Author Topic: geometric series (Read 1465 times) Tweet Share

/0 · « **on:** December 21, 2008, 06:15:23 pm »

Consider the sum $S_n = 1+\frac{x}{2}+\frac{x^2}{4}+...+ \frac{x^{n-1}}{2^{n-1}}$

a) Find the possible values of x for which the infinite sum exists. Denote this sum by S.
b) Find the values of x for which $S=2S_{10}$

I did a) alright, but I'm getting some wacky answers for b) :/

chemboy · « **Reply #1 on:** December 21, 2008, 06:35:21 pm »

Just to let you know, sequences has been removed from the specilaist course.

Collin Li · « **Reply #2 on:** December 21, 2008, 07:21:26 pm »

$r = \frac{x}{2}$

$S = \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x}$ , when $0 \leq r \leq 1 \implies 0 \leq x \leq 2$

$S_{10} = 1 + \frac{x}{2} + \frac{x^2}{4} + \ldots + \frac{x^9}{2^9}$

Consider: $P = S\left(\frac{x^{10}}{2^{10}}\right) = \frac{x^{10}}{2^{10}} + \frac{x^{11}}{2^{11}} + \ldots$

Since $S_{10} = S - P = S\left(1 - \frac{x^{10}}{2^{10}}\right)$

When $S = 2S_{10}$

$\implies 1 = 2\left(1 - \frac{x^{10}}{2^{10}}\right)$

$\implies 0 = 1 - \frac{x^{10}}{2^9}$

$\implies 2^9 = x^{10}$

$\implies x = 2^{\frac{9}{10}}$

Check that $0 < x < 2$ , which it is, so this is a solution.

/0 · « **Reply #3 on:** December 21, 2008, 07:30:20 pm »

Wow, smart coblin

Mao · « **Reply #4 on:** December 21, 2008, 08:04:41 pm »

$-2\le x < 2$

I am fairly sure

/0 · « **Reply #5 on:** December 21, 2008, 08:12:54 pm »

According to the answers the domain is -2 < x < 2, very close lol

Collin Li · « **Reply #6 on:** December 21, 2008, 08:16:19 pm »

Ah yes, I meant $|r| < 1$ , so $-1 < r <1$ .

$\implies -2 < r < 2$

dekoyl · « **Reply #7 on:** December 21, 2008, 08:33:47 pm »

Quote from: chemboy on December 21, 2008, 06:35:21 pm

Just to let you know, sequences has been removed from the specilaist course.

Has it?

I was told that although it's not specifically covered, it is still important to have a good understanding of sequences and series.

Mao · « **Reply #8 on:** December 21, 2008, 08:54:30 pm »

oh duh, i'm an idiot

1 - 1 + 1 - 1 + 1 - 1 + 1 ..... is not convergent

orsel · « **Reply #9 on:** December 22, 2008, 02:24:40 am »

Quote from: dekoyl on December 21, 2008, 08:33:47 pm

Quote from: chemboy on December 21, 2008, 06:35:21 pm
Just to let you know, sequences has been removed from the specilaist course.

Has it?

I was told that although it's not specifically covered, it is still important to have a good understanding of sequences and series.

Nah, its completely unnecessary. I skipped it and did fine.

Also, I have not seen a single question in all the past exams I did that relied on series.

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