A) A Dirac delta function is not a function, but it can be obtained as a limit o

Question

A)

A Dirac delta function is not a function, but it can be obtained as a limit of a sequence of functions. For example, consider the sequence

delta_n(x) = n for -1/(2n) < x < 1/(2n), but delta_n(x) = 0 outside that range. Verify that, in the limit n approaches infinite, Integral( dx*delta(x)*f(x)) from -infinite to positive infinite approaches f(0) for any continuous function f(x).

B)

Consider the discontinuous function f(x) = 0 for x (does not equal) 0, f(0) = 1. With the denition of the delta function from part A, do we have Integral( dx*delta(x)*f(x)) from -infinite to positive infinite = f(0)? Explain your answer.

Explanation / Answer

(A) => (int_{-infty}^{infty} delta_n(x) cdot f(x) dx) because (delta_n(x)) is zero except (-1/2n, 1/2n) so, => (lim_{n -> infty} int_{rac{-1}{2n}}^{rac{1}{2n}} delta_n(x) cdot f(x) dx) in the limit of n-->(infty) the following approx. to the above integral would be correct. => (lim_{n -> infty} f(0) int_{rac{-1}{2n}}^{rac{1}{2n}} delta_n(x) dx) = f(0) (B) the above shown derivation is valid only the continuous function around x=0 . But as we modeled dirac delta function we can also model the discontinuous functions. hence the outcome would still be valid. (for example, take f(x) = 0 every where except (-1/2n, 1/2n) where it has value 1) => (lim_{n -> infty} f(0) int_{rac{-1}{2n}}^{rac{1}{2n}} delta_n(x) dx) = f(0)

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