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

In summary, the proof of Taylor's theorem using the expression \( p(x) = q(x) - \frac{(b-a)^n}{(b-x)^n} q(a) \) illustrates the connection between polynomial approximations and limits of function behavior. This approach highlights how the difference between the function and its polynomial approximation diminishes as the interval shrinks, thereby reinforcing the significance of Taylor polynomials in approximating continuous functions. It emphasizes the role of the remainder term in estimating the accuracy of the approximation, providing a clear framework for understanding convergence and error in polynomial expansions.
  • #1
PLAGUE
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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)?
 
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  • #2
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?
 
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  • #3
PLAGUE said:
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
[tex]p(x)=q(x)−\frac{(b−a)^n}{(b−x)^n}q(a)[/tex]
where x is included in denominator. p(a)=0. p(b) diverges.
Is the above LaTex formula same as your textbook ?
 
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  • #4
anuttarasammyak said:
I am not familiar with the formula you quote
[tex]p(x)=q(x)−\frac{(b−a)^n}{(b−x)^n}q(a)[/tex]
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:
 

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  • #5
fresh_42 said:
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?
 

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  • #6
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
 
  • #7
PLAGUE said:
Here is the proof:
Upside down
[tex]p(x)=q(x)−\frac{(b−x)^n}{(b−a)^n}q(a)[/tex]? Then p(a)=p(b)=0.
 
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  • #8
anuttarasammyak said:
Upside down
[tex]p(x)=q(x)−\frac{(b−x)^n}{(b−a)^n}q(a)[/tex]? Then p(a)=p(b)=0.
 

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  • #9
1700371273198.png

Is this the part you quote ? It coincides with #7 where x is in numerator.
 
  • #10
yes it is
 

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

What is Taylor's theorem?

Taylor's theorem is a fundamental result in calculus that provides an approximation of a function as a polynomial. It states that a function can be expressed as a Taylor series around a point, which is a sum of terms calculated from the values of the function's derivatives at that point. The theorem is useful for approximating functions using polynomials, especially when the function is smooth and can be differentiated multiple times.

What does the expression p(x) = q(x) - ((b-a)^n/(b-x)^n)q(a) represent?

The expression p(x) = q(x) - ((b-a)^n/(b-x)^n)q(a) is a form used in the proof of Taylor's theorem. Here, q(x) represents a polynomial approximation of a function near a point, and the term ((b-a)^n/(b-x)^n)q(a) adjusts this approximation to account for the behavior of the function at the endpoints a and b. This expression helps to demonstrate how closely the polynomial can approximate the function by comparing it to the function's value at specific points.

Why is it important to prove Taylor's theorem using this specific form?

Proving Taylor's theorem using the form p(x) = q(x) - ((b-a)^n/(b-x)^n)q(a) is important because it provides a clear framework for understanding the convergence of the polynomial approximation to the actual function. This specific form highlights the relationship between the polynomial and the function's behavior at the endpoints, allowing for a more rigorous analysis of the approximation's accuracy and the remainder term in the Taylor series expansion.

What role do the endpoints a and b play in this proof?

The endpoints a and b play a crucial role in the proof of Taylor's theorem because they are the points at which the function is evaluated and compared to the polynomial approximation. By considering the values of the function at these endpoints, the proof can demonstrate how well the polynomial captures the function's behavior in the interval [a, b]. The choice of endpoints also influences the remainder term, which quantifies the difference between the actual function and its polynomial approximation.

How does this form relate to the concept of convergence in calculus?

This form relates to the concept of convergence in calculus by illustrating how the polynomial approximation approaches the actual function as the degree of the polynomial increases. The expression shows that as the polynomial degree increases, the term ((b-a)^n/(b-x)^n)q(a) becomes more significant in determining the accuracy of the approximation. This relationship is central to understanding convergence in Taylor series, as it highlights the conditions under which the series converges to the function within the specified interval.

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