Differentiating and Integrating Power Series
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference your lecture notes and
the relevant chapters in a textbook/online resource.
EXPECTED SKILLS:
• Know (i.e. memorize) the Maclaurin series for ex , sin x and cos x. Algebraically manipulate these series expansions, as well as other given power series expansions, to form
new expansions.
• Differentiate and integrate power series expansions term-by-term.
• Use a series expansion to approximate an integral to some specified accuracy.
PRACTICE PROBLEMS:
1. Confirm that
d x
(e ) = ex by differentiating the Maclaurin series for ex term-by-term.
dx
d x
d
x 2 x3 x4
(e ) =
+
+
+ ...
1+x+
dx
dx
2!
3!
4!
2x 3x2 4x3
=0+1+
+
+
+ ...
2!
3!
4!
x 2 x3
+
+ ...
=1+x+
2!
3!
= ex .
2. Recall that the Maclaurin series for ex converges to ex for all real numbers x. It can
be shown (in a complex analysis course) that this convergence holds for any complex
number as well. Based on this fact, use the Maclaurin series for ex , sin x, and cos x to
prove Euler’s Formula:
eix = cos x + i sin x,
where i is the imaginary number with the property i2 = −1 (and thus i3 = −i, i4 = 1,
etc).
1
Since ex = 1 + x +
x2 x3
+
+ . . . , we have
2!
3!
(ix)2 (ix)3 (ix)4 (ix)5 (ix)6 (ix)7 (ix)8
+
+
+
+
+
+
+ ...
2!
3!
4!
5!
6!
7!
8!
x2
x3 x4
x5 x6
x7 x8
= 1 + ix −
−i +
+i −
−i +
+ ...
2!
3!
4!
5! 6! 7!
8!
x3 x5 x7
x2 x4 x6 x8
+
−
+
+ ... + i x −
+
−
+ ...
= 1−
2!
4!
6!
8!
3!
5!
7!
= cos x + i sin x
eix = 1 + ix +
Note: Recall that Euler’s formula was useful when solving second order linear homogeneous differential equations with constant coefficients, specifcally for when the
characteristic equation had complex roots.
3. The purpose of this problem is to find the Maclaurin series for arctan x. If we attempt
to take successive derivatives of arctan x the computation becomes unpleasant rather
quickly (try it if you want). Here is a simpler alternative.
(a) Find the Maclaurin series for
2
3
4
1 + x + x + x + x + ... =
1
.
1−x
∞
X
xk . See Convergence of Taylor Series #5.
k=0
(b) Replace x in part (a) with the appropriate quantity to obtain the Maclaurin series
1
for 1+x
2.
If we replace x in
1
1−x
with −x2 we get
So the Maclaurin series for
1
1−(−x2 )
=
1
.
1+x2
1
is
1 + x2
1 + (−x2 ) + (−x2 )2 + (−x2 )3 + . . . = 1 − x2 + x4 − x6 + . . . =
∞
X
(−1)k x2k
k=0
(c) Integrate the answer in part (b) term-by-term to obtain the Maclaurin series for
arctan x.
Z
1
We know that
= arctan x + C.
2
Z 1+x
x3 x5 x7
2
4
6
+
−
+ . . . − C.
So arctan x = (1 − x + x − x + . . .) dx − C = x −
3
5
7
∞
X
(−1)k x2k+1
Since arctan(0) = 0, we have C = 0, and therefore arctan x =
.
2k
+
1
k=0
2
2
4. Find the first four nonzero terms of the Maclaurin series for f (x) = e(x ) arctan x
by multiplying the Maclaurin series of the factors. See the previous problem for the
Maclaurin series for arctan x.
Since ex = 1 + x +
x2
2!
+
x3
3!
2
+ . . . , we have e(x ) = 1 + x2 +
x4
2!
+
x6
3!
+ ...
So
e
(x2 )
x4 x6
x 3 x5 x7
arctan x = 1 + x +
+
+ ...
x−
+
−
+ ...
2!
3!
3
5
7
5
3 4 3
x
x
x
x
2
2
+ (x )(x) + (1)
+ (x ) −
+
(x)
= (1)(x) + (1) −
3
5
3
2!
5 4 3 6 7
x
x
x
x
x
2
+ (x )
+
−
+
(x) + . . .
+ (1) −
7
5
2!
3!
3!
2
11
2
= x + x3 + x5 + x7 + . . .
3
30
35
2
5. Consider the function f (x) = sin x cos x.
(a) Find the first three nonzero terms of the Maclaurin series for f (x) by multiplying
the Maclaurin series of the factors.
2
2
x − x3 + x5 − . . .
3
15
(b) Confirm your answer in part (a) by using the trigonometric identity
sin 2x = 2 sin x cos x.
1
1
(2x)3 (2x)5
2
2
sin x cos x = sin 2x =
2x −
+
− . . . = x − x3 + x5 − . . .
2
2
3!
5!
3
15
6. Find the first three nonzero terms of the Maclaurin series for f (x) = tan x by performing a long division on the Maclaurin series for sin x and cos x.
1
2
x + x3 + x5 + . . .; Detailed Solution: Here
3
15
7. Use the result in #6 to find the first three nonzero terms of the Maclaurin series for
f (x) = sec2 x.
d
d
1 3
2 5
2
2
sec x =
(tan x) =
x + x + x + . . . = 1 + x2 + x4 + . . .
dx
dx
3
15
3
Z
8. Use a Maclaurin series to approximate
0
3
1
cos(x2 ) dx to four decimal-place accuracy.
∞
∞
X
X
(−1)k x2k
(−1)k x4k
, so cos (x2 ) =
.
(2k)!
(2k)!
k=0
k=0
1
Z 1
∞
∞
k 4k+1 X
X
(−1)
x
(−1)k
cos(x2 ) dx =
Thus
,
=
(2k)!(4k
+
1)
(2k)!(4k
+
1)
0
k=0
k=0
0
Z 1
cos(x2 ) dx.
which is an alternating series that converges to S =
cos x =
0
Thus |S − sn | < an+1 , where sn is the n-th partial sum and ak =
1
.
(2k)!(4k+1)
For four decimal-place accuracy, we want
an+1 =
1
≤ 0.00005.
(2n + 2)!(4n + 5)
(2n + 2)!(4n + 5) ≥ 20, 000
Now for n = 2 : (6!)(13) < 20, 000 and for n = 3 : (8!)(17) > 20, 000.
So we want the 3-rd partial sum s3 .
Z
1
2
cos(x ) dx ≈
Therefore
0
3
X
k=0
(−1)k
1
1
1
=1−
+
−
.
(2k)!(4k + 1)
(2!)(5) (4!)(9) (6!)(13)
1
Z
arctan(x2 ) dx to two decimal-place accuracy.
9. Use a Maclaurin series to approximate
0
See problem #3 for the Maclaurin series for arctan x.
Z 1
4
X
1
1
1
1
1
(−1)k
2
arctan(x ) dx ≈
= −
+
−
+
.
(2k
+
1)(4k
+
3)
3
(3)(7)
(5)(11)
(7)(15)
(9)(19)
0
k=0
4