Computational Methods Exercise 20.1 | Basic Mathematics Solution [NEB UPDATED]

Exercise 20.1

1) Apply the method of successive bisection to find the

a) Square root of 3 within 2 places of decimal in (1,2)

Solution:

Let x be the square root of 3.

Let f(x) = x² – 3

f(1) = 1 – 3 = 2 = -ve

f(2) = (2)² – 3 = 1 = +ve

f(1.5) = (1.5)² – 3 = -0.75

So, the root lies between 1.5 and 2

Or, $\frac{{1.5 + 2}}{2}$ = 1.75

f(1.75) = (1.75)² = 0.0625

So, the root lies between 1.5 and 1.75.

Or, $\frac{{1.5 + 1.75}}{2}$ = 1.625.

f(1.625) = (1.625)² – 3 = 0.359375.

So, the root lies between 1.625 and 1.75.

Or, $\frac{{1.625 + 1.75}}{2}$ = 1.71875

So, f(1.71875) = (1.71875)² – 3 = -0.045898

So, the root lies in between 1.71875 and 1.75

So, $\frac{{1.71875 + 1.75}}{2}$ = 1.734375

Or, f(1,734375) = (1.734375)² – 3 = 0.0080566 which is small enough.

So, the required square root of 3 = 1.734375.

= 1.73.

So, the square root of 3 is 1.73.

 

b) Square root of 123 within 2 places of decimal in (11, 12)

Solution:

Let x be the square root of 123.

Let f(x) = x² – 123

Or, f(11) = 121 – 123 = -2

And f(12) = 144 – 123 = 21

Or, $\frac{{11 + 12}}{2}$ = 11.5

Or, f(11.5) = (11.5)² – 123 = 9.25

So, the root lies in between 11 and 11.5

Or, $\frac{{11 + 11.5}}{2}$ = 11.25.

Or, f(11.25) = (11.25)² – 123 = 3.5625.

So, the root lies in between 11 and 11.25

Or, $\frac{{11 + 11.25}}{2}$ = 11.125

Or, $\frac{{11 + 11.125}}{2}$ = 11.0625

Or, f(11.0625) = (11.0625)² – 123 = -0.62109

So, the root lies between 11.0625 and 11.125

Or, $\frac{{11.0625 + 11.125}}{2}$ = 11.09375.

Or, f(11,09375) = (11.09375)² – 123 = 0.071289

So, the root lies between 11.0625 and 11.09375.

So, $\frac{{11.0625 + 11.09375}}{2}$ = 11.078125.

Or, f(11.078125) = (11.078125)² – 123 = -0.275146.

So, the root lies in between 11.078125 and 11.09375

So, $\frac{{11.078125 + 11.09375}}{2}$ = 11.0859.

Or, f(11.8059) = (11.8059)² – 123 = -0.10282.

So, the root lies between 11.0859 and 11.09375.

Or, $\frac{{11.0859 + 11.09375}}{2}$ = 11.089825.

Or, f(11.089825) = (11.089825)² – 123 = -0.01578

So, the root = 11.089825.

So, the square root of 123 = 11.09.

c) The approximate value of $\sqrt 2$ within an error of 10^-3

Solution:

Let the square of 2 be x.

Then x² – 2 = 0.

Let f(x) = x² – 2

Or, f(1) = 1 – 2 = -1 < 0

Or, f(2) = 4 – 2 = 2 > 0

Or, f(1).f(2) = -1.2 = -2 < 0.

So, the root lies between 1 and 2.

Or, $\frac{{1 + 2}}{2}$ = 1.5

Or, f(1.5) = (1.5)² – 2 = 0.25 > 0

So, the root lies between 1 and 1.5.

Or, $\frac{{1 + 1.5}}{2}$ = 1.25.

Or, f(1.25) = (1.25)² – 2 = 0.4375 < 0.

SO, the root lies in between 1.25 and 1.5.

Or, $\frac{{1.25 + 1.5}}{2}$ = 1.375.

Or, f(1.375) = (1.375)² – 2 = 0.109375 < 0.

So, the root lies between 1.375 and 1.5.

Or, $\frac{{1.375 + 1.5}}{2}$ = 1.4375.

Or, f(1.4375) = (1.4375)² – 2 = -0.6640625 > 0.

So, the root lies in between 1.375 and 1.4375.

Or, $\frac{{1.375 + 1.4375}}{2}$ = 1.40625.

Or, f(1.40625) = (1.40625)² – 2 = -0.2246.

So, the root les in between 1.40625 and 1.4325

Or, $\frac{{1.40625 + 1.421875}}{2}$ = 1.4140625.

Or, f(1.4140625) = (1.4140625)² – 2 = 0.000427.

Whose absolute value is less than 10^-3.

So, the required square root = 1.4141.

d) The value of $\sqrt[3]{{10}}$ with error not more than 0.03

Solution:

Let the cube of 10 be x.

Then x³ – 10 = 0

Let f(x) = x³ – 10

Or, f(2) = (2)³ – 10 = - 2 < 0.

Or, f(3) = (3)³ – 10 = 17 > 0.

Or, f(2).f(3) = -2 * 17 = -34 < 0.

So, the root lies between 2 and 3.

Or, $\frac{{2 + 3}}{2}$ = 2.5

Or, f(2.5) = (2.5)² – 10 = 5.625 > 0.

So, the root lies between 2 and 2.5.

Or, $\frac{{2 + 2.5}}{2}$ = 2.25.

Or, f(2.1875) = (2.1875)³ – 10 = 0.4675 > 0.

The root lies between 2.125 and 2.1875.

Or, f(2.25) = (2.25)³ – 10 = 1.3906 > 0.

So, the root lies between 2 and 2.25.

Or, $\frac{{2.125 + 2.25}}{2}$ = 2.1875.

Or, $\frac{{2 + 2.25}}{2}$ = 2.125.

Or, f(2.125) = (2.125)³ – 10 = -0.40429 < 0.

So, the root lies between 2.125 and 2.25.

Or, $\frac{{2.125 + 2.1875}}{2}$ = 2.15625.

Or, f(2,15625) = (2.15625)³ – 10 = 0.025299 which is less than 0.03

So, the root is 2.15625.

2) Apply the method of successive bisection to find the root of the equation:

a) x² +x-4=0 in (1, 2) correct to two places of decimals.

Solution:

Let f(x) = x² + x – 4.

Or, f(1) = 1 + 1 – 4 = - 2 < 0.

Or, f(2) = 4 + 2 – 4 = 2 > 0

Or, f(1).f(2) = -2 * 2 = -4 < 0.

So, the root lies between 1 and 2.

Or, $\frac{{1 + 2}}{2}$ = 1.5

Or, f(1.5) = (1.5)² – 4 = -0.25 < 0

So, the root lies between 1.5 and 2.

Or, $\frac{{1.5 + 2}}{2}$ = 1.75.

Or, f(1.75) = (1.75)² + 1.75 – 4 = 0.8125.

So, the root lies between 1.5 and 1.75.

Or, $\frac{{\left( {1.5 + 1.75} \right)}}{2}$ = 1.625.

Or, f(1.625) = (1.625)² + 1.625 – 4 = 0.265625.

So, the root lies in between 1.5 and 1.625.

Or, $\frac{{1.5 + 1.625}}{2}$ = 1.5625

f(1.5625) = (1.5625)² + 1.5625 – 4 = 0.0039025.

So, the root is 1.5626.

b) 2x²-x-5=0 correct to 4 places of decimals within an error of 0.05.

Solution:

 

c) x³-4x+1 = 0 lying between 1 and 2 correct to two places of decimal.

Solution:

Let f(x) = x³ – 4x + 1

Or, f(1) = 1 – 4 + 1 = -2.

And f(2) = 8 – 8 + 1 = 1.

Or, $\frac{{1 + 2}}{2}$ = 1.5

Or, f(1.5) = (1.5)³ – 4 * 1.5 + 1 = -1.625

So, the root lies between 1.5 and 2.

Or, $\frac{{1.5 + 2}}{2}$ = 1.75.

Or, f(1.75) = (1.75)³ – 4 * 1.75 + 1 = -0.640625.

So, the root lies between 1.75 and 2.

Or, $\frac{{1.75 + 2}}{2}$ = 1.875.

Or, f(1.875) = (1.875)³ – 4 * 1.875 + 1 = 0.0917968.

So,  the root lies in between 1.75 and 1.875.

Or, $\frac{{1.75 + 1.875}}{2}$ = 1.8125.

Or, f(1.8125) = (1.8125)³ – 4 * 1.8125 + 1 = -0.295654.

So, the root lies in between 1.825 and 1.875.

Or, $\frac{{1.8125 + 1.875}}{2}$ = 1.84375.

Or, f(1.84375) = (1.84375)³ – 4 * 1.84375 + 1 = -0.1073.

So, the root lies between 1.84375 and 1.875.

Or, $\frac{{1.84375 + 1.875}}{2}$ = 1.859375.

Or, f(1.859375) = (1.859375)³ – 4 * 1.84375 + 1 = -0.00912585.

So, the root is 1.859375.

So, the required root is 1.86.

d) x³-x-4=0 lying between 1 and 2 correct to three places of decimal.

Solution:

Let f(x) = x² – x – 4

Or, f(1) = 1 – 1 – 4 = -4.

And f(2) = 8 – 2 – 4 = 2.

Or, $\frac{{1 + 2}}{2}$ = 1.5

Or, f(1.5) = (1.5)³ – (1.5) – 4 = -2.125.

So, the root lies between 1.5 and 2.

Or, $\frac{{1.5 + 2}}{2}$ = 1.75

Or, f(1.75) = (1.75)³ – 1.75 – 4 = -0.390625.

So, the root lies in between 1.75 and 2.

Or, $\frac{{1.75 + 2}}{2}$ = 1.875.

Or, f(1.875) = (1.875)³ – 1.875 – 4 = 0.7167968.

So, the root lies between 1.75 and 1.875.

Or, $\frac{{1.75 + 1.875}}{2}$ = 1.8125.

Or, f(1.8125) = (1.8125)³ – 1.8125 – 4 = 0.1418.

So, the root lies in between 1.75 and 1.8125

Or, $\frac{{1.75 + 1.8125}}{2}$ = 1.78125.

Or, f(1.78125) = (1.78125)³ – 1.78125 – 4 = -0.129608.

So, the root lies between 1.78125 and 1.8125.

Or, $\frac{{1.78125 + 1.8125}}{2}$ = 1.796875.

Or, f(1.796875) = (1.796875)³ – 1.796875 – 4.

= 0.00480.

So, the root is 1.796875 = 1.797.

e) x²-2x-5= 0 in (2, 3) correct to three places of decimal.

[Note: In textbook mistakenly x³-2x-5= 0 is given but it is x²-2x-5= 0]

Solution:

f(x) = x² – 2x – 5

Or, f(2) = 8 – 4 – 5 = -1

And f(3) = 27 – 6 – 5 = 16.

Or, $\frac{{2 + 3}}{2}$ = 2.5.

Or, f(2.5) = (2.5)³ – 2 * 2.5 – 5 = 5.625.

So, the root lies between 2 and 2.5.

Or, $\frac{{2 + 2.5}}{2}$ = 2.25.

Or, f(2.25) = (2.25)³ – 2 * 2.25 – 5 = 1.890625.

Or, f(2.25) = (2.25)³ – 2 * 2.25 – 5 = 1.890625.

The root lies between 2 and 2.25.

Or, $\frac{{2 + 2.25}}{2}$ = 2.125.

Or, f(2.125) = (2.125)³ – 2 * 2.125 – 5 = 0.345703.

So, the root lies between 2 and 2.125.

Or, $\frac{{2 + 2.125}}{2}$ = 2.0625.

Or, f(2.0625) = (2.0625)³ – 2 * 2.0625 – 5 = -0.351318.

So, the root lies between 2.0625 and 2.125.

Or, $\frac{{2.0625 + 2.125}}{2}$ = 2.09375.

Or, f(2.09375) = (2.09375)³ – 2 * 2.09375 – 5.

= 0.00894164.

So, the root = 2.09375 = 2.094.

f) 3x-2^x=0 correct to 3 places of decimals with error less than 0.005.

Solution:

 

g) e^x-2x-1= 0 correct to 2 places of decimals within an accuracy of 10^-2.

Solution:

Let f(x) = e^x – 2x – 1

Or, f(1) = 2.71828 – 2 * 1 – 1 = -0.28172.

Or, f(2) = 7.38906 – 2 * 2 – 1 = 2.38906.

Or, f(1).f(2) = -ve.

So, the root lies between 1 and 2.

Or, $\frac{{1 + 2}}{2}$ = 1.5

Or, f(1.5) = 4.48169 – 2 * 1.5 – 1 = 0.48169.

So, the root lies between 1 and 1.5.

Or, $\frac{{1 + 1.5}}{2}$ = 1.25

Or, f(1.25) = e^1.25 – 2 * 1.25 – 1 = -0.009657.

So, the root = 1.25.

h) 2x-log_ex = 7 in (4, 5) with error less than 0.06.

Solution:

Let f(x) = 2x – log_ex – 7

Or, f(4) = 2 * 4 – log4 – 7 = -0.3863 < 0.

Or, f(5) = 2 * 5 – log5 – 7 = 1.3905 > 0.

Or, f(4).f(5) = -0.3863 * 1.3905 = -ve.

So, the root lies between 4 and 5.

Or, $\frac{{4 + 5}}{2}$ = 4.5

Or, f(4.5) = 2 * 4.5 – log4.5 – 7 = 0.4959 > 0.

So, the root lies between 4.0 and 4.5.

Or, $\frac{{4.0 + 4.5}}{2}$ = 4.25.

Or, f(4.25) = 2 * 4.25 – log4.25 – 7 = 0.05308 < 0.06.

So, the root = 4.25.

 

3.a) Show that the equation f(x) = x³ - 18 = 0 has no negative root and one positive root, and find the positive root correct to 3 places of decimal in the interval (2, 3). Use method of bisection.

Solution:

f(x) = x³ – 18

Listing the sign of terms of f(x), we have,

+ $ - $

There is only one change. So, there is one positive root.

Again, f(-x) = (-x)³ – 18 = -x³ – 18.

There is no change in sign. So, there is no negative root.

Or, f(x) = x³ – 18.

Or, f(2) = (2)³ – 18 = -10.

And f(3) = (3)³ – 18 = 9.

Or, $\frac{{2 + 3}}{2}$ = 2.5.

f(2.5) = (2.5)³ – 18 = -2.375.

So, the root lies between 2.5 and 3.

Or, $\frac{{2.5 + 3}}{2}$ = 2.75.

Or, f(2.75) = (2.75)³ – 18 = 2.796875.

So, the root lies between 2.5 and 2.75

Or, $\frac{{2.5 + 2.75}}{2}$ = 2.625.

Or, f(2.625) = (2.625)³ – 18 = 0.0087989.

The root lies between 2.5 and 2.625.

Or, f(2.5625) = (2.5625)³ – 18 = -1.173583.

So, the root lies between 2.5625 and 2.625.

Or, $\frac{{2.5625 + 2.625}}{2}$ = 2.59375.

Or, f(2.59375) = (2.59375)³ – 18 = -0.550445556.

So, the root lies in between 2.59375 and 2.625.

Or, $\frac{{2.59375 + 2.625}}{2}$ = 2.609375.

Or, f(2.609375) = (2.609375)³ – 18 = -0.2331886.

So, the root lies between 2.609375 and 2.625

Or, $\frac{{2.609375 + 2.625}}{2}$ = 2.6171875.

Or, f(2.617185) = (2.617185)³ – 18 = -0.073128223.

So, the root lies between 2.6171875 and 2.625.

Or, $\frac{{2.6171875 + 2.625}}{2}$ = 2.62109375.

Or, f(2.62109375) = (2.62109375)³ – 18 = 0.007212

So, the root is 2.62109 = 2.621.

b) Show that the equation f(x) = 2x³-5x+2=0 has one negative root and one positive root between 1 and 2. Using method of bisection, find a positive root of the equation with error less than 10^-1.

Solution:

f(x) = 2x³ – 5x + 2 = 0.

There are two sign change. Thus, there are two positive roots.

Or, f(-x) = 2(-x)³ – 5(-x) + 2 = -2x³ + 5x + 2.

There is only one change in sign, So, there is only one negative root.

Or, f(1) = 2 * 1 – 5 * 1 + 2 = -1 < 0.

Or, f(2) = 2 * 8 – 5 * 2 + 2 = 8 > 0.

Or, f(1) .f(2) = -1 * 8 = -8 < 0.

So, One positive root lies between 1 and 2.

Or, $\frac{{1 + 2}}{2}$ = 1.5.

Or, f(1.5) = 2 * (1.5)³ – 5 * 1.5 + 2 = 1.25 > 0.

So, the root lies in between 1.125 and 1.5

Or, $\frac{{1 + 1.125}}{2}$ = 1.125.

Or, f(1.125) = 2 * (1.125)³ – 5 * 1 * 1.125 + 2 = -0.7774 < 0.

So, the root lies between 1.125 and 1.5.

Or, $\frac{{1.125 + 1.5}}{2}$ = 1.3125.

Or, f(1.3125) = 2 * (1.3125)³ – 5 * 1.3125 + 2

= 0.04052 which is less than 10^-1.

So, the required root = 1.3125.

 

4. a) How many iteration (bounds) do you need to get the root if you start with a = 2 and b =3 and the tolerance is 10^-4 ?

Solution:

$\frac{{\left| {{\rm{b}} - {\rm{a}}} \right|}}{{{2^{\rm{i}}}}}$< 10^-4.

à$\frac{{3 - 2}}{{{2^{\rm{i}}}}}$< 10^-4.

à 2^-i< 10^-4.

à -i.log2 < -4.

à i >$\frac{4}{{{\rm{log}}2}}$ = $\frac{4}{{0.3010}}$ = 13.29.

So, 14 iterations ensure approximate root.

b) If (0) = -1 and f(8) = 1, how many iteration of bisections of the interval will be required to find an approximation to the root of f(x) accurate to 0.25?

Solution:

 

c) Find the minimum number of iterations required for the solution of the equation x²-3x-8=0 in the interval (2, 3) using bisection method within an accuracy of 10^-3. Also, find an approximate root correct to 4 decimal places within an accuracy of 10^-2.

Solution:

 f(x) = x² – 3x – 8 = 0.

Or, a = 2, b = 3, ԑ₁ = 10^-3.

Or, $\frac{{\left| {{\rm{b}} - {\rm{a}}} \right|}}{{{2^{\rm{i}}}}}$< ԑ_i.

Or, $\frac{{3 - 2}}{{{2^{\rm{i}}}}}$< 10^-3.

Or, 2ⁱ> 10³.

Or, i.log₁₀2 > 3log₁₀10 = 3.

Or, i >$\frac{3}{{{{\log }_{10}}2}}$ = 9.9658 $ \approx $ 10.

So, the minimum number of iteration = 10.

Or, f(2) = (2)³ – 3 * 2 – 8 = -6 < 0.

Or, f(3) = (3)³ – 3 * 3 – 8 = 10 > 0.

So, f(2).f(3) = -6 * 10 = -60 < 0.

So, the root lies between 2 and 3.

Or, $\frac{{2 + 3}}{2}$ = 2.5

Or, f(2.5) = (2.5)³ – 3 * 2.5 – 8 = 0.125.

The root lies between 2 and 2.5.

Or, $\frac{{2 + 2.5}}{2}$ = 2.25.

Or, f(2.25) = (2.25)³ – 3 * 2.25 – 8 = -3.3594 < 0

The root lies between 2.25 and 2.5.

Or, $\frac{{2.25 + 2.5}}{2}$ = 2.375.

Or, f(2.375) = (2.375)³ – 3 * 2.375 – 8 = -1.7285 < 0

So, the root lies between 2.375 and 2.5.

Or, $\frac{{2.375 + 2.5}}{2}$ = 2.4375.

Or, f(2.4375) = (2.4375)³ – 3 * 2.4375 – 8 = -0.83002 < 0.

So, the root lies between 2.4375 and 2.5.

Or, $\frac{{2.4375 + 2.5}}{2}$ = 2.46875

or, f(2.46875) = (2.46875)³ – 3 * 2.46875 – 8 = -0.35989 < 0.

So, the root lies between 2.484375 and 2.5.

Or, $\frac{{2.46875 + 2.5}}{2}$ = 2.484375

Or, f(2.484375) = (2.484375)³ – 3 * 2.484375 – 8 = -0.1192 < 0.

So, the root lies between 2.484375 and 2.5.

Or, $\frac{{2.484375 + 2.5}}{2}$ = 2.4921875.

= 0.0024.

So, the required root = 2.4922.