Recent content by ELog

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    Does the Series Involving a Function with Continuous Third Derivative Converge?

    I found the problem in the form I presented, though the derivative was with respect to x (as HoI pointed out). If I expand f as a taylor series, I think I see the fun of which you speak. EDIT: All we know about f is that its third derivative is continuous. Does this effect the taylor series (Is...
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    Does the Series Involving a Function with Continuous Third Derivative Converge?

    This problem has been bothering me for some time. Any thoughts or insights are greatly appreciated. Consider a function, f, with continuous third derivative on [-1,1]. Prove that the series \sum^{\infty}_{n=1} (nf(\frac{1}{n})-nf(-\frac{1}{n}) - 2\frac{df}{dn}(0)) converges. Thanks in...
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    Integration of (e^x)dx/ (e^(2x) +5e^(x) + 6)

    You're rushing into partial fractions too quickly. First substitute u=e^x so that du=e^xdx. Partial fractions will then yield the correct integration.
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    Solving 2nd Order Diff Eq in Quantum Mechanics

    So I was trying a few quantum mechanics problems and encountered this diff eq: \frac{\hbar^2}{2m}\frac{\partial^2}{\partialx^2}\psi(x) + \frac{1}{2}kx^2\psi(x) = E\psi(x) I put it into the form: \frac{\partial^2}{\partialx^2}\psi(x) + (\frac{2mE}{\hbar^2} - \frac{m}{\hbar^2}kx^2)\psi(x) = 0...
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    Is the Riemann Hypothesis the Key to Predicting Prime Numbers?

    Yes, that is what your teacher was referring to. However, there is no "solution to the equation." The man who developed a lot of it, riemann, basically suggested that besides the negative even integers (-2, -4, -6, etc.) that all numbers that make his zeta-function ,\zeta(s), equal 0 are complex...
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    Differential and derivatives [HELP]

    Work (A) = Force (F) * Displacement (r) If the displacement is not a straight line, then we can break the path up into a series of straight lines, and so the work is then the sum of all F*r for each r, or segment of the path. Letting this segment lengths approach zero, the sum becomes an...
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