Taylor Series, working backwards

In summary: Thanks for catching that!If you put u=(-1/3)*x then it looks like your series is 2*(u^0+u^1+u^2+u^3+...). Look familiar, enuf hints?It actually doesn't. I am still flummoxed. :|It actually doesn't. I am still flummoxed. :|
  • #1
tinylights
18
0

Homework Statement


Okay, first there is an explanation of the Taylor Series equation. This I don't have a problem with. Then, we have this:

Consider the power series 2 - (2/3)x + (2/9)x^2 - (2/27)x^3. What rational
function does this power series represent?

Homework Equations

/

The Attempt at a Solution



I basically am just stuck at the end. I do understand how this function relates to the Taylor Series equation and wrote a detailed explanation of what parts represent a, i, f(a), f'(a), etc. However, after solving and presenting this information:

f(a) = 2
f'(a) = -2/3
f''(a) = 4/9
f'''(a) = -12/27

I don't know how to sort of reverse it and find the relevant function. I know that the numerator is being multiplied by -1, then -2, then -3, and the denominator is being multiplied by 3, but how does one take that information and turn it into equations?

I don't want the answer - I just want a hint in the right direction, or a method by which I can find it. Thank you guys.
 
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  • #2
Think about what function you series represents. Start by writing your series as a sum which will help you recognize any common Maclaurin series in your sum.
 
  • #3
tinylights said:

Homework Statement


Okay, first there is an explanation of the Taylor Series equation. This I don't have a problem with. Then, we have this:

Consider the power series 2 - (2/3)x + (2/9)x^2 - (2/27)x^3. What rational
function does this power series represent?

Homework Equations

/

The Attempt at a Solution



I basically am just stuck at the end. I do understand how this function relates to the Taylor Series equation and wrote a detailed explanation of what parts represent a, i, f(a), f'(a), etc. However, after solving and presenting this information:

f(a) = 2
f'(a) = -2/3
f''(a) = 4/9
f'''(a) = -12/27

I don't know how to sort of reverse it and find the relevant function. I know that the numerator is being multiplied by -1, then -2, then -3, and the denominator is being multiplied by 3, but how does one take that information and turn it into equations?

I don't want the answer - I just want a hint in the right direction, or a method by which I can find it. Thank you guys.

If you put u=(-1/3)*x then it looks like your series is 2*(u^0+u^1+u^2+u^3+...). Look familiar, enuf hints?
 
  • #4
It actually doesn't. I am still flummoxed. :|
 
  • #5
tinylights said:
It actually doesn't. I am still flummoxed. :|

Look up "geometric series".
 
  • #6
Okay, so - a geometric sequence is represented by (1+r+r^2+r^3+r^4...) and can be expressed in the form 1/(1-r), so basically I have 2(1+u+u^2+u^3+u^4), which can equivalently be expressed in the form 2/(1-u). U = 1/3x, so we end up with 2/(1-1/3x)?

The answer seems to check out so I'll go with it. Thanks very much!
 
  • #7
tinylights said:
Okay, so - a geometric sequence is represented by (1+r+r^2+r^3+r^4...) and can be expressed in the form 1/(1-r), so basically I have 2(1+u+u^2+u^3+u^4), which can equivalently be expressed in the form 2/(1-u). U = 1/3x, so we end up with 2/(1-1/3x)?

The answer seems to check out so I'll go with it. Thanks very much!

Your series alternates signs. u=(1/3)x doesn't. How should you change it?
 
Last edited:
  • #8
Ahhh, you're right! Realized that as I was checking my work. It should be -1/3x.
 

FAQ: Taylor Series, working backwards

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is used to approximate a function by using its derivatives at a single point.

How do you find the coefficients of a Taylor series?

The coefficients of a Taylor series can be found by taking the derivatives of the function at the point of expansion and plugging them into the formula for the Taylor series.

Why is working backwards in Taylor series important?

Working backwards in Taylor series allows us to find the original function from its Taylor series representation. This is useful for simplifying complex functions and solving differential equations.

What is the relationship between Taylor series and Maclaurin series?

A Maclaurin series is a special case of a Taylor series where the point of expansion is at x=0. This means that the coefficients of a Maclaurin series can be found by simply taking the derivatives of the function at x=0.

How do you use Taylor series to approximate a function?

To approximate a function using Taylor series, we can use a finite number of terms in the series to get an approximation of the function. The more terms we use, the more accurate the approximation will be.

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