7.5 Practice Problems
Problem 7.5.1. Show that \(\displaystyle \int ^{\infty }_{x}\frac {e^{-t}}{t}\,dt\) has the asymptotic expansion given in Example 7.1.5, and prove the stated bound on the remainder \(R_n\). Where to start: integrate by parts repeatedly, keeping track of the remainder integral rather than discarding it.
Problem 7.5.2. Show that the series in that example diverges for every fixed \(x\), and determine how many terms give the best accuracy at \(x=10\). Where to start: the terms are \(n!/x^{n+1}\); find where the ratio of consecutive terms passes through \(1\).
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Solution. The ratio of consecutive terms is \((n+1)/x\), which exceeds \(1\) once \(n>x-1\), so the terms decrease only while \(n<x-1\) and grow thereafter without bound: the series diverges for every \(x\).
At \(x=10\) the smallest term is at \(n\approx 9\), so about nine or ten terms give the best attainable accuracy, and the error there is roughly the size of that term, \(9!/10^{10}\approx 3.6\times 10^{-5}\). Adding further terms makes the approximation worse — behaviour with no analogue for convergent series, and the reason an asymptotic expansion must always be truncated deliberately.
Problem 7.5.3. Use Watson’s lemma to find the first two terms of the large-\(x\) behaviour of \[\int ^{\infty }_{0}\frac {e^{-xt}}{1+t}\,dt .\] Where to start: expand \((1+t)^{-1}\) as a geometric series about \(t=0\) and apply Theorem 7.2.1 term by term.
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Solution. Near the origin \((1+t)^{-1} = 1 - t + t^{2} - \cdots \), so \(a_n = (-1)^{n}\) and \(\lambda _n = n\). Theorem 7.2.1 gives \[\int ^{\infty }_{0}\frac {e^{-xt}}{1+t}\,dt \ \sim \ \sum ^{\infty }_{n=0}\frac {(-1)^{n}\,n!}{x^{n+1}} = \frac {1}{x} - \frac {1}{x^{2}} + \frac {2}{x^{3}} - \cdots ,\] the first two terms being \(1/x - 1/x^{2}\). Note this is the expansion of Example 7.1.5 again, which it should be: the substitution \(t\mapsto t-x\) turns one integral into the other.
Problem 7.5.4. Apply Laplace’s method to \(\displaystyle \int ^{\pi /2}_{0}e^{x\cos t}\,dt\) and obtain the leading behaviour as \(x\to \infty \). Note that the maximum lies at an endpoint and explain what changes. Where to start: at an endpoint maximum only half the Gaussian is integrated, so the result carries a factor \(\tfrac 12\) — and here \(h'(0)=0\) as well, so the parabolic approximation still applies.
Problem 7.5.5. Derive Stirling’s formula in the form \(\ln n! = n\ln n - n + \tfrac 12\ln (2\pi n) + O(1/n)\), and use it to estimate \(\ln (100!)\). Compare with the exact value.
Problem 7.5.6. Use Stirling’s formula to show that \[\binom {2n}{n}\ \sim \ \frac {4^{n}}{\sqrt {\pi n}} .\] Where to start: write the binomial coefficient as \((2n)!/(n!)^{2}\) and apply the formula to each factorial.
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Solution. By Theorem 7.4.1, \[\binom {2n}{n} = \frac {(2n)!}{(n!)^{2}} \sim \frac {\sqrt {4\pi n}\,(2n/e)^{2n}}{\left [\sqrt {2\pi n}\,(n/e)^{n}\right ]^{2}} = \frac {\sqrt {4\pi n}\;4^{n}n^{2n}e^{-2n}}{2\pi n\;n^{2n}e^{-2n}} = \frac {4^{n}}{\sqrt {\pi n}} .\] This is the estimate behind the central term of the binomial distribution, and dividing by \(4^{n}\) gives the probability of exactly \(n\) heads in \(2n\) tosses as \(\left (\pi n\right )^{-1/2}\) — which decays, though slowly, and is the reason a fair coin tossed many times almost never gives exactly half heads.
Problem 7.5.7. Explain why the convergent Taylor series for \(\operatorname {erf}(x)\) is unsuitable for computing \(\operatorname {erf}(5)\), and why the divergent asymptotic series for \(\operatorname {erfc}(x)\) is suitable. Where to start: estimate the largest term of the Taylor series at \(x=5\) and compare it with the value of the function.
Problem 7.5.8. State precisely why an asymptotic expansion does not determine the function it represents uniquely, by showing that \(e^{-x}\) has the asymptotic expansion \(\sum 0\cdot x^{-n}\) as \(x\to \infty \). What does this imply about recovering \(f\) from its expansion? Where to start: \(e^{-x}\) is smaller than every power of \(1/x\) for large \(x\).
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