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Empty
 one year ago
Fun integral problem
Empty
 one year ago
Fun integral problem

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Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx\] Evaluate

freckles
 one year ago
Best ResponseYou've already chosen the best response.0like to make a conjecture \[(f(x))'_a^b \cdot g(x)(g(x))'_a^b f(x)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0oops missing pluggin a and b into those one things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SturmLiouville operators are selfadjoint, meaning $$\langle Lf,g\rangle=\langle f,Lg\rangle$$ over a suitable Hilbert space

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[[(f(x))' g(x)]_a^b[(g(x))' f(x)]_a^b\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can verify this by just integrating by parts, although in this case it is not as simple as there are no suitable boundary conditions: $$\begin{align*}\int_a^b f''(x) g(x)\, dx&=f'(x)g(x)\bigg_a^b \int_a^b f'(x) g'(x)\, dx\\&=\bigg[f'(x)g(x)f(x)g'(x)\bigg]_a^b+\int_a^b f(x) g''(x)\, dx\end{align*}$$

perl
 one year ago
Best ResponseYou've already chosen the best response.1$$\large \int\limits_a^b f''(x)g(x)dx +\int\limits_b^a g''(x)f(x)dx \\= \large \int\limits_a^b f''(x)g(x)dx \int\limits_a^b g''(x)f(x)dx \\= \large \int\limits_a^b [f''(x)g(x) g''(x) f(x)]dx $$

freckles
 one year ago
Best ResponseYou've already chosen the best response.0oh integration by parts awesome @oldrin.bataku

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, since $$\int_b^a f(x)g''(x) \, dx=\int_a^b f(x)g''(x)\, dx$$ the remaining integral cancels out and we have $$\bigg[f'(x) g(x)f(x)g'(x)\bigg]_a^b$$

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Without using IBP (Which is admittedly better since it's systematic and doesn't rely on tricks) we can us this cheap trick is if you can recognize: \[\frac{d}{dx}[f'(x)g(x)f(x)g'(x)] = f''(x)g(x)f(x)g''(x) \] Similarly you can use this sort of trick to integrate: \[\int \vec v \times \vec v '' d t = \vec v \times \vec v' \] Fun stuff :D
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