Specialist's Methods Thread

[b^3]

I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?

Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.

EDIT: Dc or someone can probably provide a more concise wording for that last part.

[Special At Specialist]

Ahh okay... so what is the difference between R --> R and R^2 --> R^2? Is there ever anything else I have to learn, like R^3 --> R^3?

[dc302]

Ahh okay... so what is the difference between R --> R and R^2 --> R^2? Is there ever anything else I have to learn, like R^3 --> R^3?

R --> R means from the real line to the real line

R^2 --> R^2 means from the plane to the plane

R^3 --> R^3 means from 3-d space to 3-d space

[Special At Specialist]

Okay cool. Will functions always be written as R^2 --> R^2? Because I could swear I remember a function given the rule f:R-->R or some weird notation like that. And it was a 2D cartesian plane graph.

[dc302]

I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?

Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.

EDIT: Dc or someone can probably provide a more concise wording for that last part.

Yeah it's hard to explain without calling upon higher maths but think of it as a map of australia. Draw a path from melbourne to sydney.

This path is represented by a function. Now what happens if you 'stretch' your map so that it;s twice as wide? The path you have seems to have changed to a new function. This is what you are doing. You have a function on the plane, and you are basically changing the geometric construction of the plane and thus obtaining a new function.


edit:

Okay cool. Will functions always be written as R^2 --> R^2? Because I could swear I remember a function given the rule f:R-->R or some weird notation like that. And it was a 2D cartesian plane graph.

It should always be R^2 -> R^2 if it's a function from the plane to the plane.

[b^3]

I think special at spesh is thinking about the domain restrictions for functions with the R->R part.

i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x²

[Special At Specialist]

I think special at spesh is thinking about the domain restrictions for functions with the R->R part.

i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x²

Yeah, that's the one

I think I get it. You're converting a function from a different dimension into a 2D cartesian plane. Would it work in all transformation situations, or only in this case? Also, what does the R^2 --> R^2 mean?

Not exactly, they are still in the same x-y plane. We are just applying a transformation to the original. The R^2->R^2 means that we are mapping a plane of the real set of numbers onto a plane of the real set of numbers.

EDIT: Dc or someone can probably provide a more concise wording for that last part.

This path is represented by a function. Now what happens if you 'stretch' your map so that it;s twice as wide? The path you have seems to have changed to a new function. This is what you are doing. You have a function on the plane, and you are basically changing the geometric construction of the plane and thus obtaining a new function.

Oh, so the function is just being moved or squished, but it never actually changes?

[dc302]

I think special at spesh is thinking about the domain restrictions for functions with the R->R part.

i.e. when you write a function out, f:Domain->R,f(x)=blah
e.g. f:R->R, f(x)=x²

Ah yes, this may have confused him because it is only a convenience the way we represent a function on a single graph. The way it would supposed to be represented is to draw two real lines, and any point on the first real line is x, and has an image f(x) on the second real line.

When you look at a 2-d plane that's representing a function from R to R, what you're looking at is actually called the 'graph of f':

g : R --> R^2 (or graph(f)), where g is defined by g(x) = (x, f(x))

An example of an actual function from the plane to the plane is:

f(x,y) = (2x,3y)

Oh, so the function is just being moved or squished, but it never actually changes?

It's changing along with the change you make to the whole plane, so it IS still changing with respect to the original function.

[Special At Specialist]

I sort of see what you're saying...
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?

[dc302]

I sort of see what you're saying...
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?

For a sure-fire way of doing it:

1. Find x in terms of x' and y', y in terms of x' and y' from the information provided. (remember x, y are the coords from the original plane, and x', y' are from the image of the original plane.

2. Substitute x and y (written in terms of x' and y') into the equation of the function.

3. Rearrange till you get the new function, in x' and y', in the desired form. ie y' = ax'^2 + bx' + c

4. Write down your solution but now change x' and y' back to x and y.

[Special At Specialist]

I sort of see what you're saying...
But with the transformations problem: is it always okay to convert a y' = x' problem into a y = x problem? Or only under certain conditions?

For a sure-fire way of doing it:

1. Find x in terms of x' and y', y in terms of x' and y' from the information provided. (remember x, y are the coords from the original plane, and x', y' are from the image of the original plane.

2. Substitute x and y (written in terms of x' and y') into the equation of the function.

3. Rearrange till you get the new function, in x' and y', in the desired form. ie y' = ax'^2 + bx' + c

4. Write down your solution but now change x' and y' back to x and y.

Ahh now that explains everything. Thank you! I'm going to attempt a few more of those problems to see if I really get the hang of it...

[Special At Specialist]

I spent literally 2 hours on this problem and still can't solve it:


It looks like such a simple algebra/logarithms question, but I don't know what to do! The best I could do was find a few solutions to c in terms of logx(y) or in terms of logx/y (y) or something along those lines. I don't know how to get rid of the x's and y's

[brightsky]

x^a = y^b
x^(a/b) = y
sub into second equation:
x^a = (x^(a/b)/x)^c
x^a = (x^(a/b - 1))^c
x^a = x^(c(a/b)-1))
a = c(a-b)/b
c = ab/(a-b)

[Special At Specialist]

x^a = y^b
x^(a/b) = y
sub into second equation:
x^a = (x^(a/b)/x)^c
x^a = (x^(a/b - 1))^c
x^a = x^(c(a/b)-1))
a = c(a-b)/b
c = ab/(a-b)

Very elegant solution and it didn't even require logarithms. Thank you!

[Special At Specialist]

I don't understand these composite functions...

Suppose that:
f(x) has a domain of [-10, 10]
f(x) has a range of [-20, 15]
g(x) has a domain of (-12, 8]
g(x) has a range of (-2, 11)

What is the domain and range of f(g(x))?
What is the domain and range of g(f(x))?