Simpson's Rule: Derivation and Error Bound

We shall now apply our approximation theorem to the computation of definite integrals. This time we shall use a polynomial of third degree to approximate the function we are to integrate. But first let us prove a result about integrals of polynomials:

Theorem 1

Let $f (x)$ be a polynomial of degree $\leq 5$ . Let $m$ be the midpoint of the interval $a \leq x \leq b$ ( $m = \frac{1}{2} (a + b)$ ). Then

\int_{a}^{b} f (x) d x = \frac{b - a}{6} (f (a) + 4 f (m) + f (b)) - \frac{(b - a)^{5}}{2880} f^{(i v)} (m) \cdot

PROOF. Observe first of all that the assertion is linear in $f$ . i.e., if it is true for $f (x)$ and for $g (x)$ , then it is true for $f (x) + g (x)$ and for $c f (x)$ , where $c$ is a constant. Therefore it will be true for all polynomials of degree $\leq 5$ , if we can show it is true for $1$ , $x$ , $x^{2}$ , $x^{3}$ , $x^{4}$ , $x^{5}$ , or if we can show it for $1$ , $(x - m)$ , $(x - m)^{2}$ , $(x - m)^{3}$ , $(x - m)^{4}$ , $(x - m)^{5}$ . For

\begin{aligned} x^n &= ((x-m)+m)^n \\ &= (x-m)^n + n(x-m)^{n-1}m + \dots + n(x-m)m^{n-1} + m^n \end{aligned}

is a polynomial of degree $n$ in $(x - m)$ for any $n$ ; thus any polynomial in $x$ of degree $\leq 5$ is also a polynomial of degree $\leq 5$ in $(x - m)$ .

First let $f (x) = (x - m)^{i}$ , where $i$ is odd: $i = 1, 3, 5$ . Then

\begin{aligned} \int_a^b (x-m)^i \, dx &= \left. \frac{1}{i+1} (x-m)^{i+1} \right|_a^b \\ &= \frac{1}{i+1} \left( (b-m)^{i+1} - (a-m)^{i+1} \right)\\ &= \frac{1}{i+1} \left( \left( \frac{b-a}{2} \right)^{i+1} - \left( \frac{a-b}{2} \right)^{i+1} \right) \cdot \end{aligned}

But $i + 1$ is even, so

{(\frac{b - a}{2})}^{i + 1} = {(\frac{a - b}{2})}^{i + 1},

and

\int_{a}^{b} (x - m)^{i} d x = 0 \cdot

On the right side, we have

\frac{b - a}{6} ((a - m)^{i} + 4 (m - m)^{i} + (b - m)^{i}) - \frac{(b - a)^{5}}{2880} f^{(i v)} (m) \cdot

If $i = 1$ or $3$ , the fourth derivative of $(x - m)^{i}$ is $0$ . If $i = 5$ , the fourth derivative is $5! (x - m)$ , and $f^{(i v)} (m) = 0$ . Thus the right side reduces to

\frac{b - a}{6} ((a - m)^{i} + (b - m)^{i}) = \frac{b - a}{6} ({(\frac{a - b}{2})}^{i} + {(\frac{b - a}{2})}^{i}) \cdot

Since $i$ is odd,

{(\frac{a - b}{2})}^{i} = - {(\frac{b - a}{2})}^{i},

and the right-hand side is $0$ . So in this case, both sides of the equation are $0$ , and we are done. It remains to treat $f (x) = 1, (x - m)^{2}$ and $(x - m)^{4}$ .

In the first two cases, $f^{(i v)} (x) = 0$ for all $x$ , so we may omit the last term on the right. For $f (x) = 1$ ,

\int_{a}^{b} 1 d x = b - a,

while

\frac{b - a}{6} (f (a) + 4 f (m) + f (b)) = \frac{b - a}{6} \cdot 6 = b - a,

Thus the assertion is true for $f (x) = 1$ . For $f (x) = (x - m)^{2}$ :

\begin{aligned} \int_a^b (x-m)^2 \, dx &= \left. \frac{1}{3} (x-m)^3 \right|_a^b \\ &= \frac{1}{3} \left( \left( \frac{b-a}{2} \right)^3 - \left( \frac{a-b}{2} \right)^3 \right)\\ &= \frac{1}{3} \left( \left( \frac{b-a}{2} \right)^3 + \left( \frac{b-a}{2} \right)^3 \right) \\ &= \frac{(b-a)^3}{12} \cdot \end{aligned}

Meanwhile,

\begin{aligned} \frac{b-a}{6} (f(a) + 4f(m) + f(b)) &= \frac{b-a}{6} \left( (a-m)^2 + (b-m)^2 \right)\\ &= \frac{b-a}{6} \left( \left( \frac{a-b}{2} \right)^2 + \left( \frac{b-a}{2} \right)^2 \right) \\ &= \frac{(b-a)^3}{12} \\ &= \int_a^b f(x) \, dx \cdot \end{aligned}

Finally, let $f (x) = (x - m)^{4}$ . Then $f^{(i v)} (x) = 24$ for all $x$ .

\int_{a}^{b} f (x) d x = \frac{1}{5} ({(\frac{b - a}{2})}^{5} - {(\frac{a - b}{2})}^{5}) = \frac{1}{80} (b - a)^{5} \cdot

On the other hand,

\begin{aligned} \frac{b-a}{6} (f(a) &+ 4f(m) + f(b)) - \frac{(b-a)^5}{2880} f^{\rm (iv)}(m)\\ &= \frac{b-a}{6} \left( \left( \frac{a-b}{2} \right)^4 + \left( \frac{b-a}{2} \right)^4 \right) - \frac{(b-a)^5}{2880} \cdot 24\\ &= \frac{(b-a)^5}{48} - \frac{(b-a)^5}{120} \\ &= \frac{(b-a)^5}{80} \\ &= \int_a^b f(x) \, dx \cdot \end{aligned}

Thus the assertion is true for all $(x - m)^{i}$ , $i \leq 5$ , and therefore for all polynomials of degree $\leq 5$ .

Now suppose we are given a function $f (x)$ having four derivatives, and suppose we have found a polynomial $P (x)$ of degree $3$ such that: $P (a) = f (a)$ , $P (b) = f (b)$ , $P (m) = f (m)$ , P'(m) = f'(m). (We shall show later how $P (x)$ can be constructed. Then $f (x) - P (x)$ has at $x = a$ and at $x = b$ a zero of multiplicity at least one, and at $x = m$ a zero of multiplicity at least $2$ . Therefore we take the $n$ of the approximation theorem to be $4$ , and $H (x) = (x - a) (x - m)^{2} (x - b)$ . Then

f (x) - P (x) = \frac{H (x)}{4!} (f^{(i v)} (ξ) - P^{(i v)} (ξ)), a < ξ < b \cdot

But $P^{(i v)} (x) = 0$ , since $P (x)$ is of degree $3$ . Now $H (x) \leq 0$ whenever $a \leq x \leq b$ , and if Max. denotes the maximum of $f^{(i v)} (x)$ for $a \leq x \leq b$ , Min. its minimum, we have

\frac{H(x)}{24} \cdot \text{Max.} \le f(x) - P(x) \le \frac{H(x)}{24} \cdot \text{Min.} \tag{**}

Since $H (x)$ is a polynomial of degree $\leq 5$ , and since $H (a) = H (b) = H (m) = 0$ , we see by our theorem that

\int_{a}^{b} H (x) d x = - \frac{(b - a)^{5}}{2880} \cdot H^{(i v)} (m) = - \frac{(b - a)^{5}}{2880} \cdot 24.

Taking the integrals in the inequality (**), we have

- \frac{(b - a)^{5}}{2880} \cdot Max. \leq \int_{a}^{b} f (x) d x - \int_{a}^{b} P (x) d x \leq - \frac{(b - a)^{5}}{2880} \cdot Min.

If we assume, as we already have in discussing its maximum and minimum, that $f^{(i v)}$ is continuous for $a \leq x \leq b$ , then by the theorem we quoted in Chapter 12,

\int_{a}^{b} f (x) d x - \int_{a}^{b} P (x) d x = - \frac{(b - a)^{5}}{2880} f^{(i v)} (ξ)

for some $ξ$ , $a \leq ξ \leq b$ . Now since $P^{(i v)} (x) = 0$ , we know by our theorem that

\begin{aligned} \int_a^b P(x) \, dx &= \frac{b-a}{6} (P(a) + 4P(m) + P(b)) \\ &= \frac{b-a}{6} (f(a) + 4f(m) + f(b)). \end{aligned}

Finally, we have Simpson's rule:

\int_{a}^{b} f(x) \, dx = \frac{b-a}{6} (f(a) + 4f(m) + f(b)) - \frac{(b-a)^5}{2880} f^{\rm (iv)}(\xi) \tag{*}

for some $ξ$ , $a ≦ ξ ≦ b$ .

The way in which this rule is used is by using the first term on the right as an approximation to

\int_{a}^{b} f (x) d x .

The second term gives an estimate of the error introduced by making this approximation. The error is at most

\frac{(b - a)^{5}}{2880} \cdot | Max . | or \frac{(b - a)^{5}}{2880} \cdot | Min . |,

whichever is larger.

Finally, we must show that a polynomial $P (x)$ can always be found satisfying our requirements. Let $Q = (a, f (a))$ , $R = (b, f (b))$ , $S = (m, f (m))$ . Then we can find a straight line $y = A x + B$ through $Q$ and $R$ (here $A$ and $B$ are simply numbers). Set

g (x) = A x + B + C (x - a) (x - b),

where $C$ is so chosen that $g (m) = f (m)$ . This is always possible because $(m - a) (m - b) \neq 0$ . We still have $g (a) = f (a)$ , $g (b) = f (b)$ . Now set

P (x) = g (x) + D (x - a) (x - b) (x - m) .

Then

P'(m) = g'(m) + D(m-a)(m-b).

Since $(m - a) (m - b) \neq 0$ , we can choose the constant $D$ so that P'(m) = f'(m). Then $P (a) = g (a) = f (a)$ , $P (b) = g (b) = f (b)$ , $P (m) = g (m) = f (m)$ and P'(m) = f'(m); moreover, $P (x)$ is of degree $3$ . Therefore $P (x)$ is our required polynomial. (Note that in applying Simpson's rule, one need not compute anything about $P (x)$ .)

EXERCISES

Exercise 1. 1) Use Simpson's rule to compute

\int_{0}^{1} \frac{d x}{1 + x^{2}};

compare with the result you get by direct integration.

Exercise 2.

Compute $\int_{- 1}^{1} e^{- x^{2}} d x :$ a) for $a = - 1, b = 1$ ;
b) by using the fact that it is $2 \int_{0}^{1} e^{- x^{2}} d x$ and using $a = 0, b = 1$ . Compare your estimates of error in each case.

Exercise 3.

Use Simpson's rule to approximate, with error, $\int_{- 1 / 2}^{1 / 2} \frac{d x}{\sqrt{1 - x^{4}}} \cdot$

No direct means is known of finding indefinite integrals in Exercises 2 and 3; therefore approximate methods assume great importance in these cases. The integral in Exercise 2 is called the error integral, and in Exercise 3 is an elliptic integral.

Simpson's Rule

EXERCISES

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