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mcastillo356

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- TL;DR Summary
- Is there something about basic analysis I am missing?

Hi PF, stucked with this proof:

Taylor's Formula

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

Taylor's Theorem

If the ##(n+1)##st-order derivative, ##f^{(n+1)} (t)##, exists for all ##t## in an interval containing ##a## and ##x##, and if ##P_{n} (x)## is the ##n##th-order Taylor polynomial for ##f## about ##a##, that is,

##P_{n} (x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^n##.

then the error ##E_{n}(x)=f(x)-P_{n}(x)## in the approximation ##f(x)\approx{P_{n}(x)}## is given by

##E_{n}(x)=\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##

where ##s## is some number between ##a## and ##x##. The resulting

formula

##f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^{n}+\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##, for some ##s## between ##a## and ##x##,

is called Taylor's formula with Lagrange remainder; the Lagrange remainder term is the explicit formula given above for ##E_{n}(x)##.

(Note that the error term (Lagrange remainder) in Taylor's formula looks just like the next term in the Taylor polynomial would look if we continued the Taylor polynomial to include one more term (of degree ##n+1##), except that the derivative ##f^{(n+1)}## is not evaluated at ##a## but rather at some (generaly unknown) point ##s## between ##a## and ##x##. This makes it easy to remember Taylor's formula.)

PROOF 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##.

Also note that the case ##n=1## is just the error formula for linearization given in the previous theorem

We will complete the proof for higher ##n## using mathematical induction.(...). Suppose, therefore, that we have proved the case ##n=k-1##, where ##k\geq{2}## is an integer. Thus, we are assuming that if ##f## is any function whose ##k##th derivative exists on an interval containing ##a## and ##x##, then

##E_{k-1}(x)=\dfrac{f^{(k)}(s)}{k!}(x-a)^k##

where ##s## is some number between ##a## and ##x##. Let us consider the next higher case ##n=k##. As in the proof of Theorem 11 (previous), we assume ##x>a## (the case ##x<a## is similar) and apply the Generalized Mean-Value Theorem to the functions ##E_{k}(t)## and ##(t-a)^{k+1}## on ##[a,x]##. Since ##E_{k}(a)=0##, we obtain a number ##u## in ##(a,x)## such that

##\dfrac{E_{k}(x)}{(x-a)^{k+1}}=\dfrac{E_{k}(x)-E_{k}(a)}{(x-a)^{k+1}-(a-a)^{k+1}}=\dfrac{E'_{k}(u)}{(k+1)(u-a)^{k}}##

Now

##E'_{k}(u)=\dfrac{d}{dt}\left(f(t)-f(a)-f'(a)(t-a)-\dfrac{f''(a)}{2!}(t-a)^2-\ldots-\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}\right)\Bigg | _{t=u}##

##=f'(u)-f'(a)-f''(a)(u-a)-\ldots-\dfrac{f^{(k)}(a)}{(k-1)!}(u-a)^{k-1}##

This last expression is just ##E_{k-1}(u)## for the function ##f'## instead of ##f##. By the induction assumption it is equal to

##\dfrac{(f')^{(k)}(s)}{k!}(u-a)^{k}=\dfrac{f^{(k+1)(s)}}{k!}(u-a)^{k}##

for some ##s## between ##a## and ##u##. Therefore,

##E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}##

We have shown that the case ##n=k## of Taylor's Theorem is true if the case ##n=k-1## is true, and the inductive proof is complete

Question: why doesn't differentiate properly ##\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}## at what I find the

Attempt: it has to be a lack of induction proof knowledge; or basic calculus understanding.

PS: I'm going to post with no preview.

Edited to improve LaTeX: Big | for differentiation

Taylor's Formula

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

Taylor's Theorem

If the ##(n+1)##st-order derivative, ##f^{(n+1)} (t)##, exists for all ##t## in an interval containing ##a## and ##x##, and if ##P_{n} (x)## is the ##n##th-order Taylor polynomial for ##f## about ##a##, that is,

##P_{n} (x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^n##.

then the error ##E_{n}(x)=f(x)-P_{n}(x)## in the approximation ##f(x)\approx{P_{n}(x)}## is given by

##E_{n}(x)=\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##

where ##s## is some number between ##a## and ##x##. The resulting

formula

##f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^{n}+\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##, for some ##s## between ##a## and ##x##,

is called Taylor's formula with Lagrange remainder; the Lagrange remainder term is the explicit formula given above for ##E_{n}(x)##.

(Note that the error term (Lagrange remainder) in Taylor's formula looks just like the next term in the Taylor polynomial would look if we continued the Taylor polynomial to include one more term (of degree ##n+1##), except that the derivative ##f^{(n+1)}## is not evaluated at ##a## but rather at some (generaly unknown) point ##s## between ##a## and ##x##. This makes it easy to remember Taylor's formula.)

PROOF 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##.

Also note that the case ##n=1## is just the error formula for linearization given in the previous theorem

We will complete the proof for higher ##n## using mathematical induction.(...). Suppose, therefore, that we have proved the case ##n=k-1##, where ##k\geq{2}## is an integer. Thus, we are assuming that if ##f## is any function whose ##k##th derivative exists on an interval containing ##a## and ##x##, then

##E_{k-1}(x)=\dfrac{f^{(k)}(s)}{k!}(x-a)^k##

where ##s## is some number between ##a## and ##x##. Let us consider the next higher case ##n=k##. As in the proof of Theorem 11 (previous), we assume ##x>a## (the case ##x<a## is similar) and apply the Generalized Mean-Value Theorem to the functions ##E_{k}(t)## and ##(t-a)^{k+1}## on ##[a,x]##. Since ##E_{k}(a)=0##, we obtain a number ##u## in ##(a,x)## such that

##\dfrac{E_{k}(x)}{(x-a)^{k+1}}=\dfrac{E_{k}(x)-E_{k}(a)}{(x-a)^{k+1}-(a-a)^{k+1}}=\dfrac{E'_{k}(u)}{(k+1)(u-a)^{k}}##

Now

*(Troublesome step):*##E'_{k}(u)=\dfrac{d}{dt}\left(f(t)-f(a)-f'(a)(t-a)-\dfrac{f''(a)}{2!}(t-a)^2-\ldots-\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}\right)\Bigg | _{t=u}##

##=f'(u)-f'(a)-f''(a)(u-a)-\ldots-\dfrac{f^{(k)}(a)}{(k-1)!}(u-a)^{k-1}##

This last expression is just ##E_{k-1}(u)## for the function ##f'## instead of ##f##. By the induction assumption it is equal to

##\dfrac{(f')^{(k)}(s)}{k!}(u-a)^{k}=\dfrac{f^{(k+1)(s)}}{k!}(u-a)^{k}##

for some ##s## between ##a## and ##u##. Therefore,

##E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}##

We have shown that the case ##n=k## of Taylor's Theorem is true if the case ##n=k-1## is true, and the inductive proof is complete

Question: why doesn't differentiate properly ##\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}## at what I find the

*difficult, troublesome step*?; why is not ##\dfrac{f^{(k+1)}(a)}{k!}(t-a)^{k}##?Attempt: it has to be a lack of induction proof knowledge; or basic calculus understanding.

PS: I'm going to post with no preview.

Edited to improve LaTeX: Big | for differentiation

Last edited: