A Short Functional Equation Problem

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Күн бұрын

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@ktuluflux 2 ай бұрын

It felt like we were going in circles but it worked out nicely in the end! Cool problem and a neat idea to use a constant inside. Higher powers may get messy though.

@RGP_Maths 2 ай бұрын

Me, looking at the thumbnail: Intuitively, f and g are both linear functions, define some coefficients, quickly solve, but I bet Dr Barker will also prove they were linear. Me, watching the video: Yep! Well done again, Dr B.

@ronbannon 2 ай бұрын

Again, excellent job. Another gem to share with my students. Thank you!

@nathanisbored 2 ай бұрын

Functional equation videos like this are my favorite, especially when you are allowed to assume the function is sufficiently “nice” (analytical or something)

@DrBarker 2 ай бұрын

It's nice to be able to do things like differentiate, and can be frustrating if we don't know that a function is differentiable. At the same time, I find it very elegant when there is a solution which only uses very simple operations like substitution.

@nathanisbored 2 ай бұрын

@@DrBarker I misspoke! I meant to say I prefer when you **arent** allowed to assume that (I typoed and said the opposite of what I meant 🤦). I completely agree with you!

@KarlFredrik 2 ай бұрын

Solved it by assuming it's linear relations. Nice to see how it's done without guessing.

@TechToppers 2 ай бұрын

You didn't solve it if you make unproved claims

@joeistead 2 ай бұрын

Suppose we have some value A for which g(A) = 0. Then f(g(A)+y) = f(y) = A + 2y + 3. But then using the original definition of f given in the problem, f(g(x) + y) = 2(g(x)+y) + A + 3 and we can get g(x) = x/2 - A/2. Using these closed form solutions for f and g to solve for g(f(x)+y), the A terms cancel out and you're left with the desired result.

@Notthatkindofdr 2 ай бұрын

How do you know such a value of A exists?

@joeistead 2 ай бұрын

@@Notthatkindofdr I didn't. It's a bit circular to find the actual value after the fact. I questioned how rigorous I should be about a YT proof. I settled on 'conversational'... so there are holes as you point out :)

@boguslawszostak1784 2 ай бұрын

@@Notthatkindofdr g(x) is linear (You can prove it)

@micknamens8659 2 ай бұрын

Shorter path: When we look at the y handling in f(g(x)+y)=x+2y+3 if follows that f doubles it's argument (beside adding some constant c). Because of f(g(x))=x+3 this also means that g devides its argument by 2 (beside adding some constant d, which has to equal (3-c)/2 ). The expression g(f(x)+y) is hence 1/2*((2x+c)+y)+(3-c)/2=x+y/2+3/2

@irhzuf 2 ай бұрын

Can't find a neat counterexample for you, but f(x) could not satisfy f(x+y) = f(x) + f(y) so you have to prove something to get there

@micknamens8659 2 ай бұрын

@@irhzuf please be more specific about which step you mean. I don't state that f(x+y)=f(x)+f(y)

@boguslawszostak1784 2 ай бұрын

@@irhzuf In school-level mathematics, the term "linear function" is used more loosely for both forms, y=ax and y=ax+b so for y=ax+b. I prefer the term affine rather than linear., f(g(x)+y)=x+2y+3 for y=-g(x) : f(g(x)-g(x))=x+2(-g(x))+3 f(0)=(x-2g(x))+c for any x It means x-2g(x) is constant C and 2g(x) =C+x g(x) =x/2+C/2 {means that g is affine: devides its argument by 2 and then adds some constant } for y=0, f(g(x))=x+3=f(x/2+C/2) for any x for t=x/2+C/2 we get x=2t-C 2t-C+3=f(t) { means that f is affine: doubles it's argument and then adds some constant } g(f(x)+y)=g(2x-C+3+y)=(2x-C+3+y)/2+C/2=(2x+y+3)/2

@Khashayarissi-ob4yj 2 ай бұрын

With luck and more power to you. hoping for more videos.

@MrGeorge1896 2 ай бұрын

Another method which is even easier: f(g(x) + y) = x + 2y + 3 Using a substitution: g(x) + y = x y = x - g(x) f(x) = x + 2x - 2g(x) + 3 = 3x + 3 - 2g(x) 2g(x) = 3x + 3 - f(x) Using another substitution: x = f(x) + y f(x) = x - y 2g(f(x) + y) = 3x + 3 - x + y = 2x + y + 3 g(f(x) + y) = (2x + y + 3) / 2 Edit: In order to not get people confused: The "y" in the second substitution is not identical to the "y" in the original equation. You could replace this y with an arbitrary variable name e.g. "n" and end up with g(f(x) + n) = (2x + n + 3) / 2 instead...

@dipankarbanerjee1130 2 ай бұрын

Woowww!!! Thank you thank you ❤

@yitanjang 2 ай бұрын

One should be wary of such abuse of notation.

@yitanjang 2 ай бұрын

You are already wrong from the first substitution. If x from lhs and rhs of the substitution is the same, it is basically asserting g(x) = x-y, which is clearly not. Try again using a different letter, say g(x) + y = u

@MrGeorge1896 2 ай бұрын

@@yitanjang I you watch the video carefully you will noticed that Dr Barker did (virtually) the same. He set x to an arbitrary value (zero) while I set y to to an arbitrary term (x - g(x)). This method works fine but there are indeed some caveats (corner cases) which are out of scope here.

@yitanjang 2 ай бұрын

@MrGeorge1896 Ah, i reread your solution, and indeed you are not wrong. That is on my behalf. the substitution was y = x -g(x) which is completely fine to do. I'm really sorry for the fuss

@sohaib_mer- 2 ай бұрын

what i did is point out how since y isn't involved in the first equation being only an addition, i presumed the function to be linear : f(x)= 2x f(g(x)+y) = 2(g(x) +3) +2y = x+2y thus 2g(x)+3= x -> g(x) = (x-3)/2

@sohaib_mer- 2 ай бұрын

though mine is a bit of cheat , yours definitely solidified the method, great vid

@Homayoun197250 2 ай бұрын

Nice problem. Thank you.

@aymantimjicht173 2 ай бұрын

We supposed that f and defined in R. Df = Dg = R. Thank you for the answer.

@minhhainguyen1979 2 ай бұрын

❤❤❤❤❤. Thanks a lot.

@mashalrazavi579 2 ай бұрын

Tnx for nice equations

@joeistead 2 ай бұрын

Yet another solution: start by noting that from the definition of f, we get f(g(x)+Ky) = x + 2Ky + 3. Intuitively, as you move along the axis by y units, the function f rises by 2y units. This tells us that f is a line, so we can write f(x) = ax+b. Then again using intuition, the only way that f(g(x)+y) could be a line is if g is also a line, so we try g(x) = cx+d and then set about trying to find a,b,c,d. We can solve for a, c, and b+2d by using our closed form f and g to compute f(g(x)+y), which was also given as x + 2y + 3. Plugging all of this into g(f(x)+y) yields the desired result.

@DrBarker 2 ай бұрын

I like the initial argument for why f must be linear! I would write this formally e.g. by setting x = 0, so we start with f(g(0) + y) = 2y + 3, then replacing y by y - g(0), we get f(y) = 2(y - g(0)) + 3 = 2y - 2g(0) + 3 = 2y + constant, so f is linear.

@boguslawszostak1784 2 ай бұрын

@@DrBarker Wjat about this? f(g(x)+y)=x+2y+3 for y=-g(x) : f(g(x)-g(x))=x+2(-g(x))+3 f(0)=(x-2g(x))+c for any x It means x-2g(x) is constant C and 2g(x) =C+x g(x) =x/2+C/2 {means that g is affine: devides its argument by 2 and then adds some constant } for y=0, f(g(x))=x+3=f(x/2+C/2) for any x for t=x/2+C/2 we get x=2t-C 2t-C+3=f(t) { means that f is affine: doubles it's argument and then adds some constant } g(f(x)+y)=g(2x-C+3+y)=(2x-C+3+y)/2+C/2=(2x+y+3)/2

@maklovitz 2 ай бұрын

Id like to see this kind of problem but with double nested function, e.g. f(g(h(x) + y)))

@DrBarker 2 ай бұрын

This is a really interesting idea! I wonder if we'd need to include 3 variables in order for there to be a unique solution from a single functional equation?

@boguslawszostak1784 2 ай бұрын

f(g(h(x) + y)))=ax+by+c for y=-h(x) f(g(0)))=ax-bh(x)+c=C h(x)=(ax+c-C)/b for any x for y=0 f(g(h(x))))=ax+c=f(g((ax-C+c)/b)) for t=(ax-C+c)/b x = (b t - c + C)/a f(g(t)=a (b t - c + C)/a+c=bt+C So, you only know that the composition of functions f and g is an affine function, and you don’t even know whether the function ff is one-to-one. For example, g(x)=atan(x) and f(x)=3tan⁡(x)+4 give an affine function when composed."

@boguslawszostak1784 2 ай бұрын

@@DrBarker The problem is not that we have too few variables, but rather that there are no conditions that the composition of functions f and g should satisfy.

@Nishaan478 2 ай бұрын

can you some bmo1/2 problems please

@aisolutionsindia7138 2 ай бұрын

think f,g are unique as well: f(x) is 2x and g(x) is (x+3)/2

@bartoloeldelaflauta8571 2 ай бұрын

what about f(x)=2x+3 and g(x)=x/2

@aisolutionsindia7138 2 ай бұрын

@@bartoloeldelaflauta8571 yes there was an error in my calculation.. the correct form seems to be f(x)=2x+k g(x)=x/2+3/2-k/2

@AlperenK. 2 ай бұрын

Ooooooh nice!

@MrRyanroberson1 2 ай бұрын

f(g(x)+y-g(x)) = x+2(y-g(x))+3; f(y)-2y = x-2g(x)+3; either side must be constant relative to the other, so choose x=0, y=0. f(0)=3-2g(0). Also g(x) = (x-f(0)+3)/2. g(f(x)+y) = f(x)/2+y/2-f(0)/2+3/2. f(x)-x = f(0), so f(x)-f(0) = x, and we get g(f(x)+y) = (x+y-f(0)+3)/2. For g, -2g(0)=y-2g(y); g(y)-g(0) = y/2: g(f(x)+y)=y/2+f(x)/2+g(0); x-f(0)+3=f(x)+2g(0); 2f(0)+2g(0)=3, f(0) must be 0, making g(0) 3/2. g(x)=x/2+3/2 and f(x)=x

@MrRyanroberson1 2 ай бұрын

I might have dropped a factor of 2 somewhere, i see now that f and g are at least double these.

@tgeofrey 2 ай бұрын

Very hard operators

@Ricardo_S 2 ай бұрын

I don't think what I'm going to write is very rigorous, but oh well... f(g(x)+y)=x+2y+3=2(x÷2+y+3÷2)= 2(((x+3)÷2)+y) g(x)=(x+3)÷2 2(g(x)+y) f(x)=2x g(f(x)+y)=((2x+y+3)÷2)

@Notthatkindofdr 2 ай бұрын

Where did the g(x)=(x+3)÷2 come from? How do you know that's true?