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 Functional feast (Posted on 2020-10-17)
Let f(x) is an odd function on R, f(1)=1 and f(x/(x-1))=xf(x) for all x<0. Find the value of

f(1)f(1/100)+f(1/2)f(1/99)+f(1/3)f(1/98)+...+f(1/50)f(1/51).

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 4.0000 (1 votes)

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 A start | Comment 1 of 3
f(1/(n+1)) = f((-1/n)/(-(n+1)/n)) = f((-1/n)/(-1/n - 1)) just algebra
= -(1/n)f(-1/n)                 see rule above
= (1/n)f(1/n)                    because an odd function

Then,

f(1) = 1
f(1/2) = 1/1 * f(1)    = 1
f(1/3) = 1/2 * f(1/2) = 1/2
f(1/4) = 1/3* f(1/3)  = 1/6
f(1/5) = 1/4 * f(1/4) = 1/24

In general, if n is a positive integer
f(1/n) = 1/(n-1)!                                    provable by induction

I do not see an obvious way to do the requested sum, so this is as far as I go

 Posted by Steve Herman on 2020-10-17 17:04:55

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