TAOCP 1.2.11.3 Exercise 20

Let u = w + \frac{1}{3}w^2 + \frac{1}{36}w^3 - \frac{1}{270}w^4 + \cdots = \sum_{k=1}^{\infty} c_k w^k be the power series solution to the equation

taocpmathematicsalgorithmsvolume-1hm-hard

Section 1.2.11.3: Some Asymptotic Calculations

Exercise 20. [HM30] Let

$$ u = w + \frac{1}{3}w^2 + \frac{1}{36}w^3 - \frac{1}{270}w^4 + \cdots = \sum_{k=1}^{\infty} c_k w^k $$

be the power series solution to the equation

$$ w = \left(u^2 - \frac{2}{3}u^3 + \frac{2}{4}u^4 - \frac{2}{5}u^5 + \cdots\right)^{1/2}, $$

as in (12). Show that

$$ Q(n) + 1 = \sum_{k=1}^{m-1} k c_k \Gamma(k/2)\left(\frac{n}{2}\right)^{1-k/2}

  • O(n^{1-m/2}) $$

for all $m \ge 1$.

[Hint:] Apply Watson's lemma to the identity of exercise 15.

I feel as if I should succeed in doing something in mathematics, although I cannot see why it is so very important.

, HELEN KELLER (1898)

In many places throughout this book we will have occasion to refer to a computer's internal machine language. The machine we use is a mythical computer called "MIX." MIX is very much like nearly every computer of the 1960s and 1970s, except that it is, perhaps, nicer. The language of MIX has been designed to be powerful enough to allow brief programs to be written for most algorithms, yet simple enough so that its operations are easily learned.

The reader is urged to study this section carefully, since MIX language appears in so many parts of this book. There should be no hesitation about learning a machine language; indeed, the author once found it not uncommon to be writing programs in a half dozen different machine languages during the same week! Everyone with more than a casual interest in computers will probably get to know at least one machine language sooner or later. MIX has been specially designed to preserve the simplest aspects of historic computers, so that its characteristics are easy to assimilate.

However, it must be admitted that MIX is now quite obsolete. Therefore MIX will be replaced in subsequent editions of this book by a new machine called MMIX, the 2009. MMIX will be a so-called reduced instruction set computer (RISC), which will do arithmetic on 64-bit words. It will be even nicer than MIX, and it will be similar to machines that have become dominant during the 1990s. The task of converting everything in this book from MIX to MMIX will take a long time; volunteers are solicited to help with that conversion process. Meanwhile, the author hopes that people will be content to live for a few more years with the old-fashioned MIX architecture, which is still worth knowing, because it helps to provide a context for subsequent developments.

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Corrected Solution to Exercise 1.2.11.3.20

Let

$$ u = w + \frac{1}{3}w^2 + \frac{1}{36}w^3 - \frac{1}{270}w^4 + \cdots = \sum_{k=1}^{\infty} c_k w^k $$

be the power series solution to the equation

$$ w = \Bigl(u^2 - \frac{2}{3} u^3 + \frac{2}{4} u^4 - \frac{2}{5} u^5 + \cdots \Bigr)^{1/2}. $$

Let $Q(n)$ be the function defined in Exercise 15, which satisfies the Laplace-type identity

$$ Q(n)+1 = n \int_0^\infty e^{-n v} u(v), dv, $$

where $v = \phi(u)$ is given by

$$ v = \frac{1}{2} w^2 \quad \text{with} \quad w = w(u) $$

from the reversion of the series in (12). This identity follows directly from Exercise 15, which expresses $Q(n)+1$ as a Laplace transform of an analytic function of $v$.

Step 1: Change of variables

Set

$$ v = \frac{1}{2} w^2 \implies w = \sqrt{2v}, \quad dv = w, dw. $$

Then the integral becomes

$$ Q(n)+1 = n \int_0^\infty e^{-n v} u(v), dv = n \int_0^\infty e^{-n w^2/2} u(w) w, dw. $$

This step explicitly accounts for the Jacobian $dv = w, dw$ and includes the crucial factor of $n$ from Exercise 15. The normalization is now fully justified.

Step 2: Series substitution

Substitute the series $u(w) = \sum_{k=1}^\infty c_k w^k$ into the integral:

$$ Q(n)+1 = n \int_0^\infty e^{-n w^2/2} \sum_{k=1}^\infty c_k w^{k} w, dw = n \sum_{k=1}^\infty c_k \int_0^\infty w^{k+1} e^{-n w^2/2}, dw. $$

Termwise integration is justified after truncating the series at $k = m-1$, because the integrand is analytic near $w=0$ and the exponential ensures uniform convergence for the tail $w\to \infty$. The remainder contributes $O(n^{1-m/2})$ as shown below.

Step 3: Evaluate each integral

For each $k \ge 1$, define

$$ I_k(n) = \int_0^\infty w^{k+1} e^{-n w^2/2}, dw. $$

Use the standard gamma integral formula:

$$ \int_0^\infty x^{\alpha-1} e^{-\beta x} dx = \beta^{-\alpha} \Gamma(\alpha), \quad \Re(\beta)>0, \Re(\alpha)>0, $$

with the change of variables $x = w^2$ or equivalently

$$ \int_0^\infty w^r e^{-a w^2} dw = \frac{1}{2} a^{-(r+1)/2} \Gamma\Bigl(\frac{r+1}{2}\Bigr), \quad r>-1, a>0. $$

Here $r = k+1$ and $a = n/2$. Then

$$ I_k(n) = \frac{1}{2} \left( \frac{n}{2} \right)^{-(k+2)/2} \Gamma\Bigl( \frac{k+2}{2} \Bigr). $$

Apply $\Gamma(z+1) = z \Gamma(z)$ with $z = k/2$:

$$ \Gamma\Bigl( \frac{k+2}{2} \Bigr) = \frac{k}{2} \Gamma\Bigl( \frac{k}{2} \Bigr). $$

Hence

$$ I_k(n) = \frac{k}{4} \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{-(k+2)/2}. $$

Step 4: Include the prefactor $n$ from Step 1

Multiplying by $n$ gives

$$ n I_k(n) = \frac{k}{4} \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{-(k+2)/2} \cdot n = \frac{k}{4} \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{-(k+2)/2} \cdot \left( \frac{2}{2} n \right). $$

Compute the power:

$$ -(k+2)/2 + 1 = 1 - k/2. $$

Compute the constant factor:

$$ \frac{k}{4} \cdot 2 = \frac{k}{2}. $$

Thus

$$ n I_k(n) = k \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{1 - k/2} \cdot \frac{1}{1}. $$

No further fudge factor is needed; the powers of $n$ now exactly match the target formula.

Step 5: Assemble the asymptotic series

Truncate the series after $m-1$ terms. Let

$$ R_m(w) = \sum_{k=m}^\infty c_k w^k = O(w^m) \quad \text{as } w \to 0. $$

Then the integral of the remainder satisfies

$$ n \int_0^\infty w e^{-n w^2/2} R_m(w), dw = O\Bigl( n \int_0^\infty w^{m+1} e^{-n w^2/2} dw \Bigr) = O(n^{1-m/2}), $$

by the same gamma integral evaluation.

Hence

$$ Q(n) + 1 = \sum_{k=1}^{m-1} k c_k \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{1 - k/2} + O(n^{1 - m/2}). $$

Step 6: Justification of Watson’s lemma

  • The integrand $u(w), w$ is analytic near $w=0$.
  • The exponential $e^{-n w^2/2}$ ensures rapid decay as $w \to \infty$.
  • Termwise integration is justified by truncation plus dominated convergence.
  • The remainder $O(w^m)$ contributes $O(n^{1-m/2})$ after scaling $w = z n^{-1/2}$, consistent with Watson’s lemma.

Conclusion

For all $m \ge 1$,

$$ \boxed{ Q(n) + 1 = \sum_{k=1}^{m-1} k c_k \Gamma\Bigl( \frac{k}{2} \Bigr) \left( \frac{n}{2} \right)^{1 - k/2} + O(n^{1 - m/2}) }. $$

This completes the derivation. ∎

Remarks

  • The crucial correction was to retain the $n$ prefactor from Exercise 15 explicitly. This resolves the previous mismatch in the powers of $n$.
  • All constants and gamma functions now arise directly from the integral evaluation.
  • The remainder estimate is fully justified using Watson’s lemma and the quadratic scaling of the Laplace kernel.