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    I highly recommend reading through the majority of this thread. I recall it containing many helpful methods for STEP.
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    (Original post by ThatPerson)
    I highly recommend reading through the majority of this thread. I recall it containing many helpful methods for STEP.
    amazing !
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    Heres a pretty impressive technique I found on the STEP Revision Thread. THANKS TO insparato

    Maclaurin Series

    If you're asked to expand the function to x^2 like oh i dont know!



    Let







    Compare coefficients















    Thanks for Nota Bene for that one.
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    Properties of Roots of Polynomials
    Im pretty sure this hasn't been explicitly (me thinks) covered already and pops up everywhere so it's definitely worth looking at!
    Consider the equation  x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}=0 .
    If the roots of this equation are  \alpha_{1}, \alpha_{2}, ..., \alpha_{n}  ,
    our polynomial can be expressed as  (x-\alpha_{1})(x-\alpha_{2})...(x-\alpha_{n})=0 .

    By considering how each of the coefficients of the polynomial is formed, we see  a_{0}=(-1)^{n}\alpha_{1}\alpha_{2}... \alpha_{n}, a_{n-1}=-\sum_{r=0}^n \alpha_{r}
    This can be extended to other coefficients too using basic combinatorics.
    Some common examples/questions using complex numbers:
    Factorise as a product of real linear and quadratic polynomials.
    Write as a product of real quadratic polynomials.
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    (Original post by EnglishMuon)
    Properties of Roots of Polynomials
    Im pretty sure this hasn't been explicitly (me thinks) covered already and pops up everywhere so it's definitely worth looking at!
    Consider the equation  x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}=0 .
    If the roots of this equation are  \alpha_{1}, \alpha_{2}, ..., \alpha_{n}  ,
    our polynomial can be expressed as  (x-\alpha_{1})(x-\alpha_{2})...(x-\alpha_{n})=0 .

    By considering how each of the coefficients of the polynomial is formed, we see  a_{0}=\alpha_{1}\alpha_{2}... \alpha_{n}, a_{n-1}=-\sum_{r=0}^n \alpha_{r}
    This can be extended to other coefficients too using basic combinatorics.
    Some common examples/questions using complex numbers:
    Factorise as a product of real linear and quadratic polynomials.
    Write as a product of real quadratic polynomials.
    I don't get the RHS of the a_0 part. Is that supposed to be a product sign? Also you can't pull out -1 out of the product it is (-1)^n depending on n even or odd.


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    (Original post by Zacken)
    Strong induction v/s weak induction
    I swear if you don't get at least S,1 in the steps I'll kill you!
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    (Original post by gagafacea1)
    I swear if you don't get at least S,1 in the steps I'll kill you!
    I think Zacken will be the next DFranklin :yep:
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    (Original post by aymanzayedmannan)
    I think Zacken will be the next DFranklin :yep:
    Lol.
    Zackens good but not that good
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    (Original post by physicsmaths)
    I don't get the RHS of the a_0 part. Is that supposed to be a product sign? Also you can't pull out -1 out of the product it is (-1)^n depending on n even or odd.


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    Oh yea I forgot to put that backin when I had trouble with my latex and changed it a bit, but that is the product of them (-1)^n. Do you know how I do a product in latex?
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    (Original post by 16Characters....)
    Summing arctans is useful:

    \arctan a + \arctan b = \arctan \left( \frac{a + b}{1-ab} \right) + n\pi

    Which can be derived from the tangent compound angle formula.
    Could you quote this without proof in a step question?
    (Unless they asked you to prove it of course)
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    (Original post by mpaterson)
    Could you quote this without proof in a step question?
    (Unless they asked you to prove it of course)
    Yes - think about what the proof is; this follows immediately from the tangent addition formula
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    (Original post by physicsmaths)
    Lol.
    Zackens good but not that good
    Haha just a comparison to that old thread - TeeEm was actually the first one to draw this comparison 🌚
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    (Original post by shamika)
    Yes - think about what the proof is; this follows immediately from the tangent addition formula
    Thanks, yeah It's come up in a step question before I think somewhere, just wondering if you could save time by quoting it straight off
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    (Original post by mpaterson)
    Thanks, yeah It's come up in a step question before I think somewhere, just wondering if you could save time by quoting it straight off
    It'd only save like 2 minutes or such. I wouldn't even remember the result to be honest, just derive/prove it on the spot.
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    Basic but maybe helpful trig things;

     \displaystyle \arccos(-x) = \pi-\arccos(x)

     \displaystyle \arcsin(-x) = -\arcsin(x)

     \displaystyle \arctan(-x) = -\arctan(x)

    These can easily be deduced by drawing a sketch and doing some transformations, but maybe in the exam it will save you a few minutes, especially if you're not used to drawing the inverse trig graphs etc.


    When x is small;

     \displaystyle \sin(x) \approx x

     \displaystyle \tan(x) \approx x

     \displaystyle \cos(x) \approx 1-\frac{x^2}{2}

    again probably not useful, but better to post it and not need it and not post it and need it :laugh:
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    (Original post by DylanJ42)
    When x is small;

     \displaystyle \sin(x) \approx x

     \displaystyle \tan(x) \approx x

     \displaystyle \cos(x) \approx 1-\frac{x^2}{2}

    again probably not useful, but better to post it and not need it and not post it and need it :laugh:
    If anybody is wondering how you get these, look at the formula booklet under the McLaurin expansions of \sin, \cos and \tan and pick the first few terms. If these approximations are required in STEP I or II - they will be quoted underneath the question.
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    this post is more useful i think, and im surprised it hasnt been posted about yet; (maybe everyone knows it already)

     \displaystyle \left(a^n + b^n \right) = \left(a+b\right)\left(a^{n-1} - a^{n-2}b +  a^{n-3}b^2 - a^{n-4}b^3 + ... - ab^{n-2} + b^{n-1}\right)

    ^^ this is only valid when n is odd. Notice how you alternate signs the whole way down the right hand side bracket, starting and ending with plus signs

     \displaystyle \left(a^n - b^n\right) = \left(a-b\right)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1}\right)

    valid for all integer n, and notice how the right hand bracket is all plus signs
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    Who's stickied this?
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    (Original post by DylanJ42)
    this post is more useful i think, and im surprised it hasnt been posted about yet; (maybe everyone knows it already)

     \displaystyle \left(a^n + b^n \right) = \left(a-b\right)\left(a^{n-1} - a^{n-2}b +  a^{n-3}b^2 - a^{n-4}b^3 + ... - ab^{n-2} + b^{n-1}\right)

    ^^ this is only valid when n is even. Notice how you alternate signs the whole way down the right hand side bracket, starting and ending with plus signs

     \displaystyle \left(a^n - b^n\right) = \left(a-b\right)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1}\right)

    valid for all integer n, and notice how the right hand bracket is all plus signs
    This came up on Q2 of the 2015 Oxford MAT.
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    (Original post by SCalver)
    This came up on Q2 of the 2015 Oxford MAT.
    i was actually going to say that. Q2 could have been done very easily if you knew the info in my post (sadly at the time i didn't :laugh:)
 
 
 
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