How Does the Delta Function and Its Derivative Interact with Shifted Functions?

AI Thread Summary
The discussion focuses on proving the relationship between the delta function and its derivative when interacting with shifted functions. The key equation to prove is that the integral of the derivative of the delta function multiplied by a shifted function equals the negative derivative of that function evaluated at the shift. The solution involves applying integration by parts and the integral definition of the delta function. The steps show that the integral of δ'(x) * f(x-a) simplifies to f(-a) - f'(-a). This illustrates the interaction between the delta function, its derivative, and shifted functions in the context of distribution theory.
gimak
Messages
56
Reaction score
2

Homework Statement


Prove the following
'()( − ) = −′()
-∞δ'(x)*f(x-a) = -f'(a)

Homework Equations


-∞δ'(x-a)*f(x) = f(a)

The Attempt at a Solution


[/B]
-∞ δ'(x)*f(x-a) = ∫δ(x)*f(x-a)dx-∫f'(x-a)*δ(x)dx = f(-a) - f'(-a)
Went from 1st to second by integration by parts
Used integral definition of delta function to go to 3rd part
 

Attachments

  • upload_2017-7-14_22-37-57.png
    upload_2017-7-14_22-37-57.png
    53.1 KB · Views: 498
Physics news on Phys.org

The Attempt at a Solution



1) ∫-∞ ∞δ'(x)*f(x-a)
2) = ∫δ(x)*f(x-a)dx-∫f'(x-a)*δ(x)dx
3) = f(-a) - f'(-a)
Went from 1st to second by integration by parts
Used integral definition of delta function to go to 3rd part

Made above easier to read
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top