STEP Prep Thread 2017

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Well, after doing some STEP problems about complex mappings (the mobius transformation?), it mapped circles and lines to either a circle or a line. And I thought to myself, why not consider lines as circles too? They have the same style of definition in the complex plane. That'd save me the effort of calling them different things. So I looked it up, and found some guys on the internet who agreed with me (https://en.wikipedia.org/wiki/Generalised_circle)

Do you think you could refer me to those STEP problems?

Btw, I love your profile picture. :-)

Reply 201

username1162972

Inversion.
Coolest thing to come out of geometry.
Can solve some very hard problems in matter of lines.

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Reply 202

FriendlyPenguin

Original post by Injective

Do you think you could refer me to those STEP problems?

Probably not - but I'd guess its STEP 3, and there will likely be some of them in the old set (1987-1990s) of STEP papers.

Btw, I love your profile picture. :-)

Are you suggesting that some people do not like Glorious Leader???

Reply 203

Injective

Original post by Mathemagicien

Probably not - but I'd guess its STEP 3, and there will likely be some of them in the old set (1987-1990s) of STEP papers.

Are you suggesting that some people do not like Glorious Leader???

I've noticed it in the old set too, part of the old specification I think. Cheers will have a look.

And not at all... :tongue:

(edited 7 years ago)

Reply 204

Zacken

Original post by injective

i've noticed it in the old set too, part of the old specification i think. Cheers will have a look.

And not at all... :tongue:

1996, iii

Reply 205

drandy76

Original post by Zacken

1996, iii

Did you know that off head, or did you use the search database?

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Reply 206

username1162972

Original post by Injective

Lol, btw I know he is joking. :wink:

I was hoping you were lol. Otherwise
2017 is Kms.

Reply 207

How to be wise

Original post by Zacken

1996, iii

Dude, did u get into Kings?

Reply 208

Zacken

Original post by drandy76

Did you know that off head, or did you use the search database?

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search database doesn't work for me idk why

Original post by STRANGER2

Dude, did u get into Kings?

Yeah

Reply 209

drandy76

Original post by Zacken

search database doesn't work for me idk why

My boy Siklos don't rate you no more

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Reply 210

drandy76

Original post by physicsmaths

I was hoping you were lol. Otherwise
2017 is Kms.

Was I joking? Who knows....

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Reply 211

girlwonder17

http://files.physicsandmathstutor.com/download/Maths/STEP/Papers/2004%20STEP%201.pdf

For question 1

For the last part iii you use that:

x^3 = 99+-70(sqrt2)

But I only know that x = 3-2(sqrt2) from ii

So how am I to assume that it works out for x=3+2(sqrt2)

Reply 212

IrrationalRoot

Original post by girlwonder17

It's pretty intuitive that it works for the positive solution as well. It's very easy to quickly verify this in your head; just think about how the signs work out when you cube the binomial.

Reply 213

zetamcfc

Original post by girlwonder17

Spoiler

Reply 214

IrrationalRoot

Original post by zetamcfc

Spoiler

I don't think this answers their question, since they got that far but were wondering what the justification was for the result of (ii) in the case of the positive solution.

Reply 215

zetamcfc

Original post by IrrationalRoot

I don't think this answers their question, since they got that far but were wondering what the justification was for the result of (ii) in the case of the positive solution.

It follows from working out the +ve case.

Reply 216

IrrationalRoot

Original post by zetamcfc

It follows from working out the +ve case.

Well ideally you shouldn't 'work out' the +ve case like you did in (ii); you can just deduce it from the -ve case.

Reply 217

girlwonder17

Original post by IrrationalRoot

It's pretty intuitive that it works for the positive solution as well. It's very easy to quickly verify this in your head; just think about how the signs work out when you cube the binomial.

I suppose you are right yea; but I would probably do the whole part b for the positive sign.

Am I right in thinking that this would work for any similar type of cubic.

thx :smile:

Original post by zetamcfc

Spoiler

I get that thnx but I don't get how you are just supposed to figure out that it works out to be positive, like irrational said check the sign I guess, just seems not like i am being complete idk

To make it clear the 99-70(sqrt 2) makes sense it is the 99+70(sqrt 2 ) that is not implied in previous parts.

Also unabomber?! He is cute, I agree with his views on technology, but killed people :/

Spoiler

(edited 7 years ago)

Reply 218

girlwonder17

STEP seems exhausting.

Its not that the content is hard (yet at least), just stretches your understanding.

Although that paper I 2004 doesn't seem too hard.

Reply 219

IrrationalRoot

Original post by girlwonder17

I suppose you are right yea; but I would probably do the whole part b for the positive sign.

Am I right in thinking that this would work for any similar type of cubic.

thx :smile:

I get that thnx but I don't get how you are just supposed to figure out that it works out to be positive, like irrational said check the sign I guess, just seems not being complete idk

To make it clear the 99-70(sqrt 2) makes sense it is the 99+70(sqrt 2 ) that is not implied in previous parts.

Also unabomber?! He is cute, I agree with his views on technology, but killed people :/

Spoiler

Doing part (ii) again is unnecessary; you definitely don't want to waste time making unnecessary justifications in STEP - it should all be as concise as possible.

The 'looking at signs method' can easily be shown to work with all such cubes of binomials, you can quickly prove this is you like since it's very easy.

It seems that you think this isn't rigorous enough, but it really does just follow from the fact the magnitude of each term in the binomial expansion will clearly be the same as before, but the signs are switched every other term due to the powers of -1 (and the rational and irrational terms alternate).