# Numerical Analysis Questions and Answers – Taylor’s Method

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Taylor’s Method”.

1. What will be y(0.1) for y′=2x+y, y(0) = -1, with step length 0.1 using Taylor Series method?
a) -1.0948
b) 1.8451
c) 4.2455
d) 3.6851

Explanation: Given,
y′=2x+y, y(0)=-1, h=0.1, y(0.1)=?
y′=2x+y
y′′=2+y′
y′′′=y′′
y′′′′=y′′′
Now substituting, we get
y0′=2×0+y0=-1
y0′′=2+y0′=1
y0′′′=y0′′=1
y0′′′′=y0′′′=1
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2 /2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=-1+0.1⋅(-1)+((0.1)2/2!)⋅(1)+((0.1)3/3!)⋅(1)+((0.1)4/4!)⋅(1)+…
=-1-0.1+0.005+0+0+…
=-1.0948
∴y(0.1)=-1.0948

2. What will be y(0.1) for y′=(x+y)/2, y(0) = -1, with step length 0.1 using Taylor Series method?
a) -0.619
b) -0.611
c) -0.614
d) -1.0487

Explanation: Given,
y′=(x+y)/2, y(0)=-1, h=0.1, y(0.1)=?
y′=(x+y)/2
y′′=(2+2y′)/4
y′′′=8y′′/16
y′′′′=128y′′′/256
Now substituting, we get
y0′=(x0+y0)/2=-0.5
y0′′=(2+2y0′)/4=0.25
y0′′′=(8y0′′)/16=0.125
y0′′′′=(128y0′′′)/256=0.0625
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2 /2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=-1+0.1⋅(-0.5)+((0.1)2/2!)⋅(0.25)+((0.1)3/3!)⋅(0.125)+((0.1)4/4!)⋅(0.0625)+…
=-1-0.05+0.0013+0+0+…
=-1.0487
∴y(0.1)=-1.0487

3. What will be y(0.1) for y′=y, y(0) = 1, with step length 0.1 using Taylor Series method?
a) 1.4722
b) 1.1052
c) 1.4721
d) 1.4727

Explanation: Given,
y′=y, y(0)=1, h=0.1, y(0.1)=?
y′=y
y′′=y′
y′′′=y′′
y′′′′=y′′′
Now substituting, we get
y0′=y0=1
y0′′=y0′=1
y0′′′=y0′′=1
y0′′′′=y0′′′=1
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(1)+((0.1)2/2!)⋅(1)+((0.1)3/3!)⋅(1)+((0.1)4/4!)⋅(1)+…
=1+0.1+0.005+0+0+…
=1.1052
∴y(0.1)=1.1052

4. What will be y(0.1) for y′=-y, y(0) = 1, with step length 0.1 using Taylor Series method?
a) 4.1245
b) 2.9512
c) 0.9048
d) 1.7532

Explanation: Given,
y′=-y, y(0)=1, h=0.1, y(0.1)=?
y′=-y
y′′=-y′
y′′′=-y′′
y′′′′=-y′′′
Now substituting, we get
y0′=-y0=-1
y0′′=-y0′=1
y0′′′=-y0′′=-1
y0′′′′=-y0′′′=1
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(-1)+((0.1)2/2!)⋅(1)+((0.1)3/3!)⋅(-1)+((0.1)4/4!)⋅(1)+…
=1-0.1+0.005+0+0+…
=0.9048
∴y(0.1)=0.9048

5. What will be y(0.1) for y′=-tan(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) -1
b) 1
c) 2
d) -2

Explanation: Given,
y′=-tan(x), y(0)=1, h=0.1, y(0.1)=?

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y′=-tan(x)
y′′=-sec2(x)
y′′′=-2sec2(x)tan(x)
y′′′′=-4sec2(x)tan2(x)-2sec4(x)
Now substituting, we get
y0′=-tan(x0)=0
y0′′=-sec2(x0)=0
y0′′′=-2sec2(x0)tan(x0)=0
y0′′′′=-4sec2(x0)tan2(x0)-2sec4(x0)=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(0)+((0.1)2/2!)⋅(0)+((0.1)3/3!)⋅(0)+((0.1)4/4!)⋅(0)+…
=1+0+0+0+0+…
=1
∴y(0.1)=1

6. What will be y(0.1) for y′=-log(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) 1.3799
b) 1.3791
c) 1.3794
d) Undefined

Explanation: Given,
y′=-log(x), y(0)=1, h=0.1, y(0.1)=?
y′=-log(x)
y′′=-1/x
y′′′=1/x2
y′′′′=-2/x3
Now substituting, we get
y0′=-log(x0)=Undefined
So further solution is not possible.

7. What will be y(0.1) for y′=-exp(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.8948
b) 2.2354
c) 3.1258
d) 4.2596

Explanation: Given,
y′=-exp(x), y(0)=1, h=0.1, y(0.1)=?
y′=-exp(x)
y′′=-ex
y′′′=-ex
y′′′′=-ex
Now substituting, we get
y0′=-ex0p(x0)=-1
y0′′=-ex0=-1
y0′′′=-ex0=-1
y0′′′′=-ex0=-1
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(-1)+((0.1)2/2!)⋅(-1)+((0.1)3/3!)⋅(-1)+((0.1)44!)⋅(-1)+…
=1-0.1-0.005+0+0+…
=0.8948
∴y(0.1)=0.8948

8. What will be y(0.1) for y′=x 2 +y 2 , y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.9091
b) 2.3256
c) 1.2454
d) 1.1115

Explanation: Given,
y′= x2 +y2 , y(0)=1, h=0.1, y(0.1)=?
y′= x2+y2
y′′=2x+2yy′
y′′′=2+2y′2 +2yy′′
y′′′′=6y′y′′+2yy′′′
Now substituting, we get
y0′=x20+y20=1
y0′′=2×0+2y0y0′=2
y0′′′=2+2y0′2+2y0y0′′=8
y0′′′′=6y0′y0′′+2y0y0′′′=28
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(1)+((0.1)2/2!)⋅(2)+((0.1)3/3!)⋅(8)+((0.1)4/4!)⋅(28)+…
=1+0.1+0.01+0.0013+0+…
=1.1115
∴y(0.1)=1.1115

9. What will be y(0.1) for y′=-ln(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) Undefined
b) 1.62
c) 2.45
d) 2.56

Explanation: Given,
y′=-ln(x), y(0)=1, h=0.1, y(0.1)=?
y′=-ln(x)
y′′=-1/x
y′′′=1/x2
y′′′′=-2/x3
Now substituting, we get
y0′=-ln(x0)=Undefined
So further solution is not possible.

10. What will be y(0.1) for y′=sin(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.9091
b) 2.3256
c) 1.005
d) 1.1115

Explanation: Given,
y′= sin(x), y(0)=1, h=0.1, y(0.1)=?
y′=sin(x)
y′′=cos(x)
y′′′=-sin(x)
y′′′′=-cos(x)
Now substituting, we get
y0′=sin(x0)=0
y0′′=cos(x0)=1
y0′′′=-sin(x0)=0
y0′′′′=-cos(x0)=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(0)+((0.1)2/2!)⋅(1)+((0.1)3/3!)⋅(0)+((0.1)4/4!)⋅(0)+…
=1+0+0.005+0+0+…
=1.005
∴y(0.1)=1.005

11. What will be y(0.1) for y′=cos(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) 1.0998
b) 2.3256
c) 1.2454
d) 1.1115

Explanation: Given,
y′= cos(x), y(0)=1, h=0.1, y(0.1)=?
y′=cos(x)
y′′=-sin(x)
y′′′=-cos(x)
y′′′′=sin(x)
Now substituting, we get
y0′=cos(x0)=1
y0′′=-sin(x0)=0
y0′′′=-cos(x0)=-1
y0′′′′=sin(x0)=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(1)+((0.1)2/2!)⋅(0)+((0.1)3/3!)⋅(-1)+((0.1)4/4!)⋅(0)+…
=1+0.1+0+0+0+…
=1.0998
∴y(0.1)=1.0998

12. What will be y(0.1) for y′=sec(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.9091
b) 2.3256
c) 1.2454
d) 1.1

Explanation: Given,
y′= sec(x), y(0)=1, h=0.1, y(0.1)=?
y′=sec(x)
y′′=sec(x)tan(x)
y′′′=sec(x)tan2(x)+sec3(x)
y′′′′=sec(x)tan3(x)+5sec3(x)tan(x)
Now substituting, we get
y0′=sec(x0)=1
y0′′=sec(x0)tan(x0)=0
y0′′′=sec(x0)tan2(x0)+sec3(x0)=0
y0′′′′=sec(x0)tan3(x0)+5sec3(x0)tan(x0)=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅((1)+(0.1)2/2!)⋅(0)+((0.1)3/3!)⋅(0)+((0.1)4/4!)⋅(0)+…
=1+0.1+0+0+0+…
=1.1
∴y(0.1)=1.1

13. What will be y(0.1) for y′=cosec(x), y(0) = 1, with step length 0.1 using Taylor Series method?
a) Undefined
b) 2.3256
c) 1.2454
d) 1.1115

Explanation: Given,
y′= cosec(x), y(0)=1, h=0.1, y(0.1)=?
y′= cosec(x)
y′′=- cosec(x)cot(x)
y′′′= cosec(x)cot2(x)+cosec3(x)
y′′′′=-cosec(x)cot3(x)-5cosec3(x)cot(x)
Now substituting, we get

y0′= cosec(x0)=Undefined
So further solution is not possible.

14. What will be y(0.1) for y′=2x+3, y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.9091
b) 2.3256
c) 1.2454
d) 1.31

Explanation: Given,
y′=2x+3, y(0)=1, h=0.1, y(0.1)=?
y′=2x+3
y′′=2
y′′′=0
y′′′′=0
Now substituting, we get
y0′=2×0+3=3
y0′′=2=2
y0′′′=0=0
y0′′′′=0=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(3)+((0.1)2/2!)⋅(2)+((0.1)3/3!)⋅(0)+((0.1)4/4!)⋅(0)+…
=1+0.3+0.01+0+0+…
=1.31
∴y(0.1)=1.31

15. What will be y(0.1) for y′=x, y(0) = 1, with step length 0.1 using Taylor Series method?
a) 0.9091
b) 2.3256
c) 1.005
d) 1.1115

Explanation: Given,
y′= x, y(0)=1, h=0.1, y(0.1)=?
y′=x
y′′=1
y′′′=0
y′′′′=0
Now substituting, we get
y0′=x0=0
y0′′=1=1
y0′′′=0=0
y0′′′′=0=0
Putting these values in Taylor Series, we have
y1=y0+hy0′+(h2/2!)y0′′+(h3/3!)y0′′′+(h4/4!)y0′′′′+…
=1+0.1⋅(0)+((0.1)2/2!)⋅(1)+((0.1)3/3!)⋅(0)+((0.1)4/4!)⋅(0)+…
=1+0+0.005+0+0+…
=1.005
∴y(0.1)=1.005

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