Solving f(x-f(y))=1-x-y

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[thatdude_]

take y=0 and x=f(0), then f(x-f(y))=f(0)=1-f(0), hence f(0)=1/2. Now again let y=0 and x arbitrary, then f(x-f(y))=f(x-1/2)=1-x. Setting x=t+1/2 we find f(t)=1/2-t.

[GirishManjunathMusic]

Given:
f(x – f(y) = 1 – x – y
To findv
f(x)

Substituting y = 0, x is free:
f(x – f(0)) = 1 – x

Substituting z = x – f(0)
f(z) = 1 – z – f(0)

Substituting z = 0:
f(0) = 1 – f(0)
2f(0) = 1
f(0) = ½

Back-substituting:
f(z) = 1 – z – ½
f(z) = ½ – z.

Resubstituting z = x:
f(x) = ½ – x.


Checking:
f(x – f(y)) = ½ – x + f(y) = ½ – x + ½ – y = 1 – x – y

[BlaqRaq]

I like how you explain your methods. Nothing is absolutely perfect but you come pretty close 😊.
Keep doing your work, I’m sure I’m not the only one enjoying the exercises.
(Cheers! From an old school learner)

[raystinger]

My method wasn't the simplest one, but it worked. Also, I assumed f is invertible since the composite f(x-f(y)) is linear.
So first, I plugged some values.
x=1 => f(1-f(y)) = -y
x=1 & y=-y => f(1-f(-y)) = y => Therefore 1-f(-x) is the inverse of the function f.
y=1 => f(x-f(1)) = -x
y=1 & x=-x => f(-x-f(1)) = x => Therefore -x-f(1) is also the inverse of the function f.
Therefore -x-f(1) = 1-f(-x)
=> f(-x) = 1 + f(1) + x
=> f(x) = 1 + f(1) - x
So now we plug some more values:
x=0 => f(0) = 1 + f(1)
y=0 => f(x - f(0)) = 1 - x
=> f(x - 1 - f(1)) = 1 - x
x=1 => f(-f(1)) = 0
Now using f(x) = 1 + f(1) - x
x=-f(1) => f(-f(1)) = 1 + f(1) + f(1) = 0
=> f(1) = -1/2
=> f(x) = 1/2 - x

[astraff]

How do we know that this is the only solution ? I mean we have'nt look for other solution so is there any other solution and how do we proof it?

[richardfarrer]

I split the difference between your solutions.

Set x = f(y) to get f(0) = 1- f(y) - y.

Set f(0) = -c+1 to get c = f(y) + y

Then substitute back to get c = 1/2

[armacham]

Set x = 0
You get: f(-f(y)) = 1 - y
multiply both sides by -1:
-f(-f(y)) = y - 1
define g(x) = -f(x)
so you have: g(g(x)) = x - 1
assuming g(x) is a first-degree polynomial, you can set g(x) = ax + b
then solve for a and b
it's not hard to do
and once you find g(x), it's super easy to find f(x)

[rachidtanan]

We have f(x)=-x+1-c
You can put x=0 and then you find c=1/2

[seanfraser]

My solution before watching the video:

Let m = f(0)
Then m = f(f(y)-f(y))
= 1-f(y)-y
Thus f(y) = 1-y-m
So m = f(0) = 1-m, thus m = 1/2
So (changing y to x) we have
f(x) = 1/2 - x

[elederiruzkin]

Very similar to:
f[x+f(y)]=x+y+1
Solved september 6th, 2021,
says the YT algorithm.

[markosv]

Is it correct to set x=y+f(y)? Very convenient.

[AliBarisa]

If f(x)= -x + k, then f(y)= -y + m where m and k are constants. Why did you assume that m = k in your explanation in the vedio? Where u said f(y) = - y +k

[SuperYoonHo]

Wow!Slovenia?!!!
SinQ/CosQ a 1x10^6 sir!

[haliltokmakmatgeo]

EMEKLERİNİZE SAĞLIK KOLAYLIKLAR DİLİYORUM HOCAM

[VictorStarukhin]

how about a proof that there are no other solutions?

[ЯрикБибик-що]

From russia pls
f (0;+inf) : x (0;+inf)
f(x+f(y+xy))= (y+1)f(x+1) -1