Math Is Fun Forum

He. Right. So I try doing that... and it gets pretty complicated quickly.
I get:

.

So, yes i did know that. But when I try to simplify its... not easy?

You're right.  I was looking in the textbook. And it said use the ratio test but looking at the reason WHY i see that root test is the same. And as you point out above is CLEARLY easier.

So my 2nd question is this:
If I graph the summands for n=3,6,9, and 12. I get:
http://render-2.snapfish.com/render2/is=Yup6aQQ%7C%3Dup6RKKt%3Axxr%3D0-qpDPfRt7Pf7mrPf
rj7t%3DzrRfDUX%3AeQaQxg%3Dr%3F87KR6xqpxQQ0QxJPPxQ0oxv8uOc5xQQQGloQnaanGPqpfVtB
%3F*KUp7BHSHqqy7XH6gX0QPGe%7CRup6GQP%7C/of=50,590,369
(dont think i can add pics to this)

Then the question asks, "do you expect convergence at either or both of the endpoints of the interval of convergence of the infinite series and prove it."

I'm not sure what this is asking?
Is it asking if the summands approach 0 when |x| equals the radius of convergence? or...?

Okay, new question (similar i suppose)... If i calculated the limits of the r(n) and p(n) from above using Mathematica and I get an interval, what exactly does that mean?

For

r(n) gives no response
p(n) gives interval from 1/2 to 2.

For

r(n) gives interval from 0 to infinity.
p(n) gives interval from 1/3 to 1/2.

Could I possible say...

The mean value theorem states "Given a function f that is differentiable at all points strictly between a and x and continuous at all points on the closed interval from a to x, there
exists a real number c strictly between a and x such that [f(x)-f(a)]/x-a = f'(c)."

Thus we can prove the question by the contrapositive. Assume f(b)=f(a) and show that there must be an f'(x)=0. By the mean value theorem we have:
[f(b)-f(a)]/b-a = 0/(b-a) = 0 = f'(x) for x strictly between a and b.

Really? I guess I'm not seeing it.
Rolle's says, "Let f be a function that is continuous on [a,b] and differentiable on (a,b) AND for which f(a)=f(b)=0. There exists at least one c, a<c<b for which f'(c)=0."

So i see that I'm assuming f(a)=f(b) but they dont necessarily have to equal 0?
This proof is killing me.

I also kind of feel like using the Intermediate Value Theorem... but that's not in this section of the book.
( what is: Generalized Mean Value Theorem, Cauchy's Remainder Theorem, L'Hospital's Rule
and Darboux's Theorem.)

Gotcha mathsyperson thanks!

So because h(0) is less than 0 and h(1) is greater than 0 then at some point in the interval of (0,1) we will have a value,c, so that h(c) = 0. And thus, since h(x)=f(x)-g(x) this point is shared by f(x) and g(x)!

Thank you!!!

Actually... for #1 (the verify by substitution part) is this correct:

n choose r =  n! / [r!(n-r)!]

Applied to left side:

2n choose (n+1) = (2n)! / [(n+1)! (2n-(n+1)!]
= (2n)! / [(n+1)!(n-1)!]

2n choose n = (2n)! / [(n!(2n-n)!]
= (2n)! / [n!n!]

Then
(2n)! / [(n+1)!(n-1)!] + (2n)! / [n!n!]
= (2n)! {1/[(n+1)!(n-1)!] + 1/(n!n!)}

The common denominator will be (n+1)!n!
= (2n)! {n/[(n+1)!n!] + (n+1)/[(n+1)!n!]
= (2n)! {(n+n+1) / [(n+1)!n!]}
= (2n)! {(2n+1) / [(n+1)!n!]}
= (2n+1)! / [(n+1)!n!]

Multiply the top and bottom by (2n+2)
= (2n+2)! / [(2n+2)(n+1)!n!]}
= (2n+2)! / [2(n+1)(n+1)!n!]
= (1/2)(2n+2)! / [(n+1)!(n+1)!]
= (1/2) (2n+2 choose n+1)

?