Official TSR Mathematical Society

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  1. Daniel Freedman's Avatar
    1341 05-02-2009  21:50
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    Re: Official TSR Mathematical Society
    Let \alpha be a root of x^4 - x^3 +x^2 - x + 1 .

    Find the value of \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1
  2. Oh I Really Don't Care's Avatar
    1342 05-02-2009  22:11
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    Let \alpha be a root of x^4 - x^3 +x^2 - x + 1 .

    Find the value of \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1
    x^4 - x^3 +x^2 - x + 1 = 0 = (x^2 + x^{-2}) - (x + x^{-1}) + 1

    2cos(2\theta) - 2cos(\theta) + 1 = 0 \Rightarrow cos(\theta) = \frac{1+\sqrt{5}}{4} \Rightarrow \theta = \frac{\pi}{5}

    Then with the use of the useful e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1 = e^{i\pi 4} - e^{i \pi 3} + e^{i\pi 2} - e^{i\pi} + 1 = 5
    Last edited by Oh I Really Don't Care; 09-02-2009 at 19:58.
  3. Daniel Freedman's Avatar
    1343 05-02-2009  22:21
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    Re: Official TSR Mathematical Society
    (Original post by DeanK22)
    x^4 - x^3 +x^2 - x + 1 = 0 = (x^2 + x^{-2}) - (x + x^{-1}) + 1

    2cos(2\theta) - 2cos(\theta) + 1 = 0 \Rightarrow cos(\theta) = \frac{1+\sqrt{5}}{4} \Rightarrow \theta = \frac{\pi}{5}

    Then with the use of the useful e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^{5} + 1 = e^{i\pi 4} - e^{i \pi 3} + e^{i\pi 2} - e^{i\pi} + 1 = 1
    Your answer doesn't match the one I have.

    EDIT: You edited.
    Last edited by Daniel Freedman; 05-02-2009 at 22:45.
  4. Oh I Really Don't Care's Avatar
    1344 05-02-2009  22:26
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    I don't completely follow what you've done, and your answer doesn't match the one I have.
    ... erm. Sorry. I added up incorrectly. The answer should not be one. I think I should be get 5 (will confirm shortly).
  5. Daniel Freedman's Avatar
    1345 05-02-2009  22:27
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    Re: Official TSR Mathematical Society
    (Original post by DeanK22)
    ... erm. Sorry. I added up incorrectly. The answer should not be one. I think I should be get 5 (will confirm shortly).
    Yes, 5 is right.

    Although, I prefer the method I have.
  6. Oh I Really Don't Care's Avatar
    1346 05-02-2009  22:31
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    Yes, 5 is right.

    Although, I prefer the method I have.
    hmm. I like using this result; z^n + z^{-n} = 2cos(n\theta)

    ... Out of curiosity how did you do it?
  7. Daniel Freedman's Avatar
    1347 05-02-2009  22:38
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    Re: Official TSR Mathematical Society
    (Original post by DeanK22)
    hmm. I like using this result; z^n + z^{-n} = 2cos(n\theta)

    ... Out of curiosity how did you do it?
    Your method is actually pretty nice

    Alternatively,

    \\ x^5+1 = (x+1)(x^4-x^3+x^2-x+1) \\ \\ \implies \alpha^5 + 1 = (\alpha+1)(\alpha^4 - \alpha^3 + \alpha^2 - \alpha + 1) = 0 \\ \\ \implies \alpha^5 = -1 \\ \\ \implies \alpha^{20} - \alpha^{15} + \alpha^{10} - \alpha^5 + 1 = 1-(-1)+1-(-1)+1=5
  8. Swayum's Avatar
    1348 05-02-2009  22:52
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    Your method is actually pretty nice

    Alternatively,

    x^5+1 = (x+1)(x^4-x^3+x^2-x+1)
    What made you think to consider x^5 + 1 in the first place?
  9. SimonM's Avatar
    1349 05-02-2009  22:54
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    Re: Official TSR Mathematical Society
    (Original post by Swayum)
    What made you think to consider x^5 + 1 in the first place?
    Geometric progressions, equalling zero... looks like roots of unity to me
  10. Daniel Freedman's Avatar
    1350 05-02-2009  22:55
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    Re: Official TSR Mathematical Society
    (Original post by Swayum)
    What made you think to consider x^5 + 1 in the first place?
    From the question, knowing something about \alpha^5 would be very useful.

    EDIT: Geometric progressions, as Simon said.
  11. Daniel Freedman's Avatar
    1351 06-02-2009  14:17
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    Re: Official TSR Mathematical Society
    I'll give credit to the source of this afterwards, as the solution is on there too.

    For all real x, evaluate \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}
    Last edited by Daniel Freedman; 06-01-2010 at 14:09.
  12. Lysdexia's Avatar
    1352 06-02-2009  15:09
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    Re: Official TSR Mathematical Society
    Here's a question I've always wanted an answer to. It's probably very simple for most of you; I'm only a lowly A2 Maths student!

    If there are an infinite amount of numbers, and therefore an infinite amount of numbers between 1 and 2, how is it possible to get from 1 to 2? That implies a beginning and an end.
  13. Daniel Freedman's Avatar
    1353 06-02-2009  15:12
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    Re: Official TSR Mathematical Society
    (Original post by Lysdexia)
    Here's a question I've always wanted an answer to. It's probably very simple for most of you; I'm only a lowly A2 Maths student!

    If there are an infinite amount of numbers, and therefore an infinite amount of numbers between 1 and 2, how is it possible to get from 1 to 2? That implies a beginning and an end.
    What do you mean by "get from 1 to 2"?
  14. Lysdexia's Avatar
    1354 06-02-2009  15:21
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    Re: Official TSR Mathematical Society
    That 1 is the beginning of a series of an infinite amount of numbers and that 2 somehow causes an abrupt end to this series.
  15. Oh I Really Don't Care's Avatar
    1355 06-02-2009  15:28
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    Re: Official TSR Mathematical Society
    (Original post by Lysdexia)
    That 1 is the beginning of a series of an infinite amount of numbers and that 2 somehow causes an abrupt end to this series.
    You are pondering over a question that makes absolutely no sense - 2 is a symbol specifying an amount yet we could have arbitarly chosen another amount [i.e. in our conventional view of numbers we could have gone from 1 object to three objects with no 2 objects represented by a whole number].
  16. Oh I Really Don't Care's Avatar
    1356 06-02-2009  15:35
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    I'll give credit to the source of this afterwards, as the solution is on there too.

    For all real x, evaluate \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}
    never done such a limit with any trig or infact never seen one until now.

    some kind of tip?
  17. Daniel Freedman's Avatar
    1357 06-02-2009  15:40
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    Re: Official TSR Mathematical Society
    (Original post by DeanK22)
    never done such a limit with any trig or infact never seen one until now.

    some kind of tip?
    Presumably you've writted out the first few terms and aren't getting anywhere?

    A good idea with summations is to try and express the general term in a nicer form that you can sum.
  18. Lysdexia's Avatar
    1358 06-02-2009  16:14
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    Re: Official TSR Mathematical Society
    (Original post by DeanK22)
    You are pondering over a question that makes absolutely no sense - 2 is a symbol specifying an amount yet we could have arbitarly chosen another amount [i.e. in our conventional view of numbers we could have gone from 1 object to three objects with no 2 objects represented by a whole number].
    I thought it would be something like that!
  19. SimonM's Avatar
    1359 06-02-2009  17:14
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    Re: Official TSR Mathematical Society
    (Original post by Daniel Freedman)
    I'll give credit to the source of this afterwards, as the solution is on there too.

    For all real x, evaluate \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r}
    Very nice question. My first though was **** me that looks bitchy

    Spoiler:
    Show


    \displaystyle \sin^3 x =\frac{1}{4} \left ( 3 \sin x - \sin 3x \right )

    Thefore we have

    \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r} = \frac{1}{4} \sum_{r=0}^{\infty} \frac{3 \sin 3^r x - \sin 3^{r+1} x}{3^r} = \frac{3}{4} \sin x

  20. thomaskurian89's Avatar
    1360 06-02-2009  17:18
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    Re: Official TSR Mathematical Society
    (Original post by SimonM)
    Very nice question. My first though was **** me that looks bitchy

    Spoiler:
    Show


    \displaystyle \sin^3 x =\frac{1}{4} \left ( 3 \sin x - \sin 3x \right )

    Thefore we have

    \displaystyle \sum_{r=0}^{\infty} \frac{\sin^3{(3^r x)}}{3^r} = \frac{1}{4} \sum_{r=0}^{\infty} \frac{3 \sin 3^r x - \sin 3^{r+1} x}{3^r} = \frac{3}{4} \sin x

    Telescopes nicely.

    Offtopic: Is University of Leicester good for math? I just got an offer.
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