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I require a polynomial $p(x)$ such that $\left|p(x) - \int_0^x \cos{(t^2)} dt\right| < \frac{1}{10!}$ for all $x \in [-1, 1]$. I know that I should probably use the fact that if $m\leq f^{n+1} {(t)} \leq M$ for $t$ in an interval containing the point $a$, then $m \frac{(x-a)^{(n+1)}}{(n+1)!} \leq E_n(x)\leq M \frac{(x-a)^{(n+1)}}{(n+1)!}$ for $x > a$ and $m \frac{(a-x)^{(n+1)}}{(n+1)!} \leq (-1)^{(n+1)}E_n(x)\leq M \frac{(a-x)^{(n+1)}}{(n+1)!}$ for $x < a$, where $E_n(x)$ is the Taylor remainder. These facts combined with the Taylor series for $\cos {w}$ followed by the appropriate substitution should give me the desired polynomial, although I keep getting stuck. Any clarification would be helpful.

Thanks.

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    "general, upper-level undergraduate mathematics knowledge." - then yes, you're only limited to Taylor. Start by looking at $x=\pm1$...2012-07-25

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Write down the ordinary Taylor expansion of $\cos(t^2)$ about $t=0$. This is done by recalling that $\cos w=1-\frac{w^2}{2!}+\frac{w^4}{4!}-\frac{w^6}{6!}+\cdots $ and substituting $w=t^2$. Now integrate term by term from $0$ to $x$.

Note that the series we get is, for $|x|\le 1$, an alternating series: the terms alternate in sign, decrease in absolute value, and approach $0$.

So the error made in cutting off somewhere is less, in absolute value, than the first "neglected" term.

That criterion works efficiently, and does not require the complicated manipulations and notation that you are using.

Remark: The various expressions for the remainder are important theoretical* tools. They are often much less useful as practical tools for making estimates.

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    Ah, that makes sense. Thank you for the help, Andre.2012-07-25