Substitution: the last integration rule
Lecture 7 read the product rule backwards and got integration by parts. What does the chain rule give, read backwards?
Lecture 7 claimed that every integration rule is a differentiation rule read backwards, and got integration by parts out of the product rule. The chain rule reverses too, and that exhausts the supply.
If $F'=f$, then differentiating $F(g(x))$ by the chain rule gives $F'(g(x))g'(x)$, so $$ \int f\bigl(g(x)\bigr)g'(x)\,dx = F\bigl(g(x)\bigr) + C\,. \tag{8.1} $$ In practice one writes $u=g(x)$ and $du = g'(x)\,dx$, turning the left-hand side into $\int f(u)\,du$.
The signal: an inside function and its derivative
This example is the app's (the lecture's first board example). Find $\displaystyle\int 2x\,e^{x^{2}}\,dx.$
Try it yourself first. Then open one step at a time.
The signal. The integrand contains an inside function, $x^{2},$ and its derivative $2x.$ That pairing is what makes (8.1) apply, with $g(x)=x^{2}$ and $f(u)=e^{u}.$
Substitute. Put $u=x^{2},$ so that $du=2x\,dx.$ The whole integrand is used up: $$ \begin{aligned} \int 2x\,e^{x^{2}}\,dx&=\int e^{u}\,du\\ &=e^{u}+C\,. \end{aligned} $$
Back to $x,$ and check. The answer is $e^{x^{2}}+C.$ By the chain rule its derivative is $e^{x^{2}}\cdot 2x,$ the integrand.
Guess first. The example after the question is the same integral. Decide which line is right before you read it.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
To evaluate $$\int_{0}^{2} x\,(1 + x^{2})^{4}\,dx$$ a student substitutes $u = 1 + x^{2}$.
Which line is right?
- Why is that answer tempting? Both errors at once: $dx$ replaced by $du$ as though they were the same thing, and the limits left in $x.$ The first is the commonest mistake with substitution (Watch out 8.3).
- Why is that answer tempting? Limits carried over correctly — and then $dx$ replaced by $du.$ They differ by $g'(x)=2x,$ so the factor $\tfrac12$ is missing and the answer is out by a factor of two.
- Why is that answer tempting? The differential handled correctly, and then the limits left in $x.$ A $u$-integrand with $x$-limits computes a number that answers no question at all.
$u = 1 + x^{2}$ gives $du = 2x\,dx,$ so $x\,dx = \tfrac12\,du.$
And the limits move with the variable: $x = 0$ gives $u = 1,$ and $x = 2$ gives $u = 5.$ So $$ \begin{aligned} \int_{1}^{5} \tfrac12 u^{4}\,du &= \Big[\tfrac{1}{10}u^{5}\Big]_{1}^{5}\\ &= \tfrac{3125 - 1}{10} = 312.4\,. \end{aligned} $$ Two things to get right. B gets the differential wrong, C the limits, and A both.
The differential and the limits
Evaluate $\displaystyle\int_{0}^{2} x\,(1+x^{2})^{4}\,dx.$
Try it yourself first. Then open one step at a time.
The substitution. The inside function is $g(x)=1+x^{2},$ and its derivative $2x$ appears in the integrand up to the factor $2$ — so substitution will work. Put $u=1+x^{2},$ so that $du = 2x\,dx$ and hence $x\,dx = \tfrac12du.$
The limits. The limits move with the variable: $x=0$ gives $u=1,$ and $x=2$ gives $u=5.$ Therefore $$ \begin{aligned} &\int_{0}^{2} x(1+x^{2})^{4}\,dx\\ &= \int_{1}^{5} \tfrac12 u^{4}\,du = \Bigl[\tfrac{1}{10}u^{5}\Bigr]_{1}^{5}\\ &= \frac{3125-1}{10} = 312.4\,. \end{aligned} $$
Two things had to be got right — the differential and the limits — and they are independent, so it is worth checking them separately.
That is the last integration rule you will be given. Parts came from the product rule and substitution from the chain rule, and there is no third differentiation rule that reverses into anything useful. Everything harder from here is a matter of recognising which of the two to try.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
With $u=x^{3}+1,$ what is $$\int_{0}^{1} x^{2}\,(x^{3}+1)^{2}\,dx\;?$$
- Why is that answer tempting? This is $\int_{1}^{2}u^{2}\,du{:}$ $x^{2}\,dx$ replaced by $du.$ But $du=3x^{2}\,dx,$ so $x^{2}\,dx=\tfrac13du,$ and the factor $\tfrac13$ is missing.
- Why is that answer tempting? This is $\int_{0}^{1}\tfrac13u^{2}\,du{:}$ the differential is right, but the limits are still the $x$-limits. At $x=0$ the new variable is $u=1,$ and at $x=1$ it is $u=2.$
- Why is that answer tempting? This is $\int_{0}^{1}u^{2}\,du{:}$ both errors at once, the factor $\tfrac13$ lost and the limits left in $x.$
$du=3x^{2}\,dx,$ so $x^{2}\,dx=\tfrac13du;$ and $x=0,1$ give $u=1,2.$ Hence $$ \begin{aligned} &\int_{0}^{1} x^{2}(x^{3}+1)^{2}\,dx\\ &=\int_{1}^{2}\tfrac13u^{2}\,du =\Bigl[\tfrac19u^{3}\Bigr]_{1}^{2}\\ &=\frac{8-1}{9}=\frac79\,. \end{aligned} $$
Infinite totals
Both $1/x$ and $1/x^{2}$ fall to zero as $x\to\infty.$ Is the area under each, from $1$ out to infinity, finite?
Both remaining pieces of single-variable calculus ask the same question: what does it mean to add up infinitely many things, and when is the answer finite? Neither requires a new operation. In both cases one adds up a finite amount and then takes a limit, exactly as in Lecture 5.
So an improper integral is an ordinary definite integral followed by a limit, and the sum of an infinite series is the limit of its partial sums.
Guess first. Answer the question below before you read the example after it, which computes both integrals.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Both $\dfrac1x$ and $\dfrac{1}{x^{2}}$ tend to $0$ as $x \to \infty$.
Which of $\displaystyle\int_{1}^{\infty}\frac{dx}{x}$ and $\displaystyle\int_{1}^{\infty}\frac{dx}{x^{2}}$ converge?
- Why is that answer tempting? Treats ‘the integrand tends to zero’ as sufficient. It is necessary and nothing more — the same necessary-versus-sufficient trap as the extreme value theorem in Lecture 6, wearing new clothes.
- Why is that answer tempting? An unbounded base does not force an unbounded area, provided the height falls away fast enough. The exponential density in statistics has an infinite base and an area of exactly 1.
- Why is that answer tempting? Backwards. Past x = 1 the function 1/x lies above 1/x², so if exactly one of them diverges it must be that one — no calculation needed to rule this out.
Both integrands vanish at infinity. One area is infinite and the other is exactly 1. So tending to zero decides nothing at all — it is the rate that decides, and the dividing line is sharp: $$ \int_{1}^{\infty} x^{-a}\,dx \text{ converges} \iff a > 1\,. $$
Two tails that both vanish
Both $1/x$ and $1/x^{2}$ fall to zero as $x\to\infty.$ Do $\int_{1}^{\infty}dx/x$ and $\int_{1}^{\infty}dx/x^{2}$ converge?
Try it yourself first. Then open one step at a time.
Integrate to $b$ and let $b$ grow: $$ \begin{aligned} \int_{1}^{b}\frac{dx}{x} &= \Bigl[\ln x\Bigr]_{1}^{b} = \ln b\\ &\longrightarrow\; \infty\,, \\ \int_{1}^{b}\frac{dx}{x^{2}} &= \Bigl[-\tfrac1x\Bigr]_{1}^{b}\\ &= 1-\frac1b \;\longrightarrow\; 1\,. \end{aligned} $$
The first diverges and the second converges, to exactly $1.$ Both integrands vanish at infinity, so vanishing decides nothing; what separates them is the rate.
Note also that $\ln b$ grows extremely slowly and still without limit: slow growth is not bounded growth.
$\int_{1}^{\infty}dx/x$ diverges; $\int_{1}^{\infty}dx/x^{2}$ converges, to $1.$
The same manoeuvre works for a sum, and for a geometric series there is a trick that evaluates it exactly.
For $|k|<1$, $$ a + ak + ak^{2} + ak^{3} + \cdots = \frac{a}{1-k}\,, $$ and for $|k|\ge1$ (with $a\ne0$) the series diverges.
Write $s_{n} = a+ak+\cdots+ak^{n-1},$ multiply by $k,$ and subtract: every term in the middle cancels, leaving $s_{n}-ks_{n} = a-ak^{n},$ so $$ s_{n} = a\,\frac{1-k^{n}}{1-k}\,. $$ If $|k|<1$ then $k^{n}\to0,$ and $s_{n}\to a/(1-k).$
The finite sum in the “why” is worth keeping on its own. For any $k\ne1$ the first $n$ terms add up to $a(1-k^{n})/(1-k),$ also when $|k|>1;$ only the limit needs $|k|<1.$ The next section and the worked example below use both.
A ratio that depends on x
This example is the app's: Fact 8.7 with a ratio that depends on $x,$ so that the answer is a set of $x.$ Nothing in it is new.
(a) For which $x$ does $$\sum_{n=1}^{\infty}\Bigl(\frac{x-3}{2}\Bigr)^{n}$$ converge, and what is its sum? For which $x$ is the sum $1$?
(b) For which $x$ does $$\sum_{n=1}^{\infty}\Bigl(\frac{\sqrt{x}}{3}\Bigr)^{n}$$ converge?
Try it yourself first. Then open one step at a time.
(a) The ratio. Written out, the series is $k+k^{2}+k^{3}+\cdots$ with $k=\dfrac{x-3}{2}{:}$ a geometric series whose first term is $a=k$ (the sum starts at $n=1$) and whose ratio is $k.$
(a) Where it converges. By Fact 8.7, exactly when $|k|<1{:}$ $$ \begin{aligned} &\Bigl|\frac{x-3}{2}\Bigr|<1\\ &\iff -2<x-3<2\\ &\iff 1<x<5\,. \end{aligned} $$ An absolute value below a number is a double inequality: both sides count. At $x=1$ and $x=5$ the ratio is $-1$ and $1,$ and the series diverges.
(a) The sum. For $1<x<5,$ $$ \begin{aligned} \frac{a}{1-k}&=\frac{k}{1-k}\\ &=\frac{(x-3)/2}{1-(x-3)/2}\\ &=\frac{x-3}{5-x}\,. \end{aligned} $$ It is $1$ when $k=1-k,$ that is $k=\tfrac12,$ so $x-3=1$ and $x=4.$ That lies in $(1,5),$ so the series really does converge there.
(b) A root: the domain first. $\sqrt{x}$ needs $x\ge0,$ and then $k=\sqrt{x}/3\ge0.$ So $|k|<1$ is just $\sqrt{x}<3,$ that is $x<9$ (both sides are non-negative, so squaring keeps the direction). The series converges exactly for $0\le x<9{:}$ the left end belongs, because at $x=0$ every term is $0;$ the right end does not, because there $k=1.$
(a) For $1<x<5,$ with sum $\dfrac{x-3}{5-x};$ the sum is $1$ at $x=4.$
(b) For $0\le x<9.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
For which $x$ does the series $$\sum_{n=0}^{\infty}(x-2)^{n}$$ converge?
- Why is that answer tempting? Only one side of $|k|<1$ was used. At $x=0$ the ratio is $k=-2,$ and the terms $1,-2,4,-8,\dots$ grow without bound.
- Why is that answer tempting? The endpoints are where $|k|=1,$ and there the series diverges: at $x=3$ it is $1+1+1+\cdots,$ and at $x=1$ the partial sums jump between $1$ and $0.$
- Why is that answer tempting? The expression $1/(1-k)$ makes sense for every $k\ne1,$ but it is the sum only when $|k|<1.$ The formula does not decide convergence; Fact 8.7 does.
The series is geometric with first term $1$ and ratio $k=x-2.$ By Fact 8.7 it converges exactly when $|x-2|<1,$ that is $-1<x-2<1,$ or $1<x<3.$ Then its sum is $\dfrac{1}{1-(x-2)}=\dfrac{1}{3-x}.$
$$ \begin{aligned} &\int_{1}^{\infty} x^{-a}\,dx \text{ converges} \iff a>1\,, \\ &\sum_{n=1}^{\infty} n^{-p} \text{ converges} \iff p>1\,. \end{aligned} $$ Geometric decay — shrinking by a constant factor — always converges. Power decay converges only above the exponent $1$.
So the two halves of this subsection have the same boundary, and $1/x$ lies exactly on it: both $\int_{1}^{\infty}dx/x$ and $\sum 1/n$ diverge, while $\int_{1}^{\infty}dx/x^{2}$ and $\sum1/n^{2}$ converge. That is not a coincidence; the two criteria can be derived from each other.
Present value: perpetuities, annuities and loans
A bond pays $100 at the end of every year, for ever. Infinitely many payments: is it worth infinitely much?
The first use of a geometric series is money that arrives later. Guess first. Answer the question below before you read the example after it, which prices the same bond.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A bond pays $100 at the end of every year, for ever. The interest rate is 5% a year.
What is it worth today?
- Why is that answer tempting? Infinitely many payments, yes — but they shrink by a constant factor, and a geometric series with $|k| < 1$ has a finite sum. This is precisely the intuition that Fact 8.7 corrects.
- Why is that answer tempting? The value if a payment also arrives today. Using $a = 100$ instead of $a = 100/1.05$ slips one extra payment into the sum — worth exactly one year's interest, and the standard error in present-value work.
- Why is that answer tempting? $2000/1.05{:}$ the whole stream discounted one year too many. The first payment arrives at the end of year 1, not year 2.
The payment at the end of year $n$ is worth $100/1.05^{n}$ today. So the value is a geometric series with $$a = \frac{100}{1.05}, \qquad k = \frac{1}{1.05}$$ and $|k| < 1,$ so it converges: $$ \begin{aligned} \frac{a}{1-k} &= \frac{100/1.05}{0.05/1.05}\\ &= \frac{100}{0.05} = 2000\,. \end{aligned} $$
A perpetuity is worth the payment divided by the interest rate. Which is exactly the amount you would have to deposit to live off the interest.
A perpetuity
A bond pays $\text{\textdollar}100$ at the end of every year, for ever. The interest rate is $5\%.$ What is it worth today?
Try it yourself first. Then open one step at a time.
Money arriving later is worth less now: at $5\%,$ a payment of $P$ arriving in $n$ years has present value $P/1.05^{n},$ since that is the sum which, invested today, grows to $P$ by then. The bond is therefore worth $$ \frac{100}{1.05} + \frac{100}{1.05^{2}} + \frac{100}{1.05^{3}} + \cdots, $$ a geometric series with first term $a = 100/1.05$ and ratio $k=1/1.05<1.$
By Fact 8.7 it converges, to $$ \begin{aligned} \frac{a}{1-k} &= \frac{100/1.05}{\,0.05/1.05\,}\\ &= \frac{100}{0.05} = 2000\,. \end{aligned} $$ So a perpetuity is worth the payment divided by the interest rate — which is exactly the deposit needed to live off the interest without touching the capital.
The commonest error here is taking $a=100$ rather than $100/1.05.$ That quietly inserts a payment arriving today, and gives $2100$ instead: one extra year's interest, from one misplaced index.
$\text{\textdollar}2000,$ the payment divided by the interest rate: $100/0.05.$
Nothing in Example 8.8 needed the payments to go on for ever. Stop them after $n$ years and the same series is a finite one, which Fact 8.7 already summed on the way past. That single observation is all there is to the mathematics of annuities and loans, and it is worth writing down because almost every financial quantity an economist meets is one of these two formulas.
Recap Words for money over time
| word | what it means here |
|---|---|
| interest rate $r$ | per period, as a decimal: $5\%$ is $r=0.05$ |
| present value | what a later payment is worth today |
| discount factor | $k=1/(1+r){:}$ one period's delay multiplies value today by $k$ |
| perpetuity | the same payment every period, for ever |
| annuity | the same payment every period, for $n$ periods |
| instalment | one of the equal payments that repay a loan |
| principal | the amount borrowed |
| balance | what an account holds at a given moment |
Guess first. A loan is repaid by a finite number of payments. Which equation fixes the payment? Answer before you read the Fact below.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
You borrow 2 000 000 kr for 30 years at 3%, and repay it in 30 equal annual instalments of $a$ kr, the first a year from now.
Which equation determines $a$?
- Why is that answer tempting? Interest ignored altogether — the payments are treated as if timing did not matter. Gives $67\,000$ kr against a true $102\,000.$
- Why is that answer tempting? The perpetuity of the question above, applied to a loan that ends. This is the payment that services the interest for ever without repaying a krona of principal — a real instrument, but not a mortgage. The tempting one, and worth naming.
- Why is that answer tempting? One payment compounded forward instead of thirty discounted back. Confuses future value with present value, and loses the sum entirely.
Thirty payments, each discounted back to today, is a finite geometric series — the one summed in the “why” of Fact 8.7: $$a\,\frac{1-(1+r)^{-n}}{r} = L$$
Solving, $a \approx 102\,000$ kr a year. Over thirty years that is about 3.06 million repaid on a two million loan.
Let $n \to \infty$ and this becomes the perpetuity $a/r$ of the question above. The finite sum is what makes the loan end.
Write $r$ for the interest rate per period and $k = 1/(1+r)$ for the discount factor. A payment of $a$ at the end of each of the next $n$ periods has total present value $$ \begin{aligned} \mathrm{PV} &= \frac{a}{1+r} + \frac{a}{(1+r)^{2}} + \cdots\\ &\qquad + \frac{a}{(1+r)^{n}}\\ &= a\,\frac{1-(1+r)^{-n}}{r}\,. \end{aligned} $$ A loan of $L$ repaid by $n$ such payments is the statement that these are the same sum of money seen from the two ends: $$ \begin{aligned} &a\,\frac{1-(1+r)^{-n}}{r} = L\\ &\qquad\Longleftrightarrow\qquad a = L\,\frac{r}{1-(1+r)^{-n}}\,. \end{aligned} $$
Both are Fact 8.7 with $a\mapsto a/(1+r)$ and $k=1/(1+r).$ The finite sum gives $$ \frac{a}{1+r}\cdot\frac{1-k^{n}}{1-k}, \qquad 1-k = \frac{r}{1+r}, $$ and the two factors of $1+r$ cancel. Letting $n\to\infty$ returns the perpetuity $a/r$ of Example 8.8, as it must.
A mortgage
Two million kronor is borrowed for $30$ years at $3\%,$ repaid in equal annual instalments. What is the instalment?
Try it yourself first. Then open one step at a time.
By Fact 8.9 with $L=2\,000\,000,$ $r=0.03$ and $n=30,$ $$ \begin{aligned} a &= 2\,000\,000\cdot\frac{0.03}{1-1.03^{-30}}\\ &\approx 102\,000 \text{ kr a year.} \end{aligned} $$
Over thirty years that is about $3.06$ million kronor repaid on a two million loan. The extra million is not a fee; it is what the lender gives up by not having the money, and it is exactly the gap the discount factor accounts for.
Two errors are worth naming. Dividing the loan by the term, $2\,000\,000/30 \approx 67\,000,$ ignores interest entirely and is far too small. Using the perpetuity formula $a = rL = 60\,000$ is smaller still: it is the payment that services the interest for ever without ever repaying the principal — a real financial instrument, but not a mortgage. The finite sum is what makes the loan end.
About $102\,000$ kr a year (more precisely $102\,038.52$ kr).
The first year a balance passes a target
This example is the app's: the finite sum of Fact 8.7 and the logarithm step of Method 2.11, joined. Nothing in it is new. At the end of every year you deposit $10\,000$ kr in an account that pays $8\%$ a year. Let $S_{n}$ be the balance just after the $n$-th deposit. Show that $$S_{n}=125\,000\,\bigl(1.08^{n}-1\bigr)\,,$$ and find the first year in which the balance exceeds $500\,000$ kr.
Try it yourself first. Then open one step at a time.
The sum the words describe. Just after the $n$-th deposit, the first deposit has earned interest for $n-1$ years, the second for $n-2,$ …, and the last for none: $$ \begin{aligned} S_{n}&=10\,000\bigl(1+1.08+\cdots\\ &\qquad+1.08^{n-1}\bigr)\,. \end{aligned} $$
The finite geometric sum. First term $a=10\,000,$ ratio $k=1.08,$ and $n$ terms. The finite sum in the “why” of Fact 8.7 holds for every $k\ne1,$ also for $k>1{:}$ $$ \begin{aligned} S_{n}&=10\,000\cdot\frac{1-1.08^{n}}{1-1.08}\\ &=\frac{10\,000}{0.08}\bigl(1.08^{n}-1\bigr)\\ &=125\,000\,\bigl(1.08^{n}-1\bigr)\,. \end{aligned} $$
The inequality. $S_{n}>500\,000$ means $1.08^{n}-1>4,$ that is $1.08^{n}>5.$
The logarithm (Method 2.11). $\ln$ is increasing, so taking $\ln$ of both sides keeps the direction: $$ \begin{aligned} &n\ln1.08>\ln5\\ &\iff n>\frac{\ln5}{\ln1.08}\\ &\qquad\approx\frac{1.6094}{0.0770}\approx20.9\,. \end{aligned} $$ We may divide by $\ln1.08$ without turning the inequality, because $\ln1.08>0.$
The smallest whole $n,$ and a check. The first whole number above $20.9$ is $21.$ Check against the year before: $S_{20}\approx457\,620<500\,000$ and $S_{21}\approx504\,229>500\,000.$
In year $21{:}$ just after the $21$st deposit the balance is $S_{21}\approx504\,229$ kr, while $S_{20}\approx457\,620$ kr.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A fund pays $3000$ kr at the end of every year for $10$ years, and nothing after that. The interest rate is $4\%.$ Which expression is its present value?
- Why is that answer tempting? This is the perpetuity, payments for ever. The fund stops after ten years, so the finite sum of Fact 8.9 is needed: $75\,000$ kr against about $24\,333.$
- Why is that answer tempting? Interest ignored: a payment in ten years is treated as worth as much as one now. That gives $30\,000,$ more than the true value.
- Why is that answer tempting? This is the balance after ten years if every payment were saved at $4\%{:}$ each payment compounded forward instead of discounted back. A value in ten years is not a value today.
By Fact 8.9 with $a=3000,$ $r=0.04$ and $n=10{:}$ $$ \begin{aligned} \mathrm{PV}&=3000\cdot\frac{1-1.04^{-10}}{0.04}\\ &\approx24\,333\text{ kr.} \end{aligned} $$ It lies below both $30\,000$ (no interest) and $75\,000$ (for ever), as it must.
A second variable
A firm doubles both its capital and its labour. Does its output double?
Single-variable calculus ends here. A firm uses capital and labour; a household's demand for a good depends on its price and on income. One input was never enough.
A way to read the table (the app's). Walk along a row: $y$ is fixed, and the values $3,8,15,24$ are just a function of $x$ alone, $x^{2}+2x.$ Walk up a column: $x$ is fixed, and $6,8,10$ is a function of $y$ alone, $4+2y,$ rising by exactly $2$ — which is $x$ — per step. A function of two variables is, in every direction we walk, a function of one.
The workhorse for the rest of the course is the Cobb–Douglas function $F(x,y)=Ax^{a}y^{b}$ with $A,a,b>0,$ met already in Lecture 2 one variable at a time. As a production function, $x$ and $y$ are inputs and $F$ is output.
Guess first. Before the example, answer the question below for a Cobb–Douglas function.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's output is $$F(K, L) = 10\,K^{1/2}L^{1/3}$$
It doubles both inputs. Output is multiplied by:
- Why is that answer tempting? Assumes constant returns to scale. That holds only when the exponents sum to 1 — the special case Cobb and Douglas happened to fit in 1927, not a law of nature.
- Why is that answer tempting? Doubles the output once for each input doubled. The factors of 2 do not multiply — they get raised to the exponents first.
- Why is that answer tempting? Adds 2^(1/2) and 2^(1/3). Multiplying the inputs multiplies the output; the exponents add, the factors do not.
Substitute and collect: $$ \begin{aligned} &F(2K, 2L)\\ &= 10\,(2K)^{1/2}(2L)^{1/3}\\ &= 2^{1/2}\,2^{1/3}\cdot 10\,K^{1/2}L^{1/3}\\ &= 2^{5/6}F(K,L) \end{aligned} $$ and $2^{5/6} \approx 1.78.$
The exponents sum to $\tfrac12 + \tfrac13 = \tfrac56 < 1,$ so doubling both inputs less than doubles output: decreasing returns to scale. Scaling by $t$ multiplies output by $t^{a+b},$ always.
Returns to scale
A firm's output is $F(K,L)=10K^{1/2}L^{1/3}.$ It doubles both inputs. By what factor does output rise?
Try it yourself first. Then open one step at a time.
Substitute and collect the constants: $$ \begin{aligned} &F(2K,2L)\\ &= 10\,(2K)^{1/2}(2L)^{1/3}\\ &= 2^{1/2}2^{1/3}\cdot 10K^{1/2}L^{1/3}\\ &= 2^{5/6}F(K,L)\,, \end{aligned} $$ and $2^{5/6}\approx1.78.$ Doubling both inputs less than doubles output.
The reason is visible in the exponents: $\tfrac12+\tfrac13=\tfrac56<1.$ In general, scaling both inputs by $t$ multiplies $Ax^{a}y^{b}$ by $t^{a+b},$ so $a+b>1$ gives increasing returns to scale, $a+b=1$ constant returns, and $a+b<1$ decreasing returns. It is the first thing anyone asks of a production function, and it takes one glance.
By $2^{5/6}\approx1.78{:}$ decreasing returns to scale, since $\tfrac12+\tfrac13<1.$
Partial derivatives
Hold $y$ still. What is left of $f(x,y)$ — and what is its derivative?
Nothing else changes. Every rule from Lectures 3 to 7 — product, quotient, chain, the lot — applies unaltered, because while $y$ stands still there is genuinely only one variable moving. The only new thing is the symbol $\partial$ and what it reminds you of. The subscript notation $f'_{1}$ and $f'_{2}$ says which argument rather than which letter, which matters as soon as functions are composed; expect both notations in the exercises.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Let $f(x,y)=x^{2}y^{3}+4y^{2}.$ What is $\dfrac{\partial f}{\partial x}$?
- Why is that answer tempting? The term $4y^{2}$ was differentiated with respect to $y.$ While $x$ moves, $4y^{2}$ is a constant, and its derivative is $0$ (Watch out 8.16).
- Why is that answer tempting? Both variables were allowed to move: $3x^{2}y^{2}$ is the derivative of $x^{2}y^{3}$ with respect to $y.$ A partial derivative moves one variable only.
- Why is that answer tempting? Both factors of $x^{2}y^{3}$ were differentiated and multiplied. With $y$ fixed, $y^{3}$ is a constant factor and stays as it is.
Hold $y$ fixed. Then $x^{2}y^{3}$ is the constant $y^{3}$ times $x^{2},$ with derivative $2xy^{3},$ and $4y^{2}$ is a constant, with derivative $0.$ So $$\frac{\partial f}{\partial x}=2xy^{3}\,.$$ For comparison, $\partial f/\partial y=3x^{2}y^{2}+8y.$
Two slices through one point
Left, the values of $f$ across the plane as shading, with a crosshair at the point the sliders choose; the blue line is “$y$ held”, the green line “$x$ held”. Right (below, on a phone), the two slices: what you see standing on each line.
- Drag the two sliders. Both slice plots are ordinary graphs with ordinary tangents; their slopes are the two partial derivatives at the point.
- Press the button with the formula until it shows the Cobb–Douglas function. Drag: both slices rise, and both flatten — two diminishing marginal products at once.
- Press it once more, for $x^{2}-y^{2},$ and put the crosshair at the origin. Both slopes are zero, yet one slice has a minimum there and the other a maximum. Lecture 11 is about exactly that.
Both partials, three times
Find both partial derivatives of (a) $z=e^{x/y};$ (b) $z=\ln(x^{2}+y^{2});$ (c) $F(K,L)=AK^{a}L^{b}.$
Try it yourself first. Then open one step at a time.
(a) By the chain rule, with $y$ constant, $\partial z/\partial x = \tfrac1y e^{x/y};$ and with $x$ constant, the inner function $x/y$ differentiates to $-x/y^{2},$ so $\partial z/\partial y = -\tfrac{x}{y^{2}}e^{x/y}.$
(b) $\displaystyle\frac{\partial z}{\partial x} = \frac{2x}{x^{2}+y^{2}},$ and by the symmetry of the formula in $x$ and $y,$ $\displaystyle\frac{\partial z}{\partial y} = \frac{2y}{x^{2}+y^{2}}.$
(c) The power rule, twice: $F'_{K} = aAK^{a-1}L^{b}$ and $F'_{L}=bAK^{a}L^{b-1}.$ Both simplify usefully: $$ F'_{K} = \frac{aF}{K}\,,\qquad F'_{L} = \frac{bF}{L}\,. $$
(a) $\tfrac1ye^{x/y}$ and $-\tfrac{x}{y^{2}}e^{x/y}.$ (b) $\dfrac{2x}{x^{2}+y^{2}}$ and $\dfrac{2y}{x^{2}+y^{2}}.$ (c) $F'_{K}=aF/K$ and $F'_{L}=bF/L.$
Part (c) has a payoff worth the whole example. Rearranging $F'_{K}=aF/K$ gives $$ \frac{K}{F}\cdot F'_{K} = a\,, $$ and the left-hand side is the elasticity of output with respect to capital: the percentage change in output per one per cent change in capital. Multiplying $F'_{K}$ by $K/F$ converts a rate measured in units into one measured in percentages, which is what makes it comparable across firms and currencies. For a Cobb–Douglas function the elasticities are the exponents themselves, readable off the formula with no calculation at all — which is why economists write production functions this way.
Guess first. A partial derivative has been measured. Can it tell you what happens when both inputs change? Answer before you read the Watch out below.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
For a firm's output $F(K,L)$ it is measured that $$F_K'(100, 50) = 5$$
The firm now raises capital from $100$ to $110$ and labour from $50$ to $55$. Output rises by:
- Why is that answer tempting? Uses 5 x 10 = 50 and ignores labour entirely. Two separate objections: the partial holds L fixed, and ten units is not a small change.
- Why is that answer tempting? Reads F′_K = 5 as ‘output rises by 5’. That would be right for one extra unit of capital with labour unchanged — and neither condition holds here.
- Why is that answer tempting? 5 x 100 — multiplies by the level of K rather than by the change in K. A derivative is a rate per unit of change, not per unit of stock.
A partial derivative answers exactly one question: how does $f$ change when this one variable moves and the others stand still?
Here labour moves too, and $F_K'$ is silent about that. Nothing in the given number tells you what labour contributes.
And even with labour held fixed, $5 \times 10 = 50$ is a linear approximation stretched over ten units, while $F_K'$ itself changes as $K$ rises — the same caution as Lecture 4.
The economist's phrase for the first objection is ceteris paribus. A partial derivative is what makes it precise.
Differentiating twice
A partial derivative is again a function of two variables. Differentiate it again — with respect to which variable?
Each partial derivative is again a function of two variables, so it can be differentiated again — with respect to either variable. There are therefore four second-order partial derivatives, and this course needs all four: they are what Lecture 11 uses to classify stationary points in two variables.
Read $\dfrac{\partial^{2}f}{\partial y\,\partial x}$ from the right: first $\partial/\partial x,$ then $\partial/\partial y.$ Each of the four second-order partial derivatives is an ordinary second derivative of a one-variable slice, or the slope of a slope taken in the other direction.
If the second-order partial derivatives of $f$ are continuous, then $$ \begin{aligned} &f''_{12} = f''_{21}\,, \qquad\text{that is}\\ &\frac{\partial^{2}f}{\partial y\,\partial x} = \frac{\partial^{2}f}{\partial x\,\partial y}\,. \end{aligned} $$ This holds for essentially every function met in economics.
That is a theorem and not a triviality. Read economically it says that the effect of capital on the marginal product of labour equals the effect of labour on the marginal product of capital — there is no obvious reason why that should be so, and it is. The result reappears throughout economics under other names, so it is worth filing as something proved rather than as a convention of notation.
Four second-order partials, and a check
Find all four second-order partial derivatives of $z=x^{3}y+x^{2}y^{2}.$
Try it yourself first. Then open one step at a time.
First the two first-order ones, each with the other variable held constant: $$ \begin{aligned} \frac{\partial z}{\partial x} &= 3x^{2}y+2xy^{2}\,, \\ \frac{\partial z}{\partial y} &= x^{3}+2x^{2}y\,. \end{aligned} $$
Differentiating each of those with respect to each variable in turn, $$ \begin{aligned} z''_{11} &= 6xy+2y^{2}\,,\\ z''_{12} &= 3x^{2}+4xy\,,\\ z''_{21} &= 3x^{2}+4xy\,,\\ z''_{22} &= 2x^{2}\,. \end{aligned} $$
The two mixed partials agree, as Fact 8.21 promises. Computing both and checking they match is a cheap and effective way of catching an algebra slip.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Let $z=x^{3}y^{2}+5x.$ What is $\dfrac{\partial^{2}z}{\partial x^{2}}$?
- Why is that answer tempting? This is $\bigl(\partial z/\partial x\bigr)^{2},$ the square of the first derivative. The second derivative differentiates $\partial z/\partial x$ once more (Watch out 8.20).
- Why is that answer tempting? This is the mixed partial $z''_{12}{:}$ the second differentiation was with respect to $y.$
- Why is that answer tempting? Adds $z''_{22}=2x^{3}.$ Each second-order partial is one function on its own; $\partial^{2}z/\partial x^{2}$ is the one with $x$ twice.
First $\partial z/\partial x=3x^{2}y^{2}+5,$ with $y$ held fixed. Differentiating that again with respect to $x$ gives $$\frac{\partial^{2}z}{\partial x^{2}}=6xy^{2}\,.$$ The other three are $z''_{12}=z''_{21}=6x^{2}y$ and $z''_{22}=2x^{3}.$
What you should take away
- Substitution is the chain rule backwards: with $u=g(x)$ and $du=g'(x)dx,$ $\int f(g(x))g'(x)dx = \int f(u)\,du.$ The signal is an inside function whose derivative also appears. Change the limits with the variable, and never swap $dx$ for $du.$
- An infinite total is a limit of finite ones — integrate to $b$ and let $b\to\infty,$ or add $n$ terms and let $n\to\infty.$ If the limit is finite it converges.
- Whether the tail tends to zero is not the question; how fast is. $\int_{1}^{\infty}x^{-a}dx$ and $\sum n^{-p}$ both converge exactly above the exponent $1,$ and $1/x$ is on the boundary in both. Geometric decay always converges; $\sum1/n$ does not, though its terms tend to zero.
- $a+ak+ak^{2}+\cdots = a/(1-k)$ for $|k|<1,$ proved by multiplying by $k$ and subtracting. A perpetuity paying $P$ for ever at rate $r$ is worth $P/r$ — the deposit that lives off its own interest.
- A partial derivative holds the other variable still, and then every rule from Lectures 3 to 7 applies unchanged. For $F=AK^{a}L^{b},$ $F'_{K}=aF/K,$ so $(K/F)F'_{K}=a{:}$ the elasticities of a Cobb–Douglas function are its exponents.
- A partial derivative is ceteris paribus made precise. It says nothing at all about what happens when the other variable moves too.
- There are four second-order partials, and for essentially every function met in practice the mixed ones agree: $f''_{12}=f''_{21}.$ Computing both is a cheap check on the algebra.
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — which line is right after substituting
To evaluate $\int_{0}^{2}x(1+x^{2})^{4}dx$ with $u=1+x^{2},$ which line is right?
(A) $\int_{0}^{2}u^{4}du;$
(B) $\int_{1}^{5}u^{4}du;$
(C) $\int_{0}^{2}\tfrac12u^{4}du;$
(D) $\int_{1}^{5}\tfrac12u^{4}du.$
Answers (B) and (C) each get exactly one of the two steps wrong, and (A) gets both. Which step does each of them miss?
One way to think about it
(D). $du=2x\,dx,$ so $x\,dx=\tfrac12du;$ and the limits move with the variable, $x=0,2$ to $u=1,5$ (Example 8.4). The value is $312.4.$
(B) gets the differential wrong: the factor $\tfrac12$ is missing, and it gives $624.8,$ twice the answer. (C) gets the limits wrong: a $u$-integrand with $x$-limits, giving $3.2,$ a number that answers no question. (A) gets both steps wrong (giving $6.4$).
Discussion question 2 — two tails, one ruled out on sight
Both $1/x$ and $1/x^{2}$ tend to $0.$ Which of $\int_{1}^{\infty}dx/x$ and $\int_{1}^{\infty}dx/x^{2}$ converge, and can you rule out one of the wrong answers without computing anything?
One way to think about it
Only $\int_{1}^{\infty}dx/x^{2}$ converges, to $1;$ the area under $1/x$ out to $b$ is $\ln b,$ which grows without limit (Example 8.6).
The answer “only $\int_{1}^{\infty}dx/x$” can be ruled out on sight. For $x\ge1$ we have $1/x\ge1/x^{2},$ so the area under $1/x$ is at least the area under $1/x^{2}.$ If the larger area were finite, the smaller one would be finite too; so they cannot be that way round.
Discussion question 3 — the value of a bond that pays for ever
A bond pays $\text{\textdollar}100$ at the end of every year for ever, at $5\%.$ What is it worth today?
(A) infinitely much;
(B) $\text{\textdollar}2000;$
(C) $\text{\textdollar}2100;$
(D) $\text{\textdollar}1905.$
One way to think about it
(B). The payments are worth $100/1.05,$ $100/1.05^{2},\dots$ today: a geometric series with $a=100/1.05$ and $k=1/1.05,$ whose sum is $100/0.05=2000$ (Example 8.8).
(A) forgets that the payments shrink by a constant factor, so the sum is finite. (C) uses $a=100,$ which counts a payment today that the bond never makes. (D) is $2000/1.05{:}$ every payment discounted one year too many.
Discussion question 4 — when doubling doubles
Output is $F(K,L)=10K^{1/2}L^{1/3}$ and both inputs double. Output is multiplied by about $1.78.$ What would the exponents have to satisfy for the answer to be exactly $2$?
One way to think about it
Their sum must be $1.$ Doubling both inputs multiplies $AK^{a}L^{b}$ by $2^{a+b},$ and $2^{a+b}=2$ exactly when $a+b=1{:}$ constant returns to scale (Example 8.14). For instance $10K^{1/2}L^{1/2}$ doubles its output when both inputs double. With $a+b=\tfrac56$ the factor is $2^{5/6}\approx1.78.$
Discussion question 5 — two objections to the obvious answer
It is measured that $F'_{K}(100,50)=5.$ The firm raises $K$ to $110$ and $L$ to $55.$ By how much does output rise? There are two separate objections to the obvious answer $50$ — find both.
One way to think about it
First: $F'_{K}$ holds labour fixed, and labour moves too. The number says nothing about what the extra labour adds (Watch out 8.18).
Second: even with labour fixed, $5\times10=50$ is a linear approximation stretched over ten units, while $F'_{K}$ changes as $K$ rises (Watch out 4.16).
A concrete case: $F(K,L)=100\sqrt{K}+20L$ has $F'_{K}=50/\sqrt{K},$ which is $5$ at $K=100.$ With $L$ fixed, raising $K$ to $110$ adds $100(\sqrt{110}-10)\approx48.8,$ not $50;$ and raising $L$ to $55$ adds $100$ more. The total is about $148.8.$ Answering it in general needs both partial derivatives and a linear approximation in two variables, which comes later in the course.
Discussion question 6 — necessary, not sufficient, once more
Why is $\sum 1/n$ the example that makes “necessary but not sufficient” stick, when the extreme value theorem and the inflection condition are the same point?
One way to think about it
Because here the counterexample is a single object you can check by hand. The terms $1/n$ visibly tend to zero, and the grouping in the “why” of Watch out 8.12 visibly pushes the partial sums past any number, half at a time. The condition holds and the conclusion fails, with nothing hidden.
The pattern has come round several times: $f'(c)=0$ does not make $c$ an extreme point, $f''(c)=0$ does not make it an inflection point (Watch out 7.8), and now terms tending to zero do not make a series converge. A necessary condition shortlists; it never confirms.
Discussion question 7 — why compute both mixed partials
For $z=x^{3}y+x^{2}y^{2},$ verify that $z''_{12}=z''_{21}$ by computing both. Why is this a useful check even when you believe the theorem?
One way to think about it
From $z'_{1}=3x^{2}y+2xy^{2},$ differentiating with respect to $y$ gives $z''_{12}=3x^{2}+4xy.$ From $z'_{2}=x^{3}+2x^{2}y,$ differentiating with respect to $x$ gives $z''_{21}=3x^{2}+4xy.$ They agree (Example 8.22).
The two routes go through different first derivatives. A slip in either $z'_{1}$ or $z'_{2}$ almost always makes the two results disagree, so agreement is cheap evidence that both first derivatives are right — and disagreement says at once that one of them is not.