Editing Euler–Maclaurin formula (section)

== Proofs ==

=== Derivation by mathematical induction ===
We outline the argument given in Apostol.<ref name=":0">{{Cite journal| first = T. M. | title = An Elementary View of Euler's Summation Formula | journal = [[The American Mathematical Monthly]] | volume = 106 | date = 1 May 1999 | issue = 5 | pages = 409–418  | issn = 0002-9890| last = Apostol | doi = 10.2307/2589145| author-link= Tom M. Apostol| publisher =    Mathematical Association of America| jstor =    2589145}}</ref>

The [[Bernoulli polynomials]] {{math|''B<sub>n</sub>''(''x'')}} and the periodic Bernoulli functions {{math|''P<sub>n</sub>''(''x'')}} for {{math|''n'' {{=}} 0, 1, 2, ...}} were introduced above.

The first several Bernoulli polynomials are
<math display=block>\begin{align}
  B_0(x) &= 1, \\
  B_1(x) &= x - \tfrac{1}{2}, \\
  B_2(x) &= x^2 - x + \tfrac{1}{6}, \\
  B_3(x) &= x^3 - \tfrac{3}{2}x^2 + \tfrac{1}{2}x, \\
  B_4(x) &= x^4 - 2x^3 + x^2 - \tfrac{1}{30}, \\
         &\,\,\,\vdots
\end{align}</math>

The values {{math|''B<sub>n</sub>''(1)}} are the [[Bernoulli numbers]] {{math|''B''<sub>''n''</sub>}}.  Notice that for {{math|''n'' ≠ 1}} we have
<math display=block>B_n = B_n(1) = B_n(0),</math>
and for {{math|''n'' {{=}} 1}},
<math display=block>B_1 = B_1(1) = -B_1(0).</math>

The functions {{math|''P''<sub>''n''</sub>}} agree with the Bernoulli polynomials on the interval {{math|[0,&nbsp;1]}}  and are [[periodic function|periodic]] with period 1.  Furthermore, except when {{math|''n'' {{=}} 1}}, they are also continuous. Thus,
<math display=block> P_n(0) = P_n(1) = B_n \quad \text{for }n \neq 1.</math>

Let {{math|''k''}} be an integer, and consider the integral
<math display=block> \int_k^{k + 1} f(x)\,dx = \int_k^{k + 1} u\,dv,</math>
where
<math display=block>\begin{align}
   u &= f(x), \\
  du &= f'(x)\,dx, \\
  dv &= P_0(x)\,dx & \text{since }P_0(x) &= 1, \\
   v &= P_1(x).
\end{align}</math>

[[integration by parts|Integrating by parts]], we get
<math display=block>\begin{align}
 \int_k^{k + 1} f(x)\,dx &= \bigl[uv\bigr]_k^{k + 1} - \int_k^{k + 1} v\,du \\
                         &= \bigl[f(x)P_1(x)\bigr]_k^{k + 1} - \int_k^{k+1} f'(x)P_1(x)\,dx \\
                         &= B_1(1)f(k+1)-B_1(0)f(k) - \int_k^{k+1} f'(x)P_1(x)\,dx.
\end{align}</math>

Using {{math|''B''<sub>1</sub>(0) {{=}} −{{sfrac|1|2}}}}, {{math|''B''<sub>1</sub>(1) {{=}} {{sfrac|1|2}}}}, and summing the above from {{math|''k'' {{=}} 0}} to {{math|''k'' {{=}} ''n'' − 1}}, we get
<math display=block>\begin{align}
\int_0^n f(x)\, dx &= \int_0^1 f(x)\,dx + \cdots + \int_{n-1}^n f(x)\,dx  \\
&= \frac{f(0)}{2}+ f(1) + \dotsb + f(n-1) + \frac{f(n)}{2} - \int_0^n f'(x) P_1(x)\,dx.
\end{align}</math>

Adding {{math|{{sfrac|''f''(''n'') − ''f''(0)|2}}}} to both sides and rearranging, we have
<math display=block> \sum_{k=1}^n f(k) = \int_0^n f(x)\,dx + \frac{f(n) - f(0)}{2} + \int_0^n f'(x) P_1(x)\,dx.</math>

This is the {{math|''p'' {{=}} 1}} case of the summation formula.  To continue the induction, we apply integration by parts to the error term:
<math display=block>\int_k^{k+1} f'(x)P_1(x)\,dx = \int_k^{k + 1} u\,dv,</math>
where
<math display=block>\begin{align}
   u &= f'(x), \\
  du &= f''(x)\,dx, \\
  dv &= P_1(x)\,dx, \\
   v &= \tfrac{1}{2}P_2(x).
\end{align}</math>

The result of integrating by parts is
<math display=block>\begin{align}
  \bigl[uv\bigr]_k^{k + 1} - \int_k^{k + 1} v\,du &= \left[\frac{f'(x)P_2(x)}{2} \right]_k^{k+1} - \frac{1}{2}\int_k^{k+1} f''(x)P_2(x)\,dx \\
                  &= \frac{B_2}{2}(f'(k + 1) - f'(k)) - \frac{1}{2}\int_k^{k + 1} f''(x)P_2(x)\,dx.
\end{align}</math>

Summing from {{math|''k'' {{=}} 0}} to {{math|''k'' {{=}} ''n'' &minus; 1}} and substituting this for the lower order error term results in the {{math|''p'' {{=}} 2}} case of the formula,
<math display=block>\sum_{k=1}^n f(k) = \int_0^n f(x)\,dx + \frac{f(n) - f(0)}{2} + \frac{B_2}{2}\bigl(f'(n) - f'(0)\bigr) - \frac{1}{2}\int_0^n f''(x)P_2(x)\,dx.</math>

This process can be iterated.  In this way we get a proof of the Euler–Maclaurin summation formula which can be formalized by [[mathematical induction]], in which the induction step relies on integration by parts and on identities for periodic Bernoulli functions.