Find all functions f:R→R that satisfies xf(y)−yf(x)=f(yx)for all x,y∈R.
Problem
Source: 2010 Indonesia TST stage 2 test 1 p1
Tags: algebra, functional, functional equation
16.12.2020 10:01
2010 Indonesia TST stage 2 test 1 p1 wrote: Find all functions f:R→R that satisfies xf(y)−yf(x)=f(yx)for all x,y≠0∈R. P(1x,y)⟹f(y)x−yf(1x)=f(xy) P(1,y)⟹f(y)−yf(1)=f(y)⟹f(1)=0 P(x,1)⟹xf(1)−f(x)=f(1x)⟹f(x)+f(1x)=0 f(y)x+yf(x)=f(xy) P(x,1y)⟹xf(1y)−f(x)y=f(1xy) −xf(y)−f(x)y=−f(xy)⟹xf(y)+f(x)y=f(xy) f(y)x+yf(x)=f(xy)=xf(y)+f(x)y⟹f(y)x2−1x=f(x)y2−1y⟹f(x)=x2−1xc, where c is just a constant by fixing y. Plugging this back into our original equation, we have xy2−1yc−yx2−1xc=(yx)2−1yxcxy2−1y−yx2−1x=((yx)2−1)xyxy2−1y−yx2−1x=y2x−xyx2(y2−1)−y2(x2−1)=y2−x2This is true, note also that f(1)=12−11=0, therefore f(x)=x2−1xc∀x≠0∈R. Answer. f(x)=x2−1xc∀x≠0∈R, where c is just an arbitrary constant.
16.12.2020 12:34
parmenides51 wrote: Find all functions f:R→R that satisfies xf(y)−yf(x)=f(yx)for all x,y∈R. Let P(x,y) be the assertion xf(y)−yf(x)=f(yx) Let a=2f(2)3 and b=f(0) P(1,1) ⟹ f(1)=0 and so P(x,1) ⟹ f(1x)=−f(x) P(1x,2) ⟹ 3a2x+2f(x)=f(2x) P(12,x) ⟹ 12f(x)+3a2x=f(2x) Subtracting, we get f(0)=b and f(x)=a(x−1x)∀x≠0, which indeed fits, whatever are a,b∈R
17.12.2020 01:28
Nice FE The only solution is of the form f(y)=y2−1yc, where c is an arbitrary real number. Let P(x,y) be the given assertion. From P(1,y) we get that f(1)=0, from this fact we can easily get from P(x,1) that −f(x)=f(1x). This implies that f(x)+f(1x)=0, plugging x=−1 into what we got we get that f(−1)=0. From P(x,−1) we get that f(x)=f(−1x)=−f(1x). This implies that f is an odd function. Adding P(x,y) and P(1x,y) we have that: (x+1x)f(y)=f(yx)+f(xy)Since we have that P(x,y)=P(1x,1y), we get that: y2−1yf(x)=x2−1xf(y)now set x=2 now we have that: f(y)=y2−1y2f(2)3This implies that f(y)=y2−1yc, plugging this in we get that this is the solution.
18.12.2020 09:39