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=== Asymptotic analysis === Arguably the most important application of the Bernoulli numbers in mathematics is their use in the [[Euler–Maclaurin formula]]. Assuming that {{mvar|f}} is a sufficiently often differentiable function the Euler–Maclaurin formula can be written as{{sfnp|Graham|Knuth|Patashnik|1989|loc=9.67}} : <math>\sum_{k=a}^{b-1} f(k) = \int_a^b f(x)\,dx + \sum_{k=1}^m \frac{B^-_k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R_-(f,m).</math> This formulation assumes the convention {{math|1=''B''{{su|p=−|b=1}} = −{{sfrac|1|2}}}}. Using the convention {{math|1=''B''{{su|p=+|b=1}} = +{{sfrac|1|2}}}} the formula becomes : <math>\sum_{k=a+1}^{b} f(k) = \int_a^b f(x)\,dx + \sum_{k=1}^m \frac{B^+_k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R_+(f,m).</math> Here <math>f^{(0)}=f</math> (i.e. the zeroth-order derivative of <math>f</math> is just <math>f</math>). Moreover, let <math>f^{(-1)}</math> denote an [[antiderivative]] of <math>f</math>. By the [[fundamental theorem of calculus]], : <math>\int_a^b f(x)\,dx = f^{(-1)}(b) - f^{(-1)}(a).</math> Thus the last formula can be further simplified to the following succinct form of the Euler–Maclaurin formula : <math> \sum_{k=a+1}^{b} f(k)= \sum_{k=0}^m \frac{B_k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R(f,m). </math> This form is for example the source for the important Euler–Maclaurin expansion of the zeta function : <math> \begin{align} \zeta(s) & =\sum_{k=0}^m \frac{B^+_k}{k!} s^{\overline{k-1}} + R(s,m) \\ & = \frac{B_0}{0!}s^{\overline{-1}} + \frac{B^+_1}{1!} s^{\overline{0}} + \frac{B_2}{2!} s^{\overline{1}} +\cdots+R(s,m) \\ & = \frac{1}{s-1} + \frac{1}{2} + \frac{1}{12}s + \cdots + R(s,m). \end{align} </math> Here {{math|''s''<sup>{{overline|''k''}}</sup>}} denotes the [[Pochhammer symbol|rising factorial power]].{{sfnp|Graham|Knuth|Patashnik|1989|loc=2.44, 2.52}} Bernoulli numbers are also frequently used in other kinds of [[asymptotic expansion]]s. The following example is the classical Poincaré-type asymptotic expansion of the [[digamma function]] {{math|''ψ''}}. :<math>\psi(z) \sim \ln z - \sum_{k=1}^\infty \frac{B^+_k}{k z^k} </math>
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