Math Is Fun Forum

let w=z-1. then:
(z+1)/(z-1)=(w+2)/w=1+2/w=1+2w*/|w|. If this is imaginary, 2Re(w)/|w|=-1. But Re(w)/|w| is equal to Cos(Arg(w)).

You are wrong at the underlined equation. For distinct x and y we dont have x*y=e. Only x*x=e and y*y=e.

To prove that G is commutative, with two elements x and y start with x*y, and try to arrive at y*x.
edit: Altough I see now, that a direct way from xy to yx seems hard to find, theres another way to do it. Hint: Since G is a group, we know that G is closed under multiplication. Thus xy is in G and yx is in G.

Remember the formula c(n,k)=c(n-1,k)+c(n-1,k-1).

yea but obv 0<1/(1+n^2)<1/n^2 so |a_n|<|b_n| for all n. Imo you should do this calculation to show that the sum converges, not just state it without motivation.

I meant that -pi/2<A,B,C<pi/2 is imposed by arctans range, then A,B,C>0 is because x,y,z>0.

But we must be careful if we just let A=arctanx, B=arctany, C=arctanz, x=tanA, y=tanB, z=tanC and consider A,B,C<pi, tanA,tanB,tanC>0 and forgets that we started with arctan and only use the positivity of x,y,z. Because now for example angles between -pi/2 and -pi would satisfies these inequalities since tan is positive in theat interval, so these problems are not completely equivalent.

First, we must define the new Sine and Cosine. The definition for the ordinary sin(x) and cos(x) are defined (the geometric definitions, some books start by defining them as their infinite series) as:
Cos(a) is the x coordinate of a point (x,y) on the unit circle, where a is the angle from the x axis in the positive direction. See following link for a better explanation:
http://en.wikipedia.org/wiki/Sine#Unit-circle_definitions

So I thought we could define the new "Cosine" and "Sine" in a similar way. Let c(a) be the new Cosine function, and s(a) be the new Sine function, which will be the x respectively y coordinate for a point (x,y)=(c(a),s(a)) lying on our "new unit circle", defined by |x|^n+|y|^n=1, and a is the same angle as in the definition above. If n=2, we get the functions Cos(a) and Sin(a), as expected. Now we want a good expression for this, and my first thought was to try to describe it with the ordinary trigonometric functions. If we draw a right angle triangle, where the hypotenuse is from the origin to a point on the "unit circle". The attached picture shows an example for n=3. Now x and y are related as |x|^n+|y|^n=1, and y/x=tan(a). Solving this system of equations for x=c(a) in the following way yields:
y=tan(a)*x.
substituting into |x|^n+|y|^n=1:
|x|^n+|tan(a)*x|^n=1
|x|^n+|tan(a)|^n*|x|^n=1
|x|^n*(1+|tan(a)|^n)=1
x=c(a)=±1/(1+|tan(a)|^n)^(1/n)
where the ± sign is determined from the quadrant the point is in. First and fourth quadrant yields +, second and third yields -. Similar for y=s(a).

Hopefully this made my post clearer

edit: a better formula would probably be:
c(a)=±|Cos(a)|/(|Cos(a)|^n+|Sin(a)|^n)^(1/n)
since it will be defined for a=pi/2+pi*k, which is obtained by multiplying nominater and denominator by |Cos(a)|
edit2: heck, now we doesnt even need the ± sign, since |Cos(a)| is multiplied by a ± depending on which quadrant the point is in by the same rule as c(a), so i think the formula ends at:

We could maybe define a new type of sine function, s(A), as the y-component of a point (x,y) satisfying x^n+y^n=1, and same with the new Cosine function, c(A), as the x-component (A being the orinary angle in radians, altough this doesnt have the same "geometric intrepetation" in our "new unit circle"). With n=2 we get the ordinary Sine and Cosine. Now consider a point p=  (x,y)=(c(A),s(A)) satisfying x^n+y^n=1. we have that x=CotA*y, which yields:
(y*CotA)^n+y^n=1
s^n(A)=y=1/(Cot^nA+1)
and similary we get:
c^n(A)=1/(Tan^nA+1)
altough im not sure how to define the "unit circle", since if n is odd, we will not have any symetry and particulary no solutions for x,y<0. Maybe we should consider the equation |x|^n+|y|^n=1 instead, maybe makes more sense.
Edit: after plotting in mathematica, I see that we obviously must consider |x|^n+|y|^n=1, otherwise we get some ridiculous function. This would give:

Cos2x+Cos3x=0
Cos2x=-Cos3x
Cos2x=Cos(pi-3x)
1)2x=3x-pi->x=pi->cosx=-1
2)2x=pi-3x->x=pi/5
looks like something is wrong with your equation...

x=1 is one solution, and that it is the only one can be realized by investigating the derivative.

I havent fully read your solution, but a hint for another solution would be to assume WLOG that

(I get the same result as you btw, ~half will be 0, ~half will be 1 to get maximum)

http://www3.hi.is/~pjo1/division%20by%20zero.jpg

http://halshop.files.wordpress.com/2007/03/phpw9jvl0pm.jpg

hint: let a+b=x and take the squares and cubes of it to get two equations, and use these equations to solve for x.