Question

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

Answer 1

add a "t" variable to each x and y...

Answer 2

[(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?

Answer 3

is homo..and degree 1

Answer 4

(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

Answer 5

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

Answer 6

thanks i got the first one wasn't sure about the second!^_^

Answer 7

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....

Answer 8

do you know how to show that eulers theorem holds for these equations?

Answer 9

...breif me on eulers thrm ...whats it state again?

Answer 10

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

Answer 11

this is from my notes but im having a hard time understanding and applying it....

Answer 12

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....

Answer 13

fi(x) seems to mean [f(x)]^i right?

Answer 14

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 :)

Answer 15

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....:)