A Functional Equation Problem

[@kehrierg]

I got there a different way. Since the goal number was f(72) and 72 prime factorizes into powers of 2 and 3, I thought to work out f(2) and f(3) and then build up to f(72). I used 108 = 18×6 along with the givens to deduce f(6) = 4/3. Then, 18=6×3, along with the givens, leads to conclude f(3)=-1/3. Then using 6=3×2, we get f(2)=2/3. Since I have f(18) already and since 72=4×18, I decided to find f(4), by writing f(4) = 2f(2) + 2f(2) = 4f(2), so f(4)= 8/3. Then f(72) = 4f(18) + 18f(4) = 56.

[@RGP_Maths]

Yes, very similar to my strategy. I defined a=f(2), b=f(3), found from the given information that 9a+12b=2 & 108a+108b=36, solved simultaneously to get 3a=2, 3b=-1, then found f(8)=8, f(9)=-2, so f(72)=72-16 = 56.

[@alipourzand6499]

This functional equation is very interesting since if we consider that x and y are functions, f will act as differential operator and the following equation can be used for the general method: f(x^n) = nx^(n-1) f(x) Another sunny friday!!!

[@wesleydeng71]

Question. Can we prove that this function is consistent, meaning, for example, there are many ways to calculate f(72) but all approaches will reach the same result?

[@jursamaj]

I suspect it is. Using the video's values for 108 and 36, I derived f(3)=-1/3. Putting this together with f(2)=2/3, I got to f(6) via 2*3, and got the same 4/3 that Dr Barker got via f(108) and f(18). I went on to derive several values, leading to some with up to 4 ways to get one value from pairs of other values, and it was always consistent. However, I don't think f() can be defined in any meaningful way over the reals, or even the rationals, as you can *only* derive values for numbers of the form (2^a)*(3^b) [note: a and b *can* be any integer divided by any power of 2, but I haven't delved far into *that* rabbit hole]. In any case, the graph of known values of f() is a weird exponential sawtooth wave. For what it's worth, here are the 3 equations I got for proceeding from known values of the function to other values: f(x),f(y)->f(xy): f(xy)=xf(y)+yf(x) f(x),f(y)->f(x/y): f(x/y)=(f(x)-(x/y)f(y))/y f(x)->f(√x): f(√x)=f(x)/2√x It follows from the 2nd equation and any x=y>0 that f(1)=0.

[@jursamaj]

Followup: For {a, an integer divided by a power of 2}, f(2^a)=(2^a)*a/3. No, you can't use a=log2(3) to get values for powers of 3. I tried that. Likewise, for {b, an integer divided by a power of 2}, f(3^b)=-(3^b)*b/9. Putting these together, f((2^a)(3^b))=2^a*-(3^b)*b/9+3^b*(2^a)*a/3=(2^a)(3^b)(a/3-b/9). I still don't see any way to escape these powers of 2 and 3 into the reals generally. Especially since the sign of the result is determined by (a-b/3), and in principle, you could form any x from different real powers of 2 and 3, thus yielding different signs for the same function input. That's why (a=log2(3), b=0) yield a different result from (a=0, b=1).

[@zyklos229]

You mean, if the functional equation leads to a well-defined function? Isn't this a math proof like "solution exists AND is unique", where in the end, you not even get the solution 🤕 I would try with first terms of 2D Taylor Series + go to bed and cry

[@jursamaj]

@@zyklos229 What Wesley meant by consistent is that you could calculate f(72)=2f(36)+36f(2) or f(72)=3f(24)+24f(3), and do those yield the same value. I haven't proven it, but it appears that it does. What I found is that the functional equation is only valid for a certain set of numbers, not all real numbers. You'd never be able to use Taylor series on it.

[@YouTube_username_not_found]

Indeed! One has to prove that the function defined in the problem exist and then find all functions. There could be some other functions with the same properties but each give a different value for f(72). By the way, there was a comment that proved the existence and uniqueness of f.

[@MasterofTheDucks]

Is that a new shirt?

[@DrBarker]

Well-spotted! 😂😂😂

[@MasterofTheDucks]

@@DrBarker thanks

[@ShubhayanKabir]

Let G(x) = f(x)/x If we divide the given identity throughout by xy we get, f(xy)/xy = f(x)/x + f(y)/y => G(xy) = G(x)+G(y) G(18) = 1/9 G(108) = 1/3 G(108) = G(18)+G(6) => G(6) = G(108)-G(18) = 2/9 Similarly, G(3) = G(18)-G(6) = -1/9 G(2) = 1/3 G(72) = 2G(2) + G(18) = 7/9 F(72) = G(72) *72 = 7/9 * 72 = 56

[@RGP_Maths]

@@ShubhayanKabir Beautiful solution! Thank you.

[@bjornfeuerbacher5514]

G(xy) = G(x)+G(y) should imply that G(x) is a logarithm... but I don't manage to figure out the base?!

[@ShubhayanKabir]

​@@bjornfeuerbacher5514that is not necessarily true, for one G(x) would have had to be a differentiable function if it were to be logarithmic, which in this case isn't mentioned in the first place.

[@PrimordialOracleOfManyWorlds]

doc, you are the super rock star of math. i am subbed.

[@swarley2500]

Thank you. A lot of competitions tend to include these types of problems, but it’s difficult to find resources and solutions.

[@isavenewspapers8890]

Who else immediately thought of the product rule for differentiation? It doesn't make sense in this context, since f maps numbers to numbers and not functions to functions, but there is a striking resemblance.

[@9WEAVER9]

Woah. That's appealing. But what if x and y are parametric functions of another variable t ?

[@cubed3942]

I noticed a rule that shows an even more striking resemblance to differentiation. f(x^n)= n*x^(n-1)*f(x) weird huh

[@isavenewspapers8890]

⁠​⁠​⁠@@9WEAVER9You mean "what if they were"? Because as things are right now, I'm pretty sure they're not. The first equation is just saying, "If you give me any pair of numbers x and y in this function's domain, you'll get a true equality here." Speaking of which, I'm not exactly sure what the domain and codomain of this function are supposed to be. The set of integers? The set of real numbers? The set of complex numbers? I mean, if you want a surjection from A to A^2, where A is one of those sets, you can certainly manage. Said surjection might have to be weird as hell, but it can exist. For a surjective function g defined by g(t) = (x(t), y(t)), you can get any pair of values for x(t) and y(t) you want by just plugging in the right value for t. As for my answer to your question... well, I'm not sure what to say. This is one of those cases where a question beginning with "what if" leaves the intent unclear to me. Like, I don't know what you wanted.

[@isavenewspapers8890]

@@cubed3942 You mean "what if they were"? Because as things are right now, I'm pretty sure they're not. The first equation is just saying, "If you give me any pair of numbers x and y in this function's domain, you'll get a true equality here." Speaking of which, I'm not exactly sure what the domain and codomain of this function are supposed to be. The set of integers? The set of real numbers? The set of complex numbers? I mean, if you want a bijection from A to A^2, where A is one of those sets, you can certainly manage. Said bijection might have to be weird as heck, but it can exist. For a bijective function g defined by g(t) = (x(t), y(t)), you can get any pair of values for x(t) and y(t) you want by just plugging in the right value for t. As for my answer to your question... well, I'm not sure what to say. This is one of those cases where a question beginning with "what if" leaves the intent unclear to me. Like, I don't know what you wanted.

[@LillianRyanUhl]

See my comment; it's not just a resemblance though it is a bit of an esoteric formulation

[@lukevideckis2260]

easy to follow

[@elephantdinosaur2284]

Nice video. It's pretty nice to explore what happens in Z (mod n) where you get different solutions. Also how did you pick the numbers? With numbers like 18, 36, 72, 108 I thought the answer was leaning towards 54 not 56 haha.

[@DrBarker]

When picking the numbers, it was important that it was actually possible to get from the first two numbers to the target numbers. The numbers 18, 108, and 72 are all made up of 2s and 3s, which is a nice way of ensuring that the problem is solvable. From there, it was a matter of choosing values for f(18) and f(108) which give a nice answer, and also give fairly nice numbers along the way for f(36) etc.

[@mathmachine4266]

f(x) isn't continuous, unfortunately. If it was, it would be of the form f(x)=C*x*ln(x), where C is some constant. In this case, though, f(x) is defined for all x=2^a*3^b, where a and b are rational numbers. Specifically, f(x)=x*(3a-b)/9 How I obtained this result: First, the continuous form, f(x)=C*x*ln(x). Simply, I realized that f(x²)=2xf(x). Then that f(x³)=x²f(x)+xf(x²)=x²f(x)+2x²f(x)=3x²f(x). Eventually, I noticed that f(x^n)=n*x^(n-1)*f(x). I then proved it by induction, f(x^(n+1))=xf(x^n)+x^nf(x)=nx^nf(x)+x^nf(x)=(n+1)x^nf(x). From there, if you try applying in reverse to find f(√(x)), f(³√(x)), or applying it in reverse and then forward to find f(x^(p/q)) for integer p and q, you'll find this relation holds for all rational powers of x. Therefore, if f(x) was continuous, this relation would have to hold for all real powers of the given x. So, if f(x^n)=nx^(n-1)*f(x), then, applying x^n=y, f(y)=log_x(y)*x^(log_x(y)-1)*f(x) = ln(y)/ln(x)*y/x*f(x). If f really is continuous, then this works for all positive x, and so this just boils down to f(y)=y*ln(y)*C. Or, in terms of x, f(x)=C*x*ln(x). However, f(x) isn't continuous, since there's no C which satisfies both f(18)=2 and f(108)=36. However, the fact still remains that, if you know f(x), you know f(x^q) for any rational q. f(2)=2/3, so f(2^a)=C*2^a*a. a=1, 2/3=C*2*1, C=1/3, f(2^a)=a*2^a/3. f(6)=3f(2)+2f(3), 4/3=2+2f(3), f(3)=-1/3. f(3)=-1/3, so f(3^b)=C*3^b*b. b=1, -1/3=C*3*1, C=-1/9, f(3^b)=-b*3^b/9 f(2^a*3^b)=2^a*f(3^b)+3^b*f(2^a)=2^a*(-b*3^b/9)+3^b*(a*2^a/3)=2^a*3^b*(a/3-b/9)=2^a*3^b*(3a-b)/9. So, for all x=2^a*3^b with rational a and b, f(x)=x(3a-b)/9.

[@firstnamelastname307]

sorry to ask : what do you mean by : a functional equation ? apart from that : great content as usual !

[@DrBarker]

Thank you! That basically means an equation which we might solve to find a function, like how f(xy) = f(x) + f(y) has a solution f(x) = log(x), rather than the usual solving an equation to find the value of a variable like 2x + 1 = 5.

[@Latronibus]

Working sorta systematically (elementary functional equation problems are never especially systematic), f(72)=8f(9)+9f(8)=48f(3)+18f(4)+36f(2)=48f(3)+108f(2), so it suffices to get those two. By factoring 108 as 18x6, the two givens get you f(6); by factoring 18 as 6x3, one of the givens together with what you just got gets you f(3); by factoring 6 as 3x2, the two things you just got get you f(2). Then you just back substitute.

[@stenzenneznets]

Does this function f(x) have a graph?

[@mistycremo9301]

I'm no trained mathematician, but I'm looking into it. I suspect that the domain is only verifiable over 18^a x 108^b, which has some very interesting holes in it

[@KSignalEingang]

As other commenters have pointed out, from the information given we can only find values where x and y are both of the form (2^n × 3^m). I'm postulating here that n & m have to be integers, but maybe I'm missing some clever trick that would allow us to extend things further [edit: this is not correct, we can for example pretty easily determine what f(√x) is if we know f(x) but I believe we're still constrained to a very limited subset of the real numbers, in which case what follows is still valid]... Assuming that, you could never uniquely determine, say, f(5). Or to put it another way, there are infinitely many functions that would fit the initial statement, each with their own values for all of the primes apart from 2 and 3. So, as far as I can see, f(x) is not well defined for most x, and is not analytic (by which I mean, continuous and differentiable). Your graph would probably consist of a selection of discrete points at certain rational values of x [edit: plus the square roots of those values, plus the roots of those roots, etc], with very large jumps in the value of f(x) between neighboring values of x.

[@zyklos229]

Not sure, if the "I know how to factorize integers" really gives all solutions. Have the impression this is a system of equations in the variables, with f in a special role, being able to build more equations. Depending on the angle which depends on what. Seems to be something between interpolation and ordinary differential equations, plus the algebraic part that x and y is interchangeable. Fascinating, also the fact that no matter how you divide 18 f.e. it MUST fit the equation, the trick is, to always get rid of at least one f on the rhs to determine another f, question is if you always reach the the end of the way 🤔

[@juandesalgado]

Fun, thanks! Question: regarding your generic method at the end of the video, suppose the initial conditions were, instead, f(18) = 2 and f(72) = 56, and you were asked for f(108) (the solution should be 36). The problem is that the system of equations now gives fractional exponents - that is, 108 = 18^(5/4) . 72^(1/4). I'm stuck trying to continue from there (when using a generic method of solution, not a special manipulation for this specific initial case). Ideas?

[@DrBarker]

We could find f(18^5 * 72) =: f(x), then find f(sqrt(x)) and f(sqrt(sqrt(x))) like how we went from f(36) to f(6). I think this approach can be generalised to finding nth roots, e.g. to find the nth root of x = a^n, breaking it down into lots of steps like a^n = a^{n - 1} * a, a^{n - 1} = a^{n - 2} * a, ...

[@juandesalgado]

@@DrBarker Thanks, will give it a try!

[@SuperCityscan]

If g(x) is f(x)/x Then the equation simplifies to g(xy) = g(x) + g(y) The only function I know who does this is logarithm so g(x)=a*logb(x) is an answer where b is the base of the logarithm so f(x) = a*x*logb(x), but we can get rid of "a" if we choose a different base so only f(x) = x*logb(x) should be good which is just f(x) = x*ln(x)*(1/ln(b)) and that makes it just f(x) = c*x*ln(x) but let's just use "a" again We know f(18) = 2 and f(108) = 36 so supposedly 2 = a*18*ln(18) so a = 0.0384, but 36=a*108*ln(108) makes a = 0.0711 So are there more functions that have the g(xy) = g(x) + g(y) behavior or there are other ways to combine logarithm functions to have more than one variable?

[@Notthatkindofdr]

Yes there are infinitely many functions satisfying g(xy) = g(x) + g(y) if you assume the Axiom of Choice, but all the non-logarithmic ones are discontinuous everywhere. See en.wikipedia.org/wiki/Cauchy%27s_functional_equation.

[@9WEAVER9]

Dr. Barker for president 2024

[@mekbebtamrat817]

No he has better things to do. Fermat became a mathematician for a reason

[@RGP_Maths]

Pierre de Fermat was a lawyer (and a mathematician in his spare time). Having another job didn't stop him achieving a great deal in mathematics!

[@9WEAVER9]

​@@mekbebtamrat817 Dr. Barker would use his math and skills to do great things to make more time for wonderful math

[@mathunt1130]

Can you solve the equation? You know f(x^2)=xf(x). From this f(0)=0.

[@KSignalEingang]

Assuming you're still referring to the system of equations in the video and not a whole other problem, your extrapolation isn't quite right. It should be f(x²) = 2xf(x) Also, while the definition given implies f(1) = 0, since f(x) = f(x) + xf(1), it does not imply the same for f(0), which *could* be 0, but could just as easily be infinity, since either would satisfy f(x × 0) = xf(0) for all x.

[@ald6980]

Everything is ok but I am afraid that function f does not exist. You can calculate f(x^n), than f(x^(1/n)) and f(x^(m/n)). And get a problem: f(x^(kn/n))!=f(x^k).

[@sil1235]

Many such functions f exist, for example f(x)=x*(v2(x)/3-v3(x)/9). Here v2(x),v3(x) are exponents of 2 (resp. 3) in prime factorization of x. That is, if the domain of f are integers (would probably work on rationals too using padic valuation).

[@ald6980]

Exotic domains should be declared :)))

[@Notthatkindofdr]

Why do you think those two expressions in the last line are not equal? Doing the calculations you indicated shows that f(x^r)=rx^(r-1)f(x) for all rational r and all x. So both sides of f(x^(kn/n))=f(x^k) are equal to kx^(k-1)f(x). I am not seeing the "problem".

[@maxvangulik1988]

f(18)=2 18=6•3 3f(6)+6f(3)=2 6=2•3 2=6f(3)+3(2f(3)+3f(2)) 12f(3)+9f(2)=2 f(108)=36 108=6•18 36=12+18(2f(3)+3f(2)) 4=6f(3)+9f(2) 2=12f(3)+9f(2) 2=-6f(3) f(3)=-1/3 f(2)=2/3 f(6)=-2/3+2=4/3 f(36)=2•6•f(6)=48/3=16 f(72)=32+24=56

[@samyachakraborty263]

we are getting these for free

[@9WEAVER9]

Attention span is the cost to ride

[@Pierumi]

Bo'a of' math equations

[@sil1235]

Overkill solution. Dividing the equation by x*y for non-zero x,y gives f(xy)/xy = f(y)/y+f(x)/x, so g(x)=f(x)/x satisfies the Cauchy's logarithmic functional equation g(x*y)=g(x)+g(y). If we restrict on positive integers only then it can be shown that the solution must be of form g(1)=0 and g(x)=v_p1(x)*r(p1)+...+v_pk(x)*r(pk) where r is any function on positive integers and v_p(x) is an exponent of p in prime factorization of x. So, f(x)=x*(v_p1(x)*r(p1)+...+v_pk(x)*r(pk)) and it only matters how the function behaves on primes. In your two cases we have 18=2^1*3^2 and so 2=f(18)=18*(r(2)+2*r(3)). Similarly from 108=2^2*3^3 we have 36=f(108)=108*(2*r(2)+3*r(3)). This system of two linear equations in two variables solves to r(2)=1/3, r(3)=-1/9. Finally, 72=2^3*3^2 and so f(72)=72*(3*(1/3)+2*(-1/9))=56. Ofc there are infinitely many functions f based on how they behave on other primes, but their values are same on numbers 2^a*3^b, and that is all we need for this problem.

[@Notthatkindofdr]

Assuming the Axiom of Choice, there are infinitely many functions f defined on the real numbers which satisfy the functional equation and given conditions. As you and @mathmachine4266 have pointed out, the given conditions determine the value of f(x) for x=2^a*3^b where a and b are rational, but for other real values of x there are infinitely many other possibilities. That is because Cauchy's logarithmic functional equation has infinitely many possible solutions by using the Axiom of Choice; however, all but one of these solutions are nowhere continuous, including the one in the video. See en.wikipedia.org/wiki/Cauchy%27s_functional_equation.

[@incription]

now try f(f(x)) = x^2 + c

[@YouTube_username_not_found]

Go to the following video and you will find in the comment made by md2perpe the general method to solve the general problem of finding an f that satisfies f(f(x)) = g(x) where g is an arbitrary given function. Video title: A Curious Functional Equation | Math Olympiads Channel: SyberMath

[@jinhuiliao1137]

(uv)'=uv'+vu'