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I have noticed that, in the theory of divergent series and divergent integrals, [tex] \sin ( \infty ) = 0[/tex].
For example
[itex]\sum_{n=0}^\infty \ (-1)^n = 1 - 1 + 1 - 1 ... = \frac{1}{1-(-1)} = \frac{1}{2}[/itex]
but
[itex]\sum_{n=0}^k \ (-1)^n = \frac{ - (-1)^{n} - 1}{1-(-1)}[/tex]
and taking the limit as [tex]k \rightarrow \infty[/tex] (use Euler's identity to put the negative power in trigonometric form)
[tex]- \lim_{\substack{k \rightarrow \infty}} \frac{(\cos( \pi k ) + i * \sin ( \pi k )) - 1}{2} = 1/2[/tex]
and hence that [itex]\sin ( \infty ) = \cos ( \infty ) = 0[/itex] (Equate real and imaginary parts).
What I would like to know is why this theory (that sine is zero at infinity) is not generally accepted even though it allows for such things as the solution to
[tex] \lim_{\substack{x \rightarrow \infty}} \frac{\sin ( x ) + x}{x}[/tex]
by use of l'Hospital's rule. Is there a good reason or is it simply too good of an example of a function that diverges by oscillation?
Please forgive any mistakes as this is the first time I've used LaTEX.
For example
[itex]\sum_{n=0}^\infty \ (-1)^n = 1 - 1 + 1 - 1 ... = \frac{1}{1-(-1)} = \frac{1}{2}[/itex]
but
[itex]\sum_{n=0}^k \ (-1)^n = \frac{ - (-1)^{n} - 1}{1-(-1)}[/tex]
and taking the limit as [tex]k \rightarrow \infty[/tex] (use Euler's identity to put the negative power in trigonometric form)
[tex]- \lim_{\substack{k \rightarrow \infty}} \frac{(\cos( \pi k ) + i * \sin ( \pi k )) - 1}{2} = 1/2[/tex]
and hence that [itex]\sin ( \infty ) = \cos ( \infty ) = 0[/itex] (Equate real and imaginary parts).
What I would like to know is why this theory (that sine is zero at infinity) is not generally accepted even though it allows for such things as the solution to
[tex] \lim_{\substack{x \rightarrow \infty}} \frac{\sin ( x ) + x}{x}[/tex]
by use of l'Hospital's rule. Is there a good reason or is it simply too good of an example of a function that diverges by oscillation?
Please forgive any mistakes as this is the first time I've used LaTEX.
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