Math 104 Section 101 Notes for 2013/11/19-26
November 26, 2013
Remark 1. These notes cover the material of the classes dedicated to approxi-
1 Linear Approximation
Using derivatives, we can ▯nd good approximations for how functions behave
near a poinp where it is easy to ▯nd their value. For example, suppose we would
like to ▯nd 101. We know it should be close to 10, but it is a little higher
since 101 > 100 and the square root is an increasing function, and the question
is how much higher than 10 it is.
In cases like this, we can apply the derivative formula in reverse. Since
0 f(x) ▯ f(a)
f (a) = lim
x!a x ▯ a
it follows that when x is \close to" a, we have
f(x) ▯ f(a)
f (a) ▯
x ▯ a
Now we get rid of the fraction and obtain f(x) ▯ f(a) ▯ f (a)(x ▯ a) and then
moving f(a) to the other side, we have
f(x) ▯ f(a) + f (a)(x ▯ a)
De▯nition 2. Let f(x) be a di▯erentiable function, and suppose that x ▯ a is
small. Given f(a) and f (a), we call f(a) + f (a)(x ▯ a) the linear approxi-
mation of f(x).
We will postpone the question of how good the approximation after some
Example 3. pet us ▯nd an approximate value for 101. We have a = 100;x =
101;f(x) = x;f (x) = 1=2 x. So f(a) = 10 and f (a) = 1=20. Therefore, by
the linear approximation formula, we have
p 1 1
101 ▯ 10 + (101 ▯ 100) = 10 + = 10:05
Checking this on a calculator gives101 ▯ 10:04988, a di▯erence of 0:00012.
pxample 4. Let us ▯nd an approximate value forp 30, using the knowledge that
532 = 2. We have a = 32;x = 30;f(x) = 5x;f (x) = 1=5x 4=5. So f(a) = 2
and f (a) = 1=(5 ▯ 32 ) = 1=(5 ▯ 16) = 1=80. Therefore we have
5 1 1
30 ▯ 2 + (30 ▯ 32) = 2 ▯ = 1:975
Checking this on a calculator gives 530 ▯ 1:97435, a di▯erence of 0:00065.
Example 5. Let us ▯nd an approximate value for e 0:1. We know that e = 1, so
x 0 x 0
we use a = 0, f(x) = e , f (x) = e , and f(a) = f (a) = 1. We use x = 0:1. So
e0:1▯ 1+1▯(0:1▯0) = 1:1. Checking this on a calculator gives e 0:1▯ 1:10517,
a di▯erence of 0:00517. Observe that this is much larger than the di▯erence
observed in the previous two examples, even though here x ▯ a = 0:1 whereas
in the previous examples it was 1 and 2.
Example 6. Let us ▯nd an approximate value for ln1:2. We know that ln1 = 0
and if f(x) = lnx then f (x) = 1=x; we have a = 1;x = 1:2;f(a) = 0;f (a) = 1.
So ln1:2 ▯ 0 + 1 ▯ (1:2 ▯ 1) = 0:2. The actual answer, based on a calculator, is
ln1:2 ▯ 0:18232, a di▯erence of 0:01768, again a relatively large di▯erence.
Note that in the case of e:1, the approximation is an underestimate, whereas
in all other cases it is an overestimate. This can be seen to be a matter of
concavity. Consult the following graph:
There are tangents at two points. At the point on the left, the curve is concave
down, and the tangent is above the graph. At the point on the right, the curve
is concave up, and the tangent is below the graph. Just as the process of taking
the derivative is analogous to approximating the tangent line by secant lines, so
is linear approximation analogous to approximating secant lines with tangent
lines. It is generally true, and can be proven, that if f (a) > 0 then near
a linear approximation will give underestimates, and if f (a) < 0 then linear
approximation will give overestimates. If a is an in
ection point, then linear
approximation will give underestimates when f is concave up between a and x
and overestimates when it is concave down.
It remains to be seen how good the approximation is. We have,
2 Proposition 7. Let f(x) be twice di▯erentiable, and let M be the maximum
value of jf (x)j for x between c and a. Then
0 (c ▯ a) 2
jf(c) ▯ (f(a) + f (a)(c ▯ a))j ▯ M
Proof. (For completeness) We will assume that c > a and that f(c) ▯ f(a) +
f (a)(c ▯ a); the other cases are similar. We put M = maxff (x) : a ▯ x ▯ cg.
We will compare the function f(x) with a quadratic function based on f(a),
f (a), and M; the quadratic function will necessarily have worse approximation.
Lemma 8. Let f(x) and g(x) be di▯erentiable, with f(a) = g(a), and f (x) ▯
g (x) whenever a ▯ x ▯ b. Then f(x) ▯ g(x) whenever a ▯ x ▯ b.
Proof. The function g(x) ▯ f(x) satis▯es g(a) ▯ f(a) = 0 and g (x) ▯ f (x) ▯ 0 0
when a ▯ x ▯ b. Then g(x)▯f(x) is an increasing function when a ▯ x ▯ b, so
g(x) ▯ f(x) ▯ 0 when a ▯ x ▯ b, or in other words f(x) ▯ g(x).
Note that we can apply the lemma twice: if f(x) and g(x) are twice di▯er-
entiable, f(a) = g(a) and f (a) = g (a), and f (x) ▯ g (x), then by the lemma
f (x) ▯ g (x) and then again f(x) ▯ g(x). So let g(x) = M(x▯a) =2+f (a)(x▯ 2 0
a) + f(a). We have g(a) = f(a) since the terms with x ▯ a vanish; we also have
g (x) = M(x ▯ a) + f (a) so g (a) = f (a). Finally, g (x) = M ▯ f (x) for 00
a ▯ x ▯ c by the de▯nition of M. So by applying the lemma twice, we get that
g(c) ▯ f(c).
Now, f(c)▯(f(a)+f (a)(c▯a)) ▯ 0 by assumption. In the other direction,
f(c) ▯ g(c) = M(c▯a) =2+f (a)(c▯a)+f(a), so f(c)▯(f(a)+f (a)(c▯a)) ▯ 0
M(c ▯ a) =2 as required.
Let us return to the four examples above. In the case of 101, we have
f (x) = ▯1=4x 3=2 , and this attains its maximum absolute value when x is
minimized, i.e. at x = a = 100; we compute M = jf (a)j = 1=(4 ▯ 100 ) =
1=4000. We also have x▯a = 1. So the error bound is M(x▯a) =2 = 1=8000 = 2
0:000125. This is quite close to the actual error of 0:00012. In the case of ln1:2,
we have f (x) = ▯1=x , and this against attains its maximum absolute value
when x is minimized, i.e. at x = a = 1; we compute M = jf (a)j = 1, and then
(x ▯ a) = 0:2, so the error bound is 1 ▯ 0:2 =2 = 0:02, which is also fairly close
to the actual error of 0:01768.
Unfortunately, in the other two cases the maximum absolute value of f is
not at a, which means the computations can get more involved. In the case of
p5 00 9=5
30, we have f (x) = ▯4=25x , whose maximum absolute value is when x
is minimized, i.e. x = 30. Instead of trying to compu