University Maths for Computing (L4) Assignment Solution

This document presents a comprehensive solution to a Maths for Computing (Level 4) assignment. The solution covers various mathematical concepts relevant to computer science, including prime numbers, greatest common divisors (GCD), least common multiples (LCM), arithmetic and geometric progressions, probability theory (binomial distribution, normal distribution), vector algebra (lines, planes, resultant forces), and calculus (derivatives, integrals, and optimization problems). The assignment explores the importance of prime numbers in computing, particularly in encryption algorithms. The solutions are detailed, providing step-by-step explanations, formulas, and calculations to aid understanding. The document includes references to supporting literature.

Maths for Computing (L4)

TABLE OF CONTENTS
LO 1.................................................................................................................................................1
Part 1............................................................................................................................................1
Part 2............................................................................................................................................1
LO 2.................................................................................................................................................3
Part 1............................................................................................................................................3
Part 2............................................................................................................................................4
LO 3.................................................................................................................................................6
Part 1 ...........................................................................................................................................6
Part 2 ...........................................................................................................................................7
LO 4.................................................................................................................................................8
Part 1............................................................................................................................................8
Part 2............................................................................................................................................9
REFERENCES .............................................................................................................................12

LO 1
Part 1
Prime numbers: 2, 3, 5, 7, 11
Solution:
Greatest common divisor (GCD)
2 = 2 * 1
3 = 3 * 1
5 = 3 * 1
7 = 3 * 1
11 = 3 * 1
Thus, the GCD of prime numbers is always one.
Least common multiple (LCM):
The least common multiple of prime numbers is equivalent to their products (Kac, 2018).
Thus, for the given pair of numbers LCM is given by:
LCM: 2 * 3 * 5 * 7 * 11 = 2310
Part 2
2.1
First term (a) = 9
Tenth term (a10) = 40.5
Last term (L) = 425.5
Solution:
The expression for the nth term of AP is:
a(n) = a + (n-1) *d where d is the common difference
The expression for the tenth term can be written as :
a10 = a +(10-1)*d
40.5 = 9 + 9d
On solving this equation we get :
Common difference d = 3.5
a. Number of terms
Solution:
1

L = a + (n-1)*d
425.5 = 9 + (n-1)*3.5
Number of terms (n) = 120
B. Sum of all terms
Solution:
Sum of n terms of A.P. = (n/2)* [2a + (n-1)*d]
On substituting values we get:
S= (120/2)* [2*9 + (120-1)*3.5]
S= 60* [434.5]
S= 26070
C. 70th term of given series
Solution:
a(70) = 9 + (70-1)*3.5
70th term = 250.5
2.2
Salary in the first year (a) = £7200
annual increment (d) = £350
Solution:
Salary in the 9th year = 7200 + (9-1)*350
= £10000
Sum of salary in the first 12 years = (12/2)* [2*7200 + (12-1)*350]
= 6* [14400 + 3850]
= £109500
Salary in the 9th year = £10000
Sum of salary in the first 12 years = £109500
2.3
Range of speeds of drilling machine: 50-750 rev/ min
Solution:
2

The general expression for a GP: a, ar, ar²....
For the given problem the first speed (a) = 50
Last speed = a*(r^5) = 750
r^5 = 750 /a = 750/50 = 15
The value of common ratio r = (15)^(1/5) = 1.7188
Thus, the terms for the GP can be determined as follows:
First term = 50
Second term = ar = 50* 1.7188 = 85.94
Third term = a*(r^2) = 50* (1.7188)² = 147.71
Fourth term = a*(r^3) = 50* (1.7188)³ = 253.89
Fifth term = a*(r^4) = 50* (1.7188)^4 = 436.39
Sixth term = 750
Rounding off to the nearest whole number, the speeds of drilling machine will be: 50, 86,
148, 254, 436, 750
Importance of prime numbers within computing
The prime numbers play very important role in computing in the field of encryption. For
the secure transmission of information between system information must be protected by means
of encryption. The algorithms required for the encryption involves large prime numbers which
are also the building blocks of natural numbers. Prime numbers have property that it is easy and
very convenient to determine large prime numbers but it is extremely tough to factorise that
number back into primes. This characteristic of prime numbers is used in computer systems to
encrypt any information From the security data security concerns in modern computer networks
prime number based algorithms are highly used and desired for the data privacy and safety (Why
do we need to know about prime numbers with millions of digits?, 2018).
LO 2
Part 1
1.1
Total number of people = 500
mean (m) = 170
Standard deviation (SD) = 9
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Solution:
Upper limit = 150
Lower limit = 195
Z value = (x – m) / SD
Z value for the lower limit = (150-170)/9 = -2.22
By using the table of partial area of normalised curve:
Area for the z value of -2.22 = 0.4868 (Feller, 2015)
Similarly,
Z value for the upper limit = (195-170)/9 = +2.78
Area for the z value of +2.78 = 0.4973 (Feller, 2015)
Total area = 0.4973 + 0.4868 = 0.9841
The probability of people having height between 150 and 195 cm = 0.9841
So number of people with height in the range 150-195 cm = 500 * 0.9841 = 492
1.2
Number of man = 20
Number of women= 33
Solution
Totol number of people in the crowdd = 20 +33 = 53
Probability of selecting a man = 20/53
Probability of selecting a woman = 33/53
Part 2
2.1
Expectation of getting 4 upwards with 3 throws of dice:
Expectation is defined as the average occurrence of any event and thus is also refereed as
the probability times the number of attempts are made.
In a single throw of a dice the probability of getting 4 upwards = 1/6
For 3 throws, probability = (1/6)*3 = 0.5
2.2
4

Total number of children (n) = 4
Solution
Let probability of having a girl = p =0.5
probability of having a boy = q = (1-p) = 0.5
By using the binomial distribution theorem we can write expand the occurrence of events p and q
in n number of trials by following equation:
(p + q)^n = (q^n) + (npq^(n-1)) + p²q^(n-2)*[n(n-1)/2ǃ]
On putting n = 4 we get:
(p + q)^4 = (q^4) + (4pq³) + (6p²q²) + (4p³q) + (p^4)
Probability (0 girl) = (q^4) = 0.0625 (i)
Probability (1 girl) = (4pq³) = 0.25 (ii)
Probability (2 girl) = (6p²q²) = 0.375 (iii)
Probability (3 girl) = (4p³q) = 0.25 (iv)
Probability (4 girl) = (p^4) = 0.0625 (v)
A. Probability of at least one girl = 1 - Probability (0 girl)
= 1- 0.0625 = 0.9375
B. Probability of at least one girl and one boy = 1 - [Probability (0 girl) +Probability (0 boy)]
= 1 – [2*0.0625]
= 0.8750
Probability theory:
This theory deals with the mathematical analysis and evaluation of the random events or
phenomenons. Though outcomes of random events cannot be predicted in advance but this
theory helps to underpin the possibilities of the expected outcomes. Random events such as coins
related experiment, dice and card games, selection as well as hash functions and load balancing
events are effectively explained by the probability theory (Butler and Stephens, 2017). Hash
function is known as the function which generates numbers in a specific range when an input
string is given. Similarly, the practical applications of load balancing can also be demonstrated
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by random events of putting balls in specific number of bins. The uniform and random nature of
probability theory helps to solve these problems by formulating chances of occurrence in a
definite frame.
LO 3
Part 1
3.1
L2 = x – 2y +2 = 0
L1 = ? Point through which LI passes (1,3)
Solution
Slope of the equation L2 can be determined as:
x – 2y +2 = 0
2y = x +2
Arranging it in y=mx+c form
y = (x/2) +1
Thus slope of line L2 = m2 = 0.5
For the perpendicular line's product of slopes = -1 (Do Carmo, 2016)
Thus, m1*m2 = -1
m1*(0.5) = -1
Slope of L1 = m1 = -2
Now the equation of line L1 can be determined by following equation:
(y-y1) = m(x-x1)
x1 = 1 and y1= 3
(y-3) = m1 (x-1)
(y-3) = (-2) (x-1)
So the equation of line L1 = 2x +y -5
3.2
Equation of plane (P) = 4x +2y +4z = -7
Solution
The unit vector normal to the plane is equivalent to the gradient. Thus, for the given plane
equation unit normal vector can be calculated as follows:
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Plane P (x,y,z)= 4x +2y +4z +7 = 0
Gradient P = 4i +2j +4k
Unit normal vector to plane = (4i +2j +4k ) / √[4² +2² + 4²]
= (4i +2j +4k ) /6
Unit normal vector to plane= (4/6)i+(1/3)j+(2/3)k
Part 2
3.2
Solution
A. (RQ)' = u +2v (By triangular law of vector addition)
B. (RN )' = 1/4(RQ)' = (1/4)(u +2v)
(As M is the mid point of RQ and N is the mid point of RM thus RN is one fourth of RQ)
C. (PN)' = [ǀuǀcosҨ + ǀuǀsinҨ] + [ǀ2vǀcosҨ + ǀ2vǀsinҨ]
(Where Ҩ is the angle between vectors)
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3.5
F1 = 30N φ1 = 45°
F2 = 40N φ2 = 150°
Solution
Vector F = [magnitude F (Cos φ)] i + [magnitude F (Sin φ)] j (Murray, 2017)
Vector F1= [30Cos 45] i + [30Sin 45] j
Vector F1 = 21.21i + 21.21j
Similarly,
Vector F2= [40Cos 150] i + [40Sin 150] j
Vector F2= -34.64i +20j
The resultant force F vector = Vector F1 +Vector F2
= 21.21i + 21.21j - 34.64i +20j
Vector F= -13.43i + 40.21j
Magnitude= √[(-13.43)² +(40.21)²] = 42.39
Tan φ = 40.21/(-13.43) = -71.51°
Magnitude of resultant force= 42.39N
Angle of resultant force= -71.51 degree
LO 4
Part 1
1.1
L = 1 + 0.00005Ҩ + 0.0000004 Ҩ²
Solution
Rate of change = dL/dҨ
dL/dҨ = 0.00005 + (2* Ҩ * 0.0000004)
L (At Ҩ = 100° C) = 0.00013 metres
L (At Ҩ = 400° C) = 0.00037 metres
8

1.2
Position x = 0.5*(9.8)*t² = 4.9t²
Solution
Velocity (v) = dx/dt
= 4.9*2*t = 9.8t
At t = 2 seconds v = 19.6
Acceleration (a) = dv/dt
= 9.8
Thus, at t = 2 seconds
velocity = 19.6 m/second
acceleration = 9.8 m/second²
Part 2
2.1
∫ 3x. dx / √(2x²+1) in the range from 0 to 2
Let (2x²+1) = u
= 3 * ∫ du / (4*√u)
= (¾) ∫ u^(-1/2) du
= (¾) [ u^(-1/2) +1 / u^(-1/2)-1]
Again substituting u = (2x²+1)
(3/2) √(2x²+1) + C
Where Cis the integration constant
On putting boundary conditions we get (3/2) √(2x²+1) + C = 3
∫ 3x. dx / √(2x²+1) in the range from 0 to 2 = 3
2.2
f(x,y) = x³ -3x² -4y² +2
Solution
Partial derivative wrt x (fx)= 3x² -6x
Partial derivative wrt y (fy) = -8y
On equating fx and fy with zero, we get
x = 2 and y = 0
9

Similarly,
fxx= 6x – 6
at x= 2 fxx = 6
fyy= -8
fxy = 0
D = fxx * fyy - (fxy) ² (Benkhettou, da Cruz and Torres, 2015)
D = 6*(-8) -0
D = -48
D<0 and thus the given function has saddle point.
2.3
Z = (x-1)² + (y-2)²
Solution
Partial derivative with respect to (WRT) x = Zx = 2 (x-1)
Partial derivative WRT y = Zy = 2 (y-2)
On equating equations i and ii to zero we get:
2 (x-1) = 0 so x=1
2 (y-2) = 0 so y =2
Thus, the function Z = (x-1)² + (y-2)² has only one stationary point which is (1,2)
Nature of the stationary point:
The nature of stationary point depends upon second partial derivatives.
Zxx = 2
Zyy = 2
Zxy = 0
D = Zxx * Zyy - (Zxy) ²
On substituting values we get:
D = 2*2 – 0 = 4
D is greater than zero (Positive) so the stationary points are either local maxima or minima.
Since D>0 and Zxx > 0 the stationary points are local minima. (Benkhettou, da Cruz and
Torres, 2015)
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Illustration 1:
Surface representation of Z=(x-1)² + (y-2)²
Illustrat
ion 2: Contour of equation Z

REFERENCES
Books and Journals
Benkhettou, N., da Cruz, A.M.B. and Torres, D.F., 2015. A fractional calculus on arbitrary time
scales: fractional differentiation and fractional integration. Signal Processing, 107,
pp.230-237.
Butler, K. and Stephens, M.A., 2017. The distribution of a sum of independent binomial random
variables. Methodology and Computing in Applied Probability, 19(2), pp.557-571.
Do Carmo, M.P., 2016. Differential Geometry of Curves and Surfaces: Revised and Updated
Second Edition. Courier Dover Publications.
Feller, W., 2015. On the normal approximation to the binomial distribution. In Selected Papers
I (pp. 655-665). Springer, Cham.
Kac, M., 2018. Statistical Independence in Probability, Analysis and Number. Courier Dover
Publications.
Murray, M.K., 2017. Differential geometry and statistics. Routledge.
Online
Why do we need to know about prime numbers with millions of digits?. 2018. [Online]. Accessed
through <https://theconversation.com/why-do-we-need-to-know-about-prime-numbers-
with-millions-of-digits-89878>
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