# 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*]:
\[P(x) \ = \ \frac{(x - x_{2})(x - x_{3})...(x - x_{n})}{(x_{1} - x_{2})(x_{1} - x_{3})...(x_{1} - x_{n})}f(x_{1})\]
\[+ \ \frac{(x - x_{2})(x - x_{3})...(x - x_{n})}{(x_{2} - x_{1})(x_{2} - x_{3})...(x_{2} - x_{n})}f(x_{2})\]
\[. \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ .\]
\[+ \ \frac{(x - x_{2})(x - x_{3})...(x - x_{n-1})}{(x_{n} - x_{1})(x_{n} - x_{2})...(x_{n} - x_{n-1})}f(x_{n}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\]

We get the following approximate formula of integration :

\[\int _{ a }^{ b }{ f(x)dx } \approx \int _{ a }^{ b }{ P(x)dx } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)\]

After some computaton it takes the form \[\int _{ a }^{ b }{ f(x)dx } \approx C_{ 1 }f(x_{ 1 })+C_{ 2 }f(x_{ 2 })+...+C_{ n }f(x_{ n })\quad \quad \quad \quad \quad \quad \quad \quad (3)\]

where the coefficients \(C_{i}\) are calculated by the formula \[C_{i} = \int_{a}^{b}{\frac{(x - x_{1})...(x - x_{i-1})(x - x_{i+1})...(x - x_{n})}{(x_{i} - x_{1})...(x_{i} - x_{i-1})(x_{i} - x_{i+1})...(x_{i} - x_{n})}}dx \ \ \ \ \ (4)\]

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

\[C_{1} = C_{2} = . \ . \ . = C_{n}\]

If we denote the total value of the coefficients \(C_{1}, C_{2}, ..., C_{n} \) by \(C_{n}\), formula (3) will take the form \[\int _{ a }^{ b }{ f(x)dx } \approx C_{ n }[f(x_{ 1 })+f(x_{ 2 })+...+f(x_{ n })\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (5)\]

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*, for *t=1*, *x=b*.

Hence, \[\int_{a}^{b}{f(x)dx} = \frac{b-a}{2} \int_{-1}^{1}{f(\frac{a+b}{2} + \frac{b-a}{2}t)dt } = \frac{b-a}{2} \int_{-1}^{1}{\Phi (t)dt}\]

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
\[\int _{ -1 }^{ 1 }{ f(x)dx } =C_{ n }[f(x_{ 1 })+f(x_{ 2 })+...+f(x_{ n })\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (6)\]
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
\[f(x)=a_{ 0 }+a_{ 1 }x+a_{ 2 }x^{ 2 }+....+a_{ n-1 }x^{ n-1 }\quad \quad \quad \quad \quad \quad \quad \quad \quad (7)\]
It will be noted that
\[\int_{-1}^{1}{f(x)dx} = \int_{-1}^{1}{( a_{0}+ a_{1}x + a_{2}x^{2} + . \ . \ . \ . + a_{n-1}x^{n-1})dx}\]

\[=\begin{cases} 2(a_{ 0 }+\frac { a_{ 2 } }{ 3 } +\frac { a_{ 4 } }{ 5 } +\frac { a_{ 6 } }{ 7 } +...+\frac { a_{ n-1 } }{ n } )\quad if\quad n\quad is\quad odd\quad \\ 2(a_{ 0 }+\frac { a_{ 2 } }{ 3 } +...+\frac { a_{ n-2 } }{ n-1 } )\quad if\quad n\quad is\quad even \end{cases}\quad \quad \quad \quad \quad \quad \quad \quad \quad (8)\]

On the other hand, the sum on te right side of (6) will, on the basis of (7), be equal to \[C_{ n }[na_{ 0 }+a_{ 1 }(x_{ 1 }+x_{ 1 }+...+x_{ n })+a_{ 2 }({ x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+...+{ x }_{ n }^{ 2 })+...+a_{ n-1 }({ x }_{ 1 }^{ n-1 }+{ x }_{ 2 }^{ n-1 }+...+{ x }_{ n }^{ n-1 })]\quad \quad \quad \quad \quad \quad \quad \quad (9)\]

Equating expressions (8) and (9), we get an equation that should hold for any \(a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n-1}\): \[2(a_{0} +\frac{ a_{2}}{3} +\frac{a_{4}}{5} +\frac{a_{6}}{7} +...+\frac{a_{n-1}}{n} ) = C_{n}[na_{0}+a_{1}(x_{1}+x_{1}+...+x_{n})+a_{2}({x}_{1}^{2}+{x}_{2}^{2}+...+{x}_{n}^{2})+ \ . \ . \ .+a_{n-1}({x}_{1}^{n-1}+{x}_{2}^{n-1}+...+{x}_{n}^{n-1})]\]

Equate the coefficients of \(a_{0}, a_{1}, a_{2}, a_{3}, ..., a_{n-1}\) on the left and right sides of the equation: \[\begin{cases} 2=C_{ n }n\quad or\quad C_{ n }=\frac { 2 }{ n } \\ x_{ 1 }+x_{ 1 }+...+x_{ n }=0 \\ { x }_{ 1 }^{ 2 }+{ x }_{ 2 }^{ 2 }+...+{ x }_{ n }^{ 2 }=\frac { 2 }{ 3C_{ n } } =\frac { n }{ 3 } \\ { x }_{ 1 }^{ 3 }+{ x }_{ 2 }^{ 3 }+...+{ x }_{ n }^{ 3 }=0 \\ { x }_{ 1 }^{ 4 }+{ x }_{ 2 }^{ 4 }+...+{ x }_{ n }^{ 4 }=\frac { 2 }{ 5C_{ n } } =\frac { n }{ 5 } \\ ............... \end{cases}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (10)\]

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**:
\[\int_{-1}^{1}{f(x)dx}= \frac{2}{n}[f(x_{1}) + f(x_{2}) + ... + f(x_{n})]\]
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
\[\int_{a}^{b}{f(x)dx}=\frac{b-a}{n}[f(X_{1}) + f(X_{2}) + ... + f(X_{n})]\]

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

**Example:** Evaluate \(\int_{1}^{2}{\frac{dx}{x}}=ln(2)\)

**Solution:** First, by a change of variable, transform this integral into a new one with limits of integration -1 to 1:
\[x+\frac{1+2}{2}+\frac{2-1}{2}t = \frac{3}{2} + \frac{t}{2} = \frac{3+t}{2}\]
\[dx = \frac{dt}{2}\]

Then, \[\int_{1}^{2}{\frac{dx}{2}} = \int_{-1}^{1}{\frac{dt}{3+t}}\] Compute the latter integral by Chebyshev's formula, taking n = 3: \[\int_{-1}^{1}{f(t)dt} = \frac{2}{3}[f(0.707107)+f(0)+f(-0.707107)]\]

Since, \[f(0.707107)=\frac{1}{3+0.7070107}=\frac{1}{3.707107}=0.269752\] \[f(0)=\frac{1}{3+0}=0.333333\] \[f(-0.707107)=\frac{1}{3-0.7070107}=\frac{1}{2.292893}=0.436130\] we have, \[\int_{-1}^{1}{\frac{dt}{3+t}}=\frac{2}{3}(0.269752+0.333333+0.436130)\] \[\frac{2}{3} \ . \ 1.039215 \ = \ 0.692810 \ \approx \boxed{0.693 \ = \ ln(2)}\]

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

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/chebyshevs-formula/