Math 253
Homework due Wednesday, January 13
SOLUTIONS
For each question, you need to show your work and explain how you got your answer. Correct answers
without accompanying explanation will not earn any credit.
1. For each of the functions below, compute the indicated Taylor polynomial centered at the given point:
(a) f (x) = sin(x), 7th degree polynomial, centered at x = 0
(b) f (x) = cos(x), 6th degree polynomial, centered at x = 0
(c) f (x) =
1
1+2x ,
third degree polynomial, centered at x = 0
(d) f (x) = ln(1 + x), 6th degree polynomial, centered at x = 0
(e) f (x) =
(f) f (x) =
(g) f (x) =
1
(3+x)2 ,
1
1−x ,
√
third degree polynomial, centered at x = 0
third degree polynomial, centered at x = 3
x, third degree polynomial, centered at x = 25.
Solution: Not given here. Routine calculations.
2. For each of the functions below, compute the indicated Taylor polynomial centered at the given point:
(a) f (x) = sin2 (x), third degree polynomial, centered at x = 0
Solution: sin(x) ≈ x −
x3
6
so
h
x3 i h
x3 i
x4
x6
sin2 (x) = sin(x) · sin(x) ≈ x −
· x−
= x2 −
+
≈ x2 .
6
6
3
36
In the last step, note that we are discarding terms of degree larger than three.
(b) f (x) = sin(x4 ), twelfth degree polynomial, centered at x = 0
Solution: sin(x) ≈ x −
x3
6
so
sin(x4 ) ≈ x3 −
(x4 )3
x12
= x3 −
.
6
6
(c) f (x) = xe3x , fourth degree polynomial, centered at x = 0
Solution: ex ≈ 1 + x +
x2
2
+
e3x ≈ 1 + (3x) +
Then
x3
6
+
x4
24 ,
so
(3x)2
(3x)3
(3x)4
9
9
27
+
+
= 1 + 3x + x2 + x3 + x4 .
2
6
24
2
2
8
9
9
xe3x ≈ x + 3x2 + x3 + x4 .
2
2
(d) f (x) =
x
ex ,
third degree polynomial, centered at x = 0
Solution: Start with ex ≈ 1 + x +
x2
2
+
e−x ≈ 1 + (−x) +
x3
6 ,
then get
(−x)3
x2
x3
(−x)2
+
=1−x+
−
2
6
2
6
and finally
x
x3
−x
2
=
xe
≈
x
−
x
+
.
ex
2
(e) f (x) =
ex
cos x ,
third degree polynomial, centered at x = 0
Solution: Here it is best to do long division:
1−
x2
2
+x2
1
+x
1
+x + x2
1
− x2
2
+ 32 x3
3
+ x6
2
x
+x2
3
+ x6
3
− x2
x
x2
+ 23 x3
x2
2 3
3x
The answer is
(f) f (x) =
1
1−2x ,
ex
cos x
≈ 1 + x + x2 + 23 x3 .
third degree polynomial, centered at x = 0
Solution: Start with the binomial series
−1 2
−1 3
(1 + u)−1 ≈ 1 − u +
u +
u = 1 − u + u2 − u3 .
2
3
Then substitute u = −2x to get
(1 − 2x)−1 ≈ 1 + 2x + 4x2 + 8x3 .
√
√
3. 123 = 1728. Approximate 3 1729 using a degree two Taylor polynomial for f (x) = 3 x centered at
x = 1728. How many decimal places will stay the same if you use a degree three Taylor polynomial
instead?
Solution: Compute that
f (1728) = 12,
f 0 (1728) =
1
1
=
,
3 · 144
432
and
f 000 (1728) =
f 00 (1728) =
10
≈ 0.00000001033.
27 · 127
Page 2
−2
−2
=
5
9 · 12
2239488
The degree two Taylor approximation is then
1
−1
(x − 1728) +
(x − 1728)2
432
2239488
1
x 3 ≈ 12 +
and so
√
3
1729 ≈ 12 +
1
−1
·1−
· 1 = 12.00231437.
432
2239488
For the degree three approximation, the next term will be one sixth of 0.00000001033. So at least the
first seven digits after the decimal point will be the same as in the degree two approximation.
4. Suppose you want to compute sin(0.5) using a Taylor polynomial centered at x = 0. What is the smallest
degree Taylor polynomial you can use that will give the correct answer up to eight decimal places?
Solution: The Taylor series for sine is
sin(x) ≈ x −
x3
x5
x7
+
−
+ ··· .
6
120 5040
A calculator shows that sin(0.5) = 0.4794255386042. The degree one Taylor approximation is just 0.5.
The degree three approximation is
0.5 −
(0.5)3
= 0.4791666667
6
and the degree five approximation is
0.5 −
(0.5)3
(0.5)5
+
= 0.4794270833
6
120
and the degree seven approximation is
0.5 −
(0.5)3
(0.5)5
(0.5)7
+
−
= 0.4794255332.
6
120
5040
So the degree seven Taylor approximation is the smallest degree that gives the correct answer to eight
decimal places.
R1
5. The integral 0 x cos(x3 ) dx is hard to do by hand. Computer calculations shows that it is close to
0.440408. Use the Taylor polynomials for cos(x) to approximate the above integral by a hand-calculation:
(a) Do it using the degree 2 Taylor polynomial for cos(x);
(b) Do it again using the degree 4 Taylor polynomial for cos(x).
Solution: Use cos(x) ≈ 1 −
x2
2 .
Then cos(x3 ) ≈ 1 −
x6
2
x cos(x3 ) ≈ x −
So
Z
1
3
Z
x cos(x ) dx ≈
0
0
1
x7
x−
2
!
and
x7
.
2
1
x2
x8 1
1
7
dx =
− = −
=
= 0.4375.
2
16 2 16
16
0
If we instead use the degree four approximation cos(x) ≈ 1 −
x cos(x3 ) ≈ x −
Page 3
x2
2
x7
x13
+
2
24
+
x4
24
then we would get
and
Z
1
x cos(x3 ) dx ≈
0
1
Z
0
x7
x13
x−
+
2
24
!
1
1
x2
x8
x14 1
1
dx =
−
+
+
≈ 0.44047619.
= −
2
16 14 · 24 2 16 336
0
Note how close this is to the actual value of the integral!
6. Consider the function g(θ) =
p
1 − 4 sin2 θ. Find the degree 5 Taylor approximation to this function.
Solution: We start with sin θ ≈ θ −
θ3
6
+
θ5
120 .
Then
θ5 θ3
θ5 θ3
+
· θ−
+
6
120
6
120
4
θ
= θ2 −
+ (higher order terms)
3
sin2 θ = (sin θ)(sin θ) ≈ θ −
Then we get
4
4 sin2 θ ≈ 4θ2 − θ4 .
3
Next use
1
1
1
1
(1 − x) 2 ≈ 1 − x − x2 − x3 + · · ·
2
8
16
Substitute x = 4 sin2 θ to get
2
3
4
1
4
1
4
1
4θ2 − θ4 −
4θ2 − θ4 +
4θ2 − θ4
2
3
8
3
16
3
4
= 1 − 2θ2 − θ4 + (terms of order θ6 and higher)
3
The desired degree 5 Taylor expansion is therefore
1
(1 − 4 sin2 θ) 2 ≈ 1 −
1
8
(1 − 4 sin2 θ) 2 ≈ 1 − 2θ2 + θ4 .
3
7. The period of a pendulum of length L (measured in meters) is given by the formula
s
L
T = 2π
g
where g = 9.8m/s2 and T is measured in seconds. For L = 30 meters, T = 10.9933 seconds.
(a) Compute the degree 2 Taylor polynomial for T (L) centered at L = 30.
(b) Compute T (32) and compare it to the value of your Taylor polynomial when L = 32.
Solution: We start by computing T 0 (30) and T 00 (30). We have T (L) =
1
√2π L 2 ,
9.8
so T 0 (L) =
3
and T 00 (L) = − 2·√π9.8 L− 2 . Then
T 0 (30) = 0.183221,
T 00 (30) = −0.00305369.
The Taylor approximation is
T (L) ≈ 10.9933 + (0.183221)(L − 30) −
Then
T (32) ≈ 10.9933 + (0.183221) · 2 −
The actual value of T (32) is 11.353815.
Page 4
1
2
1
2
0.00305369(L − 30)2 .
0.00305369 · 4 = 11.353635.
1
√π L− 2
9.8
kq
8. An electric charge q sitting at spot s on the number line generates an electric field equal to E(x) = (s−x)
2
at each location x. Now suppose we have two particles with charges of q and −q sitting at s and −s,
respectively. The total electric field at spot x is then
E(x) =
kq
kq
−
.
2
(s − x)
(s + x)2
Factoring out an s2 in the denominator, this becomes
"
#
kq
1
1
E(x) = 2 ·
−
.
s
(1 − xs )2
(1 + xs )2
Use the Taylor polynomials for
x = 0).
1
(1−u)2
to obtain a degree three Taylor polynomial for E(x) (centered at
Solution: Start with the Binomial Series
−2 2
−2 3
−2
(1 + u) ≈ 1 − 2u +
u +
u = 1 − 2u + 3u2 − 4u3 .
2
2
Then substitute −u for u to get
(1 − u)−2 = (1 + (−u))−2 ≈ 1 + 2u + 3u2 + 4u3 .
So
(1 − u)−2 − (1 − u)−2 ≈ [1 + 2u + 3u2 + 4u3 ] − [1 − 2u + 3u2 − 4u3 ] = 4u + 8u3 .
Now substitute u =
x
s
to get
#
"
x
x 3
8 i
kq h
kq
= 3 · 4x + 2 x3 .
+8
E(x) ≈ 2 · 4
s
s
s
s
s
Page 5