I need to solve this questions for assignment in HEDVA(Calculus in Hebrew)
??
some answers, here:
http://www.mathhelpforum.com/math-he...ease-help.html
If, the two functions are infinitely differenciable on some open interval, then
$\displaystyle (fg)'=f'g+fg'$
$\displaystyle (fg)''=(f'g+fg')'=(f''g+f'g+f'g'+fg'')=f''g+2f'g'+ fg''$
$\displaystyle (fg)'''=(f''g+2f'g'+fg'')'=f'''g+f''g'+2f''g'+2f'g ''+f'g''+fg'''$=$\displaystyle f'''g+3f''g'+3f'g''+fg'''$
I hope you recognize the pattern.
It follows binomial coefficients.
(You can support this with induction).
Thus,
$\displaystyle (fg)^{(n)}=\sum_{k=0}^n {n \choose k}f^{(k)}g^{(n-k)}$
Thus, you can think of,
$\displaystyle \sin x\sin 2x\sin 3x=(\sin x\sin 2x)\sin 3x$
And apply the theorem (twice because you still need to do the other product).
Thus,
$\displaystyle \sum_{k=0}^n {n\choose k}(\sin x\sin 2x)^{(k)}(\sin 3x)^{(n-k)}$
Apply theorem again,
$\displaystyle \sum_{k=0}^n {n\choose k}\cdot \sum_{i=0}^k {k\choose i} (\sin x)^{(i)}(\sin 2x)^{(k-i)} \cdot (\sin 3x)^{(n-k)}$