1. ## Follow the bouncing limit

Hi everyone, I'm running into a snag on this problem and am asking for assistance. Sorry that this is a jpg, I have no idea how to use the Math Formula options.

2. Originally Posted by HeirToPendragon
Hi everyone, I'm running into a snag on this problem and am asking for assistance. Sorry that this is a jpg, I have no idea how to use the Math Formula options.

See this.

3. While it does appear that this is what I'm looking for:

$
n = (a_n+1)^n \geq 1 + na_n + \tfrac{1}{2}n(n-1)a_n^2 \geq \tfrac{1}{2}n(n-1)a_n^2
$

Can you tell me how you got the middle equation? The problem I'm having with this current book is that it does exactly what you did here. It shows me a resolution without showing any of the scratch.

4. Originally Posted by HeirToPendragon
$
n = (a_n+1)^n \geq 1 + na_n + \tfrac{1}{2}n(n-1)a_n^2 \geq \tfrac{1}{2}n(n-1)a_n^2
$
(Here $n\geq 2$).

This follows by binomial theorem: $(a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k}$

If $a,b\geq 0$ then $(a+b)^n = \sum_{k=0}^2 {n\choose k} a^k b^{n-k} + \sum_{k=2}^n {n\choose k}a^kb^{n-k} \geq \sum_{k=0}^2 {n\choose k}a^k b^{n-k} =$ $a^n + nab^{n-2}+\tfrac{1}{2}n(n-1)a^2b^{n-2}$.

5. I don't think I've ever even seen that theorem before, which means I probably don't have access to using it

*sigh*

6. Originally Posted by HeirToPendragon
I don't think I've ever even seen that theorem before, which means I probably don't have access to using it
This is the binomial theorem that you should have learned in high-school.
If you are an analysis class, surly you must have learned it.

7. It's possible. High School was 4 years ago and I didn't exactly have the best teachings (like, the sin and cos graphs were not part of my knowledge until I hit college and had to teach myself trig).

But I'll take your word for it and use that.