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.
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.
Functional equation videos like this are my favorite, especially when you are allowed to assume the function is sufficiently “nice” (analytical or something)
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.
@@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!
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 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 :)
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 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
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...
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
@@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.
@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
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
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.
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.
@@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
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?
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."
@@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.
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
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)
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.
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.
Again, excellent job. Another gem to share with my students. Thank you!
Functional equation videos like this are my favorite, especially when you are allowed to assume the function is sufficiently “nice” (analytical or something)
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.
@@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!
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.
How do you know such a value of A exists?
@@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 :)
@@Notthatkindofdr g(x) is linear (You can prove it)
Solved it by assuming it's linear relations. Nice to see how it's done without guessing.
You didn't solve it if you make unproved claims
Nicely done!
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
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
@@irhzuf please be more specific about which step you mean. I don't state that f(x+y)=f(x)+f(y)
@@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
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...
Woowww!!! Thank you thank you ❤
One should be wary of such abuse of notation.
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
@@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.
@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
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
though mine is a bit of cheat , yours definitely solidified the method, great vid
With luck and more power to you.
hoping for more videos.
We supposed that f and defined in R. Df = Dg = R.
Thank you for the answer.
Nice problem. Thank you.
❤❤❤❤❤. Thanks a lot.
Tnx for nice equations
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.
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.
@@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
Ooooooh nice!
can you some bmo1/2 problems please
Id like to see this kind of problem but with double nested function, e.g. f(g(h(x) + y)))
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?
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."
@@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.
Very hard operators
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
I might have dropped a factor of 2 somewhere, i see now that f and g are at least double these.
think f,g are unique as well: f(x) is 2x and g(x) is (x+3)/2
what about f(x)=2x+3 and g(x)=x/2
@@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
{x+x ➖ }+{4y+4y ➖}+{3+ 3➖ }={x^2+8y^2+6}=14xy^4 2^7xy^4 2^7^1xy^4 2^1^1xy^2^2 1xy^1^2 1xy^2(xy ➖ 2xy+1). f(g+g ➖)( x+x ➖)+(y+y ➖) =14xy^2 .
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)
Where did the g(x)=(x+3)÷2 come from? How do you know that's true?
Man try this , f(g(x) +y)= x+2y+3,
f(g(x) +y)=2((x+3)/2 +y )
g(x) =( x+3)/2, and f(x)=2x
g(f(x)+y)= (2x+y+3)/2
Give me some credit