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    (Original post by miketree)
    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.
    Notation does not equate to knowledge required to solve a problem. Limits and integrals over an infinite interval are definitely covered in Further Maths specifications.
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    (Original post by miketree)
    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.

    1. Agreed.

    2. I'm fairly sure it is. I agree about the functional equation stuff though.

    15. Treatment of simple limits is on some syllabuses, certainly mine.

    22. As above.

    23. I'm not sure, the individual symbols are used at a-level so I think you can work it out. Possibly a **.
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    (Original post by miketree)
    Hey can you explain why you jumped from \displaystyle\sum_{r=1}^{n} \sin{rx} to \displaystyle\sum_{r=1}^{n} e^{ix}?

    Isn't e^{ix} = \cos{x} + i\sin{x}? Hence you've lost the r and you've included an extra cos?
    Typo, I corrected it just before you posted
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    (Original post by miketree)
    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.
    There may be notation issues because of the different boards for A level but you can just ask the problem poser to clear up what they mean :dontknow:

    Surely you have a vague idea of what a limit is from differentiation/ integration and evaluating those limits don't require anything more advanced than A level knowledge (L'Hopital's rule, etc)

    Really? I thought most A level specs covered integrating to infinity as 'improper integrals' If not, evaluating the actual integral to infinity is usually quite intuitive anyway and takes 2 minutes to learn (e.g. what happens to e^(-x) as x tends to infinity?)
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    (Original post by Jkn)
    Nicely put!

    I always wondered if people were lurking in the background, quietly appreciating our solutions :lol:
    Yeah, I'm guilty of that.. :rolleyes:


    (Original post by ukdragon37)
    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
    Haha, I was just waiting for this to be said! Yeah - quite probably, I'm just not that good at maths anymore (along the lines of Asklepios etc), I'd like to learn more anyway.
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    (Original post by Jkn)
    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.

    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!
    Once again, with all due respect, an individual ready to sit an exam must know everything in that A-level specification. Additionally, he must be able to answer questions he has never seen before, hence not relying on memorisation.

    I understand this, I have not commented on any S1-4, and any M1-5 questions only those in C1-4 and FP1-3 of which I have covered.

    Whilst I do understand your comment on A-levels being taught using memorisation techniques and not understanding, I still do think the majority of these questions are wrongly titled.
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    (Original post by miketree)
    ...
    What did you think of this question then?

    \displaystyle \int_0^1 x^{-x}\,dx
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    (Original post by LewisMead)
    Notation does not equate to knowledge required to solve a problem. Limits and integrals over an infinite interval are definitely covered in Further Maths specifications.
    C1-4, FP1-3, Edexcel has no infinite interval integrals.
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    (Original post by joostan)
    Hopefully correct . . .
    Spoiler:
    Show

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

    Consider:
    S_n = \displaystyle\sum_{r=1}^{n} e^{ix} = \dfrac{e^{ix}(1-e^{inx})}{1-e^{ix}}
    This forms a geometric series - using the sum formula:
    \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}} 

\Rightarrow S_n = \dfrac{(e^{inx}+e^{ix} - e^{(n+1)x} -1)}{2-2\cos(x)}

\displaystyle \sum_{r=1}^{n} \sin(rx) = \Im(S_n) = \dfrac{\sin(x)+\sin(nx) - \sin((n+1)x)}{2-2\cos(x)}

\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)

\mathrm{Im}(S_n) =  \dfrac{1}{2}\left(\sin(nx) + \dfrac{\sin(x)(1-\cos(nx))}{1-\cos(x)} \right)

\therefore \displaystyle \sum_{r=1}^{n} \sin(rx) = \dfrac{1}{2}\left(\sin(nx) + \cot\left(\frac{x}{2}\right)(1-\cos(nx)) \right)
    A perfect solution, and from a student who hasn't even completed their A-Levels!
    (Original post by miketree)
    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.
    Oh btw, when I typed "integral" to you I meant "summation" :lol: I've been doing waaaaay too much integration lately :lol: INTEGRALMANIA!

    I agree with 1, 2, 15 and 23. They are definitely mislabelled and are all ** problems in my opinion. this is an issue that has been frequently discussed on this thread. E.g. functional equations, limits, etc... are obviously not a part of the A-Level specification and so, however straightforward the problem, all such problems should be **

    22 is * in my opinion. We have been introduced to the idea of infinity in the A-Level syllabus. Thinks like \lim_{n \to \infty} \frac{1}{n} = 0 are taught at A-Level. Note also that this notation is also required (though, anything beyond this, is not and examiners are obviously not going to be fussed about little notational errors concerning limits for this very reason).

    Its strange... I recall having this very same argument with the many of the people who post such problems about a month ago of my STEP Problem-Solving thread :rollseyes:
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    (Original post by Jkn)
    Nicely put!

    I always wondered if people were lurking in the background, quietly appreciating our solutions :lol:
    Indeed. I particularly admired your rigorous approach regarding how a \rightarrow b where a,b are distinct constants :yes:
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    (Original post by Jkn)
    A perfect solution, and from a student who hasn't even completed their A-Levels!
    :woo:
    Spoiler:
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    But I'm a lil bit keen
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    (Original post by Jkn)
    Its strange... I recall having this very same argument with the many of the people who post such problems about a month ago of my STEP Problem-Solving thread :rollseyes:
    I think I've have this argument 3 times on TSR Which is possibly why I'm being slightly curt.
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    (Original post by bananarama2)
    What did you think of this question then?

    \displaystyle \int_0^1 x^{-x}\,dx
    Hmm

    I tried using x^{-x} = e^{-x\ln{x}} then substitution but that didn't get me anywhere, how would you do it?
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    (Original post by miketree)
    Hey can you explain why you jumped from \displaystyle\sum_{r=1}^{n} \sin{rx} to \displaystyle\sum_{r=1}^{n} e^{ix}?

    Isn't e^{ix} = \cos{x} + i\sin{x}? Hence you've lost the r and you've included an extra cos?
    Remember that a+ib=c+id \iff a=c \ b=d (I leave the proof to the reader). This is why real and imaginary parts always remain the same regardless of the manipulation
    (Original post by miketree)
    Once again, with all due respect, an individual ready to sit an exam must know everything in that A-level specification.
    I eagerly await your disprove via. counterexample
    Additionally, he must be able to answer questions he has never seen before, hence not relying on memorisation.
    I don't think I've ever seen the question 1+54354+34 before...
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    (Original post by miketree)
    ...
    Whilst I do understand your comment on A-levels being taught using memorisation techniques and not understanding, I still do think the majority of these questions are wrongly titled.
    What do you mean? Closer to real mathematics? I would have thought that the solutions to hard problems would help me learn more maths, and am grateful for that, although am new here so haven't had time to look at much or work on any.
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    (Original post by miketree)
    Hmm

    I tried using x^{-x} = e^{-x\ln{x}} then substitution but that didn't get me anywhere, how would you do it?
    The solution has been posted. But think Maclaurin (or however you spell it, that never was my strong point )
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    (Original post by henpen)
    What do you mean? Closer to real mathematics?
    Actually some of it's rather complex.
    Ba dum tss
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    (Original post by Jkn)
    I don't I've ever seen the question 1+54354+34 before...
    Fortunately for me. I'd undoubtedly get that wrong.
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    (Original post by The Polymath)
    Indeed. I particularly admired your rigorous approach regarding how a \rightarrow b where a,b are distinct constants :yes:
    :pierre:
    Spoiler:
    Show
    Hahaha :lol: Are you referring to the infamous solution that launched the massive argument with LOTF on the STEP problem-solving thread? :lol: Hahahaha wasn't ​quite that bad!
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    Well everyone is coming out of the woodwork today I really should be doing chemistry. Sorry guys and gals.
 
 
 
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