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SamBridge

Is there a way to find an inverse anyway?

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So there's all sorts of equations where when you solve for their inverses, you don't get a "function", but I honestly couldn't care less if it's not a function, I just want to know the straight up relationship between y and x of the inverse and I don't care if it fails the vertical line test, so be it. I know there's some way to convert cartesian equations to polar equations, so can I convert, say, y=^6-8x^5+4(x+2)x^4-(x-4)x^3-20, convert it to a polar equation, solve for the inverse in terms of polar coordinates, and then back substitute into cartesian coordinates to get an x-y relationship with y in terms of x? I really don't mind if I have a bunch of roots, technically sqrt(x) has two outputs just on it's own and it's not a particularly unique function.

Edited by SamBridge

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I'm not absolutely certain, but I think you can convert a graph of a function to its "inverse" by reflecting it in the line x=y.

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You want to write not y(x), but x(y)?

 

If so I think this is always possible, mod what you say about uniqueness, but not necessarily in a closed form. For example, you know that there is no closed form in general for the roots of a quintic polynomial. However the roots always exist.

 

Changing coordinate systems should also be fine, if you have a genuine function then it is no problem at all.

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You want to write not y(x), but x(y)?

 

If so I think this is always possible, mod what you say about uniqueness, but not necessarily in a closed form. For example, you know that there is no closed form in general for the roots of a quintic polynomial. However the roots always exist.

 

Changing coordinate systems should also be fine, if you have a genuine function then it is no problem at all.

Well, say I have a function that's y=...some function with an inverse that's not actually a function. It would be converted to rsin(theta)=...f^-1(rcos(theta))... so is there a way to rearrange the inverse of that and have it equal the inverse of the original Cartesian function and rearrange it to get rsin(theta)=...?

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Well, say I have a function that's y=...some function with an inverse that's not actually a function. It would be converted to rsin(theta)=...f^-1(rcos(theta))... so is there a way to rearrange the inverse of that and have it equal the inverse of the original Cartesian function and rearrange it to get rsin(theta)=...?

 

 

To save you some trouble, the inverse (in your sense) of the sine function is not the cosine function it is the "angle whose sine is the given value" and often called the arcsin.

So the pair is:-

 

if y = sin(x) then x = arcsin(y)

edit the arcsin(x) is also commonly written sin-1(x) - they refer to the same thing.

 

 

Just to be clear the reciprocal of the sin is the cosecant. 1/sin(x) does not equal cosin(x) it equals cosec(x)

 

I would warn you further that what you are proposing is a tail chasing exercise IMHO.

Edited by studiot

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To save you some trouble, the inverse (in your sense) of the sine function is not the cosine function it is the "angle whose sine is the given value" and often called the arcsin.

So the pair is:-

 

if y = sin(x) then x = arcsin(y)

edit the arcsin(x) is also commonly written sin-1(x) - they refer to the same thing.

 

 

Just to be clear the reciprocal of the sin is the cosecant. 1/sin(x) does not equal cosin(x) it equals cosec(x)

 

I would warn you further that what you are proposing is a tail chasing exercise IMHO.

I have no idea how anything you just said addresses any of my questions. I already know the inverse trig functions and their reciprocals, I'm talking about something completely different than simple trig identities, I'm talking about solving for an inverse using a substitution with a polar coordinate system. If rsin(theta)=y, then in order to back substitute in a Cartesian function in terms of y I need to solve for rsin(theta) and back substitute into Cartesian functions if it's possible. The hope is that because trig functions are more manipulable and have all sorts of weird properties that you'd get a better Cartesian equation to work with after back substituting.

Edited by SamBridge

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It completely depends on what your function is. Bear in mind that if you can't write a closed form for the inverse of the function then this approach won't really help unless you know something special about what happens when you apply a polar transformation to its inverse. It is entirely usual for the inverse of a function to exist but not be expressible in a closed form.

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