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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{equation*} xJ''+J'+xJ=0 \end{equation*}\] \[\begin{align*} J(x)&=\frac2\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ J'(x)&=\frac2\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)+\,\mathrm du\\ \\ J''(x)&=\frac{2}\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du\\ \\ \end{align*} \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align*} &xJ''+J'+xJ\\ &=\frac{2x}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du+\frac{2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du+\frac{2x}\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x\sin^2(u)\cos\left(x\sin (u)\right)\sin(u)\sin\left(x\sin (u)\right)+x\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x(1\sin^2(u))\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ \end{align*} \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1i think i need to find [?]\[\begin{equation*}\frac{\partial }{\partial u}\Big[\quad?\quad\Big]=x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]

malevolence19
 one year ago
Best ResponseYou've already chosen the best response.0I wish I was in this class :/

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1am i thinking about this problem the right way?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1hmm wolfram tells me \[\begin{equation*}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]=x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1I am meant to be able to work that bit out in my head?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1its just the product rule in reverse but is kind hard to make that step

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.2The work appears correct to me here. I can't say I'd be able to get that antiderivative easily though, in my head... lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1the wolfram stepbystep solution starts by assuming we can tell the factors in the product rule, after we have worked out these , the problem is simple, but i have trouble seeing the factors

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{align*}&=\frac{2}\pi\int\limits_0^{\pi/2}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]\,\mathrm du\\ &=\frac{2}\pi\cos(u)\sin\left(x\sin(u)\right)\Big_0^{\pi/2}\,\mathrm du\\ &=\frac{2}\pi\left[\cos(\tfrac\pi2)\sin\left(x\sin(\tfrac\pi2)\right)\cos(0)\sin\left(x\sin(0)\right)\right]\,\mathrm du\\ \\ &=0\\\end{align*}\]

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.2\[ \begin{align} \frac{\partial}{\partial u} \left( \cos u \; \sin ( x \sin u ) \right) &= \frac{\partial}{\partial u} \left( \cos u \right) \sin (x \sin u) + \cos u \frac{\partial}{\partial u} \left( \sin ( x \sin u ) \right) \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin (x \sin u) } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ x \cos u \cos (x \sin u) } \\ \text{Let } k = x \sin u; \quad k' = x \cos u. \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin k } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ k' \cos k } \end{align} \] I think it looks a little clearer when you replace the weird composition of functins and find the derivative of the inner function / replace that...

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1that is a good method letting k= x sin u , because it is common to both terms thanks. P.S. \(\neg\) is not the same as\(\)

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.2ooh, I see what that is now. I kept thinking \neg was a negative sign and the bent end was just a weird latex thing. i remember seeing it in something else now.. lol :P You're welcome, and thanks!

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1negation is used in logic and sets, for example if the universal set is U={1,2,3,4,5,6,7,8} and a subset is S={1,2} the negation of S ¬S={3,4,5,6,7,8}

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0In settheory it's called a *complement* rather than a negation, which is typically used in logic.
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