sequence question

PvtBillPilgrim · #1 $Old$ November 14th, 2007, 11:11 AM

Could someone show me what to do here:
For each positive integer n, let fn be the function defined by
fn(x) = x^2n + x^(2n−1) + ... + x^2 + x + 1. Let mn be the minimum of fn on the interval [−1,0]. Prove that the sequence (mn) converges to a limit A and then find A.

Thank you for your help.

ThePerfectHacker · #2 $Old$ November 14th, 2007, 11:34 AM

Quote:

Originally Posted by PvtBillPilgrim $View Post$

Could someone show me what to do here:
For each positive integer n, let fn be the function defined by
fn(x) = x^2n + x^(2n−1) + ... + x^2 + x + 1. Let mn be the minimum of fn on the interval [−1,0]. Prove that the sequence (mn) converges to a limit A and then find A.

Thank you for your help.

We can wrtite $x^{2n}+x^{2n-1}+...+x+1 = \frac{1-x^{2n+1}}{1-x} \mbox{ if }x\not = 1 \mbox{ and }2n+1 \mbox{ if } x=1$ . So the mimmum occurs for zero for all $n$ .

I think you missted something because it is not a very interesting problem.

PvtBillPilgrim · #3 $Old$ November 14th, 2007, 02:56 PM

Is what you did correct?

f_n+1(x) - f_n(x) is decreasing (I believe). This should imply that (m_n) is decreasing. And it is bounded by [-1,0] so it must be convergent.

As n goes to infinity, f_n(x) goes to 1/(1-x) on (-1,0]. How do I find the limit A though (to what (m_n) converges)?

Thanks for your help.

ThePerfectHacker · #4 $Old$ November 14th, 2007, 07:20 PM

Quote:

Originally Posted by PvtBillPilgrim $View Post$

Is what you did correct?

f_n+1(x) - f_n(x) is decreasing (I believe). This should imply that (m_n) is decreasing. And it is bounded by [-1,0] so it must be convergent.

As n goes to infinity, f_n(x) goes to 1/(1-x) on (-1,0]. How do I find the limit A though (to what (m_n) converges)?

Thanks for your help.

I make a minor mistake. I did it on the interval [0,1] you wanted on the interval [-1,0]. But can you use the same approach and do the problem yourself?

PvtBillPilgrim · #5 $Old$ November 14th, 2007, 07:29 PM

Well do you know what the limit A is?

I think it may be 1/2 just looking at my limit above as n goes to infinity. Does this sound right?

ThePerfectHacker · #6 $Old$ November 14th, 2007, 07:39 PM

I did not do the limit. I told you these polynomial can be written as $f(x) = \frac{1-x^{2n+1}}{1-x}$ . Now how do you find the minimum? Find the derivative and make it zero. Can you do that? That will be your sequence.