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

AI Thread Summary
The discussion revolves around the functional equation f(x+y) = f(x)f(y) and the condition f(1) + f(2) = 5, leading to the query about finding f(-1). Participants explore the implications of the function being non-zero and continuous, concluding that f(x) must be positive due to f(x)f(-x) = f(0) = 1, which prohibits f(x) from being zero. They also discuss the possibility of discontinuous functions satisfying the equation, suggesting that while f(x) could take various forms, the exponential function f(x) = b^x is a common solution. Ultimately, the consensus is that if f is continuous, then f(-1) is uniquely determined, while discontinuity allows for multiple potential solutions.
littlemathquark
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Homework Statement
##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##
Relevant Equations
##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##
I can show that ##f(x) \ge 0## but I can not show ##f(x) \ne 0##
##x=y=t/2## then ##f(t) =f^2(t/2)\ge 0##
 
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littlemathquark said:
Homework Statement: ##f(x+y) =f(x) +f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##
Relevant Equations: ##f(x+y) =f(x) +f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##

I can show that ##f(x) \ge 0## but I can not show ##f(x) \ne 0##
##x=y=t/2## then ##f(t) =f^2(t/2)\ge 0##
How? I doubt both. Apply the rule to ##x=x+0## and ##0=x+(-x).##
 
fresh_42 said:
How? I doubt both.
I am sorry, correct question must be ##f(x+y) =f(x) f(y) ##
 
littlemathquark said:
I am sorry, correct question must be ##f(x+y) =f(x) f(y) ##
And ##f(1)\cdot f(2)=5## or is this still the addition in which case it is difficult to say anything about sums?
 
fresh_42 said:
And ##f(1)\cdot f(2)=5## or is this still the addition in which case it is difficult to say anything about sums?
This is still addition, ##f(1)+f(2)=5##
 
Yes, but I could show that ##f(x)\gt 0## in this question.
 
##0=x+(-x) ## and ##f(0)=f(x+(-x))=f(x)f(-x)=1## so for every ##x## ##f(x) \ne 0##
If ##y=0## ##f(x) =f(x) f(0)## so ##f(0)=1##
 
littlemathquark said:
##0=x+(-x) ## and ##f(0)=f(x+(-x))=f(x)f(-x)=1## so for every ##x## ##f(x) \ne 0##
If ##y=0## ##f(x) =f(x) f(0)## so ##f(0)=1##
If ##f(x)\cdot f(-x)=1## then all values are invertible, aren't they?
 
  • #10
Yes, but ##f(x) =0##
 
  • #11
littlemathquark said:
Yes, but ##f(x) =0##
##1=f(0)=f(x)\cdot f(-x).## Hence, ##f(x)## cannot be zero. It has an inverse ##f(-x)=f(x)^{-1}.##

I would argue that ##f## is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have ##f(a)<0## and ##f(b)>0## then there must also be a ##c## such that ##f(c)=0## which we already ruled out.
 
  • #12
We assume f(x)=e^ax and get a by solving the quadratic equation
Y^2 +Y-5=0 where Y=e^a.
 
  • #13
Btw., we only want to know what ##f(-1)## is. So why don't you use ##(-1)^{-1}=-1##?
 
  • #14
fresh_42 said:
##1=f(0)=f(x)\cdot f(-x).## Hence, ##f(x)## cannot be zero. It has an inverse ##f(-x)=f(x)^{-1}.##

I would argue that ##f## is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have ##f(a)<0## and ##f(b)>0## then there must also be a ##c## such that ##f(c)=0## which we already ruled out.
İs f is not continuous, can we find another ##f## function that satisfies conditions of the given problem and different from exponantial function? And can you give me an example function?
 
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  • #15
anuttarasammyak said:
We assume f(x)=e^ax and get a by solving the quadratic equation
Y^2 +Y-5=0 where Y=e^a.
Why do you assume ##f(x)=e^{ax}?##
 
  • #16
Because we are familiar with the relation
e^{a(x+y)}=e^{ax}\ e^{ay} or
b^{x+y}=b^{x}\ b^{y} where ##b=e^a \ > 0##. It seems almost true to me that this is the only function which stasify this addition-multiplication relation.

BTW
f(2)=f(1+1)=f(1)^2
leads to the quadratic equation of f(1) WITHOUT the assumption. After getting f(1)
f(1)f(-1)=f(1-1)=f(0)
and
f(x)=f(x+0)=f(x)f(0)
f(0)=1 or f(x)= 0 for any x. f(1)+f(2)=5 confirms the former. Thus we get f(-1).
 
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  • #17
fresh_42 said:
Btw., we only want to know what ##f(-1)## is. So why don't you use ##(-1)^{-1}=-1##?
I can solve the problem by using your suggestion, thank you. But I want to understand link between f and continuous.
 
  • #18
anuttarasammyak said:
It seems almost true to me that this is the only function which stasify this addition-multiplication relation.
f(x)f(-x)=f(0)=1
except meaningless "For all x f(x)=0" so f(x) > 0
f(n)=f(1)^n
f(1)=f(\frac{1}{m}+ \frac{1}{m}+\frac{1}{m}+...)=f(\frac{1}{m})^m
f(\frac{1}{m})=f(1)^{\frac{1}{m}}
f(\frac{n}{m})=f(1)^{\frac{n}{m}}=b^{\frac{n}{m}}
So it is true for rational numbers. Any irrational numbers is expressed as limit of rational number sequence so
f(x)=b^x
 
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  • #19
anuttarasammyak said:
f(x)f(-x)=f(0)=1
except meaningless "For all x f(x)=0" so f(x) > 0
f(n)=f(1)^n
f(1)=f(\frac{1}{m}+ \frac{1}{m}+\frac{1}{m}+...)=f(\frac{1}{m})^m
f(\frac{1}{m})=f(1)^{\frac{1}{m}}
f(\frac{n}{m})=f(1)^{\frac{n}{m}}=b^{\frac{n}{m}}
So it is true for rational numbers. Any irrational numbers is expressed as limit of rational number sequence so
f(x)=b^x
I think ##f## must be continuous at least one point if we use limit.
 
  • #20
Yes. y=b^x is continuous everywhere.
 
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  • #21
littlemathquark said:
Homework Statement: ##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##
Relevant Equations: ##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##

I can show that f(x)≥0 but I can not show f(x)≠0
f(x)f(-x)=f(0)=1 prohibits f(x)=0.
 
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  • #22
Thank you, fresh_42 wrote the same fact.
 
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  • #23
Though the function is continuous, We do not have to use continuity to prove f(x) is positive.
 
  • #24
A continuous function is one that you can draw in one line without interception, roughly speaking. So if we have a point ##(a,f(a))## with ##f(a)>0 ## and a point ##(b,f(b))## with ##f(b)<0## then the line that represents the function, i.e. its graph, has to cross the ##x##-axis at some point ##x=c## and there we have ##f(c)=0,## if ##f## can be drawn in one line without interception.

Otherwise, if we allow gaps/jumps in our line, we can e.g. define
$$
f(x)= \begin{cases}
-4 &\text{ if } x<-1\\
1 &\text{ if } -1 \le x\le 1\\
4 &\text{ if } x>1
\end{cases}
$$
This does not fulfill the first condition but it should give you a sense of what might happen if we allow gaps. Things would be a lot more complicated. I think the first condition can be used to prove that this cannot happen, but it's not obvious.

That's basically the difference between a continuous and a discontinuous function. A continuous function means that if we wobble at some point, we stay within a neighborhood of the function value there, we don't have jumps. A discontinuous function could be anything.
 
  • #25
fresh_42 said:
A continuous function is one that you can draw in one line without interception, roughly speaking. So if we have a point ##(a,f(a))## with ##f(a)>0 ## and a point ##(b,f(b))## with ##f(b)<0## then the line that represents the function, i.e. its graph, has to cross the ##x##-axis at some point ##x=c## and there we have ##f(c)=0,## if ##f## can be drawn in one line without interception.

Otherwise, if we allow gaps/jumps in our line, we can e.g. define
$$
f(x)= \begin{cases}
-4 &\text{ if } x<-1\\
1 &\text{ if } -1 \le x\le 1\\
4 &\text{ if } x>1
\end{cases}
$$
This does not fulfill the first condition but it should give you a sense of what might happen if we allow gaps. Things would be a lot more complicated. I think the first condition can be used to prove that this cannot happen, but it's not obvious.

That's basically the difference between a continuous and a discontinuous function. A continuous function means that if we wobble at some point, we stay within a neighborhood of the function value there, we don't have jumps. A discontinuous function could be anything.
So if ##f## is continuous function, then the answer ##f(-1)## is unique. But ##f## is not continuous it's not unique. Am I right?
 
  • #26
littlemathquark said:
So if ##f## is continuous function, then the answer ##f(-1)## is unique. But ##f## is not continuous it's not unique. Am I right?
I am uncertain. I think we can prove that ##f## has to be continuous by using the functional equation. So your claim would become void and trivially true. But in general, yes. The variety of discontinuous functions is considerably larger than continuous functions, whatever larger means technically.
 
  • #27
fresh_42 said:
##1=f(0)=f(x)\cdot f(-x).## Hence, ##f(x)## cannot be zero. It has an inverse ##f(-x)=f(x)^{-1}.##

I would argue that ##f## is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have ##f(a)<0## and ##f(b)>0## then there must also be a ##c## such that ##f(c)=0## which we already ruled out.

From <br /> f(a) = f(\tfrac12 a + \tfrac12 a) = f(\tfrac 12 a)^2 it follows that if f(a) &lt; 0 then f(\tfrac12 a) is imaginary.
 
  • #28
Say f(a)=0 ,
f(x)=f(x-a)f(a)=0
for any x. This is a plain function which we do not expect.
 
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  • #29
just a remark, which emphasizes that the arguments should avoid continuity.
Since R is a vector space over Q, using the axiom of choice it has a basis containing 1, and π, so one can write R as a direct sum of Q-vector subspaces Q+W, where W contains π. Then define f:R-->R, or f:Q+W-->R to be f(a,b) = e^a.π^b. This satisfies the functional equation but is not continuous, since f(3.14) = e^(3.14), ..., f(3.14159) = e^(3.14159), etc..., but f(π) = π^π ≠ e^π.
 
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  • #30
anuttarasammyak said:
Yes. y=b^x is continuous everywhere.
I believe if the base b is positive, isn't it?

OP: What is your domain, the Real numbers?
 
  • #31
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.
 
  • #32
note: post 29 proves that f(x) = b^x is not the only function satisfying the functional equation.
 
  • #33
mathwonk said:
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)?
 
  • #34
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.
 
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  • #35
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.)
 
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  • #36
Can we find pairwise function that satisfies problem conditions?
 
  • #37
BTW would you show your answer of f(ー1) in order to avoid spoiler of homework?
 
  • #38
anuttarasammyak said:
BTW would you show your answer of f(ー1) in order to avoid spoiler of homework?
##f(-1)=\frac{1+\sqrt{21}}{10}##
 
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  • #39
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).
 
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  • #40
Sorry that's typo, I mean piece wice function.
 
  • #41
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.
 
  • #42
mathwonk said:
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 ?
 
  • #43
@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.
 
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  • #44
So we should know your choice representing irrational numbers, e.g. ##\pi## among ##\pi##+1,##\pi##+2,…
 
  • #45
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)
 
  • #46
Now I know you are not saying about all irrational numbers but r+s##\pi##. f(e), f(##\sqrt{2}##) should be considered independently.
 
  • #47
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).
 
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  • #48
f(x+y)=f(x)f(y) &gt;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 ?
 
  • #49
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  • #50
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. ]
 
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