# 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 \([a,b]:\)

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

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

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

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

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

\[\begin{align} x&=\frac{1+2}{2}+\frac{2-1}{2}t \\ &= \frac{3}{2} + \frac{t}{2} \\ &= \frac{3+t}{2}\\\\ \Rightarrow dx &= \frac{dt}{2}. \end{align}\]

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

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

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}\]

**Cite as:**Chebyshev's Formula.

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