1. ## Please check my work

Hello !!

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

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

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

Substituting $\displaystyle x = -\frac{1}{6}x$
And taking 2 in the binomial series gives

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

Gives

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

as

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

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

is valid for

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

Thanks Macca

2. Originally Posted by macca101
Hello !!

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

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

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

Substituting $\displaystyle x = -\frac{1}{6}x$
And taking 2 in the binomial series gives

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

Gives

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

as

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

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

is valid for

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

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

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

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

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

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

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

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

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

$\displaystyle -1<b<1\Longrightarrow -1<2a-a^2<1\Longrightarrow -1<\neg\frac{1}{3}x-$$\displaystyle (\neg\frac{1}{6}x)^2<1 and solve from there... ~ \displaystyle Q\!u\!i\!c\!k 3. Originally Posted by Quick \displaystyle -1<b<1\Longrightarrow -1<2a-a^2<1\Longrightarrow -1<$$\displaystyle \neg\frac{1}{3}x-(\neg\frac{1}{6}x)^2<1$
and solve from there...

~ $\displaystyle Q\!u\!i\!c\!k$
You can complete the square, watch: (correct me if I'm wrong)

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

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

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

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

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

$\displaystyle -36<x^2+12x<36$

$\displaystyle -36+36<x^2+12x+36<36+36$

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

$\displaystyle \sqrt0<x+6<\sqrt{72}$

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

I hope that's right.
Actually, I know I'm right

4. How is that a Taylor series?

5. 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)

6. Originally Posted by ThePerfectHacker
How is that a Taylor series?
The Taylor series was found (hopefully) in the first part of my question.

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

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

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

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

$\displaystyle \boxed{-2<x<2}$

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

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

$\displaystyle f(x)=\frac{1}{(1+\frac{1}{6}x)^2}=1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3+.....$
You mean geomteric series.
---
I assume you want an infinite series for a function,
$\displaystyle \frac{1}{(1+1/6x)^2}$
Note the infinite geometric,
$\displaystyle \frac{1}{1-t}=1+t+t^2+t^3+...$ for $\displaystyle |t|<1$
Take derivative of both sides,
$\displaystyle \frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+...$
Substitute $\displaystyle t=-1/6x$
Thus,
$\displaystyle \frac{1}{(1+1/6x)^2}=1-1/3x+1/12x^2-...$
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
$\displaystyle a_k=(-1)^k\frac{(k+1)}{6^k}$
Thus,
$\displaystyle \left|\frac{a_{k+1}}{a_k}\right|=\frac{k+2}{6^{k+1 }}\cdot \frac{6^k}{k+1}=\frac{1}{6}\cdot \frac{k+2}{k+1}$
Its limit as, $\displaystyle k\to\infty$ is 1/6,
Therefore the reciprocal of the limit tells you the radius of convergence which is 6.
Thus,
$\displaystyle |x|<6$
Same as saying,
$\displaystyle -6<x<6$

8. Thanks For That