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Math Help - Taylor's Theorem with Lagrange's form of the remainder.

  1. #1
    Junior Member pearlyc's Avatar
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    Taylor's Theorem with Lagrange's form of the remainder.

    The question goes like this.

    Suppose that two real-valued functions C and S are defined and differentiable on the domain (-\infty,\infty) and satisfy the following properties:

    S'(\theta) = C(\theta), C'(\theta) = - S(\theta), S(0) = 0, C(0) = 1.

    (a) Use Taylor's Theorem with Lagrange's form of the remainder to prove that for all \theta \in R,

     1 - \frac{\theta^2}{2} \leq C(\theta) \leq 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24}

    (b) Deduce that the function C(\theta) has no zeros in the interval [0,1.4] but at least one zero in the interval [1.4,1.6]. Let \lambda denote the smallest such zero. Prove that S(\lambda) = 1, C(\theta + 2\lambda) = -C(\theta) and S(\theta + 2\lambda) = -S(\theta)
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    Quote Originally Posted by pearlyc View Post
    The question goes like this.

    Suppose that two real-valued functions C and S are defined and differentiable on the domain (-\infty,\infty) and satisfy the following properties:

    S'(\theta) = C(\theta), C'(\theta) = - S(\theta), S(0) = 0, C(0) = 1.

    (a) Use Taylor's Theorem with Lagrange's form of the remainder to prove that for all \theta \in R,

     1 - \frac{\theta^2}{2} \leq C(\theta) \leq 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24}
    These functions ought to remind you of functions called c** and s** that you have met before. That gives a motivation for the definition that follows.

    Define f(\theta) = (S(\theta))^2 + (C(\theta))^2. Its derivative is f'(\theta) = 2S(\theta)C(\theta) - 2C(\theta)S(\theta) = 0. So f is constant, and by evaluating it at 0 you see that the constant is 1. Therefore (S(\theta))^2 + (C(\theta))^2 = 1. This shows in particular that |C(\theta)|\leqslant 1 for all \theta.

    Now you can apply Taylor's Theorem with Lagrange's form of the remainder (taking the first two terms plus remainder) to see that C(\theta) = 1 - \frac{\theta^2}2C(\phi) for some \phi between 0 and \theta, and this must be \geqslant 1 - \frac{\theta^2}2. Now do the same, taking four terms of the Taylor series, plus remainder, to get the other inequality.

    Quote Originally Posted by pearlyc View Post
    (b) Deduce that the function C(\theta) has no zeros in the interval [0,1.4] but at least one zero in the interval [1.4,1.6]. Let \lambda denote the smallest such zero. Prove that S(\lambda) = 1, ...
    This follows easily from the intermediate value theorem together with the magic formula (S(\theta))^2 + (C(\theta))^2 = 1.

    Quote Originally Posted by pearlyc View Post
    ... C(\theta + 2\lambda) = -C(\theta) and S(\theta + 2\lambda) = -S(\theta).
    I don't see an easy way to get these results. You want to show that S(\lambda+\theta) = C(\theta) and C(\lambda+\theta) = -S(\theta). The only way I can see to do this is to show that C(\theta) is the sum of its Taylor series 1-\frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \ldots\,, and then to show that the function S(\lambda+\theta) is the sum of the same Taylor series.
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