Given
(1+2x)1/2(1+4x)1/4(1+6x)1/6...(1+98x)1/98(1+100x)1/100
y = -----------------------------------------------------
(1+3x)1/3(1+5x)1/5(1+7x)1/7...(1+99x)1/99(1+101x)1/101
Find y' at x = 0.
Each of the 50 terms in the numerator and each of the 50 in the denominator, if approximated by their Taylor expansion, would be (1+x).
So this would simplify to (1+x)^50 / (1+x)^50 = 1
And the derivative of constant 1 is zero.
This should be particularly accurate since the Taylor expansion is around x=0 and we are looking for y' at x=0.
Not rigorous, but I would be very surprised if it is not the answer.
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Posted by Larry
on 2025-01-09 23:14:43 |
Hand waving non-proof
