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Math Help - Continuity, series, inverse function

  1. #1
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    Red face Continuity, series, inverse function

    1. If f:\mathbb{R}\mapsto\mathbb{R} is monotonic increasing and f(f(x)) is continuous, prove that f(x) is continuous!

    2. Find the value of \sum^{\infty}_{k=1} \frac{1}{2^k}\tan (\frac{\pi}{2^{k+2}})

    3. If f:\mathbb{R}^2\mapsto\mathbb{R}^2\text{ where } f((x_1,x_2))=(x_1+2x_2,2x_1+x_2)\forall x_1,x_2\in\mathbb{R},\text{ then find } f^{-1}
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    Quote Originally Posted by GOKILL View Post
    2. Find the value of \sum^{\infty}_{k=1} \frac{1}{2^k}\tan (\frac{\pi}{2^{k+2}})
    Let f(x) = -\sum_{k=1}^\infty\ln\bigl(\cos(x/2^k)\bigr). Then f'(x) = \sum_{k=1}^\infty2^{-k}\tan(x/2^k), so we are looking for f'(\pi/4).

    To compute f(x), write \sum_{k=1}^n\ln\bigl(\cos(x/2^k)\bigr) = \ln\Bigl(\prod_{k=1}^n\cos(x/2^k)\Bigr). Then use the formula \cos x\cos y = \tfrac12\bigl(\cos(x+y) + \cos(x-y)\bigr) to get

    \cos\tfrac x2\cos\tfrac x4 = \tfrac12\bigl(\cos\tfrac14x +\cos\tfrac34x\bigr),

    \cos\tfrac x2\cos\tfrac x4\cos\tfrac x8 = \tfrac14\bigl(\cos\tfrac18x +\cos\tfrac38x + \cos\tfrac58x +\cos\tfrac78x\bigr),

    ...

    \prod_{k=1}^n\cos(x/2^k) = \frac1{2^{n-1}}\sum_{k=1}^{2^{n-1}} \cos\Bigl(\frac{2k-1}{2^n}x\Bigr).

    That last line can be viewed as a Riemann sum for \int_0^1\!\!\!\cos(xt)\,dt = \frac{\sin x}x. Therefore f(x) = -\ln\Bigl(\frac{\sin x}x\Bigr), and so f'(x) = x^{-1} - \cot x.

    Finally, f'(\pi/4) = \frac4\pi-1 \approx 0.2732.
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