In trying to understand the Laplace transform of the nth derivative of a function, i have started by calculating the transform of the first derivative:
would result in , which in the solution leads to just . This leads to my question: How do we evaluate what leads to, since we do not know what kind function f(x) is? The denominator's infinity comes from an exponential function, which goes faster towards infinity than a non-exponential, but what if f(x) were also an exponential?
If anyone could reply, i would be grateful.
I don't really know what those restrictions are, but what i wanted to know was
what happened to f(t) and e^(-pt) when t->infinity, in other words, why was only f(0) the only one remaining? I've seen that demonstration, but i do not understand the part about why just f(0) remains, i don't understand what happens to f(t)*e^(-pt) when t-> infinity.
Could you please explain in a bit more detail? I am guessing it may be because of those restrictions you mentioned, but i do not know what those are, but would be very appreciative if you would explain
I have found this explanation http://math.fullerton.edu/mathews/c2...eTheorem.1.pdf, however i do not understand what the term 'exponential order' refers to. Does it refer to e^x?
I've found another constraint in the form of , where
Is the one you provided stating the same as this one? What i understand from this one is that all positive values of the image of the original function must not exceed that exponential. The reason why the restriction on M is so that the exponential is only positive and the constraint on K is so that, well, if K were less than 0, then that would become , which i'm guessing doesn't express the conditions of the Laplace transform. Am i correct in my judgement on that?
Why this restriction must hold, from what i have read, is so that the output of the transform is a finite value. What i do not understand is how that condition assures it will be a finite value? I am trying fairly hard to understand this and not just go by some formulas, so i would greatly appreciate further feedback
I am sorry for the late reply, but i had to look around and get myself better informed before i asked any further.
This is a sufficient condition for the Laplace transform to exist.
This implies that the improper integral must converge.
Note I said it is sufficient, but not necessary. The function
Has a Laplace transform but for any see what happens as