Plotting the approximation of the Dirac delta function

Lambda96
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Homework Statement
Plot ##g^{\epsilon}(x)## for ##\epsilon=1## , ##\epsilon=\frac{1}{2}## and ##\epsilon=\frac{1}{4}##
Relevant Equations
none
Hi,

I am not sure if I have solved the following task correctly

Bildschirmfoto 2024-01-15 um 16.34.11.png


I did the plotting in mathematica and got the following

Bildschirmfoto 2024-01-15 um 16.48.02.png

Would the plots be correct? What is meant by check for normalization, is the following meant?

For the approximation for ##\epsilon > 0##, does it mean that for the area of ## 0 < x < \epsilon## the area must be 1, so for the case ##\epsilon=1## the total area would be 1, the half for ## 0 < x < \epsilon## would then be ##\frac{1}{2}##, so the normalization constant should be 2?
 
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To sketch usually means to draw by hand. Is this behavior that you expected? Look at the formula. Between minus epsilon and +epsilon, what is the function doing (does it change with x)?

Normalization refers to the area under the "curve" equal to 1 over the entire spectrum. Do these graphs satisfy that?
 
Last edited:
Lambda96 said:
Would the plots be correct?
Yes, those plots look correct.
Lambda96 said:
What is meant by check for normalization, is the following meant?
It means to check that the integral is 1.
Lambda96 said:
For the approximation for ##\epsilon > 0##, does it mean that for the area of ## 0 < x < \epsilon## the area must be 1,
No. It means that the area of ##- \epsilon < x < \epsilon ## must be 1.
Lambda96 said:
so for the case ##\epsilon=1## the total area would be 1, the half for ## 0 < x < \epsilon## would then be ##\frac{1}{2}##, so the normalization constant should be 2?
No. No further normalization is needed. The area between each graph and the x axis is already 1.
 
Thank you scottdave and FactChecker for your help 👍👍
 
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