scigal89
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In class we worked out the following
<br /> <br /> \int e^{ik(x-X)}dk=\frac{e^{ik(x-X)}}{i(x-X)}\approx \frac{sin[k(x-X)]}{x-X}<br /> <br />
by taking the real part of the solution. My teacher wants us to graph the following functions
<br /> <br /> \psi_{1} \sim \frac{sin(x)}{x}<br /> <br />
<br /> <br /> \psi_{2} \sim \frac{sin(x)}{x}-\frac{1}{2}\frac{sin(2x)}{2x}<br /> <br />
The second function, though, has a little dip in it at 0 that shouldn't be there. He says that's due to the fact that the functions aren't normalized and that we should be doing Gram-Scmidt or some other procedure to obtain the proper results. However, when I do Gram-Schmidt I get something nasty that can't possibly be correct (and most importantly, doesn't correct the problem). So what am I doing wrong?
<br /> <br /> \int e^{ik(x-X)}dk=\frac{e^{ik(x-X)}}{i(x-X)}\approx \frac{sin[k(x-X)]}{x-X}<br /> <br />
by taking the real part of the solution. My teacher wants us to graph the following functions
<br /> <br /> \psi_{1} \sim \frac{sin(x)}{x}<br /> <br />
<br /> <br /> \psi_{2} \sim \frac{sin(x)}{x}-\frac{1}{2}\frac{sin(2x)}{2x}<br /> <br />
The second function, though, has a little dip in it at 0 that shouldn't be there. He says that's due to the fact that the functions aren't normalized and that we should be doing Gram-Scmidt or some other procedure to obtain the proper results. However, when I do Gram-Schmidt I get something nasty that can't possibly be correct (and most importantly, doesn't correct the problem). So what am I doing wrong?
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