Solving Taylor Series Problem with m-th Derivative Bound

Click For Summary
SUMMARY

The discussion focuses on solving a Taylor series problem involving the m-th derivative of a function f(x) bounded by the expression |f(x)(m)| < m|11x - 28 - x²| for 4 < x < 6. The key question is determining the number of terms required in the Taylor series expansion around x = 4.5 to accurately evaluate f(5.8) to seven decimal places. The remainder theorem is essential for this calculation, specifically finding n such that the remainder is less than 0.0000001.

PREREQUISITES
  • Taylor series expansion
  • Remainder theorem in calculus
  • Understanding of derivatives
  • Quadratic functions and their properties
NEXT STEPS
  • Study the Remainder Theorem in detail
  • Learn how to calculate Taylor series expansions
  • Investigate the properties of quadratic functions, particularly |11x - 28 - x²|
  • Practice problems involving Taylor series convergence and accuracy
USEFUL FOR

Students in numerical methods, mathematicians, and anyone interested in advanced calculus concepts, particularly those working with Taylor series and error estimation.

brad sue
Messages
270
Reaction score
0
Hi ,
I have some difficulties to solve this problem. It is from my numerical methods class but the problem is about taylor series:

It is known that for 4 < x < 6, the absolute value of the m-th derivative of a certain function f(x) is bounded by m times the absolute value of the quadratic 11x-28-x2. That is, |f(x)(m)| < m|11x - 28 - x2| for m > 0. How many terms would be required in the Taylor series for f about x = 4.5 to evaluate f(5.8) accurate to 7 decimals?

Please, I need some sugestions. I am not sure if i need to use the remainder formula to find it.

Please somebody help me!

Thank you

Brad
 
Physics news on Phys.org
Yes, you should use the remainder theorem. You would find the remainder after n terms for the Taylor's series in terms of (x- 4.5)n. Of course, here x- 4.5= 5.8- 4.5= 1.3. For what n is the remainder less than 0.0000001?

What is the maximum possible value of |11x- 28- x2| for 4< x< 6?
 
ok,
thank I think i get it.

HallsofIvy said:
Yes, you should use the remainder theorem. You would find the remainder after n terms for the Taylor's series in terms of (x- 4.5)n. Of course, here x- 4.5= 5.8- 4.5= 1.3. For what n is the remainder less than 0.0000001?

What is the maximum possible value of |11x- 28- x2| for 4< x< 6?
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
4K
Replies
17
Views
10K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
17K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K