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When I see the alternating signs in the infinite series expansion of $\sin x$, I'm reminded of the inclusion-exclusion principle. Could there be any way to visualize it in such a way?

Also, is there an elementary reason why the Taylor series approximates a function? I've read the wikipedia entry but didn't understand it so it's greatly appreciated if someone can point me in the right direction.

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Interesting suggestion, and challenge to ingenuity. For example, the usual series expression for $e^{-1}$ does arise by inclusion/exclusion from the problem of finding the probability that a permutation is a derangement. –  André Nicolas Mar 26 '12 at 21:18
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On the Taylor Series point - standard texts on analysis will acquaint you with the remainder term. Part of the deal is that the remainder term goes to zero, justifying the approximation/series. For example, $f(x)=e^{-x^2}$ at x=0 is a classic series where the remainder does not tend to zero. –  Mark Bennet Mar 26 '12 at 21:27
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@Mark, are you thinking of $e^{-x^{-2}}$? –  Gerry Myerson Mar 26 '12 at 21:59
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A lot of infinite series have alternating signs, e.g., the series for $(1+x)^{-1}$. Many finite sequences come out alternating, too, as in Euler's vertices minus edges plus faces formula and its generalizations to higher dimensions. It may be too much to try to connect every alternating sequence of signs to inclusion-exclusion. –  Gerry Myerson Mar 26 '12 at 22:04
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@GerryMyerson - yes, thanks –  Mark Bennet Mar 27 '12 at 5:24

2 Answers 2

The Taylor series approximates a function because of its behavior at the origin point of the Taylor series, the point $x=a$ in the expansion $\sum b_j (x-a)^j$.

At every derivative of the Taylor series, first, second, third, etc. the value of the Taylor series at the origin point $x=a$,

is the same as the value of the derivative of the function being estimated at that point.

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This is not a complete answer, but alternating sums sometimes can be factored into products of form $\prod_{i=1}^k (1-a_i)$, which is 0 when any $a_i$ is 1.

Sine factored:

$\sin \pi x = \pi x \prod_{i=1}^{\infty} (1-x^2/n^2)$, which is $0$ when $x \in \mathbb Z$.

The inclusion-exclusion principle factored:

$\prod_{i=1}^{k} (1 - 1_{A_i}(x))$, which is $0$ when $x \in 1_{\bigcup A_i}$.

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