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calculation in proof of Poincare's Lemma

oblixps

Member
May 20, 2012
38
[tex] A_{j_{1}...j_{p}} [/tex] is a (0, p) tensor defined in a star shaped region of some point P where the coordinates [tex] x^1 = ... = x^n = 0 [/tex].

in the course of proving Poincare's lemma my book does the following: [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l}\frac{\partial(tx^l)}{\partial x^j} = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} t\delta^{l}_{j} = t\frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} [/tex].

what i'm confused about is why didn't the book do [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial (tx^l)}\frac{\partial(tx^l)}{\partial x^j} [/tex].

what happened to that t in the "denominator" of the first fraction in the chain rule?
 

oblixps

Member
May 20, 2012
38
i was looking around on google and i ran across this related result in some lecture slides.

letting g(t) = f(tx, ty) and using the chain rule:

[tex] g'(t) = (\frac{\partial f}{\partial x})(tx, ty) * x + (\frac{\partial f}{\partial y})(tx, ty) * y [/tex].

once again i am confused on why they wrote [tex] \frac{\partial f}{\partial x} [/tex] instead of [tex] \frac{\partial f}{\partial (tx)} [/tex].
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
At the risk of getting an infraction from one of our moderators, I'm going to bump this thread as the OP has waited a bit and posted more info yet hasn't been helped. If possible I'd help myself but alas, this problem is out of my league.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi oblixps, :)

[tex] A_{j_{1}...j_{p}} [/tex] is a (0, p) tensor defined in a star shaped region of some point P where the coordinates [tex] x^1 = ... = x^n = 0 [/tex].

in the course of proving Poincare's lemma my book does the following: [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l}\frac{\partial(tx^l)}{\partial x^j} = \frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} t\delta^{l}_{j} = t\frac{\partial A_{j_{1}...j_{p}}}{\partial x^l} [/tex].

what i'm confused about is why didn't the book do [tex] \frac{\partial}{\partial x^j}A_{j_{1}...j_{p}}(tx^h) = \frac{\partial A_{j_{1}...j_{p}}}{\partial (tx^l)}\frac{\partial(tx^l)}{\partial x^j} [/tex].

what happened to that t in the "denominator" of the first fraction in the chain rule?
Can you please tell me what your book is......

i was looking around on google and i ran across this related result in some lecture slides.

letting g(t) = f(tx, ty) and using the chain rule:

[tex] g'(t) = (\frac{\partial f}{\partial x})(tx, ty) * x + (\frac{\partial f}{\partial y})(tx, ty) * y [/tex].

once again i am confused on why they wrote [tex] \frac{\partial f}{\partial x} [/tex] instead of [tex] \frac{\partial f}{\partial (tx)} [/tex].
.... and the web-link where you found the above statement.

Kind Regards,
Sudharaka.
 

oblixps

Member
May 20, 2012
38
thanks for the reply.

the book i'm using is "Tensors, Differential Forms, and Variational Principles" by Lovelock and Rund. Just in case you have access to a copy, it should be on page 143.

and the lecture slides I referred to are from: http://www.math.upenn.edu/~ryblair/Math 600/papers/Lec1.pdf

the result i posted is on slide 22 and is what starts off one of the proofs.