What a function is
A circle is a perfectly good curve. Why is it not the graph of a function?
Economics and statistics are, to a remarkable extent, applied mathematics. Almost every claim an economist makes — that a tax will reduce output, that a firm should produce twenty units rather than thirty — is at bottom a statement about a function. Demand and supply, costs and production are functions; profit maximisation and consumer choice are optimisation problems; every marginal quantity is a derivative; least squares, expectation and maximum likelihood are minimisation and maximisation problems; and an economy with many goods is a system of linear equations, which is to say a matrix.
The course follows that list, in four parts, each building on the one before:
- Functions of one variable and the derivative. Which functions do we use, what do their graphs look like, and how fast do they change?
- Integration. Adding up infinitely many infinitesimal pieces.
- Functions of several variables and optimisation. Real economic problems almost never involve just one variable, and almost always involve a constraint.
- Matrices and determinants. The bookkeeping device that makes many variables manageable.
How to read a lecture in this app
Every lecture is built from the same kinds of boxes.
- A Definition says what a word means, and everything else depends on them.
- A Fact is a result worth remembering. Under most of them is a button Why is this true? with the reason.
- A Method is something to carry out with pen and paper.
- The Examples are worked examples, which are the heart of the notes. The solution opens one step at a time.
- A Watch out is a mistake made every single year.
- A Common error is a mistake that the answers of an earlier term show was really made, together with the check that catches it.
- Check yourself is a multiple-choice question, one of those voted on in class. Pick an answer before you read on.
Mathematics is not a spectator sport: keep a pen in your hand while you read, and recompute every example before you look at the solution.
The two emphasised words carry all the weight: “each” says that $f(x)$ must be defined for every $x$ in the domain, and “exactly one” says that $f$ may not be ambiguous. The second has a picture — a set of points in the plane is the graph of a function if and only if every vertical line meets it at most once, the vertical line test, which is why a circle is not one.
Function or not? Three sets of points
Which of these sets of points in the plane is the graph of a function of $x$?
(a) The line $y = 2x + 1.$
(b) The circle $x^2 + y^2 = 1.$
(c) The upper half of that circle: the points with $x^2 + y^2 = 1$ and $y \ge 0.$
Try it yourself first. Then open one step at a time.
(a) Yes. For each real number $x$ the formula gives exactly one value, $2x + 1.$ Every vertical line meets the line once. The domain is all of ${\mathbb R}.$
(b) No. The vertical line $x = 0$ meets the circle twice, at $(0, 1)$ and at $(0, -1).$ A function would have to give $x = 0$ exactly one value, and here there are two candidates.
(c) Yes. For each $x$ with $-1 \le x \le 1$ there is exactly one point of the circle with $y \ge 0,$ namely $y = \sqrt{1 - x^2}.$ This is a function with domain $D = [-1, 1].$ For $x = 2$ there is no point at all, and that is allowed: $2$ is simply not in the domain.
(a) and (c) are graphs of functions; (b) is not, because the line $x = 0$ meets the circle twice.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Which of these rules does not define a function $f\colon D\to{\mathbb R}$ on the stated domain $D$?
- Why is that answer tempting? The formula has no value at $x = 0,$ but $0$ is not in this domain. “Each” only asks for a value at every $x$ in $D.$
- Why is that answer tempting? Here $f(2) = f(-2) = 4,$ so two different $x$ share one value. That is allowed. “Exactly one” says that one $x$ may not have two values, and it says nothing about two $x$ with the same value.
- Why is that answer tempting? A constant is a function: every $x$ gets exactly one value, and it happens to be the same value each time. Its graph is a horizontal line.
Definition 1.1 asks for two things: a value for each $x$ in the domain, and exactly one value. With $D = {\mathbb R}$ the rule $f(x) = 1/x$ fails the first, since it gives no value at $x = 0.$ Removing $0$ from the domain repairs it, so a function is a rule together with its domain.
Linear functions: the slope is a rate
You are told only that every extra unit costs the same to produce. What does the cost function look like?
Suppose you are told only this about a cost function: every extra unit costs the same to produce. That single sentence pins the function down almost completely, and the resulting shape is the straight line.
If $f(x)=ax+b$, then increasing $x$ by one unit changes $f(x)$ by exactly $a$, no matter where you start: $$ \begin{aligned} &f(x+1)-f(x) \\ &\quad= \bigl(a(x+1)+b\bigr) - (ax+b) \\ &\quad= a\,. \end{aligned} $$ So $a>0$ gives a graph rising to the right, the steeper the larger $a$ is; $a<0$ one falling to the right, with steepness $|a|$; and $a=0$ the horizontal line $y=b$.
That the answer does not depend on $x$ is the defining feature of a linear function. The slope also carries units — dollars per ton, millions of trips per unit of price — and asking what they are is the quickest way to catch a mistake.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
The cost function for US Steel over 1917–1938 was estimated as $$C = 55.73\,x + 182\,100\,000$$ where $C$ is total cost in dollars per year and $x$ is tons of steel produced per year.
What does the number $55.73$ mean?
- Why is that answer tempting? This is the average cost, mistaken for the cost of one more ton. The average cost per ton here is $55.73 + 182\,100\,000/x,$ which is nowhere near $55.73.$ Keep the two apart: the same confusion comes back all through the course.
- Why is that answer tempting? This reads a rate with units (dollars per ton) as a percentage, which has no units. Ask what the units of the slope are.
- Why is that answer tempting? This swaps the slope $a$ and the intercept $b.$ The cost at $x = 0$ is the intercept, $182\,100\,000$ dollars.
$C(x+1) - C(x) = 55.73$ — that is the whole content of the number, and it is what Fact 1.3 says.
One more ton, not the average ton. The average cost at $x$ tons is $$\frac{C(x)}{x} = 55.73 + \frac{182\,100\,000}{x},$$ which is enormous at low output. It falls towards $55.73$ as $x$ grows but never reaches it.
And $182\,100\,000$ is the intercept: what the firm pays before making any steel at all.
Units are the fastest check available. The slope is dollars per ton. That alone rules out C.
Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be distinct points on a non-vertical straight line $\ell$. Then the slope of $\ell$ is $$ a = \frac{y_2-y_1}{x_2-x_1}\,, \tag{1.1} $$ independently of which two points of $\ell$ are chosen; and the line through $(x_1,y_1)$ with slope $a$ has equation $$ y - y_1 = a\,(x-x_1)\,. \tag{1.2} $$
If $y_1=ax_1+b$ and $y_2=ax_2+b,$ then $b = y_1-ax_1 = y_2-ax_2.$ Hence $$ y_2 - y_1 = a(x_2 - x_1)\,. $$ Dividing both sides by $(x_2-x_1)$ gives the desired result.
Here $x_1$ and $y_1$ are fixed numbers, the coordinates of the given point, whereas $x$ and $y$ are variables; confusing the two is the commonest error in this subsection. A line through two given points needs no separate formula: compute the slope with (1.1), use (1.2), and check by substituting the given points back in.
A demand function from two observations
Demand $D$ for a good is a linear function of its price $P.$ At a price of $\text{\textdollar}10$ demand is $200$ units; at $\text{\textdollar}15$ it is $150$ units. Find the demand function.
Try it yourself first. Then open one step at a time.
The slope. We are given two points $(P,D)=(10,200)$ and $(15,150).$ The slope is $$ \frac{150-200}{15-10} = \frac{-50}{5} = -10\,. $$
The line. So by (1.2), $D-200 = -10(P-10),$ that is, $$ D = 300 - 10P\,. $$
Check: at $P=15$ this gives $300-150=150,$ which is what we want.
What the numbers say. The slope $-10$ says that each extra dollar of price costs the seller $10$ units of sales; it is negative, as demand curves usually are. The intercept $300$ is demand at price zero, and $D=0$ at $P=30$: nobody buys at that price.
$D = 300 - 10P.$ Each extra dollar of price lowers demand by $10$ units.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A line has slope $3$ and passes through the point $(2, 5).$ Which equation describes it?
- Why is that answer tempting? The coordinates of the point have changed places. In $y - y_1 = a(x - x_1)$ the number $y_1$ stands beside $y$ and $x_1$ beside $x.$ Test: at $x = 2$ this equation gives $y = -7,$ not $5.$
- Why is that answer tempting? This takes the $y$-coordinate of the point for the intercept. The intercept is the value at $x = 0,$ and the point given has $x = 2.$ Test: at $x = 2$ this gives $y = 11.$
- Why is that answer tempting? The bracket is missing: the slope must multiply the whole of $x - x_1.$ Test: at $x = 2$ this gives $y = 9.$
Formula (1.2) with $a = 3$ and $(x_1, y_1) = (2, 5)$ gives $y - 5 = 3(x - 2),$ that is, $y = 3x - 1.$
The check costs five seconds: put $x = 2$ into your answer and see whether $y = 5$ comes out. Each of the three wrong equations fails it. In the practice problems on this formula about four answers in ten were wrong, and this substitution is the check that catches such a slip.
Two lines at once
Two equations in two unknowns describe two lines, and a pair $(x,y)$ solves the system exactly when the point lies on both. So the algebra of solving and the geometry of intersecting are the same thing, and there are only three possibilities.
A system of two linear equations in two unknowns has either
- exactly one solution — the lines cross at a single point;
- no solution — the lines are parallel and distinct;
- infinitely many solutions — the two equations describe the same line.
In particular, it is impossible for such a system to have exactly two solutions.
Before drawing anything, put both equations in the form $y=ax+b$ and compare the slopes: unequal slopes mean case (1), equal slopes mean case (2) or (3), and the intercepts decide which. So $3x+4y=2$ and $6x+8y=24$ are $y=-\tfrac34x+\tfrac12$ and $y=-\tfrac34x+3$ — parallel and distinct, no solution.
Replacing $=$ by $\le$ turns a line into a half-plane: the points with $2x+y\le 4$ are those on or below $y=-2x+4.$ The most important instance in economics is the budget set.
The budget set
A consumer has $\text{\textdollar}m$ to spend on two goods, whose prices are $\text{\textdollar}p$ and $\text{\textdollar}q$ per unit, with $p,q>0.$ Buying $x$ units of the first good and $y$ of the second requires $x\ge 0,$ $y\ge 0$ and $px+qy\le m,$ so the consumer is restricted to the budget set $$ \begin{aligned} B = \{(x,y) :{}& px+qy\le m,\\ & x\ge 0,\ y\ge 0\}\,. \end{aligned} $$ Find the slope and the intercepts of the budget line $px+qy=m.$
Try it yourself first. Then open one step at a time.
The shape. The conditions $x\ge0$ and $y\ge0$ confine us to the first quadrant, and $px+qy\le m$ then cuts out a triangle.
Slope and $y$-intercept. Solving $px+qy=m$ for $y$ gives $$ y = -\frac{p}{q}\,x + \frac{m}{q}\,, $$ so the slope is $-p/q$ and the $y$-intercept is $m/q.$
The other intercept. Setting $y=0$ gives $x=m/p.$ Hence the budget line joins $(m/p,0)$ to $(0,m/q),$ and $B$ is the triangle beneath it.
The slope is $-p/q.$ The line meets the $y$-axis at $m/q$ and the $x$-axis at $m/p.$
Draw that triangle, and keep the picture. The slope $-p/q$ has a genuinely economic meaning: to buy one more unit of good $1$ you must give up $p/q$ units of good $2,$ so $p/q$ is the opportunity cost of good $1$ measured in units of good $2$ — the market's exchange rate between the two goods. Everything else in consumer theory is built on it.
The slope of the budget line is set by the two prices alone. It says nothing about what the consumer happens to like.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A consumer's budget set is $px + qy \le m$ with $x, y \ge 0$.
The price $p$ of good $x$ doubles. Income $m$ and the price $q$ are unchanged.
What happens to the budget line?
- Why is that answer tempting? This treats a higher price like a lower income. It is true that the consumer can afford less, but look at where: a parallel shift keeps the slope $-p/q,$ and the slope has changed.
- Why is that answer tempting? The slope of the budget line is $-p/q,$ so it cannot stay the same when $p$ doubles.
- Why is that answer tempting? The right mechanism on the wrong axis. Ask: if you spend everything on $y,$ does the price of $x$ matter?
Spending everything on $y$ gives $m/q$ units — the price of $x$ never entered. So the $y$-intercept cannot move.
Spending everything on $x$ gives $m/p$ units, and $p$ doubled, so that intercept halves.
One end fixed, the other pulled in: the line pivots. The slope steepens from $-p/q$ to $-2p/q$ — of course it does, $x$ is now twice as expensive in terms of $y$.
| change | what moves |
|---|---|
| $m$ falls | both intercepts, slope fixed — parallel shift |
| $p$ rises | $x$-intercept only, slope steepens — pivot |
A price change and an income change are different animals. Almost all of consumer theory lives in that distinction.
Move the budget line
Now see it move. The sliders set the two prices $p$ and $q$ and the income $m.$
- Drag $p$ with $q$ and $m$ fixed. Which intercept moves, and which stays where it is?
- Drag $m.$ Both intercepts move, and the line stays parallel to itself.
- Press Reset and then Double p, and compare with your answer to the question above.
Quadratic functions: where is the top?
A profit rises with output at first and falls later. At which output is it largest?
Linear functions never turn around. A firm's profit, though, typically rises with output at first and falls later; a cost per unit typically falls and then rises. The simplest functions that can do this are the quadratics.
The condition $a\ne0$ is not a technicality: if $a=0$ the function is linear, and linear functions behave completely differently. Two questions about (1.3) matter: for which $x,$ if any, is $ax^2+bx+c=0;$ and where is the vertex? The second is the one that matters for economics, since profit maximisation is exactly the search for a vertex. Both drop out of a single algebraic manoeuvre.
For $a\ne 0$, $$ ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^{\!2} - \frac{b^2-4ac}{4a}\,. \tag{1.4} $$
Expand the right-hand side: $$ \begin{aligned} &a\bigl(x^2 + \tfrac{b}{a}x + \tfrac{b^2}{4a^2}\bigr) - \tfrac{b^2-4ac}{4a} \\ &\quad= ax^2 + bx + \tfrac{b^2}{4a} - \tfrac{b^2}{4a} + c \\ &\quad= ax^2+bx+c\,. \end{aligned} $$
Formula (1.4) is worth staring at, because on the right-hand side the variable $x$ occurs only once. All the difficulty of a quadratic lies in the fact that $x$ appears twice; completing the square removes it. And a squared bracket is a quantity we understand perfectly: it is never negative, and it is zero exactly when the bracket is, that is when $x=-b/2a.$
Let $f(x)=ax^2+bx+c$ with $a\ne0$, and set $x^{*} = -\dfrac{b}{2a}$.
- If $a>0$, then $f$ has a minimum at $x^{*}$, and no maximum.
- If $a<0$, then $f$ has a maximum at $x^{*}$, and no minimum.
In both cases the extreme value is $$ f(x^{*}) = c - \frac{b^2}{4a} = -\frac{b^2-4ac}{4a}\,, $$ and the parabola is symmetric about the vertical line $x=x^{*}$: for every $t$, $f(x^{*}+t) = f(x^{*}-t)$.
The extremum. In (1.4) the only part depending on $x$ is $a\bigl(x+\frac{b}{2a}\bigr)^2,$ which for $a>0$ is $\ge 0$ and vanishes precisely at $x=-b/2a,$ so the expression is smallest there; for $a<0$ it is $\le 0$ and the expression is largest there.
The symmetry. For the symmetry, put $x = x^{*}\pm t$: the bracket becomes $\pm t,$ and squaring destroys the sign.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Where does $f$ attain its maximum?
- Why is that answer tempting? This is $-b/2a$ with the minus sign of $a$ dropped: $-12/(2\cdot 2)$ instead of $-12/(2\cdot(-2)).$ It is the commonest slip with this formula.
- Why is that answer tempting? This is $-b/a$: the $2$ in the denominator is missing. The $2$ comes from completing the square, where the bracket is $x + \frac{b}{2a}.$
- Why is that answer tempting? “Opens downward” is not the same as “decreasing”. A parabola with $a < 0$ rises to its vertex and falls after it, so it has a maximum and no minimum.
Two independent checks, both faster than redoing the algebra:
Roots. $-2x^2 + 12x - 7 = 0$ has roots $3 \pm \sqrt{22}/2$. Their midpoint is $3$. Of course it is.
Sketch. $a < 0$, so it opens downward, and $f(0) = -7 < 0$ while $f(3) = 11 > 0$. The top is somewhere to the right of $0$ — which rules out $-3$ on sight.
For the value: substitute. $f(3) = -18 + 36 - 7 = 11$.
The number $\Delta = b^2 - 4ac$ in step 4 is called the discriminant of the quadratic: its sign alone says how many roots there are.
The check in step 4 is worth a second look. The two roots are $\bigl(-b+\sqrt{\Delta}\bigr)/2a$ and $\bigl(-b-\sqrt{\Delta}\bigr)/2a,$ and they differ only in the sign before the square root. In their average the square roots cancel, and what is left is $-b/2a.$ So the vertex lies midway between the roots, which is one way to remember where the $2$ and the minus sign come from. When there are no roots, the vertex is still at $-b/2a$: Fact 1.10 never mentions them.
Method 1.13 from start to finish
Analyse $f(x) = -2x^2 + 8x - 6$: its shape, its vertex, its extreme value and its roots.
Try it yourself first. Then open one step at a time.
Step 1: the shape. Here $a = -2,$ $b = 8$ and $c = -6.$ Since $a < 0$ the parabola is $\cap$-shaped, and $f$ has a maximum and no minimum.
Step 2: where. With both signs written out, $$x^{*} = -\frac{8}{2\cdot(-2)} = -\frac{8}{-4} = 2\,.$$
Step 3: the value. Substitute $x = 2$ into $f$: $$ \begin{aligned} f(2) &= -2\cdot 4 + 8\cdot 2 - 6 \\ &= -8 + 16 - 6 = 2\,. \end{aligned} $$ The largest value of $f$ is $2,$ and it is attained at $x = 2.$
Step 4: the roots. $\Delta = b^2 - 4ac = 64 - 48 = 16 > 0,$ so there are two roots: $$x = \frac{-8 \pm \sqrt{16}}{2\cdot(-2)} = \frac{-8 \pm 4}{-4}\,,$$ that is, $x = 1$ or $x = 3.$
Check: the average of the roots is $(1 + 3)/2 = 2 = x^{*},$ and $f(1) = -2 + 8 - 6 = 0.$
A $\cap$-shaped parabola with vertex $(2, 2)$: the maximum value is $2,$ at $x = 2.$ The roots are $x = 1$ and $x = 3.$
Move the parabola
The sliders set $a,$ $b$ and $c$ in $f(x) = ax^2 + bx + c.$
- Drag $b$ with $a$ and $c$ fixed. The vertex slides along, and the midpoint of the roots slides with it.
- Drag $c$ upward until the roots disappear. The line $x = -b/2a$ does not move.
- Drag $a$ through $0$ and watch the parabola turn over.
- Press Completed square to see formula (1.4) with the current numbers.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
What is the largest value of $f$?
- Why is that answer tempting? $3$ is $x^{*},$ the place where the maximum occurs. The question asks for the value, $f(3).$
- Why is that answer tempting? This is $f(-3)$: the sign of $a$ was lost in $-b/2a,$ giving $x^{*} = -3$ instead of $3.$ A largest value below $f(0) = -4$ cannot be right.
- Why is that answer tempting? This is $c - b^2/4,$ the formula for the value with the $a$ in $c - b^2/4a$ forgotten. Substituting is safer than remembering a second formula.
$a = -1 < 0,$ so there is a maximum, at $$x^{*} = -\frac{6}{2\cdot(-1)} = 3\,.$$ The largest value is what $f$ gives there: $$f(3) = -9 + 18 - 4 = 5\,.$$ Check by symmetry: $f(2) = -4 + 12 - 4 = 4$ and $f(4) = -16 + 24 - 4 = 4,$ equal and below $5.$
Quadratics at work: profit and the mean
A firm sells $Q$ units at the price $102 - 2Q$ each. How many should it make?
We do not yet have calculus, but Fact 1.11 already solves every problem in which revenue is linear in quantity and costs are quadratic.
Recap The economics words used in this section
Nothing here needs economics you have not met; these are the words.
- Quantity $Q$: how many units the firm produces and sells.
- Price $P$: what the firm receives per unit.
- Demand: how much buyers take at a given price. The more the firm wants to sell, the lower the price must be, for instance $P = 102 - 2Q.$
- Revenue $R$: the money that comes in, price times quantity: $R = PQ.$
- Cost $C$: what it costs to produce $Q$ units.
- Profit $\pi$: revenue minus cost, $\pi = R - C.$ Here $\pi$ is a name for the profit, not the number $3.14\ldots$
- Monopolist: a firm that is the only seller of its good, so that its own quantity decides the price.
- Tax per unit: an amount $\tau$ (tau) paid to the government for every unit produced. It adds $\tau Q$ to the cost.
Many of last term's comments on this lecture were about exactly these words, not about the mathematics. The mathematics is always the same two lines: write the profit as a quadratic in $Q,$ then find its vertex.
The output that maximises profit
When a firm produces and sells $Q$ units it receives the price $P=102-2Q$ per unit, and its cost of producing $Q$ units is $C = 2Q + \tfrac12 Q^2.$ Find the output that maximises profit, and the maximal profit.
Try it yourself first. Then open one step at a time.
Profit as a function of $Q.$ Revenue is price times quantity, $R = PQ = (102-2Q)Q,$ so profit is $$ \begin{aligned} &\pi(Q) = R - C \\ &\ = (102-2Q)Q - \bigl(2Q+\tfrac12 Q^2\bigr) \\ &\ = 100Q - \tfrac52 Q^2\,. \end{aligned} $$
Where the maximum is. This is a quadratic in $Q$ with $a=-\tfrac52<0$ and $b=100,$ so it has a maximum, at $$ Q^{*} = -\frac{100}{2\cdot(-\tfrac52)} = 20\,. $$
The maximal profit. Substituting, $$ \begin{aligned} \pi(Q^{*}) &= 100\cdot 20 - \tfrac52\cdot 400 \\ &= 2000-1000 = 1000\,. \end{aligned} $$
In words. The firm should produce $20$ units and will earn a profit of $1000;$ the corresponding price is $P=102-40=62.$
$Q^{*} = 20$ units, and the maximal profit is $\pi(20) = 1000.$ The price is then $62.$
One proviso applies to every such calculation. Optimising over all real numbers may return an answer the problem forbids: had the numbers given $Q^{*}<0,$ the answer would be $Q^{*}=0,$ the firm shutting down, and not the formula. Always check a formula obtained by optimising against the constraints of the problem, here $Q\ge0.$
Addition: the same calculation with letters — a monopolist
This example is not in the lecture notes; it is from the lecture. A monopolist can sell $Q$ units at the price $P = a - bQ$ per unit, and its cost is $C = \alpha Q + \beta Q^2.$ Here $a,$ $b,$ $\alpha$ (alpha) and $\beta$ (beta) are positive constants with $a > \alpha.$ Which output $Q_M$ maximises the profit?
Try it yourself first. Then open one step at a time.
Profit as a function of $Q.$ Revenue minus cost: $$ \begin{aligned} &\pi(Q) \\ &\ = (a - bQ)Q - (\alpha Q + \beta Q^2) \\ &\ = (a - \alpha)Q - (b + \beta)Q^2\,. \end{aligned} $$
A quadratic opening downward. The coefficient of $Q^2$ is $-(b + \beta) < 0$ and the coefficient of $Q$ is $a - \alpha.$ By Fact 1.11 the maximum is at $$ \begin{aligned} Q_M &= -\frac{a - \alpha}{2\cdot\bigl(-(b + \beta)\bigr)} \\[4pt] &= \frac{a - \alpha}{2(b + \beta)}\,. \end{aligned} $$ Since $a > \alpha$ this is positive, so the constraint $Q \ge 0$ is satisfied.
Read the formula. The numerator $a - \alpha$ is the gap between the highest price buyers will pay and the cost of the first units: no gap, no production. The denominator says how quickly that gap closes: $b$ because the price falls as more is sold, $\beta$ because each further unit costs more. The steeper either is, the sooner the firm stops.
Check with Example 1.14: there $a = 102,$ $b = 2,$ $\alpha = 2$ and $\beta = \tfrac12,$ so $Q_M = \dfrac{100}{2\cdot\tfrac52} = 20,$ as before.
$Q_M = \dfrac{a - \alpha}{2(b + \beta)},$ the vertex of $\pi(Q) = (a - \alpha)Q - (b + \beta)Q^2.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
The monopolist above chooses $Q_M = \dfrac{a-\alpha}{2(b+\beta)}$.
The government now imposes a tax of $\tau$ per unit produced.
By how much does the optimal output fall?
- Why is that answer tempting? This assumes that the tax goes through one for one. The units already say no: $\tau$ is money per unit, and $Q$ is a number of units.
- Why is that answer tempting? True for a tax that is one fixed sum, false for a tax per unit. A fixed sum moves the whole profit curve down without changing where its top is; a tax per unit changes the coefficient of $Q.$
- Why is that answer tempting? A half-remembered “the firm bears half”. This answer ignores $b$ and $\beta$ altogether, and how much the output falls must depend on how steep demand and cost are.
The tax makes total cost $\alpha Q + \beta Q^2 + \tau Q = (\alpha + \tau)Q + \beta Q^2$.
Nothing else changed — the same problem with $\alpha$ replaced by $\alpha + \tau$:
Why not $\tau$? The firm re-optimises: it absorbs part of the tax and passes on part, and how much depends on how steep demand and cost are. That is the $2(b+\beta)$.
Why is C wrong? A lump-sum tax really would leave $Q_M$ alone — it shifts $\pi$ down without changing its shape. A per-unit tax tilts it.
Why the mean is the “expected” value
An urn contains $T$ balls, the $t$-th carrying the number $y_t.$ One ball will be drawn at random and you must guess the number on it; call your guess $\hat y.$ Statisticians measure the badness of a guess by the mean squared error $L(\hat y) = \frac{1}{T}\sum_{t=1}^{T}(y_t-\hat y)^2.$ Which guess minimises $L$?
Try it yourself first. Then open one step at a time.
A quadratic in the guess. Expand the square and split the sum: $$ \begin{aligned} &L(\hat y) \\ &\ = \frac{1}{T}\sum_{t=1}^{T} y_t^2 - \frac{2}{T}\Bigl(\sum_{t=1}^{T} y_t\Bigr)\hat y + \hat y^{\,2} \\ &\ = \hat y^{\,2} - 2\mu_y\,\hat y + \frac{1}{T}\sum_{t=1}^{T} y_t^2\,, \end{aligned} $$ where $\mu_y = \frac1T\sum_t y_t$ is the average of the numbers.
Its vertex. This is a quadratic in the variable $\hat y$, with $a=1$ and $b=-2\mu_y,$ so by Fact 1.11 it is minimised at $$ \hat y = -\frac{b}{2a} = -\frac{-2\mu_y}{2} = \mu_y\,. $$
What it means. The best guess, in the mean-squared-error sense, is the mean. This is precisely why the mean of a random variable is called its expected value, and it is the reason squared errors are everywhere in statistics.
$\hat y = \mu_y$: the guess with the smallest mean squared error is the mean of the numbers.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
An urn holds three balls, carrying the numbers $2,$ $3$ and $10.$ Which guess $\hat y$ gives the smallest mean squared error $L(\hat y) = \tfrac13\bigl((2-\hat y)^2 + (3-\hat y)^2 + (10-\hat y)^2\bigr)$?
- Why is that answer tempting? $3$ is the middle one of the three numbers (the median). It is a natural guess, but it is not what squared errors reward: $L(3) = 50/3,$ while $L(5) = 38/3.$
- Why is that answer tempting? $6$ is halfway between the smallest and the largest number. The mean uses all three numbers, not only the two at the ends: $L(6) = 41/3,$ more than $L(5) = 38/3.$
- Why is that answer tempting? A guess at the largest number is far from the other two, and squaring makes large misses expensive: $L(10) = 113/3.$
By Example 1.15 the best guess is the mean, $$\mu_y = \frac{2 + 3 + 10}{3} = 5\,.$$ You can also see the quadratic directly: $L(\hat y) = \hat y^{\,2} - 10\,\hat y + \tfrac{113}{3},$ with vertex at $\hat y = -\dfrac{-10}{2} = 5$ and smallest value $L(5) = \tfrac{38}{3}.$
Polynomials: degree, roots and factors
$x^3 - 2x^2 - 5x + 6 = 0.$ There is no formula to reach for. Where do you start looking for a solution?
Linear and quadratic functions are the first two members of a family.
Degrees $1,$ $2,$ $3$ and $4$ are called linear, quadratic, cubic and quartic. Functions such as $5+x^{-2}$ or $1/(x^3-x+2)$ are not polynomials: negative powers and division by a non-constant are both forbidden.
Cubics are the first degree genuinely useful in economics. A realistic total cost function has a fixed cost $C(0)>0,$ then rises quickly at low output, more slowly as equipment is used fully, and steeply again at high output. A line cannot do this, and neither can a parabola, whose rate of increase changes only once; a cubic $C(Q)=aQ^3+bQ^2+cQ+d$ can, and does so provided $a>0,$ $b<0,$ $c>0,$ $d>0$ and $3ac>b^2$ — the characteristic “lazy S”.
The single most useful fact about polynomials connects roots with factors.
Let $P$ be a polynomial and $a$ a number. Then $$ x-a \text{ is a factor of } P(x) \;\Longleftrightarrow\; P(a)=0\,. $$
Dividing $P(x)$ by $x-a$ leaves a remainder of degree less than $1,$ a constant $r,$ so $P(x) = q(x)(x-a) + r$ for all $x;$ setting $x=a$ gives $P(a)=r.$ The remainder therefore vanishes exactly when $P(a)=0.$
It follows that a polynomial of degree $n$ has at most $n$ roots: each root $a$ contributes a distinct factor $x-a,$ and multiplying $n+1$ of them would already give degree $n+1.$ It may have none at all — $x^{100}+1$ is never zero.
Addition: turning points
This paragraph is not in the lecture notes; it is from the lecture. A turning point of a graph is a place where it switches from rising to falling, or back. A line has none, and a parabola has exactly one, its vertex. In general a polynomial of degree $n$ has at most $n - 1$ turning points. We cannot give the reason yet; it comes in a later lecture.
Like the bound on the number of roots, this is a ceiling and not a count: $x^{100}+1$ has degree $100,$ no roots and only one turning point.
The theorem is only useful if we can find roots. In general this is hard, but there is a good trick when the coefficients are integers.
Suppose all the coefficients $a_n,\ldots,a_1,a_0$ in (1.5) are integers. Then every integer root of $P$ is a factor of the constant term $a_0$.
Let $x$ be an integer with $P(x)=0.$ Move $a_0$ across and take out a factor of $x$: $$ x\bigl(a_nx^{n-1}+a_{n-1}x^{n-2}+\cdots+a_1\bigr) = -a_0\,. $$ The bracket is an integer, so $x$ divides $a_0.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
All coefficients are integers. Which of these can be ruled out as an integer root immediately, with no substitution at all?
- Why is that answer tempting? $1$ divides $6,$ so it is a candidate. It is in fact a root: $P(1) = 0.$
- Why is that answer tempting? Negative divisors count too. If you list only the positive factors of $a_0$ you lose half of the candidates.
- Why is that answer tempting? This is the heart of the question: $6$ divides $6,$ so it is a candidate. It just happens not to be a root. Candidate and root are different words.
An integer root must divide the constant term $a_0 = 6$. The candidates are therefore $$\pm 1,\; \pm 2,\; \pm 3,\; \pm 6$$
$4$ is not among them, so it is out before you write anything down.
$6$ is a candidate — but $P(6) = 120 \neq 0$, so it is not a root. That is the distinction the question is about:
The theorem shortens the list. It does not answer the question.
Testing the rest: $P(1) = P(3) = P(-2) = 0$. Degree three, so that is all of them, and $P(x) = (x-1)(x-3)(x+2)$.
All roots of a cubic with integer coefficients
Find all roots of $-x^3+4x^2-x-6=0.$
Try it yourself first. Then open one step at a time.
The candidates. The coefficients are integers and the constant term is $-6,$ whose factors are $\pm1,\pm2,\pm3,\pm6.$
The test. Substituting each of these eight candidates in turn shows that $-1,$ $2$ and $3$ are roots and the others are not.
Nothing is missing. A cubic has at most three roots, so we have found them all, and $$ \begin{aligned} &-x^3+4x^2-x-6 \\ &\quad= -(x+1)(x-2)(x-3)\,. \end{aligned} $$
The roots are $x = -1,$ $x = 2$ and $x = 3.$
Division with remainder, and rational functions
$17 \div 5$ is $3$ with remainder $2.$ What is $(x^4 + 3x^2 - 4) \div (x^2 + 2x)$?
Dividing one polynomial by another, which is what step (4) of Method 1.19 asks for, works exactly like long division of numbers.
Let $P$ have degree $m$ and $Q$ have degree $n$, with $m\ge n$. Then there is exactly one pair of polynomials $q$ (of degree $m-n$) and $r$ (of degree less than $n$) with $$ \begin{aligned} &P(x) = q(x)\,Q(x) + r(x)\,, \\ &\text{equivalently} \\ &\frac{P(x)}{Q(x)} = q(x) + \frac{r(x)}{Q(x)} \end{aligned} \tag{1.6} $$ wherever $Q(x)\ne0$. We call $q$ the quotient and $r$ the remainder; if $r$ is the zero polynomial, $Q$ is a factor of $P$.
With numbers: $17 = 3\cdot 5 + 2,$ and the remainder $2$ is smaller than the divisor $5.$ With polynomials “smaller” means of lower degree, and that is what tells you when to stop dividing.
Long division of polynomials
Compute $(x^4+3x^2-4)\div(x^2+2x).$
Try it yourself first. Then open one step at a time.
How to set it up. Arrange the work as for long division of numbers, asking at each step what the leading term of the divisor must be multiplied by to match the leading term of what is left. There are no terms in $x^3$ or $x$ in the numerator; write in the coefficients $0,$ so that like terms stay in line.
The division. $$ \scriptsize \begin{array}{rl} & \underline{\;x^2 - 2x + 7\;}\\[2pt] x^2+2x\,\bigl)\; & x^4+0x^3+3x^2+0x-4\\ & \underline{x^4+2x^3}\phantom{{}+3x^2+0x-4}\\ & \phantom{x^4+}-2x^3+3x^2+0x-4\\ & \phantom{x^4+}\underline{-2x^3-4x^2}\phantom{{}+0x-4}\\ & \phantom{x^4+-2x^3+}7x^2+0x-4\\ & \phantom{x^4+-2x^3+}\underline{7x^2+14x}\phantom{{}-4}\\ & \phantom{x^4+-2x^3+7x^2}-14x-4 \end{array} $$ Read it from the top: $x^2\cdot x^2 = x^4$ gives the first term $x^2$ of the quotient, and $x^2(x^2+2x)$ is subtracted; then $-2x\cdot x^2 = -2x^3;$ then $7\cdot x^2 = 7x^2.$
When to stop. The remainder $-14x-4$ has degree $1,$ less than the degree $2$ of the divisor, so we stop. Hence $x^4+3x^2-4 = (x^2-2x+7)(x^2+2x) + (-14x-4),$ or equivalently $$ \begin{aligned} \frac{x^4+3x^2-4}{x^2+2x} ={}& x^2-2x+7 \\ & - \frac{14x+4}{x^2+2x}\,. \end{aligned} $$
The quotient is $x^2 - 2x + 7$ and the remainder is $-14x - 4.$
Always check a division by multiplying back: here $(x^2-2x+7)(x^2+2x) = x^4 + 3x^2 + 14x,$ and adding the remainder $-14x - 4$ gives $x^4 + 3x^2 - 4,$ as it should. On the exam this one line is part of a complete solution.
Fact 1.22 says that every improper rational function is a polynomial plus a proper one — Example 1.23 did exactly this — and that is usually the first step in understanding one, since the polynomial part governs the behaviour for large $|x|$ while the proper part dies away. The simplest interesting case is $R(x)=a/x$ with $a>0,$ whose graph is a hyperbola; curves $xy=\text{constant}$ appear in economics as the indifference curves of a consumer with Cobb–Douglas preferences.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
$P(x) = x^3 - 3x + 4$ is divided by $x - 2.$ What is the remainder? (No long division is needed.)
- Why is that answer tempting? This is $P(-2).$ Dividing by $x - 2$ means $a = 2$ in $P(x) = q(x)(x - a) + r,$ so the number to substitute is $+2.$ Substituting $-2$ gives the remainder on division by $x + 2.$
- Why is that answer tempting? $2$ divides the constant term $4,$ so $2$ is a candidate for a root. But $P(2) = 6 \ne 0$: it is not a root, and $x - 2$ is not a factor.
- Why is that answer tempting? $4$ is the constant term, which is $P(0)$: the remainder on division by $x,$ not by $x - 2.$
The divisor $x - 2$ has degree $1,$ so the remainder is a constant $r,$ and $P(x) = q(x)(x - 2) + r$ for all $x.$ Put $x = 2$: $$r = P(2) = 8 - 6 + 4 = 6\,.$$ This is the argument behind Fact 1.17, used for a remainder that is not zero. The long division gives the same thing: $x^3 - 3x + 4 = (x^2 + 2x + 1)(x - 2) + 6.$
What you should take away
- A linear function $f(x)=ax+b$ is exactly one whose value changes by the same amount $a$ for every unit increase in $x.$ Slope through two points: $(y_2-y_1)/(x_2-x_1);$ line through one point: $y-y_1=a(x-x_1).$
- Two linear equations in two unknowns have one solution, none, or infinitely many — never two. Compare the slopes to see which.
- The budget line $px+qy=m$ has slope $-p/q$ and intercepts $m/p$ and $m/q;$ the slope is the opportunity cost of good $1$ in terms of good $2.$
- Completing the square, $ax^2+bx+c = a\bigl(x+\tfrac{b}{2a}\bigr)^{2}-\tfrac{b^2-4ac}{4a},$ makes $x$ occur only once and answers every question about a quadratic. The vertex is at $x^{*}=-b/2a,$ a minimum if $a>0$ and a maximum if $a<0;$ the extreme value is $f(x^{*}).$
- $x-a$ is a factor of $P(x)$ if and only if $P(a)=0;$ a polynomial of degree $n$ has at most $n$ roots; any integer root divides the constant term; and polynomials divide with a remainder, $P = qQ + r$ with $\deg r<\deg Q.$
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — a price equal to the cost of one more unit
A firm's cost function is $C(Q)=100+7Q$ and the price per unit sold is $\text{\textdollar}7.$ The profit function is
(A) increasing in $Q;$
(B) decreasing in $Q;$
(C) constant in $Q;$
(D) not enough information.
One way to think about it
(C). Revenue is $7Q,$ so the profit is $$\pi(Q) = 7Q - (100 + 7Q) = -100\,.$$ Every unit brings in $7$ dollars and costs $7$ dollars more to make, so selling more changes nothing: the profit is $-100$ at every output, a loss equal to the fixed cost. The slope of the revenue line and the slope of the cost line are equal, and the profit is the difference of two parallel lines.
Discussion question 2 — doubling the income, the prices, or all three
What happens to the budget set when $m$ doubles? When both prices double? When $m$ and both prices all double?
One way to think about it
Look at the two intercepts $m/p$ and $m/q$ and at the slope $-p/q.$
$m$ doubles. Both intercepts double and the slope is unchanged: the budget line shifts outward, parallel to itself. The new triangle has twice the base and twice the height, so four times the area.
Both prices double. Both intercepts halve, and the slope $-2p/2q = -p/q$ is unchanged: a parallel shift inward, exactly as if the income had been halved.
All three double. Nothing happens: $2px + 2qy \le 2m$ is the same condition as $px + qy \le m.$ Only the relative sizes of income and prices matter.
Discussion question 3 — which way the budget line turns
If the price $p$ of good $1$ falls, the budget line
(A) shifts outward parallel to itself;
(B) rotates outward about the point $(0,m/q);$
(C) rotates outward about the point $(m/p,0);$
(D) is unchanged.
One way to think about it
(B). If the consumer spends everything on good $2,$ the price of good $1$ does not matter: the intercept $(0, m/q)$ stays where it is. Spending everything on good $1$ now buys more, so the intercept $m/p$ on the $x$-axis moves outward. One end fixed and the other moving out: the line rotates outward about $(0, m/q),$ and its slope $-p/q$ becomes less steep. A parallel shift, as in (A), is what a change of income does.
Discussion question 4 — a tax per unit in Example 1.14
In Example 1.14, the government levies a tax of $\tau$ per unit produced. What happens to $Q^{*},$ and does the firm pass the whole tax on?
One way to think about it
The tax adds $\tau Q$ to the cost, so the profit becomes $$ \begin{aligned} \pi(Q) &= 100Q - \tfrac52 Q^2 - \tau Q \\ &= (100 - \tau)Q - \tfrac52 Q^2\,. \end{aligned} $$ It is the same quadratic with $b = 100$ replaced by $100 - \tau,$ so $$Q^{*} = \frac{100 - \tau}{5} = 20 - \frac{\tau}{5}\,.$$ Output falls by $\tau/5$ (as long as $\tau \le 100;$ beyond that the firm shuts down, $Q^{*} = 0$).
The price is $P = 102 - 2Q^{*} = 62 + \tfrac25\tau.$ So the price rises by only two fifths of the tax: the firm passes on $40\,\%$ of it to the buyers and bears the rest itself. How much is passed on depends on how steep the demand and the cost are, which is what the denominator $2(b + \beta)$ in the monopolist's formula records.
Discussion question 5 — two equal values fix the axis of symmetry
The quadratic $f(x)=ax^2+bx+c$ has $f(1)=f(5).$ Then its axis of symmetry is
(A) $x=2;$
(B) $x=3;$
(C) $x=4;$
(D) impossible to say without knowing $a,b,c.$
One way to think about it
(B). By Fact 1.11 a parabola is symmetric about $x = x^{*},$ so two points with the same value lie equally far from the axis, one on each side. The axis is therefore midway between them: $x^{*} = (1 + 5)/2 = 3.$
The algebra agrees: $f(1) = f(5)$ says $a + b + c = 25a + 5b + c,$ that is, $b = -6a,$ and then $-b/2a = 3$ whatever $a$ and $c$ are. So (D) is wrong: the two equal values are enough.
Discussion question 6 — a cubic is fixed by four points
True or false: two polynomials of degree $3$ that agree at four different points must be equal.
One way to think about it
True. Call them $P$ and $Q$ and look at the difference $D(x) = P(x) - Q(x).$ It is a polynomial of degree at most $3,$ and it is zero at the four points where $P$ and $Q$ agree. But a polynomial of degree $n$ has at most $n$ roots, so a polynomial of degree $1,$ $2$ or $3$ cannot have four; and a non-zero constant has none. The only possibility left is that $D$ is zero everywhere, that is, $P = Q.$
The same argument shows that a line is fixed by two points and a parabola by three.