Taylor series of this function

In summary, In this conversation, a student is trying to find the Taylor series for a function at a specific point and is having trouble doing so. Another student provides helpful advice on simplifying the expression.
  • #1
Kuzu
15
0
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
 
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  • #2
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.
 
  • #3
I tried to substitute x=u+2 but that didn't make it easier. Still too much work with the derivatives.

[tex]\frac{x2+2x+1}{(x-6)2(x+2)}[/tex]

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

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

and

"Find the taylor series of f(x) = [tex]\frac{x2+2x+1}{(x-2)2(x+1)}[/tex] 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.
 
  • #4
Kuzu said:
I tried to substitute x=u+2 but that didn't make it easier. Still too much work with the derivatives.

[tex]\frac{x2+2x+1}{(x-6)2(x+2)}[/tex]

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

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

and

"Find the taylor series of f(x) = [tex]\frac{x2+2x+1}{(x-2)2(x+1)}[/tex] 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.
 
  • #5
Hey thanks for the advice!
I used partial fractions like this:

[tex]\frac{A}{x+2}+\frac{B}{x-6}+\frac{B}{(x-6)^2}[/tex]

I found A=1/64, B=189/192 and C=49/8
some strange numbers.. is this right?
 
  • #6
Kuzu said:
Hey thanks for the advice!
I used partial fractions like this:

[tex]\frac{A}{x+2}+\frac{B}{x-6}+\frac{B}{(x-6)^2}[/tex]

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.
 
  • #7
Thanks to both of you!
 
  • #8
Kuzu said:
Thanks to both of you!

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

1. What is a Taylor series and how is it related to a function?

A Taylor series is a representation of a function as an infinite sum of terms. It is derived from the derivatives of the function evaluated at a specific point, and it allows for approximating the value of the function at nearby points.

2. How do you find the Taylor series of a function?

The Taylor series of a function can be found by using the formula:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
where a is the point at which the series is centered and the derivatives are evaluated.

3. What is the significance of the remainder term in a Taylor series?

The remainder term in a Taylor series represents the difference between the actual value of the function and the approximated value using a finite number of terms in the series. It helps to quantify the accuracy of the approximation.

4. Can you use a Taylor series to find the value of a function at a point outside its domain?

No, a Taylor series can only be used to approximate the value of a function at points within its domain. It is not valid for points outside the domain of the function.

5. What are some applications of Taylor series in real-world problems?

Taylor series can be used in a variety of fields, including physics, engineering, and finance. They are commonly used to approximate complex functions and to solve differential equations. In finance, they are used to calculate interest rates and in engineering, they can help with optimization problems.

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