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# STEP 2016 Solutions Watch

1. (Original post by farryharnworth)
I was not. Was there another farry there?
Ok. Never mind then.

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2. (Original post by farryharnworth)
I was not. Was there another farry there?
3. (Original post by farryharnworth)
In a simultaneous effort to learn a bit of LaTeX and to rub in my own face just how stupid I was to miss out some of the questions, I've typed up the questions and solutions for STEP II (credit is given, even if I've extended, abridged, corrected, or tweaked solutions).

I figured you guys may as well have it. Give me a shout if I've messed something up.
I think its marvelous - definitely should be attached in the OP. Thanks for sharing!

There is one very small thing that you have missed out of question (8i) - I hesitate to mention it, since I don't think its worth much bother to add it in, and you have done a great job otherwise on the LaTex - which is that, for our succinct notation to work, we just need to clarify what happens when m=1; that is, we should define

But this is a very nice resource for future STEPpers (and makes my solutions look a lot neater than they were! ), so thank you again for sharing!
4. (Original post by Mathemagicien)
I think its marvelous - definitely should be attached in the OP. Thanks for sharing!

There is one very small thing that you have missed out of question (8i) - I hesitate to mention it, since I don't think its worth much bother to add it in, and you have done a great job otherwise on the LaTex - which is that, for our succinct notation to work, we just need to clarify what happens when m=1; that is, we should define

But this is a very nice resource for future STEPpers (and makes my solutions look a lot neater than they were! ), so thank you again for sharing!
Thank you, you're too kind

You're absolutely right, I did miss that. It does seem logical to me that it would be true (and a quick google turns up this, which is a bit vague, but seems to support the convention), but of course it's not worth being anything but 100% clear in a STEP exam. I've updated it - can't get it to stick in an edit, so here it is with minor correction.
Attached Images
5. STEP II 2016.pdf (357.6 KB, 75 views)
6. (Original post by farryharnworth)
Give me a shout if I've messed something up.
Q3, the LHS is the sum of non-negative terms, not positive terms.

Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both.

Another edit: you seem to be missing the "write down for four events" in Q12.
7. (Original post by Zacken)
Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both
8. (Original post by Mathemagicien)
There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
9. (Original post by Zacken)
There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
Damn, I didn't read that bit

I agree
10. (Original post by Mathemagicien)
Damn, I didn't read that bit

I agree
(Original post by Zacken)
There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?
z(1) = 1 follows because y_n(1) = 1 and z = 2y_n^2 - 1 etc but surely that's all you'd have to say? It's given that there is one solution satisfying y_k(1) = 1 for each k, and you've provided one for k = 2n, so z = y_2n.
11. (Original post by StrangeBanana)
z(1) = 1 follows because y_n(1) = 1 and z = 2y_n^2 - 1 etc but surely that's all you'd have to say? It's given that there is one solution satisfying y_k(1) = 1 for each k, and you've provided one for k = 2n, so z = y_2n.
Yeah, was just saying that there's probably a mark allocated for stating it.
12. (Original post by Zacken)
Q3, the LHS is the sum of non-negative terms, not positive terms.

Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both.
Right you are; sorted.

(Original post by Zacken)
Another edit: you seem to be missing the "write down for four events" in Q12.
Not sure how I managed that - whoops.

Thank you very much.
Attached Images
13. STEP II 2016.pdf (357.9 KB, 109 views)
14. (Original post by farryharnworth)
Right you are; sorted.

Not sure how I managed that - whoops.

Thank you very much.
Cheers, great work!
15. Are we allowed to post Step III solutions yet?
16. (Original post by riquix)
Are we allowed to post Step III solutions yet?
Yes

Edit: no
17. (Original post by riquix)
Are we allowed to post Step III solutions yet?
I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm
18. (Original post by 13 1 20 8 42)
I believe you have to wait 24 hours after the start time of the exam, i.e. you have to wait until 2pm
Ah right
20. (Original post by Shrek1234)
No discussion until 2pm (BST)

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21. 2pm **** sake
22. saving this space for question 7

don't steel it :<

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