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Desperately Difficult Derivative (Posted on 2005-03-30) Difficulty: 4 of 5
If f(x)=sin(x), find the value of its 2001st derivative, at x=0.

See The Solution Submitted by Old Original Oskar!    
Rating: 3.8000 (5 votes)

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Solution re: Simple solution (spoiler alert!) | Comment 4 of 9 |
(In reply to Simple solution (spoiler alert!) by ajosin)

I've rewritten ajosin's solution to be more easily understood, as it took me quite a while to understand that.

Since sin z = z - z^3/3! + z^5/5! + ... + (-1)^(n+1) z^(2n-1)/(2n-1)!, in particular

sin x^3 = x^3 - (x^3)^3/3! + (x^3)^5/5! + ... + (-1)^(n+1) (x^3)^(2n-1)/(2n-1)!

where the generic term is

(-1)^(n+1) * (x^3)^(2n-1) / (2*n - 1)


(-1)^(n+1) * (x^(6*n-3)) / (2*n - 1)

Each time the derivative is taken, the exponent of x decreases by one. Eventually it gets to be 1, and then zero, after which it remains zero, and the derivative of the particular term is then zero also.  In those cases of non-zero exponent, the evaluation of the term at x=0 will result in the term being zero.  So we are concerned only with the term where the exponent of x has just turned zero.

Since the exponent of x is 6*n-3-d, where d is the level of derivative taken, we want 6*n-3 = d, where d in this instance is 2001. This solves to n = 334, so the term whose 2001st derivative concerns us is (-1)^(334+1) x^2001/667!. That 2001st derivative is -2001!/667!.

BTW, that amounts to about  -9.84416159176 * 10^4142.

  Posted by Charlie on 2005-03-31 15:09:28
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