Taylor Polynomials and Numerical Analysis

mynorka
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Homework Statement


Use a Taylor Polynomial about pi/4 to approximate cos(42){degrees} to an accuracy of 10^-6.

*To get an accuracy of 10^-6, use the error term to determine an nth Taylor Polynomial to use.

Homework Equations


x = 45 or pi/4, x0 = 42 or 7pi/30

cos(x) = Pn(x) + Rn(x)

Polynomial Term - Pn(x) = ∑f^(k)(x-x0)^k/(k)!

Error Term - Rn(x) = f^(n+1)(ζ(x))(x-x0)^(n+1)/(n+1)!

The Attempt at a Solution


(pi/60)^n/(n+1)! < ((-60/pi)*10^-6)/pi

^I get stuck at this part. I'm supposed to solve for n, but the left-hand side of this inequality confuses me.
 
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mynorka said:

Homework Statement


Use a Taylor Polynomial about pi/4 to approximate cos(42){degrees} to an accuracy of 10^-6.

*To get an accuracy of 10^-6, use the error term to determine an nth Taylor Polynomial to use.

Homework Equations


x = 45 or pi/4, x0 = 42 or 7pi/30

cos(x) = Pn(x) + Rn(x)

Polynomial Term - Pn(x) = ∑f^(k)(x-x0)^k/(k)!

Error Term - Rn(x) = f^(n+1)(ζ(x))(x-x0)^(n+1)/(n+1)!

The Attempt at a Solution


(pi/60)^n/(n+1)! < ((-60/pi)*10^-6)/pi

^I get stuck at this part. I'm supposed to solve for n, but the left-hand side of this inequality confuses me.

What are you doing? We have ##|\text{error}| \leq (\pi/60)^{n+1}/(n+1)!##, and this must not exceed ##10^{-6}##.
 
this is pretty tough to read. perhaps put it in latex and then we can check it out
 
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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