# Finding the image of a multivariate function

• docnet
In summary, the conversation discusses a problem involving finding the image of a map and simplifying it. Different approaches are suggested, including using polar coordinates and the Weierstraß substitution. It is determined that the image exists for positive u and positive and negative v, and that the best solution would involve manipulating u and v to give x2 + y2 = 1. The conversation also touches on the difficulty of finding a single neat equation to describe the image and the asymptotic behavior of the image.
docnet
Gold Member
Homework Statement
Find the image of the unit disk under this transformation.
Relevant Equations
Please see below.

Hi all,

I am trying to figure out a way to simplify this problem to give the image of the map. I have not seen this function before and I am having trouble figuring out what the image should come out as.

I have tried graphing u and v separately as a function of x and y in R3 and both surfaces go to infinite z within the domain of our unit disc. This means u and v go to infinity in the UV plane.

What we know is the image exists for positive u and for positive and negative v.

The best and neatest solution would be obtained by manipulating u and v to give x2 + y2 = 1

I tried computing simple choices like u2+v2 that do not yield a solution.

Does anyone know what the image is off the top of their head?

Thank you.

Delta2
Ever considered polar coordinates ?

BvU said:
Ever considered polar coordinates ?
I don't think that solves the problem because we still have two independent parameters, whether we call them ##x,y## or ##\varphi,\psi##. I would attempt concentric circles: ##x^2+y^2=r^2## with ##0\leq r< 1##. Also the Weierstraß substitution comes to mind, but then again we have two angles.

I mean, it is already described, so which form of description are you heading for?

It's homework - we need an attempt from the OP first. To get going post #2 might at least lead to an attempt.
And yes, the image is 2D.

BvU said:
Ever considered polar coordinates ?
I tried substituting x = rcosθ and y = rsinθ

since our domain is defined as r= 1, we would need to solve the expression for r = 1. I have not been able to do this.

fresh_42 said:
I don't think that solves the problem because we still have two independent parameters, whether we call them ##x,y## or ##\varphi,\psi##. I would attempt concentric circles: ##x^2+y^2=r^2## with ##0\leq r< 1##. Also the Weierstraß substitution comes to mind, but then again we have two angles.

I mean, it is already described, so which form of description are you heading for?

Could you expand on the weierstraB substitution? My professor is known for giving devious problems that cannot be solved without some esoteric techniques.

What do you mean by concentric circles? I think the image on the UV plane does not include the -u quadrants.

A comprehensive description would be finding equation to describe the image, locating any singular sets that may be on the image, where the image is defined, whether orientation ispreserved or reversed, and how many times the image is covered at the corresponding complements of the singular set, if there are any.

You know ##x^2+y^2<1##, the unit disc. This can be filled out by concentric circles ##x^2+y^2=r^2## with radius ##r<1##. This way you reduce the two parameters to one parameter. Now you can use polar coordinates and see where the circles are mapped to. The Weierstraß substitution is a half tangent substitution used to transform trigonometric integrals into polynomial integrals. It can also be used the other way around. You can look it up on Wikipedia.

fresh_42 said:
You know ##x^2+y^2<1##, the unit disc. This can be filled out by concentric circles ##x^2+y^2=r^2## with radius ##r<1##. This way you reduce the two parameters to one parameter. Now you can use polar coordinates and see where the circles are mapped to. The Weierstraß substitution is a half tangent substitution used to transform trigonometric integrals into polynomial integrals. It can also be used the other way around. You can look it up on Wikipedia.

I understand. I think the difficulty comes from the fact we can't seem to solve for u or v as a function of theta alone. It's difficult to find a single neat equation to describe the image.

Trying your method, the graph of u and v shows u diverges to + infinity at integer multiples of pi, while v diverges to + / - infinity at integer multiples of pi for r=1. The following is a graph with r = .9.

Whish confirms our expectations.So the solution is

Delta2
I don't really understand. Given what's written, how do you determine if, say, (4,7) is in the image? It seems like all they've done is discover that some points in the top right and bottom right corner need to be in the image, and we still don't know what the rest of it looks like.

Even if the picture is right, what is the asymptomatic behavior? Does the image tighten in on some line, or something like ##v=\sqrt{u}## as u goes to infinity?

Office_Shredder said:
I don't really understand. Given what's written, how do you determine if, say, (4,7) is in the image? It seems like all they've done is discover that some points in the top right and bottom right corner need to be in the image, and we still don't know what the rest of it looks like.

Even if the picture is right, what is the asymptomatic behavior? Does the image tighten in on some line, or something like ##v=\sqrt{u}## as u goes to infinity?

We don't know if 4,7 is in the image from what is written, as this is not a complete solution.

We know from plotting points (0,0), (1,0), (0,1), and (1,1) are in the image, the geometry of the image is still unknown.

To determine if an arbitrary point (u,v) is in the image, we require solving for the inverse of map F. Finding the inverse map can be a diabolical problem that involves a lot of guess-work, trial and error, and luck. Finding the inverse map of F is equivalent to finding the implicit function of the image. If we have the inverse map, with variables x and y expressed solely in terms of u and v, we could simply compute x^2 + y^2 = 1 to find the implicit function of the image. It would be a neat solution.

I think you make a good point, that the precise asymptotic behavior could be determined. We have not learned how to do this in class yet.

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I'll revisit this problem tomorrow and see if I can find explicit equations for the image using the concentric circles idea by fresh_42. If we are lucky we should be able to locate a set of curves that give a better idea of the image.

fresh_42 said:
I would attempt concentric circles
Or some other family, e.g. circles radius r centred at (r-1,0), 0<r<1.
Office_Shredder said:
asymptomatic
Autocompleted, or just virus-free?

SammyS and Delta2
haruspex said:
Or some other family, e.g. circles radius r centred at (r-1,0), 0<r<1.
I tried this and ended up with circle equations: ##x+1=p=r\cdot \cos \varphi\, , \,s=r\cdot\sin \varphi .##

docnet said:
Does anyone know what the image is off the top of their head?

I think the image should be the right half plane ##u>0.## This mapping looks like the usual conformal equivalence from the disk to the half plane in ##\mathbb{C}## written out in cartesian coordinates (I didn't check this though).

Anyway it should be clear that the ##u## coordinate of any point in the image can be any positive number, so try to show for fixed ##u## that you can also make ##v## arbitrary. I think you can either do this directly or with the trig arguments suggested above. If you work out what this function is as a holomorphic map, you can probably even write down an inverse.

fresh_42 said:
I tried this and ended up with circle equations: x+1=p=r⋅cos⁡φ,s=r⋅sin⁡φ.

haruspex said:
Or some other family, e.g. circles radius r centred at (r-1,0), 0<r<1.

The issue may be that our domain is the unit disk defined as {x2 + y2 <1} and taking from a different subset of the xy-plane such as an offset disc will map to different points on the uv- plane.
Infrared said:
I think the image should be the right half plane u>0. This mapping looks like the usual conformal equivalence from the disk to the half plane in C written out in cartesian coordinates (I didn't check this though).

Anyway it should be clear that the u coordinate of any point in the image can be any positive number, so try to show for fixed u that you can also make v arbitrary. I think you can either do this directly or with the trig arguments suggested above. If you work out what this function is as a holomorphic map, you can probably even write down an inverse.

It seems that the image does not fill the right half-plane. An arbitrary point (u,v) corresponds to a point (x,y) and our map does not allow v to be arbitrary with respect to u, or vice versa.

Is holomorphic mapping a complex analysis technique? It is beyond my experience or knowledge at the moment.

Here is a graph I have been staring at for a few days. It is a graph of the equations in our map, and it is only a representation of it, or a way to visualize it. Since we can only work in three dimensions, we substitute u and v with z and use the two colors to establish the fourth dimension. In this graph, each pair of the z coordinates of the two surfaces correspond to a point (u, v) on the uv plane. We choose the point (u,v) by looking at the points on the surfaces above a single point (x, y) (taken from the unit disc, because it is our domain), take their z coordinates and and map them mentally in the uv plane. We observe the singularities at the point (x, y) = (-1, 0). This graph tells us everything about the image except for the asymptotic behavior.

I also re-visited the concentric circles method. As we observed earlier, finding the curves in the UV plane as implicit equations in Θ is difficult. We try to do this by using trigonometric substitutions for u and v and solve for Θ, and set them equal to each other to get rid of Θ. Let r = .5

From the equation u, we have the following. (ignore the 2pi n term, as it does not matter to us).

Θ =

From the equation v, here is what we have for Θ (replace x with Θ).

So the implicit equation for the curve {r=.5} under our map is the equality between these two expressions.

I would really try my substitutions (##w=x+1## followed by polar coordinates ##y=r\sin\varphi## and ##w=r\cos \varphi## and applying the restrictions ##|x|,|y|,x^2+y^2<1##). Your graphics are three dimensional, but the result is two dimensional. I think it is a Möbius transformation at its core, even though we have non linear terms.

docnet
docnet said:
The issue may be that our domain is the unit disk defined as {x2 + y2 <1} and taking from a different subset of the xy-plane such as an offset disc will map to different points on the uv- plane.
Unless I have blundered, the family of circles I defined spans exactly the same domain. Imagine an inverted cone with circles marked at horizontal planes. Instead of looking straight down from above, imagine looking down a line tangent to one side of the cone.

docnet
The equations get a lot simpler if we substitute ##x=r\cos(\theta)-1##, ##y=r\sin(\theta)##. This makes the domain the unit disc centred at (1,0):
##u=2\frac{\cos(\theta)}r-1##
##v=2\frac{\sin(\theta)}r##
Writing u'=u+1, in the (u',v) plane the mapping becomes an inversion (https://en.m.wikipedia.org/wiki/Inversive_geometry) in the circle radius √2, centred at the origin.

Edit: I hadn't noticed @fresh_42 already suggested the same substitution.

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docnet
fresh_42 said:
I would really try my substitutions (##w=x+1## followed by polar coordinates ##y=r\sin\varphi## and ##w=r\cos \varphi## and applying the restrictions ##|x|,|y|,x^2+y^2<1##). Your graphics are three dimensional, but the result is two dimensional. I think it is a Möbius transformation at its core, even though we have non linear terms.
haruspex said:
The equations get a lot simpler if we substitute ##x=r\cos(\theta)-1##, ##y=r\sin(\theta)##. This makes the domain the unit disc centred at (1,0):
##u=2\frac{\cos(\theta)}r-1##
##v=2\frac{\sin(\theta)}r##
Writing u'=u+1, in the (u',v) plane the mapping becomes an inversion (https://en.m.wikipedia.org/wiki/Inversive_geometry) in the circle radius √2, centred at the origin.

Wow, thank you haruspex and fresh_42 for the insightful explanations. I have never heard of inversive geometry before, so it is a first for me.

The following is my understanding of this geometry. This map inverses every point in our offset unit disc with respect to the reference circle of radius √2 centered at (1,0) in the uv plane. The image includes everything outside of an offset circle of radius 2 centered at (1,0). Our domain has radius 1 => 1 x 2 = (√2)2, so the boundary of our domain will get mapped to the boundary of our image, and the image goes to infinity in all directions, where the points at infinity correspond to the center of our domain.

Another question I had is, if we compute the composite map of this map with it self, will it simply give x=x and y=y, because applying the inversion twice gives the identity transformation?

docnet said:
This map inverses every point in our offset unit disc with respect to the reference circle of radius √2 centered at (1,0) in the uv plane.
No, it's at the origin in the (u',v) plane, so at (-1,0) in the (u,v) plane.

docnet
The image also excludes the left half of the UV plane, as the map is undefined there. Which is odd, because the determinant of the matrix of partial derivatives is nonzero for the unit disk, indicates there is no singular set.

## 1. What is the purpose of finding the image of a multivariate function?

The image of a multivariate function is the set of all possible output values that the function can produce for a given set of input values. It is important to find the image of a function in order to understand the range of values that the function can take on and to determine the behavior of the function.

## 2. How do you find the image of a multivariate function?

To find the image of a multivariate function, you need to plug in different combinations of input values and observe the corresponding output values. This can be done by creating a table of values or by graphing the function and identifying the range of values on the y-axis.

## 3. What is the difference between the domain and the image of a multivariate function?

The domain of a multivariate function is the set of all possible input values, while the image is the set of all possible output values. In other words, the domain represents the independent variables of the function, while the image represents the dependent variables.

## 4. Can a multivariate function have more than one image?

Yes, a multivariate function can have more than one image. This means that for a given set of input values, the function can produce multiple output values. This is known as a many-to-one function.

## 5. Why is it important to consider the image of a multivariate function when analyzing it?

Considering the image of a multivariate function allows us to understand the behavior and range of values of the function. It can also help us identify any patterns or relationships between the input and output values, which can be useful in making predictions or solving problems.

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