Looking at the series expansion of $\tan x$, it appears as the monotonically increasing function which we know isn't true. So, how can we describe it's periodicity from it's Maclaurin series?
Why $\tan x$ is not infinity at all the points > $\frac{\pi}{2}$
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functional-analysis
trigonometry
taylor-expansion
periodic-functions
trigonometric-series
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3The McLauin series of the tangent has a restricted interval of convergence. So it doesn't really make sense to talk about periodicity here...See this link:https://proofwiki.org/wiki/Taylor_Series_Expansion_for_Tangent_Function. – 2017-02-03
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0For a simpler example of the same phenomenon, take the Maclaurin series for $1/(1-x)$. Looks monotonically increasing too, but for $x>1$ the function is negative. It's not safe to draw conclusions about a function by looking at its Taylor series outside its interval of convergence. – 2017-02-03
2 Answers
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The Taylor series of a smooth function at a point need not converge everywhere; even if it does, it may not equal the function everywhere! In this case, the Taylor series at $0$ converges and agrees with tangent on the connected component of the graph containing the origin; but this isn't the whole graph! So the rest of the tangent function is invisible to the Taylor series at zero.
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0Thanks for the link. Never knew there existed such a thing. – 2017-02-03
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I believe that it's because the function is discontinuous (and undifferentiable) at $x= \pm \frac{ \pi}{2}$. Since Taylor series rely only on continuity and looks at the derivatives at a single point, in theory it should rapidly diverge to infinity after $x= \pm \frac{ \pi}{2}$ (not sure if it actually does).
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1MIght want to change to $\pm \frac{\pi}{2}$ – 2017-02-03