I was in the middle of answering this question that asked how to get the Taylor series of $cos(x)$ centered at $a=\frac{\pi}{3}$. I was reaching the point where I was about to write $(-1)^n$ making the it an alternating series, when I began to realize why calculus II courses only talk about the Maclaurin series of $cos(x)$. In the Maclaurin series when $a=0$, the function alternates signs every iteration using $(-1)^n$, and has the pattern of $\{-,+,-,+,-,+\}$. However, when the function is centered at $a=\frac{\pi}{3}$, the series has the pattern $\{+,-,-,+,+,-,-,+,+,...\}$
What form of $(-1)^{kn}$ can create a sequence with this kind of pattern?
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$\begingroup$ I guess this behaviour is due to $\sin$ and $\cos$ terms rather than to $(-1)^n$, isn'it? $\endgroup$– 7raiden7May 11, 2014 at 19:13
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2 Answers
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For sequences of period 4, one has to rely on the fourth roots of unity, that is, to linear combinations of the sequences of general term $1$, $\mathrm i^n$, $(-1)^n$ and $(-\mathrm i)^n$. In your case, starting at $n=1$, the sequence is $$ \frac{1-\mathrm i}2\cdot\mathrm i^n+\frac{1+\mathrm i}2\cdot(-\mathrm i)^n. $$ Another way to express the same sequence is to use cosine and sine functions with period 4, in your case, still starting at $n=1$, one would consider $$ \sqrt2\cdot\cos\left(n\frac\pi2-\frac\pi4\right). $$
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$\begingroup$ If an expanded Taylor series is what we would use to estimate $cos(x)$, is it "legal" to use $cos(x)$ in its closed form? Wouldn't that be finding the value of $cos(x)$ by using a series that requires knowing the value of $cos(x)$? $\endgroup$– EphraimMay 11, 2014 at 19:27
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$\begingroup$ There is no Taylor series involved here, whether of the cosine function or otherwise. The second formulation in my post only uses the values of $\cos(\pi/4)$, $\cos(3\pi/4)$, $\cos(5\pi/4)$ and $\cos(7\pi/4)$. $\endgroup$– DidMay 11, 2014 at 19:33
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$\begingroup$ The original reason for my question was to figure out how to express the Taylor series of the function $cos(x)$ centered at $a=\frac{\pi}{3}$. The first part of your answer looks like it would work no matter what. What I was asking in my comment was whether or not $\sqrt2\cdot\cos\left(n\frac\pi2-\frac\pi4\right)$ can be used to write a taylor series for the function $cos(x)$ $\endgroup$– EphraimMay 11, 2014 at 19:39
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$$(-1)^{\lceil n/2\rceil}$$ where $\lceil x\rceil$ is the ceiling function, i.e. the smallest integer $\ge x$.
answered May 11, 2014 at 19:22