STEP Tricks and Hacks Thread

Is this "long loop" terminology standard? I've never heard of it before.

First time I've seen "loop" used in the context of proofs is in this thread :smile:

. So I thought "long loop" just had to be invented. We just need to invent a notion of homotopy of proof loops and we're done for Easter.

The first time I saw it was in one of those god-awful linear algebra proofs about the equivalence of "Rank A = n" and "unique solutions of Ax=b" and about 1/2 a dozen other things. It struck me as too clever-clever at the time, and I'm not sure that I really believed the proof - it looked like stuff had been missed out.

Yup, I've never liked it. If there's an equivalence between two facts, then I like to see it demonstrated directly - for the simple reason that you see better why the equivalence is true. Sometimes it may be difficult to prove a direct equivalence, and going through an intermediate lemma is required; but this should be done only when there's little alternative.

Reply 181

atsruser

11

Original post by Gregorius

First time I've seen "loop" used in the context of proofs is in this thread :smile:

. So I thought "long loop" just had to be invented.

Yes, I think it's a pretty nice term - it ought to be made more popular.

We just need to invent a notion of homotopy of proof loops and we're done for Easter.

That would be pretty cool - you could prove the equivalence of two sets of proofs. A "proof loop" isn't a very continuous thing though - don't homotopies only work when stuff is continuous? Still, I await your upcoming paper (and subsequent Fields Medal, of which I claim, ooh, 5%, for suggesting a use of the concept) with interest.

Yup, I've never liked it. If there's an equivalence between two facts, then I like to see it demonstrated directly - for the simple reason that you see better why the equivalence is true. Sometimes it may be difficult to prove a direct equivalence, and going through an intermediate lemma is required; but this should be done only when there's little alternative.

I agree - however, it's probably the only sensible approach in those "long list of N equivalent properties" proofs - you'd have N(N-1)/2 proofs to do otherwise.

Reply 182

joostan

13

Original post by shamika

If any STEP question required that it would be much too difficult I think. It's a very neat trick though, useful for algebra courses at uni!

I agree, but @Lucasium was asking about a loop method that was beyond FM, and I couldn't really think of one.

Original post by Zacken

Wait, this is boggling me - how does this work? You get logical equivalence between a number of statements just by proving implications from one to another? :redface:

As Shamika said - once it's explained to you it seems quite obvious.

Original post by physicsmaths

This is very strong. Damn
Posted from TSR Mobile

Original post by Student403

That's pretty cool

I'm glad you liked it, but as others have said, it's not often used except in theorems requiring proof that a long sequence of statements are equivalent.
I've not seen many questions that this is applicable to, and certainly none in STEP, but it's something to look out for :smile:

(edited 8 years ago)

Reply 183

Insight314

7

Original post by joostan

I agree, but @Insight314 was asking about a loop method that was beyond FM, and I couldn't really think of one.

It was @Lucasium, not me. :biggrin:

Reply 184

joostan

13

Original post by Insight314

It was @Lucasium, not me. :biggrin:

That is my bad - being dumb :frown:

Reply 185

shamika

16

Original post by joostan

I agree, but @Lucasium was asking about a loop method that was beyond FM, and I couldn't really think of one.

Don't get me wrong, I think it's a cool trick (my mind was blown when I first saw it in lectures, but I think at the time I was just overly keen :biggrin:

), just wanted to remind people that they shouldn't worry if they haven't seen it.

Hope Cambridge is going well :smile:

Reply 186

Insight314

7

@joostan

In which Cambridge college are you studying? :smile:

Reply 187

joostan

13

Original post by Insight314

@joostan
In which Cambridge college are you studying? :smile:

Spoiler

Original post by shamika

Don't get me wrong, I think it's a cool trick (my mind was blown when I first saw it in lectures, but I think at the time I was just overly keen :biggrin:

), just wanted to remind people that they shouldn't worry if they haven't seen it.

Hope Cambridge is going well :smile:

Thanks.

Oh definitely don't want folks worrying - for me it was a tip from a (then) second year regarding a problem we were given.

Reply 188

Abscissa

6

This idea shows up here:
http://msekce.karlin.mff.cuni.cz/~krajicek/mendelson.pdf
page 18, 1.8

Reply 189

Tau

10

Original post by Zacken

Oh, and somebody remind me to do a post about multiplying by one creatively.

Did you ever get round to making this post? I'd be very interested :colondollar:

Reply 190

Zacken

OP

22

Original post by Tau

Did you ever get round to making this post? I'd be very interested :colondollar:

Nope, never did. Maybe I'll do it one of these days. :tongue:

Reply 191

username1162972

18

Original post by Zacken

Nope, never did. Maybe I'll do it one of these days. :tongue:

Thank you very much sir

Posted from TSR Mobile

Reply 192

atsruser

11

Original post by Zacken

Nope, never did. Maybe I'll do it one of these days. :tongue:

I'd like to see it too. I can only think of two tricks that fit this description:

1. Finding equivalent fractions e.g.

\frac{2}{3} = \frac{2}{3} \times 1 = \frac{2}{3} \times \frac{n}{n} = \frac{2n}{3n}

and variants thereof.

2. Changing units e.g.

1 \text{ km} = 1000 \text{ m} \Rightarrow 1 = \frac{1000 \text{ m}}{1 \text{ km}}

so

5.2 \text{ km} = 5.2 \text{ km} \times 1 = 5.2 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 5200 \text{ m}

Reply 193

Zacken

OP

22

Original post by atsruser

I'd like to see it too. I can only think of two tricks that fit this description:

Was mainly thinking about it as an integration trick, like:

\displaystyle \int \frac{x^2}{(x \sin x + \cos x)^2} \, \mathrm{d}x = \int \frac{x}{\cos x} \times \frac{x \cos x}{(x\sin x + \cos x)^2} \, \mathrm{d}x

which immediately renders it vulnerable to IBP since

Unparseable latex formula:

z = x \sin x + \cos x \Rightarrorw z' = x\cos x

.

Stuff along those lines.

Reply 194

IrrationalRoot

20

Original post by atsruser

...

Main use I can think of is rationalising denominators/dividing by complex numbers. This probably falls under finding equivalent fractions but is actually a useful technique.

Reply 195

Melanie Leconte

2

Original post by Duke Glacia

1. Adding zero creatively (i) (Zacken)
2. Trigonometrical symmetry and substitutions for integration (Zacken)
3. Trigonometrical functions as shifts of others (Zacken)
4. Implicit differentiation and logarithmic differentiation (Zacken)
5. Adding zero creatively (ii) (Zacken)
6. Algebra shenanigans with 'it suffices' (IrrationalRoot)
7. Discriminants as a useful tool for inequalities and in mechanics (Duke Glacia)
8. Weierstrass subs (Shamika)
9. Inequalities and injectivity (Zacken)
10. Summing arc-tangents (16Characters...)
11. L'Hopitals rule for limits (Joostan)
12. Factorising special polynomials and techniques to do so (Aymanzayenmannan)
13. Strong induction (Zacken)

Hopefully we can grow this into a useful and extensive collection of tricks and techniques - feel free to share your own!

So to start with Zacken posted a amazing techniques which could have saved me a lot of time and effort.

OP written by Zacken

Thank you so much for this :smile:

Reply 196

Major-fury

6

Original post by Zacken

I'd just like to showcase an astounding example of the adding zero creatively trick (even if it's been quoted already by Duke in the OP), it featured heavily in STEP II, 2005, Q8 for example and the examiners report and official solutions didn't even mention it, they decided to mention ridiculous things like hyperbolic trigonometrical substitutions instead of a simple trick.

So what we have is:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3y^2}{(1+x^2)^{5/2}}\end{equation*}

The seperable stuff is easy and you all know how to integrate

y^{-2}

, but the RHS is slightly more tricky, how would you integrate

x^3/(1+x^2)^{5/2}

- what I'd do:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{x^3 +x - x}{(1+x^2)^{5/2}} = \frac{x(1+x^2) - x}{(1+x^2)^{5/2}} = \frac{x}{(1+x^2)^{3/2}} - \frac{x}{(1+x^2)^{5/2}} \end{equation*}

Whereby integrating is very standard now, since

(1+x^2)' = 2x

. Oh, and somebody remind me to do a post about multiplying by one creatively.

Hi there, I was looking through the tricks and hacks and was wondering what the multiplying by one creatively was. If possible could you show me please. Thank you.

Reply 197

IrrationalRoot

20

Original post by Major-fury

Hi there, I was looking through the tricks and hacks and was wondering what the multiplying by one creatively was. If possible could you show me please. Thank you.

You've probably done it before but just haven't noticed it was 'multiplying by one creatively': consider rationalising the denominator of a fraction:

\dfrac{1}{\sqrt{3}}= \dfrac{1}{\sqrt{3}} \times 1=\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}= \dfrac{\sqrt{3}}{3}

.

Another example is proving that differentiability implies continuity; this involves both(!) creatively adding zero and creatively multiplying by 1:

\displaystyle \lim_{h \to 0} f(a+h) = \displaystyle \lim_{h \to 0} (f(a+h)-f(a)+f(a))= [br][br]\displaystyle \lim_{h \to 0} \left(\dfrac{f(a+h)-f(a)}{h} \times h+f(a)\right)=f'(a) \times 0 +f(a)=f(a)

as required.

In most cases the general idea is to force a desired expression into the picture.

Reply 198

CauchyIntegral

2

Potentially useful trick for STEP. It can help to know the sin and cos half angle identities which can be proven using the double angle formulae

\sin(\frac{x}{2})=\pm\sqrt\frac{1-\cos(x)}{2}

\cos(\frac{x}{2})=\pm\sqrt\frac{1+\cos(x)}{2}

Can be useful to spot if you need to simplify a result. One example is STEP I 1998 Q6

Reply 199

cxs

14

only 10 days till STEP. Write the last posts before STEP and then leave my laptop, plunge into the final mocks.
I'll illustrate this trick (well it is more likely to be a skill.) by 4 questions. topic? I'd like to call it
Manipulation Of Variables

Level1
compare different expressions of a Irrelevant variable
2008.s3.5.png

2008,S3,Q5
Many induction methods could lead to what we need to prove. But have you ever thought about why?I mean,
if you are not told what we need to find, then how can we find it?
A good starting point is to write seperate x---as x is irrelevant of the sequence T.
So, rearranging, we have [T_n+1(x)+T_n-1(x)]/2T_n(x)=x
similiarly,[T_n(x)+T_n-2(x)]/2T_n-1(x)=x
now we find 2 ways, then trating x as constant, compare 2 expressions, the result soon follow.

level2
seperate variables
1990.A.7.png

1990, furtherA,Q7
p+1=ke^-p
The graph looks terrible. Obviously, we can not easily find which k satisfies that they are tangential.
A normal way is perhaps to defferentiate, find the tangent to the point, and put it into p+1---looong work
But there's a very simple way.
As we need to find set of k, why not express k in terms of p?
so, seperate variables.
Rearrange we have k=(p+1)*e^p
Now it looks easy. As p can vary (which means the intersection can vary), so we just need to find the range of
this expression, and the value of the turning point of p is when there's 1 root. For 2 roots or 1-- draw a graph
of this expression, you'll get everything.

STEP Tricks and Hacks Thread

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you