Homogeneous function of degree 0

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JacksonBrukes

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Hello I have stumbled upon a task that I have tried solving for some time, but I am stuck.
There is a task in my book that asks me if this function has homogeneous degree 0 (yes or no, and why), picture of this function is below.
I have had easier tasks that in my book about finding homogeneous degree , but this one stands out.
If anyone could help me it would be great!

Is this function homogeneous degree 0?
Capture.PNG

Thank you!
 
JacksonBrukes said:
Hello I have stumbled upon a task that I have tried solving for some time, but I am stuck.
There is a task in my book that asks me if this function has homogeneous degree 0 (yes or no, and why), picture of this function is below.
I have had easier tasks that in my book about finding homogeneous degree , but this one stands out.
If anyone could help me it would be great!

Is this function homogeneous degree 0?
View attachment 14558

Thank you!
Click to expand...
Is the following an EXACT replica of your question:

Is this function homogeneous degree 0?

If NOT please provide verbatim replica of the question.

Also please tell us:

What is the definition of homogeneous function (please cite example)?

What is the definition of homogeneous degree (please cite example)?
 
JacksonBrukes said:
Hello I have stumbled upon a task that I have tried solving for some time, but I am stuck.
There is a task in my book that asks me if this function has homogeneous degree 0 (yes or no, and why), picture of this function is below.
I have had easier tasks that in my book about finding homogeneous degree , but this one stands out.
If anyone could help me it would be great!

Is this function homogeneous degree 0?
View attachment 14558

Thank you!
Click to expand...
Evidently (as in your title but not in the question) you meant homogeneous of degree 0. That is defined, for example, in https://en.wikipedia.org/wiki/Homogeneous_function

A presentation with only two variables is given at https://www.mathsisfun.com/calculus/homogeneous-function.html

You may want to show us how the definition was stated for you, if it was different from these.

Now please show us where you had trouble applying the definition you were given.
 
Yes the question is, is the function homogeneous of degree 0.
The definition that I use in my book is a function is homogeneous if f(tx, ty) = tk f(x,y) for all t > 0.
An easy example would to be:
We have a function f(x,y) =x^2 + y^2

f(tx,ty) = (tx)^2 +(ty)^2 =t^2x^2 + t^2y^2 = t^2(x^2 + y^2)

by looking at the result we see that f(tx,ty) = t^2f(x)
which means that the functions is homogeneous of degree 2.
 
Dr.Peterson said:
Evidently (as in your title but not in the question) you meant homogeneous of degree 0. That is defined, for example, in https://en.wikipedia.org/wiki/Homogeneous_function

A presentation with only two variables is given at https://www.mathsisfun.com/calculus/homogeneous-function.html

You may want to show us how the definition was stated for you, if it was different from these.

Now please show us where you had trouble applying the definition you were given.
Click to expand...

Thank you Dr. Peterson, sorry for not being clear enough. I have posted a reply that explain the definition.
Just by the look at this function it does not look like it is homogeneous of degree 0. I have read through your sources and they were useful, thank you.
The problem I have with this function is that it includes subtraction and division, which I am not sure how to handle (what I am allowed to do), the examples in the sources show only multiplication and addition.
 
Please show what you tried for the given function. I don't think subtraction and division change things as much as you think. You're allowed to do anything algebra permits, and I imagine you have plenty of experience with algebra!

Often the way to learn something new is to boldly try the things you are not sure of, rather than wait for someone else to tell you that you can. If what you try is wrong, you will find out soon enough.
 
I can add that the process for your problem will start the same as the example you gave: substitute, expand, and factor. That isn't affected by subtraction. Then you will have to deal with the division; that amounts to simplifying a fraction.
 
JacksonBrukes said:
Yes the question is, is the function homogeneous of degree 0.
The definition that I use in my book is a function is homogeneous if f(tx, ty) = tk f(x,y) for all t > 0.
An easy example would to be:
We have a function f(x,y) =x^2 + y^2

f(tx,ty) = (tx)^2 +(ty)^2 =t^2x^2 + t^2y^2 = t^2(x^2 + y^2)

by looking at the result we see that f(tx,ty) = t^2f(x)
which means that the functions is homogeneous of degree 2.
Click to expand...
Yes, that function is homogeneous of degree 2. The function you originally gave was \(\displaystyle f(x)= \frac{x^3y+ x^4- y^4}{x^4+ x^2y^2- xy^3}\). Replacing x with xt and y with yt that becomes \(\displaystyle \frac{(x^3t^3)(yt)+ (x^4t^4)- (y^4t^4)}{x^4t^4+ (x^2t^2)(y^2t^2)- (xt)(y^3t^3)}= \)\(\displaystyle \frac{t^4(x^3y+ x^4- y^4)}{t^4(x^4+ x^2y^2- xy^3)}\).

Now what is the degree of homogeneity of that function?
 
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HallsofIvy said:
Yes, that function is homogeneous of degree 2. The function you originally gave was \(\displaystyle f(x)= \frac{x^3y+ x^4- y^4}{x^4+ x^2y^2- xy^3}\). Replacing x with xt and y with yt that becomes tex]\frac{(x^3t^3)(yt)+ (x^4t^4)- (y^4t^4)}{x^4t^4+ (x^2t^2)(y^2t^2)- (xt_)(y^3t^3}= \Frac{t^4(x^3y+ x^4- y^4)}{t^4(x^4+ x^2y^2- x^3}[/tex].

Now what is the degree of homogeneity of that function?
Click to expand...

Thank you for the reply. When you put it this way it makes it clear that the homogeneity of that function is 4 :D So the division if this function made it seem like it was much harder then it actually. Thank you once again to all that helped me.

PS (that may seem like a stupid question but just to be 100% sure)
HallsofIvy said:
{t^4(x^3y+ x^4- y^4)}{t^4(x^4+ x^2y^2- x^3}
Click to expand...
there should xy^3 at the end right? (it was just a mistype I assume)
 
Dr.Peterson said:
No, the function is not homogeneous of degree 4. I think you stopped too soon; there's another step after what HallsofIvy showed you.
Click to expand...
Well this is awkward, I was so happy to think it was the answer.
I have been thinking about it, and came up to this. I have to divide t so that it becomes 1, then I figured out that 1 is the same as t^0 so it is homogeneous of degree 0. Is this also incorrect?
 
JacksonBrukes said:
Well this is awkward, I was so happy to think it was the answer.
I have been thinking about it, and came up to this. I have to divide t so that it becomes 1, then I figured out that 1 is the same as t^0 so it is homogeneous of degree 0. Is this also incorrect?
Click to expand...
Yes, of course that is correct. Just look at your definition which is homogeneous of degree k if f(tx, ty) = t^k f(x,y) for all t > 0. It did not say \(\displaystyle \dfrac{t^k}{t^k}f(x,y)\) which by the way equals \(\displaystyle \dfrac{t^m}{t^m}f(x,y)\). That is \(\displaystyle \dfrac{t^4}{t^4}f(x,y)\) = \(\displaystyle \dfrac{t^7}{t^7}f(x,y)\) This is exactly why your method would not work as you (incorrectly) get the degree to be 4 and 7 and......
 
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