There was one time that the answer was a quadratic function. I remember noticing that a linear function did not satisfy the functional equation (I don't recall whether a constant satisfied it or not). Also, there is a linear case that works, although it is a degenerate case for when B = 0.
It is not immediately obvious that g(x+y) = g(x)g(y) => g(x) = C^x. In fact, we have g(x) = a exp(L(x)) where L is a Q-linear function on R. Plugging the result back in the formula relating f and g ensures that this function is increasing, and thus R-linear.
Absolutely. Just like in the first case, the fact that f is increasing must be used here. This piece of the puzzle is classic but very much important and not obvious!
♡Functional Equations by Titu Andreescu ♡functional equations A problem solving approach book by B.J. Venkatachala ●Functional equations in mathematical olympiads by Amir Hossein Parvardi ●Introduction to functional equations by Costas Efthimiou ●Functional Equations and how to solve them by Christopher G. Small ●Lectures on functional Equations and their applications volume 19 by J. Aczèl
Merry Christmas to you and your familiy, Michael. Many thanks for your inspiring maths work on your TH-cam channels. Even for mathematicians very enjoyable. My fav. topic is all about the zeta function (and L-functions / modular forms). 🎄 Chris
This question was previously asked in 2004 Antalya Math Contest in Turkey to 2nd year high school students. It is nice to see a question here that I've just recently solved.
I had a hard time getting what he meant with "x=1 and y=x" until I got it : he actually does "x=1 and y=y" and then uses y as a dummy variable, renaming it x.
The solutions can't really be combined into one. If you try by setting C=1 in the second solution, you end up dividing by zero, and that's a no-no. Even if we ignore the division problem since it's part of the constant coefficient anyway, we still can't choose a value of C that makes f(x)=A⋅(Cˣ-1)→f(x)=Ax. Also, you probably should have pointed out that f(1) is a free variable. For reference, solution 2: g(x)=Cˣ, f(x)=A⋅(Cˣ-1) (with solution 1: g(x)≡1, f(x)=Ax).
The first substitution I did was x=dy. f(y+dy)-f(y)=g(y)f(dy). As soon as you get an equation that looks like this, you are looking to evaluate a limit, and with some tinkering and an application of LHR (and renaming variables) you get g(x)=f’(x)/f’(0). Since we know g(0) is 1, that means that f’ must either be a constant or a multiple of itself, which leads us to the exponential pair and the linear and constant pair quite nicely (I came up with B*a^x-B, but that’s essentially the same thing with a renamed coefficient where my B is the multiplicative inverse of yours)
At about 5:00 you set x=1, y=x. I'm not very practiced at functional equations, so not sure what is happening there. If y=x and x=1, then isn't y=1? Maybe a different notation would be a bit more understandable, or some intermediate steps to include more detail to arrive at those substitutions. Always interesting videos!
8:11 it's not exactly obvious that f(x) = Ax is the only solution. But we can prove that if there is some value k s.t. f(k) != Ak, it will lead to contradiction.
I assumed f is C2, then what you can get is 1) [f(x+y)-f(y)]/x = f(x)/x*g(y) -> g(s) = f'(s)/f'(0). So the two functions are related. 2) [f(x+y)-f(x)]/y =f(x)[g(y)-1]/y+f(y)/y -> f'(x) = f''(0)/f'(0) f(x) + f'(0). You get either f(x)=Ax+B or f(x) = C(1-exp(rx)). As you demonstrated, both lead to valid solutions
Isn't continuity required to prove these are all the functions? I think it all comes to the functional equation f(x+y)=f(x)+f(y), right? I think if we have continuity then we can prove it has to be linear function. However without continuity I think there could be function that is partially linear, so for given real r f(q*r)=a_r*q*r for all rational qs. a_r could be different for different r but if r_1-r_2 is rational then a_r_1=a_r_2. And I think we could prove such functions exist by taking one number from each class of equivalance relation (where r1 and r2 are equivalent when their difference is rational) and we set different slopes for first 2 numbers and we use them for defining set and then for when we have all of chosen representatives of these classes in well ordered chain of length continuum, we take the representative only when it can't be written as finite rational combination q1*r1+q2*r2+... taken at some previous step and we define function using first slope. If it can then it should already be well defined, because then it has only one form and so we define using our formula. Since we have chosen 2 different slopes at first, the final definition is not a simple linear function. So I think continuity if needed.
You need something more than f(x+y) = f(x) + f(y). Continuity works but monotony too. If f(q) = Aq on the rationals and f is increasing, then for any real number x, f(x) >= sup {f(q) | q rational less than x} = Ax and in the other direction f(x)
There's no need as long as f is increasing: In case 2, g=1+Bf is increasing since B>0, and therefore g(x+y)=g(x)g(y) indeed implies that g is an exponential. I agree that it should be pointed out at some point in the proof
@@ezequielangelucci1263 with x=y=0 you get f(0)=0 then let A=f(1), by very brief inductions you can show that f(n)=An for all integers n let r be a rational number, there exists integers p,q such that r=p/q then by induction you can also show that f(qr)=qf(r). You get this equality : Ap=f(p)=f(q*r)=qf(r) therefore f(r)=Ar for all rationals r. Finally, since Q is dense in R and f is strictly increasing, all real numbers can be approached by rationnal sequences and thus f(x)=Ax holds for all real numbers Conversely, if there exists A in R such that f(x)=Ax for all x in R, you also get that f(x+y)=f(x)+f(y)
@@ezequielangelucci1263 any real number r can be expressed as a limit of a monotonic increasing sequence x_n of rational numbers. Assuming f is increasing, we get f(x_n)
@@justanotherman1114 I think this is still assuming without proof that f(x) = Ax for all rational x, and because I think that part is pretty important for understanding why this property has to hold, I'll give a proof: First of all, for any integer n and number x, f(n x) = f(x + x + ... + x) = f(x) + ... + f(x) = n f(x). (it works similarly for negative n, but you get the idea) Let x, y be in Q with x=a/b and y=c/d, and f(x) = A x, f(y) = B y for these specific x and y. Then there exists the number z = 1/(b d) with x = ad z, y = bc z. Let f(z) = C z. From all of this, it follows that: A x = f(x) = f(ad z) = ad f(z) = ad C z = C x (A - C) x = 0. Analogously, (B - C) y = 0. If x, y =/= 0, this implies A = C = B. Now define A := f(1), implying f(1) = A 1. The previous logic yields that for any rational x =/= 0, f(x) = A x. You also get f(x) = f(x+0) = f(x) + f(0) and therefore, f(0) = 0 = A 0. This proves that there exists an A for which all rational numbers x have f(x) = A x. I don't think this idea is immediately obvious, but Michael Penn was probably just using it as already known theory rather than something you can think up on the spot.
Alternative solution. Assume we can prove the differentiability of f and g. Partial differentiate the equation with respect to y, and get f'(x+y)=f(x)g'(y)+f'(y). Setting y=0 gets us f'(x)-g'(0)f(x)=f'(0). This is a standard first order linear ordinary differential equation of f with respect to x. We are done.
You can also get it from the definition of the derivative. You should be more careful with continuity and differentiability than I am doing here, but if you move f(y) to the LHS and divide by x you get [f(x + y) - f(y)]/x = f(x)/x g(y), and if you take the limit as x -> 0 you get the limit definition of the derivative of x at y (on the LHS) and at 0 (on the RHS). This becomes: f'(y) = f'(0)g(y), which is nice because that means that g is just proportional to the derivative of f. Let's call that proportionality constant m = 1/f'(0) so that g(x) = mf'(x) Plugging back into the original equation: f(x + y) = mf(x)f'(y) + f(y) Let's build a symmetric equation by swapping x and y f(x + y) = mf(y)f'(x) + f(x), and since the LHS are equal we can equate the RHS mf(x)f'(y) + f(y) = mf(y)f'(x) + f(y), moving things around we get to m [f'(x) - 1]/f(x) = m [f'(y) - 1]/f(y), we can now fix y to see that the RHS is a constant (call it a) and the LHS has only terms that depend on x. We then get the differential equation mf' - af = 1, this 1st order ODE can easily be solved with the initial conditions f(0)= 0 and f'(0) = 1/m to get: f(x) = B(c^x - 1) and g(x) = c^x (where c = a/m)
it is important that f is increasing in the case where f(x+y)=f(x)+f(y) we concluded that f(x) is linear because if there is no other conditions there are non-trivial solutions and these non-trivial solutions can only be constructed using axiom of choice. (Very deep.)
Oh, finally, the solution is neither a constant nor a linear function.
There was one time that the answer was a quadratic function. I remember noticing that a linear function did not satisfy the functional equation (I don't recall whether a constant satisfied it or not). Also, there is a linear case that works, although it is a degenerate case for when B = 0.
www.youtube.com/@user_math2023
It is not immediately obvious that g(x+y) = g(x)g(y) => g(x) = C^x.
In fact, we have g(x) = a exp(L(x)) where L is a Q-linear function on R. Plugging the result back in the formula relating f and g ensures that this function is increasing, and thus R-linear.
That's an excellent observation, but in g(x) = a exp(L(x)) the constant a must be equal to 1 in this case.
Absolutely. Just like in the first case, the fact that f is increasing must be used here. This piece of the puzzle is classic but very much important and not obvious!
I love functional equations. What are the best books to learn more about them? Could someone please suggest me? Thank you
www.youtube.com/@user_math2023
Would also like to know more about them :)
♡Functional Equations by Titu Andreescu
♡functional equations A problem solving approach book by B.J. Venkatachala
●Functional equations in mathematical olympiads by Amir Hossein Parvardi
●Introduction to functional equations by Costas Efthimiou
●Functional Equations and how to solve them by Christopher G. Small
●Lectures on functional Equations and their applications volume 19 by J. Aczèl
@@carlobenedetti2407
Your welcome bro 😊
Merry Christmas to you and your familiy, Michael. Many thanks for your inspiring maths work on your TH-cam channels. Even for mathematicians very enjoyable. My fav. topic is all about the zeta function (and L-functions / modular forms). 🎄 Chris
13:36 Good Place To Stop
13:37 Leet
This question was previously asked in 2004 Antalya Math Contest in Turkey to 2nd year high school students. It is nice to see a question here that I've just recently solved.
I had a hard time getting what he meant with "x=1 and y=x" until I got it : he actually does "x=1 and y=y" and then uses y as a dummy variable, renaming it x.
The solutions can't really be combined into one. If you try by setting C=1 in the second solution, you end up dividing by zero, and that's a no-no. Even if we ignore the division problem since it's part of the constant coefficient anyway, we still can't choose a value of C that makes f(x)=A⋅(Cˣ-1)→f(x)=Ax. Also, you probably should have pointed out that f(1) is a free variable. For reference, solution 2: g(x)=Cˣ, f(x)=A⋅(Cˣ-1) (with solution 1: g(x)≡1, f(x)=Ax).
The first substitution I did was x=dy. f(y+dy)-f(y)=g(y)f(dy). As soon as you get an equation that looks like this, you are looking to evaluate a limit, and with some tinkering and an application of LHR (and renaming variables) you get g(x)=f’(x)/f’(0). Since we know g(0) is 1, that means that f’ must either be a constant or a multiple of itself, which leads us to the exponential pair and the linear and constant pair quite nicely (I came up with B*a^x-B, but that’s essentially the same thing with a renamed coefficient where my B is the multiplicative inverse of yours)
f is not given to be continuous.
As a Lithuanian, I feel proud for this problem.
At about 5:00 you set x=1, y=x. I'm not very practiced at functional equations, so not sure what is happening there. If y=x and x=1, then isn't y=1? Maybe a different notation would be a bit more understandable, or some intermediate steps to include more detail to arrive at those substitutions. Always interesting videos!
8:11 it's not exactly obvious that f(x) = Ax is the only solution. But we can prove that if there is some value k s.t. f(k) != Ak, it will lead to contradiction.
Yeah, it is not obvious, but it is well know fact, called the Cauchy functional equation
case 1 is a limit of case 2 pretty cool.
I assumed f is C2, then what you can get is
1) [f(x+y)-f(y)]/x = f(x)/x*g(y) -> g(s) = f'(s)/f'(0). So the two functions are related.
2) [f(x+y)-f(x)]/y =f(x)[g(y)-1]/y+f(y)/y -> f'(x) = f''(0)/f'(0) f(x) + f'(0). You get either f(x)=Ax+B or f(x) = C(1-exp(rx)). As you demonstrated, both lead to valid solutions
Isn't continuity required to prove these are all the functions? I think it all comes to the functional equation f(x+y)=f(x)+f(y), right? I think if we have continuity then we can prove it has to be linear function. However without continuity I think there could be function that is partially linear, so for given real r f(q*r)=a_r*q*r for all rational qs. a_r could be different for different r but if r_1-r_2 is rational then a_r_1=a_r_2. And I think we could prove such functions exist by taking one number from each class of equivalance relation (where r1 and r2 are equivalent when their difference is rational) and we set different slopes for first 2 numbers and we use them for defining set and then for when we have all of chosen representatives of these classes in well ordered chain of length continuum, we take the representative only when it can't be written as finite rational combination q1*r1+q2*r2+... taken at some previous step and we define function using first slope. If it can then it should already be well defined, because then it has only one form and so we define using our formula. Since we have chosen 2 different slopes at first, the final definition is not a simple linear function. So I think continuity if needed.
The strictly increasing condition on f ends up being enough to show that f must be fully linear.
You need something more than f(x+y) = f(x) + f(y). Continuity works but monotony too. If f(q) = Aq on the rationals and f is increasing, then for any real number x, f(x) >= sup {f(q) | q rational less than x} = Ax and in the other direction f(x)
@@swenji9113 Yes, burk314 already pointed it out.
You should ask g to be continuous
There's no need as long as f is increasing: In case 2, g=1+Bf is increasing since B>0, and therefore g(x+y)=g(x)g(y) indeed implies that g is an exponential. I agree that it should be pointed out at some point in the proof
Can we write the last solution as f(x) = a^x + b? Looks a little cleaner?
Amazing, could you do a video about the "obvious" f(x+y) = f(x) + f(y) ?
Just put g(x)=1 in the defining property, and you'll get this equality
@@deweiter nono, my question is about why f(x+y) = f(x) + f(y) means f(x) = Ax
@@ezequielangelucci1263 with x=y=0 you get f(0)=0
then let A=f(1), by very brief inductions you can show that f(n)=An for all integers n
let r be a rational number, there exists integers p,q such that r=p/q
then by induction you can also show that f(qr)=qf(r). You get this equality : Ap=f(p)=f(q*r)=qf(r) therefore f(r)=Ar for all rationals r.
Finally, since Q is dense in R and f is strictly increasing, all real numbers can be approached by rationnal sequences and thus f(x)=Ax holds for all real numbers
Conversely, if there exists A in R such that f(x)=Ax for all x in R, you also get that f(x+y)=f(x)+f(y)
@@ezequielangelucci1263 any real number r can be expressed as a limit of a monotonic increasing sequence x_n of rational numbers. Assuming f is increasing, we get f(x_n)
@@justanotherman1114 I think this is still assuming without proof that f(x) = Ax for all rational x, and because I think that part is pretty important for understanding why this property has to hold, I'll give a proof:
First of all, for any integer n and number x, f(n x) = f(x + x + ... + x) = f(x) + ... + f(x) = n f(x). (it works similarly for negative n, but you get the idea)
Let x, y be in Q with x=a/b and y=c/d, and f(x) = A x, f(y) = B y for these specific x and y. Then there exists the number z = 1/(b d) with x = ad z, y = bc z. Let f(z) = C z.
From all of this, it follows that:
A x = f(x) = f(ad z) = ad f(z) = ad C z = C x (A - C) x = 0. Analogously, (B - C) y = 0. If x, y =/= 0, this implies A = C = B.
Now define A := f(1), implying f(1) = A 1. The previous logic yields that for any rational x =/= 0, f(x) = A x. You also get f(x) = f(x+0) = f(x) + f(0) and therefore, f(0) = 0 = A 0.
This proves that there exists an A for which all rational numbers x have f(x) = A x.
I don't think this idea is immediately obvious, but Michael Penn was probably just using it as already known theory rather than something you can think up on the spot.
You can switch x and y and compare to original equation.
I found it. It's right before the opening parenthesis.
Alternative solution. Assume we can prove the differentiability of f and g. Partial differentiate the equation with respect to y, and get f'(x+y)=f(x)g'(y)+f'(y). Setting y=0 gets us f'(x)-g'(0)f(x)=f'(0). This is a standard first order linear ordinary differential equation of f with respect to x. We are done.
Capo
i love functional equations i want to kiss them
You can also get it from the definition of the derivative. You should be more careful with continuity and differentiability than I am doing here, but if you move f(y) to the LHS and divide by x you get [f(x + y) - f(y)]/x = f(x)/x g(y), and if you take the limit as x -> 0 you get the limit definition of the derivative of x at y (on the LHS) and at 0 (on the RHS). This becomes:
f'(y) = f'(0)g(y), which is nice because that means that g is just proportional to the derivative of f. Let's call that proportionality constant m = 1/f'(0) so that g(x) = mf'(x)
Plugging back into the original equation:
f(x + y) = mf(x)f'(y) + f(y)
Let's build a symmetric equation by swapping x and y
f(x + y) = mf(y)f'(x) + f(x), and since the LHS are equal we can equate the RHS
mf(x)f'(y) + f(y) = mf(y)f'(x) + f(y), moving things around we get to
m [f'(x) - 1]/f(x) = m [f'(y) - 1]/f(y), we can now fix y to see that the RHS is a constant (call it a) and the LHS has only terms that depend on x. We then get the differential equation
mf' - af = 1, this 1st order ODE can easily be solved with the initial conditions f(0)= 0 and f'(0) = 1/m to get:
f(x) = B(c^x - 1) and g(x) = c^x (where c = a/m)
it is important that f is increasing in the case where f(x+y)=f(x)+f(y) we concluded that f(x) is linear because if there is no other conditions there are non-trivial solutions and these non-trivial solutions can only be constructed using axiom of choice. (Very deep.)
If anyone is interested: en.wikipedia.org/wiki/Cauchy%27s_functional_equation
We have f(0) = 0 being contingent on x=0, y=y, now he uses this with x=x, y=0. That can't be valid.