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Math Help - Trigonometric "regression".

  1. #1
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    Trigonometric "regression".

    Hello.
    I have just encountered this problem. I want to find a cosine function with the graph represented on the picture below.

    I have found the following:
    \rm{A} \times \cos(cx + \phi) + d

    2 \times \cos(\frac \pi 2x + \phi)
    d=0

    But I am unable to find \phi. So then my question would be: How do I find it, please?

    In advance, thank you for spending time and efforts on my problems :]
    Attached Thumbnails Attached Thumbnails Trigonometric "regression".-graph_cos.png  
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by MatteNoob View Post
    Hello.
    I have just encountered this problem. I want to find a cosine function with the graph represented on the picture below.

    I have found the following:
    \rm{A} \times \cos(cx + \phi) + d

    2 \times \cos(\frac \pi 2x + \phi)
    d=0

    But I am unable to find \phi. So then my question would be: How do I find it, please?
    I guess that \phi\in (-\pi,\pi] ?

    One possiblity to find the value of \phi is noting that for x_0=-\frac{2}{\pi}\phi one has 2\cos\left(\frac{\pi}{2}x_0+\phi\right)=2\cos\left  (-\frac{\pi}{2}\frac{2}{\pi}\phi+\phi\right)=2. As \phi\in (-\pi,\pi] we also have x_0\in[-2,2). Using the graph, one can solve 2\cos \left( \frac{\pi}{2}x_0+\phi\right)=2 for x_0\in[-2,2) and derive the value of \phi.
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    I guess that \phi\in (-\pi,\pi] ?

    One possiblity to find the value of \phi is noting that for x_0=-\frac{2}{\pi}\phi one has 2\cos\left(\frac{\pi}{2}x_0+\phi\right)=2\cos\left  (-\frac{\pi}{2}\frac{2}{\pi}\phi+\phi\right)=2. As \phi\in (-\pi,\pi] we also have x_0\in[-2,2). Using the graph, one can solve 2\cos \left( \frac{\pi}{2}x_0+\phi\right)=2 for x_0\in[-2,2) and derive the value of \phi.
    Thank you for your reply, flyingsquirrel.

    The task doesn't say anything about \phi \in \langle -\pi, \, \pi ] however, the correct answer should be that \phi = \frac \pi 4

    I am not sure how you found x_0 = -\frac 2\pi, so please feel free to elaborate on that. :]

    Furthermore, I didn't understand how you found that:
    2\cdot \cos\left(-\frac\pi 2 \cdot \frac 2\pi \phi + \phi\right) = 2 (What I mean is, how do you know it is equal to two?)
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by MatteNoob View Post
    I am not sure how you found x_0 = -\frac 2\pi, so please feel free to elaborate on that. :]
    I don't know if you made a typo or misread but x_0=-\frac{2}{\pi}{\color{red}\phi}. I chose this value because the link between x_0 and \phi is simple hence once you know x_0 you can easily find \phi. I've no other jusification, I just noted this.

    Furthermore, I didn't understand how you found that:
    2\cdot \cos\left(-\frac\pi 2 \cdot \frac 2\pi \phi + \phi\right) = 2 (What I mean is, how do you know it is equal to two?)
    2 \cos\left(-\frac\pi 2 \cdot \frac 2\pi \phi + \phi\right) =2 \cos\left(-\phi + \phi\right) =2 \cos 0= 2
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  5. #5
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    @ flyingsquirrel

    Oh, yes, sorry about that typo, hehe. Thank you for the explanation, it is much more clear now. If you have other methods, please feel free to educate me on those too. :]

    Again, thank you very much for your time and help, I appreciate it!
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by MatteNoob View Post
    If you have other methods, please feel free to educate me on those too. :]
    Another method would be using inverse trigonometric functions.

    (we still work with \phi\in(-\pi,\pi])

    Using the graph, we know that 2\cos\left(\frac{\pi}{2}\cdot 0+\phi\right) \approx 1.39 (y-intercept) hence \cos \phi \approx \frac{1.39}{2}=0.70. This gives us two values for \phi : either \phi\approx\arccos 0.70 either \phi\approx-\arccos 0.70. To determine which of the two values is the right one, let's use another point. For example, we know that 0=2\cos\left( \frac{\pi}{2}\cdot 0.5+\phi \right) (x-intercept for x=0.5) hence \phi has to satisfy the equation 0=\cos\left(\frac{\pi}{4}+\phi\right) which is only possible if \phi\approx-\arccos 0.70=-46^{\circ}.
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  7. #7
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    @ flyingsquirrel

    Thank you very much for the other method. :]

    If you still have some time, I'd like to go back to the x_0= \frac 2\pi \phi thing. I am sorry if I come across as thick, but as you have probably noted from my signature, I have some issues regarding terminology etc.

    Quote Originally Posted by flyingsqurirrel
    I don't know if you made a typo or misread but . I chose this value because the link between and is simple hence once you know you can easily find . I've no other jusification, I just noted this.
    The problem is that I don't understand how you found x_0 to be -\frac 2\pi

    Is this because it would cancel out the period for the function and then set it to zero? Let me give you an example of what I mean.

    If f(x) = 5\cos\left(\frac \pi 4x + \phi \right)
    would: x_0 = -\frac 4\pi \phi which gives us:
    5\cos\left(\frac{\pi}{4}\cdot -\frac{4}{\pi}\phi + \phi\right) = 5\cos(0) = \underline{5}

    So then, when we know that x\in [-5,\, 5] and \phi \in [-\pi,\, \pi\rangle ?

    If so, how do you know that \phi \in [-\pi, \, \pi \rangle ?

    Sorry about the lengthy and messy post, but I can't help being curious and craving understanding here, hehe.
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by MatteNoob View Post
    The problem is that I don't understand how you found x_0 to be -\frac 2\pi

    Is this because it would cancel out the period for the function and then set it to zero?
    Yes ! (I should have written this in one of my previous posts...)
    Let me give you an example of what I mean.

    If f(x) = 5\cos\left(\frac \pi 4x + \phi \right)
    would: x_0 = -\frac 4\pi \phi which gives us:
    5\cos\left(\frac{\pi}{4}\cdot -\frac{4}{\pi}\phi + \phi\right) = 5\cos(0) = \underline{5}
    OK

    So then, when we know that x\in [-5,\, 5] and \phi \in [-\pi,\, \pi\rangle ?

    If so, how do you know that \phi \in [-\pi, \, \pi \rangle ?
    I'm not sure to understand what you mean. Anyway : the cosine function is 2\pi-periodic hence without the condition \phi \in (-\pi, \, \pi ], \phi could be any number \frac{\pi}{2}+2k\pi,\,k\in\mathbb{Z}. As we only want one value of \phi, we add the condition \phi\in(-\pi, \, \pi ]. This gives us a unique solution : \frac{\pi}{2}. Note that one can choose any interval as long as its "length" is 2\pi : we could have set \phi\in[42,42+2\pi)... (usually one takes \phi in (-\pi,\pi] or in [0,2\pi))

    Then, from the condition \phi\in(-\pi,\pi] we can find the interval in which x_0=-\frac{4}{\pi}\phi lies :

    We know that -\pi<\phi \leq \pi

    Multiply the inequalities by -1 :   \pi>-\phi\geq -\pi

    Multiplying by \frac{4}{\pi} we get  \frac{4}{\pi}\cdot \pi > -\frac{4}{\pi}\phi \geq -\frac{4}{\pi}\cdot \pi

    Simplifying, it gives us :  4>x_0\geq 4

    In other words : x_0\in[-4,4). (I should also have explained this in one of my previous posts )
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