##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##

[anuttarasammyak]

Yes, it is.( cf. post #16).
The relation f(x+y)=f(x)f(y) does not limit b. Other additional conditions may confine b to real positive.

 

[mathwonk]

note: post 29 proves that f(x) = b^x is not the only function satisfying the functional equation.

 

[WWGD]

note: post 29 proves that f(x) = b^x is not the only function satisfying the functional equation.

I saw f(3.14), but your function seems to take two inputs as f(a,b)?

 

[mathwonk]

sorry for the confusing notation. I have chosen a vector subspace W of the reals containing π and such that every real has a unique expression as a sum a+b, where a is in the subspace Q spanned by the real number 1, and where b belongs got the subspace W. Then if x is any real number, we can write x = a+b (uniquely with a in Q and b in W), and then my function f takes x to e^a.π^b. In particular, an element x in Q goes to e^x, and an element x in W goes to π^x, i.e. for x in Q we have a =x and b =0, and for x in W we have a=0 and b = x.

This choice of W also gives a vector space isomorphism from R to the vector space Q+W defined as pairs (a,b) with a in Q and b in W, and I used that isomorphism to represent the real number x by the pair (a,b). I guess that was confusing. In that notation 3.14 was represented as (3.14, 0), and π was represented as (0, π).

The isomorphism takes Q+W-->R, by sending (a,b) in Q+W, to a+b in R. This definition of Q+W as a set of pairs, is called an "exterior" direct sum, and is usually written with a circle around the plus sign, but I can't do that. Some people write Q+W to mean the "interior" sum, meaning simply the subspace of R consisting of all sums a+b of elements a in Q and b in W. That would be the image of my map above Q+W-->R.

In general, if E,F are subspaces of a vector space V, then V is the direct sum of E and F if and only if either of these two things is true, (iff both are true):
1) Every vector x in V is expressible uniquely as a sum x =a+b, with a in E and b in F.
2) The map E+F-->V from the exterior direct sum E+F to V, taking (a,b) to a+b, is an isomorphism.

 

[mathwonk]

It seems about all one can say about a map f:R-->R that satisfies f(x+y) = f(x).f(y), is that (it is either identically zero or always positive, and if not zero, then) for all real x and all rational r, that f(rx) = [f(x)]^r. If f is also continuous, then it seems that f(xy) = [f(x)]^y for all real x,y, hence f(x) = f(1.x) = [f(1)]^x, for all real x. so then f(x) = b^x where b = f(1).

If f is not continuous then f(rx) = a^r, with a = f(x), when f is restricted to the one - dimensional Q-subspace Q.x, namely all rational multiples of x, but the choice of a can thus differ for different subspaces.

This restriction is not quite yet written as f(y) = b^y, but if x is any non- zero element of the subspace, hence a basis, then every element y of the subspace Q.x, can be uniquely written as y = rx for r rational, hence we may write y/x = r, and hence, if a = f(x), then f(y) = a^(y/x), for every y in Q.x. Hence the restriction has the form f(y) = a^(cy), where c = 1/x is a constant real number. And since a^(cy) = (a^c)^y, this means that f(y) does equal b^y, if we take b = a^(1/x) = f(x)^(1/x).

Thus if f satisfies the functional equation, then on any one dimensional Q-subspace of R, spanned say by x, we have f(y) = b^y, where b = [f(x)]^(1/x). But we can write R as a direct sum of an uncountable number of one- dimensional Q-subspaces, and it seems f can have a different b in every one of them. (I don't know if that forces the b's to be different in any two different one diml subspaces.)

 

[littlemathquark]

Can we find pairwise function that satisfies problem conditions?

 

[anuttarasammyak]

BTW would you show your answer of f(ー1) in order to avoid spoiler of homework?

 

[littlemathquark]

BTW would you show your answer of f(ー1) in order to avoid spoiler of homework?

$f(-1)=\frac{1+\sqrt{21}}{10}$

 

[mathwonk]

I don't know what you mean by a "pairwise" function. But I believe I have argued that the functional equation shows that the values of f are determined on all rationals once you know the value at a single non zero rational, but the value of f at a non rational point can still be essentially anything (positive).

 

[littlemathquark]

Sorry that's typo, I mean piece wice function.

 

[mathwonk]

I suppose you mean by that a function with different definitions on two different subsets of the reals, and that is exactly what I defined in post 29, where I used the fact that every real number x can be written uniquely as a sum of two numbers, x = a+b, with rational a, and b chosen from a complementary Q- vector subspace W of the reals. Then I used the function e^x for rational x, and the function π^x for x in W. And for more general numbers of form x = a+b, with rational a, and b in W, you multiply them together getting f(x) = f(a+b) = f(a).f(b) = e^a.π^b.

 

[anuttarasammyak]

since f(3.14) = e^(3.14), ..., f(3.14159) = e^(3.14159), etc..., but f(π) = π^π ≠ e^π.

You say f($\pi$)=f($\pi$-3+3)=$\pi$^($\pi$-3)e^3 ?

 

[mathwonk]

@anuttarasammyak: by the definition I gave, f(π-3) = π^π.e^(-3). i.e. to evaluate f at x, first write x as a sum of elements of Q and of W, x= a+b, then f(x) = e^a.π^b. in your example, x = -3 + π, so f(x) = e^(-3).π^π.

 

[anuttarasammyak]

So we should know your choice representing irrational numbers, e.g. $\pi$ among $\pi$+1,$\pi$+2,…

 

[mathwonk]

by the axiom of choice, i.e. zorn's lemma, the Q-vector space R has a ( uncountable) Q- linear basis containing 1 and π. (just take a maximal Q-independent set of reals containing 1 and π.) Such a basis cannot be explicitly described, but it does contain 1 and π. So although I cannot tell you the value of f at most irrational points, e.g. I do not know the value of f(sqrt(2)), [unless I also require the basis to contain sqrt(2)], I can definitely tell it to you for numbers of form r+sπ, with r,s both rational. nonetheless the function f is well defined everywhere in terms of this basis. and I can define it on points of this basis almost at will, restricting the values to be positive.

As I simplified it, I took W to be the Q-subspace spanned by all basis elements except 1. Since I did not specify any other elements of W except (rational multiples of) π, I do not know the value of f at any elements except those of form r + sπ, for rational, r,s; but for those, f(r+sπ) = e^r.π^(sπ)

 

[anuttarasammyak]

Now I know you are not saying about all irrational numbers but r+s$\pi$. f(e), f($\sqrt{2}$) should be considered independently.

 

[mathwonk]

yes, that is why the example I gave is not so elementary. the fact that there exists a Q-basis of R at all, is quite abstract. to define an f such as I wanted, one cannot just divide the reals into rational and irrationals, one has to choose a Q-linear direct sum decomposition of R, and this cannot be done explicitly. so although my direct sum expression R = Q+W can be proved to exist, using zorn's lemma, hence the resulting f also is proven to exist, still it is not so easy to evaluate.

if I want to know the value of some such f also at sqrt(2), I note that 1, sqrt(2), and π are Q-linearly independent, hence I can find a decomposition R = Q+W where W contains both π and sqrt(2), and then my previous f will have value f(sqrt(2)) = π^(sqrt(2)).

or I could look at R as a direct sum R = Q+Q.π + V, where sqrt(2) belongs to V, and I can define f so that f(r) = e^r for rational r, and f(sπ) = π^(sπ) for rational multiples of π, and f(x) = 2^x for x in V, hence f(sqrt(2)) = 2^(sqrt(2)). this f will have f(7+π+3.sqrt(2)) = e^7.π^π.2^(3.sqrt(2)).
but now I don't know the value of f at sqrt(3).

 

[anuttarasammyak]

f(x+y)=f(x)f(y) >0
g(x):=\log f(x)
g(x+y)=g(x)+g(y)
g(0)=0,g(x)=-g(-x)
It seems almost sure that
g(x)=cx
Here we do not have to distinguish x is rational or irrational. Your discussion still works here ?

 

[mathwonk]

not only is it not almost surely true that g(x) must equal cx, rather it is certainly false. My example (or log of it, as you very perceptively observe), is a counterexample to this hope. in fact to find my example, I first searched for counterexamples to this additive statement. here is a link:
'https://math.stackexchange.com/questions/1488031/group-of-automorphisms-of-mathbbr

 

[mathwonk]

Interestingly, since you mention f(e), if we express R as a direct sum R = Q + Q.π + W, for some complementary Q-linear subspace W, and set f(x) = 2^x for x in Q+Qπ, but f(x) = 3^x for x in W, then we have no idea of the value of f(e)! I.e. we do not know how to express e as r + sπ + y, with y in W. In fact mathematicians apparently do not even know whether e+π is rational or not, so it is possible that here f(e) = 2^e.

In particular, it is not known whether 1, π and e are Q- independent, so we cannot prove there is a Q-basis of R that contains 1,π and e, hence we cannot prove the existence of a function f satisfying our functional equation and having arbitrary values at 1, π, and e, although we can do so for 1,π, and sqrt(2).

[I did not know this until tonight while searching for confirmation of the opposite (false) statement that I was about to make. ]