Taylor Series Approximation

  • Thread starter Kinetica
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  • #1
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Homework Statement



Hi!

I have a couple of problems on Taylor Series Approximation.

For the following equations, write out the second-order Taylor‐series approximation.
Let x*=1 and, for x=2, calculate the true value of the function and the approximate value given by the Taylor series approximation.

f(x)=(x0.5-1)/0.5
f(x)=(x-1)2


The Attempt at a Solution



I was able to do the approximation at x=1 for the first problem. However, for the second problem, the second derivative yields a constant meaning that I cannot plug in x=1. What should I do in this case?

I do not understand the following direction:

"for x=2, calculate the true value of the function and the approximate value given by the Taylor series approximation."

Please help.

P.S. I am sorry - I don't know how to paste formulas within this message, thus, I have to attach a Word document.
 

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Answers and Replies

  • #2
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have you learned about binomial series? If you have you can solve the second problem with it without even thinking much if you know about binomial series and just get the right answer that way.

(1 + x)^p = summation from k=0 to ∞ of (p chose k)x^k

wait nvm i don't think that will work
oh actually ya i think you can manipulate it so that way it will

oh my bad you want the power series not taylor series...
so than just do from k=0 to k=2

sorry i forgot how to do the Taylor series and just recalled the summation for (1+x)^p as a power series. You get to the point were you just do these problems so much it's just best to remeber some of the easier ones...

I'm kinda rusty on this stuff lol

look up power series of (1+x)^p as a power series and just do from k=0 to k=2
you can also just put the equation in wolfram alpha to get the answer =O
note that you still have to figure out why the answer is correct =P

P.S. I think there's some other way to do this that I forgot about

but if f(x)=constant
then f(2) = constant
it's not a problem
 
Last edited:
  • #3
Ray Vickson
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Homework Statement



Hi!

I have a couple of problems on Taylor Series Approximation.

For the following equations, write out the second-order Taylor‐series approximation.
Let x*=1 and, for x=2, calculate the true value of the function and the approximate value given by the Taylor series approximation.

f(x)=(x0.5-1)/0.5
f(x)=(x-1)2


The Attempt at a Solution



I was able to do the approximation at x=1 for the first problem. However, for the second problem, the second derivative yields a constant meaning that I cannot plug in x=1. What should I do in this case?

I do not understand the following direction:

"for x=2, calculate the true value of the function and the approximate value given by the Taylor series approximation."

Please help.

P.S. I am sorry - I don't know how to paste formulas within this message, thus, I have to attach a Word document.

Why do you say you can't plug in x = 1 in the second case? What prevents you from substitution x = 1 in the formula f''(x) = 2 (for all x)?

RGV
 
  • #4
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Why do you say you can't plug in x = 1 in the second case? What prevents you from substitution x = 1 in the formula f''(x) = 2 (for all x)?

RGV

So you mean that given that f"(x)=2, I can say that f"(1)=2?
 
  • #5
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Yes.
You may imagine it by thinking it is 2 independently by x.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
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So you mean that given that f"(x)=2, I can say that f"(1)=2?

Of course. That is what we mean when we say f''(x) = 2 for all x. It holds for x = 1 or x = 943.76/sqrt(pi) or x = 10 million, or whatever.

RGV
 
  • #7
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Oh, I see. Thanks.

Do you know how do I calculate the true value of the function and the approximate value given by the Taylor series approximation for x=2?
 
  • #8
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Oh, I see. Thanks.

Do you know how do I calculate the true value of the function and the approximate value given by the Taylor series approximation for x=2?

You're given the formulas for both functions, and you presumably have your Taylor polynomials. Just evaluate these functions at the appropriate values.
 
  • #9
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I got it! Thanks!
 

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