How to expand inverse function

In summary: I think the question for the OP has been settled. The rest of the discussion is a very good example how language can be read differently and why accuracy is so important!In summary, Inverse functions are differentiable if and only if they are given explicit definitions.
  • #1
kent davidge
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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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  • #2
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.
 
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  • #3
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}##.
 
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  • #4
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##
 
  • #5
WWGD said:
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 ...
kent davidge said:
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##.
 
  • #6
fresh_42 said:
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?
 
  • #7
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.
 
  • #8
WWGD said:
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.
 
  • #9
fresh_42 said:
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##?
 
  • #10
Why? ##\sqrt{x}## isn't differentiable at ##x=0##, so what?
 
  • #11
But it is invertible there. EDIT: Why don't we take the discussion elsewhere to avoid disrupting this one further?
 
  • #12
WWGD said:
But it is invertible there.
But who cares?
kent davidge said:
If instead, I'm given ##f^{-1}(x)##
means to me: Consider ##f^{-1}\, : \,x \longmapsto +\sqrt{x}##.
Given a function, not given the fact that ##f## is invertible. These are two different statements. I admit that the OP could be read differently, but I read it as if ##f^{-1}## was explicitly given. And in this case I can write
fresh_42 said:
$$f^{−1}(x_0+v)=f^{−1}(x_0)+J_{x_0}f^{−1}⋅v+\tilde r(v)$$
with ##x_0 \in (0,\infty)## in case of ##f^{-1}(x)=\sqrt{x}##, i.e. ##x_0 \neq 0##.
EDIT: Why don't we take the discussion elsewhere to avoid disrupting this one further?
I think the question for the OP has been settled. The rest of the discussion is a very good example how language can be read differently and why accuracy is so important! In the end the whole debate is about what is meant by ##f^{-1}## is given: as a function in its own right (my interpretation) or as the inverse function of ##f## (your interpretation). But what should is given mean if you only relate to the existence?
 

Related to How to expand inverse function

1. How do you find the inverse of a function?

To find the inverse of a function, you can switch the x and y variables and then solve for y. The resulting equation will be the inverse function.

2. Can all functions have an inverse?

No, not all functions have an inverse. For a function to have an inverse, it must pass the horizontal line test, meaning that every horizontal line intersects the graph of the function at most once.

3. What is the domain and range of an inverse function?

The domain of an inverse function is the range of the original function, and the range of an inverse function is the domain of the original function.

4. How do you graph an inverse function?

To graph an inverse function, you can reflect the original function's graph over the line y=x. This will result in the inverse function's graph.

5. Can you expand an inverse function?

Yes, you can expand an inverse function by using the inverse function property, which states that the composition of a function and its inverse is equal to the identity function. This can be used to simplify and expand inverse functions.

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