 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sum Functions Determination (Posted on 2011-02-19) Determine all possible real valued functions f, with x being a real number belonging to (0,1), such that:

ĀĀĀĀĀĀĀĀĀĀf(x) + f(1/(1-x)) = 2(1 - 2x)/(x(1 -x))

Note: The interval related symbols are in consonance with this article.

ĀĀĀĀĀĀĀĀĀĀ

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) An approach (spoiler) Comment 1 of 1
Let g(x) = 2(1 - 2x)/(x(1 -x))

Use three transformations:

a) x = y

then f(y) + f(1/(1-y)) = g(y)

b) x = (1/(1-y))

then f(1/(1-y)) + f((y-1)/y) = g(1/(1-y))

c) x= ((y-1)/y)

then f((y-1)/y) + f(y) = g((y-1)/y)

Add a and c, and subtract b, to get
2f(y) = g(y) + g((y-1)/y) - g(1/(1-y))

Thus, there is one and only one function that  works:
f(x) =  (g(x) + g((x-1)/x) - g(1/(1-x)))/2
where g(x) =  2(1 - 2x)/(x(1 -x))

I am too lazy to simplify the above, but I look forward to the answer.

Actually, I am not too lazy to remove the 2.

f(x) = h(x) + h((x-1)/x) - h(1/(1-x))
where h(x) =  (1 - 2x)/(x(1 -x))

 Posted by Steve Herman on 2011-02-19 14:13:15 Please log in:
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