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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

Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your Online Lessons are recorded. Make the most out of your live session, then play it back after

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 Online Lessons. In my Online Lessons, 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 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.

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.

The **general solution** y_{GS}(t) to a differential equation in the form ay''(t)+by'(t)+cy(t)=f(t) is the sum of the **complementary function** y_{CF}(t) and a **particular solution** y_{PS}(t).

First we find the complementary function, which is the general solution to the equation y''(t)+y(t)=0, so the right hand side is zero. The*characteristic equation* associated with this equation is r^{2}+1=0, which has solutions r=±i. Hence y_{CF}(t) = Acos(t)+Bsin(t) where A and B are arbitrary constants.

Next we find a particular solution. We guess based on the right hand side that a particular solution has the form y_{PS}(t)=Ce^{2t}, where C is a constant to be determined.Differentiating twice we find that y_{PS}''(t)=4Ce^{2t}, so y_{PS}''(t)+y_{PS}(t)_{ }= 4Ce^{2t} + Ce^{2t} = 5Ce^{2t}.But since y''(t)+y(t)=5e^{2t}, this means that 5Ce^{2t} must equal 5e^{2t}, so we see that C=1, so y_{PS}(t)=e^{2t}.

Therefore y_{GS}(t) = y_{CF}(t) + y_{PS}(t) = Acos(t) + Bsin(t) + e^{2t}.

First we find the complementary function, which is the general solution to the equation y''(t)+y(t)=0, so the right hand side is zero. The

Next we find a particular solution. We guess based on the right hand side that a particular solution has the form y

Therefore y

For n = 1, the sum is given by (1/2)(1)(1+1), which gives 1, the expected result. We now assume that the statement is true for some k. If we look at k+1, the sum is given by 1 + 2 + ... + k + (k+1). Since we have assumed that 1 + 2 + ... + k = (1/2)(k)(k+1), this can be rewritten as (1/2)(k)(k+1) + (k+1). Simplifying this gives (1/2)(k+1)(k+2), which is the required result. If the statement is true for n = k, we have shown it to be true for n = k + 1. Since the statement is true for n = 1, it is shown to be true for all n >= 1.

First we check that this is true for n=1: S_{1} = 1/(1x3) which is equal to n/(2n+1) for n=1 therefore S_{n }= n/(2n+1) is true for n = 1. Next assume that it is true for n=k. S_{k} = k/(2k+1). Now using this assumption we check that it is true for n=k+1: S_{k+1} = S_{k}+ 1/(2(k+1) - 1)(2(k+1)+1). Rearranging this and substituting in k/(2k+1) for S_{k} we get S_{k+1} = (k+1)/(2k+3) which is consistent with the original formula. Therefore we can say that since S_{n} = n/(2n+1) is true for n=1 and whenever it is true for n=k it is also true for n=k+1, it is true for all integer values of n larger than or equal to 1.

We know that z=re^(i*theta) from the definition of the exponential form of a complex number. Hence it follows that: z^4=(re^(i*theta))^4=r^4*e^(4*i*theta) We can find z^4 by converting 8(3^0.5+i) (cartesian form) into exponential form by finding the modulus and argument of this: (I will do the working of this question on the whiteboard when asked) z^4=16e^i*(π/6 + 2Kπ) , where K is any integer. We have needed to add 2Kπ to account for the arbitrary number of rotations; any integer K can vary the argument by 2π K times however since this is a full rotation this argument will still represent the same complex number. We know that z^4=(re^(i*theta))^4=r^4*e^(4*i*theta) hence we can compare coefficients: r^4=16 implies r=2 z^4=(re^(i*theta))^4=r^4*e^(4*i*theta) 4*theta = π/6+2Kπ impllies theta = π/24 + 0.5Kπ We are not given an interval for the argument so we assume the standard interval (-π,π) and find all arguments of each root within this interval by considering all the possible values of K: k=0 case theta=1/24π k=1 case theta=13/24π k=-1 case theta =-11/24π k=-2 case theta = -23/24π Hence we can conclude the four complex roots in exponential form are: 2e^(i*1/24π), 2e^(i*13/24π), 2e^(i*-11/24π), 2e^(i*-23/24π)

Answered by George G.

Studies Mathematics with Computer Science at Southampton

This is a fairly typical example of a question from the Further Maths syllabus.

We wish to demonstrate that for all integers n greater than or equal to 4, n! > n^{2} .

We briefly recap that n! is the product of all numbers from 1 up to n. For example 4! = 4x3x2x1 = 24.

As with any induction question, there are four steps that we will work through in turn. They are the base case, the induction hypothesis, the inductive step, and finally the conclusion.

We begin with the base case. Here we select the minimum integer satisfying the conditions, in this case n=4, and demonstrate that the inequality holds true. Clearly 4! = 24 > 4^{2} = 16

The second step is to assume our induction hypothesis.

For this, we simply assume that the inequality holds for n=k, for any valid integer k.

That is, we assume k! > k^{2}.

We will now use this in step three to show that IF the inequality holds true for n=k, then it also holds for n=k+1.

For step three we consider the left hand side of the inequality, for n=k+1, and show that it is indeed larger than the right hand side for the same n:

(k+1)! = (k+1) x k x (k-1) x (k-2) x.....x 3 x 2 x 1 = (k+1) x k! This is a key observation!

(k+1)! = (k+1) x k!

(k+1) x k! > (k+1) x k^{2 } By our Induction Hypothesis

(k+1) x k^{2} = k^{3} + k^{2}

k^{3}+k^{2} > k^{2} + 2k + 1 Because k^{3} > 2k+1 for k>4

k^{3}+k^{2}> (k+1)^{2}

Now linking up the chain of inequalities from beginning to end yields (k+1)! > (k+1)^{2} Which is the original inequality with n=k+1.

Hence our inductive step holds. Practically this means that if we know the inequality holds for n=k, then it also holds for n=k+1

The final step is to note that since our base case (n=4) holds. And our inductive step (n=k+1) holds, it follows that n=5 holds. Since n=5 holds, and n=k+1 holds, sp does n = 6, we can repeat this indefinitely, showing that any n greater than or equal to 4 satisfies the given inequality.

We wish to demonstrate that for all integers n greater than or equal to 4, n! > n

We briefly recap that n! is the product of all numbers from 1 up to n. For example 4! = 4x3x2x1 = 24.

As with any induction question, there are four steps that we will work through in turn. They are the base case, the induction hypothesis, the inductive step, and finally the conclusion.

We begin with the base case. Here we select the minimum integer satisfying the conditions, in this case n=4, and demonstrate that the inequality holds true. Clearly 4! = 24 > 4

The second step is to assume our induction hypothesis.

For this, we simply assume that the inequality holds for n=k, for any valid integer k.

That is, we assume k! > k

We will now use this in step three to show that IF the inequality holds true for n=k, then it also holds for n=k+1.

For step three we consider the left hand side of the inequality, for n=k+1, and show that it is indeed larger than the right hand side for the same n:

(k+1)! = (k+1) x k x (k-1) x (k-2) x.....x 3 x 2 x 1 = (k+1) x k! This is a key observation!

(k+1)! = (k+1) x k!

(k+1) x k! > (k+1) x k

(k+1) x k

k

k

Now linking up the chain of inequalities from beginning to end yields (k+1)! > (k+1)

Hence our inductive step holds. Practically this means that if we know the inequality holds for n=k, then it also holds for n=k+1

The final step is to note that since our base case (n=4) holds. And our inductive step (n=k+1) holds, it follows that n=5 holds. Since n=5 holds, and n=k+1 holds, sp does n = 6, we can repeat this indefinitely, showing that any n greater than or equal to 4 satisfies the given inequality.

Use demoivre's theorem (cosx+isinx)^n=cosnx+isinnx, put n=5 gives (cosx+isinx)^5 =cos5x +isin5x. Expand the LHS to give (let c=cos(x) and s=sin(x)); c^5+5c^4 is +10c^3 (is)^2 + 10c^2 (is)^3 + 5c(is)^4 + (is)^5.
Now compare the real parts of each side; cos5x= c5 -10c^3 s^2 + 5c s^2. Put s^2=1-c^2; cos5x = c^5 - 10 c^3 [1 - c^2] + 5 c [1 - 2 c^2 + c^4)] . Simplify to give: cos 5x= 16 (cos x)^5 -20 (cos x)^3 + 5 cos x

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Our Further Maths A Level Tutors will give you that focused, individual attention which could make all the difference to your grade. Gaps in your learning, careless mistakes and ineffective time-management during the exams can all be overcome with the expert guidance of a Further Maths A-Level Tutor. Your Maths A Level Tutor will also ensure you have plenty of past papers to complete, as practice, repetition and excellent exam technique are key to Maths A Level success.