But a2x2cannot be negative for x > 0 and positive for x < 0, as required; we cannot improve upon the tangent line approximation by using a second degree polynomial.. We need at least a t
Trang 1y
x
x
y = x
y = x
f (x) = sin x
f (x) = sin x
–1 1
.1 1
1
(a)
(b) sin 1
Figure 30.1
Refining the tangent line approximation:Any polynomial approximation, Pk(x), of sin x about x = 0 certainly ought to be as good a local approximation as is the tangent line approximation, P1(x) = x Therefore, like sin x the graph of Pk(x)must pass through (0, 0) and must have a slope of 1 at x = 0 This means that
Pk(0) = 0 and
Pk(0) = 1
Pk(x) = a0+ a1x + a2x2+ · · · + akxk so Pk(0) = a0= 0
Pk(x) = a1+ 2a2x + 3a3x2+ · · · + kakxk−1 so Pk(0) = a1= 1
Therefore, Pk(x)is of the form x + a2x2+ a3x3+ · · · + akxk
Note that sin x lies below the tangent line for x > 0 and above the tangent line for x < 0 Therefore the approximation sin x ≈ x must be decreased for x > 0 and increased for x < 0
in order to improve upon it
Second Degree Approximation
P2(x)must be of the form x + a2x2, where a2is a constant But a2x2cannot be negative for x > 0 and positive for x < 0, as required; we cannot improve upon the tangent line approximation by using a second degree polynomial We need at least a third degree polynomial to improve upon the tangent line approximation
Trang 2Before moving on, let’s look at the second degree polynomial approximation from a geometric viewpoint If P2(x) = a0+ a1x + a2x2is to be the best parabolic approximation
to f (x) = sin x about x = 0, then it must satisfy the following three conditions
It has the same value as sin x at x = 0 P2(0) = f (0)
It has the same slope as sin x at x = 0 P2(0) = f(0)
It has the same concavity as sin x at x = 0 P2(0) = f(0) Each one of these conditions determines the value of one coefficient of P2(x) The first two result in a0= 0 and a1= 1, respectively The second derivative of sin x at x = 0 is zero
d2
dx2sin x
x=0
dx cos x
x=0
= − sin x
x=0
= 0 The “parabola” must have a second derivative of zero; consequently, it is not a parabola at all
Third Degree Approximation
To determine the coefficients of the third degree polynomial of best fit, we require that the polynomial, P3(x) = a0+ a1x + a2x2+ a3x3, and f (x) = sin x agree at x = 0 and that each nonzero derivative of the polynomial is equal to the corresponding derivative of sin x
at x = 0 These four conditions determine the four coefficients
P3(0) = f (0)
P
3(0) = f(0)
P3(0) = f(0)
P3(0) = f(0)
We have already demonstrated that the first two conditions result in a0= 0, a1= 1 As
an exercise, show that the third and fourth conditions require that a2= 0 and a3= −16, respectively
P3(x) = 0 + x + 0x2−1
6x
3
6x
3
Notice that the −16x3 term is negative for x > 0 and positive for x < 0, providing an appropriate adjustment to the tangent line approximation (See Figure 30.2.)
y
x
2π –2π
P1(x) = x
P3(x) = x –
f (x) = sin x
x3 6
Figure 30.2
EXERCISE 30.1 Let f (x) = sin x and P3(x) = a0+ a1x + a2x2+ a3x3 Calculate P(x), P(x), P(x),
f(x), f(x), f(x), and evaluate each at x = 0 Show that the four conditions given above
Trang 3determine a0, a1, a2, and a3, respectively and that
a0= 0, a1= 1, a2= 0, and a3= −1
6. Using the third degree polynomial to approximate sin 0.1 gives
sin 0.1 ≈ P3(0.1) = 0.1 −(0.1)
3
6000= 0.09983
This matches the calculator estimate of sin 0.1 to six decimal places The actual value is a bit larger than P3(0.1)
Higher Degree Approximations
To find the kth degree polynomial approximation we require that Pk(x)and sin x agree at
x = 0 and each nonzero derivative of the polynomial matches that of sin x
The condition that the fourth derivatives agree ends up meaning that a4= 0, so we will proceed directly with the fifth degree polynomial
Let P5(x) = a0+ a1x + a2x2+ a3x3+ a4x4+ a5x5 Requiring that all nonzero deriva-tives of P5(x)match the derivatives of f (x) = sin x at x = 0 means the following conditions must be satisfied
P5(0) = f (0)
P5(0) = f(0)
P
5(0) = f(0)
P5(0) = f(0)
P5(4)(0) = f(4)(0) where f(p)denotes the pth derivative of f
P5(5)(0) = f(5)(0)
As an exercise, show that these conditions determine a0, a1, a2, a3, a4, and a5, respectively, and that
P5(x) = x − 1
6x
3
120x
5
EXERCISE 30.2 Let f (x) = sin x and P5(x)be the fifth degree polynomial given above Show that the six
conditions stated mean that
a0= f (0), a1= f(0), a2=f
(0)
(0)
(4)(0)
(5)(0)
where n! = n · (n − 1) · · · 3 · 2 · 1 Conclude that
a0= 0, a1= 1, a2= 0, a3=−1
3! = −1
120. Using the fifth degree polynomial approximation to sin x to approximate sin(0.1) gives sin 0.1 ≈ P5(0.1) = 0.1 − (0.1)
3
5
12 · 106≈ 0.0998334166 This agrees with the 10 decimal places given for sin 0.1
Trang 4x
2π –2π
P1(x) = x
P3(x) = x –
f (x) = sin x
x3 6
P5(x) = x – x +
3
6
x5 120
Figure 30.3
The graphs of sin x, P1(x), P3(x), and P5(x)are given in Figure 30.3
EXERCISE 30.3 Using a computer or graphing calculator, graph f (x) = sin x, P1(x) = x, P3(x) = x −x3!3,
and P5(x) = x −x3!3 +x5!5 Zoom in around x = 0 and observe how well each polynomial approximates the values of sin x near x = 0 Try to guess formulas for P7(x), P9(x), and
P11(x) Graph these as well and decide how much confidence you have in your answers Below is a table of values given to 10 decimal places
−0.2 −0.1986693308 −0.2 −0.1986666667 −0.1986693333
−0.1 −0.0998334166 −0.1 −0.0998333333 −0.0998334167
OBSERVATIONSFrom the graphical and numerical data gathered we observe that
i for a fixed x near zero, the higher the degree of the polynomial approximation the better its value approximates that of sin x, and
ii the higher the degree of the polynomial, the further away from zero the approximation
is reasonable
NOTEIn all the work we’ve done with sin x, x must be in radians, not in degrees.dxd sin x = cos x only for x in radians
Taylor Polynomial Approximations
In the previous example we constructed polynomial approximations to f (x) = sin x around
x = 0 by choosing the coefficients of the polynomial such that the polynomial and all its nonzero derivatives matched f and its corresponding derivatives at x = 0 This method of constructing polynomial approximations to a function f about a number x = b in its domain
is remarkably useful
Trang 5Let f be a function whose first n derivatives exist at x = b For the sake of simplicity,
we begin with the case b = 0
D e f i n i t i o n
The nth degree polynomial, Pn(x), that is equal to f (0) when evaluated at x = 0 and whose first n derivatives are equal to those of f (x) when evaluated at x = 0 is called
the nth degree Taylor polynomial generated by f at x = 0 The polynomial is said
to be centered at x = 0, or expanded about x = 0.
More generally, we can expand a function about x = b using a polynomial in powers of (x − b)
Pn(x) = a0+ a1(x − b) + a2(x − b)2+ a3(x − b)3+ · · · + an(x − b)n
D e f i n i t i o n
The nth degree polynomial in powers of (x − b) that is equal to f (b) when evaluated
at x = b and whose first n derivatives match those of f (x) at x = b is called the nth
degree Taylor polynomial generated by f at x = b We refer to b as the center of
the polynomial
When evaluated at its center, a Taylor polynomial is equal to the value of its generating function Our hope is that for x near the center the value of the polynomial is close to the value of the function.1
We now turn our attention to computing Taylor polynomials In the next section we will look at the accuracy of Taylor polynomial approximations, and subsequently will see what we get by allowing the degree of the Taylor polynomial to increase without bound
Computing a Taylor Polynomial Centered at x = 0
Suppose f and its first n derivatives exist at x = 0 We want to find constants a0, a1, a2, ,
ansuch that
Pn(x) = a0+ a1x + a2x2+ · · · + anxn
is the Taylor polynomial generated by f about x = 0
We impose the following (n + 1) conditions; each enables us to solve for one coeffi-cient
Pn(0) = f (0)
P
n(0) = f(0)
Pn(0) = f(0)
Pn(0) = f(0)
Pn(n)(0) = f(n)(0)
(30.1)
1
Trang 6In short, Pn(k)(0) = f(k)(0) for k = 0, 1, , n.2
We begin by finding the first n derivatives of Pn(x)
Pn(x) = a0+ a1x + a2x2 + a3x3 + a4x4 + · · · + anxn
Pn(x) = a1 + 2a2x + 3a3x 2 + 4a4x3 + · · · + nanxn−1
P
n(x) = 2 · a2+ 3 · 2 · a3x + 4 · 3 · a4x2
+ · · · + n(n − 1) · anxn−2
P
n (x) = 3 · 2 · a3 + 4 · 3 · 2 · a4x + · · · + n(n − 1)(n − 2) · anxn−3
Pn(4)(x) = 4 · 3 · 2 · a4 + · · · + n(n − 1)(n − 2)(n − 3) · anxn−4
Pn(n−1)(x) = (n − 1)(n − 2)(n − 3) · · · 3 · 2 · an−1+ n(n − 1)(n − 2) · · · 3 · 2 · anx
Pn(n)(x) = n(n − 1)(n − 2) · · · 3 · 2 · an
Next we evaluate each expression at x = 0
Pn(0) = a0
Pn(0) = a1
Pn(0) = 2 · a2
Pn(0) = 3 · 2 · a3= 3!a3
Pn(4)(0) = 4!a4
Pn(n−1)(0) = (n − 1)!an−1
Pn(n)(0) = n!an
We summarize: Pn(k)(0) = k!akfor k = 0, 1, , n Returning to (30.1), the original (n + 1) conditions, we obtain3
k!ak= f(k)(0) for k = 0, 1, , n
Solving for ak, the coefficient of xk, we obtain ak=f(k)k!(0)for k = 0, 1, , n We summa-rize our result
The nth degree Taylor polynomial generated by f (x) at x = 0 is given by
Pn(x) = f (0) + f(0)x +f
(0)
2
(0) 3! + · · · +f
(n)(0)
n
That is, Pn(x) =
n
k=0
f(k)(0)
k
This work behind us, we compute the nth degree Taylor polynomial generated by f about x = 0 as follows
2 Here we use the convention that P 0
(x) = P (x).
3 Recall: 0! = 1 and f (0) (x) = f (x).
Trang 71 Compute the first n derivatives of f
Be alert to the possibility of patterns emerging You improve your chances of noticing patterns by not multiplying out For instance, 5 · 4 · 3 · 2 is easier to recognize as 5! than is120
2 Evaluate f and each of its derivatives at x = 0.
3 The coefficient of xkis the constantf(k)k!(0)
EXAMPLE 30.2 Find the nth degree Taylor polynomial generated by exat x = 0
SOLUTION
Pn(x) = f (0) + f(0)x + f
(0)
2
+ · · · +f
(n)(0)
n
The derivative of exis ex; therefore f(k)(0) = e0= 1 for k = 0, 1, 2, , n Thus
Pn(x) = 1 + x +x
2
2! +x
3
3! + · · · + x
n
Graphs of exand several of its Taylor polynomials are shown in Figure 30.4 Note that
P1(x) = 1 + x is simply the tangent line approximation to exat x = 0
y
y
x
x
P4
P2
P3
P4
P2
P1
P1(x)
P1(x)
P1(x) = 1 + x
P2(x) = 1 + x +
P3(x) = 1 + x +
P4(x) = 1 + x +
P2(x)
P3(x)
P4(x)
e x
P3
f (x) = e x
f (x) = e x
15 10 5
5
–5 1
1
(a)
(b)
Magnification around x = 0
x2
2!
x2
2!
x3
3!
x2
2!
+
x3
3! + + x4!4
Figure 30.4
On the following page is a table of values produced using Taylor polynomials for ex Values are given to nine decimal places
Trang 8x ex P1(x) P2(x) P3(x) P4(x) P5(x)
0.1 1.105170918 1.1 1.105 1.105166666 1.105170833 1.105170917 0.2 1.221402758 1.2 1.22 1.221333333 1.221400000 1.221402667 0.5 1.648721271 1.5 1.625 1.645833333 1.6484375 1.648697917
EXAMPLE 30.3 Find the 8th degree Taylor polynomial generated by f (x) = cos x about x = 0.
SOLUTION P8(x) = f (0) + f(0)x +f2!(0)x2+ · · · +f(8)8!(0)x8
f(8)(x) = cos x f(8)(0) = 1
P8(x) = 1 −x
2
2! +x
4
4! −x
6
6! +x
8
8!
Notice that the coefficients of all the odd power terms are zero This makes sense; cosine is
an even function Analogously, the coefficients of all even power terms in the expansion of sin x about x = 0 are zero since sin x is an odd function
The graph of cos x and some of its Taylor polynomials centered at x = 0 are given in Figure 30.5 (Graph them yourself and you can zoom in around x = 0.)
P2(x) = 1 –
P4(x) = 1 –
P6(x) = 1 –
P2(x) P6(x)
f (x) = cos x
P8(x)
x2
2!
x2
2!
x4
4!
x2
2!
+
x4
4!
+ – x6!6
P8(x) = 1 – x2!2 x4
4!
6!
8! +
2π –2π
y
x
Figure 30.5
Computing a Taylor Polynomial Centered at x = b
Suppose we want to approximate ln 1.2 using a Taylor polynomial We can’t use a Taylor polynomial for f (x) = ln x expanded about x = 0 because neither f nor any of its deriva-tives exist at x = 0 We can, however, either center the Taylor polynomial for ln(1 + x) at
x = 0 or work with the Taylor polynomial for ln x expanded about x = 1 We will do the latter First we will look at how to compute a Taylor polynomial centered at x = b Recall that the nth degree Taylor polynomial for f (x) at x = b is an nth degree polynomial in powers of x − b,
Pn(x) = a0+ a1(x − b) + a2(x − b)2+ · · · + an(x − b)n,
Trang 9such that the values of Pn(x) and its nonzero derivatives are equal to those of f (x) when evaluated at x = b That is, the coefficients a0, a1, a2, , anare determined by the conditions
Pn(k)(b) = f(k)(b) for k = 0, 1, 2, , n (30.2)
As an exercise, calculate the first n derivatives of Pn(x) (Don’t multiply out (x − b)k; use the Chain Rule.) Evaluating these derivatives at a = b, conclude that
Pn(k)(b) = k!ak This result, together with (30.2) enables us to solve for ak, k = 0, 1, 2, , n
ak=f
(k)(b) k!
The nth degree Taylor polynomial generated by f (x) around x = b is given by
Pn(x) = f (b) + f(b)(x − b) +f
(b) 2! (x − b)2+ · · · +f
(n)(b) n! (x − b)n That is,4Pn(x) =
n
k=0
f(k)(b) k! (x − b)k
Given a particular function, f , and center, b, we compute the first n derivatives of f , evaluate each at x = b, and usef(k)k!(b)as the coefficient of (x − b)k
Note:
The equation of the line tangent to f at x = b is of the form y − y1= m(x − x1), where (x1, y1) = (b, f (b)) and m = f(b) The equation is therefore y = f (b) + f(b)(x − b); this is P1(x), as expected
When b = 0 we’re back to the Taylor polynomial centered at x = 0
EXAMPLE 30.4 (a) Find the nth degree Taylor polynomial for ln x centered at x = 1.
(b) Use P5to estimate ln(1.2)
SOLUTION (a) Pn(x) = f (1) + f(1)(x − 1) + f2!(1)(x − 1)2+f3!(1)(x − 1)3+ · · · +
f (n) (1) n! (x − 1)n
4 In order to use this summation notation we must adopt the convention that (x − b) 0 = 1 even if x = b.
Trang 10Compute derivatives of ln x, looking for a pattern.
f(5)(x) = 4 · 3 · 2 · x−5 f(5)(1) = 4 · 3 · 2 = 4!
f(n)(x) = (−1)n+1(n − 1)!x−n f(n)(1) = (−1)n+1(n − 1)!
Pn(x) = 0 + 1(x − 1) +−1
2!(x − 1)2+2!
3!(x − 1)3+−3!
4! (x − 1)4+
· · · +(−1)
n+1(n − 1)!
Pn(x) = (x − 1) −(x − 1)
2
3
4
4 + · · · + (−1)n+1(x − 1)
n
More compactly,
Pn(x) =
n
k=0
f(k)(1) k! (x − 1)k=
n
k=0
(−1)k+1(k − 1)!
=
n
k=0
(−1)k+1 (k − 1)!
k · (k − 1)!(x − 1)
k
=
n
k=0
(−1)k+1(x − 1)
k
(b)
ln x ≈ P5(x) = (x − 1) −(x − 1)
2
3
4
5
5
ln 1.2 ≈ P5(1.2) = 0.2 − (0.2)
2
3
4
5
Compare this with the actual value of ln 1.2; it matches for the first four decimal places
Aside: Dealing with Factorials and Alternating Signs
Factorials: Parentheses are important
(2n)! = (2n) · (2n − 1) · (2n − 2) · · · 3 · 2 · 1
= 2n · (2n − 1)!
2n + 1! = 2n + 1