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clairaut
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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 don't 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.
[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 don't 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.
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