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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

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I always look up the board my students are taking so the Online Lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

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Find the cube roots of 2^{1/2}cis(pi/4)

2^{1/2}cis(pi/4+ 2pi *k), for every integer k

By DeMoivre:

2^{1/2*1/3}cis((pi/4+ 2pi *k)/3)^{3}2^{1/6}cis(pi/12+ 2/3*pi *k)

Taking k=0,1,2 gives the three cube roots:z_{1}= 2^{1/6}cis(pi/12) (k=0)z_{2}= 2^{1/6}cis(3pi/4) (k=1)z_{3}= 2^{1/6}cis(17pi/12) (k=2)

2

By DeMoivre:

2

Taking k=0,1,2 gives the three cube roots:z

In order to understand where that annoying "+C" term comes from that is so easily forgotten from the end of the solution, let's start with the general thought process of solving an indefinite integral. When facing such a question, that is, "calculate the (indefinite) integral of this and that function", we think of the problem as follows. What is that function that if I differentiate I get the function I have to integrate? For example, let's say we want to find the indefinite integral of the function F(x) = x, which is a straight line. In our example the question then becomes "What is the function whose derivate is F(x) = x?".

If you have practised the rules of differentiation enough, then this is simple, right? If I differentiate G(x) = 0.5 * x^{2} I get G'(x) = 2 * 0.5 * x = x = F(x), which is our solution! But what if I differentiate G(x) = 0.5 * x^{2} + 1? I get G'(x) = 2 * 0.5 * x + 0 = x = F(x), which is also the solution! How about G(x) = 0.5 * x^{2 }- 47? It leads to the same result. So how can we bring all the possible solutions to the question under one roof? This is where the "+C" term comes from: if we say that the solution is G(x) = 0.5 * x^{2} + C (where C is a constant), then our solution still remains G'(x) = F(x) = x, regardless whether C is 0 or -98549 or pi. In a nutshell, the "+C" term is there to express that we don't know what the exact is solution is, and writing it this way describes all the possible solutions to the question. Lastly, note that you don't have to stick to "+C"; you can use all kinds of notation instead as long as you make it clear that it is a constant.

If you have practised the rules of differentiation enough, then this is simple, right? If I differentiate G(x) = 0.5 * x

First, we must evaluate what is given in the question. As it can be seen, the expression indicates that the problem consists of a first-order differential equation. We are also given the values of x and their respective y value. These indicate that the problem should be integrated and then solved to obtain the value for the integration constant. Finally, we must calculate the value of y for when x = 2.
Following these steps, the differential equation can be integrated to give y = 1/2*10*exp(2x) - 4*x + C. We are given that y = 6 when x = 0, thus the value of C is calculated as C = 6 - 5*exp(0) = 1. Thus the general expression of y is y = 5*exp(2x) - 4*x + 1. Substituting in the value of x = 2 gives y(2) = 5*exp(2*2) - 4*2 + 1 = 5*exp(4) - 7.

First we need to be clear how to construct a proof by induction. It has two main parts, the **base case** and the **inductive step**. Let *P(n) *denote that the statement is true for a fixed n. Our base case is to check that *P(0)* is true. Our inductive step is to show that if we assume *P(n) *is true, then we can derive that *P(n+1)* is true. This usually involves rearranging one side of *P(n+1)* untill we see one side of *P(n)* in some part of it then substituting the other side of *P(n)* and rearranging our way to the unseen side of *P(n+1)*. Finally we must remember to **conclude** the proof, drawing together what we have shown to show the statement is true in general. The conclusion is pretty much the same for any proof of this type.
The **base case** is simpler. Substitute *n=0 *into the LHS, then the RHS and check the inequality holds. To complete this part you need to make sure you understand the values of *2*^{0} and *0!*. For the **inductive step** assume *P(n)*. Write down this statement in full. Somewhere else write the statement for *P(n+1)* and remember it is our goal to derive this inquality. Starting with the LHS of *P(n+1)* argue your way to the RHS remembering to respect the inequality, so each new line should be greater or equal to the last. To complete this part it would be helpful to write out the meaning of some parts of the statement such as *(2(n+1))!* and *(n+1)! *^{2}. Our **conclusion** is that if *P(n)* is true then so is *P(n+1)*, and since *P(0) *is true, the statement is true for evey integer *n>=0*.

Answered by Connor F.

Studies Computer Science with Mathematics BSc at Manchester (Computer Science)

Unlike if we had just x in the numerature, here the chain rule will do us no good. Integration by parts in this case also is little help - we end up with expressions at least as difficult as we started with. For example:
Int x^{2}/(1+x^{2})dx = x^{2}tan^{-1}x - Int 2x tan^{-1}x dx
In this case, a very simple approach is to simply add and subtract 1 from the denominator:
Int x^{2}/(1+x^{2})dx = Int (x^{2}+1-1)/(1+x^{2})dx= Int (1+x^{2})/(1+x^{2}) - 1/(1+x^{2})dx = Int 1 - 1/(1+x^{2})dx
From here, do you know know you to solve the problem?

Euclid's algorithm is really useful to be able to, firstly, see if two numbers are co-prime, in other words to see if they share any common factors, but also to find solutions to equations. Say we have two integers that satisfy:
32x + 24y = 16
Then we use Euclid's algorithm to first calculate the greatest common divisor (gcd) of 32 and 24. Hopefully, the method of this is ok? So we get gcd(32,24) = 8. Now, we can reverse what we did to get our solutions to the equation above. But don't forget that we had the equation equal to 16, not 8. This is often used in exams to trip up students, so look out for that.

Answered by Abby R.

Studies Mathematics at Leeds

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