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Math Help - some problems ??

  1. #1
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    some problems ??

    I need to solve this questions for assignment in HEDVA(Calculus in Hebrew)
    ??
    Attached Thumbnails Attached Thumbnails some problems ??-untitled-2.gif  
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  2. #2
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  3. #3
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    If, the two functions are infinitely differenciable on some open interval, then
    (fg)'=f'g+fg'
    (fg)''=(f'g+fg')'=(f''g+f'g+f'g'+fg'')=f''g+2f'g'+  fg''
    (fg)'''=(f''g+2f'g'+fg'')'=f'''g+f''g'+2f''g'+2f'g  ''+f'g''+fg'''= f'''g+3f''g'+3f'g''+fg'''
    I hope you recognize the pattern.
    It follows binomial coefficients.
    (You can support this with induction).
    Thus,
    (fg)^{(n)}=\sum_{k=0}^n {n \choose k}f^{(k)}g^{(n-k)}
    Thus, you can think of,
    \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,
    \sum_{k=0}^n {n\choose k}(\sin x\sin 2x)^{(k)}(\sin 3x)^{(n-k)}
    Apply theorem again,
    \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)}
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