The Phoenix Rises
Calculus
Level
pending
Let f(x) = (sin(x^3))/x. How many "trailing zeros" appear in the 602-nd derivative of f at x = 0. Here, the number of "trailing zeros" is the number of times ten appears as a factor (e.g., 10000 has four "trailing zeros").
The answer is 99.
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1 solution
The Taylor series expansion of sin u is given by u - (u^3)/(3!) + (u^5)/5! - ...
By substituting u = x^3 into the series expansion above and dividing each term by x, we find that the k-th term of the Taylor series for f at x = 0 is given by:
((-1)^k)*(x^(6k+2))/(2k+1)!
But, we also know that the coefficient of the m-th term of the Taylor series expansion at x = 0 is the m-th derivative at 0 divided by m!. So, if A is the 602-nd derivative of f at x = 0, we find (by taking k=100 in the formula above):
A / 602! = ((-1)^100)/201!
A = 602! / 201!
The number of trailing zeros of 602! is given by:
int(602/5) + int(602/25) + int(602/125) = 120+24+4 = 148.
Similarly, there are 49 trailing zeros in 201!. Therefore, the number of trailing zeros in the quotient is 148 - 49 = 99.