Author Topic: Interesting questions (spesh) (Read 11828 times) Tweet Share

pi · « **Reply #120 on:** November 29, 2010, 01:14:25 pm »

Quote from: iNerd on November 29, 2010, 12:22:22 pm

Quote from: golden on November 29, 2010, 12:20:58 pm
I know why I put it up now, what does where k E Z mean?
k is an element of Z where Z represents the set of integers...I think.

That's right, k is an integer (a 'whole' numbers). Integers are now represented by Z, but in some old textbooks, were noted as J.

Mao · « **Reply #121 on:** November 30, 2010, 08:11:05 pm »

Quote from: brightsky on November 27, 2010, 07:49:15 pm

9. b) i) $1 + \frac{x}{2} + \frac{x^2}{4} + ... = \sum^{\infty}_{n=1} \frac{x^{n-1}}{2^{n-1}} = \frac{2}{2 -x} = S$

Hence the infinite sum exists for all $x \in \mathbb{R} \not \{2\}$

Not quite.. $\sum_{n=1}^{\infty}r^n = \frac{1}{1-r}$ is only true for $|r|<1$ , thus $\sum^{\infty}_{n=1} \frac{x^{n-1}}{2^{n-1}}=\sum^{\infty}_{n=1} \left(\frac{x}{2}\right)^{n-1}$ only converges for $\left| \frac{x}{2} \right| < 1$ , the rest is fairly trivial.

golden · « **Reply #122 on:** December 01, 2010, 09:57:04 am »

I'm in some sense new to this. Could someone please explain it to me?

/0 · « **Reply #123 on:** December 01, 2010, 10:13:12 am »

If you have a geometric series $a+ar+ar^2+ar^3+...+ar^n$ , the sum is

$\sum_{n=0}^m ar^n= a\frac{1-r^{m+1}}{1-r}$

If $|r|>1$ , then as $m \to \infty$ , $r^{m+1}$ will blow up, so the expression will be undefined. In the case where $r=1$ , the expression is also undefined as we have $\frac{0}{0}$ However, if $|r|<1$ , then as $m \to \infty$ , $r^{m+1} \to 0$ , so we have:

$\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$ , $|r|<1$ .

golden · « **Reply #124 on:** December 01, 2010, 10:25:31 am »

Quote from: /0 on December 01, 2010, 10:13:12 am

If you have a geometric series $a+ar+ar^2+ar^3+...+ar^n$ , the sum is

$\sum_{n=0}^m ar^n= a\frac{1-r^{m+1}}{1-r}$

If $|r|>1$ , then as $m \to \infty$ , $r^{m+1}$ will blow up, so the expression will be undefined. In the case where $r=1$ , the expression is also undefined as we have $\frac{0}{0}$ However, if $|r|<1$ , then as $m \to \infty$ , $r^{m+1} \to 0$ , so we have:

$\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$ , $|r|<1$ .

I see, and if it was $\frac{1}{99 - r}$
It would have been |r| < 99 right?

/0 · « **Reply #125 on:** December 01, 2010, 10:32:15 am »

It depends on the geometric series in question. After you have identified 'r' and 'a' in your geometric series, you can use the formula $\sum_{n=0}^\infty ar^n=\frac{a}{1-r}$ only if $|r|<1$ .

e.g.
With $1+\frac{x}{2}+\frac{x^2}{4}+...$

$a=1$ , and $r=\frac{x}{2}$ , so you can find a closed form for the infinite series only if $|r|=|\frac{x}{2}|<1$ . If this condition is not satisfied there doesn't exist a closed form for the infinite series.

i.e. $1+\frac{x}{2}+\frac{x^2}{4}+...=\frac{1}{1-\frac{x}{2}}$ is only valid for $|\frac{x}{2}|<1$ .

See if you can find when the following series converge, and their closed form if they do:
$1-1+1-1+...$
$x+2x^2+4x^3+8x^4+...$
$x-y+(x^2y^2-xy^3)+(x^3y^4-x^2y^5)+...$

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Author Topic: Interesting questions (spesh) (Read 11828 times) Tweet Share

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