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

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Taylor's formula relates the error in a Taylor approximation to the Mean-Value Theorem (MVT), demonstrating that the case of n=0 in Taylor's formula mirrors MVT's assertion about the existence of a point where the derivative equals the slope of the secant line. The discussion explores how the MVT can provide insights into error estimation for Taylor approximations, particularly for functions like f(x) = e^x. It highlights that for quadratic functions, the point of tangency is always the midpoint between x and a, while for higher-order functions, the error is influenced by delta and higher-order terms. Participants express a desire for further clarification on these concepts, particularly regarding error estimation in more complex functions. Understanding these relationships enhances the comprehension of approximation methods in calculus.
mcastillo356
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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
 
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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.
 
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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
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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