MHB Calculation in proof of Poincare's Lemma

[oblixps]

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

in the course of proving Poincare's lemma my book does the following: \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}.

what I'm confused about is why didn't the book do \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}.

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

 

[oblixps]

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:

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

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

 

[Jameson]

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]

Hi oblixps, :)

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

in the course of proving Poincare's lemma my book does the following: \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}.

what I'm confused about is why didn't the book do \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}.

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:

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

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

... and the web-link where you found the above statement.

Kind Regards,
Sudharaka.

 

[oblixps]

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.