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I had to find the Taylor series for the function $f(x)=\cos(x)$ centred at $a=\frac{\pi}{4}$.

I found the pattern but the only part I'm missing is the sign. Since the series is centred at $\frac{\pi}{4}$, no value of $f^{\{n\}}(a)$ is equal to zero, and the pattern is $+, -, -, +, +, -, -, +$.

I have checked the answer that my teacher put in the document, but he just wrote

$$f(x) = \sum_{n=0}^{\infty}{\text{sign} \frac{\sqrt{2}}{2(n!)}}\left(x-\frac{\pi}{4}\right)^n$$ here $\text{sign} =+--++--++--+ \cdots$

which I find rather disappointing...

Is there a mathematical way to insert the sign pattern into the series, similarly to $(-1)^n$ for a normal alternating series?

Blue
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Thibaut B.
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  • you could split up the sum into 2 alternating series – Jens Renders May 13 '20 at 13:44
  • Sometimes readability is to be preferred imho – Manlio May 13 '20 at 13:56
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    Notice that that's the sign pattern for cosine as we cycle counterclockwise through the quadrants. This suggests considering the $\pi/4$ angles in each quadrant: $$\sqrt{2}\cos\left(\frac\pi4+n\frac\pi2\right)=\sqrt{2}\cos\left((2n+1)\frac\pi4\right)$$ where the multiplied $\sqrt{2}$ makes the terms unit-sized ... and, conveniently, helps simplify the rest of the expression in the summation. – Blue May 13 '20 at 14:08

2 Answers2

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What about $$\frac{\cos(\pi/4+n\pi/2)}{\cos(\pi/4)}$$

b00n heT
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  • You start with something simple: $+ - - + + - -+\cdots$, and you turn it into something that makes your readers scratch their heads and reach for pencil and paper... – TonyK May 13 '20 at 14:10
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    Agreed. Although my answer wants to be geometric and is well represented by Blue's comment: in the sense that the looked-for sign sequence is $+--+$, which is exactly the sequence in which cosine changes sign as we let the angle increase. – b00n heT May 13 '20 at 14:49
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One simple method that clearly shows your intentions is: $$(-1)^{\left\lfloor\frac{n+1}{2}\right\rfloor}$$ Mathematically cleaner, but not so transparent: $$(-1)^{\frac12n(n+1)}$$ Having said that, your teacher's method is easier to read than either of these solutions.

TonyK
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