Series help

• Jun 21st 2006, 04:40 AM
chancey
Series help
I have this formula:
${c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

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

From n = 1 to n = 10
• Jun 21st 2006, 05:35 AM
ThePerfectHacker
Quote:

Originally Posted by chancey
I have this formula:
${c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

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

From n = 1 to n = 10

Not really sure what you are asking.
• Jun 21st 2006, 05:45 AM
Soroban
Hello, Malay!

Edit: You're right! . . . *blush*
. . . . I'll correct it now.

Quote:

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

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

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

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

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

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

• Jun 21st 2006, 06:20 AM
malaygoel
Quote:

Originally Posted by Soroban
Hello, chancey!

I had to baby-talk my way through it . . .

[size=3]
Let $c_1 = a$, the first term.

Then crank out the first few terms:

$c_1\;=\;a$

$c_2\;=\;\frac{a^2}{2}$

But someone check my work . . . please!

Hello Soroban!
It should be
$(c_2)^2\;=\;\frac{a^2}{2}$
Keep Smiling

Malay
• Jun 21st 2006, 06:24 AM
malaygoel
Quote:

Originally Posted by chancey
I have this formula:
${c_n}^2 = \frac{1}{2}{c_{n-1}}^2$

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

From n = 1 to n = 10

${c_n}^2 = \frac{1}{2}{c_{n-1}}^2$
I observed that
$\frac{c_n}{c_{n-1}}=\frac{1}{\sqrt{2}}$
Hence the given sequence is a G.P. with common ratio $\frac{1}{\sqrt{2}}$

Keep Smiling
Malay
• Jun 21st 2006, 12:58 PM
chancey
I need the series to do this:

Example (not the actual formula, changed it so the numbers were easier)
$c_1 = 1$

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

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

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

$c_n = ?$

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 $n_{10}$ or do I have to do it 10 times
• Jun 21st 2006, 01:19 PM
ThePerfectHacker
As explained in the previous discussions this sequence is geometric, thus, if,
$c_1=a$
then,
$c_n=(1/\sqrt{2})^{n-1}a$
or,
$c_n=(-1/\sqrt{2})^{n-1}a$
Where $a$ is the initial term.
Note, it is not possible to find $a$, it can be anything number.
• Jun 21st 2006, 01:27 PM
Soroban
Hello, chancey!

Quote:

I need the series to do this:

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

$c_1 = 1$

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

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

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

$c_n = \:?$

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 $n_{10}$
or do I have to do it 10 times
You're expected to be familiar with Arithmetic Series and Geometric Series by now
. . and to be able to "eyeball" a sequence and determine its general form.

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

How long does it take for you to see that the denominators are doubled each time?
With a little thought, we see that the $n^{th}$ denominator is: $2^{n-1}$
So the general term is: . $c_n\:= \:\frac{1}{2^{n-1}}$
• Jun 21st 2006, 10:30 PM
chancey
Quote:

Originally Posted by Soroban
Hello, chancey!

You're expected to be familiar with Arithmetic Series and Geometric Series by now
. . and to be able to "eyeball" a sequence and determine its general form.

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

How long does it take for you to see that the denominators are doubled each time?
With a little thought, we see that the $n^{th}$ denominator is: $2^{n-1}$
So the general term is: . $c_n\:= \:\frac{1}{2^{n-1}}$

Of course I can see that, that was just an example, im not really that interested in the answer, but the process is what I want to know.

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

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

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