago
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  28. experimentX
    • 3 years ago
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    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.

  29. UnkleRhaukus
    • 3 years ago
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    have i done something wrong?

  30. experimentX
    • 3 years ago
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    no it's okay ... just advised for simplicity. |dw:1358071243367:dw|

  31. experimentX
    • 3 years ago
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    |dw:1358071379099:dw|

  32. UnkleRhaukus
    • 3 years ago
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    \[\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function (x) \newcommand\erfi[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{-u^2}}{u}} % Error function integral (x) \begin{align*} \erf x &=\erfi x\\ &=\frac2{\sqrt\pi}\intl0x{\sum_{n=0}^{\infty } \frac{\left(-u^2\right)^n}{n!}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\intl0x{ u^{2n}}u\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\left.\frac{ u^{2n+1}}{2n+1}\right|_0^x\\ &=\frac2{\sqrt\pi}\sum_{n=0}^{\infty }\frac{(-1)^n}{n!}\frac{ x^{2n+1}}{2n+1}\\ &=\frac2{\sqrt\pi}\left\{x-\frac{x^3}3+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\dots\right\}\\ \end{align*} \]

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