5.6: Summary and further problems
- Page ID
- 59099
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Upon meeting a complicated problem, divide it into a big part—the most important effect—and a correction. Analyze the big part first, and worry about the correction afterward. This successive-approximation approach, a species of divide-and-conquer reasoning, gives results automatically in a low-entropy form. Low-entropy expressions admit few plausible alternatives; they are therefore memorable and comprehensible. In short, approximate results can be more useful than exact results.
Problem 5.42 Large logarithm
What is the big part in \(ln(1 + e^{2})\)? Give a short calculation to estimate \(ln(1 + e^{2}\)) to within 2%.
Problem 5.43 Bacterial mutations
In an experiment described in a Caltech biology seminar in the 1990s, researchers repeatedly irradiated a population of bacteria in order to generate mutations. In each round of radiation, 5% of the bacteria got mutated. After 140 rounds, roughly what fraction of bacteria were left unmutated? (The seminar speaker gave the audience 3 s to make a guess, hardly enough time to use or even find a calculator.)
Problem 5.44 Quadratic equations revisited
The following quadratic equation, inspired by [29], describes a very strongly damped oscillating system.
\[s^{2} + 10^{9}s + 1 = 0. \label{5.54} \]
a. Use the quadratic formula and a standard calculator to find both roots of the quadratic. What goes wrong and why?
b. Estimate the roots by taking out the big part. (Hint: Approximate and solve the equation in appropriate extreme cases.) Then improve the estimates using successive approximation.
c. What are the advantages and disadvantages of the quadratic-formula analysis versus successive approximation?
Problem 5.45 Normal approximation to the binomial distribution
The binomial expansion
\[(\frac{1}{2} + \frac{1}{2})^{2n} \label{5.55} \]
contains terms of the form
\[f(k) \equiv\left(\begin{array}{c}
2 n \\
n-k
\end{array}\right) 2^{-2 n} \label{5.56} \]
where \(k = −n . . . n\). Each term f(k) is the probability of tossing \(n − k\) heads (and \(n + k\) tails) in \(2n\) coin flips; \(f(k)\) is the so-called binomial distribution with parameters \(p = q = 1/2\). Approximate this distribution by answering the following questions:
a. Is \(f(k)\) an even or an odd function of \(k\)? For what \(k\) does \(f(k)\) have its maximum?
b. Approximate \(f(k)\) when \(k ≪ n\) and sketch \(f(k)\). Therefore, derive and explain the normal approximation to the binomial distribution.
c. Use the normal approximation to show that the variance of this binomial distribution is \(n/2\).
Problem 5.46 Beta function
The following integral appears often in Bayesian inference:
\[f(a,b) = \int_{0}^{1} x^{a}(1-x)^{b}dx, \label{5.57} \]
where \(f(a − 1, b − 1)\) is the Euler beta function. Use street-fighting methods to conjecture functional forms for \(f(a,0), f(a,a)\), and, finally, \(f(a,b)\). Check your conjectures with a high-quality table of integrals or a computer-algebra system such as Maxima.