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How to find minimum turning points

  1. May 25, 2014 #1
    I have a function:

    f(x) = Asin2(x) + Bcos2(x) + Csin(2x)

    and I want to find the minimum turning point(s). To start with, I calculated:

    f'(x) = (A - B)sin(2x) + 2Ccos(2x)

    Therefore, turning points occur when f'(x)=0, or:

    tan(2x) = 2C / (B - A)

    To find the minima, I then want to look at the 2nd derivative:

    f''(x) = 2(A - B)cos(2x) - 4Csin(2x)

    So, as far as I understand, minima will occur when both f'(x)=0 and f''(x)>0. So the second inequality can be expressed as:

    2(A - B)cos(2x) > 4Csin(2x)

    Now I'm having trouble knowing how to proceed. Since sin and cos can be positive or negative depending on x, I can't see a neat way of really understanding where the minima will be. Could anyone help?
  2. jcsd
  3. May 25, 2014 #2


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    Best to put [tex]
    f(x) = \frac12(A + B) + \frac12(B - A)\cos(2x) + C \sin(2x)
    = \frac12(A + B) + R\sin(2x + \alpha)
    using [tex]
    R\sin(2x + \alpha) = R\sin(\alpha)\cos(2x) + R\cos(\alpha)\sin(2x).
    The minima then occur when [itex]\sin(2x + \alpha) = -1[/itex].
  4. May 25, 2014 #3
    Okay that's really clever... it's taken me a while to figure out how that works.

    So, if I have [itex]0 \leq x < 2\pi[/itex], I think taking the arcsin would give:[tex]\alpha+2x=\frac{3\pi}{2}[/tex]Then, given how we've defined [itex]\alpha[/itex], I feel like we can say:[tex]\tan\alpha = \frac{(B-A)}{2C}[/tex]Putting those things together, I get:[tex]x=\frac{3\pi}{4}-\frac{1}{2}\tan^{-1}\left\{\frac{(B-A)}{2C}\right\}[/tex]Hmm... but given that tan(x) repeats itself twice on the interval I'm interested in [itex](0 \leq x < 2\pi)[/itex] ... I feel like I've just reintroduced ambiguity.

    I'm a bit out of my depth, so I've crunched some numbers with some computer simulations... and this x seems to sometimes give the minimum, sometimes the maximum. I feel like I need to somehow constrain (B - A)/2C ... but at the moment I'm somewhere on a slippery slope to proving 1 = 0...
  5. May 25, 2014 #4
    I've looked more closely at my problem and have determined three further constraints:[tex]A\geq0\\B\geq0\\C\sin(2x)\geq0[/tex]Imposing these constraints seems to provide a unique solution in my computer simulations... but I'm not really certain why. Can anyone offer any insight?
  6. May 26, 2014 #5


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    That is to be expected, because [itex]f(x)[/itex] has a period of [itex]\pi[/itex], not [itex]2\pi[/itex]. Thus there are two minima in [itex]0 \leq x \leq 2\pi[/itex].
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