- #1
Turion
- 145
- 2
Please ignore this thread.
Last edited:
Laplace Transform - Learn & Calculate Transforms Easily
- Thread starter Turion
- Start date
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Homework Help Overview
The discussion revolves around the Laplace Transform, specifically the integral definition and its evaluation. Participants are exploring the application of Theorem 4.3.3 and considering different approaches to compute the transform.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the integral form of the Laplace Transform and suggest splitting it into segments for evaluation. There are questions regarding the implications of separating cases based on whether n is even or odd. Some participants also express uncertainty about the correctness of a solutions manual.
Discussion Status
The discussion is ongoing, with participants offering different perspectives on the evaluation of the Laplace Transform. There is a suggestion to reference Theorem 4.3.3, indicating a potential direction for further exploration, though no consensus has been reached.
Contextual Notes
Some participants note the absence of specific theorems in the initial problem statement, which may affect the approach to the solution. There is also a mention of a solutions manual that may contain inaccuracies.
- #2
Turion
- 145
- 2
The correct solution:
- #3
Turion
- 145
- 2
Actually, I think the solutions manual is wrong. Please ignore this thread.
- #4
brmath
- 337
- 34
It would have helped if you stated Theorem 4.3.3. That said, we could try it without any special theorem. We have
##Lf = \int_0^{\infty} e^{-st}f(t)dt ##
which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.
Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.
Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.
##Lf = \int_0^{\infty} e^{-st}f(t)dt ##
which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.
Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.
Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.
- #5
Turion
- 145
- 2
brmath said:
Sorry for wasting your time, but I already posted:
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