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A Summer of Maths (ASoM) 2016

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    (Original post by Gregorius)
    How about showing that

     \displaystyle \sin(x + iy) = \sin{x} \cosh{y} + i \cos{x}\sinh{y}

    and working from there?
    Sorry if I'm wrong but shouldn't it be sin(x)cosh(y) — icos(x)sinh(y)
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    (Original post by Insight314)
    I know you are talking about me. No problem, I will stop tagging people.
    Course he talking about you. U tag more people then there are in Cambridge. Lol


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    (Original post by physicsmaths)
    Course he talking about you. U tag more people then there are in Cambridge. Lol


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    Lol.I actually only tagged one person there (Gregorius) since he knows his stuff. I can't see the problem of tagging one person, so not sure why he wanted to spank me for tagging only one person; I would understand if he told me off earlier when I tagged tons of people.
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    (Original post by krishdesai7)
    Sorry if I'm wrong but shouldn't it be sin(x)cosh(y) — icos(x)sinh(y)
    No?

    Name:  ImageUploadedByStudent Room1467222115.069167.jpg
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    Just out of curiosity, how regularly and for how long do you mathmos exercise?
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    (Original post by MathsCoder)
    Just out of curiosity, how regularly and for how long do you mathmos exercise?
    I'm in my job(gym) around 5 times a week, somewhat serious powerlifter


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    (Original post by drandy76)
    I'm in my job(gym) around 5 times a week, somewhat serious powerlifter


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    Quite cool!
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    (Original post by MathsCoder)
    Just out of curiosity, how regularly and for how long do you mathmos exercise?
    2x/week gym
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    (Original post by MathsCoder)
    Just out of curiosity, how regularly and for how long do you mathmos exercise?
    never

    fat ****
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    (Original post by MathsCoder)
    Just out of curiosity, how regularly and for how long do you mathmos exercise?
    I do exercise papers every day, for a few hours

    (Original post by drandy76)
    I'm in my job(gym) around 5 times a week, somewhat serious powerlifter


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    A powerlifting mathematician? :eyeball:
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    (Original post by krishdesai7)
    Sorry if I'm wrong but shouldn't it be sin(x)cosh(y) — icos(x)sinh(y)
    Nope. Take a look at this entry in the DLMF for all the identities.
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    (Original post by Insight314)
    All right, gonna try that, after I finish eating though. However, is it possible to show that it has infinite solutions using my method?
    I can't say, as I don't see how you're going to proceed from where you got
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    (Original post by MathsCoder)
    Just out of curiosity, how regularly and for how long do you mathmos exercise?
    I get up from my desk and walk over to the coffee machine at least twice a day.
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    (Original post by Gregorius)
    I can't say, as I don't see how you're going to proceed from where you got
    Using your method, I got up to having to show that  \sin x cosh y = 2 where x, y satisfy z = x+iy, has infinitely many solutions. :/
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    (Original post by Insight314)
    Using your method, I got up to having to show that  \sin x sech y = 2 where x, y satisfy z = x+iy, has infinitely many solutions. :/
    Easier than that. Think about \sin{x}\cosh{y} = 2. Choose x so that \sin(x + n \pi) = 1. Now work out what y must be,
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    (Original post by Gregorius)
    Easier than that. Think about \sin{x}\cosh{y} = 2. Choose x so that \sin(x + n \pi) = 1. Now work out what y must be,
    Whoops, I meant sin x cosh y in my reply, but you saw that.
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    Can I use the fact that sinx = (e^ix-e^-ix)/2i and solve directly for sinx = 2 to prove that it has infinite solutions
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    (Original post by Mathemagicien)
    I do exercise papers every day, for a few hours



    A powerlifting mathematician? :eyeball:
    Probably a better powerlifter tbh


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    (Original post by wlog54)
    Can I use the fact that sinx = (e^ix-e^-ix)/2i and solve directly for sinx = 2 to prove that it has infinite solutions
    You can certainly set off in this direction - the equation turns into a quadratic in  e^{iz} , which you can solve so show you get a solution. But the difficulty then is to show that you get an infinity of them (using periodicity). I suspect that you end up going in circles proving the periodicity.

    The advantage of the method I suggested is that it gets you the infinity of solutions due to the periodicity of the real function sin(x), which I assume you are allowed to quote.

    Take a look at this plot of sin(z) in the complex plane to get your intuitions fired up!
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    (Original post by Gregorius)
    You can certainly set off in this direction - the equation turns into a quadratic in  e^{iz} , which you can solve so show you get a solution. But the difficulty then is to show that you get an infinity of them (using periodicity). I suspect that you end up going in circles proving the periodicity.

    The advantage of the method I suggested is that it gets you the infinity of solutions due to the periodicity of the real function sin(x), which I assume you are allowed to quote.

    Take a look at this plot of sin(z) in the complex plane to get your intuitions fired up!
    I see where your avatar comes from!
 
 
 
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