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I am trying to rewrite sine series in a way to get rid of the factorial.

$$\begin{array}{c c c c c} sin(x) & x & -\ {x^3 \over 3!} & +\ {x^5 \over 5!} & -\ {x^7 \over 7!}\\ & T_0 & T_1 & T_2 & T_3\\ \hline \end{array}$$ If we number the terms from $0$ onward. I chose $3rd$ term over $2nd$ term to further simplify.

$$ \begin{align} {T_3 \over T_2} & = {-{x^7 \over 7!} \over {x^5 \over 5!}} \\ {T_3 \over T_2} &= -{x^7 \over 7!}\ \cdot {5! \over x^5 } \\ T_3 & = -T_2 \ . \ {x^2 \over 7 \ \cdot 6} \end{align} $$

But I wanna generalize it for $i$ so I eventually end up with something like this:

$$ ^{(*)} \ \ T_i = -\ T_{i-1} \cdot {x^2 \over (2i+1) \ 2i} $$ so that I can write it down programmatically in a recurrent algorithm: $$ T = -T * (x*x)/(2i+1)/2i $$ (-- from now on, it's magic that I'm not sure is correct --)

Problem is, when I get to the $$ T_i = - T_{i-1} \cdot {x^{2i+1} \over (2i+1)!}\ \cdot {(2i-1)! \over x^{2i-1} } \\ \text{for $i = 3$ from the $T_3$} $$ I don't know how to simplify from here. I know I need to end up with $^{(*)}$, but don't know how.

Ideas?


Solution from Jacob's hint:

$$ \begin{align} T_i & = -\ T_{i-1} \cdot {x^{(2i+1)-(2i-1)} \over (2i+1)!}\ \cdot {(2i-1)! \over 1 } \\ T_i & = -\ T_{i-1} \cdot {x^{2} \over (2i+1)((2i+1)-1)!}\ \cdot {(2i-1)! \over 1 } \\ T_i & = -\ T_{i-1} \cdot {x^{2} \over (2i+1)(2i)!}\ \cdot {(2i-1)! \over 1 } \\ T_i & = -\ T_{i-1} \cdot {x^{2} \over (2i+1)(2i)(2i-1)!}\ \cdot {(2i-1)! \over 1 } \\ T_i & = -\ T_{i-1} \cdot {x^{2} \over (2i+1)(2i)}\ \cdot {1 \over 1 } \end{align} $$

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Just a remark: typesetting $X.-Y$ without parentheses looks quite badly (and many people consider it wrong and undefined). You can try X\cdot\left(-Y\right) with the output: $$X\cdot\left(-Y\right)$$ –  tohecz Dec 2 '12 at 14:13

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up vote 5 down vote accepted

Hint: $(2i+1)!=(2i+1)(2i)(2i-1)!$ and $x^{2i+1}/x^{2i-1}=x^2$.

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Now I realized how easy it is! Going to update with your solution. –  DavidL Dec 2 '12 at 12:35

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