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confinement
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I am trying to show that a certain sum rigorously tends to a Reimann integral, but I have a certain sticking point that you will find below.
S = \sum_{n = 1}^{\infty} c_n \phi_n
<br /> \phi_n = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L})<br />
In this case the c_n can be assumed to be such that the series converges. Basically I want to show that the limit of S as L goes to infinity is a Reimann integral by definition. We have:
\lim_{L\to\infty}\sum_{n = 1}^{\infty} \sqrt{\frac{2}{L}} c_n \sin(\frac{n \pi x}{L})
consider the function sin(k), and define a partition of the positive half of the real line by the points n pi / L. Then the mesh of the partition is pi / L. To deal with the constants define c_n = c(n pi x / L) = c(k_n). Then we have:
\lim_{L\to\infty}\sum_{n = 1}^{\infty} c(k_n)\sin(k_n) \sqrt{\frac{2}{L}}
This is almost a Reimann integral by definition, we have an infinite sum of values of a function multiplied by something small, and the mesh of the partition of points is going to zero, the problem is that the mesh of the partition is \Delta k_n = \pi / L, not the \sqrt{\frac{2}{L}} that I have. The constant factor is obviously not a problem, but I cannot pretend that L and sqrt(L) are the same for this purpose. Is there anyway to recover, or does this 'almost integral' correspond to anything?
S = \sum_{n = 1}^{\infty} c_n \phi_n
<br /> \phi_n = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L})<br />
In this case the c_n can be assumed to be such that the series converges. Basically I want to show that the limit of S as L goes to infinity is a Reimann integral by definition. We have:
\lim_{L\to\infty}\sum_{n = 1}^{\infty} \sqrt{\frac{2}{L}} c_n \sin(\frac{n \pi x}{L})
consider the function sin(k), and define a partition of the positive half of the real line by the points n pi / L. Then the mesh of the partition is pi / L. To deal with the constants define c_n = c(n pi x / L) = c(k_n). Then we have:
\lim_{L\to\infty}\sum_{n = 1}^{\infty} c(k_n)\sin(k_n) \sqrt{\frac{2}{L}}
This is almost a Reimann integral by definition, we have an infinite sum of values of a function multiplied by something small, and the mesh of the partition of points is going to zero, the problem is that the mesh of the partition is \Delta k_n = \pi / L, not the \sqrt{\frac{2}{L}} that I have. The constant factor is obviously not a problem, but I cannot pretend that L and sqrt(L) are the same for this purpose. Is there anyway to recover, or does this 'almost integral' correspond to anything?
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