Thank you SO MUCH for the important part you played in helping Adam get the A level grades he needed to get to the course of his choice in Newcastle. To anyone contemplating getting Adams support - just do it :-) - having a helpful, knowledgable, friendly, same sort of age chap to ask about topics you need a bit more time on to get clarity - is invaluable. And Adam is reliable - flexible - friendly and supportive. Thanks and best of luck. Brenda - Mum to the Adam :-) - this Adam tutored.

Brenda, Parent from berks

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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.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the 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.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

To use the integrating factor method your first order DE must be of the form dy/dx + f(x)y =g(x), where f(x) and g(x) are any functions that depend only on x. lets say f(x)=3x^2 and g(x)=2, (If I feel the tutee would like a greater understanding I would leave f(x) and g(x) arbitrary). Now we define our integrating factor to be e^(integral of x^2) = e^(x^3). Now we multiply our DE by this integrating factor and notice that by using the product rule backwards we get d(e^(x^3)y)/dx =2e^(x^3). (explain this step in more detail by actually computing the left hand side and showing it is equal to what we had beforehand). Now we can use standard integration and rearranging methods to find and equation of y in terms of x. I would now go through more examples with the tutee and progress on to observing them when they try and answer a problem without my help.

Answered by Ryan A.

Studies Mathematics at Warwick

If you have a claim which says something about every element in a list of elements with each element depending on previous elements, induction might be a useful starting point. In your exams, that "list of elements" is probably going to be numbers that form a sequence, something like the Fibonnaci sequence (1,1,2=1+1,3=2+1,5=3+2,8=5+3,...), which is obviously a list of numbers, and obviously each number depends on the numbers prior to it (since the nth element is the sum of the previous two elements) or a series (which is just a sequence hidden in different notation), but the idea of listing things you want to prove something about in such a way that a thing in that list depends on the things before it is quite powerful; I still use that method in my degree for proving things about quite complex objects like certain types of compression codes, so it's a good tool to have at hand.
The reason listing things like this is useful is because induction relies on two things: that the claim is true for your first element; and that if the claim is true for all elements up to the nth element, then the claim is true for the (n+1)th element (which means it's true for the (n+2)th element, and the (n+3)th element, and so on). The way it's often described it like a chain of dominos: you tip over the first one by showing that the claim is true for the first element, and you show that the dominos all knock each other over by proving that truth up to n implies truth for n+1.

Answered by Steven R.

Studies MMath Mathematics at Oxford, Balliol College

How do I express complex numbers in the form re^{iθ}? Complex numbers have a general form of a+ib, i being imaginary. To answer this it is best shown on an argand diagram. Plot your complex number, then the angle at which the complex expression is plotted is your theta and its distance is "r".
Example 3+4i
Using pythagaros we find "r": (3^{2}+4^{2})^{1/2}=5
Using trigonometry we find "θ": tan^{-1}(4/3)=0.927 Rad
Therefore 3+4i is represented as 5e^{0.927i}

We start by finding the determinant of this matrix using the cofactor method (I.e for each entry in the matrix, cross out the row and column in which it resides and find the determinant of the matrix formed, then multiply by the entry itself and sum each result). The next step is to find the matrix of cofactors using the same crossing out method as before, but this time not multiplying by the original entry, and forming a matrix of the results instead of summing them. Finally we transpose this matrix and multiply each entry by the reciprocal of the determinant.
**REMEMBER: When finding cofactors, starting from the top left and moving row by row, each 'odd' cofactor is multiplied by +1 and each 'even' cofactor by -1

Answered by Reece R.

Studies Mathematics MMath G103 at Durham

Euler's Formula is: e^{ix} = cos(x) + isin(x)
This identity comes from the Maclaurin expansion of the exponential function. The resulting maclaurin series is a power series in x with odd terms having a factor of i. Seperating the odd and even terms, the odd terms give isin(x) and the even terms give cos(x).

Answered by Luke K.

Studies Physics at Bristol

It is possible to solve this question using integration by parts. However, we note that sin(x) is an odd function, meaning that sin(-x) = -sin(x). Thus x^{2}sin(x) is also an odd function. This means that the area under x^{2}sin(x) from 0 to pi is equal to the area under x^{2}sin(x) from -pi to 0. Hence the integral of x^{2}sin(x) between -pi and pi is 0.

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