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    (Original post by IceKidd)
    Do you know the percentage of people that have been invited to interview this year for maths and joint courses ( i imagine they have all been decided even if not all applicants have received them yet)?
    This is by no means official stats, but I think for the three CompSci courses it will work out to about 45% interviewed. We saw a big increase in the number of applications for CompSci courses this year, so the % being interviewed has fallen a bit this time around for us.

    OxfordMathsDept might be able to help you on the Maths Institute figures. (Quoted below to get her attention.)

    (Original post by OxfordMathsDept)
    OxfordMathsDept
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    Hi guys. I just wondered how people tend to approach the 'Prove that...' questions. I sometimes just don't know where to start or how to get into the problem.
    ie) Prove there are infinite numbers of primes
    ie) Prove that for all multiples of 9, the digits add up to 9.

    Thanks in advance
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    I'm new here so hello to all!
    I have just received an interview in Mathematics and Philosophy. I am from Poland and (according to Univ's webpage) am required to attend my interview at Oxford. Unfortunately because of a personal matter I would prefer to schedule a skype interview. My question is - will the request for skype interview hurt my application?
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    (Original post by ftball22)
    Hi guys. I just wondered how people tend to approach the 'Prove that...' questions. I sometimes just don't know where to start or how to get into the problem.
    ie) Prove there are infinite numbers of primes
    ie) Prove that for all multiples of 9, the digits add up to 9.

    Thanks in advance
    i) Try proving there are a finite no of primes, and derive a contradiction ( this is an extremely elegant proof by Euclid )
    ii) just consider the decimal expansion as sums of powers of 10 multiplied individually by integers from 0 to 9. Then use modular arithmetic writing 10 as 1 mod 9, or find a cunning factor of 9 out of the expansion ( rewrite 10^k as (10^k-1)+1 in general and expand). Technically they only add upto 9 in some cases, eg consider 999. It should read they add upto a multiple of 9 in base 10 to be precise.
    In geometry there is a nice technique called proof by inversion if you're into that as well as inductive, contradiction and construction proofs in general. Sometimes if you're stuck it just requires a moment to look at the problem from a different angle. For example, 'How many possible scores are there for a test where each question is graded 0 to 10 and the scores are such that the marks for a later question are never higher than a previous question?'. It falls out using a bar chart and analysing distances. The moment you make the connection between the binomial theorem the solution comes out quite easily it just requires a bit of creativity.
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    (Original post by IceKidd)
    Do you know the percentage of people that have been invited to interview this year for maths and joint courses ( i imagine they have all been decided even if not all applicants have received them yet)?
    (Original post by Oxford Computer Science Dept)
    OxfordMathsDept might be able to help you on the Maths Institute figures. (Quoted below to get her attention.)
    Attention obtained!

    For Maths and our joint schools (Stats & Phil) it's looking like we're interviewing around 54% of applicants.
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    (Original post by Noble.)
    Prepare to meet the Flynn.
    Sorry if I haven't understood it correctly lol, but I doubt he's as good as you I'm guessing.
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    (Original post by NewtonsApple)
    i) Try proving there are a finite no of primes, and derive a contradiction ( this is an extremely elegant proof by Euler )
    Sorry to be a pedant, but I'm pretty sure Euclid devised that proof to his own theorem a few thousand years before Euler was born.
    I do love Euler's proof, though. It may not be as simple but it's very beautiful.
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    (Original post by souktik)
    Sorry to be a pedant, but I'm pretty sure Euclid devised that proof to his own theorem a few thousand years before Euler was born.
    I do love Euler's proof, though. It may not be as simple but it's very beautiful.
    Oh right I always muddle up the two. *facepalm*
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    (Original post by NewtonsApple)
    Oh right I always muddle up the two. *facepalm*
    Happens to me too.
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    In general for solving a recurrence relation, e.g x'(n+1) = 3(x'n)^2 -1 is it a valid method to assume that x'n converges and then solve the equation using x'(n+1) = x'n? I think it wouldn't work necessarily for example a relation can have equal x'n values even before the convergence point, but how exactly would you write a formal solution to this problem? We could approximate the solution using approximate methods for Difference Equations, but I wonder if there is a valid solution which works without extremely good approximations.
    Appreciate the help!
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    (Original post by NewtonsApple)
    In general for solving a recurrence relation, e.g x'(n+1) = 3(x'n)^2 -1 is it a valid method to assume that x'n converges and then solve the equation using x'(n+1) = x'n? I think it wouldn't work necessarily for example a relation can have equal x'n values even before the convergence point, but how exactly would you write a formal solution to this problem? We could approximate the solution using approximate methods for Difference Equations, but I wonder if there is a valid solution which works without extremely good approximations.
    Appreciate the help!
    No, it's not a valid method in general. But if you can prove that it's a convergent sequence, setting x'(n+1)=x'n is okay; it'll help you find the convergence point. But that is only for sequences that are convergent in the first place! Take your own example with the additional condition x'(0)=1. Then the sequence is clearly unbounded.
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    (Original post by souktik)
    No, it's not a valid method in general. But if you can prove that it's a convergent sequence, setting x'(n+1)=x'n is okay; it'll help you find the convergence point. But that is only for sequences that are convergent in the first place! Take your own example with the additional condition x'(0)=1. Then the sequence is clearly unbounded.
    Even if a sequence is convergent it can have equal x'n values before the convergence point.
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    (Original post by NewtonsApple)
    Even if a sequence is convergent it can have equal x'n values before the convergence point.
    If you can prove the sequence x'n is strictly increasing or decreasing then it works
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    (Original post by NewtonsApple)
    If you can prove the sequence x'n is strictly increasing or decreasing then it works
    Eg if the general formula for x'n is 1/n, x'(n+1) = ((x'n)^-1+1)^-1 and if you set x'n =x'(n+1) you get x'n =0 which works since x'n is strictly decreasing.
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    (Original post by NewtonsApple)
    Even if a sequence is convergent it can have equal x'n values before the convergence point.
    I'm not sure what you're trying to say. I can look at a sequence <t_n>, where t_0=0, t_1=1, t_2=1 and for n \ge 3, t_n=1-\frac{1}{n}. It converges to 1, and we have t_1=t_2=1 as well. Even if <t_n> converges to t, it is possible to have t_r=t_{r+1}=t for finite r. Is this what you're saying?
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    (Original post by souktik)
    I'm not sure what you're trying to say. I can look at a sequence , where t_0=0, t_1=1, t_2=1 and for n \ge 3, t_n=1-\frac{1}{n}. It converges to 1, and we have t_1=t_2=1 as well. Even if converges to t, it is possible to have t_r=t_{r+1}=t for finite r. Is this what you're saying?
    You could get a better example though. For example, a sequence is convergent if eventually convergent so we can have a non monotonic function which touches the convergence point at successive x'n points before n tends to infinity. Define a sequence x'n = 0 for odd n, 1/n*sin ((pi/3)*n) for even n. Then we have the whole sequence converges to 0, but at n=6,7 we have a solution to x'(n+1) = x'n.

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    Q. Prove (n!)! is divisible by n!^(n-1)!

    Does anyone know how to prove this?
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    (Original post by kapur)
    Q. Prove (n!)! is divisible by n!^(n-1)!

    Does anyone know how to prove this?
    First prove that n! divides the product of any n consecutive natural numbers.
    Now note that (n!)! is the product of n!=n.(n-1)! consecutive natural numbers. Hence proved.

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    (Original post by Oxford Computer Science Dept)
    This is by no means official stats, but I think for the three CompSci courses it will work out to about 45% interviewed. We saw a big increase in the number of applications for CompSci courses this year, so the % being interviewed has fallen a bit this time around for us.
    So is it right to say around 1 in 2 interviewees will be rejected (if ~20% get accepted)?
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    (Original post by TSR561)
    So is it right to say around 1 in 2 interviewees will be rejected (if ~20% get accepted)?
    I thought it was 1 in 3 :confused:

    Sent from my GT-N7100 using Tapatalk
 
 
 
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