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 Sep24 awarded Autobiographer Jul21 revised Show solution to ODE's fourier series is a series of sines only edited body Jul21 comment Show solution to ODE's fourier series is a series of sines only Then I don't get it: The ODE itself has nothing to do with this result? It all comes strictly from the boundary values? Jul21 comment Show solution to ODE's fourier series is a series of sines only @AndrewD I guess this is theorem: en.wikipedia.org/wiki/… I guess we can't use the theorem since $u$ doesn't necessarily has a bounded variation. Jul21 comment Show solution to ODE's fourier series is a series of sines only @AndrewD There was actually another segment in the question asking us to explain why we cannot apply that theorem to conclude 2 but since I didn't know the theorem or its terms to quote, I didn't bring it in. Could you give a link to the theorem in question? We are also hinted that we should derive 2 by developing the coefficients of the Fourier series. Jul21 asked Show solution to ODE's fourier series is a series of sines only Jul19 accepted Find roots of $3z^{100} - e^z$ in the unit disc. Jul15 awarded Teacher Jul14 revised Find roots of $3z^{100} - e^z$ in the unit disc. circle->disc Jul14 answered Find roots of $3z^{100} - e^z$ in the unit disc. Jul14 revised Find roots of $3z^{100} - e^z$ in the unit disc. correction from the comments Jul14 comment Find roots of $3z^{100} - e^z$ in the unit disc. @AntonioVargas oh boy, what a blunder. Thanks for setting me straight Jul14 comment Find roots of $3z^{100} - e^z$ in the unit disc. @AntonioVargas, on $0$, $e^z=1$ is greater than $3z^{100}=0$ and on $1$ for example, $3z^{100}=3$ and $e^z=e<3$ Jul14 asked Find roots of $3z^{100} - e^z$ in the unit disc. Jul11 accepted Show that $\sum_{k=1}^{n}a_ke^{2 \pi ikx}$ has a root in $\left[ 0,1 \right]$ Jul11 awarded Commentator Jul11 comment Show that $\sum_{k=1}^{n}a_ke^{2 \pi ikx}$ has a root in $\left[ 0,1 \right]$ D'oh! I poisoned the internet Jul11 asked Show that $\sum_{k=1}^{n}a_ke^{2 \pi ikx}$ has a root in $\left[ 0,1 \right]$ Jul11 comment Show that the complex closed line integral $\oint\frac{\mathrm{d}z}{p(z)}$ is $0$ ($p$ is polynomial) Yeah, the fact that the roots are distinct is relevant to the rest of the question which I didn't present here. I just didn't want to leave it out in case it somehow is needed Jul10 accepted Show that the complex closed line integral $\oint\frac{\mathrm{d}z}{p(z)}$ is $0$ ($p$ is polynomial)