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Degree: Physics (Bachelors) - Durham University
My name is Sam Crawford and I'm currently in my first year studying Maths and Physics at Durham University.
I love tutoring people, so much so infact that I taught my friend AS-Level Maths over the summer once just for fun (he did offer to pay me at one point). I also did a year as a voluntary teaching assistant in a maths class at my secondary school.
I also 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.
I find that the best way to solve most problems is to draw pictures whenever possible and encourage anyone I'm teaching to do the same.
I also find that the ability to ask questions regularly to gauge the understanding of the student is one of the greatest things about one-to-one tuition because it means I can tailor the session to the student's precise needs.
If you have any questions at all, feel free to send me a message using the 'WebMail' on this site. If you want to choose me as your tutor then you can book a free 15 minute 'Meet-the-Tutor' session, also on this site, so we can talk about any problems you are having/what you would like me to teach you.
I hope to meet you soon!
|Chemistry||A Level||£20 /hr|
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
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For this explanation we will use the following example from a 2013 exam paper:
If u1= 2 and un+1=(5un-3)/(3un-1), then prove that un=(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 uk=(3k+1)/(3k-1)
We must then prove that it is also true for n = (k+1), we start by finding uk+1 using the original formula:
uk+1 = (5uk-3)/(3uk-1) = (5*(3k+1)/(3k-1) - 3)/(3*(3k+1)/(3k-1) - 1) = ... = (3k+4)/(3k+2)
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:
uk+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, uk+1 should now be in the form that we want to prove for unor 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:
u1 = 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!see more
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] + csee more