Evaluate the Taylor series and find the error at a given point

In summary: I think the Taylor polynomial is at x0 = 1 and you then evaluate it at x = 1/√e. This gets you terms in powers of x - 1.
  • #1
tompenny
15
3
Homework Statement
Evaluate the Taylor series and find the error at a point
Relevant Equations
$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$
And want to approximate it using Taylor at the point $$\frac{1}{\sqrt e} $$
I have the following function
$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$
And want to approximate it using Taylor at the point ##\frac{1}{\sqrt e} ##

I also want to decide (without calculator)whether the error in the approximation is smaller than ##\frac{1}{25} ##

The Taylor polynomial is:
$$f\left(x\right)\approx P\left(x\right) = 1+x+\frac{1}{2}x^{2}$$

I evaluate the function at the point: $$f\left(- \frac{1}{2}\right)=e^{- \frac{1}{2}}$$

I evaluate the series at the point: $$P\left(- \frac{1}{2}\right)=\frac{5}{8}$$

Method. 1.
The error I get is $$E=\left|f\left(- \frac{1}{2}\right) - P\left(- \frac{1}{2}\right)\right|=\frac{5}{8} - e^{- \frac{1}{2}}\approx 0.018$$

Method. 2.
Using the formula $$E=\left|\frac{f'''(c)(c-a)^3}{6}\right|$$
where c is: $$-\frac{1}{2}<c<0$$
I get the error to $$E=\left|\frac{e^{- \frac{1}{2}}(-\frac{1}{2})^3}{6}\right|=\frac{1}{48\sqrt(e)}\approx 0.013$$

So my questions:
1. Which one of my methods is correct(if any)?
2. How do I show that the error is smaller than 1/25 without using a calculator?

Many thanks for any input on this matter:)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
tompenny said:
Homework Statement:: Evaluate the Taylor series and find the error at a point
Relevant Equations:: $$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$
And want to approximate it using Taylor at the point $$\frac{1}{\sqrt e} $$

I have the following function
$$f^{(0)}\left(x\right)=f\left(x\right)=e^{x}$$
And want to approximate it using Taylor at the point ##\frac{1}{\sqrt e} ##

I also want to decide (without calculator)whether the error in the approximation is smaller than ##\frac{1}{25} ##

The Taylor polynomial is:
$$f\left(x\right)\approx P\left(x\right) = 1+x+\frac{1}{2}x^{2}$$
No, that's the Maclaurin polynomial of degree 2. For a Taylor polynomial at the point in question, the terms would be powers of ##x - \frac 1 {\sqrt e}##.
Is there some reason you're limiting the Maclaurin polynomial to 2nd degree terms?

Here's the Maclaurin series for ##e^x## -- ##1 + x + \frac {x^2}{2!} + \frac{x^3}{3!} + \dots##.
A Taylor series for the same function about, say x = 1 would look like this: ##a_0 + a_1\frac{x -1}{1} + a_2\frac{(x-2)^2}{2!} + \dots##.
tompenny said:
I evaluate the function at the point: $$f\left(- \frac{1}{2}\right)=e^{- \frac{1}{2}}$$
No. ##f(-1/2)## has nothing to do with your problem.
tompenny said:
I evaluate the series at the point: $$P\left(- \frac{1}{2}\right)=\frac{5}{8}$$

Method. 1.
The error I get is $$E=\left|f\left(- \frac{1}{2}\right) - P\left(- \frac{1}{2}\right)\right|=\frac{5}{8} - e^{- \frac{1}{2}}\approx 0.018$$

Method. 2.
Using the formula $$E=\left|\frac{f'''(c)(c-a)^3}{6}\right|$$
where c is: $$-\frac{1}{2}<c<0$$
I get the error to $$E=\left|\frac{e^{- \frac{1}{2}}(-\frac{1}{2})^3}{6}\right|=\frac{1}{48\sqrt(e)}\approx 0.013$$

So my questions:
1. Which one of my methods is correct(if any)?
2. How do I show that the error is smaller than 1/25 without using a calculator?
Your 1st method is heading down the wrong track, as pointed out. The second looks to be closer to what you need to do.

##\frac 1 {\sqrt e} \approx .60##, so it's between 1/2 and 3/4. You could calculate the error by picking a number c to make the error in your formula smaller than .04.
tompenny said:
Many thanks for any input on this matter:)
 
  • Like
Likes DaveE
  • #3
Mark, I think he's trying to take the Taylor series around x=0 to evaluate ##1/\sqrt{e}## by plugging in -1/2 into the series.

I think the first part looks fine. I'm actually not sure how you would do this without a calculator in a standardized way. I suspect they just wanted you to use the triple derivative thing you wrote down. That said, note that the actual error you calculated is larger than the estimate you got from the triple derivative thing, which should never happen - the point of that formula is to construct a strict upper bound. Do you know what went wrong?

I also think you could do something dumb like note the remaining series is smaller than
$$\sum_{i=3}^{\infty} \frac{1}{3! 2^i} = \frac{1}{24}$$

That doesn't quite get you the bound they are looking for though, but you could tweak this to be tighter by considering the first couple terms more carefully.
 
  • #4
OK, but if you want to expand a series with the best accuracy (for a given number of terms) then you will want to expand it about the point of interest. This will give you a tighter bound on the accuracy of the expansion. For a Taylor's series this means terms of (x-x0) and derivatives evaluated at x0 for an expansion at x0. Your expansion was at the point x=0, but you care most about x=1/√e

Next step, review the general formula for a Taylor's expansion.

edit: BTW, you can expand the function at 0, but you will need to include more terms to get the accuracy you need compared to expansion at the point of interest.
 
  • #5
You can't use the error estimate for expansion about x=a, when you've actually expanded about x=0 (a≠0). Notice (for future reference) that the error bound is based on the next term in the series.

Also, just to save some effort - notice how nice your function is for taking derivatives. This will allow you a simpler solution that most any other function.
 
  • #6
DaveE said:
OK, but if you want to expand a series with the best accuracy (for a given number of terms) then you will want to expand it about the point of interest. This will give you a tighter bound on the accuracy of the expansion. For a Taylor's series this means terms of (x-x0) and derivatives evaluated at x0 for an expansion at x0. Your expansion was at the point x=0, but you care most about x=1/√e

This doesn't make any sense. If the goal is to evaluate f(a), how are you supposed to make a Taylor series centered around a? The first term is f(a), which we explicitly don't know.
 
  • #7
Office_Shredder said:
This doesn't make any sense. If the goal is to evaluate f(a), how are you supposed to make a Taylor series centered around a? The first term is f(a), which we explicitly don't know.
I thought the exercise was to create an approximate polynomial function at that point and evaluate the error. Yes, if you really only care about that point you don't need an expansion. I think we do know f(a). However, you certainly could expand at a different place and either be less accurate near a or include more terms.
 

1. 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 at a specific point by taking into account the function's derivatives at that point.

2. How do you evaluate a Taylor series?

To evaluate a Taylor series, you need to know the function, the point at which you want to approximate the function, and the number of terms you want to use in the series. Then, you can plug in the values into the Taylor series formula and simplify to find the approximation.

3. What is the purpose of finding the error in a Taylor series?

The error in a Taylor series represents the difference between the actual value of the function and the approximate value obtained from the series. It is important to find the error to determine the accuracy of the approximation and to know how many terms are needed to get a desired level of accuracy.

4. How do you find the error at a given point in a Taylor series?

To find the error at a given point in a Taylor series, you can use the Lagrange remainder formula, which takes into account the value of the next term in the series and the maximum value of the derivative of the function within a specific interval.

5. Can a Taylor series always accurately represent a function?

No, a Taylor series can only accurately represent a function if the function is infinitely differentiable and the series converges at the given point. Otherwise, the series will only be an approximation with a certain level of error.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
837
  • Calculus and Beyond Homework Help
Replies
4
Views
781
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
Replies
12
Views
875
  • Calculus and Beyond Homework Help
Replies
17
Views
604
  • Calculus and Beyond Homework Help
Replies
3
Views
408
  • Calculus and Beyond Homework Help
Replies
16
Views
555
  • Calculus and Beyond Homework Help
Replies
2
Views
707
  • Calculus and Beyond Homework Help
Replies
1
Views
248
  • Calculus and Beyond Homework Help
Replies
4
Views
736
Back
Top