In order to use the derivative method, it should be specified in the problem that f is differentiable, or other conditions that leads to the differentiability of f. Where I live, students encounter this type of functional equation for the first time when they're learning calculus, so the 2nd method was actually the very first thing that came to my mind when I saw the problem! By the way, I really like your content and dedication. Keep up the great work!
To be honest, the first method looks like solving by knowing the answer. And by that I mean knowing that there exists another f(x) other that x² which wouldn't really work if you didn't know the answer before solving. And the second one is actually really good, it took me kinda some time to work that just out algebraically without any calculus. Even tho the solution could be simplified by getting f(0) = 0 via considering f(0+0) = 2f(0)+0, still nice method Nice vid btw
The video should have specified the domain of f to be R, and that f is continuous. This is because, since the video did not specify this, the video's conclusion is technically incorrect. For example, let f(x) = g(x) + h(x). Hence g(x + y) + h(x + y) = g(x) + h(x) + g(y) + h(y) + 2·x·y. From here, we can set g and h, without loss of generality, so that g(x + y) = g(x) + g(y), and h(x + y) = h(x) + h(y) + 2·x·y, and for such g, h, f will satisfy the equation. Here, notice that h satisfies the same equation f satisfies. h could be x |-> x^2, but it could be any other function satisfying the equation. So immediately concluding that h = x |-> x^2 is the only solution for h is not necessarily justified. Also, it is true that g(x + y) = g(x) + g(y) is solved by g(x) = A·x, but these are not the only solutions. They would be the only solution if g were continous, but an stipulation that g is continuous was never given for the problem.
Hi, do you care to enlightenment me about Cauchy's Functional equation, I searched on Google, looked on Quora and MSE. But I'm not sure what are the necessary conditions about an additive function for us to conclude that is of the form f(x)=kx. My teacher said that the only differentiable solution f(x+y)= f(x) + f(y) is f(x)=kx and he will prove that for us when we study functional equations chapter. I am curious whether it is enough that f is continuous (and defined everywhere) for this conclusion. I know that if it is continuous at any single point, you can prove that it is continuous everywhere.
I'm only asking you this because I see you in the comments of maths videos enlightening and correcting people all the time. And a request, pls try to dumb down your answer to high school level 😁. If that long comment made it unclear, my question is whether the condition of continuity (and domain all real numbers) is enough to conclude that f(x)=kx is the only solution, or do we require Differentiability as well (I know a proof using Differentiability, but not with just continuity)
I know the proof without continuity that f(0)= 0 and f(x)= xf(1) for rational numbers x, but I have know idea how to extend this to all real numbers using continuity (in the proof I assume that f is defined for all real numbers)
@@anshumanagrawal346 Continuity at a single point is one condition, as you said. Alternatively, you can also conclude f(x) = k·x if f is strictly monotonic and nonzero. By strictly monotonic, I mean that x < y implies f(x) < f(y) or f(x) > f(y). There are two other alternative conditions. Instead, if you know that f is bounded on some interval, you may conclude f(x) = k·x. Bounded means roughly that in that interval, it does not go to infinity or anything, there is a "largest value" f takes. And the other condition is pretty hard to explain simple layman terms. It is called "Lebesgue measurable". Explaining what that means requires some measure-theory context.
If you write f(x)= g(x)+h(x) where g(x) is even and h(x) is odd, then you can actually prove that g(x)=x^2 (by plugging in -x for y in the original expression) [edit: oh, you'll also need to plug in 0's for both x and y to see that f(0)=0]
@@trollface9903 h(x) isn't 0. See the "first method" in the video for the rest of this argument. My point was just that the first method didn't have to be an ansatz.
I use another method: First, note that if you plug in y=0, then you get that f(0) = 0, and we will use this fact later. Now, notice that if you set x and y to be equal then you get the equation f(2x) = 2x^2 + 2f(x), now call n to be a new variable and we see that f(n) = f(2*n/2) = 2(n/2)^2 + 2f(n/2) = 2(n/2)^2 + 2f(2*n/4) = ... By making this process infinitely many times and simplifying you get f(n) = (n^2)/2 + (n^2)/4 + ... + lim a->inf 2^a*f(n/2^a) In one hand, the sum part its just n^2*(1/2 + 1/4 + ...) = n^2, in the other hand, the limit is a inf*0 limit so we can transform it into a 0/0 limit and then use l'Hôpital's rule, lim a->inf 2^a*f(n/2^a) = lim a->inf f(n/2^a)/2^(-a) = lim a->inf (f'(n/2^a)*n*(2^(-a))')/(2^(-a))' = lim a->inf f'(n/2^a)*n = f'(0)*n (as n is finite) so therefore f(n) = n^2 + k*n QED.
Notice that in this step, we immediately assume differentiability. Because of our original equation, we can find that f(0) = f(0+0) = f(0)+f(0)+2*0*0 or f(0)=0 Now, look at f(h)/h and write it as (f(h)-f(0))/(h-0) If we let h approach 0, notice that we just get f'(0). Because we assumed differentiability, we know this limit exists.
@@angelmendez-rivera351 in my opinion it wasn't. Just like how technically assuming continuity isn't justified either. But since we already did assume differentiability, that limit of f(h)/h brings no problems anymore
@@skylardeslypere9909 yeah, in the first method too, it's not correct to conclude that h(x)= ax, unless it was given in the question that f is continuous
I see a lot of comments complaining the problem did not specify f is continuous. That’s part of the problem and easy solvable. 1. f(0)=? We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0) f(0)=0. 2. f continuous function? I.e. , lim {f(x+h)}=? when h->0 We calculate the limit on both sides of the original equation when y->0 => lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0 lim {f(x+y)}=f(x)+0+0, when y->0 lim {f(x+y)}=f(x) when y->0 f is continuous
If you set y = -x, you get f(0) = f(x) + f(-x) - 2x^2. Setting x = 0 will yield f(0) = 0, so you can further rearrange the equation to get f(x) + f(-x) = 2x^2. Assuming we're dealing with real numbers only, we can draw the conclusion that f(x) is symmetric about the y-axis, since swapping x with -x doesn't change the equation. From there, we can replace f(-x) with f(x) and solve to get f(x) = x^2. That said, this strategy misses the "ax" term. I suppose once you realize you're dealing with a quadratic, you could make the case to expand it to to all equations of the form x^2 + ax + b, remember that f(0) = 0, and realize that x^2 + ax for any constant "a" satisfies, but that's more of a guess-and-check approach.
There's a big problem, you never justified that f(x) = x^2 is the only solution, or even that it is the only type(or family) of solutions. And we can obviously guess that x^2 works without any work
The first time I saw that, I tried various things without success. Just now, my mind promptly substituted x² for f(x). I'd recently been doing a few (x + y)² terms.
Same here, Since I can’t sleep, I saw this one two and solved it in 10 seconds mentally before watching the video. I did also not watch the video to the end because the method seems too long and just based on guessing. Differentiating twice on both sides of the equation over x with y as parameter independent of x: => f’(x+y) = f’(x) + 2y => f”((x+y)= f”(x) for any y => f”(x) is constant (=2a) => f(x)= ax^2 + bx + c From original equation, x=y=0 we get => f(0)=2f(0) => c=f(0)=0 Now we use the result in the original equation to find a and b a(x+y)^2 +b(x+y) = (ax^2 + bx) * (ay^2 + by) + 2xy => 2axy= 2xy => a=1, b is any number => solution: f(x)= x^2+bx
Actually we could have stop at the first derivative of f over x already. with f'(x+y)=f'(x) + 2y for any x and y =>f'(0+y)=f'(0)+2y =>f'(y)=2y+f'(0) =>f(y)=y^2+f'(0)*y+c (c=0 as calculated above) => f(x)=x^2+bx
@@klopkerna3562 thank you for the question. f is continuous function. 1. f(0)=? We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0) f(0)=0. 2. f is continuous function: We calculate the limit on both sides of the equation when y->0 => lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0 lim {f(x+y)}=f(x)+f(0)+0} when y->0 lim {f(x+y)}=f(x) when y->0 f is continuous. 3. f(x)=? f(x+y)=f(x)+f(y)+2xy f(x+y)-f(x)=f(y)+2xy [ f(x+y)-f(x)]/y=[f(y)+2xy]/y with y≠0 [ f(x+y)-f(x)]/y=f(y)/y+2x Now let’s calculate the limit on both sides when y->0. lim {[f(x+y)-f(x)]/y}=lim {f(y)/y+2x} when y->0 We remind of the definition of the derivative of a function f(x) as being f’(x)= lim {[f(x+h)-f(x)]/h} when h->0; and With f(0)=0 as noted above lim {[f(x+y)-f(x)]/y}=lim {[f(0+y)-f(0)]/y+2x]} when y->0 f’(x)=f’(0)+2x f(x)=x^2+f’(0)*x+ c Since f(0)=0, c=0 Thus: f(x)=x^2+f’(0)*x
ok my solution is iterative f(x)=f(x-1)+f(1) + 2x = f(x-2) + 2f(1) + 2x + 2(x-1) etc. you get a row: f(x)= f(x-x) +xf(1)+ 2(x + x-1+...+1)= 0 + xf(1) + 2 x(x-1)/2 = x² - x + xf(1) what corresponds your solution if f(1)= 1+a
Sorry, i don't speak english. Tanto el primer como el segundo método parten de la base que f es diferenciable, ya que h debe ser diferenciable como función de Cauchy, y para garantizar que h es diferenciable, ya que h(x)=f(x)-x^2, y x^2 es diferenciable, basta que lo sea f para que lo sea h. En el segundo método, necesitas que el límite lim(h->0) (f(h)/h) exista, y una forma de tener eso es tener que f es diferenciable. Un saludo
Мне кажется, что ваши решения не достаточно строгие, но ответ всегда совпадает Возможно я не прав К примеру, чтобы доказать это уравнение, я использовал 7 страниц (маленьких, размером А5)
I got an answer f(x)=x^2, but I have no idea how to verify this answer. What I did was setting x=0, then x=1 and x=y and a little deduction afterwards. This led me to conclusion that f(x)=ax^2+bx+c and then that c=0, a=1 and b=0. I'm not sure though, this is the first functional equation I tried and got an answer at least. Let's see. : )
@@aashsyed1277 You are right. I found f(0) to be 0 and then f(1) to be 1, which would imply that f(x)=x^2. But I've made a mistake, because f(1)=b+1. At least, that's what I go. It doesn't give any information about b though, because I don't know f(1).
I got this one in my math book ,I found the answer in 2 different ways but still I felt insecure about the answer. Is weird how easy the answer appears and it feels like there would be infinite answers.
For the even part fe(x) , substitution of y=0: f(x) =f(x)+f(0),f(0)=0, but again if y=-x then f(0)=0=f(x) +f(-x) - 2x^2. Then fe(x) =x^2. For the odd part fo(x), fo(x+y) =1/2[f(x+y)-f(-x-y)] =fo(x) +fo(y) (after few semplifications), so that's true only for linear function fo(x) =ax. But f(x) =fe(x) +fo(x), so f(x) =x^2 +ax. What do you think?
@@skylardeslypere9909 Yes, but the video never talked about f being continuous. The video never even specified the domain, which is technically a mistake, since by not specifying the domain, it makes a multitude of solutions possible.
@@TheEternalVortex42 There are not. They can only be proven to exist via the axiom of choice, which is nonconstructive, or via some at least as strong axiom. I imagine there may be weaker axioms too, but if they exist, then they are also nonconstructive.
There is a big mistake, you didn't give the initial condition that f is continuous (it probably has to be differentiable for this to work but I'm not sure about that). If you don't add the constraint of continuity there are other possible solution
Often in calc we assume a general solution is of the form f(x)*h(x), but here you assumed it is of the form f(x)+h(x). I guess you solved it with calculus first!
Based on the question raised by a viewer on my previous proposed solution in another comment, I thought of this other way to solve the problem: 1. f(0)=? We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0) f(0)=0. 2. f is continuous function: We calculate the limit on both sides of the equation when y->0 => lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0 lim {f(x+y)}=f(x)+f(0)+0} when y->0 lim {f(x+y)}=f(x) when y->0 f is continuous 3. f(x)=? One way is presented in my other comment. Here another way: f(x+y)=f(x)+f(y)+2xy f(x+y)-f(x)=f(y)+2xy [ f(x+y)-f(x)]/y=[f(y)+2xy]/y with y≠0 [ f(x+y)-f(x)]/y=f(y)/y+2x Now let’s calculate the limit on both sides when y->0. lim {[f(x+y)-f(x)]/y}=lim {f(y)/y+2x} when y->0 We remind of the definition of the derivative of a function f(x) as being f’(x)= lim {[f(x+h)-f(x)]/h} when h->0; and With f(0)=0 as noted above lim {[f(x+y)-f(x)]/y}=lim {[f(0+y)-f(0)]/y+2x]} when y->0 f’(x)=f’(0)+2x f(x)=x^2+f’(0)*x+ c Since f(0)=0, c=0 Thus: f(x)=x^2+f’(0)*x
Thanks for the problem. I did it like that: Let y=x Then derivative goes:(1) f'(2x)-f'(x)=2x Then f(x)=ax^2 +bx +c Then i put f'(x)=2ax+b into (1) and get a=1 which means f(x)=x^2 +bx +c and then i simply write the equation given to me first and get that c=0 and then i saw that no matter what b is, ill get the same results( there is a better name for this but idk it in English).
Okay, commenting before I watch video's solution. I managed to get as far as being able to prove that if f(x) is continuous at x = 0, then f(x) = x^2 EDIT:... And clearly even that was wrong. argh... and ugh, I just spotted my error for that. Was a stupid thing where I lost track of a division in one spot. argleblargle. Just a result of how I was working it out and stupidly forgot to write out a division. blaaaaaaa.
En écrivant f(2x), puis f(3x)....on finit par « voir », puis démontrer par récurrence que f(nx)= nf(1) + n(n-1)x^2 Ensuite, avec x=1, f(n)= nf(1) + n(n-1) On extrapole de n à x, on développe, et on a bien f(x)= x^2 +ax
In order to use the derivative method, it should be specified in the problem that f is differentiable, or other conditions that leads to the differentiability of f. Where I live, students encounter this type of functional equation for the first time when they're learning calculus, so the 2nd method was actually the very first thing that came to my mind when I saw the problem!
By the way, I really like your content and dedication. Keep up the great work!
That's right! Thank you!
Good job. However: Plugging in y=0 in the original eq. would give you f(0)=0.
This would have made the determiniation of k in the end much easier.
Why lim when h tends to 0 of f(h) /h = a?
Some condition, e.g. continuity of f, is assumed in 1st method .
Differentiability of f at x=0 is assumed in 2nd method.
Very good approach .You are the ocean of knowledge , dear professor .
Many thanks 💖
To be honest, the first method looks like solving by knowing the answer. And by that I mean knowing that there exists another f(x) other that x² which wouldn't really work if you didn't know the answer before solving.
And the second one is actually really good, it took me kinda some time to work that just out algebraically without any calculus. Even tho the solution could be simplified by getting f(0) = 0 via considering f(0+0) = 2f(0)+0, still nice method
Nice vid btw
Cauchy's equation saves the day😉
(Only if h was increasing or continuous or the other conditions for Cauchy's equations😅)
So additive condition can't state that function is linear, am I right?
It does show it is linear on Q only. On R you need some other conditions
The video should have specified the domain of f to be R, and that f is continuous. This is because, since the video did not specify this, the video's conclusion is technically incorrect. For example, let f(x) = g(x) + h(x). Hence g(x + y) + h(x + y) = g(x) + h(x) + g(y) + h(y) + 2·x·y. From here, we can set g and h, without loss of generality, so that g(x + y) = g(x) + g(y), and h(x + y) = h(x) + h(y) + 2·x·y, and for such g, h, f will satisfy the equation. Here, notice that h satisfies the same equation f satisfies. h could be x |-> x^2, but it could be any other function satisfying the equation. So immediately concluding that h = x |-> x^2 is the only solution for h is not necessarily justified. Also, it is true that g(x + y) = g(x) + g(y) is solved by g(x) = A·x, but these are not the only solutions. They would be the only solution if g were continous, but an stipulation that g is continuous was never given for the problem.
That's right
Hi, do you care to enlightenment me about Cauchy's Functional equation, I searched on Google, looked on Quora and MSE. But I'm not sure what are the necessary conditions about an additive function for us to conclude that is of the form f(x)=kx. My teacher said that the only differentiable solution f(x+y)= f(x) + f(y) is f(x)=kx and he will prove that for us when we study functional equations chapter. I am curious whether it is enough that f is continuous (and defined everywhere) for this conclusion. I know that if it is continuous at any single point, you can prove that it is continuous everywhere.
I'm only asking you this because I see you in the comments of maths videos enlightening and correcting people all the time. And a request, pls try to dumb down your answer to high school level 😁. If that long comment made it unclear, my question is whether the condition of continuity (and domain all real numbers) is enough to conclude that f(x)=kx is the only solution, or do we require Differentiability as well (I know a proof using Differentiability, but not with just continuity)
I know the proof without continuity that f(0)= 0 and f(x)= xf(1) for rational numbers x, but I have know idea how to extend this to all real numbers using continuity (in the proof I assume that f is defined for all real numbers)
@@anshumanagrawal346 Continuity at a single point is one condition, as you said. Alternatively, you can also conclude f(x) = k·x if f is strictly monotonic and nonzero. By strictly monotonic, I mean that x < y implies f(x) < f(y) or f(x) > f(y).
There are two other alternative conditions. Instead, if you know that f is bounded on some interval, you may conclude f(x) = k·x. Bounded means roughly that in that interval, it does not go to infinity or anything, there is a "largest value" f takes. And the other condition is pretty hard to explain simple layman terms. It is called "Lebesgue measurable". Explaining what that means requires some measure-theory context.
If you write f(x)= g(x)+h(x) where g(x) is even and h(x) is odd, then you can actually prove that g(x)=x^2 (by plugging in -x for y in the original expression)
[edit: oh, you'll also need to plug in 0's for both x and y to see that f(0)=0]
You still need to show h(x) is 0
@@trollface9903 h(x) isn't 0.
See the "first method" in the video for the rest of this argument. My point was just that the first method didn't have to be an ansatz.
I use another method:
First, note that if you plug in y=0, then you get that f(0) = 0, and we will use this fact later.
Now, notice that if you set x and y to be equal then you get the equation f(2x) = 2x^2 + 2f(x), now call n to be a new variable and we see that f(n) = f(2*n/2) = 2(n/2)^2 + 2f(n/2) = 2(n/2)^2 + 2f(2*n/4) = ...
By making this process infinitely many times and simplifying you get f(n) = (n^2)/2 + (n^2)/4 + ... + lim a->inf 2^a*f(n/2^a)
In one hand, the sum part its just n^2*(1/2 + 1/4 + ...) = n^2, in the other hand, the limit is a inf*0 limit so we can transform it into a 0/0 limit and then use l'Hôpital's rule, lim a->inf 2^a*f(n/2^a) = lim a->inf f(n/2^a)/2^(-a) = lim a->inf (f'(n/2^a)*n*(2^(-a))')/(2^(-a))' = lim a->inf f'(n/2^a)*n = f'(0)*n (as n is finite) so therefore f(n) = n^2 + k*n QED.
Did you show the derivability of f?
9:13
The limit of sum is sum of limit
But if and only if both limit exist and you doesn't prove
lim h to 0 f(h)/h always exist :/
Notice that in this step, we immediately assume differentiability. Because of our original equation, we can find that f(0) = f(0+0) = f(0)+f(0)+2*0*0 or f(0)=0
Now, look at f(h)/h and write it as (f(h)-f(0))/(h-0)
If we let h approach 0, notice that we just get f'(0). Because we assumed differentiability, we know this limit exists.
@@skylardeslypere9909 You are correct, but this begs the question as to whether the differentiability assumption is justified.
@@angelmendez-rivera351 in my opinion it wasn't. Just like how technically assuming continuity isn't justified either. But since we already did assume differentiability, that limit of f(h)/h brings no problems anymore
@@skylardeslypere9909 yeah, in the first method too, it's not correct to conclude that h(x)= ax, unless it was given in the question that f is continuous
(and thus h is continuous)
I see a lot of comments complaining the problem did not specify f is continuous.
That’s part of the problem and easy solvable.
1. f(0)=?
We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0)
f(0)=0.
2. f continuous function? I.e. , lim {f(x+h)}=? when h->0
We calculate the limit on both sides of the original equation when y->0
=> lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0
lim {f(x+y)}=f(x)+0+0, when y->0
lim {f(x+y)}=f(x) when y->0
f is continuous
Very impressive!
Thank you!
Your first method is so beautiful and elegant
Thank you so much 😊
We love SyberMath😊
❤️
If you set y = -x, you get f(0) = f(x) + f(-x) - 2x^2. Setting x = 0 will yield f(0) = 0, so you can further rearrange the equation to get f(x) + f(-x) = 2x^2. Assuming we're dealing with real numbers only, we can draw the conclusion that f(x) is symmetric about the y-axis, since swapping x with -x doesn't change the equation. From there, we can replace f(-x) with f(x) and solve to get f(x) = x^2.
That said, this strategy misses the "ax" term. I suppose once you realize you're dealing with a quadratic, you could make the case to expand it to to all equations of the form x^2 + ax + b, remember that f(0) = 0, and realize that x^2 + ax for any constant "a" satisfies, but that's more of a guess-and-check approach.
There's a big problem, you never justified that f(x) = x^2 is the only solution, or even that it is the only type(or family) of solutions. And we can obviously guess that x^2 works without any work
A moment of silence for this class 7, who though he could do this sum
The second method is easier to understand than the first method.
I agree! Thanks for the feedback!
I think Cauchy Function has many solutions, so maybe you can get other solutions rather than x^2+ax in method 1.
The first time I saw that, I tried various things without success.
Just now, my mind promptly substituted x² for f(x). I'd recently been doing a few (x + y)² terms.
Solving for k is much easier when you just look at f(0).
Same here, Since I can’t sleep, I saw this one two and solved it in 10 seconds mentally before watching the video. I did also not watch the video to the end because the method seems too long and just based on guessing.
Differentiating twice on both sides of the equation over x with y as parameter independent of x:
=> f’(x+y) = f’(x) + 2y
=> f”((x+y)= f”(x) for any y
=> f”(x) is constant (=2a)
=> f(x)= ax^2 + bx + c
From original equation, x=y=0 we get
=> f(0)=2f(0)
=> c=f(0)=0
Now we use the result in the original equation to find a and b
a(x+y)^2 +b(x+y) = (ax^2 + bx) * (ay^2 + by) + 2xy
=> 2axy= 2xy
=> a=1, b is any number
=> solution: f(x)= x^2+bx
Nice!
@@SyberMath 🙏
Actually we could have stop at the first derivative of f over x already. with f'(x+y)=f'(x) + 2y for any x and y =>f'(0+y)=f'(0)+2y =>f'(y)=2y+f'(0) =>f(y)=y^2+f'(0)*y+c (c=0 as calculated above) => f(x)=x^2+bx
And how do you know that f is differentiable?
@@klopkerna3562 thank you for the question. f is continuous function.
1. f(0)=?
We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0)
f(0)=0.
2. f is continuous function:
We calculate the limit on both sides of the equation when y->0
=> lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0
lim {f(x+y)}=f(x)+f(0)+0} when y->0
lim {f(x+y)}=f(x) when y->0
f is continuous.
3. f(x)=?
f(x+y)=f(x)+f(y)+2xy
f(x+y)-f(x)=f(y)+2xy
[ f(x+y)-f(x)]/y=[f(y)+2xy]/y with y≠0
[ f(x+y)-f(x)]/y=f(y)/y+2x
Now let’s calculate the limit on both sides when y->0.
lim {[f(x+y)-f(x)]/y}=lim {f(y)/y+2x} when y->0
We remind of the definition of the derivative of a function f(x) as being f’(x)= lim {[f(x+h)-f(x)]/h} when h->0; and
With f(0)=0 as noted above
lim {[f(x+y)-f(x)]/y}=lim {[f(0+y)-f(0)]/y+2x]} when y->0
f’(x)=f’(0)+2x
f(x)=x^2+f’(0)*x+ c
Since f(0)=0, c=0
Thus:
f(x)=x^2+f’(0)*x
ok my solution is iterative
f(x)=f(x-1)+f(1) + 2x = f(x-2) + 2f(1) + 2x + 2(x-1) etc. you get a row:
f(x)= f(x-x) +xf(1)+ 2(x + x-1+...+1)= 0 + xf(1) + 2 x(x-1)/2 = x² - x + xf(1) what corresponds your solution if f(1)= 1+a
I enjoy the functional equations but I never saw them before and I have a maths degree.
Me too - I've never seen them before, let alone being taught how to solve them.
Sorry, i don't speak english.
Tanto el primer como el segundo método parten de la base que f es diferenciable, ya que h debe ser diferenciable como función de Cauchy, y para garantizar que h es diferenciable, ya que h(x)=f(x)-x^2, y x^2 es diferenciable, basta que lo sea f para que lo sea h.
En el segundo método, necesitas que el límite lim(h->0) (f(h)/h) exista, y una forma de tener eso es tener que f es diferenciable. Un saludo
Мне кажется, что ваши решения не достаточно строгие, но ответ всегда совпадает
Возможно я не прав
К примеру, чтобы доказать это уравнение, я использовал 7 страниц (маленьких, размером А5)
Great sols
Thank you!
Derivative method was lit man! I didn't expect the limit defention to be used here, Ur problems r getting good n good evryday
It is standard class 12 th problem in
Application of derivative chapter
@@NikunjKumarGupta oh
@@manojsurya1005 kaunsi class mein ho aap?
Thank you! 💖
@@NikunjKumarGupta watchu sayin bro?
Nice
Thanks
Wow, nice equation squared
Thank you!
Guess work! Total guessing
I got an answer f(x)=x^2, but I have no idea how to verify this answer. What I did was setting x=0, then x=1 and x=y and a little deduction afterwards. This led me to conclusion that f(x)=ax^2+bx+c and then that c=0, a=1 and b=0. I'm not sure though, this is the first functional equation I tried and got an answer at least. Let's see. : )
B is not necessarily 0
@@aashsyed1277 You are right. I found f(0) to be 0 and then f(1) to be 1, which would imply that f(x)=x^2. But I've made a mistake, because f(1)=b+1. At least, that's what I go. It doesn't give any information about b though, because I don't know f(1).
@@snejpu2508 edit ur comment
@@aashsyed1277 no, it shows what the commenter had originally thought, and it's good to show your mistakes too
@@anshumanagrawal346 :)
How can we integrate without f prime o that means without the value f dash o?
Why do you assume f is differentiable at 0?
I should've mentioned it at the beginning
How did we know the limit of f(h)/h converges?
That's the thing, for this question to be solvable you have to be given that f has to be continuous (and in fact differentiable)
@@anshumanagrawal346 oh gotcha, forgot if that was given
I got this one in my math book ,I found the answer in 2 different ways but still I felt insecure about the answer.
Is weird how easy the answer appears and it feels like there would be infinite answers.
I can tell by inspection that the answer is f(x)=x^2, but I don't know how to prove it.
You can solve easily by splitting f(x) in even part and odd part.
How?
For the even part fe(x) , substitution of y=0: f(x) =f(x)+f(0),f(0)=0, but again if y=-x then f(0)=0=f(x) +f(-x) - 2x^2. Then fe(x) =x^2. For the odd part fo(x), fo(x+y) =1/2[f(x+y)-f(-x-y)] =fo(x) +fo(y) (after few semplifications), so that's true only for linear function fo(x) =ax. But f(x) =fe(x) +fo(x), so f(x) =x^2 +ax. What do you think?
That's simply (x + y) ² = x² + y² + 2xy That's all, folks.
Really?
😁
perfect
Nice video.
THere are other additive functions then the linear (f(x) =kx) function though
f(x)=ax is the only _continuous_ function that satisfies f(x+y)=f(x)+f(y)
@@skylardeslypere9909 Yes, but the video never talked about f being continuous. The video never even specified the domain, which is technically a mistake, since by not specifying the domain, it makes a multitude of solutions possible.
@@angelmendez-rivera351 I 100% agree
Although I don't think there are any explicit constructions of them.
@@TheEternalVortex42 There are not. They can only be proven to exist via the axiom of choice, which is nonconstructive, or via some at least as strong axiom. I imagine there may be weaker axioms too, but if they exist, then they are also nonconstructive.
There is a big mistake, you didn't give the initial condition that f is continuous (it probably has to be differentiable for this to work but I'm not sure about that). If you don't add the constraint of continuity there are other possible solution
That's right!
Great!
You can solve for y=0 in the equation, and get f(0)=0 without doing the ling math
Often in calc we assume a general solution is of the form f(x)*h(x), but here you assumed it is of the form f(x)+h(x). I guess you solved it with calculus first!
I am a Nigerian and would like to be mathematiician like our professor My name is Patrick
f(x)=x² but I could only use calculus to solve it..... 1st
👍
👍👍👍👏👏👏👏👏👏👌👌👌👌👌👌
Based on the question raised by a viewer on my previous proposed solution in another comment, I thought of this other way to solve the problem:
1. f(0)=?
We set x=y=0 in the original equation and we get f(0+0)=f(0)+f(0)
f(0)=0.
2. f is continuous function:
We calculate the limit on both sides of the equation when y->0
=> lim {f(x+y)}=lim {f(x)+f(y)+2xy} when y->0
lim {f(x+y)}=f(x)+f(0)+0} when y->0
lim {f(x+y)}=f(x) when y->0
f is continuous
3. f(x)=?
One way is presented in my other comment. Here another way:
f(x+y)=f(x)+f(y)+2xy
f(x+y)-f(x)=f(y)+2xy
[ f(x+y)-f(x)]/y=[f(y)+2xy]/y with y≠0
[ f(x+y)-f(x)]/y=f(y)/y+2x
Now let’s calculate the limit on both sides when y->0.
lim {[f(x+y)-f(x)]/y}=lim {f(y)/y+2x} when y->0
We remind of the definition of the derivative of a function f(x) as being f’(x)= lim {[f(x+h)-f(x)]/h} when h->0; and
With f(0)=0 as noted above
lim {[f(x+y)-f(x)]/y}=lim {[f(0+y)-f(0)]/y+2x]} when y->0
f’(x)=f’(0)+2x
f(x)=x^2+f’(0)*x+ c
Since f(0)=0, c=0
Thus:
f(x)=x^2+f’(0)*x
Thanks for the problem. I did it like that:
Let y=x
Then derivative goes:(1) f'(2x)-f'(x)=2x
Then f(x)=ax^2 +bx +c
Then i put f'(x)=2ax+b into (1) and get a=1 which means f(x)=x^2 +bx +c and then i simply write the equation given to me first and get that c=0 and then i saw that no matter what b is, ill get the same results( there is a better name for this but idk it in English).
Okay, commenting before I watch video's solution. I managed to get as far as being able to prove that if f(x) is continuous at x = 0, then f(x) = x^2
EDIT:...
And clearly even that was wrong. argh... and ugh, I just spotted my error for that. Was a stupid thing where I lost track of a division in one spot. argleblargle. Just a result of how I was working it out and stupidly forgot to write out a division. blaaaaaaa.
En écrivant f(2x), puis f(3x)....on finit par « voir », puis démontrer par récurrence que f(nx)= nf(1) + n(n-1)x^2
Ensuite, avec x=1, f(n)= nf(1) + n(n-1)
On extrapole de n à x, on développe, et on a bien f(x)= x^2 +ax
Bonjour, votre solution n'est valable que pour les entiers naturels.
Cauchy equation has a lot of non-continuous solutions. en.m.wikipedia.org/wiki/Cauchy%27s_functional_equation
I am a Nigerian and would like to be mathematiician like our professor My name is Patrick