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Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) := \cos^2 x,\ x\in \mathbb{R}$

1) Find the Taylor polynomial for $f$ of degree 3 at the point 0 and show that: $$\lvert (\cos^2x-(1-x^2) \vert \le \frac{1}{3}x^4, \ \forall x\in\mathbb{R} $$

2) Use 1) to show $$\lvert \int_0^\frac{1}{2} (\cos\sqrt t)^2\,dt - \frac{3}{8}\rvert\le \frac{1}{72} $$


1) I am not going to show the computation, but just the result:

The Taylor polynomial at degree 3 at 0 is given by:
$$ T_n(x) = 1-x^2$$

The second part of the question is the remainder term: $$\lvert R_n \rvert = \lvert f(x)-T_n(x) \rvert =\lvert \cos^2 x-(1-x^2) \vert \le \frac{1}{3}x^4 $$ and I know that $$\lvert R_n \rvert \le \frac{M_n}{(n+1)!}\lvert x-x_0 \rvert^{n+1}$$ Where $M_n$ is given by:

$$ M_n \ge \max {\lvert f^{n+1}(t) \rvert \lvert t\in [x_0,x]}$$ The Taylor polynomial of degree 4 is: $8(\cos^2 x-\sin^2 x)$ at 0 $\;8(\cos^2 x - \sin^2 x) = 8$

It then follows that $\frac{M_n}{(n+1)!}\lvert x-x_0 \rvert^{n+1} = \frac{8}{4!}x^4 = \frac{1}{3}x^4$

2) How am I supposed to use this to show that the integral is less than $\frac{1}{72}$?

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Let $x=\sqrt{t}$ in (1) and integrate.

$$\int_0^{1/2}(1-t)\,dt=\frac{3}{8}$$

$$\int_0^{1/2}\frac{1}{3}t^2\,dt=\frac{1}{72}$$

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  • $\begingroup$ Ah such a simple solution, thank you! $\endgroup$ – Simbörg Jun 3 '18 at 17:31

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