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 one year ago
Show that \(\operatorname{erc}(x)\) satisfies the differential equation
\[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)
 one year ago
Show that \(\operatorname{erc}(x)\) satisfies the differential equation \[\frac{\mathrm dy}{\mathrm dx}=2xy\frac2{\sqrt \pi}\] With \(y=1\) when \(x=0\)

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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \newcommand\de[2]{\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative \newcommand\pa[2]{\frac{\partial #1}{\partial #2}} % partial derivative \newcommand\erf[1]{\operatorname {erf}\left(#1\right)} % Error function erf(x) \newcommand\erfint[1]{\frac2{\sqrt \pi}\intl{0}{#1}{e^{u^2}}{u}} % Error function integral erfc(x) \newcommand\erfc[1]{\operatorname {erfc}\left(#1\right)} % Complimentary Error function Compliment (x) \newcommand\erfcint[1]{\frac2{\sqrt \pi}\intl{#1}{\infty}{e^{u^2}}u} % Complimentary Error function integral (x) \newcommand\erc[1]{\operatorname{erc}\left(#1\right)} % Normalised Error function (x) \newcommand\ercint[1]{e^{#1^2}\erfcint{#1}} % Normalised Error function integral (x) \begin{equation*} \de yx=2xy\frac2{\sqrt \pi},\qquad\qquad y(0)=1\\ \end{equation*}\]\[ \begin{align*} y(x)&=\erc x\\ &=e^{x^2}\erfc x\\ &=\ercint x\\ \\ y(0)&=\erfcint 0\\ &=1\\ \\ \de yx&=\frac2{\sqrt \pi}\left[\de{\left(e^{x^2}\right)}x\intl x\infty{e^{u^2}}u+e^{x^2}\intl x\infty {\pa {\left(e^{u^2}\right)}u}u\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{u^2}}u+e^{x^2}\left.e^{u^2}\right_x^\infty\right]\\ &=\frac2{\sqrt \pi}\left[2xe^{x^2}\intl x\infty{e^{u^2}}ue^{x^2}e^{x^2}\right]\\ &=2xe^{x^2}\frac{2}{\sqrt \pi}\intl x\infty{e^{u^2}}u\frac2{\sqrt \pi}\\ &=2xy\frac2{\sqrt \pi}\\ \end{align*} \]

Goten77
 one year ago
Best ResponseYou've already chosen the best response.0im sorry but man... im good at math but i only know stuff up to calculus 2.... anything after i havent taken yet.... best of luck

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0what is erc(x) ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0my book defines:: ___ The error function:\[\operatorname {erf}(x)=\frac{2}{\sqrt \pi}\int\limits_0^xe^{u^2}\mathrm du\]___ The complementary error function:\[\operatorname{erfc}(x)=1\operatorname {erf}(x)\] ___\[\operatorname {erc}(x)=e^{x^2}\operatorname{erfc(x)}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i guess it is the normalised complementary error function

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0ahh I see..... what's your problem? I think you've done it..

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0so i haven't made any errors?

chihiroasleaf
 one year ago
Best ResponseYou've already chosen the best response.0It seems correct..

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large {y=\operatorname{erc}(x)=e^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]\\\frac{dy}{dx}=2xe^{x^2}\left[1\frac{2}{\sqrt \pi}\int_0^xe^{u^2}\;\mathrm du\right]+e^{x^2}\left[\frac{2}{\sqrt \pi}e^{x^2}\right]\\=2xy\frac{2}{\sqrt \pi}}\]

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2it's shorter if you apply the fundamental theorem of calculus

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2you used the derivative "under the integral sign", and i appreciate looking at a different solution. i see the function \(\operatorname{efc}(x)\) as a product, so i sued the product rule for derivative. in the process, i also saw the fundamental theorem of calculus\[\frac{\mathrm d}{\mathrm dx} \int_a^x f(u)\;\mathrm du=f(x)\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0the only difference i can see your method is that you used erf where i used erfc

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0I can't say I understand the difference between derivative "under the integral sign" & the fundamental theorem of calculus
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