Hello, help please
1) let and roots of equation .
Evaluate :
2) let , prove that :
3) Find all function such that :
You can use the quadratic formula to find a and b (alpha and beta).
x=(-B+-sqrt(B^2-4AC))/2A, so x=(-pi+-sqrt(pi^2-4sqrt(2)))/2=tan(angle).
The two possible values of arctan(x) are -.4986 and -1.2032 in radians.
Simply plug these values in for a and b in the second equation.
Pick three random positive numbers x,y,z.
Define vector u=(xy,yz,xz) and vector v=(xz,xy,yz) in R^3
Note: |u|=|v|, so let d=|u|=|v|
Now we ask, which is bigger, u*v or u*u? (where * signifies vector product where appropriate)
Since, u*v=|u||v|cos(a), for a the angle between them,
u*v=d^2cos(a) and u*u=d^2cos(0)=d^2.
Since all values are positive, u and v are confined to the first (+,+,+) quadrant, the maximum angle between them is 90*=pi/2 radians.
Since cosine is a decreasing function on the interval [0,pi/2], cos(a)<cos(0), therefore u*u>u*v. You can also think of the fact that for two vectors of the same length, the dot product increases the closer they are together (the smaller the angle between them), so the maximal dot product for any vector is the dot product with itself.
Substituting, this gives us: (xy)^2+(yz)^2+(xz)^2 > yzx^2+xzy^2+xyz^2. Divide by xyz on both sides and you have your exact answer.
I KNOW THIS FEELS LIKE USING A CANON TO KILL A MOSQUITO, BUT THIS THE LEAST CONFUSING WAY I WAS ABLE TO VISUALIZE THE PROBLEM. I HOPE YOU ARE ABLE TO USE VECTOR GEOMETRY IN YOUR PROOF.
expanding gives us now put and then divide by
this question is not bad! if in we put x = 0 we'll get3) Find all function such that :
1) is onto: suppose is given. put then by
2) : by 1) there exists with then by for all let's call this result now gives us
3) : if in we put y = 0, we'll get again if in we let y = 0 and replace x with f(x) and use we'll get:
4) : by 3), for any either or now suppose and for some non-zero then by on the other hand:
which gives us or contradiction! therefore either or