LSE MA100 (Bugs fixed, September 2018)

This is the question for Lent Term week 10 of the MA100 course at the LSE. It looks at material from chapters 39.

This is the question for Lent Term week 9 of the MA100 course at the LSE. It looks at material from chapters 37 and 38.

This is the question for Lent Term week 8 of the MA100 course at the LSE. It looks at material from chapters 35 and 36.

This is the question for Lent Term week 7 of the MA100 course at the LSE. It looks at material from chapters 33 and 34.

The following is a description of parts a and b. In particular it describes the varaibles used for those parts.

This question (parts a and b) looks at optimisation problems using the langrangian method. parts a and b of the question we will ask the student to optimise the objective function f(x,y) = y + (a/b)x subject to the constraint function r^2 = (x-centre_x)^2 + (y-centre_y)^2.

The variables centre_x and centre_y take values randomly chosen from {6,7,...,10} and r takes values randomly chosen from {1,2,...,5}.

We have the ordered set of variables (a,b,c) defined to be randomly chosen from one of the following pythagorean triplets: (3,4,5) , (5,12,13) , (8,15,17) , (7,24,25) , (20,21,29). The a and b variables here are the same as those in the objective function. They are defined in this way because the minimum will occur at (centre_x - (a/c)*r , centre_y - (b/c)*r) with value centre_y - (b/c)r + (a/b) * centre_x - (a^2/bc)*r , and the maximum will occur at (centre_x + (a/c)*r , centre_y + (b/c)*r) with value centre_y + (b/c)r + (a/b) *centre_x + (a^2/bc)r. The minimisation problem has lambda = -c/(2br) and the maximation problem has lambda* = c/(2br).

We can see that all possible max/min points and values are nice rational numbers, yet we still have good randomisation in this question. :)

This is the question for Lent Term week 6 of the MA100 course at the LSE. It looks at material from chapters 31 and 32.

This is the question for Lent Term week 5 of the MA100 course at the LSE. It looks at material from chapters 29 and 30.

This is the question for Lent Term week 4 of the MA100 course at the LSE. It looks at material from chapters 27 and 28.

This is the question for Lent Term week 3 of the MA100 course at the LSE. It looks at material from chapters 25 and 26.

This is the question for Lent Term week 2 of the MA100 course at the LSE. It looks at material from chapters 23 and 24.

This is the question for Lent Term week 1 of the MA100 course at the LSE. It looks at material from chapters 21 and 22.

This is the question for week 10 of the MA100 course at the LSE. It looks at material from chapters 19 and 20.

This is the question for week 9 of the MA100 course at the LSE. It looks at material from chapters 17 and 18.

Description of variables for part b:
For part b we want to have four functions such that the derivative of one of them, evaluated at 0, gives 0; but for the rest we do not get 0. We also want two of the ones that do not give 0, to be such that the derivative of their sum, evaluated at 0, gives 0; but when we do this for any other sum of two of our functions, we do not get 0. Ultimately this part of the question will show that even if two functions are not in a vector space (the space of functions with derivate equal to 0 when evaluated at 0), then their sum could nonetheless be in that vector space. We want variables which statisfy:

a,b,c,d,f,g,h,j,k,l,m,n are variables satisfying

Function 1: x^2 + ax + b sin(cx)
Function 2: x^2 + dx + f sin(gx)
Function 3: x^2 + hx + j sin(kx)
Function 4: x^2 + lx + m sin(nx)

u,v,w,r are variables satifying
u=a+bc
v=d+fg
w=h+jk
r=l+mn

The derivatives of each function, evaluated at zero, are:
Function 1: u
Function 2: v
Function 3: w
Function 4: r

So we will define
u as random(-5..5 except(0))
v as -u
w as 0
r as random(-5..5 except(0) except(u) except(-u))

Then the derivative of function 3, evaluated at 0, gives 0. The other functions give non-zero.
Also, the derivative of function 1 + function 2 gives 0. The other combinations of two functions give nonzero.

We now take b,c,f,g,j,k,m,n to be defined as \random(-3..3 except(0)).
We then define a,d,h,l to satisfy
u=a+bc
v=d+fg
w=h+jk
r=l+mn

Description for variables of part e:

Please look at the description of each variable for part e in the variables section, first.
As described, the vectors V3_1 , V3_2 , V3_3 are linearly independent. We will simply write v1 , v2 , v3 here.
In part e we ask the student to determine which of the following sets span, are linearly independent, are both, are neither:

both: v1,v2,v3
span: v1,v1+v2,v1+v2+v3, v1+v2+v3,2*v1+v2+v3
lin ind: v1+v2+v3
neither: v2+v3 , 2*v2 + 2*v3
neither:v1+v3,v1-2*v3,2*v1-v3
neither: v1+v2,v1-v2,v1-2*v2,2*v1-v2

This is the question for week 8 of the MA100 course at the LSE. It looks at material from chapters 15 and 16.

This is the question for week 7 of the MA100 course at the LSE. It looks at material from chapters 13 and 14.

This is the question for week 6 of the MA100 course at the LSE. It looks at material from chapters 11 and 12.

This is the question for week 5 of the MA100 course at the LSE. It looks at material from chapters 9 and 10.

The following describes how we define our revenue and cost functions for part b of the question.

We have variables c, f, m, h.

The revenue function is R(q) = -c q^2 + 2mf q .
The cost function is C(q) = f q^2 - 2mc q + h .

The "revenue - cost" function is -(c+f) q^2 +2m(c+f) q - h

Differentiating, we see that there is a maximum point at m.

We pick each one of f, m, h randomly from the set {2, .. 6}, and we pick c randomly from {h+1 , ... , h+5}. This ensures that the discriminant of the "revenue - cost" function is positive, meaning there are two real roots, meaning the maximum point lies above the x-axis. I.e. we can actually make a profit.

This is the question for week 4 of the MA100 course at the LSE. It looks at material from chapters 7 and 8. The following describes how a polynomial was defined in the question. This may be helpful for anyone who needs to edit this question.

For parts a to c, we used a polynomial defined as m*(x^4 - 2a^2  x^2 + a^4 + b), where the variables "a" and "b" are randomly chosen from a set of reaosnable size, and the variable $m$ is randomly chosen from the set {+1, -1}. We can easily see that this polynomial has stationary points at -a, 0, and a. We introduced the variable "m" so that these stationary points would not always have the same classification. The variable "b" is always positive, and so this ensures that our polynomial does not cross the x-axis. The first and second derivatives; stationary points; the evaluation of the second derivative at the stationary points; the classification of the stationary points; and the axes intercepts can all be easily expressed in terms of the variables "a", "b", and "m". Indeed, this is what we did to mark the student's answers.

This is the question for week 3 of the MA100 course at the LSE. It looks at material from chapters 5 and 6. The following describes how two polynomials were defined in the question. This may be helpful for anyone who needs to edit this question.

In part a we have a polynomial. We wanted it to have two stationary points. To create the polynomial we first created the two stationary points as variables, called StationaryPoint1 and StationaryPoint2 which we will simply write as s1 ans s2 here. s2 was defined to be larger than s1. This means that the derivative of our polynomial must be of the form a(x-s1)(x-s2) for some constant a. The constant "a" is a variable called PolynomialScalarMult, and it is defined to be a multiple of 6 so that when we integrate the derivative a(x-s1)(x-s2) we only have integer coefficients. Its possible values include positive and negative values, so that the first stationary point is not always a max (and the second always a min). Finally, we have a variable called ConstantTerm which is the constant term that we take when we integrate the derivative derivative a(x-s1)(x-s2). Hence, we can now create a randomised polynomial with integers coefficients, for which the stationary points are s1 and s2; namely (the integral of a(x-s1)(x-s2)) plus ConstantTerm.

In part e we created a more complicated polynomial. It is defined as -2x^3 + 3(s1 + s2)x^2 -(6*s1*s2) x + YIntercept on the domain [0,35]. One can easily calculate that the stationary points of this polynomials are s1 and s2. Furthermore, they are chosen so that both are in the domain and so that s1 is smaller than s2. This means that s1 is a min and s2 is a max. Hence, the maximum point of the function will occur either at 0 or s2 (The function is descreasing after s2). Furthermore, one can see that when we evaluate the function at s2 we get (s2)^2 (s2 -3*s1) + YIntercept. In particular, this is larger than YIntercept if s2 > 3 *s1, and smaller otherwise. Possible values of s2 include values which are larger than 3*s1 and values which are smaller than 3*s1. Hence, the max of the function maybe be at 0 or at s2, dependent on s2. This gives the question a good amount of randomisation.

This is the question for week 2 of the MA100 course at the LSE. It looks at material from chapters 3 and 4.

This is the question for week 1 of the MA100 course at the LSE. It looks at material from chapter 2.