If x=0 is allowed, you can determine f(y)=0 immediately, as @rickostidich said. The problem would be much more interesting if x=0 is not part of the domain. e.g. "f(x+y) = xf(y) for positive real numbers x and y." Or "f(x+y) = xf(y) for all natural numbers x and y." Your work in the video shows that the only solution is still the zero function f(x)=0.
Note that, because addition is commutative, f(x + y) = yf(x) = xf(y). Putting in y =1, we get f(x) = xf(1). Using this on our original equation, we get f(1)(x + y) = xf(1)y. If f(1) = 0, then we're done. Otherwise, we have x + y = xy, which is apparently true for all x and y. This is clearly absurd. So, f(1) = 0 is the only possible situation. So, f(x) = 0.
To be clear, this is assuming that we restrict it to "positive real numbers" or something like that, so that putting in x = 0 doesn't immediately solve everything.
If we set x=0, we immediately get f(y) =o. Looking at it another way, the partial derivative of f(x+y) w.r.t x is the same as that w.r.t y. Therefore, partial_xf(x+y) = f(y) = partial_y f(x+y) = x df(y)/dy > 1/f(y) df(y)/dy = i/x, which is not possible, as the LHS is a function of y and the RHS is a function of x alone. Therefore f(x) =o.
Great video!
Set x=0, then f(y)=0, done. This is one of the very few Sybermath videos which I didn't like. 😜
😮😜😁
If x=0 is allowed, you can determine f(y)=0 immediately, as @rickostidich said.
The problem would be much more interesting if x=0 is not part of the domain. e.g. "f(x+y) = xf(y) for positive real numbers x and y." Or "f(x+y) = xf(y) for all natural numbers x and y." Your work in the video shows that the only solution is still the zero function f(x)=0.
Note that, because addition is commutative, f(x + y) = yf(x) = xf(y). Putting in y =1, we get f(x) = xf(1). Using this on our original equation, we get f(1)(x + y) = xf(1)y. If f(1) = 0, then we're done. Otherwise, we have x + y = xy, which is apparently true for all x and y. This is clearly absurd. So, f(1) = 0 is the only possible situation. So, f(x) = 0.
To be clear, this is assuming that we restrict it to "positive real numbers" or something like that, so that putting in x = 0 doesn't immediately solve everything.
Put x=y=0
f(0)=0
Then put y=0
f(x)=xf(0)=0
Hence f(x)=0
Substitute x=0 and leave y alone
F(y)=0
If we set x=0, we immediately get f(y) =o. Looking at it another way, the partial derivative of f(x+y) w.r.t x is the same as that w.r.t y. Therefore, partial_xf(x+y) = f(y) = partial_y f(x+y) = x df(y)/dy > 1/f(y) df(y)/dy = i/x, which is not possible, as the LHS is a function of y and the RHS is a function of x alone. Therefore f(x) =o.