I have this formula:

$\displaystyle {c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

How do I convert this to a series with $\displaystyle \sum$

From n = 1 to n = 10

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- Jun 21st 2006, 04:40 AMchanceySeries help
I have this formula:

$\displaystyle {c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

How do I convert this to a series with $\displaystyle \sum$

From n = 1 to n = 10 - Jun 21st 2006, 05:35 AMThePerfectHackerQuote:

Originally Posted by**chancey**

- Jun 21st 2006, 05:45 AMSoroban
Hello, Malay!

Edit: You're right! . . . *blush*

. . . . I'll correct it now.

Quote:

I have this formula: $\displaystyle (c_n)^2 \:= \:\frac{1}{2}(c_{n-1})^2$

How do I convert this to a series with $\displaystyle \sum^{10}_{n=1}$

We have: .$\displaystyle c_n\;=\;\frac{1}{\sqrt{2}}c_{n-1}$

We have a geometric series with first term $\displaystyle c_1 = a$ and common ratio $\displaystyle r = \frac{1}{\sqrt{2}}$

The $\displaystyle k^{th}$ term is: .$\displaystyle c_n\:=\:\frac{a}{2^{\frac{n-1}{2}}} $

Then: .$\displaystyle \sum^{10}_{k=1}\frac{a}{2^{\frac{n-1}{2}}} \;=\;a \cdot \frac{1 - \frac{1}{(\sqrt{2})^{10}}} {1 - \frac{1}{\sqrt{2}}} $

- Jun 21st 2006, 06:20 AMmalaygoelQuote:

Originally Posted by**Soroban**

It should be

$\displaystyle (c_2)^2\;=\;\frac{a^2}{2}$

Keep Smiling

Malay - Jun 21st 2006, 06:24 AMmalaygoelQuote:

Originally Posted by**chancey**

I observed that

$\displaystyle \frac{c_n}{c_{n-1}}=\frac{1}{\sqrt{2}}$

Hence the given sequence is a G.P. with common ratio $\displaystyle \frac{1}{\sqrt{2}}$

Keep Smiling

Malay - Jun 21st 2006, 12:58 PMchancey
I need the series to do this:

Example (not the actual formula, changed it so the numbers were easier)

$\displaystyle c_1 = 1$

$\displaystyle c_2 = \frac{1}{2} 1 = 0.5$

$\displaystyle c_3 = \frac{1}{2} 0.5 = 0.25$

$\displaystyle c_4 = \frac{1}{2} 0.25 = 0.125$

$\displaystyle c_n = ?$

Where each number in the series has to be calculated with the previous number. My question is, can that be put into a series so that i can run a formula once to find $\displaystyle n_{10}$ or do I have to do it 10 times - Jun 21st 2006, 01:19 PMThePerfectHacker
As explained in the previous discussions this sequence is geometric, thus, if,

$\displaystyle c_1=a$

then,

$\displaystyle c_n=(1/\sqrt{2})^{n-1}a$

or,

$\displaystyle c_n=(-1/\sqrt{2})^{n-1}a$

Where $\displaystyle a$ is the initial term.

Note, it is not possible to find $\displaystyle a$, it can be anything number. - Jun 21st 2006, 01:27 PMSoroban
Hello, chancey!

Quote:

I need the series to do this:

Example (not the actual formula, changed it so the numbers were easier)

$\displaystyle c_1 = 1$

$\displaystyle c_2 = \frac{1}{2}\cdot 1 = \frac{1}{2}$ . Don't use decimals!

$\displaystyle c_3 = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$

$\displaystyle c_4 = \frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}$

$\displaystyle c_n = \:?$

Where each number in the series has to be calculated with the previous number.

My question is, can that be put into a series so that i can run a formula once to find $\displaystyle n_{10}$

or do I have to do it 10 times

. . and to be able to "eyeball" a sequence and determine its general form.

This sequence is: .$\displaystyle 1,\;\frac{1}{2},\;\frac{1}{4},\;\frac{1}{8},\; \frac{1}{16},\;\hdots$

How long does it take for you to see that the denominators are*doubled*each time?

With a little thought, we see that the $\displaystyle n^{th}$ denominator is: $\displaystyle 2^{n-1}$

So the general term is: .$\displaystyle c_n\:= \:\frac{1}{2^{n-1}} $ - Jun 21st 2006, 10:30 PMchanceyQuote:

Originally Posted by**Soroban**

So:

$\displaystyle {c_n}^2 = \frac{1}{2} {c_{n-1}}^2 - 2$

$\displaystyle {c_n} = \sqrt{\frac{1}{2} {c_{n-1}}^2 - 2}$

$\displaystyle {c_n} = (\sqrt{\frac{1}{2} a^2 - 2} )^{n-1}$