Mathematics Lecture Notes

Theorem4.2Euler-Maclaurin Formula

Let $f \in C^{2p+2}[a, b]$ and $h = (b - a)/n$ . Then the composite trapezoidal rule satisfies: $T_n(f) = \int_a^b f(x)\, dx + \sum_{k=1}^p \frac{B_{2k}}{(2k)!} h^{2k} \left[f^{(2k-1)}(b) - f^{(2k-1)}(a)\right] + R_p$ where $B_{2k}$ are the Bernoulli numbers ( $B_2 = 1/6$ , $B_4 = -1/30$ , $B_6 = 1/42, \ldots$ ) and the remainder satisfies $|R_p| \leq \frac{|B_{2p+2}|}{(2p+2)!} h^{2p+2} (b-a) \|f^{(2p+2)}\|_\infty$ .

Proof

The proof uses the Bernoulli polynomials $B_k(x)$ defined by $\frac{te^{xt}}{e^t - 1} = \sum_{k=0}^\infty B_k(x) \frac{t^k}{k!}$ .

Step 1. On a single interval $[0, h]$ , apply repeated integration by parts with $B_k(x/h)$ : $\int_0^h f(x)\, dx = \frac{h}{2}[f(0) + f(h)] - \sum_{k=1}^p \frac{B_{2k}}{(2k)!} h^{2k}[f^{(2k-1)}(h) - f^{(2k-1)}(0)] + r_p$ where $r_p = -\frac{h^{2p+2}}{(2p+2)!}\int_0^h B_{2p+2}(x/h) f^{(2p+2)}(x)\, dx$ .

Step 2. Sum over all subintervals $[x_j, x_{j+1}]$ for $j = 0, \ldots, n-1$ . Interior derivative terms telescope: $f^{(2k-1)}(x_{j+1}) - f^{(2k-1)}(x_j)$ at shared nodes cancel, leaving only the boundary contributions $f^{(2k-1)}(b) - f^{(2k-1)}(a)$ .

Step 3. The left side sums to $\int_a^b f(x)\, dx$ and the trapezoidal contributions sum to $T_n(f)$ , yielding the stated formula. The remainder bound follows from $|B_{2p+2}(t)| \leq |B_{2p+2}|$ for $t \in [0,1]$ . $\square$

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ExampleRichardson Extrapolation from Euler-Maclaurin

The Euler-Maclaurin formula shows $T(h) = I + c_1 h^2 + c_2 h^4 + \cdots$ with explicit constants $c_k$ . This asymptotic expansion in even powers of $h$ is the theoretical foundation for Romberg integration: combining $T(h)$ and $T(h/2)$ eliminates the $h^2$ term, giving the Simpson-like formula $\frac{4T(h/2) - T(h)}{3} = I + O(h^4)$ . Each level of extrapolation removes two orders of error.

RemarkThe Trapezoidal Rule for Periodic Functions

When $f$ is periodic with period $b - a$ , all boundary terms $f^{(2k-1)}(b) - f^{(2k-1)}(a) = 0$ vanish. The Euler-Maclaurin formula then gives $T_n(f) = I + R_p$ for any $p$ . If $f$ is $C^\infty$ -periodic (or analytic and periodic), the trapezoidal rule converges exponentially fast: $|T_n - I| = O(e^{-cn})$ for some $c > 0$ . This remarkable phenomenon makes the trapezoidal rule the method of choice for periodic integrands.

ExampleComputing $\sum_{k=1}^n 1/k$ via Euler-Maclaurin

The Euler-Maclaurin formula also converts sums to integrals. For $\sum_{k=1}^n \frac{1}{k}$ : $\sum_{k=1}^n \frac{1}{k} = \ln n + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \frac{1}{120n^4} - \cdots$ where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant. This asymptotic expansion allows high-precision computation of harmonic sums and the constant $\gamma$ .

Math Notes

Euler-Maclaurin Summation Formula

Proof Outline

Applications