Justification for upper bound in Taylor polynomial

This is a bound on the error of the Taylor polynomial approximation.In summary, the conversation discusses a question about an upper bound on the error in a Taylor polynomial approximation. The authors propose using a value of 2 as an alternate bound and provide several justifications for this value. These include using Rolle's theorem, the fact that the exponential function is convex, and the inequality e^x < 1/(1-x).
  • #1
woe_to_hice

Homework Statement


I've been reviewing some Taylor polynomial material, and looking over the results and examples here.
https://math.dartmouth.edu/archive/m8w10/public_html/m8l02.pdf

I'm referring to Example 3 on the page 12 (page numbering at top-left of each page). The question is asking about an upper bound on the error.

Homework Equations

The Attempt at a Solution


I was able to get the result in the PDF, but in the discussion of the example, the authors mention that there is a logical issue with using an exact value of e^(0.2) while we're approximating f(x) = e^x . As an alternate bound the authors propose using f(5)(c) < 2 .

Can anyone tell me where the justification for this value of 2 comes from? Why is this a proposed value for an upper bound?
 
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  • #2
To be honest? I can't see it either. One could take ##0.2 < \frac{1}{2}## and ##f^{(5)}(c) \leq e^{0.2} < e^{0.5} < \sqrt{3} < 2## or something like
$$ f^{(5)}(c) = 1 + c + \frac{c^2}{2!}+ \frac{c^3}{3!} + \ldots < 1+\frac{1}{5}+\left( \frac{1}{5}\right)^2+ \left( \frac{1}{5}\right)^3+\ldots < 1+\frac{1}{2}+\left( \frac{1}{2}\right)^2 +\ldots = 2$$ but this is a lot of guesswork. Did they eventually had an upper bound in the previous section, or perhaps the ##K## in Rolle?

Another possibility is to use the fact (if given), that the exponential function is convex. In this case we have
$$
f^{(5)}(c) < \frac{e^0+e^1}{2} < \frac{1}{2} + \frac{3}{2} = 2
$$
 
  • #3
Thanks for replying. I will look at the Rolle's theorem possibility.
 
  • #4
woe_to_hice said:

Homework Statement


I've been reviewing some Taylor polynomial material, and looking over the results and examples here.
https://math.dartmouth.edu/archive/m8w10/public_html/m8l02.pdf

I'm referring to Example 3 on the page 12 (page numbering at top-left of each page). The question is asking about on upper bound on the error.

Homework Equations

The Attempt at a Solution


I was able to get the result in the PDF, but in the discussion of the example, they authors mention that there is a logical issue with using an exact value of e^(0.2) while we're approximating f(x) = e^x . As an alternate bound the authors propose using f(5)(c) < 2 .

Can anyone tell me where the justification for this value of 2 comes from? Why is this a proposed value for an upper bound?

For ##x > 0## we have ##e^{-x} > 1-x##, so ##e^x < 1/(1-x)##, giving ##e^{0.2} < 1/(1 - 0.2) = 1.25 < 2##.
 

FAQ: Justification for upper bound in Taylor polynomial

1. What is the purpose of finding an upper bound in Taylor polynomial?

The upper bound in Taylor polynomial is used to approximate the value of a function at a certain point. It helps to determine how close the polynomial is to the actual value of the function at that point.

2. How is the upper bound calculated in Taylor polynomial?

The upper bound is calculated using the remainder term in the Taylor polynomial formula. This term takes into account the error between the actual value of the function and the approximation given by the polynomial.

3. Can the upper bound be negative in Taylor polynomial?

No, the upper bound in Taylor polynomial is always a positive value. This is because it represents the maximum possible error between the polynomial and the actual value of the function at the given point.

4. Why is it important to have a small upper bound in Taylor polynomial?

A small upper bound indicates that the polynomial is a good approximation for the function at the given point. This means that the polynomial is close to the actual value of the function, making it more accurate and reliable.

5. Does the degree of the polynomial affect the upper bound in Taylor polynomial?

Yes, the degree of the polynomial does affect the upper bound. As the degree increases, the upper bound decreases, indicating a better approximation of the function at the given point.

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