Infinite totals, and a second variable

(a) Substitute and change the limits. Put $u=\ln x,$ so $du=\dfrac1x\,dx.$ At $x=1,$ $u=\ln1=0;$ at $x=e,$ $u=\ln e=1.$ $$ \begin{aligned} \int_{1}^{e}\frac{(\ln x)^{2}}{x}\,dx &=\int_{0}^{1}u^{2}\,du\\ &=\Bigl[\tfrac13u^{3}\Bigr]_{0}^{1}=\tfrac13\,. \end{aligned} $$

(b) The antiderivative. With $u=x^{2}+9,$ $du=2x\,dx,$ so $x\,dx=\tfrac12du$ and $$ \begin{aligned} &\int x\sqrt{x^{2}+9}\,dx\\ &=\tfrac12\int u^{1/2}\,du\\ &=\tfrac12\cdot\tfrac23u^{3/2}+C\\ &=\tfrac13\bigl(x^{2}+9\bigr)^{3/2}+C\,. \end{aligned} $$

(b) The check. By the chain rule, $$ \begin{aligned} &\tfrac13\cdot\tfrac32\bigl(x^{2}+9\bigr)^{1/2}\cdot2x\\ &=x\sqrt{x^{2}+9}\,. \end{aligned} $$

(c) With the limits in $u.$ $x=0$ gives $u=9$ and $x=4$ gives $u=25{:}$ $$ \begin{aligned} &\tfrac12\int_{9}^{25}u^{1/2}\,du\\ &=\tfrac13\Bigl[u^{3/2}\Bigr]_{9}^{25}\\ &=\tfrac13(125-27)=\tfrac{98}{3}\,. \end{aligned} $$

(c) With the answer of (b), back in $x$ and with the old limits: $$ \begin{aligned} &\Bigl[\tfrac13\bigl(x^{2}+9\bigr)^{3/2}\Bigr]_{0}^{4}\\ &=\tfrac13\bigl(25^{3/2}-9^{3/2}\bigr)=\tfrac{98}{3}\,. \end{aligned} $$ The same number, as it must be. What is never allowed is the mixture: $u$ in the integrand and $0,4$ as limits.

(a) Over the common denominator, $$ \begin{aligned} \frac1n-\frac{1}{n+1}&=\frac{(n+1)-n}{n(n+1)}\\ &=\frac{1}{n(n+1)}\,. \end{aligned} $$

(b) Almost everything cancels. $$ \begin{aligned} s_{n}&=\Bigl(1-\tfrac12\Bigr)+\Bigl(\tfrac12-\tfrac13\Bigr)\\ &\quad+\cdots+\Bigl(\tfrac1n-\tfrac{1}{n+1}\Bigr)\\ &=1-\frac{1}{n+1}\,. \end{aligned} $$ Each $-\tfrac1j$ is cancelled by the $+\tfrac1j$ of the next bracket; only the first and the last term survive. For instance $s_{1}=\tfrac12,$ $s_{2}=\tfrac12+\tfrac16=\tfrac23,$ and $1-\tfrac13=\tfrac23.$

(c) The limit. As $n\to\infty,$ $\dfrac{1}{n+1}\to0,$ so $s_{n}\to1.$ By Definition 8.5 the series converges, and its sum is $1.$

(d) Consistent. The terms satisfy $0<\dfrac{1}{n(n+1)}<\dfrac{1}{n^{2}},$ and $\sum1/n^{2}$ converges by Fact 8.11, since $p=2>1.$ The terms shrink like a power with exponent $2,$ above the borderline $1.$

(a) The ratio. The series is $k+k^{2}+\cdots$ with $k=x^{2}-3,$ a geometric series with first term $k.$ By Fact 8.7 it converges exactly when $|k|<1{:}$ $$ \begin{aligned} &-1<x^{2}-3<1\\ &\iff 2<x^{2}<4\\ &\iff \sqrt2<|x|<2\,. \end{aligned} $$

(a) Two intervals. $\sqrt2<|x|<2$ holds for $\sqrt2<x<2$ and for $-2<x<-\sqrt2.$ The set is $$(-2,-\sqrt2)\cup(\sqrt2,2)\,.$$ Between $-\sqrt2$ and $\sqrt2$ the ratio is $\le-1;$ outside $[-2,2]$ it is $>1.$

(b) The sum. $$ \begin{aligned} \frac{k}{1-k}&=\frac{x^{2}-3}{1-(x^{2}-3)}\\ &=\frac{x^{2}-3}{4-x^{2}}\,. \end{aligned} $$

(c) Sum $1.$ $\dfrac{k}{1-k}=1$ gives $k=\tfrac12,$ so $x^{2}=\tfrac72$ and $x=\pm\sqrt{7/2}\approx\pm1.87.$ Both lie in the set of (a), since $\sqrt2\approx1.41<1.87<2.$

(a) The instalment. By Fact 8.9 with $L=300\,000,$ $r=0.04,$ $n=10{:}$ $$ \begin{aligned} a&=300\,000\cdot\frac{0.04}{1-1.04^{-10}}\\ &=\frac{12\,000}{1-1.04^{-10}}\\ &\approx36\,987\text{ kr.} \end{aligned} $$

(b) The first year. The interest on $300\,000$ kr for one year is $0.04\cdot300\,000=12\,000$ kr. The rest of the first instalment, about $36\,987-12\,000=24\,987$ kr, reduces the debt, which is then about $275\,013$ kr. (Each later year the interest is smaller, because the debt is, so more of the same instalment goes to the debt.)

(c) The total. $10\cdot36\,987\approx369\,873$ kr. The difference, about $69\,873$ kr, is the interest: what the lender is paid for not having the money. It is not a fee on top; the ten instalments, each discounted back to today, add up to exactly $300\,000$ kr.

(a) The series. The dividend at the end of year $n$ is $50\cdot1.02^{n-1},$ worth $50\cdot1.02^{n-1}/1.06^{n}$ today: $$ \begin{aligned} V&=\frac{50}{1.06}+\frac{50\cdot1.02}{1.06^{2}}\\ &\quad+\frac{50\cdot1.02^{2}}{1.06^{3}}+\cdots\,. \end{aligned} $$ Each term is the one before times $\dfrac{1.02}{1.06},$ so it is geometric with $a=\dfrac{50}{1.06}$ and $k=\dfrac{1.02}{1.06}\approx0.962.$

(b) The value. $|k|<1,$ so by Fact 8.7 $$ \begin{aligned} V&=\frac{50/1.06}{1-1.02/1.06}\\ &=\frac{50/1.06}{0.04/1.06}\\ &=\frac{50}{0.04}=1250\text{ kr.} \end{aligned} $$ Without growth it is the perpetuity $50/0.06\approx833$ kr (Example 8.8). Two per cent of growth adds half as much again.

(c) When growth catches up. With growth $g$ the ratio is $k=(1+g)/1.06.$ If $g\ge6\%$ then $k\ge1,$ and by Fact 8.7 the series diverges: the discounted dividends do not shrink, and the “value” is infinite. No one pays an infinite price, so the model breaks: a dividend cannot grow at least as fast as the interest rate for ever. The formula $50/(r-g)$ that (b) suggests is meaningless there; it would even give a negative price for $g>r.$

(a) $16^{1/4}=2$ and $9^{1/2}=3,$ so $F(16,9)=4\cdot2\cdot3=24.$

(b) Scale both. $$ \begin{aligned} F(tK,tL)&=4\,t^{1/4}K^{1/4}\,t^{1/2}L^{1/2}\\ &=t^{3/4}F(K,L)\,. \end{aligned} $$ $\tfrac14+\tfrac12=\tfrac34<1{:}$ decreasing returns to scale. Doubling both inputs multiplies output by $2^{3/4}\approx1.68.$

(c) One input alone. Only the factor $K^{1/4}$ changes, so output is multiplied by $2^{1/4}\approx1.19{:}$ $F(32,9)=24\cdot2^{1/4}\approx28.5.$ Doubling one input does much less than doubling both.

(d) Constant returns need the exponents to add up to $1{:}$ $\tfrac14+b=1,$ so $b=\tfrac34.$

(a) With $D=20\,m^{1/2}p^{-1}{:}$ $$ \begin{aligned} \frac{\partial D}{\partial p}&=-\frac{20\sqrt{m}}{p^{2}}\,,\\ \frac{\partial D}{\partial m}&=\frac{10}{p\sqrt{m}}\,. \end{aligned} $$

(b) At $(5,400).$ $\sqrt{400}=20,$ so $D=\dfrac{20\cdot20}{5}=80,$ $D'_{p}=-\dfrac{400}{25}=-16$ and $D'_{m}=\dfrac{10}{5\cdot20}=0.1.$

In words: with income held at $400,$ raising the price by one unit lowers demand by about $16$ units; with the price held at $5,$ one more unit of income raises demand by about $0.1$ units. Each holds the other variable fixed: ceteris paribus.

(c) The elasticities. $$ \begin{aligned} \frac{p}{D}\,D'_{p}&=\frac{5}{80}\cdot(-16)=-1\,,\\ \frac{m}{D}\,D'_{m}&=\frac{400}{80}\cdot0.1=0.5\,. \end{aligned} $$ $D=20\,m^{1/2}p^{-1}$ has the form of a Cobb–Douglas function, so the elasticities are the exponents, $-1$ and $\tfrac12,$ as after Example 8.17: a price $1\%$ higher lowers demand by about $1\%,$ an income $1\%$ higher raises it by about $\tfrac12\%.$

(d) Two objections. $D'_{p}=-16$ holds income fixed, but income rises too, which pushes demand up; and a price change of a whole unit (from $5$ to $6,$ that is $20\%$) is not small, while $D'_{p}$ changes with $p$ (Watch out 8.18). In fact $D(6,500)=\dfrac{20\sqrt{500}}{6}\approx74.5,$ a fall of about $5.5,$ not $16.$

(a) First order. With $y$ fixed, by the product rule and the chain rule, $$ \begin{aligned} f'_{1}&=e^{xy}+x\cdot y\,e^{xy}\\ &=(1+xy)\,e^{xy}\,. \end{aligned} $$ With $x$ fixed, $f'_{2}=x\cdot x\,e^{xy}=x^{2}e^{xy}.$

(b) Differentiate $f'_{1}$ again. $$ \begin{aligned} f''_{11}&=y\,e^{xy}+(1+xy)\,y\,e^{xy}\\ &=(2y+xy^{2})\,e^{xy}\,,\\ f''_{12}&=x\,e^{xy}+(1+xy)\,x\,e^{xy}\\ &=(2x+x^{2}y)\,e^{xy}\,. \end{aligned} $$

(b) Differentiate $f'_{2}$ again. $$ \begin{aligned} f''_{21}&=2x\,e^{xy}+x^{2}y\,e^{xy}\\ &=(2x+x^{2}y)\,e^{xy}\,,\\ f''_{22}&=x^{3}e^{xy}\,. \end{aligned} $$ $f''_{12}=f''_{21},$ as Fact 8.21 says. The two came by different routes, so their agreement checks both $f'_{1}$ and $f'_{2}.$

(c) At $(1,0).$ $e^{0}=1,$ so $f''_{11}=0,$ $f''_{12}=f''_{21}=2$ and $f''_{22}=1.$