Numerical Analysis Questions and Answers – Muller’s Method

Answer: a
Explanation: Given,
Let f(x)=2x-5
Here

x	0	1	2	3
f(x)	-5	-3	-1	1

x0=2
x1=3
x2=2.5
1st iteration :
f(x0)=f(2)=(2⋅2)-5=-1
f(x1)=f(3)=(2⋅3)-5=1
f(x2)=f(2.5)=(2⋅2.5)-5=0
h1=x1-x0=3-2=1
h2=x2-x1=2.5-3=-0.5
δ1=f(x1)-f(x0)/h1=(1-(-1))/1=2
δ2=f(x2)-f(x1)/h2=(0-1)/-0.5=2
a=(δ2-δ1)/(h2+h1)=(2-2)/(-0.5+1)=0
b=a×h2+d2=(0×(-0.5))+2=2

c=f(x2)=0
x3 = x2 + \(\frac{-2c}{b ± \sqrt{b^2-4ac}}\)
x3 = x2 + \(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
= 2.5 + \(\frac{-2×0}{2+\sqrt{2^2-4×0×0}}\)
= 2.5+(0/(2+√4))
= 2.5+(0/(2+2))
= 2.5
Approximate root of the equation 2x-5=0 using Muller method is 2.5

Answer: d
Explanation: Given,
Here x²-x-1=0
Let f(x)=x²-x-1
Here

x	0	-1	-2
f(x)	-1	1	5

x0=-1
x1=0
x2=-0.5

n	x0	x1	x2	f(x0)	f(x1)	f(x2)	a	b	c	x3	ɛan
1	-1	0	-0.5	1	-1	-0.25	1	-2	-0.25	-0.618	19.0983
2	0	-0.5	-0.618	-1	-0.25	0	1	-2.2361	0	-0.618	0

Approximate root of the equation x²-x-1=0 using Muller method is -0.618.

Answer: b
Explanation: Given,
Here x³+2x²+x-9=0
Let f(x)=x³+2x²+x-9
Here

x	0	1	2
f(x)	-9	-5	9

x0=1
x1=2
x2=1.5

n	x0	x1	x2	f(x0)	f(x1)	f(x2)	a	b	c	x3	ɛan
1	1	2	1.5	-5	9	0.375	6.5	14	0.375	1.4729	1.8418
2	2	1.5	1.4729	9	0.375	0.0068	6.9729	13.3853	0.0068	1.4724	0.0343
3	1.5	1.4729	1.4724	0.375	0.0068	0	6.4452	13.3931	0	1.4724	0

Approximate root of the equation x³+2x²+x-9=0 using Muller method is 1.4724.

Answer: c
Explanation: Given,
Here tan(x)=0
Let f(x)=tan(x)
Here

x	0	1
f(x)	0	1.5574

x0=0
x1=1
x2=0.5
1st iteration :
f(x0)=f(0)=tan(0)=0

f(x1)=f(1)=tan(1)=1.5574
f(x2)=f(0.5)=tan(0.5)=0.5463
h1=x1-x0=(1-0)=1
h2=x2-x1=(0.5-1)=-0.5
δ1=f(x1)-f(x0)/h1=(1.5574-0)/1=1.5574
δ2=f(x2)-f(x1)/h2=(0.5463-1.5574)/(-0.5)=2.0222
a=(δ2-δ1)/(h2+h1)=(2.0222-1.5574)/(-0.5+1)=0.9296
b=a×h2+d2=(0.9296×(-0.5))+2.0222=1.5574
c=f(x2)=0.5463
x3 = x2 + \(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3 = x2 + \(\frac{-2c}{(b+(b)\sqrt{b^2-4ac}}\)
= 0.5 + \(\frac{-2×0.5463}{1.5574+\sqrt{(1.5574)^2-4×0.9296×0.5463}}\)
= 0.5 + \(\frac{-1.0926}{1.5574+\sqrt{0.3941}}\)
= 0.5 + \(\frac{-1.0926}{1.5574+0.6278}\)
= 0

Now,
x0=x1=1
x1=x2=0.5
x2=x3=0

2nd iteration:

f(x0)=f(1)=tan(1)=1.5574
f(x1)=f(0.5)=tan(0.5)=0.5463
f(x2)=f(0)=tan(0)=0
h1=x1-x0=0.5-1=-0.5

h2=x2-x1=0-0.5=-0.5
δ1=f(x1)-f(x0)/h1=(0.5463-1.5574)/-0.5=2.0222
δ2=f(x2)-f(x1)/h2=(0-0.5463)/-0.5=1.0926
 a=(δ2-δ1)/(h2+h1)=(1.0926-2.0222)/(-0.5±0.5)=0.9296
b=a×h2+d2=(0.9296×(-0.5))+1.0926=0.6278
c=f(x2)=0
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+ (b)\sqrt{b^2-4ac}}\)
=0+\(\frac{-2×0}{0.6278+\sqrt{0.6278^2-4×0.9296×0}}\)
=0+\(\frac{0}{0.6278+\sqrt{0.3941}}\)
=0+\(\frac{0}{0.6278+0.6278}\)
=0

Approximate root of the equation tan(x)=0 using Muller method is 0.

Answer: b
Explanation: Given,
Here -x+2^x-1.5=0
Let f(x)=-x+2^x-1.5
Here

x	0	-1	-2
f(x)	-0.5	0	0.75

x0=-1
x1=0
x2=-0.5

n	x0	x1	x2	f(x0)	f(x1)	f(x2)	a	b	c	x3	ɛan
1	-1	0	-0.5	0	-0.5	-0.2929	0.1716	-0.5	-0.2929	-1	50
2	0	-0.5	-1	-0.5	-0.2929	0	0.1716	-0.6716	0	-1	0

Approximate root of the equation -x+2^x-1.5=0 using Muller method is -1.

Answer: d
Explanation: Given,
Here 2cos(x)-x+1=0
Let f(x)=2cos(x)-x+1
Here

x	0	1	2
f(x)	3	1.0806	-1.8323

x0=1
x1=2
x2=1.5

n	x0	x1	x2	f(x0)	f(x1)	f(x2)	a	b	c	x3	ɛan
1	1	2	1.5	1.0806	-1.8323	-0.3585	-0.0693	-2.9129	-0.3585	1.3766	8.9676
2	2	1.5	1.3766	-1.8323	-0.3585	0.0095	0.054	-2.9879	0.0095	1.3797	0.2302
3	1.5	1.3766	1.3797	-0.3585	0.0095	0.0001	-0.1514	-2.9635	0.0001	1.3798	0.0019

Approximate root of the equation 2cos(x)-x+1=0 using Muller method is 1.3798.

Answer: a
Explanation: Given,
Here log(x)=0
Let f(x)=log(x)
Here

x	0	1	2
f(x)	"Undefined"	0	0.301

x0=1

x1=2
x2=1.5

1st iteration :
f(x0)=f(1)=log(1)=0
f(x1)=f(2)=log(2)=0.301
f(x2)=f(1.5)=log(1.5)=0.1761
h1=x1-x0=2-1=1
h2=x2-x1=1.5-2=-0.5
δ1=f(x1)-f(x0)/h1=(0.301-0)/1=0.301
δ2=f(x2)-f(x1)/h2=(0.1761-0.301)/-0.5=0.2499
a=(δ2-δ1)/(h2+h1)=(0.2499-0.301)/(-0.5+1)=-0.1023
b=a×h2+d2=((-0.1023)×(-0.5))+0.2499=0.301
c=f(x2)=0.1761
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
=1.5+\(\frac{-2×0.1761}{0.301+\sqrt{(0.301)^2-4×(-0.1023)×0.1761}}\)
=1.5+\(\frac{-0.3522}{0.301+\sqrt{0.1627}}\)
=1.5+\(\frac{-0.3522}{0.301+0.4033}\)

=1

Now,
x0=x1=2
x1=x2=1.5
x2=x3=1

2nd iteration:

f(x0)=f(2)=log(2)=0.301
f(x1)=f(1.5)=log(1.5)=0.1761
f(x2)=f(1)=log(1)=0
h1=x1-x0=1.5-2=-0.5
h2=x2-x1=1-1.5=-0.5
δ1=f(x1)-f(x0)/h1=(0.1761-0.301)/-0.5=0.2499
δ2=f(x2)-f(x1)/h2=(0-0.1761)/-0.5=0.3522
 a=(δ2-δ1)/(h2+h1)=(0.3522-0.2499)/(-0.5±0.5)=-0.1023
b=a×h2+d2=(-0.1023×(-0.5))+0.3522=0.4033
c=f(x2)=0
x3=x2+\(\frac{-2c}{b±\sqrt{b^2-4ac}}\)
x3=x2+\(\frac{-2c}{b+(b)\sqrt{b^2-4ac}}\)
=1+(-2)x\(\frac{0}{0.4033+\sqrt{(0.4033)^2-4×(-0.1023)×0}}\)
=1+\(\frac{0}{0.4033+\sqrt{0.1627}}\)
=1+\(\frac{0}{0.4033+0.4033}\)
=1
Approximate root of the equation log(x)=0 using Muller method is 1.

Answer: a
Explanation: Muller’s method takes a similar approach as secant method, but projects a parabola through the points, just as in secant method where we obtain a root estimate by projecting a straight line to the x-axis through two function values.

Answer: d
Explanation: Muller’s method solves equation of the form f(x)=0, where it follows an iterative approach. It takes a similar approach as Secant’s method, but projects a parabola through the points. Furthermore, instead of giving an exact root of the function, it generates the approximate value for the root.

Answer: a
Explanation: The order of convergence for Muller’s method is 1.84. It is a quantity that represent how quickly the sequence approaches the limit. As similar to Muller’s method order of convergence for Secant’s method is 1.62 and order of convergence for Newton’s method is 2.