급수 심화 전체 문제

In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number ${H}_{k}=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{k}\right).$ Recall that ${T}_{k}={H}_{k}-\ln k$ is decreasing. Compute $T=\lim_{k\to \infty }{T}_{k}$ to four decimal places. (Hint: $\frac{1}{k+1}<{\int }_{k}^{k+1}\frac{1}{x}dx$.)

Suppose a scooter can travel $100$ km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel $100{H}_{N}$ km, where ${H}_{N}=1+1/2+\cdots +1/N.$

[T] Suppose a computer can sum one million terms per second of the divergent series $\sum_{n=1}^{N}\frac{1}{n}.$ Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100.$

[T] A fast computer can sum one million terms per second of the divergent series $\sum_{n=2}^{N}\frac{1}{n\ln n}.$ Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100.$

[T] Estimates of $\sum_{n=1}^{\infty }\frac{1}{{n}^{2}}$ can be accelerated by writing its partial sums as $\sum_{n=1}^{N}\frac{1}{{n}^{2}}=\sum_{n=1}^{N}\frac{1}{n(n+1)}+\sum_{n=1}^{N}\frac{1}{{n}^{2}(n+1)}$ and recalling that $\sum_{n=1}^{N}\frac{1}{n(n+1)}=1-\frac{1}{N+1}$ converges to one as $N\to \infty .$ Compare the estimate of ${\pi }^{2}/6$ using the sums $\sum_{n=1}^{1000}\frac{1}{{n}^{2}}$ with the estimate using $1+\sum_{n=1}^{1000}\frac{1}{{n}^{2}(n+1)}.$

A legend from India tells that a mathematician invented chess for a king. The king enjoyed the game so much he allowed the mathematician to demand any payment. The mathematician asked for one grain of rice for the first square on the chessboard, two grains of rice for the second square on the chessboard, four grains of rice for the third square on the chessboard, and so on. Find an exact expression for the total payment (in grains of rice) requested by the mathematician. Assuming there are $30,000$ grains of rice in $1$ pound, and $2000$ pounds in $1$ ton, how many tons of rice did the mathematician attempt to receive?

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula ${x}_{n+1}=b{x}_{n},$ where ${x}_{n}$ is the population of houseflies at generation $n,$ and $b$ is the average number of offspring per housefly who survive to the next generation. Assume a starting population ${x}_{0}.$

Find $\lim_{n\to \infty }{x}_{n}$ if $b>1,$ $b<1,$ and $b=1.$

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula ${x}_{n+1}=b{x}_{n},$ where ${x}_{n}$ is the population of houseflies at generation $n,$ and $b$ is the average number of offspring per housefly who survive to the next generation. Assume a starting population ${x}_{0}.$

If $b=\frac{3}{4}$ and ${x}_{0}=100,$ find ${S}_{10}$ and $\lim_{n\to \infty }{S}_{n}$

Use the approximation $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the period of a pendulum having length $10$ meters and maximum angle ${\theta }_{\max }=\frac{\pi }{6}$ where $k=\sin \left(\frac{{\theta }_{\max }}{2}\right).$ Compare this with the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}.$

Suppose that a pendulum is to have a period of $2$ seconds and a maximum angle of ${\theta }_{\max }=\frac{\pi }{6}.$ Use $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}?$

[T] An equivalent formula for the period of a pendulum with amplitude ${\theta }_{\max }$ is $T({\theta }_{\max })=2\sqrt{2}\sqrt{\frac{L}{g}}{\int }_{0}^{{\theta }_{\max }}\frac{d\theta }{\sqrt{\cos \theta }-\cos ({\theta }_{\max })}$ where $L$ is the pendulum length and $g$ is the gravitational acceleration constant. When ${\theta }_{\max }=\frac{\pi }{3}$ we get $\frac{1}{\sqrt{\cos t-1/2}}\approx \sqrt{2}\left(1+\frac{{t}^{2}}{2}+\frac{{t}^{4}}{3}+\frac{181{t}^{6}}{720}\right).$ Integrate this approximation to estimate $T\left(\frac{\pi }{3}\right)$ in terms of $L$ and $g.$ Assuming $g=9.806$ meters per second squared, find an approximate length $L$ such that $T\left(\frac{\pi }{3}\right)=2$ seconds.