- **Calculus**
(*http://www.physicsforums.com/forumdisplay.php?f=109*)

- - **Finding Complicated Inverse Functions**
(*http://www.physicsforums.com/showthread.php?t=473201*)

Finding Complicated Inverse FunctionsWe have a formula for the derivative of an inverse function:
dy/dx = 1/(dx/dy). Just how useful is it? Say we want to find the inverse of a complicated function, f(x), on an interval (a,b) on which f(x) is one-to-one. Can we use integration to find such a function? Example: Say we didn't know much about the function h(x) = sin(x), but wanted to express its inverse as an integral (this was my inspiration for the idea). How could this be done? More importantly, this would apply to functions like F(x) = x*e^x. Its inverse, W(x), is important in several applications. Say I choose the branch on (0, infinity). Could I express this branch (or any other I choose) as an integral of well-defined functions? |

Re: Finding Complicated Inverse FunctionsWhat do you mean "express its inverse as an integral?"
If you want to find the derivative of an inverse, implicit differentiation(or similar variations) is sometimes the only way. To express arcsin(x) as an integral, one would find its derivative, and say arcsin(x) = ∫ darcsin(x) or ∫ sqrt(1-x ^{2})^{-1} dx.Is that what you mean, or do you want to know how to do implicit differentiation? |

Re: Finding Complicated Inverse FunctionsOH! You mean find the inverse by differentiating, the integrating the inverse of the derivative. That's a very clever idea! But, unfortunately, in the hard cases like arcsin, it won't get you anywhere.
In the case of sin(x), you would get arcsin(x). [tex]d(sinx) = cosx dx[/tex] [tex]\frac{d(sinx)}{dx} = cosx[/tex] [tex]\frac{dx}{d(sinx)} = secx = \frac{1}{\sqrt{1 - x^2}}[/tex] The rightmost being the most common form of the derivative of arcsin(x) [tex]\int \frac{dx}{d(sinx)} = \int secx = \int \frac{1}{\sqrt{1 - x^2}} = arcsin(x) + C[/tex] A facepalm would be appropriate here. :tongue: |

Re: Finding Complicated Inverse FunctionsQuote:
Once I can find inverses of one-to-one functions, I can restrict the domains of any differentiable functions until they are one-to-one on an interval, and work from there. There is one method I can think of, which involves the intersection of a curve with a line. say you have [tex]y = f(x)[/tex]. Switch the x and y so that [tex]x = f(y)[/tex]. Now, x is a function of the independent variable y, which produces an "inverse graph". A line parallel to the y axis would now be considered "horizontal" in this context. Find the intersection(s) [tex]f(y) = c[/tex] where x = c. x is the argument of the inverse function of f(x). The values where f(y) = c are the values of the inverse function. This is only a numerical method, and will give several values for inverses of functions that are not one-to-one. My hope is that it could shed some light on my greater problem. Thanks for reading my post. Any help will be appreciated. :biggrin: |

Re: Finding Complicated Inverse FunctionsHave you learned Taylor Series? It seems to accomplish what you're attempting. Not to be a cynic, but I'm pretty sure that when trying to apply this to functions with non-elementary inverses you'll end up just integrating to that inverse(ei the non-elementary inverse will be the only function with the derivative in question). It does have application in finding inverses of functions with elementary inverses, but I, for one, would prefer to stay as far away from integrating (3x
^{2}+2x+1)^{-1} as possible. |

Re: Finding Complicated Inverse FunctionsI like the idea of series, but I don't think they would be applied in that way exactly. Integrating (3x^2 + 2x +1)^-1 would not give the inverse of f(x) = x^3 + x^3 + x.The existing formula requires that the inverse be known to find the derivative. Once you have the derivative, you can integrate it to find the inverse (a moot point). I'm wondering if I can sidestep this fact.
What is known about finding inverses? What are some examples of common and advanced procedures? This might be a launching point. |

All times are GMT -5. The time now is 10:18 PM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums