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$$\sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n}}{(2n)!}) * \sum_{n=0}^{\infty} C_{n}x^{n} = \sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n+1}}{(2n+1)!})$$

How to find coefficients until the $x^{7} $ ? It is basically sin and written in form of sums but i'm not sure how should i use this form to calculate coefficients from this factor sum. How should i precede?

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  • $\begingroup$ It might be a good idea to have the second summation index be something other than $n$. It is clear what you mean, but... $\endgroup$ – Mark Fischler Nov 21 '16 at 23:50
  • $\begingroup$ The Cauchy product of two series should help. $\endgroup$ – hamam_Abdallah Nov 22 '16 at 0:13
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Take the equation $$ (1-x^2/2+x^4/24-x^6/720+\dots)(C_0+C_1x+\dots+C_7x^7+\dots)=x-x^3/6+x^5/120-x^7/5040+\dots $$ and expand out the left side. When you collect the powers of $x$, you get $$ C_0+C_1x+(C_2-C_0/2)x^2+(C_3-C_1/2)x^3+\dots=x-x^3/6+x^5/120-x^7/5040+\dots $$ On the left side, I only went up to $x^3$, you will need to go up to $x^7$.

Now, since these two power series are equal, their coefficients must be equal, so you get $C_0=0$, $C_1=1$, $C_2-C_0/24=0$, $C_3-C_1/2=-1/6$, etc. Solve.

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Observe that

$$\sum_{n=0}^{+\infty}C_nx^n=\tan(x)$$

and $C_0=C_2=C_4=C_6=0$.

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