In analysis I, my professor wrote that:
$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} + o(x^4)$
I would like to know why that's true.
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Taylor series for $\cos$. – Workaholic Mar 09 '16 at 21:37
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BTW, your professor could have written $o(x^5)$ instead of $o(x^4).$ – Dave L. Renfro Mar 09 '16 at 21:42
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1Heck, he could have written $O(x^6)$. – zahbaz Mar 09 '16 at 21:43
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1She could have written $O(x^7)$, but she would have been wrong. – marty cohen Mar 09 '16 at 21:58
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Look up Taylor/Maclaurin series:
The idea is that we can represent many "nice" functions in terms of a power series.
That is a series of the form $$\sum_{n=0}^{\infty} a_n(x-x_0)^n$$
It turns out that the series representation for $\cos (x)$ about $x=0$ is
$1-\frac{x^2}{2!}+\frac{x^4}{4!}+\text{higher terms}$.
Chris
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Convergent Taylor series and finite Taylor polynomials are related but distinct concepts and shouldn't be muddled like this. See http://math.stackexchange.com/questions/1308992/why-doesnt-a-taylor-series-converge-always – Ian Mar 09 '16 at 21:43
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i'll post this as an answer because i can't comment yet.
The expression given by The Professor is a taylor expansion. A particular class indeed, called Maclaurin expansion. In this link
you'll see a video explaining the thing.
