/sci/ - Science & Math

>>11514409
Not sure but
f(a)+f(0)=f(f(a))
f(0)+f(a)=f(f(0)+a)
Thus
f(f(a))=f(f(0)+a)
Assuming f is injective,
f(x)=f(0)+x
f'(x) = 1
So f(x) = c + x is a solution for any c.

>>11514409
Also,
f(f(a)+b) = f(a+f(b))
Assuming injectivity
f(a)+b=a+f(b)
f(a)-f(b)=a-b
f(a)-f(0)=a
which yields the same result

Using degrees no polynomial of degree 2 or higher is going to work.

>>11514592
Proof? I think I've got the following:
deg(f(x)+f(x)) = deg(2f(x)) = deg(f(x))
But on the other hand,
deg(f(f(x)+x)) = deg(f(x)) * deg(f(x)+x) = deg(f(x))^2 assuming deg(f(x)) != deg(x) = 1.
deg(f(x)) != deg(f(x))^2 unless 0 or 1.

>>11514611
Correction: Assuming deg(f(x)) > deg(x) = 1.

f(x)= integer n such that x is in [n, n+1) works

If f(x) is differentiable for all x, then f(x)=0 and f(x)=x+c are the only solutions

For any constant b:
f(x)+f(b)=f(f(x)+b)
f(x)=f(f(x)+b)-f(b)
f'(x)=f'(f(x)+b)*f'(x)

For f'(x)≠0:
f'(x)=f'(f(x)+b)*f'(x)
1=f'(f(x)+b) //f' is invariant
f'(x)=1
f(x)=x+c
Then:
f(a)+f(b)=f(f(a)+b)
(a+c)+(b+c)=((a+c)+b+c)
a+b+2c≡a+b+2c, so true for all c
So: f(x)=x+c

For f'(x)=0:
f'(x)=0
f(x)=c
Then:
f(a)+f(b)=f(f(a)+b)
c+c=c, so c=0
So: f(x)=0

>>11514747
Doesn't this assume that f(x)+b takes all values in the domain of f?

>>11514747
Differentiability is a brainlet assumption.
Assume f is continuous. I will prove that either:
a) f is the zero function.
b) f is of the form f(x)=x+f(0)
Proof:
1.f(-f(0))=0 so the image of f is (by the intermediate value theorem) some interval containing 0.
2. Assume f is not the zero function. W.l.o.g. If f(a)>0 for some a (the argument works the same for negative values).
3. f(a) + f(a) = f(f(a)+a) = 2a is in the image and similarly na is in the image for all natural numbers n, so the interval contains [0, infinity).
4. Let x>0. Take a s.t. f(a)=x. Then f(a)+f(0)=f(f(a)+0) <=> f(x)=x+f(0)
5. We want to prove that the range is whole of R:
Let b<0
Take a such that f(a)=-b.
Then f(a)+f(b)=f(f(a)+b)=f(0)
But as |b| grows, f(a) grows so f(b) must be negative for some b. Hence again by IVT we see that the range of f is the whole of R.
6. Repeating step 4 but now for all values of x shows f(x)=x+f(0).

>>11514812
Explain step 3.

>>11515285
Essentially it follows from the linearity of the image:
if a is in the image and b is in the image, then a+b is in the image.
This follows trivially from the defining formula
f(a)+f(b)=f(f(a)+b)
Now because the reals are an archimedean field, if a>0, na gets arbitrarily large for big natural numbers n. As 0 and na are in the image, by the assumption that f is continuous and the intermediate value theorem, every number in between them is in the image. That is, [0,na] is in the image. Hence [0,infinity) is in the image.

>>11515296
I see, you meant 2f(a)=f(a)+f(a)=f(f(a)+a). What you had before was incorrect.

>>11515353
You are correct.

>>11514689
So the floor function basically?

>>11515404
Yes. Or a floor plus a constant integer. I'm sure you can find even more noncontinuous examples.

>>11515406
>>11515404
ceiling function also works

>>11515412
>>11515406
>>11515404
Also for any real a>0, you could take f(x)= a*floor(x/a)

Generating function for iterates of 0 under f is -f(0)x/(2x-1).

>>11515504
U foken wot m8?
f^n(0)=nf(0)

>>11514409
bump

>>11515533
Why are you bumping? I already solved the problem here
>>11514812

>>11515528
Yeah I was mistaken.

>>11515535
Only in the continuous case.

If I let f(0)=1 and f(1)=0, is there a way to fill anything else?

>>11514409
General solution:

Let k be a positive integer.

f( k f(a) + a ) = f( f(a) + (k-1)f(a) + a ) = f(a) + f( (k-1)f(a) + a ) = ... = (k-1)f(a) + f( f(a) + a ) = (k-1)f(a) + f(a) + f(a) = (k+1)f(a)

f( k f(a) + a ) = (k+1)f(a)

>>11515733
that's not a solution

>>11515733
Typo: k is any integer

>>11515791
It's a solution, it's just very general.

>>11515800
The problem asks you to determine all functions f(x) which satisfy the equation. You just manipulated the equation to get another equation but didn't solve the problem.

>>11515800
It's a property, but I'm not sure it's a complete description.

>>11515805
My answer gives all functions, starting only with a point (a, f(a)) and generating all other necessary points. You don't like the form of the function but that doesn't mean it isn't a function.

>>11515807
It is.

>>11515827
>>11515816
Please explain how it gives all functions. Give a description for the most general function.

>>11515816
if your answer is an equation (which it is), then unless it's in the form

f(x) = expression not involving f

you ain't solved nothing pal

>>11515816
>>11515827
You can't just choose orbits independently using your rule. For example, if f(0)=1 then f(k)=k+1 for integers k. We don't have f(1/2) defined yet, so take f(1/2)=1/2. Then we find that f(1)=1, a contradiction.

>My answer gives all functions
>How? That's your job to figure out.

>>11515837
Functions can be well-defined even if not given as an expression in terms of elementary operations. Recursion, applied correctly, can also be used to define a function.

>>11515833
>Please explain how it gives all functions.
Choose a point, say (0,1). Then the solution is of the form f(k) = k+1 where k is any integer.

>Give a description for the most general function.
I already did. Would it help you if I instead wrote

Let f(a) = c for some a. Let k be any integer.

Then f(kc+a) = (k+1)c

>>11515857
But this only defines the function on the subset {kc+a, k integer}, not the whole reals.

>>11515857
>>11515838
You have the same problem even if you only look at functions on the integers. Taking f(0)=2 gives f(2k)=2(k+1) but you can't choose f(1)=1 because that will imply f(0)=0.

>>11515838
I never said you can choose orbits independently.

>>11515862
So?

>>11515869
See >>11515874

>>11515874
Without a description of which choices are possible you haven't really given all possible functions.

>>11515838
>>11515869
You can't choose a new point (b,f(b)) if f(a)-a =/= f(b)-b

>>11515888
See >>11515896

>>11515816
all you have done is you have shown that f(a) (which cannot even be chosen arbitrarily!!) determines the values on the set

{a + k*f(a) | k is an integer }

literally nothing else

>>11515918
This.

>>11514812
>f(-f(0))=0
Would it have killed you to say something like "let a = 0 and let b = -f(0)"?

>>11515918
>f(a) (which cannot even be chosen arbitrarily!!)
Of course it can.

>literally nothing else
There is not much else to say. If you want to choose more orbits then add a new point (b, f(b)) such that f(a)-a = f(b)-b.

>>11515998
Prove that last assertion.

>>11516012
It's trivial.

If k1 f(a) + a = k2 f(b) + b then we need (k1+1)f(a) + (k2+1)f(b)

Rearrange the first equation for k1 and then sub into the second to get f(a)-a = f(b)-b.

So if we want the function to cover all reals we simply take each x for which f(x) is undefined and make f(x) = f(a)-a+x.

Then that means f( k f(x) + x ) = (k+1)f(x). And the process repeats.

>>11516046
Typo: If k1 f(a) + a = k2 f(b) + b then we need (k1+1)f(a) = (k2+1)f(b)

>>11514409
In what class do you do stuff like this. I like it

>>11516057
putnam

>>11515998
so your solution is:

f is any function such that R can be
decomposed into disjoint sets in the form {x + k*f(x) | k is an integer}, indexed by the pairs (x,f(x)) such that they all satisfy f(x) - x = f(y) - y.

>>11516061
Yes.

>>11516057
In an Analysis of Algorithms CS class I had to solve recursive (math) functions to solve for computational complexity of recursive algorithms. (Big-O notation)

Differential equations also kinda had stuff like this, with differential functions defined in terms of themselves

>>11516061
Oh but you forgot that specifically f( k f(a) + a ) = (k+1)f(a), in other words f(x) = x-a+f(a) where (x-a)/c is an integer.

>>11516100
Typo: where (x-a)/f(a) is an integer.

Ok so assuming a and b are real numbers then for the equation given to be valid f needs to have all the real numbers as domain.

Now let's take two cases.
Case 1: f varies
Set [math]b=0 [/math] then
[eqn]
f(a)+f(0) = f(f(a))
[/eqn]
Now set [math] f(0) =c , f(a) =x [/math] so we get
[eqn]
f(x) = x + c
[/eqn]
Case 2: f does not vary and is a constant c. Then
[eqn]
f(f(a)+b) = f(a)+f(b)\implies c = c +c \implies f(x) = c =0
[/eqn]

>>11516122
Dumb and wrong

So to summarize in a pleasing form, the solution is, for some chosen point (a,b):

f(x) = x-a+b where (x-a)/b is an integer

And you can increase that domain to anywhere where (x-c)/(c-a+b) is an integer, where c is any real number.

Pretty nice.

>>11516122
To add to this. If f(0) = 0 then functions that satisfy
f(f(x)) = f(x) are also valid solutions. Floor,ceiling function,round. f(0) = 0 and f(x) =c is also true etc

>>11516195
Also wrong. You are really dumb

>>11516195
Wrong. f(x)=c is not a solution for nonzero c.

>>11516195
Assume f(x)=c for c≠0:
f(a)+f(b)=f(f(a)+b)
-> c+c=c
-> c=0, a contradiction

>>11516195 #
Assume f(x)=c for c≠0:
f(a)+f(b)=f(f(a)+b)
-> c+c=c
-> c=c-c
-> c=0, a contradiction

>>11516147
I forgot that b=0 is a special case and is its own solution. You can also choose a new point (c,d) such that k1 b + a =/= k2 d + c for all integers k1 and k2. That is how you get step functions as a solution.

bump.

>>11516147
The domain should be all the real numbers. You didn't solve it.

>>11514409
2f(0) = f(f(0))

>>11517495
>The domain should be all the real numbers.
It can be all the real numbers, it doesn't have to be. Example: f(0)=0 is a solution.

>>11517495
Also you didn't read the whole post or at least didn't understand it. It can be extended to all real numbers.

Since you appear mathematically challenged, I'll explain the answer in an even simpler way: the solution is points on a line that are evenly spaced, or points on multiple lines that are parallel and spaced out all in the same manner. Over the entire reals this gives you either a straight line or a step function.

Bump

Haven't read the thread, but smells like foundations with the Axioms of Choice
might make it so that there exist many solutions.
e.g. compare

https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation