Corrected exercises: differentials and linear approximation (201-103)

$f'(x)=3x^{2}-2$, so $f'(2)=10$ and $dy=10\times 0.5=5$. Pitfall: confusing $dy$ with the actual change, $\Delta y=f(2.5)-f(2)=10.625-4=6.625$; the differential is only an estimate of it.

$f'(x)=\frac{6}{2\sqrt{x}}=\frac{3}{\sqrt{x}}$, so $f'(36)=\frac{3}{6}=0.5$ and $dy=0.5\times 3=1.5$.

$f'(x)=-\frac{200}{x^{2}}$, so $f'(20)=-\frac{200}{400}=-0.5$ and $dy=-0.5\times(-0.4)=0.2$. Pitfall: the sign. $x$ DECREASES and $\frac{200}{x}$ is decreasing, so $y$ INCREASES: $dy$ is positive.

$f(x)=\sqrt{x}$, $a=64$, $dx=-1$: $f'(64)=\frac{1}{2\times 8}=\frac{1}{16}$, so $\sqrt{63}\approx 8-\frac{1}{16}=7.9375$. The calculator gives $7.9373$: the estimate is too high by two ten-thousandths, because the tangent lies above a concave curve. Pitfall: starting from $49$, much farther away, which gives $7+\frac{14}{14}=8$.

$f(x)=\sqrt[3]{x}$, $a=27$, $dx=1$: $f'(x)=\frac{1}{3\sqrt[3]{x^{2}}}$, so $f'(27)=\frac{1}{3\times 9}=\frac{1}{27}$ and $\sqrt[3]{28}\approx 3+\frac{1}{27}\approx 3.0370$. The calculator gives $3.0366$.

$f(x)=\frac{1}{x}$, $a=4$, $dx=0.02$: $f'(4)=-\frac{1}{16}$, so $\frac{1}{4.02}\approx 0.25-\frac{0.02}{16}=0.25-0.00125=0.24875$. Pitfall: forgetting the minus sign in the derivative of $\frac{1}{x}$ and ADDING $0.00125$, when dividing by a larger number gives less.

$f(x)=x^{5}$, $a=2$, $dx=-0.02$: $f'(2)=5\times 2^{4}=80$, so $(1.98)^{5}\approx 32+80\times(-0.02)=32-1.6=30.4$. The exact value is $30.4316$: the gap comes from the curvature, which stays small because $dx$ is small.

For $y=x^{n}$, $dy=nx^{n-1}dx$, so $\frac{dy}{y}=n\,\frac{dx}{x}$: the exponent multiplies the relative error. For $y=x^{3}$: $3\times 2=6$ %. For $y=x^{-\frac{1}{2}}$: $\left|-\frac{1}{2}\right|\times 4=2$ %. Pitfall: keeping the sign of the exponent, when a margin of error is expressed in absolute value.

$\ln y=\ln u+2\ln v$, so $\frac{dy}{y}=\frac{du}{u}+2\frac{dv}{v}$. In the worst case: $0.8+2\times 1.2=3.2$ %. Pitfall: forgetting the exponent of $v$ and finding $2$ %.

$\frac{dy}{y}=\frac{du}{u}-\frac{dv}{v}$. In the worst case the two errors work in opposite directions, so the margin is $2+0.6=2.6$ %. Pitfall: subtracting the margins and announcing $1.4$ %, which assumes that the two errors always cancel out.

$U(900)=140\sqrt{900}=140\times 30=4{,}200$ users. Writing $U(b)=140\,b^{1/2}$: $U'(b)=70\,b^{-1/2}=\dfrac{70}{\sqrt{b}}$, so $U'(900)=\dfrac{70}{30}\approx 2.333$ users per thousand dollars.

With $db=40$: $dU=2.333\times 40\approx 93.33$ users. The actual change is $\Delta U=U(940)-U(900)=140\sqrt{940}-4{,}200=140\times 30.659\,4-4{,}200\approx 4{,}292.32-4{,}200=92.32$ users. The differential is the change along the TANGENT, the actual change is the change along the CURVE; the figure shows the gap between the two.

The error is $93.33-92.32\approx 1.01$ users, about $1.1$ % of the actual change. On a budget decision, being off by one user out of ninety-two is perfectly acceptable: this is what justifies using the differential rather than the exact calculation.

With $db=100$: $dU=2.333\times 100\approx 233.33$, whereas $\Delta U=140\sqrt{1{,}000}-4{,}200\approx 4{,}427.19-4{,}200=227.19$. The error rises to $6.14$ users, about six times the previous error for a step only two and a half times larger. This is no coincidence: the error of linear approximation grows like the SQUARE of the step, and $2.5^{2}=6.25$, which fits.

It OVERESTIMATES, in both cases. The reason is concavity: $U''(b)=-35\,b^{-3/2}<0$, so the curve is concave down and lies BELOW its tangent. The general rule is easy to remember and holds for the whole chapter: if $f''<0$, the tangent lies above and the differential overestimates; if $f''>0$, the tangent lies below and the differential underestimates. For a budget, this means the linear estimate is always a little optimistic about the number of users gained.

$D(8)=71.2+8-0.05\times 64=71.2+8-3.2=76.0$ and $D'(t)=1-0.1t$, so $D'(8)=0.2$. The linearization is $L(t)=D(8)+D'(8)(t-8)=76+0.2\,(t-8)$. It is exactly the equation of the tangent line, and it is the phrase “at the current rate of $0.2$ points per year” written in symbols.

$L(9)=76+0.2=76.2$ %. The exact value is $D(9)=71.2+9-0.05\times 81=71.2+9-4.05=76.15$ %. The error is $0.05$ points, negligible next to the precision of a graduation statistic.

$L(12)=76+0.2\times 4=76.8$ %. The exact value is $D(12)=71.2+12-0.05\times 144=71.2+12-7.2=76.0$ %. The error is $0.8$ points, sixteen times larger, and this time it is of the same order as the changes you are trying to measure: the forecast is unusable.

Because the error of a linear approximation grows like the SQUARE of the step. Here the step goes from 1 to 4 years, so the error is multiplied by $4^{2}=16$. You can check this exactly on a parabola: the gap between $D$ and its tangent at 8 is $-0.05\,(t-8)^{2}$, that is $-0.05$ for $t=9$ and $-0.80$ for $t=12$. For a quadratic function, the linearization error is therefore known in advance, which is not the case in general.

No, never. A line of slope $0.2$ is strictly increasing, so $L$ has no maximum and no change of direction: it predicts a graduation rate that rises forever, which is wrong as early as $t=10$ and absurd well beyond, since it passes 79 % at $t=24$ and would eventually exceed 100 %. A linearization cannot see what CURVATURE carries, that is, no maximum, no ceiling and no turning point. That is the price of its simplicity, and it is why any sentence of the type “at the current rate, in fifteen years” must be read as an extrapolation, like those of the first chapter.

By logarithmic differentiation, $\dfrac{dR}{R}=\dfrac{dp}{p}+\dfrac{dq}{q}$. Since the two errors can have any sign, the WORST CASE adds their absolute values: $\left|\dfrac{dR}{R}\right|\le 1.5+2.5=4$ %. Read “at worst” and not “on average”: if the two errors have opposite signs they partly cancel, but you cannot count on it.

$\dfrac{d(Y/P)}{Y/P}=\dfrac{dY}{Y}-\dfrac{dP}{P}$, and the worst case again adds the absolute values: $1+0.5=1.5$ %. The minus sign in the formula does not shrink the margin, because a negative error on $P$ combined with a positive error on $Y$ produces precisely the accumulation.

$\dfrac{dA}{A}=2\,\dfrac{dr}{r}$, the exponent 2 coming out through logarithmic differentiation. The margin is therefore $2\times 3=6$ %. A power MULTIPLIES the relative error by its exponent, which is why you never measure an area or a volume by measuring a length alone when you can do otherwise.

$R=24\times 540=12{,}960$ dollars, and the margin is $4$ % of that amount, that is $0.04\times 12{,}960\approx 518$ dollars. You would therefore write “$12{,}960$ dollars, give or take $518$ dollars”, and take care not to comment on a 200 dollar gap from one month to the next.

No. It is the RATES OF CHANGE that subtract, because they have a known sign; the UNCERTAINTIES have no sign, so their absolute values add. The difference is fundamental and fits in one sentence: for a speed, $\dfrac{x'}{x}-\dfrac{y'}{y}$; for an uncertainty, $\left|\dfrac{dx}{x}\right|+\left|\dfrac{dy}{y}\right|$. Confusing the two leads you to announce a precision you do not have, which is the most serious fault of a quantitative report.

$(1.04)^{5}\approx 1.216\,653$, so $V(0.04)=\dfrac{1{,}000}{1.216\,653}\approx 821.93$ dollars.

Write $V(r)=1{,}000\,(1+r)^{-5}$, so $V'(r)=-5{,}000\,(1+r)^{-6}=\dfrac{-5{,}000}{(1+r)^{6}}$. At $r=0.04$: $(1.04)^{6}\approx 1.265\,319$, so $V'(0.04)\approx -3{,}951.6$ dollars per unit of rate. The unit is puzzling and has to be read correctly: a rise of one FULL point of rate, that is $dr=0.01$, would cost about 39.52 dollars.

With $dr=0.002\,5$: $dV=-3{,}951.6\times 0.002\,5\approx -9.88$ dollars. The exact value is $V(0.042\,5)=\dfrac{1{,}000}{(1.042\,5)^{5}}\approx 812.12$ dollars, a real drop of $9.81$ dollars. The estimate is off by 7 cents out of nearly ten dollars, which is more than enough for a trading floor as for an exam.

$\dfrac{V'(r)}{V(r)}=\dfrac{-5{,}000\,(1+r)^{-6}}{1{,}000\,(1+r)^{-5}}=\dfrac{-5}{1+r}$, that is $\dfrac{-5}{1.04}\approx -4.808$ per unit of rate. For $dr=0.002\,5$: $-4.808\times 0.002\,5\approx -0.012\,02$, that is $-1.20$ %. This agrees with $-0.012\,02\times 821.93\approx -9.88$ dollars. The number 5 that appears in the numerator is the TERM of the investment: the farther away the maturity, the more sensitive the value is to the rate, which portfolio managers call duration.

Because a bond already issued promises a FIXED amount: if rates rise, the same amount can now be obtained by investing less money today, so the right to that amount is worth less. Present value and rate always move in opposite directions, and the calculation says so through the negative sign of $V'$, without any financial interpretation being needed.

$W(0.80)=\dfrac{0.80}{0.20}=4$ service times. $W'(\rho)=\dfrac{1}{(1-\rho)^{2}}$, so $W'(0.80)=\dfrac{1}{0.04}=25$. The linearization is $L(\rho)=4+25\,(\rho-0.80)$.

$L(0.85)=4+25\times 0.05=5.25$, whereas $W(0.85)=\dfrac{0.85}{0.15}\approx 5.667$. The error is $0.417$, about $7.4$ % of the exact value. Already, over five percentage points, the approximation loses its value.

$L(0.90)=4+2.5=6.5$ against $W(0.90)=9$: error of $2.5$, that is $28$ %. $L(0.95)=4+3.75=7.75$ against $W(0.95)=19$: error of $11.25$, that is $59$ %. The approximation does not degrade, it collapses: it announces eight service times where there will be nineteen.

$W'(\rho)=(1-\rho)^{-2}$, so $W''(\rho)=2\,(1-\rho)^{-3}=\dfrac{2}{(1-\rho)^{3}}$, and $W''(0.80)=\dfrac{2}{0.008}=250$. This huge value foretells precisely what question c) finds: the linearization error is approximately $\dfrac{1}{2}W''\,(\Delta\rho)^{2}$, that is $125\times 0.05^{2}=0.31$ for the first step, the right order of magnitude. For the last step, the estimate itself fails, because $W''$ is no longer constant on the interval: we are approaching the asymptote.

The forecast is $7.75$ service times instead of 19, so it is off by more than eleven, and it errs on the LOW side: the curve is convex, $W''>0$, so it lies above its tangent. This is the most dangerous direction of error for a public service, the one that reassures: the manager forecasts a queue two and a half times shorter than it will be. The rule to remember fits in one sentence: near an asymptote, any sentence beginning with “at the current rate” is wrong, and it is wrong on the optimistic side.

$u=\dfrac{24{,}800}{412{,}000}\approx 0.060\,194$, that is $6.019$ %.

$du=\dfrac{dU\,A-U\,dA}{A^{2}}=\dfrac{dU}{A}-\dfrac{U\,dA}{A^{2}}$. The two errors have unknown signs, so the worst case adds the absolute values: $|du|\le \dfrac{|dU|}{A}+\dfrac{U\,|dA|}{A^{2}}$. It is once again the rule of the chapter: for a quotient, rates of change subtract, UNCERTAINTIES add.

$\dfrac{400}{412{,}000}\approx 0.000\,971$ and $\dfrac{24{,}800\times 2{,}000}{412{,}000^{2}}=\dfrac{49{,}600{,}000}{169{,}744{,}000{,}000}\approx 0.000\,292$. The sum is about $0.001\,263$, that is $0.126$ percentage points. The rate must therefore be written $6.019$ % give or take $0.126$ points, that is, somewhere between $5.893$ % and $6.145$ %.

No, or at least not as it stands. The announced drop, $0.175$ points, is of the same ORDER OF MAGNITUDE as the margin of a single survey, $0.126$ points. Now a change compares TWO surveys, each carrying its own margin: the uncertainty on the difference is therefore larger still, of the order of $0.25$ points in the worst case. The observed drop is perfectly compatible with no change at all, and a headline about falling unemployment is not supported by the numbers.

The rule fits in one sentence: never comment on a change smaller than the margin of error. It has two practical consequences. First, every published rate must come with its margin, otherwise the reader can do nothing with it. Second, a change that falls under the margin must not be called zero, which would be another unsupported claim, but NOT MEASURABLE by the survey as it was conducted. Increasing the sample size is then the only way to settle the matter, and exercise 7 shows how to set the precision to demand.

By logarithmic differentiation, $\dfrac{dB}{B}=\dfrac{dc}{c}+\dfrac{dn}{n}$. In the worst case, $\left|\dfrac{dB}{B}\right|\le\left|\dfrac{dc}{c}\right|+\left|\dfrac{dn}{n}\right|$, so the condition is that the SUM of the two relative precisions must not exceed 1 %.

$\left|\dfrac{dn}{n}\right|\le 1-0.4=0.6$ %. The precision budget, so to speak, is shared: every tenth of a point granted to the unit cost is taken away from the headcount.

$0.006\times 12{,}400=74.4$, so the headcount must be known to within 74 people. This is a concrete survey constraint, which translates into a sample size or the quality of an administrative file.

With the exponent $1.2$: $\dfrac{dB}{B}=\dfrac{dc}{c}+1.2\,\dfrac{dn}{n}$. The condition becomes $1.2\left|\dfrac{dn}{n}\right|\le 0.6$ %, so $\left|\dfrac{dn}{n}\right|\le 0.5$ %, which makes $0.005\times 12{,}400=62$ people. An exponent greater than 1 AMPLIFIES the error on the headcount, so it tightens the requirement.

On the one whose COEFFICIENT is the largest in the sum of relative errors, here the headcount, which weighs $1.2$ against 1 for the unit cost. The general rule, valid for $B=c^{\alpha}n^{\beta}$, is that the contribution of each datum to the error of the result is its exponent times its own relative precision. Note that this rule never mentions the SIZE of the quantities: whether the unit cost is 12 dollars or 12,000, it weighs in the error exactly as much as its exponent says. Concentrating the measurement effort on the big number rather than on the high exponent is the most frequent methodological mistake in a research budget.

FALSE. $dy=f'(a)\,dx$ is the change along the TANGENT, whereas $\Delta y=f(a+dx)-f(a)$ is the change along the CURVE. The two coincide only if $f$ is affine, or in the limit as $dx$ tends to 0. Exercise 1 puts a number on the gap: $93.33$ against $92.32$. Correct statement: $dy$ APPROXIMATES $\Delta y$, with an error that grows like the square of the step.

FALSE. Relative errors ADD UP: $2+2=4$ %. Exercise 3 shows it. One way to convince yourself without a formula: raising both factors by 2 % multiplies the product by $1.02^{2}=1.040\,4$, so about 4 %. Correct statement: the relative error of a product is the sum of the relative errors of its factors.

FALSE, and this is the subtlest confusion of the chapter. It is the RATES OF CHANGE that subtract, because they have a known sign; UNCERTAINTIES have none, so their absolute values add. Correct statement: $\dfrac{d(x/y)}{x/y}=\dfrac{dx}{x}-\dfrac{dy}{y}$, but the worst-case margin is $\left|\dfrac{dx}{x}\right|+\left|\dfrac{dy}{y}\right|$.

FALSE. The quality of the approximation depends neither on the size of $f$ nor on that of $f'$: it depends on the CURVATURE $f''$ and on the size of the step, the error being approximately $\dfrac{1}{2}|f''|\,(\Delta x)^{2}$. Exercise 5 proves it: $W$ is only 4 at $\rho=0.80$, a modest value, and the approximation collapses there because $W''=250$. Correct statement: linear approximation is good when the step is small compared with the scale on which the curvature varies.

FALSE, it is the opposite. If $f''>0$, the curve is CONVEX, so it turns upward and lies ABOVE its tangent: the differential underestimates. If $f''<0$, the curve is concave and lies below: the differential overestimates. Exercise 1 illustrates the second case, with a square root, and exercise 5 the first, with the queue. Correct statement: $f''>0$ puts the tangent below the curve, $f''<0$ puts it above.