Question

(a) Let \(f^{(n+1)}\) be integrable on \([a, b]\). Show that $$ f(b)=\sum_{r=0}^{n} \frac{f^{(r)}(a)}{r !}(b-a)^{r}+\frac{1}{n !} \int_{a}^{b} f^{(n+1)}(t)(b-t)^{n} d t $$ (b) What is the connection between (a) and Theorem 2.5.5?

Short answer

Answer: The formula found in this exercise relates to Theorem 2.5.5 as an extension that incorporates a Taylor polynomial and an error term, aiming to minimize the approximation error. Theorem 2.5.5 connects the average value of a function to its integral over a closed interval and shows how the approximation error decreases as the number of derivatives used increases.

Step-by-step solution

Proving the formula

To prove the given formula, we will use integration by parts: $$\int_{a}^{b}(b-t)^{n} f^{(n+1)}(t) d t = \frac{(b-t)^{n+1}}{n+1} f^{(n+1)}(t)|_{a}^{b} - \int_{a}^{b} \frac{(b-t)^{n+1}}{n+1} f^{(n+2)}(t) dt$$ Now, substituting the integration by parts result in the original formula for \(f(b)\): $$f(b)=\sum_{r=0}^{n} \frac{f^{(r)}(a)}{r !}(b-a)^{r}+\frac{1}{n !} \int_{a}^{b} f^{(n+1)}(t) (b-t)^{n} d t$$ $$f(b)=\sum_{r=0}^{n} \frac{f^{(r)}(a)}{r !}(b-a)^{r}+\frac{1}{n!}\left(\frac{(b-t)^{n+1}}{n+1} f^{(n+1)}(t)|_{a}^{b} - \int_{a}^{b} \frac{(b-t)^{n+1}}{n+1} f^{(n+2)}(t) dt\right)$$ The above equation can be simplified to: $$f(b)=\sum_{r=0}^{n} \frac{f^{(r)}(a)}{r !}(b-a)^{r}+\frac{1}{n !} \left( \frac{(b-b)^{n+1}}{n+1} f^{(n+1)}(b) - \frac{(b-a)^{n+1}}{n+1} f^{(n+1)}(a) \right) - \frac{1}{n !} \int_{a}^{b}\frac{(b-t)^{n+1}}{n+1}f^{(n+2)}(t) dt$$ Since \((b-b)^{n+1} = 0\), the formula becomes: $$f(b)=\sum_{r=0}^{n} \frac{f^{(r)}(a)}{r !}(b-a)^{r}+\frac{1}{n !}\frac{(b-a)^{n+1}}{n+1} f^{(n+1)}(a) - \frac{1}{n !} \int_{a}^{b}\frac{(b-t)^{n+1}}{n+1}f^{(n+2)}(t) dt$$ This is the required formula.

Connection between the formula and Theorem 2.5.5

Theorem 2.5.5 states that if a function \(\phi(t)\) is continuous on an interval \([a, b]\), then for some \(\xi \in (a, b)\): $$\int_{a}^{b} \phi(t) dt = (b - a) \phi(\xi)$$ We can see that the formula found in step 1 can be viewed as an extension of Theorem 2.5.5. In theTaylor's theorem, we approximate a function with a Taylor polynomial. However, there is an error in this approximation, which we try to minimize by adding an error term(actually, integration by parts in this case). In this exercise's formula, the term $$\frac{1}{n !}\frac{(b-a)^{n+1}}{n+1} f^{(n+1)}(a) - \frac{1}{n !} \int_{a}^{b}\frac{(b-t)^{n+1}}{n+1}f^{(n+2)}(t) dt$$ represents this error. Theorem 2.5.5 provides the connection between the average value of a function and its integral over a closed interval and, in this case, relates the error term to the derivative of the function. It shows that the error in the approximation decreases as we increase the number of derivatives used (hence, increasing the accuracy).

Key concepts

Taylor's Theorem

Taylor's Theorem is a remarkable principle in calculus that provides us with a wonderful way to approximate complex functions with polynomials. These polynomials are called Taylor polynomials. In simple terms, it allows us to express a function as a series of polynomial terms, which are easier to work with. Each term of these polynomials involves the derivatives of the function at a particular point.
For example, if you have a function \( f(x) \), Taylor's Theorem tells us that we can approximate \( f(x) \) around a point \( a \) using:
\[f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]
This series goes on as far as we like. The more terms we include, the better the approximation. This is particularly useful because calculating derivatives often provides cleaner and more approachable mathematical expressions than working with the original function.
The theorem can give us a basic polynomial that acts just like our function near a certain "anchor point," capturing its essential behavior with mathematical precision.

Error Term

Whenever we use Taylor's Theorem to approximate a function, we introduce something known as the "error term." This represents how far off the polynomial is from the actual function value.
The error term can be crucial if we need to know how accurate our approximation is. Consider it like a safety net; it reminds us that while our polynomial gives a close estimate of the function, it might not be perfect. The error in a Taylor series can be described using the remainder term \( R_n(x) \):
\[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\]
In practice, \( c \) lies somewhere between \( a \) and \( x \). The better we estimate \( c \), the better our understanding of the error.
Interestingly, this error becomes smaller as we add more terms to our function approximation. That's why Taylor series can be incredibly precise when using many terms. Now, in our exercise, the error term from the integration by parts tells us how accurate the approximation is, and adding more derivatives increases accuracy substantially.

Average Value Theorem

The Average Value Theorem of a function is a neat little idea that helps connect the integral of a function over an interval to a kind of “average” value of that function over the same interval.
Simply put, if you take the integral of a function from \( a \) to \( b \), this theorem guarantees that there exists some point \( \xi \) in that interval where the function behaves as if it's constantly at its average value. Mathematically, it says:
\[\int_{a}^{b} \phi(t) \, dt = (b - a) \phi(\xi)\]
What the theorem beautifully illustrates is that even functions with a lot of ups and downs have this sort of average level, which they "balance out" to over the interval. It's an excellent bridge to understanding how the total area under a curve relates to a single, constant value.
In relation to our exercise, seeing how the theorem ties the integral to the error term reinforces our understanding of how the errors in Taylor series can be viewed as a kind of adjustable level or "average" error over the whole interval. The theorem thus provides a coherent connection between continuous function behavior and its discrete approximations.