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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.


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If you're restricted only to Taylor polynomials, then you will have to take a lot of terms to reach your bound. Here's a simplifying suggestion you might want to consider: since Taylor polynomials, be design, work best for arguments near the expansion point, the points where the error will be greatest would be near the boundaries of the interval you're considering. Try looking into how to minimize the error at the ends. – J. M. Jul 25 '12 at 17:43
On the other hand, if this is actual work, and not a textbook assignment you're working on, you can certainly do better than the Taylor polynomial (e.g. using Chebyshev series). – J. M. Jul 25 '12 at 17:44
This is a problem from a practice master's entrance exam that tests general, upper-level undergraduate mathematics knowledge. As such, the expected answer most likely does not involve Chebyshev series. That being said, a solution is a solution. There is a similar problem, however, that uses a Taylor approximation in Ch.7 of Apostol Vol. 1. Thank you, J.M., for the suggestion about the end points. – user36387 Jul 25 '12 at 18:11
"general, upper-level undergraduate mathematics knowledge." - then yes, you're only limited to Taylor. Start by looking at $x=\pm1$... – J. M. Jul 25 '12 at 18:14
up vote 2 down vote accepted

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. – user36387 Jul 25 '12 at 18:45

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