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    (Original post by Zacken)
    PTSD (post traumatic step disorder) to 16, II, Q3
    Acronym inside an acronym. I like it.

    I wonder if we can go deeper.


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    (Original post by IrrationalRoot)
    Nah I just derive it in the exam if I can't remember. Also e^x is quite a tricky one to remember, so I usually derive that too just to be on the safe side.
    I thought every remembers it at the sum of the even function cosh and the odd function sinh?


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    (Original post by sweeneyrod)
    I do usually remember it, I think I just wrote it down wrong. (And technically cos is even, sine is odd!)
    Nah it's for diff stuff, sin goes to cos, cos goes to -sin, hence the odd


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    Did you guys know that the first four letters of Stephen Siklos' name make up the word STEP?

    Coincidence? I think not.

    Edit: 1, 2, 3, 4. Yay!

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    (Original post by Insight314)
    Did you guys know that the first three letters of Stephen Siklos' name make up the word STEP?

    Coincidence? I think not.


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    Four* letters, you made a typo


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    (Original post by drandy76)
    Four* letters, you made a typo


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    I actually didn't make a typo. I miscounted. I hope Camb isn't watching.


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    STEP III 2015:
    Spoiler:
    Show
    I thought this was a weird paper, due to the three questions on proof that weren't on any normal A-level stuff at all (not complaining, because they were quite nice).

    2 - pretty straightforward
    5 - also pretty straightforward, except I couldn't get the if part at the start of (ii)
    6 - gave myself 2 marks for an attempt at half of (i)
    7 - nice question
    8 - would have been an easy full if I hadn't for some reason differentiated sin as -cos and vice versa. Instead it was an easy half.
    12 - (i) fairly easy, although a bit dodgy that you didn't seem to have to prove that Rn's p.g.f. was always G(x). Didn't make much progress in (ii), and looking at the mark scheme it looked horrendous
    13 - drew a little graph, gave up

    Overall 73, a solidish 1. Not bad, although it was a bit odd I wouldn't mind a paper like it this year.
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    (Original post by Insight314)
    I actually didn't make a typo. I miscounted. I hope Camb isn't watching.


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    You counted 3 letters because triangles have 3 sides illuminati confirm
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    (Original post by Insight314)
    I actually didn't make a typo. I miscounted. I hope Camb isn't watching.


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    My boy SS just dmed me, soz but you've been relocated to London met



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    (Original post by sweeneyrod)
    You counted 3 letters because triangles have 3 sides illuminati confirm
    Wait a minute.

    1. STEP is made up of three letters.
    2. STEP is also made up of three papers - I, II, III.

    However, STEP is not made up of three letters, so by using proof by contradiction it has been proven that STEP is also not made up of three papers. STEP IV is being secretly hidden by Siklos confirmed. QED Illuminati controls STEP confirmed.

    Try to refute my logic, you mortal peasants.

    Pro-tip: You can't.


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    (Original post by drandy76)
    My boy SS just dmed me, soz but you've been relocated to London met



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    I am an optimist: at least it's not Oxford.


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    (Original post by Insight314)
    Wait a minute.

    1. STEP is made up of three letters.
    2. STEP is also made up of three papers - I, II, III.

    However, STEP is not made up of three letters, so by using proof by contradiction it has been proven that STEP is also not made up of three papers. STEP IV is being secretly hidden by Siklos confirmed. QED Illuminati controls STEP confirmed.

    Try to refute my logic, you mortal peasants.

    Pro-tip: You can't.


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    Using the principles of strong induction (thank you zacin bas$d God)
    We can consider the initial case, where step has 3 letters.
    1-S
    2-T
    3-E
    4-P
    .......... Oh my gawd, step 4 confirmed


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    (Original post by drandy76)
    Using the principles of strong induction (thank you zacin bas$d God)
    We can consider the initial case, where step has 3 letters.
    1-S
    2-T
    3-E
    4-P
    .......... Oh my gawd, step 4 confirmed


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    The 4th paper is called STEP S.

    It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

    You need SSS in STEP I-III to qualify and the paper is taken in September.

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    (Original post by jneill)
    The 4th paper is called STEP S.

    It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

    You need SSS in STEP I-III to qualify and the paper is taken in September.

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    dont u mean...

    STEPtember???
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    (Original post by drandy76)
    Using the principles of strong induction (thank you zacin bas$d God)
    We can consider the initial case, where step has 3 letters.
    1-S
    2-T
    3-E
    4-P
    .......... Oh my gawd, step 4 confirmed


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    Using proof by exhaustion:

    Case 1: STEP is a subset of Stephen Siklos.
    Case 2: STEP I, II, III are elements of the subset STEP.

    Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED

    My logic is irrefutable.


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    (Original post by jneill)
    The 4th paper is called STEP S.

    It's sat by applicants who want to skip Part IA so they can fasttrack the Tripos.

    You need SSS in STEP I-III to qualify and the paper is taken in September.

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    I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)


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    (Original post by sweeneyrod)
    dont u mean...

    STEPtember???
    That would be silly...

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    (Original post by drandy76)
    I'm so excited to be the first half of the day before I get a follow back on my way home from work (spammed suggestive text and couldn't stop because sentence was interesting)


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    That made so much sense!


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    (Original post by Insight314)
    Using proof by exhaustion:

    Case 1: STEP is a subset of Stephen Siklos.
    Case 2: STEP I, II, III are elements of the subset STEP.

    Since STEP I, II, III are subsets of STEP, and due to the fact that STEP is a subset of Stephen Siklos, then there must be a distinct subset of elements of the set Stephen Siklos that are members of it. Therefore, using proof by exhaustion, the proposition that Stephen Siklos is hiding STEP IV is true. QED

    My logic is irrefutable.


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    Gonna take a page from my boy ramanujan and do a proof by intuition:
    Consider how cool primes are(very)
    Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too

    Where's my field medal?


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    (Original post by drandy76)
    Gonna take a page from my boy ramanujan and do a proof by intuition:
    Consider how cool primes are(very)
    Consider that they're are 3 step papers, a prime, it follows that Siklos would only work in prime number of step papers so there MUST be a FIFTH step paper too

    Where's my field medal?


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    Using direct proof:

    Axiom 1: Stephen Siklos is the set with subset of STEP.
    Axiom 2: STEP is a subset with elements I, II, III; it's cardinality is a non-fixed positive integer, solely controlled by Stephen "the madman" Siklos.

    By considering the above axioms, we can see that the cardinality of set Stephen Siklos is 13. Since 1 + 3 = 4, then through direct proof we have proven that the new cardinality of the set STEP, as outlined in Axiom 2, is one more than its cardinality over the real elements of the family. Therefore, there must be a STEP IV. QED





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