Vector Identity (del operator)

AI Thread Summary
The discussion focuses on the transformation of the expression p^ (1/m)∇p to (m/(m+1))∇p^(m+1/m). Participants highlight the use of the chain rule in vector calculus, specifically noting that ∇p^a equals a p^(a-1)∇p. By selecting an appropriate value for 'a', the transformation can be simplified effectively. The conversation emphasizes the clever application of calculus principles to achieve the desired result. Overall, the exchange illustrates a practical approach to understanding vector identities.
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i am completely lost as to how to go from
p^ \frac{1}{m}∇p
to
\frac{m}{m+1} ∇p^\frac{m+1}{m}
 
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In general, the chain rule should tell you that,

\nabla p^a = a p^{a-1} \nabla p

Choose an appropriate value for a and see what happens.
 
oh wow...nice trick
what-you-did-there-i-see-it.thumbnail.jpg
 
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