Sections 10.1–3Taylor Polynomials and Series
Math E-16
December 19, 2007
Announcements
I Matthew Leingang ([email protected]) at yourservice
I Homework for next time:I 10.1: #2,4,5,7,8,10-13,18,26-28,34I 10.2: #1,6,16,17,21,22,28,32,40I 10.3: #8-12,20,22,36
Taylor PolynomialsMotivationDerivationExamples
Taylor SeriesDefinitionPower Series and the Convergence IssueFamous Taylor SeriesNew Taylor Series from Old
Motivation
I What is sin(91◦)?
I What is√
4.001?
How does your calculator work?
Suppose you ask your calculator for sin(31◦).
I Does it construct a right triangle with one angle equal to 31◦,then measure opposite-over-hypotenuse?
I Does it construct a unit circle, measure the arc length equalto 31◦, then use the vertical coordinate?
I No, it uses a polynomial approximation. Deep down,calculators can only add and multiply anyway.
Constant Approximation
If f is continuous at a, then
f (x) ≈ f (a)
when x is “close to” a.
Example
I√
4.001 ≈√
4 = 2
I sin(91◦) ≈ sin(90◦) = 1
But we should be able to do better.
Linear Approximation
How can we approximate a function with a line?
DefinitionLet f be a function and a apoint at which f isdifferentiable. Then the linearapproximation to f at a is
L(x) = f (a) + f ′(a)(x − a)
•
Linear Approximation
How can we approximate a function with a line?
DefinitionLet f be a function and a apoint at which f isdifferentiable. Then the linearapproximation to f at a is
L(x) = f (a) + f ′(a)(x − a)
•
Example
Estimate these by linear approximation.
(i)√
4.001
(ii) sin(91◦)
Solution (i)
We use f (x) =√
x, a = 4. Then f ′(4) = 14 , so
L(x) = 2 + 14(x − 4)
This means √4.001 ≈ 2 + 1
4 ·1
1000 = 2.00025
Notice(2.00025)2 = 4.001000063
Example
Estimate these by linear approximation.
(i)√
4.001
(ii) sin(91◦)
Solution (i)
We use f (x) =√
x, a = 4. Then f ′(4) = 14 , so
L(x) = 2 + 14(x − 4)
This means √4.001 ≈ 2 + 1
4 ·1
1000 = 2.00025
Notice(2.00025)2 = 4.001000063
Solution (ii)
We use f (x) = sin x, a = 90◦ = π/2. Then f (a) = 1, f ′(x) = cos x,so f ′(a) = 0. This means
L(x) = 1
so the linear approximation is no better than the constant one.
Quadratic Approximation
How can we approximate a function with a parabola?
We arelooking for a function
Q(x) = c0 + c1(x − a) + c2(x − a)2
with Q(a) = f (a), Q ′(a) = f ′(a), Q ′′(a) = f ′′(a). What arec0, c1, c2 in terms of f ?
I Since Q(a) = c0, we need c0 = f (a).
I SinceQ ′(x) = c1 + 2c2(x − a) =⇒ Q(a) = c1
we need c1 = f ′(a).
I SinceQ ′′(x) = 2c2 = Q(a)
we need c2 = 12 f ′′(a).
Quadratic Approximation
How can we approximate a function with a parabola? We arelooking for a function
Q(x) = c0 + c1(x − a) + c2(x − a)2
with Q(a) = f (a), Q ′(a) = f ′(a), Q ′′(a) = f ′′(a). What arec0, c1, c2 in terms of f ?
I Since Q(a) = c0, we need c0 = f (a).
I SinceQ ′(x) = c1 + 2c2(x − a) =⇒ Q(a) = c1
we need c1 = f ′(a).
I SinceQ ′′(x) = 2c2 = Q(a)
we need c2 = 12 f ′′(a).
Quadratic Approximation
How can we approximate a function with a parabola? We arelooking for a function
Q(x) = c0 + c1(x − a) + c2(x − a)2
with Q(a) = f (a), Q ′(a) = f ′(a), Q ′′(a) = f ′′(a). What arec0, c1, c2 in terms of f ?
I Since Q(a) = c0, we need c0 = f (a).
I SinceQ ′(x) = c1 + 2c2(x − a) =⇒ Q(a) = c1
we need c1 = f ′(a).
I SinceQ ′′(x) = 2c2 = Q(a)
we need c2 = 12 f ′′(a).
DefinitionLet f be a function and a apoint at which f is twicedifferentiable. Then thequadratic approximation tof at a is
Q(x) = f (a)+f ′(a)(x−a)+12 f ′′(a)(x−a)2
•
Example
Estimate these by quadratic approximation.
I√
4.001
I sin(91◦)
Solution (i)
We use f (x) =√
x, a = 4. Then f ′(4) = 14 and f ′′(4) = − 1
32 , so
Q(x) = 2 + 14(x − 4)− 1
64(x − 4)2
This means
√4.001 ≈ 2 + 1
4 ·1
103 − 164
1106 = 2.000249984
This is the same answer my TI-83 gives for√
4.001.
Example
Estimate these by quadratic approximation.
I√
4.001
I sin(91◦)
Solution (i)
We use f (x) =√
x, a = 4. Then f ′(4) = 14 and f ′′(4) = − 1
32 , so
Q(x) = 2 + 14(x − 4)− 1
64(x − 4)2
This means
√4.001 ≈ 2 + 1
4 ·1
103 − 164
1106 = 2.000249984
This is the same answer my TI-83 gives for√
4.001.
Solution (ii)
We use f (x) = sin x, a = 90◦ = π/2. Then f (a) = 1, f ′(a) = 0,and f ′′(x) = − sin x, so f ′′(a) = −1. This means
Q(x) = 1− 12(x − π)2
sosin(91◦) ≈ 1− 1
2
(π
180
)2 ≈ 0.9998476913
My TI-83 hassin(91◦) = 0.9998476952
which this agrees with up to the ninth place.
In General
How can we approximate a function with a polynomial of degree n?
DefinitionLet f be a function and a a point at which f is n timesdifferentiable. The Taylor Polynomial of degree n for f centeredat a is
Pn(x) = f (a) + f ′(a)(x − a) +f ′′(a)
2!(x − a)2 + · · ·+ f (n)(a)
n!(x − a)n
=n∑
k=0
f (k)(a)
k!(x − a)k
(Convention: f (0)(x) = f (x))
In General
How can we approximate a function with a polynomial of degree n?
DefinitionLet f be a function and a a point at which f is n timesdifferentiable. The Taylor Polynomial of degree n for f centeredat a is
Pn(x) = f (a) + f ′(a)(x − a) +f ′′(a)
2!(x − a)2 + · · ·+ f (n)(a)
n!(x − a)n
=n∑
k=0
f (k)(a)
k!(x − a)k
(Convention: f (0)(x) = f (x))
Taylor PolynomialsMotivationDerivationExamples
Taylor SeriesDefinitionPower Series and the Convergence IssueFamous Taylor SeriesNew Taylor Series from Old
Take it to the Limit
DefinitionLet f be a function and a a point at which f is infinitelydifferentiable. The Taylor Series of for f centered at a is
T (x) = f (a) + f ′(a)(x − a) +f ′′(a)
2!(x − a)2+
· · ·+ f (n)(a)
n!(x − a)n + . . .
=∞∑
k=0
f (k)(a)
k!(x − a)k
Example
Compute the Taylor Series for f (x) = ln x centered at 1.
Solution
f (x) = ln x f (1) = 0
f ′(x) = x−1 f ′(1) = 1
f ′′(x) = −x−2 f ′′(1) = −1
f ′′′(x) = 2x−3 f ′′′(1) = 2
f (4)(x) = −6x−3 f (4)(1) = −6
. . . . . . . . . . . .
f (k)(x) = (−1)k+1(k − 1)!x−k f (k)(1) = (−1)k+1(k − 1)!
So
T (x) =∞∑
k=1
(−1)k+1(k − 1)!
k!(x − 1)k =
∞∑k=1
(−1)k+1
k(x − 1)k
Example
Compute the Taylor Series for f (x) = ln x centered at 1.
Solution
f (x) = ln x f (1) = 0
f ′(x) = x−1 f ′(1) = 1
f ′′(x) = −x−2 f ′′(1) = −1
f ′′′(x) = 2x−3 f ′′′(1) = 2
f (4)(x) = −6x−3 f (4)(1) = −6
. . . . . . . . . . . .
f (k)(x) = (−1)k+1(k − 1)!x−k f (k)(1) = (−1)k+1(k − 1)!
So
T (x) =∞∑
k=1
(−1)k+1(k − 1)!
k!(x − 1)k =
∞∑k=1
(−1)k+1
k(x − 1)k
Caution
The infinite sum is dangerous!
I Sometimes it gives very good approximations quickly
I Sometimes it gives good approximations slowly
I Sometimes it doesn’t give anything.
To see this, let Pn be the nth degree Taylor Polynomial off (x) = ln x centered at 1. For which x is T (x) = lim
n→∞Pn(x)
meaningful?
n Pn(12) Pn(2) Pn(3)
1 -0.5 1. 2.02 -0.625 0.5 0.03 -0.666667 0.833333 2.666674 -0.682292 0.583333 -1.333335 -0.688542 0.783333 5.066676 -0.691146 0.616667 -5.67 -0.692262 0.759524 12.68578 -0.69275 0.634524 -19.31439 -0.692967 0.745635 37.5746
10 -0.693065 0.645635 -64.8254
n Pn(12) Pn(2) Pn(3)
10 -0.693065 0.645635 -64.825420 -0.693147 0.668771 -34359.730 -0.693147 0.676758 -2.3593×107
40 -0.693147 0.680803 -1.81712×1010
50 -0.693147 0.683247 -1.49113×1013
60 -0.693147 0.684883 -1.27387×1016
70 -0.693147 0.686055 -1.11899×1019
80 -0.693147 0.686936 -1.00322×1022
90 -0.693147 0.687622 -9.13584×1024
100 -0.693147 0.688172 -8.42274×1027
n Pn(12) Pn(2) Pn(3)
100 -0.693147 0.688172 -8.42274×1027
200 -0.693147 0.690653 -5.34752×1057
300 -0.693147 0.691483 -4.52171×1087
400 -0.693147 0.691899 -4.30016×10117
500 -0.693147 0.692148 -4.36161×10147
600 -0.693147 0.692315 -4.60801×10177
700 -0.693147 0.692433 -5.00727×10207
800 -0.693147 0.692523 -5.55436×10237
900 -0.693147 0.692592 -6.25895×10267
1000 -0.693147 0.692647 -7.14101×10297
Observations
I Pn(12) converges to f (1
2) = − ln 2
I Pn(2) converges to f (2) = ln 2, but more slowly
I Pn(3) does not converge at all!
So in examining this process of approximation by polynomials, wehave to be a little bit careful about what numbers we plug in.
Definition (cf. §9.5)
A power series centered at a is a sum of constants times powersof (x − a):
c0 + c1(xa) + c2(x − a)2 + · · ·+ cn(x − a)n + . . .
Theorem
For a given power series∞∑
n=1
cn(x − a)n there are only three
possiblities:
1. There is a number R such that the series converges when|x − a| < R and diverges when |x − a| > R.
2. The series converges for all x (R =∞)
3. The series converges only when x = a (R = 0).
R is called the radius of convergence of the power series.
Why radius?
An open interval is kind of like a one-dimensional circle:
a− R a a + R
convergence on (a− R, a + R)divergence on (−∞, a− R) and (a + R,∞)at the endpoints—???
Why radius?
An open interval is kind of like a one-dimensional circle:
a− R a a + R
convergence on (a− R, a + R)
divergence on (−∞, a− R) and (a + R,∞)at the endpoints—???
Why radius?
An open interval is kind of like a one-dimensional circle:
a− R a a + R
convergence on (a− R, a + R)
divergence on (−∞, a− R) and (a + R,∞)
at the endpoints—???
Why radius?
An open interval is kind of like a one-dimensional circle:
a− R a a + R
convergence on (a− R, a + R)divergence on (−∞, a− R) and (a + R,∞)
at the endpoints—???
Example
Compute the radius of convergence of the power series
f (x) =∞∑
k=0
xk
Solution
This is a geometric series. We know it converges to1
1− xwhen
|x | < 1, and not when x = 1 or x = −1. So
I The radius of convergence is 1
I The interval of convergence is the open interval (−1, 1)
Example
Compute the radius of convergence of the power series
f (x) =∞∑
k=0
xk
Solution
This is a geometric series. We know it converges to1
1− xwhen
|x | < 1, and not when x = 1 or x = −1. So
I The radius of convergence is 1
I The interval of convergence is the open interval (−1, 1)
Famous Taylor Series
Example
Compute Taylor series centered at zero for the following functions:
I ex
I sin x
I cos x
I (1 + x)p
Example
Compute the Taylor series centered at zero for f (x) = ex
Solution
f (x) = ex f (0) = 1
f ′(x) = ex f ′(0) = 1
f ′′(x) = ex f ′′(0) = 1
f ′′′(x) = ex f ′′′(0) = 1
. . . . . . . . . . . .
f (k)(x) = ex f (k)(0) = 1
So
T (x) =∞∑
k=0
xk
k!
Example
Compute the Taylor series centered at zero for f (x) = ex
Solution
f (x) = ex f (0) = 1
f ′(x) = ex f ′(0) = 1
f ′′(x) = ex f ′′(0) = 1
f ′′′(x) = ex f ′′′(0) = 1
. . . . . . . . . . . .
f (k)(x) = ex f (k)(0) = 1
So
T (x) =∞∑
k=0
xk
k!
FactThe Taylor series for the function f (x) = ex converges for all x toex .
The convergence is because the factorials k! grow much fasterthan the exponentials xk . It’s a little more work to say that itconverges to ex .
FactThe Taylor series for the function f (x) = ex converges for all x toex .
The convergence is because the factorials k! grow much fasterthan the exponentials xk . It’s a little more work to say that itconverges to ex .
Example
Compute the Taylor series centered at zero for f (x) = sin x .
Solution
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 (4)(x) = cos x f (4)(0) = 1
And repeat! So
T (x) =∞∑
k=0k odd
±xk
k!=∞∑
m=0
(−1)mx2m+1
(2m + 1)!= x − x3
3!+
x5
5!− · · ·
This turns out to converge for all x to sin x.
Example
Compute the Taylor series centered at zero for f (x) = sin x .
Solution
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 (4)(x) = cos x f (4)(0) = 1
And repeat! So
T (x) =∞∑
k=0k odd
±xk
k!=∞∑
m=0
(−1)mx2m+1
(2m + 1)!= x − x3
3!+
x5
5!− · · ·
This turns out to converge for all x to sin x.
Example
Compute the Taylor series centered at zero for f (x) = cos x .
Solution
f (x) = cos x f (0) = 1
f ′(x) = − sin x f ′(0) = 0
f ′′(x) = − cos x f ′′(0) = −1
f ′′′(x) = sin x f ′′′(0) = 0
f (4)(x) = sin x f (4)(0) = 0
And repeat! So
T (x) =∞∑
k=0k even
±xk
k!=∞∑
m=0
(−1)mx2m
(2m)!= 1− x2
2!+
x4
4!− · · ·
This turns out to converge for all x to cos x.
Example
Compute the Taylor series centered at zero for f (x) = cos x .
Solution
f (x) = cos x f (0) = 1
f ′(x) = − sin x f ′(0) = 0
f ′′(x) = − cos x f ′′(0) = −1
f ′′′(x) = sin x f ′′′(0) = 0
f (4)(x) = sin x f (4)(0) = 0
And repeat! So
T (x) =∞∑
k=0k even
±xk
k!=∞∑
m=0
(−1)mx2m
(2m)!= 1− x2
2!+
x4
4!− · · ·
This turns out to converge for all x to cos x.
Example (The Binomial Series)
Compute the Taylor series centered at zero for f (x) = (1 + x)p,where p is any number (not just a whole number).
Solution
f (x) = (1 + x)p f (0) = 1
f ′(x) = p(1 + x)p−1 f ′(0) = p
f ′′(x) = p(p − 1)(1 + x)p−2 f ′′(0) = p(p − 1)
f ′′′(x) = p(p − 1)(p − 2)(1 + x)p−3 f ′′′(0) = p(p − 1)(p − 2)
. . . . . . . . . . . .
So
T (x) = 1 + px +p(p − 1)
2x2 + . . .
=∞∑
k=0
p(p − 1)(p − 2) · · · (p − k + 1)
k!xk
Example (The Binomial Series)
Compute the Taylor series centered at zero for f (x) = (1 + x)p,where p is any number (not just a whole number).
Solution
f (x) = (1 + x)p f (0) = 1
f ′(x) = p(1 + x)p−1 f ′(0) = p
f ′′(x) = p(p − 1)(1 + x)p−2 f ′′(0) = p(p − 1)
f ′′′(x) = p(p − 1)(p − 2)(1 + x)p−3 f ′′′(0) = p(p − 1)(p − 2)
. . . . . . . . . . . .
So
T (x) = 1 + px +p(p − 1)
2x2 + . . .
=∞∑
k=0
p(p − 1)(p − 2) · · · (p − k + 1)
k!xk
New Taylor Series from Old
Big Time Theorem
We can integrate and differentiate power series, and the ROC stays
the same: If f (x) =∞∑
k=0
ck(x − a)k has radius of convergence R,
that is, if
f (x) =∞∑
k=0
ck(x − a)k when |x − a| < R,
then
f ′(x) =∞∑
k=1
kck(x − a)k−1 when |x − a| < R,
and∫f (x) dx =
∞∑k=0
1
k + 1ck(x − a)k+1 + C when |x − a| < R.
This is really saying two things:
1. The power series which is differentiated (or integrated)term-by-term has the same radius of convergence as theoriginal power series.
2. It converges to the thing we want: the derivative orantiderivative of f
The other big theorem
If f has any power series representation near a, then it is equal tothe Taylor Series.
Example
Compute the Taylor series centered at 0 for arctan x .
Solution
First, we’ll find the Taylor Series for1
1 + x2. It’s geometric:
1
1 + x2=
1
1− (−x2)=∞∑
k=0
(−x2)k =∞∑
k=0
(−1)kx2k .
This converges when∣∣x2∣∣ < 1 ⇐⇒ |x | < 1, so the ROC is 1.
Now arctan is the antiderivative:
arctan x =
∫ ∞∑k=0
(−1)kx2k dx =∞∑
k=0
(−1)k
∫x2k dx =
∞∑k=0
(−1)kx2k+1
(2k + 1)
And the ROC of this 1, too.
Example
Compute the Taylor series centered at 0 for arctan x .
Solution
First, we’ll find the Taylor Series for1
1 + x2. It’s geometric:
1
1 + x2=
1
1− (−x2)=∞∑
k=0
(−x2)k =∞∑
k=0
(−1)kx2k .
This converges when∣∣x2∣∣ < 1 ⇐⇒ |x | < 1, so the ROC is 1.
Now arctan is the antiderivative:
arctan x =
∫ ∞∑k=0
(−1)kx2k dx =∞∑
k=0
(−1)k
∫x2k dx =
∞∑k=0
(−1)kx2k+1
(2k + 1)
And the ROC of this 1, too.
Cool result
This means if
∞∑k=0
(−1)k
(2k + 1)= 1− 1
3+
1
5− 1
7+ · · ·
converges (and it does), it converges to
arctan(1) =π
4
Example
I Compute the Taylor series centered at 0 for f (x) = x7 sin(x3).
I Find f (2008)(0).
Solution
x7 sin(x3) = x7∞∑
m=0
(−1)m(x3)2m+1
(2m + 1)!= x7
∞∑m=0
(−1)mx6m+3
(2m + 1)!
=∞∑
m=0
(−1)mx6m+10
(2m + 1)!
Example
I Compute the Taylor series centered at 0 for f (x) = x7 sin(x3).
I Find f (2008)(0).
Solution
x7 sin(x3) = x7∞∑
m=0
(−1)m(x3)2m+1
(2m + 1)!= x7
∞∑m=0
(−1)mx6m+3
(2m + 1)!
=∞∑
m=0
(−1)mx6m+10
(2m + 1)!
To find f (2008)(0), note
∞∑m=0
(−1)mx6m+10
(2m + 1)!=∞∑
k=0
f (k)(0)xk
k!
Equating the coefficients of x2008 will get us what we want. If6m + 10 = 2008, then m = 333. So
f (2008)(0)
2008!=
(−1)333
667!
and thus
f (2008)(0) = −2008!
667!
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