Tangent lines and Taylor polynomials
A theme in calculus is that of approximating a nonlinear function by a
linear one.
Goal =⇒ to approximate the graph of f near a with a line �(x) that
has the same value and the same slope as f does at a
The tangent line to f at a satisfies these conditions. Using the point-slope
form of a line, we can write down an equation for it:
y − y0 = m(x − x0 )
�(x) f (a) f � (a)
a
Thus, the equation for the tangent line �(x) is
�(x) = f (a) + f �(a)(x − a).
This is a function of x — f (a), f �(a), and a are numbers!
Remark For values of x near a, the tangent line approximates the
original function. In fact, it is the unique linear function that has the
same function value f (a) and derivative f �(a) at the point a, so in this
sense, it is the best linear approximation to f at a.
Example Find the tangent line to f (x) =
We have f �(x) =
line is
1
√
,
2 x
√
x near a = 100.
so f (100) = 10 and f �(100) =
�(x) = 10 +
Math 30G - Prof. Kindred - Lecture 7
1
20 .
Thus, the tangent
1
(x − 100).
20
Page 1
Example Estimate
√
103 (without a calculator).
Using our equation for the tangent line from the previous example, we
have
√
1
103 = f (103) ≈ �(103) = 10 + (103 − 100) = 10.15 .
20
√
(Note that the actual value is 103 is 10.1489...)
Question
Can we get a better approximation?
Not with a line, but we can do better if we approximate using a quadratic.
Consider
P2(x) = f (a) + f �(a)(x − a) + C(x − a)2
� �� �
�
��
� � �� �
2nd-order
�(x)=P1 (x)
linear approx.
quadratic
improvement
where C is to be determined.
We have
• P2(a) = f (a)
• P2� (a) = f �(a) since P2� (x) = f �(a) + 2C(x − a)
• We want P2��(a) = f ��(a), and since P2��(x) = C, we need to set
f ��(a)
C=
.
2
Thus,
f ��(a)
P2(x) = f (a) + f (a)(x − a) +
(x − a)2
2
is the best quadratic approximation to f at a.
�
Math 30G - Prof. Kindred - Lecture 7
Page 2
Question
Then let
What about the best cubic approximation to f at a?
f ���(a)
P3(x) = P2(x) +
(x − a)3.
� 3! ��
�
cubic improvement
Exercise: check that P3 and f have the same value at a, as well as the
same 1st, 2nd, and 3rd derivatives at a.
Let’s generalize this to a kth order polynomial.
Definition The kth order Taylor polynomial of f (x) at the point a is
f ��(a)
f (k)(a)
2
Pk (x) = f (a) + f (a) +
(x − a) + · · · +
(x − a)k .
2!
k!
This kth degree polynomial has the same value and first k derivatives as
f does at x = a.
�
Example Approximate sin x near x = 0 by a 3rd-order Taylor poly.
We have
so
f (x) = sin x =⇒ f (0) = 0,
f �(x) = cos x =⇒ f �(0) = 1,
f ��(x) = − sin x =⇒ f ��(0) = 0,
f ���(x) = − cos x =⇒ f ���(0) = −1,
f ��(0)
f ���(0)
2
P3(x) = f (0) + f (0) +
(x − 0) +
(x − 0)3
2!
3!
1
= 0 + 1 · x + 0 · x2 − · x3
6
�
x3
= x−
6
Math 30G - Prof. Kindred - Lecture 7
Page 3
Note that because of the pattern of the derivatives of sin x, we can see
that
x3 x5 x7
P7(x) = x −
+
−
3!
5!
7!
How good are these approximations?
Question
How close is Pn(x) to f (x)?
Define En(x) = f (x) − Pn(x). (We can think of this as the “error” or
“remainder”.)
Theorem (Taylor’s theorem). If f (x) is of class C n+1, then for each
x, there exists a c ∈ (a, x) such that
f (n+1)(c)
En(x) =
(x − a)n+1.
(n + 1)!
While this is an exact formula for the error, the value of c is unknown.
However, we do know that a < c < x so this helps us bound En(x).
3
Example For sin x near the point 0, we found P3(x) = x − x6 . Using
Taylor’s theorem, we have
� (4)
� �
�
� f (c)
�
�
�
sin
c
4�
4�
�
�
|E3(x)| = �
(x − 0) � = �
x
4!
4! �
| sin c| 4
=
|x|
24
|x|4
≤
since | sin c| ≤ 1.
4!
Math 30G - Prof. Kindred - Lecture 7
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Now, let’s check this error bound for the approximation of sin 1. We have
13
sin 1 ≈ 1 − = 0.83
6
and our error bound
in this case is
|E3(1)| ≤
1
= 0.0416.
24
We see that our actual error, sin 1 − 0.83 ≈ 0.0081, is less than 0.0416.
Math 30G - Prof. Kindred - Lecture 7
Page 5
3rd-order Taylor polynomial for sin(x) at 0