Laplace Transform - Learn & Calculate Transforms Easily

Turion
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Please ignore this thread.
 
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The correct solution:

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Actually, I think the solutions manual is wrong. Please ignore this thread.
 
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.
 
brmath said:
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.

Sorry for wasting your time, but I already posted:

Actually, I think the solutions manual is wrong. Please ignore this thread.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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