Question on isomorphism between addition and multiplication

In summary, the function required satisfies the property \phi(x+y)=\phi(x)\phi(y) for all x,y in the real or complex plane, where x,y can be any real or complex number. There are other functions which satisfy the equation, but they are not necessarily continuous. To find such a function, one would need to try different exponents to see if any of them satisfy the equation.
  • #1
mnb96
715
5
Hello,
I want to find a family of functions [itex]\phi:\mathbb{R} \rightarrow \mathbb{C}[/itex] that have the property: [tex]\phi(x+y)=\phi(x)\phi(y)[/tex] where [itex]x,y\in \mathbb{R}[/itex].

I know that any exponential function of the kind [itex]\phi(x)=a^x[/itex] with [itex]a\in\mathbb{C}[/itex] will satisfy this property.
Is this the only choice, or are there other functions that I am missing that satisfy the above property?
 
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  • #2
Try playing around with the formula to answer this question for yourself.

For instance the formula shows that [itex]\phi[/itex](0) = 1
 
  • #3
hello mnb96! :smile:

various ways, eg put ##\psi = ln\phi##, or what is ##\phi '(x+y)## ? :wink:
 
  • #4
Some remarks:

1) I'm not sure how you define [itex]a^x[/itex] for [itex]a\in \mathbb{C}[/itex]. You got to be careful, because those exponents usually take on multiple values and you need to choose the correct one.

2) You might want to add as an hypothesis that [itex]\varphi[/itex] is continuous. In that case, you will indeed be able to prove what you want. If [itex]\varphi[/itex] is not continuous, then there might be other functions which satisfy the equation, and those functions are very ill-behaved.
 
  • #5
micromass said:
2) You might want to add as an hypothesis that [itex]\varphi[/itex] is continuous. In that case, you will indeed be able to prove what you want. If [itex]\varphi[/itex] is not continuous, then there might be other functions which satisfy the equation, and those functions are very ill-behaved.

In the case of a real rather than complex valued function I would guess that the rule forces the function to be continuous.

- the rule implies that f(0) = 1 and f(x) >0

- the rule says that f(x) = f(x/n)^n so f(x/n) must approach 1 as n grows large. This indicates (but doesn't prove)continuity at zero. But if it is continuous at zero it is everyehere. If not, it is discontinuous everywhere.
 
  • #6
lavinia said:
In the case of a real rather than complex valued function I would guess that the rule forces the function to be continuous.

- the rule implies that f(0) = 1 and f(x) >0

- the rule says that f(x) = f(x/n)^n so f(x/n) must approach 1 as n grows large. This indicates (but doesn't prove)continuity at zero. But if it is continuous at zero it is everyehere. If not, it is discontinuous everywhere.

Well, here is a counterexample:
We know that [itex]\mathbb{R}[/itex] is a [itex]\mathbb{Q}[/itex]-vector space, so it has an (infinite) basis E. Take a particular [itex]e\in E[/itex].
Any element [itex]z\in \mathbb{R}[/itex] can be written as the finite sum

[tex]z=\sum_{x\in E} \alpha_x x[/tex]

Now define [itex]g(z)=\alpha_ee[/itex]. Then [itex]g:\mathbb{R}\rightarrow\mathbb{R}[/itex] satisfies [itex]g(x+y)=g(x)+g(y)[/itex] for all reals x and y. But it is not [itex]\mathbb{R}[/itex]-linear and thus not continuous.

Now define [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] as [itex]f(z)=2^{g(z)}[/itex]. Then this satsifies [itex]f(x+y)=f(x)f(y)[/itex] but it is not continuous.
 
  • #7
micromass said:
Well, here is a counterexample:
We know that [itex]\mathbb{R}[/itex] is a [itex]\mathbb{Q}[/itex]-vector space, so it has an (infinite) basis E. Take a particular [itex]e\in E[/itex].
Any element [itex]z\in \mathbb{R}[/itex] can be written as the finite sum

[tex]z=\sum_{x\in E} \alpha_x x[/tex]

Now define [itex]g(z)=\alpha_ee[/itex]. Then [itex]g:\mathbb{R}\rightarrow\mathbb{R}[/itex] satisfies [itex]g(x+y)=g(x)+g(y)[/itex] for all reals x and y. But it is not [itex]\mathbb{R}[/itex]-linear and thus not continuous.

Now define [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex] as [itex]f(z)=2^{g(z)}[/itex]. Then this satsifies [itex]f(x+y)=f(x)f(y)[/itex] but it is not continuous.

cool.

so exponentiate any Q but not R linear map of the reals to the reals.

So... the sequence x/n will have a constant coefficient divided by n with respect the the basis vector so that's why the function looks continuous on it.

And this means that there is a number with a coefficient bounded away from zero in any interval around zero.
 
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  • #8
Mmm...:confused: ... I am a bit confused.
Let's stick for now with the case [itex]\phi:\mathbb{R} \rightarrow \mathbb{R}[/itex].
Assuming [itex]\phi[/itex] is an isomorphism between (ℝ,+) and (ℝ+,×) that satisfies the property [itex]\phi(x+y)=\phi(x)\phi(y)[/itex], and that is continuous, we can say that:

1) [itex]\phi(0)=\phi(x-x)=\phi(x)\phi(-x)[/itex] for all [itex]x\in \mathbb{R}[/itex], thus [itex]\phi(0)=1[/itex]

2) from 1) we have that [itex]\phi(-x)=\frac{1}{\phi(x)}[/itex]

3) [itex]\phi(x)=\phi(x/2+x/2)=\phi(x/2)^2 > 0[/itex], thus [itex]\phi(x)>0[/itex]

4) [itex]\phi[/itex] must be bijective, thus [itex]\phi'(x)>0[/itex]

5) [itex]\phi(n) = \phi(1+1+\ldots+1)=\phi(1)^n[/itex] for all [itex]n\in \mathbb{Z}[/itex]

In conclusion [itex]\phi[/itex] must be a continuous positive monotonic increasing function passing through the point (0,1) and through the points [itex](n, \phi(1)^n)[/itex]. It seems clear that the only possibility is to choose [itex]\phi(x)=\phi(1)^x=a^x[/itex], although I don't know how to put it rigorously.

Now the problem is, what if [itex]\phi:\mathbb{R}\rightarrow\mathbb{C}[/itex] maps the reals to a subset of the complex numbers?
 
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  • #9
mnb96 said:
Mmm...:confused: ... I am a bit confused.
Let's stick for now with the case [itex]\phi:\mathbb{R} \rightarrow \mathbb{R}[/itex].
Assuming [itex]\phi[/itex] is an isomorphism between (ℝ,+) and (ℝ+,×) that satisfies the property [itex]\phi(x+y)=\phi(x)\phi(y)[/itex], and that is continuous, we can say that:

1) [itex]\phi(0)=\phi(x-x)=\phi(x)\phi(-x)[/itex] for all [itex]x\in \mathbb{R}[/itex], thus [itex]\phi(0)=1[/itex]

[itex]\phi[/itex]( 0 + x) = [itex]\phi[/itex](0)[itex]\phi[/itex](x) so [itex]\phi(0)=1[/itex]

4) [itex]\phi[/itex] must be bijective, thus [itex]\phi'(x)>0[/itex]

you don't know that [itex]\phi[/itex] is differentiable just because it is continuous or bijective.
In conclusion [itex]\phi[/itex] must be a continuous positive monotonic increasing function passing through the point (0,1) and through the points [itex](n, \phi(1)^n)[/itex]. It seems clear that the only possibility is to choose [itex]\phi(x)=\phi(1)^x=a^x[/itex], although I don't know how to put it rigorously.

You need to go from continuity to the conclusion
 
  • #10
if ##\phi## is differentiable, there's a very quick proof

i suspect that that proof can be adapted to the merely continuous case (but i haven't tried)
 
  • #11
lavinia said:
You need to go from continuity to the conclusion

Ok. But wouldn't it be possible to define a continuous (and monotonic increasing function) that still passes through the points [itex](n,a^n)[/itex] but does "strange things" between [itex](n,a^n)[/itex] and [itex](n+1,a^{n+1})[/itex] ?
 
  • #12
mnb96 said:
[itex]\phi[/itex] must be bijective, thus [itex]\phi'(x)>0[/itex]


The function e[itex]^{-x}[/itex] is bijective from the reals to the positive reals but its derivative is always negative.
 
  • #13
mnb96 said:
Ok. But wouldn't it be possible to define a continuous (and monotonic increasing function) that still passes through the points [itex](n,a^n)[/itex] but does "strange things" between [itex](n,a^n)[/itex] and [itex](n+1,a^{n+1})[/itex] ?

a continuous monotonically increasing function may not be everywhere differentiable although it seems right that it can only have a discrete set of kinks. See if you can find some examples.
 
  • #14
tiny-tim said:
if ##\phi## is differentiable, there's a very quick proof

i suspect that that proof can be adapted to the merely continuous case (but i haven't tried)

ln[itex]\phi[/itex] is linear over the rational numbers. If is is continuous it follows that it is linear over the reals. Easy proof.

A continuous linear map of the reals to the reals is multiplication by a constant.
 
  • #15
As micromass said for the complex case one need to wonder whether one can define a single branch of the logarithm on the values of [itex]\phi[/itex].

If ln[itex]\phi[/itex] is single valued then its projections onto the x and y-axis are linear.
 

What is isomorphism between addition and multiplication?

The isomorphism between addition and multiplication is a mathematical concept that states that the operation of addition and the operation of multiplication are related in a one-to-one correspondence. This means that for every addition operation, there exists a corresponding multiplication operation that produces the same result.

Why is isomorphism between addition and multiplication important?

Isomorphism between addition and multiplication is important because it helps us understand the underlying structure of the two operations and their relationship to each other. It also allows us to use properties and rules from one operation to solve problems in the other operation.

How do you prove isomorphism between addition and multiplication?

To prove isomorphism between addition and multiplication, you need to show that the two operations follow the same basic structure and have similar properties. This can be done through mathematical proofs and by demonstrating that the two operations produce the same results for a given set of numbers.

What are some examples of isomorphism between addition and multiplication?

One example of isomorphism between addition and multiplication is the distributive property, where a(b+c) = ab + ac. This shows that the addition operation of (b+c) has a corresponding multiplication operation of ab + ac, which produces the same result.

Another example is the commutative property, where a+b = b+a and ab = ba. This shows that the order of the numbers does not affect the result in both addition and multiplication operations.

Can isomorphism between addition and multiplication be applied to other mathematical operations?

Yes, isomorphism can be applied to other mathematical operations as well, as long as they have a similar structure and properties to addition and multiplication. For example, isomorphism can be applied to exponentiation and logarithms, as they also have a one-to-one correspondence and follow similar rules and properties.

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