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Physics 160 Homework 2

Define the function we will be using for the forward differencing as \[u=ge^{ikx_j}\] \[\frac{\partial{u}}{\partial{x}}=iku\] and the forward differencing equation as \[\Delta_x^{'}f_j=\frac{f_{j+1}-f_j}{\Delta}\] where are function \(f\) will be our functuon \(u\).

Putting this is we get \[\Delta_x^{'}u=\frac{ge^{ikx_{j+1}}-ge^{ikx_j}}{\Delta}\] define \(x_{j+1}=x_j+\Delta\) then the above equation becomes \[\Delta_x^{'}u=\frac{ge^{ikx_j+\Delta}-ge^{ikx_j}}{\Delta}\]

Factoring we get \[\Delta_x^{'}u=\frac{ge^{ikx_j}(e^{ik\Delta}-1)}{\Delta}\] and taylor expanding we get \[\approx\frac{ge^{ikx_j}(1+ik\Delta+\frac{(ik\Delta)^2}{2}-1)}{\Delta}\] Then factoring \[=\frac{ik\Delta ge^{ikx_j}(1+\frac{ik\Delta}{2})}{\Delta}\]. then \[\Delta_x^{'}u=\frac{\partial{u}}{\partial{x}}(1+\frac{ik\Delta}{2})\] So \[\Delta_x^{'}=(1+\frac{ik\Delta}{2})\frac{\partial{}}{\partial{x}}\]

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