1. ## Wordy question

Ok, I'm going to be honest here and say I don't completely understand what I'm being asked here.

f(t)=3sin(t) + 5cos(3t)

suppose that for t >= 0, the function f(t) is non-negative and that it's derivative is bounded, I.e there exists M E R such that for all t >= 0,
|df/dt|<= M.
Let F1(s) = Laplace{f(t)} and F2(s) = Laplace{f^2(t)} and suppose that these transforms are defined for positive values of s. Prove that there exist two constants A, B E R such that for all s >= 0 we have the following upper bound.

F2(s) <= (1/s)(A+BF1(s))

thank you for any help I'm really struggling on this one.

Rodregez

2. Originally Posted by Therodregez
Ok, I'm going to be honest here and say I don't completely understand what I'm being asked here.

f(t)=3sin(t) + 5cos(3t)

suppose that for t >= 0, the function f(t) is non-negative and that it's derivative is bounded, I.e there exists M E R such that for all t >= 0,
|df/dt|<= M.
Let F1(s) = Laplace{f(t)} and F2(s) = Laplace{f^2(t)} and suppose that these transforms are defined for positive values of s. Prove that there exist two constants A, B E R such that for all s >= 0 we have the following upper bound.

F2(s) <= (1/s)(A+BF1(s))

thank you for any help I'm really struggling on this one.

Rodregez
Are you sure you haven't confused two different problems here? f(x)= 3sin(t)+ 5cos(3t) is NOT "non-negative for t> 0". In particular, f(1)= 3sin(1)+ 5cos(3)= -2.4255.