6.4: Tangent roots- A daunting transcendental sum

Our farewell example, chosen because its analysis combines diverse streetfighting tools, is a difficult infinite sum.

\(\text{Find} S ≡ \sum x_{n}^{-2} \text{ where the } x_{n} \text{ are the positive solutions of } \tan x = x\)

The solutions to \(\tan x = x\) or, equivalently, the roots of \(\tan x − x\), are transcendental and have no closed form, yet a closed form is required for almost every summation method. Street-fighting methods will come to our rescue.

Pictures and easy cases

Begin the analysis with a hopefully easy case.

Question

What is the first root \(x_{1}\)?

The roots of \(\tan x − x\) are given by the intersections of \(y = x\) and \(y = \tan x\). Surprisingly, no intersection occurs in the branch of \(\tan x\) where \(0 < x < \pi/2\) (Problem 6.23); the first intersection is just before the asymptote at \(x = 3\pi/2\). Thus, \(x_{1} ≈ 3\pi/2\).

Problem 6.23 No intersection with the main branch

Show symbolically that \(\tan x = x\) has no solution for \(0 < x < \pi/2\). (The result looks plausible pictorially but is worth checking in order to draw the picture.)

Question

Where, approximately, are the subsequent intersections?

As \(x\) grows, the \(y = x\) line intersects the \(y = \tan x\) graph ever higher and therefore ever closer to the vertical asymptotes. Therefore, make the following asymptote approximation for the big part of \(x_{n}\):

\[x_{n} ≈ (n + \frac{1}{2})\pi. \label{6.18} \]

Taking out the big part

This approximate, low-entropy expression for \(x_{n}\) gives the big part of \(S\) (the zeroth approximation).

\[S ≈ \sum[\underbrace{(n + \frac{1}{2}\pi}_{≈x_{n}}]^{-2} = \frac{4}{\pi^{2}} \sum_{1}^{\infty} \frac{1}{(2n + 1)^{2}}. \label{6.19} \]

The sum \(\sum_{1}^{\infty} (2n + 1)^{−2}\) is, from a picture (Section 4.5) or from Euler–MacLaurin summation (Section 6.3.2), roughly the following integral.

\[\sum_{1}^{\infty}(2n + 1)^{-2} \approx \int_{1}^{\infty}(2n+1)^{-2} \mathrm{~d} n=-\frac{1}{2} \times\left.\frac{1}{2n + 1}\right|_{1} ^{\infty}=\frac{1}{6}\label{6.20} \]

Therefore,

\[S ≈ \frac{4}{\pi^{2}} x \frac{1}{6} = 0.067547... \label{6.21} \]

The shaded protrusions are roughly triangles, and they sum to one-half of the first rectangle. That rectangle has area 1/9, so

\[\sum_{1}^{\infty} (2n + 1)^{-2} ≈ \frac{1}{6} + \frac{1}{2} x \frac{1}{9} = \frac{2}{9}. \label{6.22} \]

Therefore, a more accurate estimate of \(S\) is

\[S ≈ \frac{4}{\pi^{2}} x \frac{2}{9} = 0.090063..., \label{6.23} \]

which is slightly higher than the first estimate.

Question

Is the new approximation an overestimate or an underestimate ?

The new approximation is based on two underestimates. First, the asymptote approximation \(x_{n} ≈ (n + 0.5)\pi\) overestimates each xn and therefore underestimates the squared reciprocals in the sum \(\sum x_{n}^{-2}\). Second, after making the asymptote approximation, the pictorial approximation to the sum \(\sum_{1}^{\infty}(2n + 1)^{−2}\) replaces each protrusion with an inscribed triangle and thereby underestimates each protrusion (Problem 6.24).

Problem 6.24 Picture for the second underestimate

Draw a picture of the underestimate in the pictorial approximation

\[\int_{1}^{\infty} \frac{1}{(2n + 1)^{2}} ≈ \frac{1}{6} + \frac{1}{2} x \frac{1}{9}. \label{6.24} \]

Question

How can these two underestimates be remedied ?

The second underestimate (the protrusions) is eliminated by summing \(\sum_{1}^{\infty} (2n + 1)^{−2}\) exactly. The sum is unfamiliar partly because its first term is the fraction 1/9 whose arbitrariness increases the entropy of the sum. Including the \(n = 0\) term, which is 1, and the even squared reciprocals \(1/(2n)^{2}\) produces a compact and familiar lower-entropy sum.

\[\int_{1}^{\infty} \frac{1}{(2n + 1)^{2}} + 1 + \int_{1}^{\infty} \frac{1}{(2n)^{2}} = \int_{1}^{\infty} \frac{1}{n^{2}}. \label{6.25} \]

The final, low-entropy sum is the famous Basel sum (high-entropy results are not often famous). Its value is \(B = \pi^{2}/6\) (Problem 6.22).

Question

How does knowing \(B = \pi^{2}/6\) help evaluate the original sum \(\sum_{1}^{\infty} (2n + 1)^{−2}\) ?

The major modification from the original sum was to include the even squared reciprocals. Their sum is \(B/4\).

\[\sum_{1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{4} \sum_{1}^{\infty} \frac{1}{n^{2}}. \label{6.26} \]

The second modification was to include the \(n = 0\) term. Thus, to obtain \(\sum_{1}^{\infty} (2n + 1)^{−2}\), adjust the Basel value \(B\) by subtracting \(B/4\) and then the \(n = 0\) term. The result, after substituting \(B = \pi^{2}/6\), is

\[\sum_{1}^{\infty} \frac{1}{(2n + 1)^{2}} = B - \frac{1}{4}B - 1 = \frac{\pi^{2}}{8} - 1 \label{6.27} \]

This exact sum, based on the asymptote approximation for \(x_{n}\), produces the following estimate of \(S\).

\[S ≈ \frac{4}{\pi^{2}} \sum_{1}^{\infty} \frac{1}{(2n + 1)^{2}} = \frac{4}{\pi^{2}} (\frac{\pi^{2}}{8} - 1) \label{6.28} \]

Simplifying by expanding the product gives

\[S ≈ \frac{1}{2} - \frac{4}{pi^{2}} = 0.094715... \label{6.29} \]

Problem 6.25 Check the earlier reasoning

Check the earlier pictorial reasoning (Problem 6.24) that \(1/6 + 1/18 = 2/9\) underestimates \(\sum_{1}^{\infty} (2n + 1)^{−2}\). How accurate was that estimate?

This estimate of \(S\) is the third that uses the asymptote approximation \(x_{n} ≈ (n + 0.5)\pi\). Assembled together, the estimates are

Because the third estimate incorporated the exact value of \(\sum_{1}^{\infty} (2n + 1)^{−2}\), any remaining error in the estimate of \(S\) must belong to the asymptote approximation itself.

Question

For which term of \(\sum {x_{n}}^{−2}\) is the asymptote approximation most inaccurate ?

As \(x\) grows, the graphs of \(x\) and \(\tan x\) intersect ever closer to the vertical asymptote. Thus, the asymptote approximation makes its largest absolute error when \(n = 1\). Because \(x_{1}\) is the smallest root, the fractional error in \(x_{n}\) is, relative to the absolute error in \(x_{n}\), even more concentrated at \(n = 1\). The fractional error in \(x_{n}^{−2}\), being −2 times the fractional error \(n\) in \(x\) (Section 5.3), is equally concentrated at \(n = 1\). Because \(x_{n}^{−2}\) is the largest at \(n = 1\), the absolute error in \(x_{n}^{−2}\) (the fractional error times \(x_{n}^{−2}\) itself) is, by far, the largest at \(n = 1\).

Question

Estimate, as a function of \(n\), the absolute error in \(x_{n}^{−2}\) that is produced by the \(n\) asymptote approximation.

With the error so concentrated at \(n = 1\), the greatest improvement in the estimate of \(S\) comes from replacing the approximation \(x_{1} = (n + 0.5)\pi\) with a more accurate value. A simple numerical approach is successive approximation using the Newton–Raphson method (Problem 4.38). To find a root with this method, make a starting guess \(x\) and repeatedly improve it using the replacement

\[x \rightarrow x - \frac{\tan x - x}{\sec^{2}x - 1}. \label{6.30} \]

When the starting guess for \(x\) is slightly below the first asymptote at \(1.5\pi\) the procedure rapidly converges to \(x_{1} = 4.4934 . . .\)

Therefore, to improve the estimate \(S ≈ 0.094715\), which was based on the asymptote approximation, subtract its approximate first term (its big part) and add the corrected first term.

\[S = S_{\text{old}} - \frac{1}{(1.5\pi)^{2}} + \frac{1}{4.4934^{2}} ≈ 0.09921. \label{6.31} \]

Using the Newton–Raphson method to refine, in addition, the \(1/x_{2}^{2}\) term gives \(S ≈ 0.09978\) (Problem 6.27). Therefore, a highly educated guess is

\[S = \frac{1}{10}. \label{6.32} \]

The infinite sum of unknown transcendental numbers seems to be neither transcendental nor irrational! This simple and surprising rational number deserves a simple explanation.

Problem 6.27 Continuing the corrections

Choose a small \(N\), say 4. Then use the Newton–Raphson method to compute accurate values of \(x_{n}\) for \(n = 1 . . . N\); and use those values to refine the estimate of \(S\). As you extend the computation to larger values of \(N\), do the refined estimates of \(S\) approach our educated guess of 1/10?

Analogy with polynomials

If only the equation \(\tan x − x = 0\) had just a few closed-form solutions! Then the sum \(S\) would be easy to compute. That wish is fulfilled by replacing \(\tan x − x\) with a polynomial equation with simple roots. The simplest interesting polynomial is the quadratic, so experiment with a simple quadratic for example, \(x^{2} − 3x + 2\).

This polynomial has two roots, \(x = 1\) and \(x = 2\); therefore \(\sum x_{n}^{−2}\), the polynomial-root sum analog of the tangent-root sum, has two terms.

\[\sum x_{n}^{-2} = \frac{1}{1^{2}} + \frac{1}{2^{2}} = \frac{5}{4}. \label{6.33} \]

This brute-force method for computing the root sum requires a solution to the quadratic equation. However, a method that can transfer to the equation \(\tan x − x = 0\), which has no closed-form solution, cannot use the roots themselves. It must use only surface features of the quadratic namely, its two coefficients 2 and −3. Unfortunately, no plausible method of combining 2 and −3 predicts that \(\sum x_{n}^{−2} = 5/4\).

Question

Where did the polynomial analogy go wrong ?

The problem is that the quadratic \(x^{2} − 3x + 2\) is not sufficiently similar to \(\tan x − x\). The quadratic has only positive roots; however, \(\tan x − x\), an odd function, has symmetric positive and negative roots and has a root at \(x = 0\). Indeed, the Taylor series for \(\tan x\) is \(x + x^{3}/3 + 2x^{5}/15 + ···\) (Problem 6.28); therefore,

\[\tan x - x = \frac{x^{3}}{3} + \frac{2x^{5}}{15} + ... \label{6.34} \]

The common factor of \(x^{3}\) means that \(\tan x − x\) has a triple root at \(x = 0\).

An analogous polynomial here, one with a triple root at \(x = 0\), a positive root, and a symmetric negative root is \((x+2) x^{3}(x−2)\) or, after expansion, \(x^{5} − 4x^{3}\). The sum \(\sum x_{n}\) (using the positive root) contains only one term and is simply 1/4. This value could plausibly arise as the (negative) ratio of the last two coefficients of the polynomial.
To decide whether that pattern is a coincidence, try a richer polynomial: one with roots at −2, −1, 0 (threefold), 1, and 2. One such polynomial is

\[(x + 2)(x + 1)x^{3}(x - 1)(x - 2) = x^{7} - 5x^{5} + 4x^{3}. \label{6.35} \]

The polynomial-root sum uses only the two positive roots 1 and 2 and is \(1/1^{2} + 1/2^{2}\), which is 5/4 the (negative) ratio of the last two coefficients.

As a final test of this pattern, include −3 and 3 among the roots. The resulting polynomial is

\[(x^{7} - 5x^{5} +4x^{3}) (x + 3)(x - 3) = x^{9} - 14x^{7} + 49x^{5} - 36x^{3}. \label{6.36} \]

The polynomial-root sum uses the three positive roots 1, 2, and 3 and is \(1/1^{2} + 1/2^{2} + 1/3^{2}\), which is 49/36 again the (negative) ratio of the last two coefficients in the expanded polynomial.

Question

What is the origin of the pattern, and how can it be extended to \(\tan x − x\)?

To explain the pattern, tidy the polynomial as follows:

\[x^{9} − 14x^{7} + 49x^{5} − 36x^{3} = −36x^{3} (1 - \frac{49}{36}x^{2} + \frac{14}{36}x^{4} - \frac{1}{36}x^{6}). \label{6.37} \]

In this arrangement, the sum 49/36 appears as the negative of the first interesting coefficient. Let’s generalize. Placing \(k\) roots at \(x = 0\) and single roots at \(± x_{1}, ± x_{2}, . . ., ± x_{n}\) gives the polynomial

\[Ax^{k}(1 - \frac{x^{2}}{x_{1}^{2}}) \frac{x^{2}}{x_{2}^{2}}) \frac{x^{2}}{x_{3}^{2}}) ... \frac{x^{2}}{x_{n}^{2}}), \label{6.38} \]

where \(A\) is a constant. When expanding the product of the factors in parentheses, the coefficient of the \(x^{2}\) term in the expansion receives one contribution from each \(x^{2}/x_{k}^{2}\) term in a factor. Thus, the expansion begins

\[Ax^{k}[1 - (\frac{1}{x_{1}^{2}} + \frac{1}{x_{2}^{2}} + \frac{1}{x_{3}^{2}} + ... + \frac{1}{x_{n}^{2}})x^{2} + ...].\label{6.39} \]

The coefficient of \(x^{2}\) in parentheses is \(\sum_{x_{n}}^{−2}\), which is the polynomial analog of the tangent-root sum.

Let’s apply this method to \(\tan x − x\). Although it is not a polynomial, its Taylor series is like an infinite-degree polynomial. The Taylor series is

\[\frac{x^{3}}{3} + \frac{2x^{5}}{15} + \frac{17x^{7}}{315} + ... = \frac{x^{3}}{3} (1 + \frac{2}{5}x^{2} + \frac{17}{105}x^{4} + ...). \label{6.40} \]

The negative of the \(x^{2}\) coefficient should be −\(\sum x_{n}^{-2}\). For the tangentsum problem, \(x − 2\) should therefore be −2/5. Unfortunately, the sum of positive quantities cannot be negative!

Question

What went wrong with the analogy?

One problem is that \(\tan x − x\) might have imaginary or complex roots whose squares contribute negative amounts to \(S\). Fortunately, all its roots are real (Problem 6.29). A harder-to-solve problem is that \(\tan x − x\) goes to infinity at finite values of \(x\), and does so infinitely often, whereas no polynomial does so even once.

The solution is to construct a function having no infinities but having the same roots as \(\tan x − x\). The infinities of \(\tan x − x\) occur where \(\tan x\) blows up, which is where \(\cos x = 0\). To remove the infinities without creating or destroying any roots, multiply \(\tan x − x\) by \(\cos x\). The polynomial-like function to expand is therefore \(\sin x − x \cos x\).

Its Taylor expansion is

\[\underbrace{\left(x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots\right)}_{\sin x}-\underbrace{\left(x-\frac{x^{3}}{2}+\frac{x^{5}}{24}-\cdots\right)}_{x \cos x} . \label{6.41} \]

The difference of the two series is

\[\sin x - x\cos x = \frac{x^{3}}{3}(1 - \frac{1}{10}x^{2} + ...). \label{6.42} \]

The \(x^{3}/3\) factor indicates the triple root at \(x = 0\). And there at last, as the negative of the \(x^{2}\) coefficient, sits our tangent-root sum \(S = 1/10\).

Multiple problems

Problem 6.28 Taylor series for the tangent

Use the Taylor series for \(\sin x\) and \(\cos x\) to show that

\[tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + ... . \label{6.43} \]

Hint: Use taking out the big part.

Problem 6.29 Only real roots

Show that all roots of \(\tan x − x\) are real.

Problem 6.30 Exact Basel sum

Use the polynomial analogy to evaluate the Basel sum

\[\sum_{1}^{\infty} \frac{1}{n^{2}} \label{6.44} \]

Compare your result with your solution to Problem 6.22.

Problem 6.31 Misleading alternative expansions

Squaring and taking the reciprocal of \(\tan x = x\) gives \(cot^{2}x = x^{−2}\); equivalently, \(cot^{2}x = x^{−2} = 0\). Therefore, if \(x\) is a root of \(\tan x − x\), it is a root of \(cot^{2}x = x^{−2}\). The Taylor expansion of \(cot^{2}x = x^{−2}\) is

\[-\frac{2}{3}(1 - \frac{1}{10}x^{2} - \frac{1}{63}x^{4} - ...). \label{6.45} \]

Because the coefficient of \(x^{2}\) is −1/10, the tangent-root sum \(S\) for \(\cot x = x^{−2}\) and therefore \(\tan x = x\) should be 1/10. As we found experimentally and analytically for \(\tan x = x\), the conclusion is correct. However, what is wrong with the reasoning?

Problem 6.32 Fourth powers of the reciprocals

The Taylor series for \(\sin x − x \cos x\) continues

\[\frac{x^{3}}{3}(1 - \frac{x^{2}}{10} + \frac{x^{4}}{280} - ...). \label{6.46} \]

Therefore find \(\sum x_{n}^{−4}\) for the positive roots of \(\tan x = x\). Check numerically that your result is plausible.

Problem 6.33 Other source equations for the roots

Find \(\sum x_{n}^{−2}\), where the \(x_{n}\) are the positive roots of \(\cos x\).