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Calculus question involving sine and relative extrema

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the function [tex]f(x) = x\sin\left( \frac{\pi}{x} \right)[/tex] has an infinite number of relative extrema.

    2. Relevant equations

    3. The attempt at a solution
    Using the product rule and chain rule to find the derivative f'(x):
    [tex] f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{x\pi}{x^2} \cos\left( \frac{\pi}{x} \right) [/tex]
    [tex] f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) [/tex]

    To try to find the relative extrema, I will determine for which values f'(x) = 0.
    [tex] \sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) = 0 [/tex]
    [tex] \sin\left( \frac{\pi}{x} \right) = \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) [/tex]
    [tex] \frac{\sin\left( \frac{\pi}{x} \right)}{\cos\left( \frac{\pi}{x} \right)} = \frac{\pi}{x} [/tex]
    [tex] \frac{\pi}{x}=\tan\left( \frac{\pi}{x} \right ) [/tex]
    Now I can replace π/x with k, if k ≠ 0:
    [tex] k=\tan(k) [/tex]
    The solution to the equation above are the points of interception between the line y = k and the curve y = tan(k). Since there is an infinite number of interceptions, the equation above has an infinite number of solutions.
    There is an infinite number of interceptions because, in the function y = tan(k), there are infinitely many vertical asymptotes, so, the line y = k will intercept the curve y = tan(k) in infinitely many points.
    So, f(x) = x sin(π/x) has an infinite number of relative extrema.

    Is this correct? Is there some important step missing in the middle?

    Thank you in advance.
     
  2. jcsd
  3. Nov 30, 2011 #2

    micromass

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    That is ok, but I have one annoying point.

    That [itex]f^\prime(x)=0[/itex] does NOT mean that there is a relative extremum at x. Indeed, look at the function [itex]f(x)=x^3[/itex]. This has derivative 0 at 0, but it is no extremum there.

    In order to be a relative extremum, you additionaly need [itex]f^{\prime\prime}(x)\neq 0[/itex].
     
  4. Nov 30, 2011 #3
    To know that, try plotting k=tan(k) and f(x)=x sin(x/pi) in the same graph.

    the sin function gives you the maximum value when sin(pi/x)=1--> pi/x=(2n+1)*pi/2 for n= 0,1,...

    So 1/x= n+1/2--> x=2/(2n+1)

    (2n+1)pi/2= tan(pi*(2n+1)/2)


    For small k the equation tank=k is an identity
     
  5. Nov 30, 2011 #4

    micromass

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    No it is never an identity for small k (only for k=0). That it is a good approximation does NOT mean that it is equal.
     
  6. Dec 1, 2011 #5
    That's true, the points where f''(x) = 0 cannot be relative extrema. In order to be a relative extremum f''(x) has to be nonzero. If f''(a) > 0, then the function is concave upwards in f(a), and, if f''(a) < 0, it's concave downwards in f(a). In f(x) = x³, f(0) is an inflection point, not a relative extremum, since f''(0) = 0 and f''(x) changes sign in x = 0.
    Calculating f''(x) using product rule and chain rule:
    [tex]f''(x)=-\frac{\pi\cos\left(\frac{\pi}{x}\right)}{x^2}-\pi \left( -\frac{\cos\left(\frac{\pi}{x}\right)}{x^2}+\frac{\pi\sin\left(\frac{\pi}{x}\right)}{x^3} \right)[/tex]
    [tex]f''(x)=-\frac{\pi^2\sin\left(\frac{\pi}{x}\right)}{x^3}[/tex]
    The expression above will be zero if and only if [itex] \sin\left(\frac{\pi}{x}\right) = 0 [/itex]. This will be true if and only if [itex] \frac{\pi}{x} = n\pi [/itex], where n is any nonzero integer (nonzero because [itex] \frac{\pi}{x} \neq 0 [/itex]).
    Since if [itex] \sin\left(\frac{\pi}{x}\right) = 0 [/itex], then [itex] \tan\left(\frac{\pi}{x}\right) = 0 [/itex], the values for which f''(x) = 0 can't be solutions to k = tan(k) where k = π/x. Therefore, all solutions to k = tan(k) are points in which [itex]f''(x) \neq 0[/itex]. Thus, they are relative extrema.
     
  7. Dec 1, 2011 #6

    micromass

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    This is correct!! You solved that very neatly! :smile:
     
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