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The Basel problem asks for the exact sum of the reciprocal squares of the positive integers. The answer isThe integral test proves that the series converges, but does not identify this exact value. The Euler evaluation of even zeta values does: the equally spaced poles of cotangent connect reciprocal integer powers to Bernoulli numbers and $\pi$. This separates three different questions: whether a sum exists, how well a partial sum approximates it, and what its exact value is.

For example, the ten-term partial sum is about $1.549768$, while the exact value is about $1.644934$. The zeta series truncation error bounds the missing amount between $1/11$ and $1/10$; it is much larger than the first omitted term $1/121$. More terms improve the approximation, but a numerical coincidence with ${\pi }^2/6$ would not prove the identity.

In On the sums of series of reciprocals by Leonhard Euler, §2 identifies the reciprocal-square sum with one-sixth of the squared circumference of a unit-diameter circle. The translation's publication note dates the presentation to 5 December 1735 and publication to 1740.^[ref] The cotangent coefficient proof of the even zeta values gives a short modern outline, while zeta series truncation error addresses the distinct approximation question.

The cosecant is $\csc z=1/\sin z$, wherever the sine denominator is nonzero. The square of the cosecant occurs in the derivative $d(\cot z)/dz=-{\csc }^{2}z$ of the cotangent.

For $w\notin \mathbb{Z}$, integrate $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ around large squares whose sides have half-integer coordinates. The cotangent is uniformly bounded on these sides and the integral tends to zero. The residue theorem gives residues $(n-w{)}^{-2}$ at the integers and $-{\pi }^2{\csc }^2(\pi w)$ at $w$, proving the identity. For $\mathrm{Im}w>0$, the geometric-series formula for the cotangent also gives ${\pi }^2{\csc }^2(\pi w)=-4{\pi }^2{\sum }_{r\ge 1}r{e}^{2\pi irw}$.

The cotangent kernel $\pi \cot (\pi z)$ has a simple pole of residue one at every integer. If a meromorphic $h$ is regular at the integers, these residues sample its integer values. Integrating around squares of half-side $N+1/2$ and applying the residue theorem converts a finite sum of samples into residues at the other poles. When the contour integral tends to zero and $h$ has only finitely many other poles, the limiting sum is the negative sum of those off-integer residues of $\pi \cot (\pi z)h(z)$. A uniform cotangent bound and sufficiently rapid decay of $h$ justify the vanishing contour term. At an integer where $h$ itself has a pole, use the full Laurent series to find the combined residue instead of substituting a divergent sample; the Basel problem calculation with $h(z)=z^{-2}$ illustrates this exception.

For $\mathrm{Im}z>0$, writing cotangent as $i(e^{2\pi iz}+1)/(e^{2\pi iz}-1)$ and expanding a geometric series gives this Fourier series. Differentiating the cotangent partial-fraction expansion and this Fourier series $k-1$ times yields

Away from integer poles,The series is locally uniformly convergent. Integrate $\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with sides $\mathrm{Re}\zeta ,\mathrm{Im}\zeta =\pm (N+1/2)$. The cotangent is uniformly bounded there, so the integral tends to zero. The residue theorem then gives $\pi \cot (\pi z)/z+{\sum }_{n\in \mathbb{Z}}(n^2-z^2{)}^{-1}=0$, which is the expansion. Differentiation generates higher integer-lattice pole sums. Setting $w=e^{2\pi iz}$ gives $\pi \cot \pi z=\pi i(w+1)/(w-1)$. On $0<|z|<1$, expansion of each paired summand gives $\pi \cot \pi z=z^{-1}-2{\sum }_{m\ge 0}\zeta (2m+2)z^{2m+1}$, relating this trigonometric function to the Riemann zeta function.

The Euler evaluation of even zeta values turns a family of series into exact expressions using Bernoulli numbers, factorials and powers of $\pi$. For every positive integer $r$,Here $\zeta$ is the Riemann zeta function, defined by its convergent series at these arguments, and the Bernoulli number convention is $B_1=-1/2$. The alternating sign combines with the sign of $B_{2r}$ to give a positive sum.^[ref]

Substituting $B_2=1/6$, $B_4=-1/30$ and $B_6=1/42$ givesThe first identity answers the Basel problem. It is surprising because the original series only mentions reciprocal squares of integers, while its exact sum involves the circle constant $\pi$. The cotangent partial-fraction expansion supplies a connection between the two: trigonometric functions encode equally spaced integer poles, whose expansion brings the reciprocal powers into view.

For an original numerical comparison, the first ten terms give ${\sum }_{n=1}^{10}{n}^{-2}\approx 1.549767731$, while ${\pi }^2/6\approx 1.644934067$. The omitted positive terms still contribute approximately $0.095166336$. For fourth powers, the ten-term partial sum is approximately $1.082036583$, only $0.000286650$ below ${\pi }^4/90\approx 1.082323234$. Both are convergent series, but the second approaches its limit much faster.

Exact evaluation is useful beyond obtaining more digits. It supplies an independent check on a numerical summation, exposes how Bernoulli numbers organize infinitely many identities, and makes the remaining series truncation error measurable without knowing any omitted terms individually. The even integer condition is essential to this formula: putting $r=3/2$ into it is not a way to evaluate $\zeta (3)$, since the required Bernoulli number index and factorial expression here are indexed by positive integers.

Historically, Leonhard Euler announced reciprocal-square and reciprocal-fourth-power evaluations in §2 of On the sums of series of reciprocals by Leonhard Euler. That paper was presented on 5 December 1735 and published in 1740, as the English translation's first-page note records. His §§3–8 develop a route through the sine power series and relationships between roots and coefficients. The cotangent calculation below is a concise modern presentation of the connection.^[ref]

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.

On any compact set avoiding the integers, writeFor all sufficiently large $|n|$, uniformly on that compact set this is $O(|n{|}^{-2})$. The summands in the higher series are $O(|n{|}^{-k})$ for $k\ge 2$. The Weierstrass M-test therefore gives locally uniform convergence away from the integer poles and permits termwise differentiation. Near an integer $m$, isolate its singular summand; all the others converge uniformly on a small neighborhood. Thus the first function has a simple pole of residue one at each integer, and the $k$th function has a pole of order $k$ there with principal part $(z-m{)}^{-k}$. There are no other singularities, so all are meromorphic functions.

Pairing positive and negative indices is legitimate because the regularized first series converges absolutely. It becomesHere is a direct derivation of its cotangent partial-fraction expansion. Fix $z\notin \mathbb{Z}$ and integrate $F(\zeta )=\pi \cot (\pi \zeta )/({\zeta }^2-z^2)$ around the square with real and imaginary coordinates $\pm R$, $R=N+1/2$, taking $N$ large enough to enclose $\pm z$. On the vertical sides $|\cot \pi \zeta |\le 1$, and on the horizontal sides it is at most $\coth (\pi R)$. Thus the cotangent is uniformly bounded and $|{\zeta }^2-z^2|\ge R^2-|z{|}^2$, so the contour integral is $O(R^{-1})\to 0$.

The residues at integers are $1/(n^2-z^2)$; the two residues at $z$ and $-z$ sum to $\pi \cot (\pi z)/z$. The residue theorem and absolute convergence of the integer residue sum consequently giveUsing the exponential definitions of sine and cosine now yields, for $w=e^{2\pi iz}$,The same locally uniform convergence gives ${\epsilon }_k^{\prime }=-k{\epsilon }_{k+1}$, including $k=1$. Hence if ${\epsilon }_k(z)={\phi }_k(e^{2\pi iz})$, the chain rule givesThis is a rational function whenever ${\phi }_k$ is, proving the requested descent for every $k$ by induction, withIn particular periodicity is a consequence of the explicit descent, rather than an assumption about the summation order.

Finally, for $|z|<1$ expand each paired summand as a geometric series:On $|z|\le r<1$ the double series converges absolutely, since its absolute sum is at most $2r{\sum }_{n\ge 1}{n}^{-2}/(1-r^2)$. Interchanging the sums and using the Riemann zeta function gives the Laurent seriesThus the $z^{-1}$ coefficient is one; the positive odd coefficients are $-2\zeta (2j+2)$, and all other coefficients vanish.

The displayed definition uses iterated Eisenstein summation in weight two: the sum in $n$ is evaluated before the sum in $m$, with the single $(0,0)$ term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.

For noninteger $w$, the cosecant partial-fraction identity isFor completeness, apply the residue theorem to $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is $O(R^{-1})$. Its residues at the integers are $(n-w{)}^{-2}$ and its residue at $w$ is $-{\pi }^2{\csc }^2(\pi w)$, proving the formula. If $\mathrm{Im}w>0$, the geometric-series expression $\cot (\pi w)=-i(1+2{\sum }_{r\ge 1}{e}^{2\pi irw})$, differentiated termwise, gives the cotangent partial-fraction Fourier kernel

Put $q=e^{2\pi iz}$ with $z$ in the complex upper half-plane. For positive $m$, this gives $-4{\pi }^2{\sum }_{r\ge 1}r{q}^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is ${\sum }_{n\ne 0}{n}^{-2}={\pi }^2/3$, by the Basel problem. The resulting series in $m,r$ does have absolute convergence, locally uniformly in $z$, so collecting the coefficient at $q^{\ell }$ is legitimate:The coefficient is the sum-of-divisors function, since $r$ runs over the positive divisors of $\ell$. ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.

Where the original expression is defined, multiplication by $\cos x$ givesThus the graph repeats with period $\pi$ and has vertical asymptotesBetween consecutive asymptotes the extended cotangent branch decreases from $+\infty$ to $-\infty$, since $y^{\prime }=-1/(\cos x+\sin x{)}^2<0$. It passes through $(k\pi ,1)$ and $(\pi /4+k\pi ,1/2)$.

It is essential to preserve the domain when simplifying a function. At $x=\pi /2+k\pi$, the original tangent is undefined. The finite limiting value is zero, so the graph has holes at $(\pi /2+k\pi ,0)$, rather than actual zeros. The simplified expression extends across those points, but the original one does not.

The original PDF specifies slope $\sqrt{3}$; the transcription's $\sqrt{-1}$ is erroneous. By part (a), $\cot (t/2)=\sqrt{3}$. Since $t/2\in (0,\pi )$ and cotangent is strictly decreasing there, the unique parameter value is $t=\pi /3$. ThereforeThe tangent line drawn in the preceding figure has the required positive slope.

Trigonometric functions turn an angle into a coordinate or a ratio of lengths. Start at $(1,0)$ on the unit circle and rotate counterclockwise through $\theta$ radians: the resulting point isThe cosine records horizontal position and the sine records vertical position. This definition continues around the whole circle, including negative angles, so the signs encode direction rather than requiring a separate rule for each right triangle.

The unit circle picture explains why these functions are useful beyond measuring triangles. The two coordinates of a point rotating at constant angular speed are periodic functions. Scaling the radius to $A$ and using $\theta =\omega t+\varphi$ gives $x=A\cos (\omega t+\varphi )$ and $y=A\sin (\omega t+\varphi )$. A circular motion becomes two simple repeating traces, connecting Euclidean geometry, simple harmonic motion, and Fourier series.

For an acute angle in a right triangle, rescaling its hypotenuse to one recovers the familiar ratios: sine is opposite side divided by hypotenuse, and cosine is adjacent side divided by hypotenuse. The tangent function is their ratio, $\tan \theta =\sin \theta /\cos \theta$, wherever the denominator is nonzero; the secant function, cosecant, and cotangent give the other reciprocal ratios.

The same names extend to complex numbers through the complex exponential:These definitions agree with the circle interpretation for real angles. For complex arguments, the values need not lie between $-1$ and $1$; the literal circle-coordinate picture is restricted to real arguments. ^[ref]