• Jul 22nd 2006, 04:24 AM
macca101
Hello !!

Give the Taylor series about 0 for f and determine a range of validity for the series

$
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}
$

Using the standard taylor series
$
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1$

Substituting $
x = -\frac{1}{6}x
$

And taking 2 in the binomial series gives

$
1+(2)*(-\frac{1}{6}x)+\frac{(2)(3)}{2!}(-\frac{1}{6}x)^2+\frac{(2)(3)(4)}{3!}(-\frac{1}{6}x)^3
$

Gives

$
1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3
$

as

$
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1$

Then
$
\frac{1}{(1+\frac{1}{6}x)^2}
$

is valid for

$
-1<-\frac{1}{6}x<1
$

Thanks Macca
• Jul 22nd 2006, 05:41 AM
Quick
Quote:

Originally Posted by macca101
Hello !!

Give the Taylor series about 0 for f and determine a range of validity for the series

$
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}
$

Using the standard taylor series
$
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1$

Substituting $
x = -\frac{1}{6}x
$

And taking 2 in the binomial series gives

$
1+(2)*(-\frac{1}{6}x)+\frac{(2)(3)}{2!}(-\frac{1}{6}x)^2+\frac{(2)(3)(4)}{3!}(-\frac{1}{6}x)^3
$

Gives

$
1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3
$

as

$
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1$

Then
$
\frac{1}{(1+\frac{1}{6}x)^2}
$

is valid for

$
-1<-\frac{1}{6}x<1
$

Thanks Macca

I'm going to try a different, simpler, approach (tell me if you see anything wrong)

we have the taylor series: $f(b)=\frac{1}{(1-b)}= 1+ b+b^2+b^3+.... for -1

and we have the series... $f(x)=\frac{1}{(1+\frac{1}{6}x)^2}$

now let's solve for: $(1+\frac{1}{6}x)^2$

but let's say $\neg\frac{1}{6}x=a$

then rewrite the problem: $(1-a)^2=1-2a+a^2=1-(2a-a^2)$

and then let's say $2a-a^2=b$

now remembering the taylor series: $f(b)=\frac{1}{(1-b)}= 1+ b+b^2+b^3+.... for -1
we see that $b$ must be between -1 and 1, let's figure out what $x$ needs to be between:

$-1 -1<2a-a^2<1\Longrightarrow -1<\neg\frac{1}{3}x-$ $(\neg\frac{1}{6}x)^2<1$
and solve from there...

~ $Q\!u\!i\!c\!k$
• Jul 22nd 2006, 05:57 AM
Quick
Quote:

Originally Posted by Quick
$-1 -1<2a-a^2<1\Longrightarrow -1<$ $\neg\frac{1}{3}x-(\neg\frac{1}{6}x)^2<1$
and solve from there...

~ $Q\!u\!i\!c\!k$

You can complete the square, watch: (correct me if I'm wrong)

$-1<-\frac{1}{3}x-(\neg\frac{1}{6}x)^2<1$

$-1<-\frac{1}{3}x-\frac{1}{36}x^2<1$

$1>\frac{1}{3}x+\frac{1}{36}x^2>-1$

$-1<\frac{1}{36}x^2+\frac{1}{3}x<1$

$-36

$-36

$-36+36

$0<(x+6)^2<72$

$\sqrt0

$\boxed{-6

I hope that's right.
Actually, I know I'm right :D
• Jul 22nd 2006, 06:27 PM
ThePerfectHacker
How is that a Taylor series?
• Jul 22nd 2006, 06:40 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
How is that a Taylor series?

I hope I didn't do anything wrong, (I haven't heard of Taylor series before so I just went off of what macca said)
• Jul 23rd 2006, 01:18 AM
macca101
Quote:

Originally Posted by ThePerfectHacker
How is that a Taylor series?

The Taylor series was found (hopefully) in the first part of my question.

$
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}=1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3+.....
$

What I was and sill am struggling with is finding the interval which the series is valid.
Quick gets it to

$
\boxed{-6<\sqrt{72}-6}
$

I have a math cad file that allows me to plot a Taylor series against the original function see below. This indicates to me the series is valid for approximately

$
\boxed{-2$

http://ianmc.bulldoghome.com/pages/i...com/taylor.jpg

I hope I am making myself clear here I'm not saying quick is wrong (far be it from me) I just don't understand how to find the interval algebraically
• Jul 23rd 2006, 08:02 AM
ThePerfectHacker
Quote:

Originally Posted by macca101
The Taylor series was found (hopefully) in the first part of my question.

$
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}=1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3+.....
$

You mean geomteric series.
---
I assume you want an infinite series for a function,
$\frac{1}{(1+1/6x)^2}$
Note the infinite geometric,
$\frac{1}{1-t}=1+t+t^2+t^3+...$ for $|t|<1$
Take derivative of both sides,
$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+...$
Substitute $t=-1/6x$
Thus,
$\frac{1}{(1+1/6x)^2}=1-1/3x+1/12x^2-...$
Good you got that right.

You are trying to find for what values this series converges.
You are going to use the generalized ratio test, with
$a_k=(-1)^k\frac{(k+1)}{6^k}$
Thus,
$\left|\frac{a_{k+1}}{a_k}\right|=\frac{k+2}{6^{k+1 }}\cdot \frac{6^k}{k+1}=\frac{1}{6}\cdot \frac{k+2}{k+1}$
Its limit as, $k\to\infty$ is 1/6,
Therefore the reciprocal of the limit tells you the radius of convergence which is 6.
Thus,
$|x|<6$
Same as saying,
$-6
• Jul 23rd 2006, 09:42 AM
macca101
Thanks For That