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STEP Tricks and Hacks Thread

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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!
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 :wink:
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<0y>0 -or- y<0 (depending on the coefficient of x2x^2 )for all values of x in a quadratic.

b2−4ac<0b^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.


Posted from TSR Mobile
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:u: m8 cnt wait to c u in real life
Original post by Duke Glacia
those were the days:u: m8 cnt wait to c u in real life


yes fam. im a delight.
@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.
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

Examples:

Spoiler

General logic:

Spoiler

Methods of Proof:

Spoiler

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. :smile:
(edited 8 years ago)
anyone have any nice resources for cubic graphs?

to help answer questions like x3+bx2+6x+c \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 :frown:
Original post by DylanJ42
anyone have any nice resources for cubic graphs?

to help answer questions like x3+bx2+6x+c \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 :frown:


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.
(edited 8 years ago)
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?
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.
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 :biggrin:
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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. :smile:


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 :tongue: ) - 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  ⟹  B  ⟺  ¬B  ⟹  ¬AA \implies B \iff \neg B \implies \neg A

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

But we also know that ¬B  ⟹  ¬A\neg B \implies \neg A is literally always true unless we have that ¬B\neg B is true and ¬A\neg A is false, this can be reworded as ¬B∧¬¬A \neg B \wedge \neg \neg A, simplifying gets us: A∧¬B A \wedge \neg B oh look... what a coincidence...
(edited 8 years ago)

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