A "spiral" in the Complex plane

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The discussion centers on the convergence of a spiral in the complex plane, specifically to the value 1+i-1/2-i/3!+1/4!+i/5!-1/6!-i/7!.... It is noted that this series can be separated into real and imaginary components, with the real part represented by the series 1-1/2+1/4!-1/6!... and the imaginary part by 1-1/3!+1/5!-1/7!.... Participants seek methods to compute these series and inquire about the values of specific sums involving factorials at x=1, which relate to the cosine and sine functions. The conversation highlights the connection between these series and the exponential function expressed as a power series. Understanding these relationships is crucial for further exploration of complex analysis.
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Homework Statement
Starting from the origin, go one unit east, then the same length north, then (1/2) of the previous length west, then (1/3) of the previous length south, then (1/4) of the previous length east, and so on. What point does this “spiral” converge to?
Relevant Equations
series sum
I understand that the "spiral" converges to 1+i-1/2-i/3!+1/4!+i/5!-1/6!-i/7!... .
It splits into two: one for Re, 1-1/2+1/4!-1/6!..., and the other for Im, 1-1/3!+1/5!-1/7!... .
Any hints on how to compute them?
 
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What are <br /> \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!} and \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} when x = 1?
 
pasmith said:
What are <br /> \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!} and \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} when x = 1?
##\cos## and ##\sin##, of course. Thanks!
 
Or, even better:
$$
\sum_{n=0}^\infty \frac{(ix)^n}{n!}
$$
 
Orodruin said:
Or, even better:
$$
\sum_{n=0}^\infty \frac{(ix)^n}{n!}
$$
Yes. Straight.
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...