Define a function
where denotes the greatest integer less than or equal to , also called the floor function.
Then, define a function
- Draw the graph of for and prove that is periodic with period 1.
- For , prove
and prove that is also periodic with period 1.
- Give a formula for in terms of the floor of .
- Find a such that
- For the constant from part (d), define
Prove that is periodic with period 1 and that if , we have,
- The graph of is as follows:
Then, we prove that is periodic with period 1.
Proof. We compute to show (where I’ve replaced Apostol’s notation with since this less likely to cause confusion, also we use the solution to this exercise in the second line)
- Proof. First, we establish the requested formula,
This was the requested formula. Next, to show is periodic with period 1, we show for any ,
Then, using the formula in the first part of (b) we have
Further, from part (a) we know is periodic with period 1, so , and we have,
by definition of . Thus, is periodic with period
- To express in terms of of we compute as follows:
where since is an integer by definition, and has period 1, so for any integer . Continuing,
Here, we know the term in the integral is zero since for all since . Then,
This is the requested expression of in terms of .
- Here we compute the integral, using the formula for we established in part (c), and solve for the requested constant ,
- First we give the proof that is periodic with period 1.
Proof. We compute,
Next, we establish the requested formula. If , we have
Let be relatively prime positive integers (i.e., they have no common factors other than 1). Then we have the formula
The sum is 0 when .
- Prove this result by a geometric argument.
- Prove this result by an analytic argument.
- Proof. We know by the previous exercise (1.11, #6) that
Further, from this exercise (1.7, #4), we know
where number of interior lattice points, and number of boundary lattice points. We also know by the formula for the area of a right triangle that
Thus, we have,
Then, to calculate we note there are no boundary points on the hypotenuse of our right triangle (since and have no common factor). (This follows since if there were such a point then for some , but then we would have divides , contradicting that they have no common factor.) Thus, . So,
- Proof. To derive the result analytically, first, by counting in the other direction we have,
Use the previous exercise (1.11, #4 parts (d) and (e)) to deduce a formula for and prove this formula is correct.
Proof. Let , then we have
Thus, there exists some with such that
Hence, . Thus,
Then, for each with we have
Then, for , we have
For any we denote the greatest integer less than or equal to by . Prove the following properties of the function :
- for any integer .
- or .
- Proof. Let for some integer . Then,
But, we defined . Thus, for any
- Proof. If , then for some . Hence, and
On the other hand, if , then let . This gives us,
- Proof. Let and , then we have
So, adding, we obtain,
- Proof. By part (c) we have,
If , then let ,
On the other hand, if , then let , and
- Proof. By part (c) we have
So, putting these together we have,
If , then, let , so
Next, if , then let , giving us,
Finally, if , then let , and we have