Mathematical Preliminaries and Error Analysis Part 2

Review Of Calculus

The integral is the other basic concept of calculus that is used extensively.The Riemann integral of the function $f$ on the interval $[a,b]$ is the following limit, provided it exists.

\[\int_{a}^{b}f\left(x\right)dx=\lim_{\max\Delta x_{i}\to0}\sum_{i=1}f\left(z_{i}\right)\Delta x_{i}\]
where the numbers $x_{0},x_{1},\cdots,x_{n}$ satisfy $a=x_{0}<x_{1}<\cdots<x_{n}=b$
and where $\Delta x_{i}=x_{i}-x_{i-1},$ for each $i=1,2,....,n,$ and $z_{i}$ is arbitrarily chosen in the interval $\left[x_{i-1},x_{i}\right]$

A function $f$ that is continuous on an interval $[a,b]$ is also Riemann integrable on $[a,b]$. This permits us to choose, for computational convenience, the points $x_{i}$ to be equally spaced in $[a,b]$ and for each $i=1,2,...,n,$ to choose $z_{i}=x_{i}$. In this case
\[\int_{a}^{b}f\left(x\right)dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=1}^{n}f\left(x_{i}\right)\]

where the numbers shown in Figure 1.5 as $x_{i}$ are $x_{i}=a+\left(i\left(b-a\right)/n\right)$

Figure 1.5

Two more basic results are needed in our study of numerical methods. The first is a generalization of the usual Mean Value Theorem for Integrals.

[Mean Value Theorem for Integrals] If $f\in C[a,b]$, $g$ is integrable on $[a,b]$ and $g(x)$ does not change sign on $[a,b]$, then there exists a number $c$ in $(a,b)$ with
\[\int_{a}^{b}f\left(x\right)g\left(x\right)dx=f\left(c\right)\int_{a}^{b}g\left(x\right)dx\]

When $g\left(x\right)\equiv1$, this result reduces to the usual Mean Value Theorem for Integrals. It gives the average value of the function $f$ over the interval $[a,b]$ as \[f\left(c\right)=\frac{1}{b-a}\int_{a}^{b}f\left(x\right)dx\]

(See Figure 1.6.)

Figure 1.6

The next theorem presented is the Intermediate Value Theorem. Although its statement is not difficult, the proof is beyond the scope of the usual calculus course.

[Intermediate Value Theorem] If $f\in C[a,b]$ and $K$ is any number between $f(a)$ and $f(b)$, then there exists a number $c$ in $(a,b)$ for which $f(c)=K$. (Figure 1.7 shows one of the three possibilities for this function and interval.)

Figure 1.7

Example 1


To show that $x^{5}-2x^{3}+3x^{2}-1=0$ has a solution in the interval $[0,1]$, consider $f\left(x\right)=x^{5}-2x^{3}+3x^{2}-1$. We have \[f\left(0\right)=-1<0\,\,\,\text{and}\,\,\,0<1=f\left(1\right)\]
and $f$ is continuous. Hence, the Intermediate Value Theorem implies a number $x$ exists, with $0<x<1$, for which $x^{5}-2x^{3}+3x^{2}-1=0$                                   $\square$

As seen in Example 1, the Intermediate Value Theorem is used to help determine when solutions to certain problems exist. It does not, however, give an efficient means for finding these solutions. This topic is considered in next Chapter.

The final theorem in this review from calculus describes the development of the Taylor polynomials. The importance of the Taylor polynomials to the study of numerical analysis cannot be overemphasized, and the following result is used repeatedly.