Linear and Quadratic Approximations

Unit 2: Applications of Differentiation
Today, we’ll be using differentiation to make approximations.
Linear Approximation
y=f(x)
y = b+a(x-x0y )
x
b = f(x0) ;
x0 ,f(x0 ( ))
a = f’(x0 )
Figure 1: Tangent as a linear approximation to a curve
The tangent line approximates f(x). It gives a good approximation near the tangent point x0.
As you move away from x0, however, the approximation grows less accurate.
f(x) ? f(x0) + f?(x0)(x ? x0)
Example 1. f(x) = ln x, x0 = 1 (basepoint)
1
f(1) = ln 1 = 0; f?(1) = = 1
x x=1
ln x
Change the basepoint:
Basepoint u0 = x0 ? 1 = 0.
? f(1) + f?(1)(x ? 1) = 0 + 1 · (x ? 1) = x ? 1
x = 1 + u =? u = x ? 1
ln(1 + u) ? u
Basic list of linear approximations
In this list, we always use base point x0 = 0 and assume that |x| << 1.
1. sin x ? x (if x ? 0) (see part a of Fig. 2)
2. cos x ? 1 (if x ? 0) (see part b of Fig. 2)
3. ex ? 1 + x (if x ? 0)
4. ln(1 + x) ? x (if x ? 0)
5. (1 + x)r ? 1 + rx (if x ? 0)
Proofs
Proof of 1: Take f(x) = sin x, then f?(x) = cos x and f(0) = 0
f?(0) = 1, f(x) ? f(0) + f?(0)(x ? 0) = 0 + 1.x
So using basepoint x0 = 0, f(x) = x. (The proofs of 2, 3 are similar. We already proved 4 above.)
Proof of 5:
f(x) = (1 + x)r; f(0) = 1
f?(0) = d (1 + x)r
x=0 = r(1 + x)r?1
x=0 = r
dx
| |
f(x) = f(0) + f?(0)x = 1 + rx