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Math Help - Taylor's Theorem

  1. #1
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    Taylor's Theorem

    Suppose f:\mathbb{R} \rightarrow \mathbb{R} is such that both f\prime and f\prime\prime exist for all x \in \mathbb{R}, so that Taylor's theorem tells us that, for each a, h\in \mathbb{R} there is a \theta such that 0<\theta<1 so that f(a+h)=f(a)+hf\prime(a)+\frac{1}{2}h^2 f\prime\prime(a+\theta h). Suppose further that on [0,2] the inequalities |f(x)|\leq 1 and |f\prime\prime(x)| \leq 1 hold. Write down the Taylor expansions of f(0) and f(2) about the point x \in [0,2], using the above form of Taylor's Theorem, with a remainder involving f\prime\prime. Hence prove that for all x \in [0,2] we have |f\prime(x)| \leq 2.

    I'm not sure about the first part, what I have is:
    f(0)=f(-h)+hf\prime(-h)+\frac{1}{2}h^2f\prime\prime(h(\theta-1)) by setting a+h=0 and
    f(2)=f(2-h)+hf\prime(2-h)+\frac{1}{2}f\prime\prime(2+h(\theta-1)) but I don't think this is correct. It says about the point x in [0,2] but I don't know how to apply this to the formula. I have no idea for the last part. Thanks!
    Last edited by worc3247; April 4th 2011 at 01:58 PM.
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    ignore this
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  3. #3
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    Taylor's expansion about x is (in a slightly rewritten form):
     f(y)=f(x)+(y-x)f\prime(x)+\frac{1}{2}(y-x)^2 f\prime\prime(\theta)
    now plug in y=2 and y=0:
     f(0)=f(x)-xf\prime(x)+\frac{1}{2}x^2 f\prime\prime(\theta_1)
     f(2)=f(x)+(2-x)f\prime(x)+\frac{1}{2}(2-x)^2 f\prime\prime(\theta_2)
    from first equation you can get e.g.
     xf\prime(x) =f(0)+f(x)+ \frac{1}{2}x^2 f\prime\prime(\theta_1) \le 1+1+2=4
    and second
     (2-x)f\prime(x) =f(2)+f(x)+ \frac{1}{2}(y-x)^2 f\prime\prime(\theta_2) \le 1+1+2=4
    add them up and you get  2f\prime(x) \le 8, i.e.  f\prime(x) \le 4. maybe i'm messing smth up, dunno... :/
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  4. #4
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    Quote Originally Posted by choovuck View Post
    Taylor's expansion about x is (in a slightly rewritten form):
     f(y)=f(x)+(y-x)f'(x)+\frac{1}{2}(y-x)^2 f\prime\prime(\theta)
    now plug in y=2 and y=0:
     f(0)=f(x)-xf'(x)+\frac{1}{2}x^2 f''(\theta_1)
     f(2)=f(x)+(2-x)f'(x)+\frac{1}{2}(2-x)^2 f''(\theta_2)
    This is an infuriating problem. I have come across variants of it many times, and it always takes me ages to get the answer to come out right. You need to do the steps of the argument in exactly the right order.

    From the first equation,
    xf'(x) = f(x) - f(0) + \frac12x^2f''(\theta_1).

    From the second equation,
    (2-x)f'(x) = f(2) - f(x) - \frac12(2-x)^2f''(\theta_2).

    Add those equations:
    2f'(x) = f(2) - f(0) + \frac12\bigl(x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr).

    Therefore
    |f'(x)| \leqslant \frac12|f(2) - f(0)| + \frac14\bigl|x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr|.

    But |f(2) - f(0)| \leqslant |f(2)| + |f(0)| \leqslant2, and

    \bigl|x^2f''(\theta_1) - (2-x)^2f''(\theta_2)\bigr| \leqslant x^2 + (2-x)^2 = 4 - 2x(2-x)\leqslant4.

    Thus |f'(x)| \leqslant 1+1=2.
    Last edited by Opalg; April 6th 2011 at 08:53 AM. Reason: corrected error
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