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can someone help me simplify the following formula? $$y = \pi(x-a)-\frac{(\pi(x-a))^3}{3!}+\frac{(\pi(x-a))^5}{5!}-\frac{(\pi(x-a))^7}{7!}+\frac{(\pi(x-a))^9}{9!}-\frac{(\pi(x-a))^{11}}{11!}+\frac{(\pi(x-a))^{13}}{13!}-\frac{(\pi(x-a))^{15}}{15!}+\frac{(\pi(x-a))^{17}}{17!}-\frac{(\pi(x-a))^{19}}{19!}+\frac{(\pi(x-a))^{21}}{21!}-\frac{(\pi(x-a))^{23}}{23!}$$

d_air
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    This is not a geometric series. Hint: consider the Taylor expansion of $\sin z$. – lulu May 15 '19 at 00:36
  • ok. I'm sorry. I was not sure about that. Yes I got this from sin formula but I'm not sure how to simplify this so it can easily be computed. – d_air May 15 '19 at 00:38
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    I doubt there's any general way to simplify it that works for all $x$. For $x$ near $a$ then this is very close to $\sin (\pi(x-a))$. – lulu May 15 '19 at 00:39
  • would that mean that If I have to plug in the values for x and a, i need to evaluate each of the 12 terms to get the value of y? – d_air May 15 '19 at 00:41
  • In general, sure. Taylor polynomials work well when the argument is near the point you are expanding around. They don't work terribly well far from the expansion point. Of course, $23!\approx 2.6\times 10^{22}$ so if $\pi(x-a)^{23}$ is small compared to that you can drop that term, and so on. – lulu May 15 '19 at 00:45
  • But why would you want to compute a Taylor polynomial for $\sin z$ for $z$ very far from $0$? What's the context for that? – lulu May 15 '19 at 00:46
  • I may not be able able to explain this fully. I'm filtering a set of x values that makes $y=0$. – d_air May 15 '19 at 00:57
  • How about if all $a$'s are removed from the equation, or if $a$ always equal to $0$, is the equation still can not be simplified? – d_air May 15 '19 at 01:33

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If you work with this limited expression, first let $t=\pi(x-a)$ to write $$y=\sum_{i=0}^{11}(-1)^n\frac {t^{2n+1}}{(2n+1)!}=t\sum_{i=0}^{11}(-1)^n\frac {t^{2n}}{(2n+1)!}$$ You have the obvious root $t=0$. For the other roots, let $t^2=u$ and then consider solving for $u$ the polynomial equation $$\sum_{i=0}^{11}(-1)^n\frac {u^{n}}{(2n+1)!}=0$$ which is not most pleasant task. There are $11$ roots but only three of them are real. Graphing, they look to be very close to $(10,40,85)$

In my opinion, the easiest way to get the accurate solutions is to use Newton method starting from these estimates. The iterates would be $$\left( \begin{array}{cc} n & u_n \\ 0 & 10.00000000 \\ 1 & 9.868296250 \\ 2 & 9.869604271 \\ 3 & 9.869604401 \end{array} \right)$$

$$\left( \begin{array}{cc} n & u_n \\ 0 & 40.00000000 \\ 1 & 39.47302741 \\ 2 & 39.47848608 \\ 3 & 39.47848665 \end{array} \right)$$ $$\left( \begin{array}{cc} n & u_n \\ 0 & 85.00000000 \\ 1 & 86.98238456 \\ 2 & 86.95292238 \\ 3 & 86.95291554 \end{array} \right)$$ Then the six non-trivial solutions for $t$ (they are quite close to $\pm \pi$, $\pm 2\pi$, $\pm 3\pi$).

Claude Leibovici
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  • I'm studying your answer. Thanks for the effort. – d_air May 15 '19 at 10:56
  • This did not completely answer my question because I am expecting for a shorter formula. As commented by lulu that there is no general formula for this, so I just have to accept that there is no perfect answer for my question. However your answer triggers me to study closely the Newton's method which helps me understand that method better and find out that its a great method to use to solve my equation. So I accepted your answer. – d_air May 18 '19 at 00:54