# I Problem when evaluating bounds...Is the result 1 or 0^0?

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1. Nov 13, 2017

### nathangeo

Consider the summation ∑,i=0,n (t^(n-i))*e^(-st) evaluated from zero to infinity.

You could break down the sum into: (t^(n))*e + (t^(n-1))*e + (t^(n-1))*e + ... + (t^(n-n))*e ; where e = e^(-st)

To evaluate this, notice that all terms will go to zero when evaluated at infinity

However, when evaluated at zero, notice the last term of the summation; when i=n : (t^(0))*e^(-st)

(t^0) is equivalent to one so we could rewrite as (1). When evaluating the last term at zero, then, we obtain (1) from the e^(-st) term. But, if you think of (1) as t^0, evaluating the last term at zero will give you 0^0, or which cannot be evaluated.

My question is, how do we handle a situation like this? I can change (t^0) to (1) and then evaluate the bounds; or leave it as it is (t^0), and find that evaluating the bound at zero creates an indeterminant value.

2. Nov 13, 2017

From what it looks like, you are evaluating an integral. If that is the case, an isolated point is in general of no consequence, and the fact that you can't evaluate $0^0$ does not affect the result. You can take the limit $t \rightarrow 0$ for the zero point.

3. Nov 13, 2017

### nathangeo

Interesting, so that evaluates to one correct? And this is an integral. Specifically the Laplace transform of [(A_n)(t^n)] so any nth root polynomial

4. Nov 13, 2017

If you are evaluating the Laplace transform of $t^n$, you might try this link: I am not sure exactly where your summation comes from.

5. Nov 13, 2017

### nathangeo

It comes through inspection. And that summation doesn't appear in the problem anyways but it's similar. What I mean is a polynomial of degree n. For example if n=3 you would have A_0+A_1(t)+A_2(t^2)+A_3(t^3)

6. Nov 13, 2017

### nathangeo

I know the Laplace transform is a linear operator but i wanted to do it all in one operation.

7. Nov 14, 2017

### Staff: Mentor

In series like this, you would typically evaluate 00 as 1, as that usually makes the function continuous (if it is well-defined around 0). If your exponent is not discrete, things can get more complicated.