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    (Original post by Zacken)
    Strong induction v/s weak induction

    However, sometimes, this isn't enough and we need to use a different kind of induction with a slight twist:
    I'm very impressed that you are familiar with this. But there's one point worth making: strong induction isn't actually "stronger" than "weak" induction since they are, in fact, logically equivalent (i.e. strong induction \Leftrightarrow weak induction), so anything that can be proved with one can be proved with the other, but of course, strong induction seems to arise naturally in some problems.
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    (Original post by atsruser)
    I'm very impressed that you are familiar with this. But there's one point worth making: strong induction isn't actually "stronger" than "weak" induction since they are, in fact, logically equivalent (i.e. strong induction \Leftrightarrow weak induction), so anything that can be proved with one can be proved with the other, but of course, strong induction seems to arise naturally in some problems.
    Definitely, from what I know we also have that 'weak' induction, 'strong' induction and the well-ordering principle are all logical equivalent and where one can be used, the other, could in principle be used, although that would be exceedingly tiresome. i.e, recasting everything into sets to use the well ordering principle would be a pain in the arse - so it's a matter of using whichever is the most convenient and aesthetically pleasing.

    Thanks for bringing this up though, it's a very good point and I'll edit my post with a link to yours!
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    (Original post by StrangeBanana)
    wtf man
    Don't tell me zackens alternative solution didn't do the same for you the first time you saw it
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    (Original post by Duke Glacia)
    Here's one idea which I found particularly useful.(Tho it may be very obvious)

    Use of Determinants to show that the y>0 -or- y<0 (depending on the coefficient of x^2 )for all values of x in a quadratic.

    b^2 -4ac<0

    Here's one Question where this may come handy.

    Whats the minimum angles to the vertical for a projectile(ball) to be realeased with speed v such that at any point of time the distance to the ball is increasing.
    I remember thatquestion. Was nice, before our interviews init.


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    Suppose you have
    y= f(x)/g(x) such that g(x) is of 2nd degree and deg(f(x))<=2
    y(g(x))=f(x)
    Then y(g(x))-f(x)=0 now a quadratic in x and use b^2-4ac to find the range of y values this curve can take! Very useful for graphs sometimes.


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    (Original post by physicsmaths)
    Suppose you have
    y= f(x)/g(x) such that g(x) is of 2nd degree and deg(f(x))<=2
    y(g(x))=f(x)
    Then y(g(x))-f(x)=0 now a quadratic in x and use b^2-4ac to find the range of y values this curve can take! Very useful for graphs sometimes.


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    Couldn't g(X) also be of 2nd degree, necessary when you rearrange you'd get a coefficient of (a-by) for the X term, which gives a quadratic in Y after using b^2-4ac


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    (Original post by drandy76)
    Couldn't g(X) also be of 1st degree, necessary when you rearrange you'd get a coefficient of (a-by) for the X term, which gives a quadratic in Y after using b^2-4ac


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    (Original post by drandy76)
    Couldn't g(X) also be of 2nd degree, necessary when you rearrange you'd get a coefficient of (a-by) for the X term, which gives a quadratic in Y after using b^2-4ac


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    Lol thats what I said.
    Either f(x) or g(x) must be of 2nd degree and the others degree can be <=2 for this to work.


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    (Original post by physicsmaths)
    Lol thats what I said.
    Either f(x) or g(x) must be of 2nd degree and the others degree can be <=2 for this to work.


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    Oh lol, haven't had my morning coffee yet soz


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    (Original post by physicsmaths)
    I remember thatquestion. Was nice, before our interviews init.


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    those were the days m8 cnt wait to c u in real life
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    (Original post by Duke Glacia)
    those were the days m8 cnt wait to c u in real life
    yes fam. im a delight.
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    Zacken can u write ur post for 'necessary and sufficient conditions'
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    (Original post by Duke Glacia)
    Zacken can u write ur post for 'necessary and sufficient conditions'
    I've got quite a lot to say for this, damn - it'll have to wait for a day when I'm less busy.
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    (Original post by Duke Glacia)
    @Zacken can u write ur post for 'necessary and sufficient conditions'

    Since Zacken is busy and I'm taking the evening off. . .

    Overview:
    Spoiler:
    Show
    The gist of this is that for two statements A andB:
    A is sufficient for B means that A\implies B. So A is said to imply B.
    This can also be written as if A then B.

    A is necessary for B means that B\implies A. So B is said to imply A.
    This can also be written as B only if A.

    Hence for A to be both necessary and sufficient forB, we must have A \iff B. In this caseA is said to be equivalent to B,logically speaking.
    This can also be written as A if and only if (iff)B.
    Examples:
    Spoiler:
    Show
    Necessary but not Sufficient:
    Let A be the statement that a number,n, is divisible by 3, and B be the statement that a number is divisible by 6.
    Then A is clearly necessary for B but certainly B is not necessary for A, take for instance 3|3 but 3\not| 6.
    Hence A is necessary, but not sufficient for B.

    Sufficient but not Necessary:
    Let A be the statement that a number,n, is divisible by 4, and B be the statement that a number is divisible by 2.
    Then Ais sufficient for B, as if 4|n for some n then certainly 2|n,but clearly A is not necessary for B.

    Necessary and Sufficient:
    Simple examples of these tend to be patently tautological, this is the best I can do. . .
    Let A be the statement that n is divisible by both 2 and 3. Let B be the statement that both 6|n.
    Then it is easily shown that A \implies B and B\implies A.
    Hence these statements are equivalent.
    General logic:
    Spoiler:
    Show
    If asked to prove a statement, one needs to be careful regarding the direction of an implication. The reason for this is that a false statement, can imply a true one. For example 0=1 \implies 0=-1, summing these, we get0=0 which is a true statement.

    It is all too easy to be a little lazy with this, for example when proving the AM-GM inequality, a common mistake by the uninitiated might be to say that:
    \dfrac{a+b}{2} \geq \sqrt{ab} \implies (a+b)^2 \geq 4ab \implies (a-b)^2 \geq 0.
    Whilst this logic is indeed valid, it does not show the desired result, why not?
    Here we have simply shown that \dfrac{a+b}{2} \geq \sqrt{ab} \implies (a-b)^2 \geq 0 but as we noted before, a false statement can imply a true one, so how do we know that the statement we started with was true?

    The simplest fix for this is to use equivalence as above, as the steps in this argument are patently reversible, so we have instead:
    \dfrac{a+b}{2} \geq \sqrt{ab} \iff (a+b)^2 \geq 4ab \iff (a-b)^2 \geq 0.
    Alternatively, one could go for a contradiction right from the off, and suppose: \dfrac{a+b}{2} &lt; \sqrt{ab}
    \implies (a+b)^2 &lt;4ab \implies (a-b)^2&lt;0, which gives us our contradiction.

    More generally one can consider a truth table, and see that if T is a true statement and F is false statement then:
    F \implies F = T
    F \implies T = T
    T \implies F = F
    T \implies T =T.
    Methods of Proof:
    Spoiler:
    Show
    The methods of proof you'll most likely run into in STEP will be inductive with the occasional odd ball thrown in there - contradiction, construction etc.
    One that is quite nice to be aware of given the current discussion is contraposition.
    That is to say A \implies B \iff \neg B \implies \neg A. Here, \neg means "not", and it might be a nice exercise in understanding to see why this is the case.
    I've probably missed something you'd like to know, or something of what Zacken was going to say. Lemme know what and I'll edit or something.
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    anyone have any nice resources for cubic graphs?

    to help answer questions like  \displaystyle x^3 +bx^2 +6x + c find values for b and c so that this curve has 3 distinct solutons, 2 distinct, 1 distinct and none etc etc

    they seem like nice questions but i can never do 'em
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    (Original post by DylanJ42)
    anyone have any nice resources for cubic graphs?

    to help answer questions like  \displaystyle x^3 +bx^2 +6x + c find values for b and c so that this curve has 3 distinct solutons, 2 distinct, 1 distinct and none etc etc

    they seem like nice questions but i can never do 'em
    As it's a cubic then it will have at least one root regardless.
    Conditions on the functional values and the type of the stationary points will give you the rest I suppose.
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    (Original post by joostan)
    As it's a cubic then it will have at least one root regardless.
    Conditions on the functional values at the stationary points will give you the rest I suppose.
    so thats really all you would do everytime? find x values for stationary points and go through all the algebra?
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    (Original post by DylanJ42)
    so thats really all you would do everytime? find x values for stationary points and go through all the algebra?
    Short of spotting a factor, possibly, I'd need to give it more thought. . .
    There is a formula for the discriminant of a cubic, and the roots too which you can use though it isn't particularly nice and I never learned it.
    It's reasonably straight forward to assess the derivative to see what types of stationary point you have and then from there easy enough to figure out how many roots you have.
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    (Original post by joostan)
    Short of spotting a factor, possibly, I'd need to give it more thought. . .
    There is a formula for the discriminant of a cubic, and the roots too which you can use though it isn't particularly nice and I never learned it.
    It's reasonably straight forward to assess the derivative to see what types of stationary point you have and then from there easy enough to figure out how many roots you have.
    ah thats fair enough

    I was hoping there would be a nice method for them, I dont really fancy a mess of bs and cs all over the page. Ill maybe try some in non exam conditions to see what they're like, I just avoid long algebra questions as much as possible in the actual mocks, mistakes are too easy to make and spotting a missed sign is a nightmare

    thank you for your help
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    (Original post by joostan)
    I've probably missed something you'd like to know, or something of what Zacken was going to say. Lemme know what and I'll edit or something.
    Aye, that was very well elucidated - thank you for that, Joostan. It had basically everything I wanted to say (except yours was properly written instead of the waffle I'd have come up with ) - just a few suggestions, you might want to add in what's wrong with the AM-GM inequality "proof" that you showed and how to fix it, this is, I find - rather important when dealing with STEP inequalities. Lots of people start from the result and show something that is obviously true, but then instead of using iff arrows or rewriting the proof in reverse they just conclude there and then.

    P.S: Nice touch adding the proof by contrapositive, I just want to add my thoughts on it for anybody who doesn't see why A \implies B \iff \neg B \implies \neg A

    We know that A \implies B is a true statement literally every single time unless we have the case that A is true and B is false, we can rewrite this as: A \implies B is a true statement unless  A \wedge \neg B.

    But we also know that \neg B \implies \neg A is literally always true unless we have that \neg B is true and \neg A is false, this can be reworded as  \neg B \wedge \neg \neg A, simplifying gets us:  A \wedge \neg B oh look... what a coincidence...
 
 
 
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