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Thread: 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 $\displaystyle (-\infty,\infty)$ and satisfy the following properties:

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

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

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

    (b) Deduce that the function $\displaystyle C(\theta)$ has no zeros in the interval [0,1.4] but at least one zero in the interval [1.4,1.6]. Let $\displaystyle \lambda$ denote the smallest such zero. Prove that $\displaystyle S(\lambda) = 1, C(\theta + 2\lambda) = -C(\theta)$ and $\displaystyle S(\theta + 2\lambda) = -S(\theta)$
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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 $\displaystyle (-\infty,\infty)$ and satisfy the following properties:

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

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

    $\displaystyle 1 - \frac{\theta^2}{2} \leq C(\theta) \leq 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24}$
    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 $\displaystyle f(\theta) = (S(\theta))^2 + (C(\theta))^2$. Its derivative is $\displaystyle f'(\theta) = 2S(\theta)C(\theta) - 2C(\theta)S(\theta) = 0$. So f is constant, and by evaluating it at 0 you see that the constant is 1. Therefore $\displaystyle (S(\theta))^2 + (C(\theta))^2 = 1$. This shows in particular that $\displaystyle |C(\theta)|\leqslant 1$ for all $\displaystyle \theta$.

    Now you can apply Taylor's Theorem with Lagrange's form of the remainder (taking the first two terms plus remainder) to see that $\displaystyle C(\theta) = 1 - \frac{\theta^2}2C(\phi)$ for some $\displaystyle \phi$ between 0 and $\displaystyle \theta$, and this must be $\displaystyle \geqslant 1 - \frac{\theta^2}2$. 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 $\displaystyle C(\theta)$ has no zeros in the interval [0,1.4] but at least one zero in the interval [1.4,1.6]. Let $\displaystyle \lambda$ denote the smallest such zero. Prove that $\displaystyle S(\lambda) = 1, ...$
    This follows easily from the intermediate value theorem together with the magic formula $\displaystyle (S(\theta))^2 + (C(\theta))^2 = 1$.

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