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Homework Help: Homogeneous function of degree n

  1. Dec 26, 2013 #1
    A function f is called homogeneous of degree n if it satisfies the equation

    [f(tx,ty,tz)]=(t^n) *[f(x,y,z)] for all t, where n is a positive integer and f has continuous second order partial derivatives".

    I dont have equation editor so let curly d=D

    I need help to show that

    (x)(D[f(tx,ty,tz)]/Dtx)+(y)(D[f(tx,ty,tz)]/Dty)+ (z)(D[f(tx,ty,tz)]/Dtz) = (n) * [f(x,y,z)]

    The hint that is given is to use the chain rule to differentiate [f(tx,ty,tz)] with respect to t.

    I can get to

    (n)(t^n-1)[f(x,y,z)]= (x) * D[f(tx,ty,tz)]/D(tx)
    +(y) * D[f(tx,ty,tz)]/D(ty)
    +(z) * D[f(tx,ty,tz)]/D(tz)

    And I only see

    x(Df/Dx)+y(Df/Dy)+z(Df/Dz) = n[f(x,y,z)]

    For n=1 and the trivial t=1

    However, this equality should work for all n and ALL t

    HELP ME.
    Last edited: Dec 26, 2013
  2. jcsd
  3. Dec 26, 2013 #2


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    Staff: Mentor

    This equation is true for all (real and positive) t, so you can use whatever you like. The right choice will give you the result you need.

    Please do not add so many pointless empty lines, they make the post hard to read.

    You can use LaTeX between $$ $$ or ## ## (inline mode):
    $$\frac{\partial f}{\partial x}$$
  4. Dec 26, 2013 #3
    You haven't answered my question...

    I'm sorry, but I can see why it would work for all positive and real t, but the correct derivation gets rid of t and only leaves n

    Again, I see why n=1 would give the
    CORRECT derivation only.

    x(Df/Dx)+y(Df/Dy)+z(Df/Dz) = n[f(x,y,z)]

    NO t^n-1

  5. Dec 26, 2013 #4


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    Staff: Mentor

    I did.

    Instead of a specific value for n (like n=1), put a specific value for t in your equation.

    For which value of t do you get tn-1=1?
  6. Dec 26, 2013 #5
    I get

    (n)(t^n-1)[f(x,y,z)]= (x) * D[f(tx,ty,tz)]/D(tx)
    +(y) * D[f(tx,ty,tz)]/D(ty)
    +(z) * D[f(tx,ty,tz)]/D(tz)

    Is different from the correct answer,

    (n) * [f(x,y,z)] = (x) * D[f(tx,ty,tz)]/D(tx)
    +(y) * D[f(tx,ty,tz)]/D(ty)
    +(z) * D[f(tx,ty,tz)]/D(tz)
  7. Dec 26, 2013 #6
    For which value of t do you get t^n-1=1?

    t= 1
  8. Dec 26, 2013 #7
    Shouldn't t be all positive real numbers and NOT only 1?
  9. Dec 26, 2013 #8


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    Staff: Mentor

    You wrote a wrong "correct answer" in post 5 (at least one that differs from the one in post 1).

    The correct answer is a special case of your result. If you set t=1 in your equation (remember: it is true for all positive t, so it is true for t=1), you get the correct answer.
  10. Dec 26, 2013 #9
    I've corrected post 1.

    I'm still confused.
  11. Dec 26, 2013 #10
    Suppose t = 10

    What happens?
  12. Dec 26, 2013 #11
    Your explanation is not very comprehensive.

    I can see that it works for t=1. Plug in and no brainer... But this derivation is not this trivial.

    I'm sorry, but how long have you been answering questions here on pf?

    Thank you
    Last edited: Dec 26, 2013
  13. Dec 26, 2013 #12


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    Staff: Mentor

    Your question is not very clear, and changing with time (so you want an answer with t inside now? What changed since you wrote the original post?).
    If everything in mathematics would be trivial, we wouldn't need mathematicians.
    1 year 9 months.
  14. Dec 26, 2013 #13
    I wrote "this derivation is NOT this trivial" to point out your trivial explanation.

    I don't mean any offense... But this derivation is not merely a plug in 1.
    Last edited: Dec 26, 2013
  15. Dec 26, 2013 #14
  16. Dec 26, 2013 #15


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    Staff: Mentor

    Plugging in t=1 perfectly removes that factor.
    I don't see the problem.
  17. Dec 26, 2013 #16
    Plugging in t=10 doesn't appear to work as it works for plugging in t=1

    How would it work for t=10?


    Somebody help me?!?!
    Last edited: Dec 26, 2013
  18. Dec 27, 2013 #17


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    Staff Emeritus
    Science Advisor
    Gold Member

    The left hand side of this equation has t's in it, but the right hand side does not. What is your full problem statement? I suspect it is
    [tex] x \frac{\partial f(x,y,z)}{\partial x} + y \frac{\partial f(x,y,z)}{\partial y} + z \frac{\partial f(x,y,z)}{\partial z} = n f(x,y,z) [/tex]

    This is important because when they suggest that you add in t and differentiate with respect to it, you have an extra variable that you are free to do whatever you want with. In particular, you can make t=1 and just hope that you get the correct answer at the end, and you don't have to worry about it working for other values of t.
  19. Dec 27, 2013 #18
    Office shredder,

    (n)(t^n-1)[f(x,y,z)]= (x) * D[f(tx,ty,tz)]/D(tx)

    +(y) * D[f(tx,ty,tz)]/D(ty)

    +(z) * D[f(tx,ty,tz)]/D(tz)

    The above should turn into

    (n)[f(x,y,z)]= (x) * D[f(x,y,z)]/D(x)

    +(y) * D[f(x,y,tz)]/D(y)

    +(z) * D[f(x,y,z)]/D(z)

    For all real t and for all positive integers n

    It is obvious and ludicrously clear that this derivation will work for the no brainer t=1 and n=1

    how does this derivation work for the rest of t OTHER THAN 1??
    Last edited: Dec 27, 2013
  20. Dec 27, 2013 #19
    Can somebody who desires a bit of a challenge derive the derivation in post 18 using t= 7 ?

    I am not interested in people looking to just find one condition that makes this derivation true.
  21. Dec 29, 2013 #20
    Never thought I would be more knowledgeable and more detailed than a physics forum tutor
  22. Dec 29, 2013 #21
    I guess I won't share also. YEAH CAPITALISM
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