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Using the tangent in place of the curve, how fast the error grows, and which way it runs.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write the linearisation of a function at a point and use it to estimate a nearby value, choose an anchor that is both exactly computable and close to the target, say from the sign of the second derivative whether the estimate is high or low, state that the error is second order in the step and use that to predict how it changes when the step changes, and propagate a relative measurement error through a computed quantity with a differential.
Near $x = a$, a differentiable function is well approximated by its tangent line: $$f(x) \approx L(x) = f(a) + f'(a)(x - a).$$ $L$ is the linearisation of $f$ at $a$, and it is the best linear approximation — the only line agreeing with $f$ in both value and slope at $a$.
In differential notation, with $dx$ a small change in the input and $dy = f'(x)\,dx$ the corresponding change along the tangent: $$\Delta y \approx dy = f'(x)\,dx.$$ $\Delta y$ is what the function actually does; $dy$ is what the tangent does; the approximation is that they are close for small $dx$.
Why it is good, and how good. For a twice-differentiable $f$, $$f(a + h) = f(a) + f'(a)h + \tfrac12 f''(\xi)h^2$$ for some $\xi$ between $a$ and $a+h$ — which is Taylor's theorem, proved from the mean value theorem two lessons from now. The error is proportional to $h^2$: halve the step and the error falls by four.
Which way the error runs. The sign of $f''$ decides it, without computing anything:
| Bend | Curve relative to tangent | Estimate |
|---|---|---|
| $f'' > 0$, concave up | above | under-estimate |
| $f'' < 0$, concave down | below | over-estimate |
Knowing the direction of an error is often worth more than a bound on its size.
Another way: picture
A curve with its tangent at $a$. Right at $a$ they touch; a little either side they have barely separated; far away the gap is obvious. Zooming in on the point of tangency makes the two indistinguishable — which is what differentiability means, and what the approximation is trading on.
Another way: steps
This is what linearisation is for outside a textbook. A quantity is measured with some uncertainty, and something is computed from it; how much uncertainty does the result carry?
If $y = f(x)$ and $x$ is uncertain by $dx$, then $y$ is uncertain by about $$dy = f'(x)\,dx.$$
Relative errors are usually what is wanted, and they are tidier. For $A = \pi r^2$: $$\frac{dA}{A} = \frac{2\pi r\,dr}{\pi r^2} = 2\,\frac{dr}{r}.$$ A $1\%$ error in the radius becomes a $2\%$ error in the area. For $V = \tfrac43\pi r^3$ it becomes $3\%$.
The general rule: for $y = x^n$, the relative error is multiplied by $|n|$. Cubing triples it, squaring doubles it, and a square root halves it — which is why measuring a quantity and taking its root is more forgiving than measuring it and cubing it, and why an experimentalist would rather compute a length from a volume than the other way round.
Anchoring where $f(a)$ is not exactly known. The whole point is to trade an unknown value for a known one. Estimating $\sqrt{26}$ by anchoring at $26$ is circular.
Anchoring too far away. $\sqrt{101}$ anchored at $1$ is a terrible estimate; anchored at $100$ it is excellent. The error grows with the square of the distance.
Using it far from the anchor and trusting it anyway. A linearisation is a local statement. There is no distance at which it is declared invalid, which is precisely why it must be used with the error term in mind.
Getting the concavity backwards. Concave up means the curve is above the tangent, so the tangent under-estimates. Draw a parabola and a tangent if it ever seems otherwise.
Confusing $\Delta y$ with $dy$. $\Delta y = f(a + dx) - f(a)$ is the true change; $dy = f'(a)\,dx$ is the tangent's change. They differ by exactly the error being discussed.
$f(x) \approx L(x)$ is often read as "near enough", which makes it useless: near enough for what? The statement has content, and the content is the error term.
Size: the error is at most $\tfrac12 \max|f''| \cdot h^2$ on the interval between the anchor and the target. Second order in the step, so it collapses as you move in and blows up as you move out.
Direction: the sign of $f''$ settles whether the estimate is high or low, and no calculation is needed to get it.
This is the difference between a working approximation and a guess. An engineer who says a linearisation is good to $0.1\%$ over a range has computed the second derivative; one who says it is "close enough" has not, and cannot say when it stops being so.
Estimate $\sqrt{101}$. Anchor at $a = 100$: $f(100) = 10$, $f'(x) = \dfrac{1}{2\sqrt x}$, so $f'(100) = \dfrac{1}{20}$.
Nearest exact point.
$L(101) = 10 + \dfrac{1}{20}(1) = 10.05$.
One step of arithmetic.
$f''< 0$, so this is an over-estimate. True value $10.0499\ldots$ — the estimate is high by $0.0001$, and the bound $\tfrac12|f''|h^2$ predicted about that.
Direction and size both foretold.
$f(x) = \sin x$ at $a = 0$: $f(0) = 0$ and $f'(0) = \cos 0 = 1$.
The anchor is exact.
$L(x) = 0 + 1 \cdot (x - 0) = x$. So $\sin x \approx x$ for small $x$ — in radians, necessarily.
This is the small-angle approximation.
At $x = 0.1$: estimate $0.1$, true $0.09983$, error $0.00017$. The error is about $|x|^3/6$ here rather than $x^2/2$, because $f''(0) = 0$ — a flat second derivative buys an extra order of accuracy, which is why this particular approximation is so unreasonably good.
$f(x) = x^3$, $f(2) = 8$, $f'(x) = 3x^2$, so $f'(2) = 12$.
Value and slope at the anchor.
$L(2.01) = 8 + 12(0.01) = 8.12$.
$f'' = 6x > 0$ at $2$, so it is an under-estimate. True: $8.120601$.
Match each approximation ingredient to its role.
| The tangent-line height at the anchor | The tangent-line slope | The horizontal displacement from the anchor | The tangent line tends to underestimate | |
|---|---|---|---|---|
| $f(a)$ | ||||
| $f'(a)$ | ||||
| $x-a$ | ||||
| $f''>0$ near $a$ |
Match each fact about a linear approximation to the conclusion it supports.
| $L$ is the tangent-line model at the anchor | The tangent estimate is an under-estimate | The tangent estimate is an over-estimate | A quadratic leading error becomes about one quarter as large | |
|---|---|---|---|---|
| At $x=a$, $L(a)=f(a)$ and $L'(a)=f'(a)$ | ||||
| $f''(x)>0$ near the anchor | ||||
| $f''(x)<0$ near the anchor | ||||
| The input step is halved |
Write the linearisation $L(x)$ of $f(x) = 4x^2$ at $x = 1$.
Answer:
To estimate $\sqrt{5 + 26}$ by linear approximation, which anchor $a$ should be used?
| Keeps tangent-line error small | Makes the linear model computable | Is circular when estimating an unknown value | Can make a poor local approximation | |
|---|---|---|---|---|
| An anchor near the target | ||||
| An anchor with known $f(a)$ and $f'(a)$ | ||||
| Using the unknown target itself as anchor | ||||
| A distant but simple anchor |
$f(x) = 8x^2$ has $f'' = 16 > 0$. Is its linearisation an over-estimate or an under-estimate?
| Tangent under-estimates nearby | Tangent over-estimates nearby | Does not imply equality away from the anchor | Does not decide approximation direction | |
|---|---|---|---|---|
| $f''>0$ | ||||
| $f''<0$ | ||||
| Tangent and graph agree at the anchor | ||||
| The sign of $f'$ |
A circle's radius is measured with a relative error of up to $1\%$. What is the largest relative error in the computed area, in per cent?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = x^2$ at $a = 2$, the linearisation's error at a step $h$ is exactly $h^2$. By what factor does the error shrink when the step is halved?
Answer:
You can linearise, estimate, and say which way and by roughly how much the estimate errs. Say in your own words why a concave-up curve is under-estimated by its tangent. Next: maxima and minima, and the search the extreme value theorem permits.
8. Your turn: estimate $(2.01)^3$ by linearising at $a = 2$, step 3
Direction predicted, then confirmed.