Taylor's Theorem: f(x) & Error Estimation

In summary, we used Taylor's theorem to find the error R_3(x) in approximating the function f(x) = (1+x)^(1/3) with a third-degree polynomial, and found that the unknown constant c lies between 0 and 2^(1/3) in order to minimize the error.
  • #1
snowing_tsw
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Homework Statement


f(x) = (1+x)^(1/3)
Use taylor's theorem to write down an expression for the error R_3(x) = f(x) - P_3(x). In what interval does the unknown constant c lie?

I can't find what value is x where c lies between x and a.

Homework Equations


in previous question, we are required to find 6^(1/3) by doing 2(1- 1/4)^(1/3)

The Attempt at a Solution



i got a= 0 and c must lie between a and x.
and R_3(x) = 10x^4/[243(1-x)^(11/3)]

i know the first boundary of c must be zero but i can't find the other boundary
 
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  • #2
of c.
Hello,

Thank you for your question. First, let's review Taylor's theorem. It states that any function f(x) that is infinitely differentiable at a point x=a can be approximated by a polynomial P_n(x) of degree n, where the error R_n(x) = f(x) - P_n(x) is given by:

R_n(x) = (1/n!) * f^(n+1)(c) * (x-a)^(n+1)

where c is some unknown constant between x and a.

In this case, we are given the function f(x) = (1+x)^(1/3) and we want to approximate it using a third-degree polynomial P_3(x). This means that n = 3, and the error R_3(x) is given by:

R_3(x) = (1/3!) * f^(4)(c) * (x-a)^4

To find the unknown constant c, we need to find the value of x where it lies between x and a. In this case, a = 0, so we need to find the value of c between 0 and x. Since we are trying to approximate the value of f(x) = (1+x)^(1/3) at x = 6^(1/3), we can set x = 6^(1/3) in the error equation to get:

R_3(6^(1/3)) = (1/3!) * f^(4)(c) * (6^(1/3) - 0)^4

So the unknown constant c lies between 0 and 6^(1/3). However, we can further narrow down this interval by finding the value of c that makes the error R_3(x) as small as possible. To do this, we need to find the maximum value of the fourth derivative of f(x) between 0 and 6^(1/3). This can be done by taking the derivative of f(x) four times and setting it equal to 0, then solving for x. The result is x = 2^(1/3).

So, the unknown constant c lies between 0 and 2^(1/3) in order to minimize the error R_3(x). I hope this helps! Let me know if you have any further questions.
 

1. What is Taylor's Theorem and why is it important?

Taylor's Theorem is a mathematical formula that allows us to approximate a function using a polynomial. It is important because it helps us to understand the behavior of a function around a specific point and provides a way to estimate the value of a function at a certain point without having to evaluate the function at every point.

2. What is the difference between Taylor's Theorem and Maclaurin's Theorem?

Taylor's Theorem is a generalization of Maclaurin's Theorem, which is a special case where the function is expanded around the point x = 0. In other words, Maclaurin's Theorem is a specific case of Taylor's Theorem where the center point is 0.

3. How is the error in Taylor's Theorem estimated?

The error in Taylor's Theorem is estimated using the remainder term, also known as the Lagrange remainder. This term takes into account the difference between the actual value of the function and the approximation using the Taylor polynomial.

4. Can Taylor's Theorem be used for all functions?

No, Taylor's Theorem can only be used for functions that are infinitely differentiable (meaning they have derivatives of all orders) around the point of expansion. Additionally, the function must be evaluated at a finite number of points.

5. How can Taylor's Theorem be applied in real-world situations?

Taylor's Theorem can be applied in fields such as physics, engineering, and economics to approximate the behavior of a function and make predictions. It can also be used in numerical analysis to improve the accuracy of numerical methods for solving equations or integrals.

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