Taylor Series, working backwards

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SUMMARY

The discussion focuses on determining the rational function represented by the power series 2 - (2/3)x + (2/9)x² - (2/27)x³ using the Taylor Series framework. The participant successfully identifies the derivatives at a point, specifically f(a) = 2, f'(a) = -2/3, f''(a) = 4/9, and f'''(a) = -12/27. By recognizing the series as a geometric series, they derive the function as 2/(1 + (1/3)x) after correcting the sign in the variable substitution.

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tinylights
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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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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.
 
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?
 
It actually doesn't. I am still flummoxed. :|
 
tinylights said:
It actually doesn't. I am still flummoxed. :|

Look up "geometric series".
 
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!
 
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:
Ahhh, you're right! Realized that as I was checking my work. It should be -1/3x.
 

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