Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?

[PLAGUE]

TL;DR Summary

I am using a book where they prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a). I have included a proof of mean value theorem where they shows how they get h(x) = f(x) – f(a) – {(f(b)-f(a))/(b-a)} (x - a). Is there any similar way to show how they come up with the p(x)?

Here is a proof of mean value theorem:

Consider a line passing through the points (a, f(a)) and (b, f(b)). The equation of the line is

y-f(a) = {(f(b)-f(a))/(b-a)} (x-a)

or y = f(a)+ {(f(b)-f(a))/(b-a)} (x-a)

Let h be a function that defines the difference between any function f and the above line.

h(x) = f(x) – f(a) – {(f(b)-f(a))/(b-a)} (x-a)

Using “Rolle’s theorem”, we have

h'(x) = f'(x) – {(f(b)-f(a))/(b-a)}

Or f(b) - f(a) = f'(x) (b - a). Hence, proved.

The source of this proof is here.
It is clear why and how they used h(x) = f(x) – f(a) – {(f(b)-f(a))/(b-a)} (x-a).

Now here is a proof of Taylor's theorem:
Consider the function $p(x)$ defined in (a, b) by p(x)=q(x)−((b−a)^n/(b−x)^n)q(a) where q(x) = f(b) - f(x) - (b - x)f'(x) - {(b-x)/2!}f''(x) -... {(b-x)^{n-1}/(n-1)!}) f^{n-1}(x)

Then they show that, p(a)=p(b)=0 and apply roll's theorem and proves Taylor's theorem. I understand this part.

I don't understand why they use, p(x)=q(x)−((b−a)^n/(b−x)^n)q(a). When proving mean value theorem, they first derived h(x) = f(x) – f(a) – {(f(b)-f(a))/(b-a)} (x-a) and then applied rolls theorem. My question is what is the derivation of p(x)=q(x)−((b−a)^n/(b−x)^n)q(a). p(x)=q(x)−((b−a)^n/(b−x)^n)q(a) and h(x) = f(x) – f(a) – {(f(b)-f(a))/(b-a)} (x-a) seems similar to me. Can we derive p(x) the same way as they did for h(x)?

 

[fresh_42]

This is very hard to read. Could you write the formulas with LaTeX, see how here ...
https://www.physicsforums.com/help/latexhelp/
Just spend a lot of ## and \dfrac{}{}. And it is Rolle's theorem. It has nothing to do with rolls.I have a derivation of Taylor by the mean value theorem of integration and integration by parts, would this help?

 

[anuttarasammyak]

TL;DR Summary: I am using a book where they prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a).

I am not familiar with the formula you quote
$$p(x)=q(x)−\frac{(b−a)^n}{(b−x)^n}q(a)$$
where x is included in denominator. p(a)=0. p(b) diverges.
Is the above LaTex formula same as your textbook ?

 

[PLAGUE]

I am not familiar with the formula you quote
$$p(x)=q(x)−\frac{(b−a)^n}{(b−x)^n}q(a)$$
where x is included in divider. p(a)=0. p(b) diverges.
Is the above LaTex formula same as your textbook ?

Here is the proof:

 

[PLAGUE]

This is very hard to read. Could you write the formulas with LaTeX, see how here ...
https://www.physicsforums.com/help/latexhelp/
Just spend a lot of ## and \dfrac{}{}. And it is Rolle's theorem. It has nothing to do with rolls.I have a deviation of Taylor by the mean value theorem of integration and integration by parts, would this help?

 

[fresh_42]

Can you tell us the location where you have the problem, in the same terms as in the pictures?

The proof basically goes like this:

• the partial sum (plus correction terms) of the Taylor series at $x=a$ minus the partial sum (plus correction terms) of the Taylor series at $x=b$,
• where the correction terms make the expression zero at both ends of the interval $[a,b]$
• such that the theorem of Rolle can be applied to find $\psi'(k)=0$
• so that with some elementary algebra we get the Taylor series of $f(x)$ developed at the point $x=a.$

I admit that this approach isn't very insightful. One would need to find a good compromise between what I have just written and what the textbook says plus all lines between the equations (1) - (4).

A different approach that uses integration instead of differentiation and the mean value theorem of integration instead of Rolle can be found here:
https://www.physicsforums.com/insights/series-in-mathematics-from-zeno-to-quantum-theory/#Functions

 

[anuttarasammyak]

Here is the proof:

Upside down
$$p(x)=q(x)−\frac{(b−x)^n}{(b−a)^n}q(a)$$? Then p(a)=p(b)=0.

 

[PLAGUE]

Upside down
$$p(x)=q(x)−\frac{(b−x)^n}{(b−a)^n}q(a)$$? Then p(a)=p(b)=0.

 

[anuttarasammyak]

Is this the part you quote ? It coincides with #7 where x is in numerator.

 

[PLAGUE]

yes it is