A Quick and Easy Functional Equation from Slovenia

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[Legoro_]

Hey Syber!
I would like to thank you for everything, all the videos and every method of solving problems. Yesterday I took one of the hardest tests here in Brazil and there was a question that you needed to factor a cubic polynomial and match the coefficients with other expression, I didn't study that at my school, but I was able to do it thanks to you, and not only that, every shortcut, mental calculations, fast thinking and creative ideas have to do with your videos. Thank you!

[DrBarker]

How about this?
Set x = y + f(y), to get
f(y) = 1 - f(y) - 2y
So rearranging, f(y) = 1/2 - y, as required.

[michaelz]

If you replace x by x + f(y) your functional equation becomes f(x) = 1 - x - (y + f(y)). So y + f(y) is independent of y, giving f(y) = c - y for some c. Plugging this into the original equation gives f(x + y - c) = 1 - x - y. Using f(x + y - c) = c- (x + y - c) = 2c -x - y now implies 2c = 1, leading to c = 1/2. So f(x) = 1/2 - x.

[f-th]

You can set x = f(y), giving you f(0) = 1 - f(y) - y, which implies f(y) = 1-y-f(0).
Now all you have to do is use this to unpack the original equation, and solving for f(0), which will be 1/2.

[kirbynrv]

Hello,
Wasn't it simpler when you get to f(x)=1-x-c to plug in x=c. Then, f(c)=0=1-c-c, so you directly get to c=1/2 without manipulating too much your equation and finally, f(x)=1-x-1/2 <=> f(x)=1/2-x.
Thank you nonetheless, I really like watching you resolve these problems (some I succeed too, and some I don't).

[liyuan-chuanli]

Step 1. Take x = f(y). We get f(0) = 1 - f(y) - y.
Stept 2. Take y = 0. We get f(0) = 1/2.
Therefore f(y) = 1/2 - y.

[adandap]

This is a variation on an answer others have posted, but set x+y = 1 and we get f(x - f(1-x) ) = 0. Since f(x)=/= 0, that means that the argument x - f(1-x) = C, or if we do x --> 1-x, then f(x) = k - x (k = 1-C). To find k, we use f(x - f(y)) = f(x - (k-y) ) = f(x+y-k) = k - (x+y-k) = 2k - x - y, so k=1/2.

[the_dpad]

I had a somewhat different approach for my solution:

f(x - f(y)) = 1 - x - y
when f(y) = x, f(0) = 1 - f(y) - y
f(y) = 1 - y - f(0)
when y = 0, f(0) = 1 - f(0)
f(0) = 1/2
f(y) = 1 - y - 1/2
f(y) = 1/2 - y

testing against the original formula:
f(x - f(y))
= f(x - (1/2 - y))
= f(x + y - 1/2)
= 1/2 - (x + y - 1/2)
= 1 - x - y

[chaosredefined]

Falls apart if we show that the function is injective.

Suppose that there is some values a and b such that f(a) = f(b). Then, consider x = a + f(0) and y = 0. That gives us f(a + f(0) - f(0)) = 1 - a - 0, or, more simply, f(a) = 1 - a. Similarly, we can get f(b) = 1 - b. So, since f(a) = f(b), we have 1 - a = 1 - b, which gives us a = b. So, if f(a) = f(b), then a = b. (aka, the function is injective)

Consider what happens when we swap the roles of x and y in the equation. This gives us f(y - f(x)) = 1 - y - x. But that's the same thing as 1 - x - y. So, f(x - f(y)) = f(y - f(x)). But, by the first thing, we can get rid of the f's, giving us x - f(y) = y - f(x). Rearranging, we get f(x) - f(y) = y - x. Putting in y = 0, we get f(x) = f(0) - x

So, f(x - f(y)) = f(x - (f(0) - y)) = f(0) - (x - (f(0) - y)) = 2f(0) - x - y. And this is equal to 1 - x - y. So, 2f(0) = 1, or f(0) = 1/2.

Therefore, the solution is f(x) = 1/2 - x

[trnfncb]

If you set first x=0 and then y=0 and compare, (assuming the function has inverse), you easily get f(x)=f(0)-x. Substituting back into the original equation, you find f(0)=1/2. Thank you for all these nice problems.

[alfreds]

If P(x;y) is f(x-f(y))=1-x-y then P(x+f(0);0) is f(x)=-x+1-f(0), 1-f(0) is a constant so we can say 1-f(0)=c that means f(x)=-x+c. Then just go to start equality and we receive that -x-y+2c=1-x-y or 2c = 1 or c=1/2 => f(x)=-x+1/2

[maxm]

Another method:
i calculated the derivative (dx) of f(x+f(y)) treating y and f(y) as a constant. d(1-x-y)/dx gives a constant -1. Thus, we know that f(x) is linear.
i.e. f(x)=mx+b
substituting this back to the original equation of f(x-f(y)) = 1-x-y, you get mx - m2y + b(1-m)= -x -y +1

establishing equality among coefficients, it's easy to see that m=-1. Then, b(1-m) = 2b = 1 -> b=0.5
Thus f(x)=-x+0.5

[amineoueslati]

We set x=f(y) we get f(0)=1-f(y)-y
since f(0) is a constant then 1-f(y)-y is a constant then f(y)+y=c f(y)=c-y
We set y=0 we get f(0)=c. Since we know that f(0)=1-f(y)-y=1-c then 1-c=c ---> c=1/2
Then f(y) =1/2 - y

[insidious_topo]

Let f(x) = mx+b.
(note: f(y) = my+b)
Then, f(x-f(y))=f(x-my-b)
= m(x-my-b) + b
= mx-(m^2)y-mb+b.
Given f(x-f(y))=1-x-y,
mx-(m^2)y-mb+b = 1-x-y
This imples that m = +/- 1 (equate coefficients in terms on x, y & b)
but we know that m = -1, as there will be no solutions for b if m were indeed 1.
Hence, -(-1)b+b = 2b = 1
=> b = 1/2.
Conclusion: f(x) = -x + 1/2

[Suppawat-dess]

f(x)=1-x-c
x->c : f(c)=1-2c
2c=1 c=1/2

[richard]

Hi bro can you help me solve this problem: How many integers y there is for x€ (1/3;3) satisfied 27^3x²+xy=(1+xy).27^9x, thank you so much i hope you'll have a nice day

[forgetittube]

Nice one!

Though, I have a simple question :: which program are you using as a “canvas/blackboard” ?

[JohnRandomness]

The function f needs a zero. There is a line, x + y = 1, where f(x - f(y)) = 0. Let a be a zero of f: f(a) = 0. Then f(x) = f(x - f(a)) = 1 - x - a. f(a) = 1 - 2a = 0, so a = 1/2. f(x) = 1 - x - 1/2 = 1/2 - x.

Check: f(x - f(y)) = 1/2 - (x - f(y)) = 1/2 - x + 1/2 - y = 1 - x - y.

[angelmendez-rivera]

Let x = y + f(y). Hence f[y + f(y) – f(y)] = 1 – [y + f(y)] – y = f(y) = 1 – 2·y – f(y). This is equivalent to 2·f(y) = 1 – 2·y. Therefore, f(y) = 1/2 – y.

[dumplet]

f is surjective by fixing x and varying y

Let k such that f(k) = 0

y = k -> f(x) = -x + 1 - k, f is of the form -x + c for some real c.

Substituting f(x) = -x + c,

f(x + y - c) = 1 - x - y

-x - y + c + c = 1 - x - y

c = 1/2

Thus, f(x) = -x + 1/2