# Homework Help: Homogeneous nonlinear function

1. Jul 7, 2007

### maxxedit

Could you give me an example of a function that satisfies scalar multiplication but not addition?

more specifically, F: R$$^2$$ -> R such that F(av)=a F(v) but F(v1 + v2) != F(v1) + F(v2)

The best thing I could come up with is F(x,y)= |x| . This obviously does not satisfy additivity, but satisfies sc. mul. for only a => 0.

2. Jul 7, 2007

### ircdan

f(v) = 2, or any other constant function.

3. Jul 7, 2007

### d_leet

This doesn't satisfy a multiplicative property like the original poster describes.

If we have f(v)=2

then af(v)=2a, but f(av)=2, so this is only true for a=1.

4. Jul 7, 2007

### matt grime

It is very important that you have R^2 mapping to R. Let's think of an example that fails, again. Suppose f is a simple poly in two variables, like f(x,y)=x^2+y^2, then f(kx,ky)=k^2f(x,y), so that won't do... but can we make it better? Next guess would be to take g(x) as sqrt(f(x)), but no, that gives |k|, not k. So can you think of a better examle? I don't know where you got the title of the thread from, but you should have thought about that.

5. Jul 7, 2007

### maxxedit

For one, we can't take sqrt because then F could send things to C.
You said R^2 to R is significant...I've been playing around with x and y, but every nonlinear function seems to break both sc.mul and addition.

6. Jul 7, 2007

### d_leet

I don't think that that was exactly what matt was getting at, if you work off of what he showed you should be able to come up with something that works with the property you want, I can think of a few that would work all based on the idea matt was trying to convey.

7. Jul 7, 2007

### maxxedit

Oh, I see. I found one!
Thanks for the guidance. Took me hours, and would have taken hours more without your help.

8. Jul 8, 2007

### cryptic26

Here is one.

f(x,y) = sqrt( x^2 + y^2) is one example. You can find several such functions.

9. Jul 8, 2007

### Dick

This function fails for a<0. Try again.

10. Jul 8, 2007

### cryptic26

No it does not. sqrt( x^2 + y^2) for (ax, ay) shall be

(+-)a * sqrt(x^2 + y^2).

For a < 0, you can find -a* sqrt(x^2+y^2) that satisfies the property

f(av) = a*f(v).

Last edited: Jul 9, 2007
11. Jul 8, 2007

### cryptic26

Alternatively, one can also have the function (x^3 + y^3) ^ (1/3).

12. Jul 9, 2007

### cshum00

Just like cryptic26 did, put a square root to the whole function or in exponential form.

F : R2 -> R

suppose that R = f(x,y,z) = x3 + 5yz
therefore, F : R2 = (x3 + 5yz)2
so to satisfy our conditions, F(x,y,z) = ((x3 + 5yz)2)1/2

or more generally
F(g(x)) = R2 -> R = [F(g(x))]1/2

I wonder if my notations and the way i wrote them is correct. :grumpy:

13. Jul 9, 2007

### matt grime

You are wrong. The square root operator always returns the positive square root. If you wish to artifically assert an 'if a>0 then, else if x<0 then...' fine, but what you wrote is still wrong.

14. Jul 9, 2007

### matt grime

this is acceptable and precisely the example i had in mind (which makes me wonder did you bother to read the posts ahead of you that actually answered the OPs questions?).

thee cube root of a^3 is always a - the choice of principal branch takes care of that automatically.

15. Jul 9, 2007

### matt grime

I don't think you've understood the question at all. R^2 means the real plane. Not the square of a function R.

16. Jul 9, 2007

### cryptic26

You seem to be confused. Why would the square root operator return only the positive square root? Besides, x can take any value (both negative an positive). There are two solutions to that equation. It is not necessary for the function to be injective.

17. Jul 9, 2007

### Dick

A relation that associates multiple values with a given (x,y) is just that. A 'relation'. Not a function. +/-sqrt(x^2+y^2) defines TWO functions. Neither of which works for all a. Your example (x^3+y^3)^(1/3) works MUCH BETTER. Why not just go with that?

18. Jul 9, 2007

### cryptic26

I understand that (x^3+y^3)^(1/3) is a better example. But, even that can have multiple roots (only that in the real plane, there shall be just one root). If need to write for the exam, then I would chose this particular one. Thanks for your explanation.

19. Jul 9, 2007

### matt grime

1. the square root function by definition only returns one value (the principal root, usually).

2. A function cannot return two possible values - that ceases to be a function

The symbol $\sqrt{y}$ does not mean 'both solutions to the equation x^2=y'. It means the principal value, which for real numbers means the positive square root.

your attempt at the function is not well defined: f(x,y)=sqrt(x^2+y^2) and then if we multiply (x,y) by a, we have to pull the correct square root out. This would require you to specify infinitely many things in a consistent way. For instance what of (3,-4) and (-3,4)? sqrt(x^2+y^2)=5 in both cases. Now which one do I choose to be 5 and which -5? Now what about (5,-12) and (-5,12)? Which of those do I choose to be 13 and which -13?

Last edited: Jul 9, 2007
20. Jul 9, 2007

### matt grime

For the umpteenth time: x^{1/3} is a unique number, it is not a multiple choice. The 27^{1/3}=3, (-27)^{1/3}=-3. That is the definition of those symbols: x^1/3 does not mean the set of cube roots of x. It means _the_ value on the principal branch.

21. Jul 9, 2007

### cryptic26

Who/What is OP? Yeah, I read the posts ahead and I don't see who actually answered the question.

Last edited: Jul 9, 2007
22. Jul 9, 2007

### cryptic26

Dick pretty much nailed it, when he explained the difference between function and relation. I understand why it is important to have unique values, when defining a function. Let me say that one more time or as you would say "the umpteenth" time.

23. Jul 9, 2007

### Dick

'OP' is 'online poster' or something like that. I.e. the (presumed) person who posted the original question. And, yes, the question was never actually answered in the thread. Before you reopened the thread I was thinking of examples like f(x,y)=y^2/x. Apparently not thinking very hard either since that isn't defined on all of R^2.

24. Jul 9, 2007

### cshum00

Yeah, i am completely confused. I re-read the question carefully and researched a little. Is the question about making a function of a vector which satisfy the associative property of the scalar multiplication and not the distributivity property right? Or not? :grumpy:

25. Jul 9, 2007

### cryptic26

cshum00, I think you already know various example of transformation from R^2 -> R. Think about the area of a rectangle. It takes in two variables (x , y) and gives one value.
Area= f(x,y) = xy
But, in order for a function to be a linear map, it has to satisfy both properties
1)f(u+v) = f(u) + f(v), where say u = (x1,y1) and v = (x2,y2).
2) f(a*u) = a*f(u).

As you can see, area is not a linear map because it fails both (1) and (2).

Area(u+v) = (x1+x2)*(y1 +y2) = x1y1+x2y2+x1y2+x2y1 (that is not the same as Area(u) +Area(v)).

Also, Area (a*u) = Area(ax1,ay1) = a^2 * x1y1 (which is not the same as a*x1y1).

The original problem requires you to think of a function that satisfies (2) but fails (1).

The function that you wrote earlier was R^3-> R as it takes 3 variables x,y and z. Squaring the equation will not make any difference because you still need 3 variables to define the vector space.

Last edited: Jul 9, 2007