Taylor series of this function

[Kuzu]

I have a homework question like this.

"Find the taylor series of the function f(x) = (x2+2x+1)/(x-6)²(x+2) at x=2"

I'm trying to simplify this expression so I can take the derivative.

I only got this far: (x+1)(x+1)/(x-6)(x-6)(x+2)

Can this be simplified more so that I can easily write the taylor series for this?

Thanks

 

[Dick]

I can't think of any way to do it easily. But if you put u=x-2 you want an expansion in powers of u. So I would substitute x=u+2 and then write your expression using partial fractions in terms of u. It's not really easy, but it's doable.

 

[Kuzu]

I tried to substitute x=u+2 but that didn't make it easier. Still too much work with the derivatives.

\frac{x²+2x+1}{(x-6)²(x+2)}

similarly I have two other questions like this, which are past years exam questions.

"Find the taylor series of f(x) = \frac{x²+2x+1}{(x-2)(x+8)} at x=-3"

and

"Find the taylor series of f(x) = \frac{x²+2x+1}{(x-2)²(x+1)} at x=1"

all three questions are very similar and I bet one of these will show on my exam this monday, so it would be big help if anyone can help me.

My prof. said there is a way to really simplify this expression. (with some substitution I guess) but he wouldn't tell me :)

I think I can handle the rest for the taylor series but I can't take any derivatives of this complicated expression.

 

[Dick]

I tried to substitute x=u+2 but that didn't make it easier. Still too much work with the derivatives.

\frac{x²+2x+1}{(x-6)²(x+2)}

similarly I have two other questions like this, which are past years exam questions.

"Find the taylor series of f(x) = \frac{x²+2x+1}{(x-2)(x+8)} at x=-3"

and

"Find the taylor series of f(x) = \frac{x²+2x+1}{(x-2)²(x+1)} at x=1"

all three questions are very similar and I bet one of these will show on my exam this monday, so it would be big help if anyone can help me.

My prof. said there is a way to really simplify this expression. (with some substitution I guess) but he wouldn't tell me :)

I think I can handle the rest for the taylor series but I can't take any derivatives of this complicated expression.

You still haven't tried using partial fractions to simplify the form. Take the function (x+3)/((x-1)(x+1)). It looks pretty bad for computing a taylor expansion. But if you use partial fractions to write it as 2/(x-1)-1/(x+1) it looks not so bad.

 

[Kuzu]

Hey thanks for the advice!
I used partial fractions like this:

\frac{A}{x+2}+\frac{B}{x-6}+\frac{B}{(x-6)^2}

I found A=1/64, B=189/192 and C=49/8
some strange numbers.. is this right?

 

[stevenb]

Hey thanks for the advice!
I used partial fractions like this:

\frac{A}{x+2}+\frac{B}{x-6}+\frac{B}{(x-6)^2}

I found A=1/64, B=189/192 and C=49/8
some strange numbers.. is this right?

That looks right, but note that 189/192 is also 63/64.

 

[Kuzu]

Thanks to both of you!

 

[stevenb]

Thanks to both of you!

I'll accept a 1% portion of the thanks and leave the appropriate 99% for our worthy mentor.