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

tompenny
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:)

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Mentor
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:)

DaveE
Staff Emeritus
Gold Member
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.

Gold Member
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.

Gold Member
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.

Staff Emeritus
Gold Member
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.