Let be two points of , be a unit vector and let be a curve passing through those points. (that is there exist such that .) Prove that the shortest path between these two points is the line.
The following facts hold:
Integrating we get that:
where is the length of the curve between the points . Taking we have
Let be a group and such that and
where is the identity element of the group. Find the order of .
We will begin stating a lemma:
Lemma: If then .
First we multiply with and from right and left respectively. Thus one can see that
Now, we use the main relation and so
By repeating the previous procedure, one can prove the result.
Using the lemma we see that and thus . Since is prime the order of will be either or .
Evaluate the contour integral .
We are applying the substitution
since the function has only one pole in the specific contour , namely .
Let be a finite subgroup of this is the group of the invertible matrices over ). If then prove that .
Let us suppose that
. We note that for every
and onto. Thus:
Thus the matrix is idempotent. thus its trace equals to its class. (since we are over which is a field of zero characteristic.) Hence
This implies that hence .
The exercise can also be found at mathematica.gr
Prove that for every constant the set
has the same volume for all continuous functions .
on the plane
is a disk of constant radius . Thus the set
is a “cylinder” which axis is the curve
and its radius is .More specifically , the set is bounded by the planes and and for every the intersection of with the plane is the disk
The area of this disk is the same with the disk
The latter one has an area of
It follows from Cavalieri’s Principal that has the same volume and that is equal to
which is the same for all continuous functions .
A somewhat visualization would be the following:
This was an exam’s question somewhere in Greece. The answer was migrated from mathematica.gr .