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Can cos(a + 90 - b)=k.a be solved?

  1. Jun 25, 2011 #1
    hi,

    quick question, is it possible to solve an equation of the type:

    cos(a + 90 - b) = k.a

    to find a where b and k are known? Do you have to solve this by iteration?

    kenneth
     
  2. jcsd
  3. Jun 25, 2011 #2

    micromass

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    Hi kennethgilpin! :smile:

    I very much doubt that your equation can be solved analytically. But I'm quite sure you can find a numerical answer. Let's define

    [tex]f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow \frac{\cos(x+90-b)}{k}[/tex]

    take [itex]x_0[/itex] arbitrary and keep applying f on it. You'll get a sequence

    [tex]x_0,f(x_0),f(f(x_0)),f(f(f(x_0))),...[/tex]

    Compute the first few terms and see if it converges. If it doesn't, then we'll have to apply more powerful methods...
     
  4. Jun 26, 2011 #3
    We can group x and 90 and use the angle sum identity of cosine to say cos(x+90-b)=cos(x+90)cos(b)+sin(x+90)sin(b). We can use the same identity to say cos(x+90=cos(x)cos(90)-sin(x)sin(90)=-sin(x) and we can use the angle sum identity of sine to say sin(x+90)=sin(x)cos(90)+sin(90)cos(x)=cos(x). It follows that cos(x+90)cos(b)+sin(x+90)sin(b)=-sin(x)cos(b)+cos(x)sin(b). Therefore, cos(x+90-b)=cos(x)sin(b)-sin(x)cos(b).

    We had to use the same identity twice, so I guess you could call it iterated.
     
  5. Jun 27, 2011 #4
    Sorry, didn't understand a thing the guys before me said xD sorry, but I'm guessing that it's not possible to find a FINITE answer as it is only one ecuation with three unknown answers.
     
  6. Jun 27, 2011 #5

    Mentallic

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    TylerH misunderstood the question, and the word you're looking for isn't finite, it's "expressible in terms of elementary functions". The solution(s) are definitely finite.
     
  7. Jun 28, 2011 #6
    Thank you for your detailed replies.

    although I was pretty handy at maths, that was 'A' level which was almost 20 years ago, so I don't understand the notation in micromasses answer.

    I'm a physiological modeller, and solved the equation by induction, i.e. by knowing my upper and lower limit for the equation I picked a value halfway, then calculated an answer for both sides of the equation, then picked a value halfway between that and so on. I'm guessing that micromass was sort of suggesting that.

    The problem is that this has to be done about 25 times (*10 iterations) and each iteration has floating point stuff that computers don't like. Having to solve things by induction is the bane of physiological modelling.

    kenneth
     
  8. Jun 28, 2011 #7

    Mentallic

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    The numerical method you're applying is quite slow, as you can tell. Applying Newton's method however will be a lot quicker.

    It's defined as follows:

    [tex]a_n=a_{n-1}-\frac{f(a_{n-1})}{f '(a_{n-1})}[/tex]

    And basically, this is how it works:

    Since you need to solve for a (usually denoted as x), you let [tex]y=f(a)=ka-\cos\left(a+90-b\right)[/tex]

    Then find its derivative, [tex]f '(a)=k+\sin\left(a+90-b\right)[/tex]

    And for the first iteration, take a reasonably good guess at what the value of a is (denoted a0), then apply the Newton's method:

    [tex]a_1=a_0-\frac{ka_0-\cos\left(a_0+90-b\right)}{k+\sin\left(a_0+90-b\right)}[/tex]

    a1 will be a better approximation, then you apply the same formula to find a2 etc.


    EDIT: it's supposed to be f ' (a), LaTeX is screwing with me
     
  9. Jun 28, 2011 #8
    Ahh, I remember this from school. Presumably it only works if you can find a derivative of the original function.
     
  10. Jun 28, 2011 #9

    Mentallic

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    Yep, but unless you have other non-elementary functions that you're also dealing with, it shouldn't be a worry.
     
  11. Jun 28, 2011 #10

    PAllen

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    One caution about Newton's method is that a bad initial choice can diverge. If you are programming this, it should be easy to find open source code for 'safe newton/raphson', where you bound that a root must be in some interval. Then, the code detects a bad guess and uses another method for one iteration, going back to Newton thereafter. As long as you can find some bounds in which a root must be present, such codes are guaranteed convergent and quite fast.

    In this case, for example, you can guarantee there is s root between [-pi,+pi].

    [EDIT: oh, you are using degrees; make that between [-180,180] ]
     
    Last edited: Jun 28, 2011
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