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Chapter 8: Problem 2

Verify that the sequence of functions \(\left\\{f_{n}(x)\right\\}_{n=1}^{\infty}\), defined by \(f_{n}(x)=\) \(\frac{n}{2} e^{-n|x|}\), approaches a delta function.

Short Answer

Expert verified
The given sequence of functions \(f_{n}(x)=\frac{n}{2} e^{-n|x|}\) does not approach a Delta function. However, the sequence of functions \(f_{n}(x)=\frac{n}{4} e^{-n|x|}\) does indeed approach a Delta function as \(n \to \infty\), meaning that it's integral is equal to 1 over the entire real line and its limit as n tends to infinity is \(0\) for \(x\neq0\) and \(\infty\) for \(x=0\).

Step by step solution

01

Understanding the Delta Function

The Dirac Delta function, often denoted as \(\delta(x)\), is not a usual function. Instead, it is a distribution. There are two crucial characteristics of delta function: 1. \(\delta(x) = 0\) for \(x\neq 0\) and \( \delta(x) = \infty \) for \(x=0\). 2. The integral of the delta function over the whole real line is equal to 1, i.e., \(\int_{-\infty}^{\infty} \delta(x)dx = 1\). We will use these properties to check if our function sequence \(f_{n}(x)\) approaches a delta function for large n.
02

Compute the Limit of the Function

We have to compute: \(\lim_{n \to \infty} f_{n}(x)\). If \(x\neq 0\), the result is straightforward, \(\lim_{n \to \infty} \frac{n}{2} e^{-n|x|}=0\). This behavior matches the first property of the delta function. If \(x=0\), our \(f_n(x)\) becomes \(\lim_{n \to \infty} \frac{n}{2}\), which tends to \(\infty\), matching the second property of the delta function.
03

Compute the Integral of the Function

Integrate \(f_{n}(x)\) over all real numbers to see if the function complies with the second property of the Delta function. The integral of \(f_{n}(x)\) from \(-\infty\) to \(\infty\) can be computed as \( \int_{-\infty}^{\infty} \frac{n}{2} e^{-n|x|} dx \). By splitting it into two parts, from \(-\infty\) to 0 and from 0 to \(\infty\), we get two integrals: \( \int_{-\infty}^{0} \frac{n}{2} e^{nx} dx + \int_{0}^{\infty} \frac{n}{2} e^{-nx} dx \). Both of these evaluate to 1, so their sum is 1 + 1 = 2, not matching the second property of the Delta function.
04

Rescaling the Function

To fix the problem noted in step 3, we rescale our function sequence \(f_{n}(x)\) by dividing it by 2. The new sequence is: \(f_{n}(x) = \frac{n}{4} e^{-n |x|}\). Now, compute the integral again, we will find that \(\int_{-\infty}^{\infty} f_{n}(x) dx = 1 \), which fulfills the property of the Delta function.

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