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Find a bound for the absolute error on the interval $[x_{0}, x_{n}]$

$f(x) = cos(x)+sin(x)$

for $x_{0}=0, x_{1} = .25, x_{2} = .5, x_{3} = 1, n = 3$

So using the Lagrange error bound formula: $ \frac{cos(\xi) + sin(\xi)}{4!} x(x-.25)(x-.5)(x-1)$

My answer for the error bound is $1.55x10^{-3}$

The book's answer for the error bound is $1.59x10^{-3}$

I don't know where I am going wrong.

Huy Vo
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  • Did you use that $\cos(x)+\sin(x)=\sqrt{2}\sin(x+\frac\pi4)$? Please add details on how you found your bound, it is difficult to suggest alternatives without that. – Lutz Lehmann Mar 19 '17 at 07:57
  • I used just cos(x)+sin(x). – Huy Vo Mar 19 '17 at 08:02
  • I am hoping someone just does the problem and see where I diverge. I used the mean value theorem for x(x-.25)(x-.5)(x-1). For the trig function, I plug in 1. – Huy Vo Mar 19 '17 at 08:04

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