f(2ˣ+1) = 4ˣ; f(x) = ? Well, what operations would you perform on 2ˣ+1 to turn it into 4ˣ? It should be easy to see that all that needs to be done is to subtract 1, and square: [(2ˣ+1) - 1]² = (2ˣ)² = 4ˣ So f(x) = (x - 1)² = x² -2x + 1 Fred
Functional equations of the type f[g(x)] = h(x) are only easy if g is right-cancellable. If g is not right-cancellable, then the equations are much more difficult, and much more interesting. In this channel, though, every video I have seen using this type of functional equation always uses an example where g is right-cancellable. The equations of the type f°g = h°f are certainly more difficult, as cancellability is not enough, but in general, these are not too difficult either if you know the functional properties of h and g.
I first substituted x with log u (base 2), then I substituted u with t-1. Finally it got simplified and I replaced t with x. My method was unnecessarily long.
You forgot the interpretation range, when the original x is real, then the original function isn't interpletable in every real number, the solution is f(x)=(x-1)^2, x>1. Sorry for my english it isn't perfect.
@@erikkonstas 1 is not in the domain of definition of f. You can continuously extend f, surely, but there is no need to declare such a continuous extension to be a solution to the equation, since _all_ extensions, continuous or not, are solutions to the equation. The only part of the domain that actually matters for the functional equation is (1, ♾). This is why domain and codomain matter when defining a function. Saying f(x) = y is not enough.
@@erikkonstas You are correct, but the type of functional equation we are dealing with is for functions f : X -> Y, where X and Y are subsets of real numbers. If we were to solve for f : X -> Y where X and Y are subsets of complex numbers instead, then you are certainly correct that this makes a difference. In this case, let g : C -> C\{1} such that g(z) = 2^z + 1, and let h : C -> C\{0} such that h(z) = 4^z. Thus, the challenge is to find all the functions f : C\{1} -> C{0} such that f°g = h. Here, notice that g has infinitely many right inverses, of the form g[m]^ρ = ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2). Thus, we expect that f[m]°g = h implies f[m] = h°g[m]^ρ, and notice that (h°g[m]^ρ)(z) = 4^{ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2)} = exp[2·ln(2)·{ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2)}] = exp[2·ln(|z - 1|) + 2·arg(z - 1)·i + 4·m·π·i] = (z - 1)^2. Therefore, f[m](z) = (z - 1)^2 for all integers m. Therefore, f[m] = f, and so f : C\{1} -> C\{0} with f(z) = (z - 1)^2 is the unique function satisfying f°g = h. Notice, though, that even though 1 is a singularity of f, it is a removable singularity. However, 1 is also irrelevant for the functional equation, which is why it is excluded from the domain.
We can also de it as follows: For x=1, f(3)= 4 = (2-1)² For x=2, f(5)= 16 = (5-1)² For x=3, f(9)= 64 = (9-1)² From the above pattern, we observe: f(x)= (x-1)²
This one could easily be done even by heart 👍 By the way when you say something like (x-1)², instead of saying the clumsy "x minus 1 quantity square" why not just simply say "x minus 1 all square"?
Brother , u are a great person , your mathematical problems are really nice and important , can u pls suggest books on number theory ? Pls , for Olympiad .
Well what he found out was f(t) but since it is a variable it doesn't matter what you call it. For the convention of functions it is named f(x) just like how we do with inverse of functions, you can use f(t) or f(u) or better yet, in the original question use y instead of x
When he says "It's not the same x" he means in f(z)=(z-1)^2 the letter z is essentially a dummy variable so we could replace it with any letter we want (he chose x as it's a nice letter for equations). It's not relating it to the equations before where we used 'x' as a letter though.
@@karnavthakur5868 You have to specify that x is a dummy variable. In fact, technically, these equations can be solved functionally, without ever using dummy variables in the algebra.
Let g : R -> (1, ♾) such that g(x) = 2^x + 1 and h : R -> (0, ♾) such that h(x) = 4^x = (2^x)^2. Notice that sqrt°h + 1 = g, which is equivalent to h = sqrt^(-1)°(g - 1), so let φ : (1, ♾) -> (0, ♾) such that φ(x) = sqrt^(-1)(x - 1) = (x - 1)^2. Hence h = φ°g. The challenge is to find some f : (1, ♾) -> (0, ♾) such that f°g = h. Since h = φ°g, f°g = φ°g. g is right- cancellable, since it is an exponential function, and is thus a bijective function. Therefore, f = φ. Therefore, f(x) = (x - 1)^2 for all x in (1, ♾).
Let y = 2^x +1 Notice (y-1) = 2^x and (2^x)^2 = 4^x So f(y) = (y-1)^2. This only works though on the domain of y. Since 2^x>0, y>1. So f(x) = (x-1)^2 for x>1, and you have no information about f(x) for x
The 1st method sucks. I thought your 2nd method would be the 1st method and optionally as a second method, if one does not immediately realize that: 4^x = (2^x)^2 = [(2^x + 1) -1]^2 == (z-1)^2, to solve for x, 2^x + 1=z using log2 and set the result in 4^x😉 I.e., x=log2(z-1) and f(z)=4^[log2(z-1)] = (z-1)^2
Ah BANANAS in action! f(x)=(x-1)/2 say, f(1) = (1-1)/2=0 f(2) = (2-1)/2=1/2 ...etc... f(🍌)=(🍌-1)/2 f( 2🍌) = (🍌+🍌)/2 - 1/2 = 🍌-1/2 this is called the _banana principle_ , very useful in inverse functions, e.g. let f(x) = (x-1)/2 what is the inverse function g(x) such that g(f(x))=f(g(x)) = x of course? f(g) = (g-1)/2 by the banana principle so x = (g -1)/2 g -1 = 2x g(x) = 2x+1 the inverse to f(x).😲
f(2ˣ+1) = 4ˣ; f(x) = ?
Well, what operations would you perform on 2ˣ+1 to turn it into 4ˣ? It should be easy to see that all that needs to be done is to subtract 1, and square:
[(2ˣ+1) - 1]² = (2ˣ)² = 4ˣ
So f(x) = (x - 1)² = x² -2x + 1
Fred
Method 1 is more like playing around.and finding a solution by coincidence. Method 2 is what you see at a glance.
Hello there Wolfgang, consider look to my channel too for similar math olympiad problems. Thanks and regards.
Functional equations where f(g(x)) = h(x) are generally easy. The mainly difficult ones are of the form f(g(x)) = p(f(x)).
I agree. Do you have an example of the second type?
Functional equations of the type f[g(x)] = h(x) are only easy if g is right-cancellable. If g is not right-cancellable, then the equations are much more difficult, and much more interesting. In this channel, though, every video I have seen using this type of functional equation always uses an example where g is right-cancellable.
The equations of the type f°g = h°f are certainly more difficult, as cancellability is not enough, but in general, these are not too difficult either if you know the functional properties of h and g.
@@SyberMath One I can think of right now is f(2x+1) = 2f²(x) - f(x)³
I used the second method, but the first is cool too!
Hi Rohan! consider look to my channel too for similar math olympiad problems. Thanks and regards.
I first substituted x with log u (base 2), then I substituted u with t-1. Finally it got simplified and I replaced t with x. My method was unnecessarily long.
Set 2^x = y and you know why.
You'll got f(y + 1) = y^2
Replace y with x - 1 to got f(x) = (x - 1)^2 and we've done. Easy 😂
Ngl, I didn't even think about that, that's good 🍺🍺🍻.
I did it in a way similar to your 2nd, i subbed y=2^x and got f(y+1)=y^2, then subbed y with x-1 and got f(x)=(x-1)^2
Yeah, this is how I did it, too. It's pretty much the same as method #2, only you & I did it in two baby steps, whereas he did it in one.
Please specify the domain and Co-domain for these types of problems. Otherwise it's just incomplete
Substitution.
Let u = 2^x + 1. Then x = lb(u-1) (lb is base-2 logarithm)
So f(u) = 4^lb(u-1)
= 2^(2lb(u-1))
= (u-1)^2
So f(x) = (x-1)^2
Yep, that's how I did it as well.
Could be another method substitute 2^x with t and do a reverse translation and then replace t with x?
Sure
great methods. really awesome
Glad you think so!
@@SyberMath :)
i first replaced x with log2(x) (log base 2 x) expressed 4 as 2² and got the equation f(x + 1) = x² again put x as x - 1 and i got f(x) = (x - 1)²
Same here
You forgot the interpretation range, when the original x is real, then the original function isn't interpletable in every real number, the solution is f(x)=(x-1)^2, x>1.
Sorry for my english it isn't perfect.
Even with complex numbers I think this doesn't define f(1).
@@erikkonstas 1 is not in the domain of definition of f. You can continuously extend f, surely, but there is no need to declare such a continuous extension to be a solution to the equation, since _all_ extensions, continuous or not, are solutions to the equation. The only part of the domain that actually matters for the functional equation is (1, ♾). This is why domain and codomain matter when defining a function. Saying f(x) = y is not enough.
@@angelmendez-rivera351 I meant if x isn't necessarily real, then 2^x could be negative as well but not 0.
@@erikkonstas You are correct, but the type of functional equation we are dealing with is for functions f : X -> Y, where X and Y are subsets of real numbers.
If we were to solve for f : X -> Y where X and Y are subsets of complex numbers instead, then you are certainly correct that this makes a difference. In this case, let g : C -> C\{1} such that g(z) = 2^z + 1, and let h : C -> C\{0} such that h(z) = 4^z. Thus, the challenge is to find all the functions f : C\{1} -> C{0} such that f°g = h. Here, notice that g has infinitely many right inverses, of the form g[m]^ρ = ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2). Thus, we expect that f[m]°g = h implies f[m] = h°g[m]^ρ, and notice that (h°g[m]^ρ)(z) = 4^{ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2)} = exp[2·ln(2)·{ln(|z - 1|)/ln(2) + [arg(z - 1) + 2·m·π]·i/ln(2)}] = exp[2·ln(|z - 1|) + 2·arg(z - 1)·i + 4·m·π·i] = (z - 1)^2. Therefore, f[m](z) = (z - 1)^2 for all integers m. Therefore, f[m] = f, and so f : C\{1} -> C\{0} with f(z) = (z - 1)^2 is the unique function satisfying f°g = h. Notice, though, that even though 1 is a singularity of f, it is a removable singularity. However, 1 is also irrelevant for the functional equation, which is why it is excluded from the domain.
Here is a very simple method
First write 4^x = (2^x)^2
Then replace x by log2(x-1)
Then we get f(x+1-1) = (x-1)^2
We get f(x) = (x-1)^2
Both methods are very nice. Thank you.
Most welcome 😊
Where is the case for x≦1 as a domain?
Method 1 is more logical 👍👍
Thanks for the feedback, bro! 🙂
we only define f for x going from 1 to +inf. So... Idk (2^x+1>1)
We can also de it as follows:
For x=1, f(3)= 4 = (2-1)²
For x=2, f(5)= 16 = (5-1)²
For x=3, f(9)= 64 = (9-1)²
From the above pattern, we observe: f(x)= (x-1)²
Nice video thanks
the first method is perfect.Thanks
Welcome 😊
Brilliant.
We can only know f(x)=(x-1)2 when x >1. We know nothing about f(x) if x
This one could easily be done even by heart 👍
By the way when you say something like (x-1)², instead of saying the clumsy "x minus 1 quantity square" why not just simply say "x minus 1 all square"?
or x minus one whole square
@@ashokkhullar6650
Exactly!
in 10 sec, just substituate (as you already learned us) y for 2^x+1 and you get f(y)=(y-1)^2 and so on...
What is the domain?
(1, ♾) (open interval)
Brother , u are a great person , your mathematical problems are really nice and important , can u pls suggest books on number theory ? Pls , for Olympiad .
I did this in seconds. Basically method 2.
Əla hllə görə təşəkkürlər.Bakıdan salamlar.
You keep saying that x is not the same x but nerver explain the difference
Well what he found out was f(t) but since it is a variable it doesn't matter what you call it. For the convention of functions it is named f(x) just like how we do with inverse of functions, you can use f(t) or f(u) or better yet, in the original question use y instead of x
When he says "It's not the same x" he means in f(z)=(z-1)^2 the letter z is essentially a dummy variable so we could replace it with any letter we want (he chose x as it's a nice letter for equations). It's not relating it to the equations before where we used 'x' as a letter though.
@@karnavthakur5868 You have to specify that x is a dummy variable. In fact, technically, these equations can be solved functionally, without ever using dummy variables in the algebra.
I used the second method.
Interesting
method 2 is cool
why not f(x)=2^(x-1)?
it seems alright
If f(x)=2^(x-1), then f(2^x+1)=2^(2^x+1-1)=2^(2^x)≠4^x=(2^2)^x
Does it worth more than 5 second?
We need a basic understanding first. I would be gr8 if u actually did that
It's basic, really is
f(x)=(x-1)^2
Let g : R -> (1, ♾) such that g(x) = 2^x + 1 and h : R -> (0, ♾) such that h(x) = 4^x = (2^x)^2. Notice that sqrt°h + 1 = g, which is equivalent to h = sqrt^(-1)°(g - 1), so let φ : (1, ♾) -> (0, ♾) such that φ(x) = sqrt^(-1)(x - 1) = (x - 1)^2. Hence h = φ°g.
The challenge is to find some f : (1, ♾) -> (0, ♾) such that f°g = h. Since h = φ°g, f°g = φ°g. g is right- cancellable, since it is an exponential function, and is thus a bijective function. Therefore, f = φ. Therefore, f(x) = (x - 1)^2 for all x in (1, ♾).
👍
Let y = 2^x +1
Notice (y-1) = 2^x and (2^x)^2 = 4^x
So f(y) = (y-1)^2.
This only works though on the domain of y. Since 2^x>0, y>1. So f(x) = (x-1)^2 for x>1, and you have no information about f(x) for x
The 1st method sucks.
I thought your 2nd method would be the 1st method and optionally as a second method, if one does not immediately realize that:
4^x = (2^x)^2 = [(2^x + 1) -1]^2
== (z-1)^2,
to solve for x, 2^x + 1=z using log2 and set the result in 4^x😉
I.e., x=log2(z-1) and f(z)=4^[log2(z-1)] = (z-1)^2
Hello
Something missing in your approach. One should prove that all real numbers can be written as 2power x+1
Ah BANANAS in action!
f(x)=(x-1)/2 say,
f(1) = (1-1)/2=0
f(2) = (2-1)/2=1/2
...etc...
f(🍌)=(🍌-1)/2
f( 2🍌) = (🍌+🍌)/2 - 1/2 = 🍌-1/2
this is called the _banana principle_ , very useful in inverse functions,
e.g. let f(x) = (x-1)/2
what is the inverse function g(x) such that g(f(x))=f(g(x)) = x of course?
f(g) = (g-1)/2 by the banana principle
so x = (g -1)/2
g -1 = 2x
g(x) = 2x+1 the inverse to f(x).😲
Brilliant.
f(x) =(x-1)^2