### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science1

### DIFFRENTIA TION AND INTEGRATION OF POWER SERIES

Consider the power series∑ ak(x− x0)k= ak+ ak(x− x0)+ ak(x− x0)2+ ⋯

∞

k=0

(1)

With a given interval of convergence I. For each x in I we may de fine a new function f by

f(x)= ∑ ak(x− x0)k

∞

k=0

(2)

EXAMPLE 1:We know that

∑ xk

∞

k=0

= 1+ x+ xk+ ⋯ = 1

1− x if| x|< 1 (3)

Thus the function 1

1−xfor| x|< 1 can be defined by

f(x)= 1

1− x= ∑ xk

∞

k=0

if| x |< 1

EXAMPLE 2: Substituting x4 for x in (3) leads to the equality

f(x)= 1

1− x4= ∑ x4k = 1+ x4+ x8+ x12 + ⋯ if| x|< 1

∞

k=0

### EXAMPLE 3 Substituting −x for x in (3) leads to the equality

f(x)= 11− (−x)= 1

1+ x∑ (−1)kxk

∞

k=0

### = 1− x+ x2− x3+ x4− ⋯ if| x|< 1

Once we see that certain function can be written as power ser ies, the next question that naturallyarises is whether such function can be differentiated and integrated The remarkable theorem

given next ensures that every function represented as a power series can be differentiated and

integrated at any x such that| x |< R . the radius of convergence Moreover, we see how the

derivative and integral can be calculated .

### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science2

Theorem 1: Let the power series ∑ akxk ∞k=0 have the radius of convergence R > 0 Let

f(x)= ∑ akxk

∞

k=0

= a0+ a1x+ a2x2+ ⋯ for| x|< R

Then for| x|< R we have the following:

(i) f (x) is continuous

(ii)The derivative f′(x)exists .and

f′(x)= d

dx a0+ d

dx a1x+ d

dx a2x2+ ⋯

= a1+ 2a2x+ 3a3x2+ ⋯ ∑ ka kxk−1 ∞k=1

(iii ) The antiderivative ∫f(x) dx exists and

∫ f(x) dx = ∫ a0 dx + ∫ a1x dx + ∫ a2x2 dx + ⋯

= a0x+ a1

x2

2 + a2

x3

3 + ⋯ + C = ∑ ak

xk+1

k+ 1

∞

k=0

+ C

Moreover, the two series ∑ ka kxk−1 ∞k=1 and ∑ ka kxk+1/(k+ 1) ∞k=0 both have the radius of

convergence R

Simply put, this theorem says that the derivative of a converging power series is the series of

derivatives of i ts terms and that the integral of a converging power series is the series of integrals

of its terms

EXAMPLE 4 From Example 3 we have

1

1+ x= 1− x+ x2− x3+ ⋯ = ∑ (−1)kxk

∞

k=0

(4)

for| x|< 1 Substituting u= x+ 1 . we have x= u− 1 and 1

1+x= 1

u. if − 1 < x< 1.then 0 <

u< 2 .and we obtain

1

u= 1− (u− 1)+ (u− 1)2− (u− 1)3+ ⋯ ∑ (−1)k(u− 1)k

∞

k=0

### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science3

### For 0< u< 2 Integration then yields

In u= ∫duu = u− (u−1)2

2 + (u−1)3

3 − ⋯ + C

Si nce In 1 = 0 . we immediately find that C = −1 . so that

In = ∑ (−1)k (u− 1)k

k+ 1

∞

k=0

(5)

For 0< u< 2 Here we have expressed the log arithmic function defined on the interval as a

power series

EXAMPLE 5: The series

f(x)= 1+ x+ x2

2!+ x3

3!+ ⋯ = ∑ xk

k!

∞

k=0

(6)

Converges for every real number x (i ∙e∙.R = ∞ ; But

f′(x)= d

dx 1+ d

dx x+ d

dx

x2

2!+ ⋯ = 1+ x2

2!+ ⋯ = f(x)

Thus f satisfies the differential equation f′= f.

f(x)= ce x (7)

For some constant c Then substituting into equations (6) and (7) yields

f(0)= 1= ce 0= c .

So that f(x)= ex We have obtained the important expansion that is valid for any real number x:

ex= 1+ x+ x2

2!+ x3

3!+ ⋯ = ∑ xk

k!

∞

k=0

(8)

For example, if we substitute the value x= 1 into (8), we obtain partial sum approximations

for e= 1+ 1+ 1

2!+ ⋯ (see Table 1) The last value ∑ xk

k!

∞k=0 is correct to five decimal places

### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science4

### EXAMPLE 6: Substituting x= −x in (8) we obtain

e−x= 1+ x+ x22!+ x3

3!+ ⋯ = ∑ (−1)k xk

k!

∞

k=0

(9)

Since this is an alternating series if x> 0 . tells us that the error| S− Sn |. In

approximating e−x for x> 0 is bounded by| an+1|= xn+1/(n+ 1)!∙† for example, to

calculate e−1 with an error of less than 10 ∙000 . we must have| S− Sn|≤ 1

(n+1)!< 0∙0001 =

1

10.000 .or (n+ 1)!> 10 ∙000 If n= 7.(n+ 1)!= 8!= 40 .320 . so that ∑ (−1)k 7k=0 /k! will

approximate e−1 correct to four decimal places We obtain

e−1≈ 1− 1+ 1

2!− 1

3!+ 1

4!− 1

5!+ 1

6!− 1

7!

= 1

2− 1

6+ 1

24 − 1

120 + 1

720 − 1

5040

= 0∙5− 0∙16667 + 0∙04167 − 0∙00833 + 0∙00139 − 0∙0002 ≈ 0∙36786

Note that e−1≈ 0∙36788 correct to five decimal places

EXAMPLE 7 Consider the series

f(x)= 1− x2

2!+ x4

4!− x6

6!+ ⋯ = ∑ (−1)k x2k

(2k )!

∞

k=0

(10 )

It is easy to see that R = ∞ since the series is absolutely convergent for every x by comparison

with the series (8) for ex [ The series (8) is larger than the series (10) since it contains the

terms xn/n! for n both even a nd odd, not just for n even , as in (10)]Differentiating we obtain

f′(x)= −x+ x3

3!+ x5

5!− x7

7!+ ⋯ = ∑ (−1)k x2k+1

(2k + 1)!

∞

k=0

(11 )

Since this series has a radius o f convergences R = ∞ . we can differentiate once more to

obtain

f′′(x)= −1+ x2

2!+ x4

4!− x6

6!+ ⋯ = ∑ (−1)k+1x2k

(2k )!

∞

k=0

= −f(x)

Thus we see that f satisfies the differential equation

f′′+ f= 0 (12 )

### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science5

### Moreover, from equations (10) and (11) we see that

f (0)= 1 and f′(0)= 0 (13 )It is easily seen that the function f(x)= cos x satisfies equation (12) together with the conditions

(13) In fact, although .We do not prove here; it is the only func tion that does so Thus we have

cos x= 1− x2

2!+ x4

4!− x6

6!+ ⋯ = ∑ (−1)k x2k

(2k )!

∞

k=0

(14 )

Since d

dx cos x= − sin x .

We obtain, from (14), the series (after multiplying bo th sides by -1)

sin x= x− x3

3!+ x5

5!− x7

7!+ ⋯ = ∑ (−1)k x2k+1

(2k + 1)!

∞

k=0

(15 )

Power series expansions can be very useful for approximate integration

EXAMPLE8 : Calculate ∫ e−t2 1

0 with an error < 0∙0001

Solution Substituting t2 for x in (9) , we find that

e−t2= 1− t2+ t4

2!+ t6

3!+ ⋯ = ∑ (−1)k t2k

k!

∞

k=0

(16 )

Then we inte grate to find that

### ∫ e−t2 x

0dt = x− x3

3!+ x5

5∙2!− x7

7∙3!+ ⋯ = ∑ (−1)kx2k+1

(2k + 1)!

∞

k=0

(17 )

The error| S− Sn |is bounded by| an−1|= | x2(n+1)−1

[2(n+1)+1](n+1)! |

In our exampl e x= 1 . so we need to choose n so that

1

(2n + 3)(n+ 1)!< 0∙0001 .or (2n + 3)(n+ 1)!> 10 .000

If n= 6 . then (2n + 3 )(n+ 1)!= (15 )(7)!= 75 .600 With this choice of n(n= 5 is too

small ), we obtain

### Lecture no .2-7 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science6

### ∫ e−t2 1

0dt = x− 1

3+ 1

5∙2!− 1

7∙3!+ 1

9∙4!− 1

11 ∙5!+ 1

13 ∙6!

≈ 1− 0∙33333 + 0∙1− 0∙02381 + 0∙00463 − 0∙00076 + 0∙00011 = 0∙74684 .

And to four de cimal places, ∫ e−t2 1

0 dt = 0∙7468

PROBL EMS :

1∙ By substituting x2 for x in (4) , find a series expansion for 1

1+x2 that is valid for| x|< 1

2. Integrate the series obtained in Problem 1 to obtain a series expansion for tan −1x

3.Use the result of Problem 2 to obtain an estimate of π that is correct to two decimal place s

4. Use the series expansion for In x to calculate the following to two decimal places of accuracy:

(a) In 0.5

In Problems 5 -13 estimate the given integral to within the given accuracy

5∙∫ e−t2 1

0

dt ;error 0∙01 6∙∫ e−t3 1

0

dt ;error < 0∙001

7∙∫ cos t2 dt ;error 0∙001

12

0

8∙∫ sin t2 dt ;error 0∙0001

12

0

### 9∙ ∫ t2e−t2 1

0dt ;error 0∙01

[ Hint: The series explanation of t2e−12 is obtained by multiplying each term of the series

expansion of e−12 by t2 ]

10 ∙∫ t5e−t5 1

0

dt ; error < 0∙0001 11 ∙∫ cos √t

1

0

dt ;error < 0∙001

12 ∙ ∫ tsin √t dt ;error < 0∙001

1

0

13 ∙∫ dt

1+ t8 ;error < 0∙0001

12

0

### 14. Find a series expansion f or xe x that is valid for all values of x

15. Use the result of Problem 14 to find a series expansion for ∫ tex dt x0