(25 marks) Consider a wave packet defined by \(\psi(x)=\int_{-\infty}^{\infty} A(k) \cos k x d k,\) where \(A(k)\) is given by
$$ A(k)=\left\{\begin{array}{ll} 1 & \text { for } 0 \leq k \leq k_{0} \\ 0 & \text { otherwise } \end{array}\right. $$
The width of \(A(k)\) is thus equal to \(\Delta k=k_{0}\). (a) Determine the function \(\psi(x)\). (b) What is the value of \(\psi(0) .\) (c) Sketch the function \(\psi(x)\). (d) The function \(\psi(x)\) is sharply peaked at \(x=0\). Define the width of the function by \(\Delta x=2\left|x_{0}\right|,\) where \(x_{0}\) is the position where \(\psi(x)\) first vanishes. Calculate the product \(\Delta x \Delta k\). [Remark: This example illustrates the concept that the widths of a wave packet \(\psi(x)\) and that of its Fourier transform \(A(k)\) are directly related. \(]\)
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(2 points) The area \(A\) of the region \(S\) that lies under the graph of the continuous function \(f\) on the interval \([a, b]\) is the limit of the sum of the areas of approximating rectangles:$$ A=\lim _{n \rightarrow \infty}\left[f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\ldots+f\left(x_{n}\right) \Delta x\right]=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x $$where \(\Delta x=\frac{b-a}{n}\) and \(x_{i}=a+i \Delta x\).The expression$$ A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{\pi}{8 n} \tan \left(\frac{i \pi}{8 n}\right) $$gives the area of the function \(f(x)=\) on...
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