I believe there are infinitely more solutions. But they are easy to represent f(x) = x + c with c in R when k = 1 then k-1 is 0 and there is no restrictions on c so any c should do the trick.
If you assume f is differentiable then there is a quick proof that f must be linear in x. Proof(ish). Since f(f(x+y)) = f(x) + f(y), this implies f(x+h) + f(0) = f(x) + f(h) for every x, h (to see this, note that x+h = (x+h) + 0 to express the initial equation in two different ways, and equate them). Rearranging this gives f(x+h) - f(x) = f(h) - f(0). Therefore taking first principles definition of the derivative, dividing the last equation by h and taking h to zero gives f’(x) = f’(0) for every x. So the derivative of f must be constant everywhere!
Great explanation but there is a little mistake at the end: when you calculate k and c in the case k=1, c can be anything since k-1=0 . If you plug into the original equation the function f(x) = x+c with c arbitrary real, you have a valid solution.
your solution is correct up until the last part. The problem is that the values that t can take on is only the image of f, meaning there is no guarantee that t can be any real number. One counter example is the function “floor of x” (the greatest integer not exceeding x). This indeed satisfies f(f(x)) = f(x) + f(0), however is not in the form f(x) = x + c. For your argument to be correct, you must prove surjectivity, which I believe is possible using continuity and the intermediate value theorem.
Here’s what I came up with: Assume f(x) =/ 0 (you have already done this case in case 1). Then, there must exist “a” such that f(a) =/ 0. Plug x=a, y=0 and we get f(f(a)) = f(a) + f(0). x=a, y=f(a) yields f(a + f(a)) = 2f(a) + f(0) and so on, meaning there exists “b” such that for all “n” in the natural numbers, f(b) = nf(a) + f(0). Now we want to extend n to the integers. Plugging in x=x, y=-x we get that f(-x) = -f(x) + f(f(0)), but f(f(0)) = 2f(0) when x=y=0 so f(-x) = -f(x) + 2f(0). Therefore we have found a way to introduce -f(a). Therefore n can be any integer. Now, using continuity, the intermediate value theorem can be applied, meaning that there exists “b” such that f(b) = x for all real numbers x, meaning you can just substitute b and get f(x) = x + f(0). Some notes: 1. I have glossed over the handling of the f(0)s in the equations to clean them up into our desired form, but they shouldn’t really be a problem. 2. the fact that f(a) =/ 0 is very crucial as it means for any real number x, there exists n in the integers such that nf(a) + f(0) =< x =< (n{+ or -}1)f(a) + f(0). If f(a) = 0 then f(0) =< x =< f(0) which is clearly not true for any real number x. 3. Continuity is what makes or breaks this proof. Without it, counterexamples like the floor functions exist.
Remark: f need not be continuous everywhere. It can be discontinuous on a set of measure zero (Lebesgue function) or on a meagre set (Bair function). It is a general question: what additional property should a function have, so that the only solution to the Cauchy equation is a linear function? One cannot drop continuity entirely, as there are counterexamples.
The solutions are f(x) = 0 and f(x) = x + c. The process of reducing the problem to a Cauchy functional equation was key to the solutions. Thank you for sharing.
@@slavinojunepri7648no it wasn't key to anything... I just rewritten formulas as f(x) = kx + t, applied to both sides, found k = 1 from f(f(0)) = 2 f(0), noticed how t cancels out when solving with k=1 And got f(x) = x + t
I did something different but there is a mistake. I let y be equal to x and i found that f[f(2x)] is equal to 2*f(x) then i let y be equal to 3*x so f[f(4*x)] is equal to f(3x)+f(x) and it's equal to 2*f(2x) too. Because of the property explained before so f(3*x)+f(x)=2*f(2*x) But doing so f(x) can be equal to every constant and x. What did I do wrong?
Thanks you so much Prime Newtons. I enjoy deeplly all your problems. In the case of funcional equations, im having troubles to understand the deep of this at all, but i will try again in home to study it. Please if you can and think thats is good idea, do more problems of this family.
Couldn't we just said that f(x) is gonna be a function like ax+b after we found f(f(x))=f(x)+f(0) which would mean that f(0) is a constant and then just found the answer normally
@@chaosredefined3834 well f(0) is constant and if f(x+y)=f(x)+C then just like in the video you can find that its in that form and its also easier because its skips a lot of steps
You could assume it’s a differentiable function and differentiate it wrt to x keeping y as a constant, then keep y=0 and solve, you can get the values of constant later on by comparison
@@donwald3436 Ok, I'm picking up what you're putting down now. The problem sais find all solutions for f not x,y. x,y are just random variables. So let's say we have f(x)+f(y)=5, and we are looking for f. so we can substitute x=y=0 into the same equation, without changing f. Since f is true for ANY and ALL values of x,y. 2f(0)=5, f(0)=2.5, for any value of x or y. We found something about the FUNCTION without knowledge of variables. Knowing f(0), we can get f(x) using f(x)+f(0)=5, f(x)=2.5 Since f is the same, f(y)=2.5 as well, making the above FUNCTIONAL equation true for all x and y. Using the functional equation (with answer) from the video: f(f(x+y))=f(x)+f(y). using f(x)=x+c (f(x)=kx+c ,k=1) f((x+y)+c)=x+c+y+c ((x+y)+c)+c=(x+y)+2c x+y+2c=x+y+2c which means this solution is correct. What the values of the variables are doesn't matter in a functional equation, because the given equation is true for ANY and ALL x,y pairs. I hope this made it clearer
Why did you not solve f(y)? Because I thought that you need to find the solutions for f(f(x+y)) and not only for f(x)….this video is not the same standard as your normal videos…
I believe there are infinitely more solutions.
But they are easy to represent
f(x) = x + c with c in R
when k = 1 then k-1 is 0 and there is no restrictions on c
so any c should do the trick.
That's what I thought as well. Given k=1, c is any real number
Yea I agree
It is correct
I agree as well
Then, where is the error? 🙂
Oh my god, that's why i love math! An amazing teacher you are! Please, continue your excellent work!
If you assume f is differentiable then there is a quick proof that f must be linear in x.
Proof(ish). Since f(f(x+y)) = f(x) + f(y), this implies
f(x+h) + f(0) = f(x) + f(h) for every x, h (to see this, note that x+h = (x+h) + 0 to express the initial equation in two different ways, and equate them).
Rearranging this gives
f(x+h) - f(x) = f(h) - f(0).
Therefore taking first principles definition of the derivative, dividing the last equation by h and taking h to zero gives
f’(x) = f’(0) for every x. So the derivative of f must be constant everywhere!
never stop teaching !!!
such an amazing teacher😭❤
Great explanation but there is a little mistake at the end: when you calculate k and c in the case k=1, c can be anything since k-1=0 . If you plug into the original equation the function f(x) = x+c with c arbitrary real, you have a valid solution.
i really really enjoy watching his teaching. So funny and entertaining and educational and .......
THANK U SIR!!.. U are an amazing teacher
Didn't understand anything but feel good after watching 😊
this is a pretty long solution. I think I have the easier one
1st case: f(x)=C. Then C=C+C which means f(x)=0 for all values of x
2nd case: f(x)>
your solution is correct up until the last part. The problem is that the values that t can take on is only the image of f, meaning there is no guarantee that t can be any real number. One counter example is the function “floor of x” (the greatest integer not exceeding x). This indeed satisfies f(f(x)) = f(x) + f(0), however is not in the form f(x) = x + c. For your argument to be correct, you must prove surjectivity, which I believe is possible using continuity and the intermediate value theorem.
Here’s what I came up with: Assume f(x) =/ 0 (you have already done this case in case 1). Then, there must exist “a” such that f(a) =/ 0. Plug x=a, y=0 and we get f(f(a)) = f(a) + f(0). x=a, y=f(a) yields f(a + f(a)) = 2f(a) + f(0) and so on, meaning there exists “b” such that for all “n” in the natural numbers, f(b) = nf(a) + f(0). Now we want to extend n to the integers. Plugging in x=x, y=-x we get that f(-x) = -f(x) + f(f(0)), but f(f(0)) = 2f(0) when x=y=0 so f(-x) = -f(x) + 2f(0). Therefore we have found a way to introduce -f(a). Therefore n can be any integer. Now, using continuity, the intermediate value theorem can be applied, meaning that there exists “b” such that f(b) = x for all real numbers x, meaning you can just substitute b and get f(x) = x + f(0).
Some notes: 1. I have glossed over the handling of the f(0)s in the equations to clean them up into our desired form, but they shouldn’t really be a problem. 2. the fact that f(a) =/ 0 is very crucial as it means for any real number x, there exists n in the integers such that nf(a) + f(0) =< x =< (n{+ or -}1)f(a) + f(0). If f(a) = 0 then f(0) =< x =< f(0) which is clearly not true for any real number x. 3. Continuity is what makes or breaks this proof. Without it, counterexamples like the floor functions exist.
f(x)=x+b
Because
((x+y)+b)+b = (x + b) + (y + b)
Remark: f need not be continuous everywhere. It can be discontinuous on a set of measure zero (Lebesgue function) or on a meagre set (Bair function). It is a general question: what additional property should a function have, so that the only solution to the Cauchy equation is a linear function? One cannot drop continuity entirely, as there are counterexamples.
Ok. Can you do a video on what a Cauchy Functional Equation is and how it’s derived? That be handy.
x+c is also a solution where c is real
Yes. Now that I think about it
The solutions are f(x) = 0 and f(x) = x + c. The process of reducing the problem to a Cauchy functional equation was key to the solutions. Thank you for sharing.
@@slavinojunepri7648no it wasn't key to anything...
I just rewritten formulas as f(x) = kx + t, applied to both sides, found k = 1 from f(f(0)) = 2 f(0), noticed how t cancels out when solving with k=1
And got f(x) = x + t
3:46 f(x+y) = f(x + y) +2f(0), Does it can be possible sir ?
I did something different but there is a mistake. I let y be equal to x and i found that f[f(2x)] is equal to 2*f(x) then i let y be equal to 3*x so f[f(4*x)] is equal to f(3x)+f(x) and it's equal to 2*f(2x) too. Because of the property explained before so f(3*x)+f(x)=2*f(2*x) But doing so f(x) can be equal to every constant and x. What did I do wrong?
I actually took notes for this video
f(f(y))= f(y)+ C , c=f(0)
Replace f(y) with x
f(x) = x+ c
f(f(x+y) = f(x)+f(y)
f(x+y+c) = x+c + y+c
x+y+c+c = x+c+y+c which is always true
f(x) = x+c
Got confused after 5 min
Then again watched it very carefully😅
How did you conclude that f(0)= h(0)?
I solved it myself and jumped to the end. Why isn't any x + c a solution?
try x=1and y=2 into f(f(x)), you can find that c must be 0
@@singcheung2362
Okay, let's try f(x) = x + 1
f(f(1+2)) = ((1+2) + 1) + 1 = 5
f(1) + f(2) = (1+1) + (2+1) = 5
It works.
@@singcheung2362
Any value works for c. Can you explain why you think c must be 0?
Thanks you so much Prime Newtons.
I enjoy deeplly all your problems.
In the case of funcional equations, im having troubles to understand the deep of this at all, but i will try again in home to study it.
Please if you can and think thats is good idea, do more problems of this family.
Hi, I wrote an explanation in the comment by donwald3436, which could help you understand the intrecacies a little bit more
@@Dshado hi fellow. I really apreciate your comment and your explanation. I will check and practice it for sure. 🤗
@@brenobelloc8617 hi, if you get stuck just messege here, i might be able to help :D
3:51 meme material
👍👍👍
Who else got lost at step 3? 😅🤦🏽
x + y just represents some input into the double composition. So, if you replace all the y terms in line 2 with x + y, you get line 3
I think step 3 means that he’s drawing a conclusion: since f(f(x)) = f(x) + f(0) and f(f(y)) = f(y) + f(0) ➡️ f(f(whatever)) = f(whatever) + f(0)
@@rayyt5566 oooooooo
@@oliverkoller5468 gotcha thx
@@collapsingwavefunction_.3356 you’re welcome : )
Couldn't we just said that f(x) is gonna be a function like ax+b after we found f(f(x))=f(x)+f(0) which would mean that f(0) is a constant and then just found the answer normally
How do you know that f(x) will be of the form ax+b?
@@chaosredefined3834 well f(0) is constant and if f(x+y)=f(x)+C then just like in the video you can find that its in that form and its also easier because its skips a lot of steps
@@ben_adel3437 If f(x+y) = f(x) + C, then f(y) = f(0) + C when x = 0, so f(y) is a constant.
@@chaosredefined3834 well yeah you made both sides constants which wouldn't give you info about the function
@@ben_adel3437 You said that we get f(x+y) = f(x) + C. For this to be true, it needs to be true for all values of x. So, it needs to be true for x=0.
The video is not clear at 3min 41 sec …the conclusion is not obvious…can you explain in more details?
He’s replacing x with x+y in the equation f(f(x)) = f(x) + f(0)
cool necklace
You could assume it’s a differentiable function and differentiate it wrt to x keeping y as a constant, then keep y=0 and solve, you can get the values of constant later on by comparison
You can't assume it is differentiable from it being continuous.
How do you justify just setting y to 0 then saying that is equivalent to the original function?
Its a functional equation so its true for ANY value of y, so he substituted a value of 0 instead, since it needs to be true for any and all values
@@Dshado Okay so I set y to 0 and x to 0 so f(f(0)) = f(0) + f(0) and somehow that is just as general as the original function?
@@donwald3436 its not as general, but it still is true. Giving the x and y values, does not make it NOT true
@@Dshado That's my point, the problem said find all solutions.
@@donwald3436 Ok, I'm picking up what you're putting down now.
The problem sais find all solutions for f not x,y. x,y are just random variables.
So let's say we have f(x)+f(y)=5, and we are looking for f.
so we can substitute x=y=0 into the same equation, without changing f. Since f is true for ANY and ALL values of x,y.
2f(0)=5, f(0)=2.5, for any value of x or y. We found something about the FUNCTION without knowledge of variables.
Knowing f(0), we can get f(x) using f(x)+f(0)=5, f(x)=2.5
Since f is the same, f(y)=2.5 as well, making the above FUNCTIONAL equation true for all x and y.
Using the functional equation (with answer) from the video:
f(f(x+y))=f(x)+f(y).
using f(x)=x+c (f(x)=kx+c ,k=1)
f((x+y)+c)=x+c+y+c
((x+y)+c)+c=(x+y)+2c
x+y+2c=x+y+2c
which means this solution is correct.
What the values of the variables are doesn't matter in a functional equation, because the given equation is true for ANY and ALL x,y pairs.
I hope this made it clearer
Why did you not solve f(y)? Because I thought that you need to find the solutions for f(f(x+y)) and not only for f(x)….this video is not the same standard as your normal videos…
You’re trying to find functions f which satisfy the equation. It doesn’t matter if you find f(x) or f(y) because they are just the same
i smart f(x) = x
Pl.try to write a little bold so that it can be followed clearly
I am from india.