Analysis: Taylor Polynomial Approximation

• Shackleford
In summary: This gives f(48^\circ)=-1. Problem 5: You set x_0=0 for 5a and changed the interval to I=[0, .2]. This change is incorrect because I explained in 4a that the function needs to be expanded about x_0=\pi/4. You need to set x_0=.266pi for 5a in order to follow the original instructions. In summary, Shackleford - according to what he has read, the maximum absolute value of the n+1 derivative is between x and x0. To find this value, he needs to
Shackleford
For #4, I'm mostly confident I did it correctly. In determining the error, we're supposed to find the maximum absolute value on an interval I. I set I = (0,2pi). Is that right?

http://i111.photobucket.com/albums/n149/camarolt4z28/4-1.png

For #5,

http://i111.photobucket.com/albums/n149/camarolt4z28/5.png

For #5, again, I'm mostly confident I did it correctly. But, again, in determining the error, we're supposed to find the maximum absolute value on an interval I. I set I = (0, .2). Is that right? I think I'm okay on b, too. I just want to make sure. I was a bit delinquent in taking notes in class when we covered this. Thanks.

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For number 4, why did you pick $x_0 = 0$ in your equation for p3?

TheoMcCloskey said:
For number 4, why did you pick $x_0 = 0$ in your equation for p3?

For 4(b)? I didn't know what to assign x0, so I picked zero. What should I use? How do I know what to use?

What did you use for the first half of this question? Why change?

TheoMcCloskey said:
What did you use for the first half of this question? Why change?

Good question. I didn't make the connection in incorporating the two problems. I have to redo the problem.

Also, am I correct in identifying the proper intervals for determining maximum absolute value of the n+1 derivative?

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Also, am I correct in identifying the proper intervals for determining maximum absolute value of the n+1 derivative?

Looks like your using the Lagrange form for error estimates. Your interval for problem 5b is correct, but review problem 4b again based on the other most recent corrections.

TheoMcCloskey said:
Looks like your using the Lagrange form for error estimates. Your interval for problem 5b is correct, but review problem 4b again based on the other most recent corrections.

For 5(a), I set x0 = 0. Is that a correct assumption? I set the interval I = [0, .2]. In general, what do I make the interval? Is it between x and x0?

For 4(a), I redid the polynomial with x0 = pi/4 and x = .266pi. What do I make the interval I for the error estimation?

You should review your notes, text, or explore the web. The interval is such that $x_0 < c < x$ (sometimes expressed as $a < \xi_L < x$).

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TheoMcCloskey said:
You should review your notes, text, or explore the web. The interval is such that $a < c < x$ (sometimes expressed as $a < \xi_L < x$).

What do you think I've been doing? The answer isn't clear to me. That's why I'm asking it.

According to what I've read, you want to find the maximum value of the derivative in between x and x0. In 4(b), it would be between pi/4 and .266pi.

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Shackleford - Calm down, take a deep breath - your answer for 5a is correct. But I guess you are having some difficulty understanding why, so I'l try again. Please be patient.

I think you know the general form of the Taylor Series so I won't waste time going through that. But I want to make sure you understand what we are really saying when we look for "a Taylor expansion about $x_0$."

The resulting expression is a local approximation of the function such that it approximates the function within an interval about the point $x_0$. By definition, the approximation is exact at $x_0$.

Also by definition, the Taylor series is an infinite sum and the function equates to this sum if it converges. When we take a truncation of this infinite series, the function is approximated because we are taking a finite number of terms.

The error term (ie, remainder term) resulting from the difference between the function and the approximation by the truncated series is typically a generalization based on certain restrictions on the function ($C^n$ continuity, $f^{(n+1)}$ exists, etc) and the Mean Value Theorem that allow us to construct the remainder term in such an abbreviated expression as the one you have been using.

Let's restate the Theorem:
Suppose $f\in C^n[a,b]$ and $f^{(n+1)}$ exists on $[a,b]$. Let $x_0\in[a,b]$. For every $x \in [a,b]$, there exist $\xi(x)$ between $x_0$ and $x$ with
$$f(x) = P_n(x)+R_n(x)$$
where
$$P_n(x) = \sum_{k=0}^n{ \frac{f^k(x_0)}{k!}\,(x-x_0)^k}$$
and
$$R_n(x)=\frac{f^{(n+1)}(\xi(x))}{(n+1)!} \, (x-x_0)^{n+1}$$

Now, the $f^{(n+1)}(\xi(x))$ term in the expression for $R_n$ is written that way since we really don't know where that point $\xi$ is since the Mean Value Theorem was used to generalize all the remaining terms (we didn't show these steps). We do know it is somewhere between $x_0$ and $x$.

Now let's look at your example problems.

Problem 4a: you recognized correctly that, since we wanted a polynomial in terms of $(x-\pi/4)$, the function needs to be expanded about $x_0=\pi/4$. Your $p_4$ expansion is correct.

Problem 4b: Continuing with the development of 4a, approximate $f(x)$ with $p_3$ and $x=48^\circ$. Since the value of $x$ is very close to $x_0=45^\circ$ (recall $45^\circ$ corresponds to $\pi/4$radians) from problem 4a, use the terms of the polynomial you developed before up to and including the third order term to construct $p_3$. The terms evaluate to powers of $(x-\pi/4)$ as before or $(48-45) \pi / 180= \pi/60$ radians.

Now, since $x=(48/180)\pi = 0.2667 \pi$ and $x_0=0.2500 \pi$, we know somewhere in that interval there is a value $\xi$ such that $f^{(n+1)}(\xi)$will yield the correct term to summarize the error in the expression for $R_n$. But we don't have to know the exact value if we just want to bound the error, that is, to find the maximum magnitude of the error will only require us to find the maximum value (or approximation) of $f^{(n+1)}(\xi)$ over the interval $x_0<\xi<x$.

For your problem, we need to bound $f^{(4)}(\xi) = \cos(\xi)$. Thus, we should use $f^{(4)}(\xi) = \cos(\xi)< \cos(\pi/4)$ since the cosine function is monotonically decreasing for increasing $x$ in this interval (ie, $\cos(x)$ gets smaller with increasing values of $x$). Hence, the error term is approximated as
$$R_3 \le \frac{\cos(\pi/4)}{4!} \, (\pi/60)^4 \approx 2.21 \times 10^{-7}$$

I know this is a bit much to swallow, but I hope it helps. Let me know if it doesn't and I try some more.

Yeah, I was able to write down the correct expression for 4(b).

For 5(a), it says denote the Taylor polynomial of degree n in powers of x for f. Of course, the general form is x - x0, so that's how I know we're looking at the function around x0 = 0.

You're right. I needed just a little bit of clarification on the theory and how it ties into the computation. Thanks. I'm stumped on the last problem, so I may post it here sometime tonight.

Do I actually have to prove anything? Just by looking at the notes and the definition, I can clearly see that f is uniformly continuous since the derivative is bounded.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

According to the notes:

Let I be a bounded interval with endpoints a and b, and let f: I → R, be continuous.

If: |f '(x)|≤ M for all x ∈ I.

Then:

1. ...

2. Let ε > 0. Then there is a δ> 0 such that |f(x2) − f(x1)| < ε whenever |x2 − x1| <δ, x1,x2 ∈ I

Def. f : I → R is uniformly continuous on I if to each ε > 0 there is a δ> 0 such that |f(x2) − f(x1)| < ε whenever |x2 − x1| <δ, x1,x2 ∈ I.

Please, disregard the previous post. The professor went over this problem today in class.

What is Taylor Polynomial Approximation?

Taylor Polynomial Approximation is a mathematical method used to approximate a function with a polynomial. It is based on the Taylor series, which is a way to represent a function as an infinite sum of terms, and allows for a polynomial to be used to approximate the function with a certain degree of accuracy.

How is Taylor Polynomial Approximation calculated?

To calculate a Taylor Polynomial Approximation, the function is first centered at a specific point, usually denoted as "a". Then, the derivatives of the function at that point are calculated. These derivatives are used to create the terms of the Taylor polynomial, and the more terms used, the more accurate the approximation will be.

What are the benefits of using Taylor Polynomial Approximation?

Taylor Polynomial Approximation allows for functions to be approximated with polynomials, which are often easier to work with mathematically. This can simplify calculations and make it easier to analyze the behavior of a function. It also allows for the approximation of functions that may be difficult to integrate or differentiate.

What is the difference between Taylor Polynomial Approximation and Taylor Series?

Taylor Polynomial Approximation is a finite sum of terms used to approximate a function, while Taylor Series is an infinite sum of terms used to represent a function. Taylor Polynomial Approximation is more useful for practical calculations, while Taylor Series is used for theoretical and mathematical purposes.

When is Taylor Polynomial Approximation most accurate?

Taylor Polynomial Approximation is most accurate when the function being approximated is smooth and well-behaved, and when a sufficient number of terms are used in the polynomial. The accuracy can also be improved by centering the polynomial at a point closer to the desired approximation point.

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