# How to expand inverse function

• I
• kent davidge

#### kent davidge

If I'm given a function ##f(x)##, say it has continuos first derivative, then I expand it as ##f(x + \Delta x) = f(x) + (df / dx) \Delta x##. If instead, I'm given ##f^{-1}(x)## how do I go about expanding it? Will this be just ##f^{-1}(x + \Delta x) = f^{-1}(x) + (df^{-1} / dx) \Delta x##?

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Well, I'd have to imagine that since an inverse function is simply another function it should hold that rules for normal functions hold for inverse functions. I'd say yes then.

kent davidge
We generally have $$f(x_0+v)= f(x_0) + J_{x_0}f\cdot v + r(v)$$
We have the same equation for ##f^{-1}(x)##, i.e.
$$f^{-1}(x_0+v)= f^{-1}(x_0) + J_{x_0}f^{-1}\cdot v + \tilde r(v)$$
So yes, you can write it this way, just be careful with ##\dfrac{df^{-1}}{dx}##.

Harperchisari and kent davidge
But I think you first need to make sure the conditions of the inverse/implicit function theorem are met, so that you know you have at least a local (differentiable) inverse. Consider the problems, e.g. of ##\sqrt x## near ##x=0##. And then maybe you can use the chain rule on ##f(f^{-1}(x))=x##

But I think you first need to make sure the conditions of the inverse/implicit function theorem are met, so that you know you have at least a local (differentiable) inverse. Consider the problems, e.g. of ##\sqrt x## near ##x=0##. And then maybe you can use the chain rule on ##f(f^{-1}(x))=x##
Not in the situation stated in the OP ...
If instead, I'm given ##f^{−1}(x)## ...
... then we can write down the Weierstraß equation simply for this given function. How to compute the derivative given the derivative of ##f## is another question. If ##f^{-1}## is given, then we do not have to bother about ##f##.

Not in the situation stated in the OP ...

... then we can write down the Weierstraß equation simply for this given function. How to compute the derivative given the derivative of ##f## is another question. If ##f^{-1}## is given, then we do not have to bother about ##f##.
How do you know the inverse is differentiable?

I can't think at this point of a counter, but I don't see how we can know for certain under the conditions in the OP.

How do you know the inverse is differentiable?
By inspection. Given ##f^{-1}## means to me, given a function ##x \longmapsto f^{-1}(x)## and not given that ##f## has an inverse. I consider it as just another function, since this was what I read from post #1. In your example with the square root, I assume that a sign is given. And ##x \longmapsto +\sqrt{x}## is differentiable on ##(0,\infty)## regardless whether it inverts ##x \longmapsto x^2## somewhere or not.

By inspection. Given ##f^{-1}## means to me, given a function ##x \longmapsto f^{-1}(x)## and not given that ##f## has an inverse. I consider it as just another function, since this was what I read from post #1. In your example with the square root, I assume that a sign is given. And ##x \longmapsto +\sqrt{x}## is differentiable on ##(0,\infty)## regardless whether it inverts ##x \longmapsto x^2## somewhere or not.
But don't we need the inverse to be defined in a neighborhood of a point, as in the case of ##\sqrt x## at ##x=0##?

Why? ##\sqrt{x}## isn't differentiable at ##x=0##, so what?

But it is invertible there. EDIT: Why don't we take the discussion elsewhere to avoid disrupting this one further?

But it is invertible there.
But who cares?
$$f^{−1}(x_0+v)=f^{−1}(x_0)+J_{x_0}f^{−1}⋅v+\tilde r(v)$$