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Implicit function on functions composed of itself

  • Thread starter chy1013m1
  • Start date
15
0
Suppose F(x, y) is C1. F(0, 0) = 0. What conditions on F will guarantee that the equation F(F(x, y), y) = 0 can be solved for y as a C1 function of x near (0, 0) ?

would it simply be dF/dy not equal 0 ?
 

Answers and Replies

112
0
Use the implicit function theorem. Seems you are on the right track, but F is a composition of F and y, so you would have to use the chain rule to right out dF/dy properly.
 

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