Really - Taylor Polynomial Approximation Error

bcjochim07
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Homework Statement



Use Taylor's theorem to obtain an upper bound of the error of the approximation. Then calculate the exact value of the error.

cos(.3) is approximately equal to 1 - (.3)^2/2! + (.3)^4/4!

Homework Equations





The Attempt at a Solution



I came up with upper bound saying

Rn = (-sinz/5!)*(.3)^5 < (.3)^5/5!

so the upper bound is (.3)^5/5! which is about .00002

But for the exact error I have no idea how to calculate it. There aren't any examples in my textbook.
 
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I think they want you to use a calculator to actually find cos(.3) then subtract the approximation to see if how well your estimate worked.
 
The only way to calculate the exact value of the error is to take the exact value of the cosine and subtract the value of the series. I think they just mean punch cos(.3) into your calculator and then subtract the series value and see that it's less than the remainder term.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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