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anonymous

  • 5 years ago

if homogeneous find the degree of homogeneity 1) (x^2+y^2)^1/2 2) (x+y)^5 + x^5-y^5 help plzz....:)

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  1. amistre64
    • 5 years ago
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    add a "t" variable to each x and y...

  2. amistre64
    • 5 years ago
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    [(tx)^2 + (ty)^2] ^ (1/2) [t^2x^2 + t^2y^2]^(1/2) [t^2(x^2 + y^2)]^(1/2) t^2^(1/2) = t right?

  3. amistre64
    • 5 years ago
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    is homo..and degree 1

  4. amistre64
    • 5 years ago
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    (tx+ty)^5 + (tx)^5-(ty)^5 (tx+ty)^5 + t^5(x^5-y^5) I dont think that left hand part is going to be homo.... youd have to factor it out and see if a single "t" variable can be pulled from it... and I dont think it can

  5. amistre64
    • 5 years ago
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    t^5x^5 +5t^4x^4ty +10txty +10txty +5txty +t^5y^5 yeah, its good.... if I did it right... its homo and t^5 degree

  6. anonymous
    • 5 years ago
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    thanks i got the first one wasn't sure about the second!^_^

  7. amistre64
    • 5 years ago
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    I used pascals triangle to determine the cooefficients for the ^5 part. and then realized that the xy variables all have exponents that equal ^5.... t^4 t^1 = t^5...and so on....

  8. anonymous
    • 5 years ago
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    do you know how to show that eulers theorem holds for these equations?

  9. amistre64
    • 5 years ago
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    ...breif me on eulers thrm ...whats it state again?

  10. anonymous
    • 5 years ago
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    23.9 Euler Theorem Consider a function f(x1, x2, x3,….,xn) that is homogeneous of degree r. Let us rewrite f(kx1, kx2, kx3,….,kxn) as f(w1, w2, w3,….,wn) where wi = kxi for i = 1, 2,..,n. By Definition 23.5.1 of the Total Differential we can write df = f1 dw1 + f2 dw2 + ….+ fi dwi +…... ……+ fn dwn. where each fi is the first order partial derivative with respect to wi. If we now divide by dk we get the following result: df = f1 dw1 + f2 dw2 + ….+ fi dwi +…... ……+ fn dwn. dk dk dk dk dk Since dwi = xi for each i = 1, 2, ….n we can refine the above result to dk df = f1 x1 + f2 x2 + ….+ fi xi +…... ……+ fn xn. * dk Recall that f(kx1, kx2, kx3,….,kxn) = kr f(x1, x2, x3,….,xn). Thus df = r k r-1 f(x1, x2, x3,….,xn). ** dk Also since each fi = f/wi it follows from the Chain Rule that f/xi = f/wi wi/xi = k f/wi *** Substituting ** into * we get r kr-1 f(x1, x2, x3,….,xn) = f1 x1 + f2 x2 + ….+ fi xi +…... ……+ fn xn Multiplying both sides by k we get r kr f(x1, x2, x3,….,xn) = kf1 x1 + kf2 x2 + ….+ kfi xi +…... ……+ kfn xn Substituting *** into the above equation we get r kr f(x1, x2, x3,….,xn) = f/x1 x1 + f/x2 x2 + ….+ f/xi xi +…... ……+ f/xn xn Hence the result in the next theorem. 23.9.1 Euler’s Theorem Assume that f(x1, x2, x3,….,xn) is a homogeneous function of degree r. Then r f(x1, x2, x3,….,xn) = f1(x) x1 + f2(x) x2 + ……….+ fi(x) xi +…..……+ fn(x) xn where x = (x1, x2, x3,….,xn). The right hand side is the sum of the products of the partial derivative fi(x) and the variable xi . The left hand side is simply r times the function f(x). In the case of n = 2 variables, Euler’s Theorem for a homogeneous function f(x1, x2) of degree r is given by r f(x1, x2) = f1(x) x1 + f2(x) x2

  11. anonymous
    • 5 years ago
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    this is from my notes but im having a hard time understanding and applying it....

  12. amistre64
    • 5 years ago
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    they use "k" instead of "t"; r is the value of the "degree". that "i" just means that with each iteration of the function we get a certain value....

  13. amistre64
    • 5 years ago
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    fi(x) seems to mean [f(x)]^i right?

  14. amistre64
    • 5 years ago
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    r f(x, y) = f^1(x) x + f^2(x) y is all I can make out of that, and dont even know if I m right :)

  15. anonymous
    • 5 years ago
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    yeah i get that just not sure how to apply it...i think it jus means i= variable in question....ok thanks for your help....:)

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