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The Proof is Trivial!

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Reply 1400
Original post by hollieeilloh
I have to disagree with you, I mean I think this thread is just a bit of fun and obviously, yeah, you have to be pretty good (understatement) at maths to do most of the problems - but hey what's the point of doing something if it's not challenging? From the looks of things, some **/*** problems are less complicated than * problems - it all just depends on what you know :smile:

I only watch this thread to have my mind blown on a regular basis, but felt the need to post. :colone:


I understand this forum is fun, but whats the point in including the title * when they can't be done with A-level knowledge.
Original post by miketree
With all due respect, STEP is not A-level knowledge. Not every person who has completed A-level maths, or A-level maths + further maths will have done STEP, because not everyone requires it.


That isn't my point, my point is only STEP requires a-level maths. These * problems only require a-level maths.
Original post by miketree
I understand this forum is fun, but whats the point in including the title * when they can't be done with A-level knowledge.


They can. Can you give an example?
Original post by Jkn

Problem 217*

Simplify r=1nsin(rx)\displaystyle \sum_{r=1}^{n} \sin(rx)

Hint:

Spoiler


Hopefully correct . . .
It may be possible to simplify further but I got bored. :redface:
Solution 217:

Consider:
Sn=r=1neirxS_n = \displaystyle\sum_{r=1}^{n} e^{irx}
This forms a geometric series - using the sum formula:
Sn=eix(1einx)1eix=eix(1einx)1eix×1eix1eix[br]Sn=(einx+eixe(n+1)x1)22cos(x)[br]r=1nsin(rx)=(Sn)=sin(x)+sin(nx)sin((n+1)x)22cos(x)[br]Im(Sn)=12(sin(x)+sin(nx)sin(nx)cos(x)cos(nx)sin(x)1cos(x))[br]Im(Sn)=12(sin(nx)+sin(x)(1cos(nx))1cos(x))[br]r=1nsin(rx)=12(sin(nx)+cot(x2)(1cos(nx)))\Rightarrow S_n = \dfrac{e^{ix}(1-e^{inx})}{1-e^{ix}} = \dfrac{e^{ix}(1-e^{inx})}{1-e^{ix}} \times \dfrac{1-e^{-ix}}{1-e^{-ix}} [br]\Rightarrow S_n = \dfrac{(e^{inx}+e^{ix} - e^{(n+1)x} -1)}{2-2\cos(x)}[br]\displaystyle \sum_{r=1}^{n} \sin(rx) = \Im(S_n) = \dfrac{\sin(x)+\sin(nx) - \sin((n+1)x)}{2-2\cos(x)}[br]\mathrm{Im}(S_n) = \dfrac{1}{2} \left(\dfrac{\sin(x)+\sin(nx) - \sin(nx)\cos(x) - \cos(nx)\sin(x)}{1-\cos(x)} \right)[br]\mathrm{Im}(S_n) = \dfrac{1}{2}\left(\sin(nx) + \dfrac{\sin(x)(1-\cos(nx))}{1-\cos(x)} \right)[br]\therefore \displaystyle \sum_{r=1}^{n} \sin(rx) = \dfrac{1}{2}\left(\sin(nx) + \cot\left(\frac{x}{2}\right)(1-\cos(nx)) \right)
(edited 10 years ago)
Original post by miketree
I understand this forum is fun, but whats the point in including the title * when they can't be done with A-level knowledge.

The can, occasionally some of them have been mis-labelled and probably should've had an extra * but they seem to be usually labelled appropriately...
Original post by miketree
With all due respect, STEP is not A-level knowledge. Not every person who has completed A-level maths, or A-level maths + further maths will have done STEP, because not everyone requires it.


Not everyone who completes A Level Maths and Further Maths or only Maths can even attempt STEP questions as they are aimed at the top students, but they are still doable with A Level knowledge as I'm able to complete questions and that's the only knowledge I have.
Original post by miketree
I understand this forum is fun, but whats the point in including the title * when they can't be done with A-level knowledge.


Yeah I can't comment on that because I'm Scottish, but I think these guys know what they're doing, maybe the applications are a bit more advanced, but I'm sure it can't be that far off. :tongue:
Reply 1407
Original post by Felix Felicis
It can be done with A-level knowledge, surely you know de Moivre's theorem?


Yes I know de Moivre's theorem.

In the OPs hint, he uses "Im", I just googled it now to know what it meant but I have not come across that in my A-level knowledge.

He then goes on to say I must use an integral, once again turning summations into integrals is not on the A-level specification.

PS. Don't want to come across as rude to you or OP.
Original post by miketree
In the OPs hint, he uses "Im", I just googled it now to know what it meant but I have not come across that in my A-level knowledge.


Which board are you on?
Original post by miketree
Yes I know de Moivre's theorem.

In the OPs hint, he uses "Im", I just googled it now to know what it meant but I have not come across that in my A-level knowledge.

He then goes on to say I must use an integral, once again turning summations into integrals is not on the A-level specification.

PS. Don't want to come across as rude to you or OP.

Ok, that's just semantics/ notation confusion, it essentially means the imaginary component of ... - you know what the imaginary component of a complex number is.

I'm not exactly sure what Jkn meant by if that was a hint to your question or whether it was a general comment about how to go about tackling the problems on this thread but my solution doesn't use any integrals, just de Moivre's theorem and trig.
(edited 10 years ago)
Original post by bananarama2
x


You called? :tongue:
Reply 1411
Original post by miketree

a) If you must have STEP knowledge for this question, it must be titled with ** because that does not "require only A-level knowledge" quoted from your original post.

STEP knowledge only differs from A-Level knowledge in that you must know the definition of an irrational number and also how to establish a proof by contradiction (though if such things are needed a question would most likely fall in the ** category)

b) I don't believe transforming a summation into an integral is in any A-level specification, apologies if I'm wrong (maybe MEI).

I can assure you that you do not need to transform the summation into an integral :smile:

I think the issue you are having is that you are assuming that questions that only require A-Level knowledge are accessible to those who have prepared for A-Level exams! Whilst such questions would not be set on A-Level exam papers, the idea is that a * solution can be derived from pieces of mathematics that are taught in a typical A-Level syllabus. I would guess that you are underestimating the sheer breadth of knowledge contained within an A-Level syllabus. Whilst teachers and exams will likely only prepare you for straightforward questions that can be done through memorisation, that does not mean that a question that requires more than just memorisation requires more knowledge than is suitable for an A-Level question, though it certainly does tend to mean it requires a greater ability to think mathematically than an A-Level question would (and, hence, why such questions would not appear on exams). The way we see it, since anyone with A-Level knowledge will be able to attempt the question (and more knowledge would not be needed or, often, help), the * rating is appropriate.

That said, many problems are mislabelled with many are annoyed about. I believe a large number of such mislabelled problems are set by (sorry guys :colondollar:) overseas students who overestimate the mathematics that is accessible to an student after having completed a typically A-Level course. Case and point: I still await LOTF's * solution to the problem that involves integrating ln(sin(x))ln(cos(x))\ln( \sin(x) ) \ln( \cos(x)) :lol:

Another thing to remember (though it is not likely relevant to any of the problems you have seen in the recent pages), there are roughly 18 modules on every exam board and so there may be pieces of mathematics in the A-Level syllabus that you are not familiar with!
Original post by joostan
You called? :tongue:


I was going to say that I always attempt the calligraphic symbols when I writing and they all end up looking like hieroglyphics.
Original post by bananarama2
I was going to say that I always attempt the calligraphic symbols when I writing and they all end up looking like hieroglyphics.


Yeah I swapped them for Im cos it looked less fancy, and I hadn't actually intended to put them in, I guessed that Im would be like ln and tan, but it didn't work out :tongue:
Original post by joostan
Yeah I swapped them for Im cos it looked less fancy, and I hadn't actually intended to put them in, I guessed that Im would be like ln and tan, but it didn't work out :tongue:

I think it looks funky :ahee:

Original post by hollieeilloh
...

We meet again Hollie :sexface:
(edited 10 years ago)
Reply 1415
Original post by bananarama2
They can. Can you give an example?


1. I understand what a continuous function is but I don't recall it coming from my A-level.
2. Notation not covered in A-level spec.
15. Limits not covered in A-level spec.
22. Integrating infinity not covered in A-level spec.
23. Notation again.

I mean I could go through the whole thread but that would be tedious and I would rather not create such a large post.
Original post by Felix Felicis
I think it looks funky :ahee:


I left one in just for lols :laugh:
Reply 1417
Original post by hollieeilloh
I have to disagree with you, I mean I think this thread is just a bit of fun and obviously, yeah, you have to be pretty good (understatement) at maths to do most of the problems - but hey what's the point of doing something if it's not challenging? From the looks of things, some **/*** problems are less complicated than * problems - it all just depends on what you know :smile:

I only watch this thread to have my mind blown on a regular basis, but felt the need to post. :colone:

Nicely put!

I always wondered if people were lurking in the background, quietly appreciating our solutions :lol: :colone:
Original post by hollieeilloh
Yeah I can't comment on that because I'm Scottish, but I think these guys know what they're doing, maybe the applications are a bit more advanced, but I'm sure it can't be that far off. :tongue:


Whenever they say it can be done with A-level knowledge, 99% of the time it can also be done with AH as well, so the questions should be accessible to you to if you can apply your knowledge in clever ways :tongue:
Reply 1419
Original post by joostan

Sn=r=1neix=eix(1einx)1eixS_n = \displaystyle\sum_{r=1}^{n} e^{ix} = \dfrac{e^{ix}(1-e^{inx})}{1-e^{ix}}


Hey can you explain why you jumped from r=1nsinrx\displaystyle\sum_{r=1}^{n} \sin{rx} to r=1neix\displaystyle\sum_{r=1}^{n} e^{ix}?

Isn't eix=cosx+isinxe^{ix} = \cos{x} + i\sin{x}? Hence you've lost the r and you've included an extra cos?

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