Mean-Value Theorem, Taylor's formula, and error estimation

Click For Summary
SUMMARY

The discussion centers on the relationship between the Mean-Value Theorem (MVT) and Taylor's formula, particularly in error estimation for Taylor approximations. The case of n=0 in Taylor's formula aligns with the MVT, illustrating that the slope of the secant line between points on the function equals the slope of the tangent at some point s between a and x. The example provided uses the function f(x)=e^x, demonstrating how to estimate the error in the approximation. The conversation highlights that for higher-order functions, error estimation relies on delta and higher-order terms.

PREREQUISITES
  • Understanding of Taylor's theorem and its applications
  • Familiarity with the Mean-Value Theorem in calculus
  • Knowledge of exponential functions, specifically f(x)=e^x
  • Basic skills in error estimation techniques in numerical analysis
NEXT STEPS
  • Study the explicit formulas for the remainder in Taylor's theorem
  • Explore error estimation methods for polynomial approximations
  • Learn about higher-order derivatives and their impact on approximation accuracy
  • Investigate the application of the Mean-Value Theorem in various mathematical contexts
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, numerical analysis, and approximation theory. This discussion is beneficial for anyone looking to deepen their understanding of Taylor's formula and error estimation techniques.

mcastillo356
Gold Member
Messages
658
Reaction score
361
TL;DR
Can't conclude any error estimation from MVT, i.e., Taylor's formula for ##n=0##
Hi, PF

Taylor's formula provides a formula for the error in a Taylor approximation ##f(x)\approx{P_{n}(x)}## similar to that provided for linear approximation.

Observe that the case ##n=0## of Taylor's formula, namely,

##f(x)=P_{0}(x)+E_{0}(x)=f(a)+\dfrac{f'(s)}{1!}(x-a)##,

is just the Mean-Value Theorem

##\dfrac{f(x)-f(a)}{x-a}=f'(s)## for some ##s## between ##a##

and ##x##

The question is: to what extent, i.e. how, MVT provides a formula for the error in a Taylor approximation?

Attempt: let ##f(x)=e^{x}##, ##x=2##, ##a=0##: for ##n=0##

1- ##\dfrac{e^{2}-1}{2}=e^{s}##
2- ##\ln{\dfrac{e^{2}-1}{2}}=s\cdot{\ln{e}}\Rightarrow##
3- ##\Rightarrow{s\approx{1.1614}}##
4- ##\therefore{f(s)\approx{3.1944}}\Rightarrow##
5- All I can conclude is that the slope of the chord line joining the points ##(0,1)## and ##(2,7.3890)## is equal to the slope of the tangent line to the curve ##y=e^x## at the point ##(1.1614,3.1944)##, so the two lines are parallel.
6- Which is here the error estimation, or how can I find out?

Greetings

geogebra-export.png

PS: I post with no preview. Twice edited
 
Last edited:
Physics news on Phys.org
MVT says there is at least one point between x and a that has a tangent egual to the slope of the line through f(x) and f(a) (the secant). For a quadratic, that point will always be exactly be the mid point between x and a (on the x axis). For higher order functions, the error depends on the delta and the higher order terms in the function.
 
  • Like
Likes   Reactions: mcastillo356
valenumr said:
MVT says there is at least one point between x and a that has a tangent egual to the slope of the line through f(x) and f(a) (the secant). For a quadratic, that point will always be exactly be the mid point between x and a (on the x axis).
Didn't know that fact for quadratics. Nice
valenumr said:
For higher order functions, the error depends on the delta and the higher order terms in the function.
Sorry, could this be explained furthermore?

Love, thanks @valenumr
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K