Zoom in far enough on a smooth curve and it looks straight. That straight line is the tangent line, and near the point of tangency it is a very good stand in for the function. This is how you can estimate 4.1 in your head, and how engineers judge how much a small measurement error affects a computed result.
Key terms
Linearization
Definition Let f be differentiable at a. The linearization of f at a is the function
L(x)=f(a)+f′(a)(x−a).
Its graph is the tangent line to the graph of f at (a,f(a)).
In plain words The linearization is the tangent line, written as a function you can evaluate.
Example Take f(x)=x at a=4. Then f(4)=2 and
f′(x)=2x1,f′(4)=41,
so
L(x)=2+41(x−4).
Tangent line approximation
Definition The tangent line approximation, or linear approximation, of f near a is
f(x)≈L(x)=f(a)+f′(a)(x−a),
used for x close to a. The error of the approximation is f(x)−L(x).
In plain words Close to a, read values off the tangent line instead of the curve. Pick a so that f(a) and f′(a) are easy to compute.
Example The linearization above gives
4.1≈2+40.1=2.025.
The true value is 2.024846…, so the error is only about 0.00015.
Differentials and actual change
Definition Let y=f(x) with f differentiable at x. The differential dx is an independent variable, any real number, and the differential dy is defined by
dy=f′(x)dx.
For an increment Δx=dx, the actual change in y is
Δy=f(x+Δx)−f(x).
The differential dy is the change along the tangent line, and Δy≈dy when Δx is small.
In plain words The increment Δy is how much the function really changes. The differential dy is how much the tangent line predicts it changes. They are close for small steps but not equal.
Example For y=x3 at x=2 with dx=Δx=0.1,
dy=3⋅22⋅0.1=1.2,Δy=2.13−23=9.261−8=1.261.
The differential misses the actual change by 0.061.
The error shrinks faster than the step
Definition A function f defined on an open interval containing a is differentiable at a if and only if there is a number m such that the error E(h)=f(a+h)−f(a)−mh satisfies
h→0limhE(h)=0,
and in that case m=f′(a). So for a differentiable function the error of the linear approximation goes to 0 faster than the increment h itself.
In plain words It is not enough that the error gets small, since any continuous function does that. The error gets small even compared with the size of the step, and among all lines through (a,f(a)) only the tangent line has this property.
Example For f(x)=x3 at a=2 we have f′(2)=12 and
E(h)=(2+h)3−8−12h=6h2+h3.
h
E(h)
E(h)/h
0.1
0.061
0.61
0.01
0.000601
0.0601
0.001
0.000006001
0.006001
The ratio
hE(h)=6h+h2
tends to 0.
Big-O and little-o notation
Definition Let E and φ be defined for h near 0, with h=0. As h→0, write
E(h)=o(φ(h))ifh→0limφ(h)E(h)=0,
where φ(h)=0 for small h=0, and write E(h)=O(φ(h)) if there are constants C>0 and δ>0 with ∣E(h)∣≤C∣φ(h)∣ whenever 0<∣h∣<δ. In this notation, f is differentiable at a exactly when
f(a+h)=f(a)+f′(a)h+o(h).
In plain words Little-o of h means negligible compared with h. Big-O of h2 means at most a fixed multiple of h2.
Example Take E(h)=6h2+h3. The limit
h→0limhE(h)=h→0lim(6h+h2)=0
shows E(h)=o(h). And for 0<∣h∣<1,
∣6h2+h3∣≤6h2+∣h∣h2≤7h2,
so E(h)=O(h2) with C=7 and δ=1.
Error bounds from the mean value theorem
Definition Let f be twice differentiable on an open interval containing a and x.
If ∣f′(t)∣≤M1 for every t between a and x, then ∣f(x)−f(a)∣≤M1∣x−a∣. This follows from the mean value theorem.
If ∣f′′(t)∣≤M2 for every t between a and x, then the error of the linear approximation satisfies
∣f(x)−L(x)∣≤2M2(x−a)2.
The second bound comes from Taylor’s theorem, which says that
f(x)−L(x)=2f′′(c)(x−a)2
for some c between a and x. Applying the plain mean value theorem twice only gives the weaker bound M2(x−a)2, without the factor 1/2.
In plain words The tangent line goes wrong only because the slope changes, and f′′ measures how fast the slope changes. A small f′′ and a small step mean a small error, and the error shrinks like the square of the step.
Example For f(x)=x near 4 the second derivative is f′′(t)=−41t−3/2, whose size decreases as t grows. For t between 4 and 4.1,
∣f′′(t)∣≤41⋅4−3/2=321.
So
∣4.1−2.025∣≤641(0.1)2=0.00015625,
and the actual error 0.000154 fits under this bound.
Check yourself
Use a linearization to estimate 38.06.
Show the answer
With f(x)=x1/3 and a=8 we get f(8)=2 and
f′(8)=31⋅8−2/3=121.
So
38.06≈2+120.06=2.005.
The true value is about 2.00499.
With x in radians, the linearization of sinx at 0 is L(x)=x. Bound the error of sin0.1≈0.1.
Show the answer
Here ∣f′′(t)∣=∣sint∣≤1, so the error is at most
21(0.1)2=0.005.
The actual error is about 0.000167, well within the bound.
A sphere’s radius is measured as 10 centimeters with a possible error of 0.05 centimeters. Estimate the possible error in its computed volume.
Show the answer
The volume is V=34πr3, so
dV=4πr2dr=4π⋅100⋅0.05=20π≈62.8.
The possible error in the volume is about 62.8 cubic centimeters.
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