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<?xml version="1.0" encoding="UTF-8"?>
<article xmlns="http://docbook.org/ns/docbook" xmlns:m="http://www.w3.org/1998/Math/MathML" version="5.0" xml:lang="en-US" role="numbered">
<info>
<title>From Basel to Gauss: Sums, Products and Integrals</title>
<author>
<orgname>Maxprograms</orgname>
</author>
<pubdate>14 November 2026</pubdate>
<copyright>
<year>2026</year>
<holder>Maxprograms</holder>
</copyright>
<abstract>
<para>A short walk through classical results of analysis: telescoping sums, the geometric
series, Euler’s product for the sine, the Gaussian integral, an eigenvalue problem and
the binomial theorem. Each result is stated, proved in a few lines and checked against
numbers.</para>
</abstract>
</info>
<section xml:id="sec-intro">
<title>Introduction</title>
<para>In 1734 Leonhard Euler found the exact value of the sum of the reciprocals of the squares
of the natural numbers, a problem that had resisted the best mathematicians for almost a
century:</para>
<informalequation xml:id="eq-basel">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mrow><m:mi>ζ</m:mi><m:mo>(</m:mo><m:mn>2</m:mn><m:mo>)</m:mo></m:mrow>
<m:mo>=</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mfrac>
<m:mn>1</m:mn>
<m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac>
<m:msup><m:mi>π</m:mi><m:mn>2</m:mn></m:msup>
<m:mn>6</m:mn>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>The series converges, because its terms behave like
<inlineequation><m:math><m:mn>1</m:mn><m:mo>/</m:mo><m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup></m:math></inlineequation>
for large <inlineequation><m:math><m:mi>n</m:mi></m:math></inlineequation>, but it converges
slowly: after <inlineequation><m:math><m:mi>n</m:mi></m:math></inlineequation> terms the
error is about
<inlineequation><m:math><m:mn>1</m:mn><m:mo>/</m:mo><m:mi>n</m:mi></m:math></inlineequation>,
as the table in <xref linkend="sec-numbers" xrefstyle="section"/> shows. This article collects
proofs of nearby facts that use the same tools, from the first year of calculus to a little
linear algebra. Equation <xref linkend="eq-basel" xrefstyle="equation"/> is the destination of the walk.</para>
</section>
<section xml:id="sec-telescoping">
<title>Telescoping sums</title>
<example role="theorem">
<title>Theorem 1</title>
<para>For every natural number
<inlineequation><m:math><m:mi>n</m:mi><m:mo>≥</m:mo><m:mn>1</m:mn></m:math></inlineequation>,</para>
<informalequation xml:id="eq-telescoping">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mi>k</m:mi><m:mo>(</m:mo><m:mi>k</m:mi><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>)</m:mo></m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac>
<m:mi>n</m:mi>
<m:mrow><m:mi>n</m:mi><m:mo>+</m:mo><m:mn>1</m:mn></m:mrow>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
</example>
<para role="proof"><emphasis>Proof.</emphasis> Each term can be split into two fractions:</para>
<informalequation xml:id="eq-partial-fractions">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mi>k</m:mi><m:mo>(</m:mo><m:mi>k</m:mi><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>)</m:mo></m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac><m:mn>1</m:mn><m:mi>k</m:mi></m:mfrac>
<m:mo>−</m:mo>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mi>k</m:mi><m:mo>+</m:mo><m:mn>1</m:mn></m:mrow>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para role="proof end">so the sum telescopes: every term cancels with the next one, and only the
first and the last survive. What remains is
<inlineequation><m:math>
<m:mn>1</m:mn><m:mo>−</m:mo>
<m:mfrac><m:mn>1</m:mn><m:mrow><m:mi>n</m:mi><m:mo>+</m:mo><m:mn>1</m:mn></m:mrow></m:mfrac>
<m:mo>=</m:mo>
<m:mi>n</m:mi><m:mo>/</m:mo><m:mo>(</m:mo><m:mi>n</m:mi><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>)</m:mo>
</m:math></inlineequation>.</para>
<para>Letting
<inlineequation><m:math><m:mi>n</m:mi><m:mo>→</m:mo><m:mi>∞</m:mi></m:math></inlineequation>
gives a sum of exactly 1. The same idea bounds the series of <xref linkend="eq-basel" xrefstyle="equation"/>: for
<inlineequation><m:math><m:mi>k</m:mi><m:mo>≥</m:mo><m:mn>2</m:mn></m:math></inlineequation>
we have
<inlineequation><m:math>
<m:mn>1</m:mn><m:mo>/</m:mo><m:msup><m:mi>k</m:mi><m:mn>2</m:mn></m:msup>
<m:mo><</m:mo>
<m:mn>1</m:mn><m:mo>/</m:mo><m:mo>(</m:mo><m:mi>k</m:mi><m:mo>(</m:mo><m:mi>k</m:mi><m:mo>−</m:mo><m:mn>1</m:mn><m:mo>)</m:mo><m:mo>)</m:mo>
</m:math></inlineequation>, and the second sum telescopes as above, so</para>
<informalequation xml:id="eq-bound">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:mfrac>
<m:mn>1</m:mn>
<m:msup><m:mi>k</m:mi><m:mn>2</m:mn></m:msup>
</m:mfrac>
<m:mo><</m:mo>
<m:mn>2</m:mn><m:mo>−</m:mo><m:mfrac><m:mn>1</m:mn><m:mi>n</m:mi></m:mfrac>
<m:mo><</m:mo>
<m:mn>2</m:mn>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>The partial sums grow and stay below 2, so the series converges.</para>
</section>
<section xml:id="sec-geometric">
<title>The geometric series</title>
<para>For a real number
<inlineequation><m:math><m:mi>r</m:mi><m:mo>≠</m:mo><m:mn>1</m:mn></m:math></inlineequation>
the partial sums of the powers of
<inlineequation><m:math><m:mi>r</m:mi></m:math></inlineequation> have a closed form:</para>
<informalequation xml:id="eq-geometric">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:apply>
<m:eq/>
<m:apply>
<m:sum/>
<m:bvar><m:ci>k</m:ci></m:bvar>
<m:lowlimit><m:cn>0</m:cn></m:lowlimit>
<m:uplimit><m:ci>n</m:ci></m:uplimit>
<m:apply><m:power/><m:ci>r</m:ci><m:ci>k</m:ci></m:apply>
</m:apply>
<m:apply>
<m:divide/>
<m:apply>
<m:minus/>
<m:cn>1</m:cn>
<m:apply><m:power/><m:ci>r</m:ci><m:apply><m:plus/><m:ci>n</m:ci><m:cn>1</m:cn></m:apply></m:apply>
</m:apply>
<m:apply><m:minus/><m:cn>1</m:cn><m:ci>r</m:ci></m:apply>
</m:apply>
</m:apply>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>To see this, multiply the sum by
<inlineequation><m:math><m:mn>1</m:mn><m:mo>−</m:mo><m:mi>r</m:mi></m:math></inlineequation>:
all the intermediate powers cancel in pairs and only
<inlineequation><m:math><m:mn>1</m:mn><m:mo>−</m:mo><m:msup><m:mi>r</m:mi><m:mrow><m:mi>n</m:mi><m:mo>+</m:mo><m:mn>1</m:mn></m:mrow></m:msup></m:math></inlineequation>
is left. When
<inlineequation><m:math><m:mrow><m:mo>|</m:mo><m:mi>r</m:mi><m:mo>|</m:mo></m:mrow><m:mo><</m:mo><m:mn>1</m:mn></m:math></inlineequation>
the power
<inlineequation><m:math><m:msup><m:mi>r</m:mi><m:mrow><m:mi>n</m:mi><m:mo>+</m:mo><m:mn>1</m:mn></m:mrow></m:msup></m:math></inlineequation>
tends to 0, and the series has a sum:</para>
<informalequation xml:id="eq-geometric-series">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:msup><m:mi>r</m:mi><m:mi>k</m:mi></m:msup>
<m:mo>=</m:mo>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mn>1</m:mn><m:mo>−</m:mo><m:mi>r</m:mi></m:mrow>
</m:mfrac>
<m:mo>,</m:mo>
<m:mspace width="1.5em"/>
<m:mrow><m:mo>|</m:mo><m:mi>r</m:mi><m:mo>|</m:mo></m:mrow>
<m:mo><</m:mo>
<m:mn>1</m:mn>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>With
<inlineequation><m:math><m:mi>r</m:mi><m:mo>=</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mn>2</m:mn></m:math></inlineequation>
this says that
<inlineequation><m:math><m:mn>1</m:mn><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mn>2</m:mn><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mn>4</m:mn><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mn>8</m:mn><m:mo>+</m:mo><m:mo>⋯</m:mo></m:math></inlineequation>
adds up to 2, the picture of a segment that is halved again and again.</para>
</section>
<section xml:id="sec-euler">
<title>Euler’s product for the sine</title>
<para>Euler’s idea was to treat the sine as a polynomial of infinite degree whose roots are the
integers. Written as a product over its roots, and scaled so that its value at 0 is 1, it
becomes<footnote><para>The product converges for every real
<inlineequation><m:math><m:mi>x</m:mi></m:math></inlineequation>. Taking
<inlineequation><m:math><m:mi>x</m:mi><m:mo>=</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mn>2</m:mn></m:math></inlineequation>
gives the product of Wallis,
<inlineequation><m:math>
<m:mi>π</m:mi><m:mo>/</m:mo><m:mn>2</m:mn><m:mo>=</m:mo>
<m:munderover>
<m:mo>∏</m:mo>
<m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mfrac>
<m:mrow><m:mn>4</m:mn><m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup></m:mrow>
<m:mrow><m:mn>4</m:mn><m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow>
</m:mfrac>
</m:math></inlineequation>. A rigorous proof of the product needs the factorization theorem
of Weierstrass.</para></footnote></para>
<informalequation xml:id="eq-product">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mfrac>
<m:mrow><m:mi>sin</m:mi><m:mo>⁡</m:mo><m:mo>(</m:mo><m:mi>π</m:mi><m:mi>x</m:mi><m:mo>)</m:mo></m:mrow>
<m:mrow><m:mi>π</m:mi><m:mi>x</m:mi></m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:munderover>
<m:mo>∏</m:mo>
<m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mrow>
<m:mo>(</m:mo>
<m:mn>1</m:mn><m:mo>−</m:mo>
<m:mfrac>
<m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup>
<m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup>
</m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>On the other hand, the Taylor series of the sine, divided by
<inlineequation><m:math><m:mi>π</m:mi><m:mi>x</m:mi></m:math></inlineequation>, gives</para>
<informalequation xml:id="eq-taylor">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mfrac>
<m:mrow><m:mi>sin</m:mi><m:mo>⁡</m:mo><m:mo>(</m:mo><m:mi>π</m:mi><m:mi>x</m:mi><m:mo>)</m:mo></m:mrow>
<m:mrow><m:mi>π</m:mi><m:mi>x</m:mi></m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mfrac>
<m:mrow>
<m:msup><m:mrow><m:mo>(</m:mo><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow><m:mo>)</m:mo></m:mrow><m:mi>k</m:mi></m:msup>
<m:msup><m:mrow><m:mo>(</m:mo><m:mi>π</m:mi><m:mi>x</m:mi><m:mo>)</m:mo></m:mrow><m:mrow><m:mn>2</m:mn><m:mi>k</m:mi></m:mrow></m:msup>
</m:mrow>
<m:mrow><m:mo>(</m:mo><m:mn>2</m:mn><m:mi>k</m:mi><m:mo>+</m:mo><m:mn>1</m:mn><m:mo>)</m:mo><m:mo>!</m:mo></m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mn>1</m:mn><m:mo>−</m:mo>
<m:mfrac>
<m:mrow><m:msup><m:mi>π</m:mi><m:mn>2</m:mn></m:msup><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup></m:mrow>
<m:mn>6</m:mn>
</m:mfrac>
<m:mo>+</m:mo>
<m:mfrac>
<m:mrow><m:msup><m:mi>π</m:mi><m:mn>4</m:mn></m:msup><m:msup><m:mi>x</m:mi><m:mn>4</m:mn></m:msup></m:mrow>
<m:mn>120</m:mn>
</m:mfrac>
<m:mo>−</m:mo>
<m:mo>⋯</m:mo>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>Compare the coefficients of
<inlineequation><m:math><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup></m:math></inlineequation>
in <xref linkend="eq-product" xrefstyle="equation"/> and <xref linkend="eq-taylor" xrefstyle="equation"/>. Multiplying out the
product, the only way to get
<inlineequation><m:math><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup></m:math></inlineequation>
is to take one factor
<inlineequation><m:math><m:mo>−</m:mo><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup><m:mo>/</m:mo><m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup></m:math></inlineequation>
and 1 from all the others, so the coefficient is
<inlineequation><m:math>
<m:mo>−</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>n</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mn>1</m:mn><m:mo>/</m:mo><m:msup><m:mi>n</m:mi><m:mn>2</m:mn></m:msup>
</m:math></inlineequation>. In the series it is
<inlineequation><m:math><m:mo>−</m:mo><m:msup><m:mi>π</m:mi><m:mn>2</m:mn></m:msup><m:mo>/</m:mo><m:mn>6</m:mn></m:math></inlineequation>.
Equating them gives <xref linkend="eq-basel" xrefstyle="equation"/>.</para>
</section>
<section xml:id="sec-gauss">
<title>The Gaussian integral</title>
<example role="theorem">
<title>Theorem 2</title>
<para>The area under the bell curve is</para>
<informalequation xml:id="eq-gauss">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:msubsup>
<m:mo>∫</m:mo>
<m:mrow><m:mo>−</m:mo><m:mi>∞</m:mi></m:mrow>
<m:mi>∞</m:mi>
</m:msubsup>
<m:msup><m:mi>e</m:mi><m:mrow><m:mo>−</m:mo><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup></m:mrow></m:msup>
<m:mspace width="0.2em"/>
<m:mi>d</m:mi><m:mi>x</m:mi>
<m:mo>=</m:mo>
<m:msqrt><m:mi>π</m:mi></m:msqrt>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
</example>
<para role="proof"><emphasis>Proof.</emphasis> Call the integral
<inlineequation><m:math><m:mi>I</m:mi></m:math></inlineequation>. The trick is to square it and
to read the product of two integrals as an integral over the plane:</para>
<informalequation xml:id="eq-plane">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:msup><m:mi>I</m:mi><m:mn>2</m:mn></m:msup>
<m:mo>=</m:mo>
<m:msub>
<m:mo>∬</m:mo>
<m:msup><m:mi mathvariant="double-struck">R</m:mi><m:mn>2</m:mn></m:msup>
</m:msub>
<m:msup>
<m:mi>e</m:mi>
<m:mrow>
<m:mo>−</m:mo><m:mo>(</m:mo>
<m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup><m:mo>+</m:mo><m:msup><m:mi>y</m:mi><m:mn>2</m:mn></m:msup>
<m:mo>)</m:mo>
</m:mrow>
</m:msup>
<m:mspace width="0.2em"/>
<m:mi>d</m:mi><m:mi>x</m:mi><m:mspace width="0.15em"/><m:mi>d</m:mi><m:mi>y</m:mi>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para role="proof">In polar coordinates the area element is
<inlineequation><m:math><m:mi>r</m:mi><m:mspace width="0.2em"/><m:mi>d</m:mi><m:mi>r</m:mi><m:mspace width="0.15em"/><m:mi>d</m:mi><m:mi>θ</m:mi></m:math></inlineequation>,
and the integral becomes elementary:</para>
<informalequation xml:id="eq-polar">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:msup><m:mi>I</m:mi><m:mn>2</m:mn></m:msup>
<m:mo>=</m:mo>
<m:msubsup>
<m:mo>∫</m:mo><m:mn>0</m:mn><m:mrow><m:mn>2</m:mn><m:mi>π</m:mi></m:mrow>
</m:msubsup>
<m:msubsup>
<m:mo>∫</m:mo><m:mn>0</m:mn><m:mi>∞</m:mi>
</m:msubsup>
<m:msup><m:mi>e</m:mi><m:mrow><m:mo>−</m:mo><m:msup><m:mi>r</m:mi><m:mn>2</m:mn></m:msup></m:mrow></m:msup>
<m:mspace width="0.2em"/>
<m:mi>r</m:mi><m:mspace width="0.2em"/><m:mi>d</m:mi><m:mi>r</m:mi><m:mspace width="0.15em"/><m:mi>d</m:mi><m:mi>θ</m:mi>
<m:mo>=</m:mo>
<m:mn>2</m:mn><m:mi>π</m:mi><m:mo>⋅</m:mo><m:mfrac><m:mn>1</m:mn><m:mn>2</m:mn></m:mfrac>
<m:mo>=</m:mo>
<m:mi>π</m:mi>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para role="proof end">Since <inlineequation><m:math><m:mi>I</m:mi></m:math></inlineequation> is
positive,
<inlineequation><m:math><m:mi>I</m:mi><m:mo>=</m:mo><m:msqrt><m:mi>π</m:mi></m:msqrt></m:math></inlineequation>.</para>
<para>Statisticians use this result every day. Dividing the bell curve by its area, and
allowing a mean <inlineequation><m:math><m:mi>μ</m:mi></m:math></inlineequation> and a standard
deviation <inlineequation><m:math><m:mi>σ</m:mi></m:math></inlineequation>, gives the density
of the normal distribution, whose total probability is 1 as it should be:</para>
<informalequation xml:id="eq-normal">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:msubsup>
<m:mo>∫</m:mo>
<m:mrow><m:mo>−</m:mo><m:mi>∞</m:mi></m:mrow>
<m:mi>∞</m:mi>
</m:msubsup>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mi>σ</m:mi><m:msqrt><m:mn>2</m:mn><m:mi>π</m:mi></m:msqrt></m:mrow>
</m:mfrac>
<m:msup>
<m:mi>e</m:mi>
<m:mrow>
<m:mo>−</m:mo>
<m:mfrac>
<m:msup><m:mrow><m:mo>(</m:mo><m:mi>x</m:mi><m:mo>−</m:mo><m:mi>μ</m:mi><m:mo>)</m:mo></m:mrow><m:mn>2</m:mn></m:msup>
<m:mrow><m:mn>2</m:mn><m:msup><m:mi>σ</m:mi><m:mn>2</m:mn></m:msup></m:mrow>
</m:mfrac>
</m:mrow>
</m:msup>
<m:mspace width="0.2em"/>
<m:mi>d</m:mi><m:mi>x</m:mi>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
</section>
<section xml:id="sec-matrix">
<title>A small eigenvalue problem</title>
<para>Linear algebra has its own sums. Take the symmetric matrix</para>
<informalequation xml:id="eq-matrix">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mi>A</m:mi>
<m:mo>=</m:mo>
<m:mrow>
<m:mo>(</m:mo>
<m:mtable>
<m:mtr><m:mtd><m:mn>2</m:mn></m:mtd><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd><m:mtd><m:mn>2</m:mn></m:mtd></m:mtr>
</m:mtable>
<m:mo>)</m:mo>
</m:mrow>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>Its eigenvalues <inlineequation><m:math><m:mi>λ</m:mi></m:math></inlineequation> are the
roots of the characteristic polynomial:</para>
<informalequation xml:id="eq-characteristic">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mrow>
<m:mi>det</m:mi><m:mo>⁡</m:mo>
<m:mo>(</m:mo><m:mi>A</m:mi><m:mo>−</m:mo><m:mi>λ</m:mi><m:mi>I</m:mi><m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:mrow>
<m:mo>|</m:mo>
<m:mtable>
<m:mtr>
<m:mtd><m:mn>2</m:mn><m:mo>−</m:mo><m:mi>λ</m:mi></m:mtd>
<m:mtd><m:mn>1</m:mn></m:mtd>
</m:mtr>
<m:mtr>
<m:mtd><m:mn>1</m:mn></m:mtd>
<m:mtd><m:mn>2</m:mn><m:mo>−</m:mo><m:mi>λ</m:mi></m:mtd>
</m:mtr>
</m:mtable>
<m:mo>|</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:msup><m:mrow><m:mo>(</m:mo><m:mn>2</m:mn><m:mo>−</m:mo><m:mi>λ</m:mi><m:mo>)</m:mo></m:mrow><m:mn>2</m:mn></m:msup>
<m:mo>−</m:mo><m:mn>1</m:mn>
<m:mo>=</m:mo>
<m:msup><m:mi>λ</m:mi><m:mn>2</m:mn></m:msup><m:mo>−</m:mo><m:mn>4</m:mn><m:mi>λ</m:mi><m:mo>+</m:mo><m:mn>3</m:mn>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>The quadratic formula gives the roots of any polynomial
<inlineequation><m:math>
<m:mi>a</m:mi><m:msup><m:mi>x</m:mi><m:mn>2</m:mn></m:msup><m:mo>+</m:mo><m:mi>b</m:mi><m:mi>x</m:mi><m:mo>+</m:mo><m:mi>c</m:mi>
</m:math></inlineequation>:</para>
<informalequation xml:id="eq-quadratic">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mi>x</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mo>−</m:mo><m:mi>b</m:mi><m:mo>±</m:mo>
<m:msqrt><m:msup><m:mi>b</m:mi><m:mn>2</m:mn></m:msup><m:mo>−</m:mo><m:mn>4</m:mn><m:mi>a</m:mi><m:mi>c</m:mi></m:msqrt>
</m:mrow>
<m:mrow><m:mn>2</m:mn><m:mi>a</m:mi></m:mrow>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>For <xref linkend="eq-characteristic" xrefstyle="equation"/> it gives
<inlineequation><m:math><m:mi>λ</m:mi><m:mo>=</m:mo><m:mo>(</m:mo><m:mn>4</m:mn><m:mo>±</m:mo><m:mn>2</m:mn><m:mo>)</m:mo><m:mo>/</m:mo><m:mn>2</m:mn></m:math></inlineequation>,
that is, <inlineequation><m:math><m:mi>λ</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:math></inlineequation>
and <inlineequation><m:math><m:mi>λ</m:mi><m:mo>=</m:mo><m:mn>3</m:mn></m:math></inlineequation>.
A check with the vector
<inlineequation><m:math>
<m:msup>
<m:mrow><m:mo>(</m:mo><m:mn>1</m:mn><m:mo>,</m:mo><m:mn>1</m:mn><m:mo>)</m:mo></m:mrow>
<m:mi>T</m:mi>
</m:msup>
</m:math></inlineequation> confirms the larger one:</para>
<informalequation xml:id="eq-eigen">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mrow>
<m:mo>(</m:mo>
<m:mtable>
<m:mtr><m:mtd><m:mn>2</m:mn></m:mtd><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd><m:mtd><m:mn>2</m:mn></m:mtd></m:mtr>
</m:mtable>
<m:mo>)</m:mo>
</m:mrow>
<m:mrow>
<m:mo>(</m:mo>
<m:mtable>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
</m:mtable>
<m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:mrow>
<m:mo>(</m:mo>
<m:mtable>
<m:mtr><m:mtd><m:mn>3</m:mn></m:mtd></m:mtr>
<m:mtr><m:mtd><m:mn>3</m:mn></m:mtd></m:mtr>
</m:mtable>
<m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:mn>3</m:mn>
<m:mrow>
<m:mo>(</m:mo>
<m:mtable>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
<m:mtr><m:mtd><m:mn>1</m:mn></m:mtd></m:mtr>
</m:mtable>
<m:mo>)</m:mo>
</m:mrow>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>In the same way
<inlineequation><m:math><m:msup><m:mrow><m:mo>(</m:mo><m:mn>1</m:mn><m:mo>,</m:mo><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow><m:mo>)</m:mo></m:mrow><m:mi>T</m:mi></m:msup></m:math></inlineequation>
is an eigenvector for
<inlineequation><m:math><m:mi>λ</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:math></inlineequation>. The
two eigenvectors are orthogonal, as they must be for a symmetric matrix.</para>
</section>
<section xml:id="sec-binomial">
<title>The binomial theorem</title>
<para>Expanding the power of a sum gives the coefficients that Pascal’s triangle lists:</para>
<informalequation xml:id="eq-binomial">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:msup>
<m:mrow><m:mo>(</m:mo><m:mi>a</m:mi><m:mo>+</m:mo><m:mi>b</m:mi><m:mo>)</m:mo></m:mrow>
<m:mi>n</m:mi>
</m:msup>
<m:mo>=</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:mrow>
<m:mo>(</m:mo>
<m:mfrac linethickness="0"><m:mi>n</m:mi><m:mi>k</m:mi></m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
<m:msup><m:mi>a</m:mi><m:mrow><m:mi>n</m:mi><m:mo>−</m:mo><m:mi>k</m:mi></m:mrow></m:msup>
<m:msup><m:mi>b</m:mi><m:mi>k</m:mi></m:msup>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>The binomial coefficient counts the subsets with
<inlineequation><m:math><m:mi>k</m:mi></m:math></inlineequation> elements of a set with
<inlineequation><m:math><m:mi>n</m:mi></m:math></inlineequation> elements:</para>
<informalequation xml:id="eq-coefficient">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mrow>
<m:mo>(</m:mo>
<m:mfrac linethickness="0"><m:mi>n</m:mi><m:mi>k</m:mi></m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow><m:mi>n</m:mi><m:mo>!</m:mo></m:mrow>
<m:mrow><m:mi>k</m:mi><m:mo>!</m:mo><m:mo>(</m:mo><m:mi>n</m:mi><m:mo>−</m:mo><m:mi>k</m:mi><m:mo>)</m:mo><m:mo>!</m:mo></m:mrow>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>Two choices of <inlineequation><m:math><m:mi>a</m:mi></m:math></inlineequation> and
<inlineequation><m:math><m:mi>b</m:mi></m:math></inlineequation> in
<xref linkend="eq-binomial" xrefstyle="equation"/> give identities that are worth remembering. With
<inlineequation><m:math><m:mi>a</m:mi><m:mo>=</m:mo><m:mi>b</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:math></inlineequation>
the coefficients add up to
<inlineequation><m:math><m:msup><m:mn>2</m:mn><m:mi>n</m:mi></m:msup></m:math></inlineequation>,
the number of all subsets:</para>
<informalequation xml:id="eq-subsets">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:mrow>
<m:mo>(</m:mo>
<m:mfrac linethickness="0"><m:mi>n</m:mi><m:mi>k</m:mi></m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:msup><m:mn>2</m:mn><m:mi>n</m:mi></m:msup>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>With
<inlineequation><m:math><m:mi>a</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:math></inlineequation> and
<inlineequation><m:math><m:mi>b</m:mi><m:mo>=</m:mo><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow></m:math></inlineequation>
the signs alternate, and for
<inlineequation><m:math><m:mi>n</m:mi><m:mo>≥</m:mo><m:mn>1</m:mn></m:math></inlineequation> the
sum is zero:</para>
<informalequation xml:id="eq-alternating">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:msup><m:mrow><m:mo>(</m:mo><m:mrow><m:mo>−</m:mo><m:mn>1</m:mn></m:mrow><m:mo>)</m:mo></m:mrow><m:mi>k</m:mi></m:msup>
<m:mrow>
<m:mo>(</m:mo>
<m:mfrac linethickness="0"><m:mi>n</m:mi><m:mi>k</m:mi></m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
<m:mo>=</m:mo>
<m:mn>0</m:mn>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
</section>
<section xml:id="sec-e">
<title>The number e</title>
<para>The binomial theorem also leads to the number
<inlineequation><m:math><m:mi>e</m:mi></m:math></inlineequation>. Take
<inlineequation><m:math><m:mi>b</m:mi><m:mo>=</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mi>n</m:mi></m:math></inlineequation>
and <inlineequation><m:math><m:mi>a</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:math></inlineequation>
in <xref linkend="eq-binomial" xrefstyle="equation"/>: the coefficient of each power grows towards
<inlineequation><m:math><m:mn>1</m:mn><m:mo>/</m:mo><m:mi>k</m:mi><m:mo>!</m:mo></m:math></inlineequation>
as <inlineequation><m:math><m:mi>n</m:mi></m:math></inlineequation> grows, and in the
limit</para>
<informalequation xml:id="eq-e-limit">
<m:math display="block">
<m:mtable width="100%" columnwidth="fit auto fit" columnalign="center center right">
<m:mtr>
<m:mtd/>
<m:mtd>
<m:mi>e</m:mi>
<m:mo>=</m:mo>
<m:munder>
<m:mi>lim</m:mi>
<m:mrow><m:mi>n</m:mi><m:mo>→</m:mo><m:mi>∞</m:mi></m:mrow>
</m:munder>
<m:msup>
<m:mrow>
<m:mo>(</m:mo>
<m:mn>1</m:mn><m:mo>+</m:mo><m:mfrac><m:mn>1</m:mn><m:mi>n</m:mi></m:mfrac>
<m:mo>)</m:mo>
</m:mrow>
<m:mi>n</m:mi>
</m:msup>
<m:mo>=</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>0</m:mn></m:mrow>
<m:mi>∞</m:mi>
</m:munderover>
<m:mfrac>
<m:mn>1</m:mn>
<m:mrow><m:mi>k</m:mi><m:mo>!</m:mo></m:mrow>
</m:mfrac>
</m:mtd>
<m:mtd class="eqnum"/>
</m:mtr>
</m:mtable>
</m:math>
</informalequation>
<para>so that
<inlineequation><m:math><m:mi>e</m:mi><m:mo>≈</m:mo><m:mn>2.71828</m:mn></m:math></inlineequation>.
Both expressions converge, but the series is far faster: its seventh partial sum already has
four correct decimals.</para>
</section>
<section xml:id="sec-numbers">
<title>Checking the numbers</title>
<para>The table lists the partial sums of the series of <xref linkend="eq-basel" xrefstyle="equation"/> and their
distance to the limit
<inlineequation><m:math><m:msup><m:mi>π</m:mi><m:mn>2</m:mn></m:msup><m:mo>/</m:mo><m:mn>6</m:mn><m:mo>≈</m:mo><m:mn>1.6449</m:mn></m:math></inlineequation>.
The error shrinks like
<inlineequation><m:math><m:mn>1</m:mn><m:mo>/</m:mo><m:mi>n</m:mi></m:math></inlineequation>,
and every row respects the bound of <xref linkend="eq-bound" xrefstyle="equation"/>.</para>
<table>
<title>Partial sums of the series for ζ(2)</title>
<tgroup cols="4">
<colspec colname="n"/>
<colspec colname="sum"/>
<colspec colname="error"/>
<colspec colname="bound"/>
<thead>
<row>
<entry><inlineequation><m:math><m:mi>n</m:mi></m:math></inlineequation></entry>
<entry><inlineequation><m:math>
<m:msub><m:mi>s</m:mi><m:mi>n</m:mi></m:msub><m:mo>=</m:mo>
<m:munderover>
<m:mo>∑</m:mo>
<m:mrow><m:mi>k</m:mi><m:mo>=</m:mo><m:mn>1</m:mn></m:mrow>
<m:mi>n</m:mi>
</m:munderover>
<m:mn>1</m:mn><m:mo>/</m:mo><m:msup><m:mi>k</m:mi><m:mn>2</m:mn></m:msup>
</m:math></inlineequation></entry>
<entry><inlineequation><m:math>
<m:msup><m:mi>π</m:mi><m:mn>2</m:mn></m:msup><m:mo>/</m:mo><m:mn>6</m:mn>
<m:mo>−</m:mo><m:msub><m:mi>s</m:mi><m:mi>n</m:mi></m:msub>
</m:math></inlineequation></entry>
<entry><inlineequation><m:math>
<m:mn>2</m:mn><m:mo>−</m:mo><m:mn>1</m:mn><m:mo>/</m:mo><m:mi>n</m:mi>
</m:math></inlineequation></entry>
</row>
</thead>
<tbody>
<row><entry>1</entry><entry>1.0000</entry><entry>0.6449</entry><entry>1.0000</entry></row>
<row><entry>2</entry><entry>1.2500</entry><entry>0.3949</entry><entry>1.5000</entry></row>
<row><entry>3</entry><entry>1.3611</entry><entry>0.2838</entry><entry>1.6667</entry></row>
<row><entry>4</entry><entry>1.4236</entry><entry>0.2213</entry><entry>1.7500</entry></row>
<row><entry>5</entry><entry>1.4636</entry><entry>0.1813</entry><entry>1.8000</entry></row>
<row><entry>10</entry><entry>1.5498</entry><entry>0.0952</entry><entry>1.9000</entry></row>
</tbody>
</tgroup>
</table>
<para>Euler’s result, the integral of Gauss and the binomial theorem are three views of the
same landscape: sums that look infinite until one finds the right way to fold them.</para>
</section>
</article>