I'd be interested in seeing various proofs that 0^0=1, if anyone has any

[Johnny5]

Well I just used L'Hopital's rule' date=' I didn't actually prove that the limit is -1, but it is plain to see. But to prove it, some would ask for an epsilon-delta proof, though I don't feel that's necessary, when you have something like L'Hopitals rule at your disposal.

 

 

I have to think about this... absolute value of ratio of consecutive terms in the series, as the index tends to infinity.

 

Let me try to interpret that formula clearly.

 

[math'] (1+x)^{1/2} = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{3/2 -k}{k}

[/math]

 

The series is going to be composed of terms.

 

Power series have the following form:

 

[math] C_0+C_1x+C_2x^2+C_3X^3+...C_nx^n+... [/math]

 

The first term of the series is C0, the second term of the series is C1x, and so on. The nth term of the series is Cn xⁿ.

 

Now, pick a term of the series at random.

 

How about n=p=999493245932459934593495

 

Forget about all other terms in the series but this one. So ignore the decimal representation of this term, and call it p.

 

So the pth term of the series is given by:

 

[math] x^p \prod_{k=1}^{k=p} \frac{3/2 -k}{k}

[/math]

 

Now, I am going to think about what you said again, very carefully.

 

The absolute value of my f(k) is the radius of convergence... being the ratio of the absolute value of consecutive terms of the series as the index k tends to infinity. hmm

 

I don't see how you get "absolute value" from anything which I've done.

 

Actually, let me let p go from 1 to 4, to see the pattern you say is there.

 

Case 1: p=1

 

[math] x^1 \prod_{k=1}^{k=1} \frac{3/2 -k}{k} = x \frac{3/2 -1}{1} = \frac{x}{2}

[/math]

 

Case 2: p=2

 

[math] x^2 \prod_{k=1}^{k=2} \frac{3/2 -k}{k} = x^2 \frac{3/2 -1}{1}\frac{3/2 -2}{2} = x^2 (\frac{1}{2}) (\frac{-1}{4})[/math]

 

Case 3: p=3

 

[math] x^3 \prod_{k=1}^{k=3} \frac{3/2 -k}{k} = x^3 \frac{3/2 -1}{1}\frac{3/2 -2}{2} \frac{3/2 -3}{3} = x^3 (\frac{1}{2}) (\frac{-1}{4})(-\frac{1}{2}) [/math]

 

[math] x^4 \prod_{k=1}^{k=4} \frac{3/2 -k}{k} = x^4 (\frac{1}{2}) (\frac{-1}{4})(-\frac{1}{2})(\frac{-5}{8}) [/math]

 

now you say... "ratio of absolute value of consecutive terms of series."

 

 

Yes. I went off and thought about it Matt. My f(k) is the ratio of consecutive terms of the series. Yes.

 

Disregard the powers of x, and focus on two consecutive coefficients of terms in the series, say the coefficient of the pth term, and the coefficient of the p+1th term. Now, take the ratio of the coefficients of the pth term, and the p+1th term, like so:

 

Coefficient of pth term:

 

[math] \prod_{k=1}^{k=p} \frac{3/2 -k}{k}

[/math]

 

Coefficient of p+1th term:

 

[math] \prod_{k=1}^{k=p+1} \frac{3/2 -k}{k}

[/math]

 

[math] \text{Ratio of two consecutive coefficients} [/math]

 

 

[math] \frac{\prod_{k=1}^{k=p+1} \frac{3/2 -k}{k}}{\prod_{k=1}^{k=p} \frac{3/2 -k}{k}} = \frac{\frac{3/2-(p+1)}{p+1}\prod_{k=1}^{k=p} \frac{3/2 -k}{k}}{\prod_{k=1}^{k=p} \frac{3/2 -k}{k}} =\frac{3/2-(p+1)}{p+1} [/math]

 

And p is arbitrary. So letting k=p+1, implying k is arbitrary as well, we have, as the ratio of two consecutive terms of the series:

 

[math] \frac{3/2-k}{k} [/math]

 

which is my f(k), exactly as you said.

 

 

SOOOOOOOOOOOOOO

 

In taking the limit, as k tends to infinity, of f(k), we are in fact performing the ratio test, precisely because f(k) is the ratio of the nth term, and the n-1th term.

 

Of course I still don't see where the absolute value bit comes in at, but I see exactly what you meant. Just the f(k) is the ratio of two consecutive terms of the series. Uh huh.

 

 

So by taking the limit as k approaches infinity of f(k), I am in effect performing the ratio test.