My name is Sam Crawford, I recently graduated with a First Class BSc in Mathematics and Physics from Durham University, and have a Masters offer for Mathematics at the University of Cambridge. I love tutoring people, so much so in fact that I taught my friend AS-Level Maths over the summer once just for fun (he did offer to pay me at one point) as well as spending a year as a voluntary teaching assistant in a maths class at my secondary school. I tend to be the go-to guy at university if someone has a problem with their homework; people often come and find me to help them out. I received 4 A*'s at A-Level in Maths, Further Maths, Physics and Chemistry as well as the gold award in both the Physics and Chemistry Olympiads, so please come to me if you would like some help with those, they are great fun! I did some of the more obscure modules in Further Maths: FP2, FP3, and FP4 (AQA), as well as studying some of the M3 module just out of curiosity.

My name is Sam Crawford, I recently graduated with a First Class BSc in Mathematics and Physics from Durham University, and have a Masters offer for Mathematics at the University of Cambridge. I love tutoring people, so much so in fact that I taught my friend AS-Level Maths over the summer once just for fun (he did offer to pay me at one point) as well as spending a year as a voluntary teaching assistant in a maths class at my secondary school. I tend to be the go-to guy at university if someone has a problem with their homework; people often come and find me to help them out. I received 4 A*'s at A-Level in Maths, Further Maths, Physics and Chemistry as well as the gold award in both the Physics and Chemistry Olympiads, so please come to me if you would like some help with those, they are great fun! I did some of the more obscure modules in Further Maths: FP2, FP3, and FP4 (AQA), as well as studying some of the M3 module just out of curiosity.

Unless I'm introducing a student to a topic for the first time I have found that starting by working through past exam questions together is by far the most efficient method. This does not mean I'll sit there and watch you fill out past papers, far from it! The questions merely act as 'jumping off points' as they allow me to quickly see the key piece of understanding the student might be missing. We then explore this gap as I present a variety of different perspectives that might help the student plug it. Every student is different, and for every issue there will be one explanation that works best for you. My job then is simply to find that explanation and express it to you clearly (and enthusiastically, I live and breathe my subject, as you'll soon find out!) I also like to encourage students to tackle set problems outside of the tutorials, and message me with any quick questions they may have. The tutoring does not have to be confined to the 60 minute sessions! I can assess work to give feedback on during sessions, or set more work hand picked for each student. Whatever works for you!

Unless I'm introducing a student to a topic for the first time I have found that starting by working through past exam questions together is by far the most efficient method. This does not mean I'll sit there and watch you fill out past papers, far from it! The questions merely act as 'jumping off points' as they allow me to quickly see the key piece of understanding the student might be missing. We then explore this gap as I present a variety of different perspectives that might help the student plug it. Every student is different, and for every issue there will be one explanation that works best for you. My job then is simply to find that explanation and express it to you clearly (and enthusiastically, I live and breathe my subject, as you'll soon find out!) I also like to encourage students to tackle set problems outside of the tutorials, and message me with any quick questions they may have. The tutoring does not have to be confined to the 60 minute sessions! I can assess work to give feedback on during sessions, or set more work hand picked for each student. Whatever works for you!

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For this explanation we will use the following example from a 2013 exam paper:

* If u _{1}= 2 and u_{n+1}=(5u_{n}-3)/(3u_{n}-1), then prove that u_{n}=(3n+1)/(3n-1) for all n>=1*

The first step of any proof by induction is to make the assumption that what we want to prove is true for a particular value n = k:

*Assume there exists k such that u _{k}=(3k+1)/(3k-1)*

We must then prove that it is also true for n = (k+1), we start by finding u_{k+1} using the original formula:

* u _{k+1 }= (5u_{k}-3)/(3*

We now want to write this in terms of k+1, in this case it is fairly straightforward but other times it may be harder to see:

*u _{k+1} = (3k+4)/(3k+2) = (3(k+1) - 3 + 4)/(3(k+1) - 3 +2) = (3(k+1)+1)/(3(k+1)-1)*

When written in terms of k+1, u_{k+1 }should now be in the form that we want to prove for u_{n}or a form that can be rearranged into that one. There is still one step left however which is CRUCIAL for this to be a proper proof by induction. We have to prove this is true for a certain value of n, in this case n = 1:

* u _{1 }= 2 = (3*1+1)/(3*1-1) therefore the assumption is true for n = 1. It is therefore true for n = 1, 2, 3, ...*

This last step is usually very simple but can often be overlooked so make sure to include it!

For this explanation we will use the following example from a 2013 exam paper:

_{1}= 2 and u_{n+1}=(5u_{n}-3)/(3u_{n}-1), then prove that u_{n}=(3n+1)/(3n-1) for all n>=1

The first step of any proof by induction is to make the assumption that what we want to prove is true for a particular value n = k:

*Assume there exists k such that u _{k}=(3k+1)/(3k-1)*

We must then prove that it is also true for n = (k+1), we start by finding u_{k+1} using the original formula:

* u _{k+1 }= (5u_{k}-3)/(3*

We now want to write this in terms of k+1, in this case it is fairly straightforward but other times it may be harder to see:

*u _{k+1} = (3k+4)/(3k+2) = (3(k+1) - 3 + 4)/(3(k+1) - 3 +2) = (3(k+1)+1)/(3(k+1)-1)*

When written in terms of k+1, u_{k+1 }should now be in the form that we want to prove for u_{n}or a form that can be rearranged into that one. There is still one step left however which is CRUCIAL for this to be a proper proof by induction. We have to prove this is true for a certain value of n, in this case n = 1:

_{1 }= 2 = (3*1+1)/(3*1-1) therefore the assumption is true for n = 1. It is therefore true for n = 1, 2, 3, ...

This last step is usually very simple but can often be overlooked so make sure to include it!

This problem is all about using integration by parts, so let's start by quoting the formula for integration by parts:

∫u*(dv/dx)dx = uv - ∫v*(du/dx)dx

To get the integral we want on the left hand side we can use the subtitutions u = dv/dx = ln(x). This means that we will have to find ∫ln(x)dx, this is also done using integration by parts:

To find ∫ln(x)dx we can use the substitutions u = ln(x) and dv/dx = 1. Using the formula above will then give us:

∫ln(x)*1dx = ln(x)*x - ∫x*(1/x)dx

= xln(x) - ∫dx = xln(x) - x = x(ln(x)-1)

Using this we can now use our original substitutions in the formula to get:

∫ln(x)*ln(x)dx = ln(x)*x(ln(x)-1) - ∫x(ln(x)-1)*(1/x)dx

= xln(x)*(ln(x)-1) - ∫(ln(x)-1)dx

= xln(x)*(ln(x)-1) - x(ln(x)-1) + x + c

Now we just have to tidy this up to get our final answer:

∫(ln(x))^2dx = x[(ln(x)+1)^2 + 1] + c

This problem is all about using integration by parts, so let's start by quoting the formula for integration by parts:

∫u*(dv/dx)dx = uv - ∫v*(du/dx)dx

To get the integral we want on the left hand side we can use the subtitutions u = dv/dx = ln(x). This means that we will have to find ∫ln(x)dx, this is also done using integration by parts:

To find ∫ln(x)dx we can use the substitutions u = ln(x) and dv/dx = 1. Using the formula above will then give us:

∫ln(x)*1dx = ln(x)*x - ∫x*(1/x)dx

= xln(x) - ∫dx = xln(x) - x = x(ln(x)-1)

Using this we can now use our original substitutions in the formula to get:

∫ln(x)*ln(x)dx = ln(x)*x(ln(x)-1) - ∫x(ln(x)-1)*(1/x)dx

= xln(x)*(ln(x)-1) - ∫(ln(x)-1)dx

= xln(x)*(ln(x)-1) - x(ln(x)-1) + x + c

Now we just have to tidy this up to get our final answer:

∫(ln(x))^2dx = x[(ln(x)+1)^2 + 1] + c