Taylor Polynomial Homework: Estimating x Range with Error < 0.01

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Homework Statement



I can either use the alternating series estimation thereom (which i don't really know) or Taylor's Inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error.

sin(x) = x - (x^3)/6 (|error| < 0.01)

Do I just start writing out the terms of the sine series? I'm not sure exactly what I'm supposed to do here.
 
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You are supposed to estimate the error in a truncation of the infinite series. As you don't seem to know either the alternating series estimate or the Taylor series remainder term, I think you will need to look at least one of them up. Can you do that? I would suggest starting with the alternating series version. It's easier.
 
thanks, Dick, I will look that up
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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