Nice solution. Just 2 little remarks: 1. It could be a little bit more efficient (at 1:46) to substitute: x = y + c. 2. Why the restriction to positive real numbers and not to R\{0} or even to C\{0} ?
With x = y + f(x) we get f(x) = x / (1 + x (x - f(x))) x f²(x) - (x² + 1) f(x) - x = 0 solving this quadratic equation for f(x): f(x) = [x² + 1 ± (x² - 1) ] / 2x Two solutions f(x) = x or f(x) = 1 / x but only the second one solves the gives function equation.
y=0 in the first equation gives f(f(x))=x. So twice the function of x gives back x, which gave me the idea of f(x)=1/x. When applied in the first equation it gave the right answer.
If you set x = 0, the RHS is zero for all y, meaning the only solution is the trivial f(x) = 0. So you need to exclude zero to get an interesting problem.
You don't know the value of c during your shift so f(y) is only determined for y bigger than c. Edit: for clarity's sake, you can only substitute y=1 iff c
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Nice solution.
Just 2 little remarks:
1. It could be a little bit more efficient (at 1:46) to substitute: x = y + c.
2. Why the restriction to positive real numbers and not to R\{0} or even to C\{0} ?
With x = y + f(x) we get f(x) = x / (1 + x (x - f(x)))
x f²(x) - (x² + 1) f(x) - x = 0 solving this quadratic equation for f(x):
f(x) = [x² + 1 ± (x² - 1) ] / 2x
Two solutions f(x) = x or f(x) = 1 / x but only the second one solves the gives function equation.
y=0 in the first equation gives f(f(x))=x. So twice the function of x gives back x, which gave me the idea of f(x)=1/x. When applied in the first equation it gave the right answer.
Great👍
this could be done with undetermined constants by guessing some form
for example f(x) = c1 + c2x + c3/x +... and finding the missing numbers
I find these types of problems a little difficult and they seem to involve a lot of trial and error without any fixed method on how to solve.
Great, should i sub your channel?
Absolutely! and watch all the videos 😍😄
Why x = 0 is not allowed? Doesnt seem obvious to me lol
Defined that way.
If you set x = 0, the RHS is zero for all y, meaning the only solution is the trivial f(x) = 0. So you need to exclude zero to get an interesting problem.
@@quantumcaffeine And this solution does not satisfy the original functional equation for
x, y ≠ 0
Because set of +integers doesn't not include 0. Zero is neither + or -
Division by zero
You don't know the value of c during your shift so f(y) is only determined for y bigger than c.
Edit: for clarity's sake, you can only substitute y=1 iff c
why do you claim f(1) is constant
Because 1 is constant.
Noice