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Math Help - Taylor series and extrema of a function

  1. #1
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    Taylor series and extrema of a function

    Hello everyone,

    if I'm asked to prove that f(x) has a local maxima for x=a (a is a given real number and f: D->R), and if I know that if f^{2n}(a)>0, (\n \in \mathbb{N^*}) \Rightarrow f has a maxima in a.Then I can use the taylor series of f evaluated in a to prove it. BUT how can I know till what degree of derivation I need to go. sure I could find the associate series of f(x) but it can be really annoying. So is there a way to find how many 2n times I need to expand me series to find what I want, because, for now if the second derivate is equal to zero I check the 4th and so on.

    Here is an example:
    f(x)=\frac{(x^2+(1+x^2)^{\frac{1}{2}})^{\frac{1}{3  }}}{1+2\sin(x)^2+\cos(x)^{2}}
    prove that f(0) is maxima.
    f^{2}(0)=0, f^{4}(0)>0
    how can I know that the 4th derivate at o is not equal to 0.

    Thanks in advance.
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  2. #2
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    For a max or min, it's necessary that the first derivative is zero, and then that the first non-vanishing derivative is an even numbered one. If that derivative is negative then the point will be a max, if it's positive then the point is a min. If the first non-vanishing derivative is an odd numbered one, then you have a point of inflection, (so it is not sufficient simply to evaluate the second, fourth, ... derivatives, you must evaluate the odd numbered ones as well). (Unless that is you know for certain that the point is definitely a max or min). Usually you will not know in advance just how far you have to keep differentiating.

    The (somewhat disgusting) function f(x) you give as an example, were you given it or did you make it up ?

    Any differentiation for this function is overkill. It's even, meaning that its graph will be symmetric about the vertical axis. All you need do is evaluate it at the origin and some near-by point and compare values.
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  3. #3
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    No I found this in a book as en exercise. Indeed this function is atrocious to derivate. I think the goal of the exercise is to show that it's easier to use the Taylor series and it is what I did and that's why I asked this.
    The way I did it is the following, 1st you found the Ts evaluated in 0 to the 4th derivate of (1+x^2)^{\frac{1}{2}} then you do it again for x^2+(1+x^2)^\frac{1}{2} then you do it for z^{\frac{1}{3}} evaluated in 1 and so on, then you put all the Ts you found together to find the one of f.
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  4. #4
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    You still have to incorporate the trig functions on the bottom line don't you ?

    The base point is the origin so you can simply write them in as the well-known Maclaurin series, but there is still work to be done after that.

    If the question asks you to use Taylor series then O.K. but if it's simply a case of checking on a max or min at x=0, (and you know that there will be a max or min at x=0 because the function is symmetric about the vertical axis), then evaluating the function at x=0 and some very small value of x is easiest.
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