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    (Original post by joostan)
    It is indeed possible to write any real number as a continued fraction.
    Sadly however there's not always a nice recurrence. Even in seemingly simple cases.
    For example, \sqrt{190} has 14 recurring integers rather than just the two in the example I gave.
    There is an algorithm for computing the continued fraction of a given number, any rational's continued fraction will necessarily terminate. For a general real number you can compute the values of the integers known as partial quotients, the corresponding fraction formed by the partial quotients will tend to the real value you're computing from.
    This gives a way to find rational approximations to real values, such as \pi. Where many of the well known approximations to \pi arise from this technique.

    There are more general cases without requirements of integers and so forth, but simple continued fractions are easier to deal with.
    Interesting, thanks for the insight! Yeah I was thinking there must be ways to generalise this to a continued fraction based off of a sequence with period n, but I imagine it could get a bit fiddly. Will need to look into it more sometime
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    Day 7 Summary
    From completing many STEP mocks, I see that my mark in the paper correlates directly with my mentality at the time- If I end up thinking "I have 1 hour left to complete another question and a half at least fully", I normally end up rushing and making silly mistakes. This then leads to general frustration and most probably not completing what I wanted to! However if I think "Oh this is interesting" and I 'do it for the maths', I end up completing questions more consistently so I will have to practice getting into this frame of mind in future!.

    I looked over a BMO2 question yesterday I completed a while ago now, and it was interesting in exploring another potential solution.
    "Suppose that p is a prime number and that there are different positiveintegers  u and v such that p^2 is the mean of u^2 and v^2.
    Prove that 2p- u - v is a square or twice a square."
    Originally I was recommended to consider  (u+v)^{2}+(u-v)^{2} which immediately gave something which looked like a partial solution, but I was also interested in looking at it interms of diaphantine equation  a^{2}+b^{2}=2c^{2} . The following article was very useful in this approach http://www.math.uconn.edu/~kconrad/b...thagtriple.pdf
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    (Original post by EnglishMuon)
    Day 7 Summary
    From completing many STEP mocks, I see that my mark in the paper correlates directly with my mentality at the time- If I end up thinking "I have 1 hour left to complete another question and a half at least fully", I normally end up rushing and making silly mistakes. This then leads to general frustration and most probably not completing what I wanted to! However if I think "Oh this is interesting" and I 'do it for the maths', I end up completing questions more consistently so I will have to practice getting into this frame of mind in future!.
    I agree very much with this, only problem is that in June when we're going in to sit an exam that will determine whether we get into the university we want or not, it's hard to get into that mindset.
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    (Original post by Zacken)
    I agree very much with this, only problem is that in June when we're going in to sit an exam that will determine whether we get into the university we want or not, it's hard to get into that mindset.
    Yep, maybe. But by that time we'll be significantly better than we will be now so hopefully we'll have the confidence to relax about them I used the method of pretending I really didnt care about the exam i was doing last year and that seemed to keep me calm.
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    Day 8 Summary
    Today I finally got some time to read up on Group Theory, in particular revision of permutations (and they seem to come up in the old STEP papers so will be useful in that sense).
    This reminded me of the nice proof of "Fermat's Little Theorem" that uses little more than Lagrange's theorem and the ideas of cyclic subgroups.
    Let's define a group  G = [1,2,...,p-1] with group operation * as multiplication. Let  a \in G such that  a^{o(a)} \equiv 1modp where p is a prime. By Lagrange's Theorem,  o(a)|o(G) , but  o(G)=p-1 hence  p-1=ko(a), k \in \mathbb{Z} .
    Therefore  a^{p-1} \equiv a^{ko(a)} \equiv (a^{o(a)})^{k} \equiv 1^{k} \equiv 1 .

    So For any prime  p ,  a^{p} \equiv amodp
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    (Original post by EnglishMuon)
    ...
    You may find x \pmod{y} useful as it gives x \pmod{y}. :yep:
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    (Original post by Zacken)
    You may find x \pmod{y} useful as it gives x \pmod{y}. :yep:
    Ah thanks, it's annoying when text is italicised when it shouldn't
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    (Original post by EnglishMuon)
    Day 8 Summary
    Today I finally got some time to read up on Group Theory, in particular revision of permutations (and they seem to come up in the old STEP papers so will be useful in that sense).
    This reminded me of the nice proof of "Fermat's Little Theorem" that uses little more than Lagrange's theorem and the ideas of cyclic subgroups.
    Let's define a group  G = [1,2,...,p-1] with group operation * as multiplication. Let  a \in G such that  a^{o(a)} \equiv 1modp where p is a prime. By Lagrange's Theorem,  o(a)|o(G) , but  o(G)=p-1 hence  p-1=ko(a), k \in \mathbb{Z} .
    Therefore  a^{p-1} \equiv a^{ko(a)} \equiv (a^{o(a)})^{k} \equiv 1^{k} \equiv 1 .

    So For any prime  p ,  a^{p} \equiv amodp
    You should look at the results called Fermat-Euler, Chinese remainder, Eulers criterion if FLT interests you


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    (Original post by physicsmaths)
    You should look at the results called Fermat-Euler, Chinese remainder, Eulers criterion if FLT interests you


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    Thank you! Its funny because my maths teacher back in the day used to ask us to explain to him why all these prime factorisation questions hold but the Chinese remainder solves them all really
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    Day 9 Summary
    Recently Ive seen some interesting/useful differential equation techniques that reduce the generic problem into something simpler to solve!
    For example, consider the situation where we drop a ball of unit mass and throw a similar ball upwards from a height d above the first ball.The only forces acting on the particles are weight and a resistance of  kv . If we want to find the distance between them at a time t, we could then formulate two 2nd order diff. equations and solve but slightly nicer is to define a new variable:
    Let  x_{1} be displacement of ball A (dropped ball) and  x_{2} be displacement of ball B (thrown ball). Define  X= x_{2}-x_{1} .
    By resolving upwards we can see
     \ddot{X} = \ddot{x_{2}} - \ddot{x_{1}}= k( \dot{x_{1}}-\dot{x_{2}})
    So   \ddot{X} = -k \dot{X} .
    But by the chain rule   \ddot{X} = \dot{X} \dfrac{d \dot{X}}{dX}
    so   \dfrac{d \dot{X}}{dX} = -k .
    We now have a first order which can solve easily to find  \dot{X} in terms of X and then integrate w.r.t. X.
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    (Original post by EnglishMuon)
    Day 9 Summary
    Recently Ive seen some interesting/useful differential equation techniques that reduce the generic problem into something simpler to solve!
    For example, consider the situation where we drop a ball of unit mass and throw a similar ball upwards from a height d above the first ball.The only forces acting on the particles are weight and a resistance of  kv . If we want to find the distance between them at a time t, we could then formulate two 2nd order diff. equations and solve but slightly nicer is to define a new variable:
    Let  x_{1} be displacement of ball A (dropped ball) and  x_{2} be displacement of ball B (thrown ball). Define  X= x_{2}-x_{1} .
    By resolving upwards we can see
     \ddot{X} = \ddot{x_{2}} - \ddot{x_{1}}= k( \dot{x_{1}}-\dot{x_{2}})
    So   \ddot{X} = -k \dot{X} .
    But by the chain rule   \ddot{X} = \dot{X} \dfrac{d \dot{X}}{dX}
    so   \dfrac{d \dot{X}}{dX} = -k .
    We now have a first order which can solve easily to find  \dot{X} in terms of X and then integrate w.r.t. X.
    Nice, you're mad good at mechanics.
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    (Original post by Marxist)
    Nice, you're mad good at mechanics.
    XD Thanks, but I wouldn't go that far. Plenty of people better than me. I just like to think and try and find nice solutions
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    Day 10 Summary
    Conditions for recurrence relations to form different sequences:
    I think every STEP Q ive seen that focuses solely on recurrence relations has involved this at some point- these are small conditions that you might overlook if you were rushing, but could cost you plenty of marks!
    In particular, consider the sequence  u_{n} defined by
     u_{2n}=pu_{2n-1}, u_{2n+1}=qu_{2n}, u_{1}=a, u_{2}=b
    From this, we can deduce that
     u_{2n}= p^{n}q^{n-1}a, u_{2n+1}= (pq)^{n-1}a

    We can then look at neccessary and sufficient conditions to make our sequence have certain characteristics:

    If our sequence is geometric, there needs to be a constant common ratio between any two consecutive terms implying  p=q .
    If we want our sequence to have a particular period (i.e. a period of  k means every k terms the terms repeat/are the same), we can make
     u_{k+1}=u_{1} .
    But beware! In doing this, we will form a polynomial of degree k (with k roots). But notice how if we want the k+1th term to equal the first, this occurs for a sequence of period 1 and other integer devisers of k so they will also be roots. We therefore pick the roots that havent yet been used in other periods.

    This pdf seems to cover everything needed on recurrence relations for STEP http://db.math.ust.hk/notes_download...ebra/ae_A8.pdf
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    (Original post by EnglishMuon)
    ...
    One very useful cheat/tip is since they usually do "find condition for period 1, period 3, period etc..." then when you get your polynomial for period etc... you should use your answers to previous parts to help factor that polynomial. You'll know that one root will be the root for period 1 and you can factor it out, then etc... very useful!
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    (Original post by Zacken)
    One very useful cheat/tip is since they usually do "find condition for period 1, period 3, period etc..." then when you get your polynomial for period etc... you should use your answers to previous parts to help factor that polynomial. You'll know that one root will be the root for period 1 and you can factor it out, then etc... very useful!
    It is indeed! How was the mock btw?
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    STEP Questions on FP1 style topics (conics and roots of polynomials)
    http://www.mathshelper.co.uk/STEP%20III%202003.pdf Question 5
    http://www.mathshelper.co.uk/STEP%20III%202005.pdf Question 3
    http://www.mathshelper.co.uk/STEP%20III%202008.pdf Question 3
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    (Original post by EnglishMuon)
    STEP Questions on FP1 style topics (conics and roots of polynomials)
    http://www.mathshelper.co.uk/STEP%20III%202003.pdf Question 5
    http://www.mathshelper.co.uk/STEP%20III%202005.pdf Question 3
    http://www.mathshelper.co.uk/STEP%20III%202008.pdf Question 3
    gracias!
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    Day 11 Summary
    Save time by using limits! This is something that comes up across nearly all integration using topics in A level and STEP, in both pure maths and applied. Basically back in the day (last year) when I was stupid, I used to spend aaages coming up with solutions to routine problems that now take only a couple of minutes at most. The only difference is that I avoid having to manually work out constants of integration at all costs, even though mark schemes normally do the long winded way.

    Consider the following differential equation,
     \dfrac{dx}{dt}= \sqrt{6(1-\dfrac{1}{(x+1)^{2}})} and at  x=0, t=0 .
    If we want to find the value of x when t=2, we could manually integrate, find our initial expression with a '+c', substitute in the initial conditions, find c, substitute t=2 in. However consider the following:

    Rearranging, we have
     \displaystyle\int^2_0 \ dt = \displaystyle\int^D_0 \dfrac{1}{ \sqrt{6(1-\dfrac{1}{(x+1)^{2}})}} \ dx .
    Since we know the initial conditions, we can just form a definite integral between time t=0 and t=2, or between x=0 and x='D' where D is a constant representing the distance moved we want to find. And this works for all initial conditions, not just both equal at 0.
    This may be very useful in STEP I 2010 Question 5 (http://www.mathshelper.co.uk/STEP%20I%202010.pdf) .... (Dont worry this doesnt spoil how to solve the question )
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    Day 12 Summary
    Factorising techniques: Here are some useful things to look out for whenever factorising any expression!

    Difference of two powers to the  n :
     a^{n}-b^{n}= (a - b) \displaystyle\sum_{i=1}^{n} a^{n-i}b^{i-1} . (This is easily proven by considering the expansion  (a^{n}-b^{n})(a - b) .

    Inspection
    Consider the horrible looking expression
     6y^{3}-y^{2}-21y+2x^{2}+12x-4xy+x^{2}y-5xy^{2}+10

    Notice how the  y^2 term cant cancel with anything, suggesting a "let y=..." may be useful. In this case, the  x^2 term will have to cancel with the  x^{2}y term, so it makes sense to let y=-2. Upon trying this, all terms can be seen to cancel so  (y+2) is a factor.

    Symmetry
    Consider question B, MAT 2002:
    http://www.mathshelper.co.uk/Oxford%...est%202002.pdf

    Notice only (iii) and (i) (RHS) has the property of symmetry between x,y,z- that is if we replace all the xs,ys, zs with each other, we should get the same expression. By comparing to the left hand side, we can check to see if these properties still hold...

    Other Hints from the Question
    If the question does not make the technique required obvious, refer to earlier (and perhaps independent looking parts of the question) and repeat anything that has been used there- usually this is the solution.
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    Day 13-15 Summary
    After attending the Warwick STEP day, here are a few interesting points I picked up on:

    One of the opening plenarys demonstrated a simple way of showing some of the logarithms properties using only basic integration techniques (which we shall assume true for this example ).
    Consider the definition of the natural log of x,  ln(x)= \displaystyle\int^x_1 \dfrac{1}{t} \ dt .
    Then
      ln(x)+ ln(y)=  \displaystyle\int^x_1 \dfrac{1}{t} \ dt + \displaystyle\int^y_1 \dfrac{1}{t} \ dt .
    By performing the substitution  u=tx on  ln(y) , we have
     ln(y) = \displaystyle\int^{xy}_x \dfrac{1}{x} \dfrac{x}{u} \ du =  \displaystyle\int^{xy}_x \dfrac{1}{t} \ dt
    hence
      ln(x)+ ln(y) = \displaystyle\int^x_1 \dfrac{1}{t} \ dt + \displaystyle\int^{xy}_x \dfrac{1}{t} \ dt = \displaystyle\int^{xy}_1 \dfrac{1}{t} \ dt = ln(xy)
    Also, by noting that (via the sub.  u= \frac{1}{t} )
     -ln(x)= - \displaystyle\int^1_{ \frac{1}{x}} \dfrac{1}{t} t^{2} \ du = \displaystyle\int^{ \frac{1}{x}}_1 \dfrac{1}{u} \ du 

= ln(x^{-1})
    The previous result implies  ln(x)-ln(y)=ln( \dfrac{x}{y})
    Similar techniques can be used to show the other common properties too.

    Perhaps one of the only things I managed to extract from the hundreds of brackets in the vector session (yet still interesting) is contradictions using linearly independent vectors:

    If we have the vector equation  \lambda \mathbf{a}+ \mu \mathbf{b} =0 (where lambda, mu are scalars), we can only say this equation implies  \mathbf{a}=0 , \mathbf{b} =0 iff a and b are linearly independent (i.e.  \mathbf{a} \cdot \mathbf{b} = 0 ) as otherwise a could be expressed in terms of b and vice versa. This can be used in proof by contradictions- e.g. showing some expression=0 iff a and b satisfy these properties.

    Example:
    Suppose we have a triangle with vertices A,B,C represented by position vectors  \mathbf{a}, \mathbf{b}, \mathbf{c} respectively. If we were to find the intersections of the perpendiculars from each vertex the midpoints of the opposite side using vector equations of lines, we would see get an expression of the form above, implying that either  \mathbf{b-a} =0 or b and a are not linearly independent. Of course as we know it is a triangle, a cant equal b so the other case must hold. ( Zacken Sorry I cant give a more concrete example, the ones Ive come up with end up effectively saying the same thing but require lots or preliminary working, so I hope the basic idea is clear!)
 
 
 
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