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 tutorials 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 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.

Here we are examining the imaginary number ‘i’, defined as the square root (sqrt) of ‘-1’.

We begin by using Euler’s identity:

e^(i*π)+1=0

e^(i*π)=-1

Since sqrt(x) is the same as x^(1/2):

(e^(i*π))^(1/2)=sqrt(-1)

Using (a^b)^c=a^(b*c) and the definition of i=sqrt(-1):

e^(i*π/2)=i

Then to achieve i^i as specified in the question:

(e^(i*π/2))^i=i^i

Using (a^b)^c=a^(b*c) again:

e^(-π/2)=i^i

Since the LHS has no imaginary part, the RHS is also real. We have proven that i^i is real and equal to e^(-π/2).

Answered by Lewis C.

Studies Natural Sciences at Durham

When differentiating cos inverse x, the typical method is to make y equal to cos inverse x.

By taking cos of both sides: x = cosy.

You can then differentiate with respect to y, obtaining that: (dx/dy) = - siny

Using our knowledge of derivatives, we now know that: (dy/dx) = -1/(siny)

From x = cosy, x^2 = (cosy)^2

= 1 - (siny)^2

(siny)^2 = 1 - x^2

siny = (1-x^2)^(1/2)

Combining this with the equation stating (dy/dx), we get:

(dy/dx) = (-1)/((1-x^2)^(1/2))

Since y is equal to the cos inverse function, this is now equal to the derivative of cos inverse x.

To solve the the problem we need to recognize what type of differentiation technique we shall be employing

y = (e^{x})^{7}

the x unction which we are diferentiating is a power of an exponential function therefore we must employ a substituion method to solve this

if u = e^{x}

therefore y = (u)^{7}

dy/du = 7(u)^{6}

we can say du/dx = e^{x}

therefore dy/dx = dy/~~du ~~ * ~~du~~/dx

dy/dx = 7(e^{x})^{6} * e^{x}

dy/dx = 7(e^{x})^{6} * e^{x}

dy/dx = 7(e^{x})^{7}

Modulus-argument form implies that we should express z in terms of its straight line distance from the origin and the angle this straight line would make with the x axis. This is expressed as z = r e^{i*theta} where r is the modulus and theta is the argument. We thus wish to express r and theta in terms of x and y.

This problem is best solved visually by considering an Argand diagram for the general complex number, z. In this way we can see that z is represented as a point x units along the x-axis and y units along the y-axis, forming a right angled triangle with the vertical and horizontal.

As with any right angled triangle, using Pythagoras, we can see that the length of the hypotenuse (I.e. the distance from the origin, the modulus) is given by r = sqrt(x^{2}+y^{2}). Similarly, using basic trigonometry we can also see that the angle between this line and the x axis (theta, the argument) is given by theta = arctan(y/x).

This means that in modulus-argument form:

z = sqrt(x^{2}+y^{2}) exp(i*arctan(y/x))

Answered by Ben H.

Studies Theoretical Physics at Birmingham

Proof by induction has three core elements to it. To start with you must prove that the statement is true for the 'basic case'. For the most part this is 1, but some questions state it is higher.

Do this by subbing 1 into the equation and ensuring that it is divisible by 3

1^3 =1

5(1)=5

1+5=6 6/3=2 Therefore divisible by three and true for 1.

Then in order to futher prove it, we are going to assume that this is true for n=k

leaving us with the equation k^3+5k=3a as it is divisible by 3.

The next stage is to prove true for n=k+1.

Do this by subbing k+1 into the original equation:

(k+1)^3 +5(k+1)

multiplying this out gives:

k^3+3k^2+3k+1+5k+5

Now we have already established that k^3+5k=3a so through rearranging, k^3=3a-5k.

Subbing this into the k+1 equation gives us:

3d+3k^2+3k-6. Each element is a multiple of three so by taking three out leaves us:

3(d+k^2+k-2) which is a multiple of three and thus divisible by three.

Then leave a concluding statement along the lines of:

'As n^3+5n is true for n=k, then it is true for n=k+1. As it is true for n=1, then it must be true for n is greater than 1'

When dealing with powers of complex numbers, always start by putting the quantity into exponential form.

** i** has a magnitude of

**i = exp(iπ/2)**

Now the expression is in exponential form, taking the square root is easy, using basic exponential math.

**sqrt(i) = (exp(iπ/2))^(1/2) = exp(iπ/4)**

This quantity has a modulus of** 1** and an argument of

**sqrt(i) = (1 + i)/sqrt(2)**

Company information

Popular requests

Other subjects

Cookies:

Are you there? â€“ We have noticed a period of inactivity, click yes to stay logged in or you will be logged out in 2 minutes

Your session has timed out after a period of inactivity.

Please click the link below to continue (you will probably have to log in again)

Every tutor on our site is from a **top UK university** and has been** personally interviewed** and ID checked. With over 7 applications for each tutor place, you can rest assured you’re getting the best.

As well as offering **free tutor meetings**, we **guarantee every tutor who has yet to be reviewed on this site,** no matter how much prior experience they have. Please let us know within 48 hours if you’re not completely satisfied and we’ll **refund you in full.**

Every time a student and parent lets us know they have enjoyed a tutorial with a tutor, one 'happy student' is added to the tutor's profile.

mtw:mercury1:status:ok

Version: | 3.25.2 |

Build: | cc5aabdedcc0 |

Time: | 2016-11-28T14:43:35Z |

Your message has been sent and you'll receive an email to let you know when responds.

Tutors typically reply within 24 hours.

Tutors typically reply within 24 hours.

Thanks , your message has been sent and we’ll drop you an email when replies. You should hear back within 24 hours.

After that, we recommend you set up a free, 15 minute meeting. They’re a great way to make sure the tutor you’ve chosen is right for you.

Thanks , your message has been sent and we’ll drop you an email when replies. You should hear back within 24 hours.

After that, we recommend you set up a free, 15 minute meeting. They’re a great way to make sure the tutor you’ve chosen is right for you.

**Limit reached ***(don't worry though, your email has still been sent)*

Now you've sent a few messages, we'd like to **give the tutors a chance to respond**.

Our team has been notified that you're waiting, but please contact us via

support@mytutor.co.uk or **drop us a call on +44 (0)203 773 6020** if you're in a rush.

**Office hours are 8am to 7pm**, Monday to Friday, and we pick up emails on weekends.

Thanks,

The MyTutor team

**Limit reached ***(don't worry though, your email has still been sent)*

Now you've sent a few messages, we'd like to **give the tutors a chance to respond**.

Our team has been notified that you're waiting, but please contact us via

support@mytutor.co.uk or **drop us a call on +44 (0)203 773 6020** if you're in a rush.

**Office hours are 8am to 7pm**, Monday to Friday, and we pick up emails on weekends.

Thanks,

The MyTutor team

**Limit exceeded ***(Your message will not be sent)*

You should have recieved a notification with your previous message, and **our team have also been notified that you're waiting.**

If you're in a rush, please contact us via support@mytutor.co.uk or **drop us a call on +44 (0)203 773 6020** .

**Office hours are 8am to 7pm**, Monday to Friday. We also pick up emails on weekends.

Thanks,

The MyTutor team

**Limit exceeded ***(Your message will not be sent)*

You should have recieved a notification with your previous message, and **our team have also been notified that you're waiting.**

If you're in a rush, please contact us via support@mytutor.co.uk or **drop us a call on +44 (0)203 773 6020** .

**Office hours are 8am to 7pm**, Monday to Friday. We also pick up emails on weekends.

Thanks,

The MyTutor team

OurÂ FurtherÂ Maths A Level TutorsÂ 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 aÂ FurtherÂ Maths A-Level Tutor.Â YourÂ Maths A Level TutorÂ 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.