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Taylor Approximation ofr $\\cos(2x)$

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Fin the fourth order Taylor Approximation of $\\cos(2x)$

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a)

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We need the derivatives of $\\cos(2x)$ up to fourth order. They are 

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\\begin{align}
\\cos^{(0)}(2x) &= \\cos(2x) \\quad   &\\cos^{(1)}(2x) &= -2\\sin(2x)\\\\
\\cos^{(2)}(2x) &= -4\\cos(2x) \\quad  &\\cos^{(3)}(2x) &= 8\\sin(2x)\\\\
\\cos^{(4)}(2x) &= 16\\cos(2x) \\quad  
\\end{align}

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Then the fourth order Taylor Approximation to the given function is 

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\\[\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right) -
2\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right) -
\\frac{4\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{2}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^2
+\\frac{8\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3!}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^3 +
\\frac{16\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{4!}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^4 =
\\\\[3mm]
\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right) -
2\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right) -
2\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^2 +
\\frac{4\\sin\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^3 +
\\frac{2\\cos\\left(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi}\\right)}{3}\\left(\\simplify[zeroTerm, fractionNumbers]{x-{r}*pi}\\right)^4
\\]

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(Note that in the answer above, the algorithm approximates sine and cosine values that appear in the polynomial we gave here)

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b)

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We need the fifth derivative for the error term. That is $\\cos^{(5)}(2x) = -32\\sin(2x)$. Then by Taylor's Theorem the error can computed by 

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\\[R_5(\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}) = \\frac{(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi - {r}*pi})}{5!}\\cos^{(5)}(2\\eta) =
\\frac{(\\simplify[fractionNumbers, zeroFactor, unitFactor]{{r2}*pi - {r}*pi})}{5!}(-32(\\sin(2\\eta)) \\quad \\\\[1mm] \\mbox{for some } \\eta\\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}\\right] \\quad (\\mbox{or }\\;
\\eta\\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}\\right])\\]

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Note that findin what $\\eta$ should is not possible at this stage. However, one can still bound the error. For that you need to find the $\\eta \\in \\left[\\simplify[fractionNumbers, unitFactor, zeroFactor]{{r}*pi}, \\simplify[fractionNumbers, unitFactor, zeroFactor]{{r2}*pi}\\right]$ such that $\\sin(2\\eta)$ is maximal.

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Find the $4$-th order Taylor Approximation to $\\cos x$ around $x=\\simplify[zeroFactor, fractionNumbers]{{r}*pi}$.

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Find an upper bound for the error between the approximation you found above and $\\cos(2x)$ at $x=\\simplify[zeroFactor, unitFactor, fractionNumbers]{{r2}*pi}$.

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