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STEP Maths I,II,III 1987 Solutions

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Out of interest, there must be some conditions that a series must meet in order for all the terms to cancel out, like in I/7?

Reply 121

ukgea

Square

Out of interest, there must be some conditions that a series must meet in order for all the terms to cancel out, like in I/7?

If a series is set as a STEP question, there's definitely a neat solution involving everything cancelling out. If a series arises in real life there most probably isn't. :smile:

Reply 122

DFranklin

Square

Out of interest, there must be some conditions that a series must meet in order for all the terms to cancel out, like in I/7?

Not really.

If I have

S(N) = \sum_1^N f(n)

, then

f(1) = S(1), f(n) = S(n)-S(n-1) (n > 1)

, so

\sum_1^N f(n) = S(1)+S(2)-S(1)+S(3)-S(2)+...+S(N)-S(N-1) = S(N)

So there always going is always going to be a way of reexpressing the sum so that all the terms cancel out.

Of course, the problem is that that expression is basically the answer you're looking for, so finding it is not always very easy...

Reply 123

Square

Righty, seems to make sense now :smile:

Reply 124

DFranklin

Dystopia

Interesting. It certainly sounds possible. Are you basing it on previous experience (perhaps with the tripos exams)?

More from apocryphal stories than anything else. Though I'm sure I've seen the occasional churlish examiners report saying something like:

[indent]despite a very similar question being asked in 27BC, the vast majority of candidates attempting this question failed to spot that 6271888543106341 factors as 238829 x 26261000729 and therefore made little progress...[/indent]

Reply 125

nota bene

I'll post up my attempt to a solution to I/7, using the approach with taking the imaginary parts...

Spoiler

Tell me if I've made some silly mistakes, quite likely...

edit: The solution should now be correct.

Reply 126

DFranklin

nota bene

Tell me if I've made some silly mistakes, quite likely...

Well, the question does ask you to express the answer in the form of a single tangent...

Reply 127

nota bene

DFranklin

Well, the question does ask you to express the answer in the form of a single tangent...

Oops, sorry. Didn't have the question when working on it, so missed that. However, I'm not sure I have ever heard a question worded that way - what does 'a single tangent' really mean? just like:

\displaystyle\frac{\sin^2(\frac{22\pi}{23})}{\sin( \frac {2\pi}{23})}=\Im e^{i(\frac{44}{23}\pi-\frac{2\pi}{23})}=\Im e^{i\frac{42}{23}\pi}=\sin(\frac{42}{23}\pi)

Reply 128

DFranklin

nota bene

Oops, sorry. Didn't have the question when working on it, so missed that. However, I'm not sure I have ever heard a question worded that way - what does 'a single tangent' really mean? just like:

\displaystyle\frac{\sin^2(\frac{22\pi}{23})}{\sin( \frac {2\pi}{23})}=\Im e^{i(\frac{44}{23}\pi-\frac{2\pi}{23})}=\Im e^{i\frac{42}{23}\pi}=\sin(\frac{42}{23}\pi)

Am I missing something? I don't think that's right:

\sin^2 x \neq \Im (e^{2ix})

.

As far as the question wording goes, as far as I remember you can get the answer in the form

\tan \theta

for a suitable

\theta

. (Possibly with a constant factor of 2 somewhere, although I don't think so).

Reply 129

nota bene

DFranklin

Am I missing something? I don't think that's right:

\sin^2 x \neq \Im (e^{2ix})

.

As far as the question wording goes, as far as I remember you can get the answer in the form

\tan \theta

for a suitable

\theta

. (Possibly with a constant factor of 2 somewhere, although I don't think so).

You're right, I'm messing up as usual.

I'll try to come up with something in closed form involving tan, although I'm not directly seeing in which step it would be nice to change to tan...

Reply 130

DFranklin

nota bene

You're right, I'm messing up as usual.

I'll try to come up with something in closed form involving tan, although I'm not directly seeing in which step it would be nice to change to tan...

Right at the end. It's just a tidying up job. You might want to think about

\sin(\pi-x) = \sin(x)

type symmetries as well...

Reply 131

nota bene

DFranklin

Right at the end. It's just a tidying up job. You might want to think about

\sin(\pi-x) = \sin(x)

type symmetries as well...

Ah!
For the first one:

Spoiler

Now to the second one...

Reply 132

generalebriety

Ahem. Check that. :p:

Reply 133

nota bene

generalebriety

Ahem. Check that. :p:

Yeah yeah, I suddenly decided to drop a factor of 1/2 :p:

Reply 134

generalebriety

nota bene

Yeah yeah, I suddenly decided to drop a factor of 1/2 :p:

Yeah yeah, I suddenly decided to fall over and break my leg. :p:

Reply 135

nota bene

So, seems like I don't know my trig identities etc.
This is what I tried

Spoiler

edit: Curse latex won't display it - it says [noparse]

\sin(\frac{2\pi}{23}k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23}k) - \sin^2 (\frac{2\pi}{23}k) \sin(\frac{2\pi}{23}) + \sin^2 (\frac{2\pi}{23}) \cos(\frac{2\pi}{23}k) - \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23}k) \cos(\frac{2\pi}{23}k)

[/noparse]

Reply 136

generalebriety

nota bene

So, seems like I don't know my trig identities etc.
This is what I tried

Spoiler

edit: Curse latex won't display it - it says [noparse]

\sin(\frac{2\pi}{23}k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23}k) - \sin^2 (\frac{2\pi}{23}k) \sin(\frac{2\pi}{23}) + \sin^2 (\frac{2\pi}{23}) \cos(\frac{2\pi}{23}k) - \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23}k) \cos(\frac{2\pi}{23}k)

[/noparse]

Was that meant to be the following?

\sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k) - \sin^2 (\frac{2\pi}{23} k) \sin(\frac{2\pi}{23}) + \sin^2 (\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k)

- \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23} k)

Reply 137

nota bene

generalebriety

Was that meant to be the following?

\sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k) - \sin^2 (\frac{2\pi}{23} k) \sin(\frac{2\pi}{23}) + \sin^2 (\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k)

- \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23} k)

Yes, except one typo of mine it shall be:

\sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k) - \sin^2 (\frac{2\pi}{23} k) \sin(\frac{2\pi}{23}) + \sin (\frac{2\pi}{23}) \cos^2 (\frac{2\pi}{23} k)

- \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23} k)

Thanks for sorting the latex, no idea why it wouldn't take my code:/ (except that it was long...)

I don't really see how this is going to be something nice in terms of tan :frown:

Reply 138

Dystopia

STEP III, Q3.

Spoiler

Nice question. :smile:

Reply 139

DFranklin

nota bene

Yes, except one typo of mine it shall be:

\sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23}) \cos(\frac{2\pi}{23} k) - \sin^2 (\frac{2\pi}{23} k) \sin(\frac{2\pi}{23}) + \sin (\frac{2\pi}{23}) \cos^2 (\frac{2\pi}{23} k)

- \sin(\frac{2\pi}{23}) \tan(\frac{2\pi}{23}) \sin(\frac{2\pi}{23} k) \cos(\frac{2\pi}{23} k)

Thanks for sorting the latex, no idea why it wouldn't take my code:/ (except that it was long...)

I don't really see how this is going to be something nice in terms of tan :frown:

I'm having a certain amount of difficulty working out what you're actually trying to do here (more me being tired than any fault on your part).

One particular problem I'm having is that your using k as the indexing variable in the sum (i.e. you have

\sum_{k=0}^{10} \{ \text{stuff} \}

), but then you sum the GP and you then have k's in your answer. I think it would be clearer (for me, and possibly for you!) if you put in the actual values (k=10 or k=11, not sure which off hand).

If you can get things down to an actual formula, I'll plug it into my computer here and see if it does at least agree numerically.

Judging by reactions, it seems this question is pretty tough for STEP I!