Chebyshev's Formula

In engineering computations, use of Chebyshev's formula of approximate integration is frequently made.

Let it be required to compute $\displaystyle \int _{ a }^{ b }{ f(x)dx }$.

Replace the integrand by the Lagrange interpolation polynominal $P(x)$ and take certain $n$ values of the function on the interval $[a.b]: \ f(x_{1}), \ f(x_{2}), ..., \ f(x_{n}),$ where $x_{1}, x_{2}, ..., x_{n}$ are any points of the interval $[a,b]:$

$$\begin{aligned} P(x) = \frac{(x - x_{2})(x - x_{3})\cdots (x - x_{n})}{(x_{1} - x_{2})(x_{1} - x_{3})\cdots (x_{1} - x_{n})}f(x_{1}) &+ \frac{(x - x_{2})(x - x_{3})\cdots (x - x_{n})}{(x_{2} - x_{1})(x_{2} - x_{3})\cdots (x_{2} - x_{n})}f(x_{2})\\\\&+\cdots + \frac{(x - x_{2})(x - x_{3})\cdots (x - x_{n-1})}{(x_{n} - x_{1})(x_{n} - x_{2})\cdots (x_{n} - x_{n-1})}f(x_{n}). &&&(1) \end{aligned}$$

We get the following approximate formula of integration:

$$\int _{ a }^{ b }{ f(x)dx } \approx \int _{ a }^{ b }{ P(x)dx }. \qquad \qquad (2)$$

After some computaton, it takes the form

$$\int _{ a }^{ b }{ f(x)dx } \approx C_{ 1 }f(x_{ 1 })+C_{ 2 }f(x_{ 2 })+\cdots +C_{ n }f(x_{ n }), \qquad \qquad (3)$$

where the coefficients $C_{i}$ are calculated by the formula

$$C_{i} = \int_{a}^{b}{\frac{(x - x_{1})\cdots (x - x_{i-1})(x - x_{i+1})\cdots (x - x_{n})}{(x_{i} - x_{1})\cdots (x_{i} - x_{i-1})(x_{i} - x_{i+1})\cdots (x_{i} - x_{n})}}\,dx. \qquad \qquad (4)$$

Formula (3) is cumbersome and inconvenient for computation because the coefficients $C_{i}$ are expressed by complex fractions.

Chebyshev posed the inverse problem: specify not the abscissae $x_{1}, x_{2}, ..., x_{n}$ but the coefficients $C_{1}, C_{2}, ..., C_{n}$ and determine the abscissae $x_{1}, x_{2}, ..., x_{n}$.

The coefficients $C_{i}$ are specified so that formula (3) should be as simple as possible for computation. This will occur when all coefficients $C_{i}$ are equal:

$$C_{1} = C_{2} = \cdots = C_{n}.$$

If we denote the total value of the coefficients $C_{1}, C_{2}, ..., C_{n}$ by $C_{n}$, formula (3) will take the form

$$\int _{ a }^{ b }{ f(x)dx } \approx C_{ n }\big[f(x_{ 1 })+f(x_{ 2 })+\cdots +f(x_{ n })\big].\qquad \qquad (5)$$

Formula (5) is, generally speaking, an approximate equation, but if $f(x)$ is a polynomial of degree not higher than $n - 1,$ then the equation will be exact. This circumstance is what permits determining the quantities $C_{n}, x_{1}, x_{2}, ..., x_{n}$.

To obtain a formula that is convenient for any interval of integration, let us transform the interval of integration $[a,b]$ into the interval $[-1, 1].$ To do this, put

$$x = \frac{a+b}{2} + \frac{b-a}{2}t,$$

then for $t = -1$ we will have $x = a,$ and for $t=1,$ $x=b.$

Hence,

$$\int_{a}^{b}{f(x)\, dx} = \frac{b-a}{2} \int_{-1}^{1}{f\left(\frac{a+b}{2} + \frac{b-a}{2}t\right)\, dt } = \frac{b-a}{2} \int_{-1}^{1}{\Phi (t)\, dt},$$

where $\Phi(t)$ denotes the function of t under the integral sign. Thus, the problem of integrating the given function $f(x)$ on the interval $[a,b]$ can always be reduced to integrating some other function $\Phi(x)$ on the interval $[-1,1].$

To summarize, the problem has reduced to choosing in the formula

$$\int _{ -1 }^{ 1 }{ f(x)\, dx } =C_{ n }\big[f(x_{ 1 })+f(x_{ 2 })+\cdots +f(x_{ n })\big] \qquad \qquad (6)$$

the numbers $C_{n}, x_{1}, x_{2}, ..., x_{n}$ so that this formula will be exact for any function $f(x)$ of the form

$$f(x)=a_{ 0 }+a_{ 1 }x+a_{ 2 }x^{ 2 }+\cdots +a_{ n-1 }x^{ n-1 }. \qquad \qquad (7)$$

It will be noted that

$$\begin{aligned} \int_{-1}^{1}{f(x)\, dx} &= \int_{-1}^{1}{\big( a_{0}+ a_{1}x + a_{2}x^{2} + \cdots + a_{n-1}x^{n-1}\big)\, dx}\\\\ &=\begin{cases} 2\left(a_{ 0 }+\frac { a_{ 2 } }{ 3 } +\frac { a_{ 4 } }{ 5 } +\frac { a_{ 6 } }{ 7 } +\cdots+\frac { a_{ n-1 } }{ n } \right)\ \ \text{ if } n \text{ is odd} \\\\ 2\left(a_{ 0 }+\frac { a_{ 2 } }{ 3 } +\cdots +\frac { a_{ n-2 } }{ n-1 } \right)\ \ \text{ if } n \text{ is even.} \end{cases}\qquad \qquad (8) \end{aligned}$$

On the other hand, the sum on te right side of (6) will, on the basis of (7), be equal to

$$C_{ n }\Big[na_{ 0 }+a_{ 1 }(x_{ 1 }+x_{ 1 }+\cdots +x_{ n })+a_{ 2 }\big({ x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+\cdots +{ x }_{ n }^{ 2 }\big)+\cdots +a_{ n-1 }\big({ x }_{ 1 }^{ n-1 }+{ x }_{ 2 }^{ n-1 }+\cdots +{ x }_{ n }^{ n-1 }\big)\Big]. \qquad \qquad (9)$$

Equating expressions (8) and (9), we get an equation that should hold for any $a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n-1}:$

$$2\left(a_{0} +\frac{ a_{2}}{3} +\frac{a_{4}}{5} +\frac{a_{6}}{7} +\cdots +\frac{a_{n-1}}{n} \right) = C_{n}\Big[na_{0}+a_{1}(x_{1}+x_{1}+\cdots +x_{n})+a_{2}\big({x}_{1}^{2}+{x}_{2}^{2}+\cdots +{x}_{n}^{2}\big)+\cdots +a_{n-1}\big({x}_{1}^{n-1}+{x}_{2}^{n-1}+\cdots +{x}_{n}^{n-1}\big)\Big].$$

Equate the coefficients of $a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n-1}$ on the left and right sides of the equation:

$$\begin{cases} 2=C_{ n }n\quad\text{ or }\quad C_{ n }=\frac { 2 }{ n } \\\\ x_{ 1 }+x_{ 1 }+\cdots +x_{ n }=0 \\\\ { x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+\cdots +{ x }_{ n }^{ 2 }=\frac { 2 }{ 3C_{ n } } =\frac { n }{ 3 } \\\\ { x }_{ 1 }^{ 3 }+{ x }_{ 2 }^{ 3 }+\cdots +{ x }_{ n }^{ 3 }=0 \\\\ { x }_{ 1 }^{ 4 }+{ x }_{ 2 }^{ 4 }+\cdots +{ x }_{ n }^{ 4 }=\frac { 2 }{ 5C_{ n } } =\frac { n }{ 5 } \\\\ \cdots \cdots \end{cases}\qquad \qquad (10)$$

From these $n-1$ equations, we find the abscissas $x_{1}, x_{2}, ..., x_{n}$. These solutions were found by Chebyshev for various values of $n.$ The following solutions are those that he found for cases when the number of intermediate points $N$ is equal to 3, 4, 5, 6, 7, 9:

Thus, on the interval $[-1,1],$ an integral can be approximated by the following Chebyshev formula:

$$\int_{-1}^{1} f(x)\, dx= \frac{2}{n}\big[f(x_{1}) + f(x_{2}) +\cdots + f(x_{n})\big],$$

where $n$ is one of the numbers 3, 4, 5, 6, 7, 9, and $x_{1}, ..., x_{n}$are the numbers given in the table. Here, $n$ cannot be 8 or any number exceeding 9, for then the system of equations (10) yields imaginary roots.

When the given integral has limits of integration $a$ and $b,$ the Chebyshev formula takes the form

$$\int_{a}^{b}{f(x)\, dx}=\frac{b-a}{n}\big[f(X_{1}) + f(X_{2}) +\cdots + f(X_{n})\big],$$

where $X_{i}=\frac{b+a}{2} + \frac{b-a}{2}x_{i}$ for $i=1, 2, ..., n$ and $x_{i}$ have the values given in the table.

Evaluate $\displaystyle{\int_{1}^{2}}{\frac{dx}{x}}=\ln(2).$

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First, by a change of variable, transform this integral into a new one with limits of integration -1 to 1:

$$\begin{aligned} x&=\frac{1+2}{2}+\frac{2-1}{2}t \\ &= \frac{3}{2} + \frac{t}{2} \\ &= \frac{3+t}{2}\\\\ \Rightarrow dx &= \frac{dt}{2}. \end{aligned}$$

Then

$$\int_{1}^{2}{\frac{dx}{x}} = \int_{-1}^{1}{\frac{dt}{3+t}}.$$

Compute the latter integral by Chebyshev's formula, taking $n = 3:$

$$\int_{-1}^{1}{f(t)\, dt} = \frac{2}{3}\big[f(0.707107)+f(0)+f(-0.707107)\big].$$

Since

$$\begin{aligned} f(0.707107)&=\frac{1}{3+0.7070107}\\&=\frac{1}{3.707107}\\&=0.269752\\\\ f(0)&=\frac{1}{3+0}\\&=0.333333\\\\ f(-0.707107)&=\frac{1}{3-0.7070107}\\&=\frac{1}{2.292893}\\&=0.436130, \end{aligned}$$

we have

\[\begin{align} \int_{-1}^{1}{\frac{dt}{3+t}}&=\frac{2}{3}(0.269752+0.333333+0.436130)\\ &=\frac{2}{3} \times 1.039215 \\ &= 0.692810 \\
&\approx \boxed{0.693 \ = \ln(2)}.\ _\square \end{align}\]