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Exercise [10.07]
http://www.roadtoreality.info/viewtopic.php?f=19&t=185
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Author:  Sameed Zahoor [ 14 Jul 2008, 10:37 ]
Post subject:  Exercise [10.07]

\xi=A\frac{\partial}{\partial{X}}+B\frac{\partial}{\partial{Y}}

\Rightarrow a\frac{\partial}{\partial{x}}+b\frac{\partial}{\partial{y}}=A(\frac{\partial{x}}{\partial{X}}\frac{\partial}{\partial{x}}+\frac{\partial{y}}{\partial{X}}\frac{\partial}{\partial{y}})+B(\frac{\partial{x}}{\partial{Y}}\frac{\partial}{\partial{x}}+\frac{\partial{y}}{\partial{y}}\frac{\partial}{\partial{y}})

\Rightarrow a\frac{\partial}{\partial{x}}+b\frac{\partial}{\partial{y}}=(A\frac{\partial{x}}{\partial{X}}+B\frac{\partial{x}}{\partial{Y}})\frac{\partial}{\partial{x}}+(A\frac{\partial{y}}{\partial{X}}+B\frac{\partial{y}}{\partial{Y}})\frac{\partial}{\partial{y}}

Comparing b/s we get,

a=A\frac{\partial{x}}{\partial{X}}+B\frac{\partial{x}}{\partial{Y}}

b=A\frac{\partial{y}}{\partial{X}}+B\frac{\partial{y}}{\partial{Y}}

By analogy,

A=a\frac{\partial{X}}{\partial{x}}+b\frac{\partial{X}}{\partial{y}}

B=a\frac{\partial{Y}}{\partial{x}}+b\frac{\partial{Y}}{\partial{y}}

Author:  Langing [ 12 May 2011, 17:21 ]
Post subject:  Re: Exercise [10.07]

Typo at end of second line at . . . B(. . .+, the partial derivative of small y with respect to small y should be with respect to large Y. Except for this, this partial derivative is correct when used below the second line.

Author:  Langing [ 23 May 2011, 18:10 ]
Post subject:  Re: Exercise [10.07]

Sameed answered the second part of Exercise 10.7 first and then found the first part by analogy. I addressed Exercise 10.7 the other way around, which was requested, 'Find A and B in terms of a and b, then use analogy to write down a and b in terms of A and B.'

I also provide enough discussion, that someone who needs elaboration has it.

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