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

mcastillo356
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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
 
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