You can do these by matrices. In this case f(x) and g(x) are represented by matrices F and G such that F=|1 -1|, F^(-1) ~ |1 1| (the matrices act on homogeneous |1 1| |-1 1| coordinates so can be multiplied by a scalar) FG = |2 1| |1 -3| G = |1 1| |2 1| |-1 1| |1 -3|
G = |3 -2| |-1 -4| equivalent to g(x) = (-3x + 2)/(x + 4)
@@pwmiles56 Maybe I was a little bit too excited about using it. I expected it to be solving a problem like this one: f(x) = (x + 1)² / (x - 1)² and f(g(x)) = (2x² - 1)² asking for g(x)...
@@MrGeorge1896 I see. There is a theory of higher order rational functions, called Cremona transformations, but it's quite difficult and not done by matrices. EDIT: For example the bilinear transformation f(z) = (az+b)/(cz+d) is also a Cremona transformation of the complex plane: f(z)=(az+b)(c*z*+d*)/|cz+d|^2 In homogeneous coordinates, with z=x+iy, this becomes f(x; y; 1) = (real quadratic in x,y; real quadratic in x,y; |cz+d|^2 i.e. another real quadratic in x,y) Because f(z) has an inverse, so does the Cremona transformation, which is its defining property.
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An interesting property of proportion that I didn't know about before! (:
G(x)=2-3x/x+4
You can do these by matrices. In this case f(x) and g(x) are represented by matrices F and G such that
F=|1 -1|, F^(-1) ~ |1 1| (the matrices act on homogeneous
|1 1| |-1 1| coordinates so can be multiplied by a scalar)
FG = |2 1|
|1 -3|
G = |1 1| |2 1|
|-1 1| |1 -3|
G = |3 -2|
|-1 -4|
equivalent to
g(x) = (-3x + 2)/(x + 4)
Nice approach especially for more complex cases with higher powers of x involved.
@@MrGeorge1896 Interesting. Do you have an example in mind?
@@pwmiles56 Maybe I was a little bit too excited about using it. I expected it to be solving a problem like this one:
f(x) = (x + 1)² / (x - 1)² and f(g(x)) = (2x² - 1)² asking for g(x)...
@@MrGeorge1896 I see. There is a theory of higher order rational functions, called Cremona transformations, but it's quite difficult and not done by matrices.
EDIT: For example the bilinear transformation
f(z) = (az+b)/(cz+d)
is also a Cremona transformation of the complex plane:
f(z)=(az+b)(c*z*+d*)/|cz+d|^2
In homogeneous coordinates, with z=x+iy, this becomes
f(x; y; 1) = (real quadratic in x,y; real quadratic in x,y; |cz+d|^2 i.e. another real quadratic in x,y)
Because f(z) has an inverse, so does the Cremona transformation, which is its defining property.
g(x) = (-3x+2)/(x+4)
Or both side -1 add karne par
f(g(x))=[g(x)-1]/[g(x)+1] = (2x+1)/(x-3) > (x+4)g(x)=2-3x > g(x)=(2-3x)/(4+x).
f(x) = (x - 1)/(x + 1)
f(x) = (x + 1 - 2)/(x + 1)
f(x) = (x + 1)/(x + 1) - 2/(x + 1)
f(x) = 1 - 2/(x + 1)
2/(x + 1) = 1 - f(x)
x + 1 = 2/(1 - f(x))
x = 2/(1 - f(x)) - 1
g(x) = 2/(1 - f(g(x))) - 1
1 - f(g(x)) = 1 - (2x + 1)/(x - 3) = (x - 3)/(x - 3) - (2x + 1)/(x - 3) = (-x - 4)/(x - 3) = (x + 4)/(3 - x)
2/(1 - f(g(x))) = 2/((x + 4)/(3 - x)) = (6 - 2x)/(x + 4)
g(x) = 2/(1 - f(g(x))) - 1 = (6 - 2x)/(x + 4) - 1 = (6 - 2x)/(x + 4) - (x + 4)/(x + 4) = (2 - 3x)/(x + 4)
Hence g(x) = (2 - 3x)/(x + 4)