Chapter 22 the Binomial Series

CHAPTER 22 THE BINOMIAL SERIES

EXERCISE 93 Page 195

1. Use Pascal's triangle to expand (x – y)7

From Pascal’s triangle on page 195 of the textbook,

(a+=++++++ x )6 a 6 6 ax 5 15 ax 42 20 ax 33 15 ax 24 6 ax 5 x 6

Thus, (a+=+++++++ x )7 a 7 7 ax 6 21 ax 52 35 ax 43 35 ax 34 21 ax 25 7 ax 6 x 7 and (xyx− )776 =+ 7()21()35()35()21()7()() xy −+ xy 52 −+ xy 43 −+ xy 34 −+ xy 25 −+−+− xy 6 y 7 i.e. (x−=− y )7 x 7 7 xy 6 + 21 xy 52 − 35 xy 43 + 35 xy 34 − 21 xy 25 + 7 xy 6 − y 7

2. Expand (2a + 3b) 5 using Pascal’s triangle.

5 From page 195 of the textbook, (a+=+++++ x) a55 ax 4 10 ax 32 10 ax 23 5 ax 4 x 5

Thus, replacing a with 2a and x with 3b gives:

5 5 4 32 23 4 5 (2ab+=+++ 3) ( 2 a) 5( 2 ab) ( 3) 10( 2 ab) ( 3) 10( 2 ab) ( 3) ++ 5( 2 ab)( 3) ( 3 b)

= 32a5++ 240 ab 4 720 ab32 + 1080 ab23 ++ 810 ab4 243 b 5

EXERCISE 94 Page 197

1. Use the binomial theorem to expand (a + 2x)4

4 (43)( ) 2( 432)( )( ) 34 (ax+ 2 ) = a43++42 a( x) a2( 2 x) + ax(2) +( 2 x) 2! 3!

= a4++8 a 3 x 24 a 22 x + 32 ax 3 + 16 x 4

2. Use the binomial theorem to expand (2 – x)6

6 (65)( ) 23(654)( )( ) (6)(5)(4)(3) (2 − x) = 26+ 6(2) 5( −+xx) (2)4( −) + (2)3(− x) + (2)24 (− x ) 2! 3! 4! (6)(5)(4)(3)(2) + (2)()()−xx56 +− 5!

= 64−+ 192x 240 x2 − 160 x 3456 +−+ 60 x 12 xx

3. Expand (2x – 3y)4

4 4 3 (43)( ) 22(432)( )( ) 3 4 (2xy− 3) =( 2 x) + 42( x) ( − 3 y) + ( 2 x) (− 3 y) + (2x)(− 3 y) +−( 3 y) 2! 3!

= 16x4−+ 96 x 3 y 216 x 22 y − 216 xy 3 + 81 y 4

5 2 4. Determine the expansion of 2x + x

5 23 2  54 22(54)( ) 3 (543)( )( ) 2 2 2x+=  ( 2 xx) + 52( )  + (2 x) + (2x)  x   xx2!  3!  x

45 (5432)( )( )( ) 22  ++(2x)  4! xx 

320 160 32 = 32xxx53+ 160 + 320 +++ xxx35

5. Expand (p + 2q)11 as far as the fifth term.

(11)( 10) 23(11)( 10)( 9) (11)(10)(9)(8) (p + 2q)11 = p11++11 p 10 ( 2 q) pq9 (2 ) +pq8 (2 ) + pq7(2 ) 4 2! 3! 4!

= p11++22 p 10 q 210 pq9 2 + 1320 pq8 3 + 5280 pq7 4

13 q 6. Determine the sixth term of 3p + 3

n nn( −−1)( n 2) to ( r − 1) terms The 6th term of (ax+ ) is axnr−−( 1) r − 1 (r −1!)

13 5 5 q (13)( 12)( 11)( 10)( 9) 13− 5 q 8 q Hence, the 6th term of 3p + is: (3p)  = 1287( 3p)  3 5! 3 3

= 34749 pq85

7. Determine the middle term of (2a – 5b)8

With power 8 there will be 9 terms in the series. The middle term will be the fifth term

(2a – 5b)8

(87)( ) 23(876)( )( ) (8)(7)(6)(5) = (2a )87+ 8(2 ab )( −+ 5) (2 ab )6( − 5 ) + (2 ab )5(− 5 ) + (2ab )44 (− 5 ) 2! 3! 4!

(8)(7)(6)(5) Hence, the fifth term is: (2ab )44 (− 5 ) = 700000ab44 4!

8. Use the binomial theorem to determine, correct to 4 decimal places:

(a) (1.003)8 (b) (0.98)7

88(87)( ) ( 876)( )( ) (a) (1.003) =( 1 + 0.003) = 1++ 8(0.003) (0.003)23 +(0.003) + ... 2! 3!

= (1+ 0.024 + 0.000252 + 0.000001512 += ...) 1.0242535...

= 1.0243, correct to 4 decimal places.

(b)

77 (76)( ) 2( 765)( )( ) 3( 7654)( )( )( ) 4 (0.98) =−=+−+( 1 0.02) 1 7( 0.02) ( −+0.02) ( −+0.02) ( −+0.02) ... 2! 3! 4!

= 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 – …

= 0.8681, correct to 4 decimal places

9. Evaluate (4.044)6 correct to 2 decimal places.

6 660.044 6 (4.044) =+=+=+( 4 0.044)  4 1 46 ( 1 0.011) 4

6(65)( ) 23( 653)( )( ) ( 7654)( )( )( ) 4 (1+=++ 0.011) 1 6( 0.011) ( 0.011) +(0.011) + (0.011) 2! 3! 4! (7)(6)(5)(4)(3) ++(0.011)5 ... 5!

= 1 + 0.066 + 0.001815 + 0.00002662 + 0.000000512435 + 0.000000003382071

= 1.067842136...

6 6 6 + 6 Hence, (4.044) = 4( 1 0.011) = 4 (1.067842136...) = 4373.88, correct to 2 decimal places

EXERCISE 95 Page 199

1 1. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (1− x )

State in each case the limits of x for which the series is valid.

1 −1(−−12)( ) 23( −−− 123)( )( ) =(1 −xx) = 1 +− ( 1)( − ) +( − x) + (− x) +... (1− x) 2! 3!

= 1++xx23 + x +... and x < 1

1 2. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (1+ x ) 2

State in each case the limits of x for which the series is valid.

1 −2(−−23)( ) 23( −−− 234)( )( ) 2 =+(1xx) =−+ 1 2 ( x) + ( x) +... (1+ x) 2! 3!

= 1−+ 2xx 323 − 4 x + ... and x < 1

1 3. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (2+ x )3

State in each case the limits of x for which the series is valid.

−3 23 1 −3 x   xx(−−34)( )  ( −−− 345)( )( )  x = (2+=x) 2−−33 1 +  = 2 1 − 3  +  + +... (2+ x )3 2  2 2!  2 3! 2

1 33 5 = 1−+xx23 − x +... 23  22 4

1 33 5 = 1−+xx23 − x +... 8 22 4

x The series is true provided < 1 i.e. x < 2 2

4. Expand 2 + x in ascending powers of x as far as the term in x3, using the binomial theorem.

State in each case the limits of x for which the series is valid.

1/2 x x x 2 + x = 21+ = 21+ = 21+ 2 2 2

Using the expansion of (1 + x)n,

1/2 23 x 1xx (1/ 2)(− 1/ 2) (1/ 2)(−− 1/ 2)( 3 / 2)  x 21+ = 2 [ ++  +... ] 2 2 2 2! 2 3! 2

xx23 x = 2 1+− + −... 4 32 128

x This is valid when < 1 i.e. x < 2 or –2 < x < 2 2

1 5. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. 13+ x

State in each case the limits of x for which the series is valid.

13  135 −− −−− 1   1 − 1 2223  222 = (1+ 3xx) 2 =+− 1 (3 ) + (3 x) + (3 x) ... 13+ x 2 2! 3!

3 27 135 = 1−+xx23 − x as far as the term in x3 2 8 16

1 The series is true provided 3x < 1 i.e. x < 3

6. Expand (2 + 3x)–6 to three terms. For what values of x is the expansion valid?

−62 −6 3x   33 xx(−−67)( )  (2+ 3x) = 2−−66 1 +  = 2 1 +− ( 6) +  to three terms 2  2 2!  2 

1 189 1 189 = 1−+ 9xx2 + ... = 1−+ 9xx2 + ... 246 64 4

3x 2 The series is true provided < 1 i.e. x < 2 3

7. When x is very small show that:

1 5 (1− 2x ) 15+ x 19 (a) ≈ 1 + x (b) ≈ 1 + 10x (c) ≈ 1 + x (1−−xx )2 (1 ) 2 (1− 3x ) 4 3 12− x 6

1 1 −−2 x =− −2 ≈+ + (a) 2 (1xx) ( 1) ( 12 x) 1 (11−−xx) ( ) 2 x ≈ 12++x ignoring the x2 term and above 2 5 ≈ 1+ x 2

(12− x) −4 (b) 4 =−−≈−+(12xx)( 13) ( 12 x)( 112 x) (13− x) ≈ 1 + 12x – 2x ignoring the x2 term and above

≈ 1 + 10x

11 15+ x − 5 2 (c) =+(15xx)23( 12 −) ≈+ 1 xx 1 + 3 (12− x) 23 25 ≈ 1++xx ignoring the x2 term and above 32 19 5 2 15+ 4 19 ≈ 1+ x += = 6 23 6 6

8. If x is very small such that x2 and higher powers may be neglected, determine the power series for

xx+−483

5 (1+ x ) 3

11 3 11 3 3 xx+−48 −−1 xx 231  =+−(4xxx)23( 8) ( 1 +=+) 5 412 813 −( 1 +x) 5 3    5 + 48   (1 x)

3 113xx  ≈ ( 4)( 81)+−−  ( x) ignoring the x2 term and above 24 38  5

xx3 x 15xx−− 5 72 x 62x ≈ 41+− − ≈ 41+ ≈ 41− 8 24 5 120 120

31x 31 ≈ 41− = 4 – x 60 15

9. Express the following as power series in ascending powers of x as far as the term in x2. State in

each case the range of x for which the series is valid.

1− x (1+−xx )3 (1 3 ) 2 (a)  (b) 1+ x (1+ x2 )

 11  13 − −− 11   1− x− xx22 2 22 (a) =−(11xx)22( +) ≈ 1 −+( −x) 1 −+ x2  1+ x  2 2!  2 2!     as far as the term in x2 xx2  xx332  xxxxx2 22 = 11−−  −+  ≈−+ 1 −+ − 28  28  28248 x2 ≈ 1−+x as far as the term in x2 2 The series is valid if x < 1

2 3 21 (1+−xx) ( 13) − (b) =+−+(1xxx)( 13)32( 1 2 ) (1+ x2 ) 21    −  33   2 x2 ≈ (1+xx)  1 −+ 2 ( −3 x) + ... 1 −+ ... as far as the term in x2 2! 2 

x2 ≈ (1+x)( 12 −− xx2 ) 1 − 2

x2 ≈ (12−xx −22 +− x 2 x) 1 − as far as the term in x2 2

xx22 ≈ (1−−xx 3122) − ≈− 1 −−xx 3 neglecting x3 terms and above 22 7 ≈ 1−−xx2 2

1 The series is valid provided 3x < 1 i.e. x < 3

EXERCISE 96 Page 201

1. Pressure p and volume v are related by pv3 = c, where c is a constant. Determine the approximate

percentage change in c when p is increased by 3% and v decreased by 1.2%.

New pressure = 1.03p = (1 + 0.03)p and new volume = 0.988v = (1 – 0.012)v

New value of c = (1 + 0.03)p(1 – 0.012) 3v3

= pv3(1 + 0.03)(1 – 0.012) 3

≈ pv3(1 + 0.03 – (3)0.012)

≈ pv3(1 + 0.03 – 0.036) ≈ pv3(1 – 0.006) i.e. 99.4% of the original value of c

Thus, the approximate percentage change in c is a reduction of 0.6%

1 2. Kinetic energy is given by mv2. Determine the approximate change in the kinetic energy when 2 mass m is increased by 2.5% and the velocity v is reduced by 3%.

New mass = 1.025m = (1 + 0.025)m and new velocity = 0.97v = (1 – 0.03)v 11 New kinetic energy = (1+ 0.025)m (1 − 0.03)22 v ≈+− mv 2 (1 0.025)(1 0.06) 22 11 ≈ mv22(1+−= 0.025 0.06)mv (0.965) 22 i.e. 96.5% of the original kinetic energy

Thus, the approximate change in kinetic energy is a reduction of 3.5%

3. An error of +1.5% was made when measuring the radius of a sphere. Ignoring the products of

small quantities, determine the approximate error in calculating (a) the volume and (b) the surface

area.

4 (a) Volume of sphere, V = π r3 3

44 4 4 New volume = ππ(1.015rr )33=+≈+ (1 0.015) 33 π r [1 3(0.015)] =+ π r3 (1 0.045) 33 3 3 = 1.045V

i.e. the volume has increased by 4.5%

(b) Surface area of sphere, A = 4π r 2

2 New surface area = 4ππ( 1+ 0.015) rr22 ≈+ 4 [1 2(0.015)] =+ 4 πr2 (1 0.03)

= 1.03A

i.e. the surface area has increased by 3.0%

4. The power developed by an engine is given by I = kPLAN, where k is a constant. Determine the

approximate percentage change in the power when P and A are each increased by 2.5% and L and

N are each decreased by 1.4%.

I = kPLAN

New power = kP(1+ 0.025) (1 −+− 0.014) LAN (1 0.025) (1 0.014)

≈ kPLAN(1+−+− 0.025 0.014 0.025 0.014) =kPLAN (1 + 0.022) i.e. 102.2% of the original power

Thus, the approximate percentage change in power is an increase of 2.2%

5. The radius of a cone is increased by 2.7% and its height reduced by 0.9%. Determine the

approximate percentage change in its volume, neglecting the products of small terms.

1 Volume of cone = π rh2 3

1 2 1 New volume = π (1+− 0.027) rh2 ( 1 0.009) ≈ π rh2 (1+− (2)0.027 0.009) 3 3

1 ≈ π rh2 (1+ 0.045) 3 i.e. 104.5% of the original volume

6. The electric field strength H due to a magnet of length 2l and moment M at a point on its axis

distance x from the centre is given by:

M 11 H = − 2(l xl−+ )22 ( xl ) 2M Show that if l is very small compared with x, then H ≈ x3

 −−22 M11 M 1 1Ml   l  H = −  = − = 11 −  −+  22l −+22l 2 2 2xl2  x  x  ( xl) ( xl) 22ll   xx11−+  xx 

M22 l  l  Ml  4  2M ≈11 +  −−  ≈  ≈ 22xl22 x  x  xl  x  x3

kT 7. The shear stress t in a shaft of diameter D under a torque T is given by: τ = π D3 Determine the approximate percentage error in calculating τ if T is measured 3% too small and D

1.5% too large.

New value of T = (1 – 0.03)T and new value of D = (1 + 0.015)D

k(1− 0.03) T kT −3 Hence, new shear stress = =(1 −+ 0.03)( 1 0.015) ππ(1+ 0.015)33DD 3

kT kT ≈(1 − 0.03)( 1 − 0.045) ≈[ 1 −− 0.03 0.045] ππDD33

kT ≈(1 −=− 0.075) τ (1 0.075) π D3 i.e. the new torque has decreased by 7.5%

8. The energy W stored in a flywheel is given by: W = kr5N2, where k is a constant, r is the radius

and N the number of revolutions. Determine the approximate percentage change in W when r is

increased by 1.3% and N is decreased by 2%.

W = kr5N2

New energy = k(1 + 0.013) 5r5 (1 – 0.02) 2N2

≈ kr5N2(1 + (5)0.013 – (2)0.02)

≈ kr5N2(1 + 0.065 – 0.04) ≈ kr5N2(1 + 0.025) i.e. 102.5% of the original volume

Thus, the approximate percentage change in energy is an increase of 2.5%

9. In a series electrical circuit containing inductance L and capacitance C the resonant frequency is

1 given by: fr = . If the values of L and C used in the calculation are 2.6% too large and 2π LC

0.8% too small, respectively, determine the approximate percentage error in the frequency.

New value of inductance = (1 + 0.026)L and new value of capacitance = (1 – 0.008)C 11 Hence, new resonant frequency = = 2ππ (1+− 0.026)LC (1 0.008) 2 (1 + 0.026) L (1 − 0.008) C

1 1 11 1 − − −− = (1+− 0.026) 2 LC2 (1 0.008) 22 2π

11 ≈(1 − 0.013)( 1 + 0.004) ≈(1 −+ 0.013 0.004) 11π 2π LC22 2 LC

≈−fr (1 0.009) i.e. the new resonant frequency is 0.9% smaller

kr 4 10. The viscosity η of a liquid is given by: η = , where k is a constant. If there is an error in r of νl +2%, in υ of +4% and I of –3%, what is the resultant error in η?

New value of r = (1 + 0.02)r, new value of ν = (1 + 0.04)ν and new value of l = (1 – 0.03)l

k(1+ 0.02)44 r kr 4 4−− 11 Hence, new value of viscosity = =++−(1 0.02) ( 1 0.04) ( 1 0.03) (1+− 0.04)νν (1 0.03)ll

kr 4 ≈+−+(1 0.08)( 1 0.04)( 1 0.03) νl

kr 4 ≈(1 +−+ 0.08 0.04 0.03) ≈+η ( 1 0.07) νl i.e. the viscosity increases by 7%

11. A magnetic pole, distance x from the plane of a coil of radius r, and on the axis of the coil, is kx subject to a force F when a current flows in the coil. The force is given by: F = , 5 (rx22+ ) where k is a constant. Use the binomial theorem to show that when x is small compared to r, kx5 kx3 then F ≈ − rr572

5 − 55 kx kx −−x2 2 F = = =+=kx( r22 x) 22 kx( r 2) 1 + 5 5 2 22 r (rx+ ) (rx22+ ) 2 

5 − x2 2 = kxr −5 1+ r 2

kx5 x2 ≈ 1+− when x is small compared to r rr522

kx5 x2 kx kx5 x2 ≈ 1− ≈ −  rr522 rr552 r 2

kx5 kx3 ≈ − rr572

5 (3dH) 12. The flow of water through a pipe is given by: G = . If d decreases by 2% and H by L 1%, use the binomial theorem to estimate the decrease in G.

New value of d = (1 – 0.02)d and new value of H = (1 – 0.01)H

5 5 1 55 1 1 (3dH) 355dH2 2 3 (1−− 0.02)22d (1 0.01) 2 H 2 Hence, new value of G = = = L 11 LL22

51 35 dH22 51   ≈ −− 1 1 (0.02)  1 (0.01)  22   L2 

(3dH )55(3dH ) ≈(1 − 0.05)( 1 − 0.005) ≈(1 −− 0.05 0.005) LL

≈−G(1 0.055) i.e. the flow G has decreased by 5.5%