Calculus by Apostol Exercise 1.7

In summary, the formula for the area of a polygon with lattice point vertices is I + B/2 - 1, where I is the number of interior points and B is the number of boundary points. This formula is valid for rectangles with sides parallel to the coordinate axes, right triangles, and parallelograms. By constructing the parallelogram as a combination of a rectangle and two right triangles, the formula can be applied to any general polygon by counting the number of interior and boundary points.
  • #1
shinobi20
271
20

Homework Statement


A point (x,y) in the plane is called a lattice point if both coordinates x and y are integers. Let P be a polygon whose vertices are lattice points. The area of P is I + B/2 - 1, where I denotes the number of lattice points inside the polygon and B denotes the number on the boundary.

(a) Prove that the formula is valid for rectangles with sides parallel to the coordinate axes. (b) Prove that the formula is valid for right triangles and parallelograms. (c) Use induction on the number of edges to construct a proof for general polygons.

Homework Equations


Area of an arbitrary polygon P
P = I + B/2 - 1
where I is the number of interior points and B is the number of boundary points

The Attempt at a Solution


I have done part (a) but stuck on part (b) and (c). I have tried to represent the parallelogram as the difference between a rectangle and triangles, but no success as of now. I know I can just assume that a parallelogram has the area same as a rectangle because I have proved that it can be constructed by a rectangle and triangles, but I want to use the equation above in order to actually show that it is equal to the area of a rectangle. Any suggestions?
 
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  • #2
Can you show that the result applies to any triangle, not just a right triangle?

Can you show by construction that any such polygon can be constructed of triangles with corners that are lattice points?

Don't forget to use the fact the vertices are lattice points.
 
  • #3
As you said, the difference between a rectangle and right triangles might be a good way to work this.
You know that a rectangle containing the parallelogram and two right triangles has area P = I + B/2 - 1.
You know that a right triangle containg A interior points and C boundary points has area Q = A + C/2 - 1.
So I suggest you just put them together, you should have
Area of parellelogram = ## I + B/2 - 1 - Q_1 -Q_2 ##.
Then by counting the number of boundary points and interior points leftover, you should have a convincing argument.
 
  • #4
DEvens said:
Can you show that the result applies to any triangle, not just a right triangle?

Can you show by construction that any such polygon can be constructed of triangles with corners that are lattice points?

Don't forget to use the fact the vertices are lattice points.
I don't get it, what is the relationship of showing that the formula applies to a general triangle and the area of a parallelogram?
 
  • #5
shinobi20 said:
I don't get it, what is the relationship of showing that the formula applies to a general triangle and the area of a parallelogram?

It goes back to the fact the vertices are lattice points. If you take any such polygon, you can build it out of triangles that have their vertices as lattice points. A parallelogram is just two such triangles. So the parallelogram will get you thinking in terms of constructing the general polygon.
 

FAQ: Calculus by Apostol Exercise 1.7

1. What is the purpose of Exercise 1.7 in "Calculus by Apostol"?

The purpose of Exercise 1.7 is to help students practice and gain a deeper understanding of the concepts and techniques learned in Chapter 1. It covers topics such as derivatives, limits, and continuity.

2. How many exercises are included in Exercise 1.7 of "Calculus by Apostol"?

Exercise 1.7 consists of 20 exercises, providing ample opportunities for students to practice and reinforce their understanding of key calculus concepts.

3. Are the exercises in Exercise 1.7 of "Calculus by Apostol" suitable for beginners?

Yes, the exercises in Exercise 1.7 are designed for students who are new to calculus. They start with basic concepts and gradually increase in difficulty, allowing beginners to build a strong foundation in calculus.

4. Can the exercises in Exercise 1.7 be solved using a calculator?

Yes, most of the exercises in Exercise 1.7 can be solved using a calculator. However, it is important for students to also understand the underlying concepts and techniques behind the calculations.

5. How can I check if my answers to the exercises in Exercise 1.7 are correct?

The solutions to the exercises in Exercise 1.7 can be found in the back of the textbook, allowing students to check their answers and identify any areas for improvement. It is also helpful to work with a study group or consult a teacher or tutor for feedback on your solutions.

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