OK...

inverse: if x=g(f(x)) then g is the inverse of f and visa versa.
f(x)^-1: if 1=f(x)*g(x) then g(x)=1/f(x)=f(x)-1

examples:

f(x) = x^2+7
inverse: x=g(f(x)) find g(x):
x = g(x^2+7) therefor:
g(x)=sqrt(x-7)

f(x)^-1 = 1/f(x) = 1/(x^2+7)=x^-2+1/7

in math, we have the tan function, the cotan function, and the arctan function. cotan = tan(x)^-1. arctan(tan(x))=x. arctan is the inverse. on a calculator =, they use the symbol "tan^-1" for the arctan function. It is not "mathematically" correct.

I hope this helps.

As for your code problem , we will need to see your code to be able to tell you what is wrong with it.

Dwiel

quote:

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Original post by Tazzel3D
f(x) = x^2+7
inverse: x=g(f(x)) find g(x):
x = g(x^2+7) therefor:
g(x)=sqrt(x-7)



f(x)^-1 = 1/f(x) = 1/(x^2+7)=x^-2+1/7

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You totally sure about this? You can''t distribute the division like that...if you want to turn a fraction with a polynomial in the denominator into a sum of fractions, you need to use the process of partial fractions. This can''t be done with 1/(x^2+7), I think...

Plus, for your example, I''ll just note that the inverse of the function f(x)=x^2+7 is actually not a function itself.

quote:

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Original post by Anonymous Poster

You totally sure about this? You can''t distribute the division like that...if you want to turn a fraction with a polynomial in the denominator into a sum of fractions, you need to use the process of partial fractions. This can''t be done with 1/(x^2+7), I think...

Plus, for your example, I''ll just note that the inverse of the function f(x)=x^2+7 is actually not a function itself.

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sorry bout that... yes, you are correct that it is not a function.. it is two: f(x)=sqrt(x-7) or f(x)=-sqrt(x-7). Also, I was thinking too quickly and am sorry about distributing the polynomial like I did.. that was wrong. Other than that, I believe, everything is correct.

nobodynews:
you can use the method that you suggested... it is basically the same thing... kind of. I was just trying to be more abstract using the def of inverse function as stated in my calc book....

hopefully the confusion is gone now!

Dwiel

quote:

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Original post by Tazzel3D

sorry bout that... yes, you are correct that it is not a function.. it is two: f(x)=sqrt(x-7) or f(x)=-sqrt(x-7). Also, I was thinking too quickly and am sorry about distributing the polynomial like I did.. that was wrong. Other than that, I believe, everything is correct.

nobodynews:
you can use the method that you suggested... it is basically the same thing... kind of. I was just trying to be more abstract using the def of inverse function as stated in my calc book....

hopefully the confusion is gone now!

Dwiel

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There''s two main ways to define an inverse function(or relation):

Algebraic: if y = f(x), then the inverse is f^-1(x) such that x = g(f(x)), like you said

Geometric: the inverse of a function(or any relation, for that matter) is the reflection of that relation over the line y=x

Something which is annoying but has to be put up with: f^-1(x) means inverse function, while according to the rules of algebra f(x)^-1 is 1/f(x).

Another note: If you know the derivative(dy/dx) of a function, the derivative of the inverse function is 1/(dy/dx).

CyberGhost,

It is really easy. If you know the triangle you are over, then just interpolate the y value between the vertices to get the height of the ground and add your height over the ground:

Yea that sounds kewl,
And I even understand it, guess I'm not as thick as I thought
Just need to check sommat tho.
As you know each cell in the vertice matrix consists of 4 vertices, in which they construct two triangles.
Looking at da vertices placement for your example I guess your talking about the left triangle,
However what would I do if I am working with the right triangle.
Does vertices order mater, as long as i keep v0v2 parallel to the x coordinate and v0v1 parallel to the z coordinate.

Would the following work or am I miss understanding something:


Is there anything I must do to the calculation for the right triangle, such as change the order of the vertices.
I hope I havent confused everyone to much, this is my first time I have done any collision detection n finding it hard to get started as there seems to be lack of tuts on the subject which are writtern in English instead of Maths.
Thanx

[edited by - CyberGhost on March 21, 2003 5:04:51 AM]