Math Problem Statement

"Find the recurrence relation satisfied by Rn, where Rn is the number of regions that a plane is divided into by n lines, if no two of the lines are parallel and no three of the lines go through the same point." answer: "Rn = n + Rn−1, R0 = 1" explain answer

Solution

To explain the recurrence relation Rn=n+Rn1,R0=1R_n = n + R_{n-1}, R_0 = 1, let’s break down the problem step by step.

Problem Overview

We need to determine the number of regions (RnR_n) that a plane is divided into by nn lines under the conditions:

  1. No two lines are parallel.
  2. No three lines are concurrent (i.e., no three lines pass through a single point).

Step-by-Step Explanation

  1. Base Case:

    • When n=0n = 0 (no lines), the entire plane is undivided and thus forms a single region: R0=1.R_0 = 1.
  2. Adding the First Line:

    • Adding one line to the plane divides the plane into two regions: R1=2.R_1 = 2.
  3. Adding the nn-th Line:

    • When we add the nn-th line to the existing n1n-1 lines, this new line intersects all n1n-1 lines already present. Since no two lines are parallel and no three lines meet at a single point, the nn-th line intersects each of the n1n-1 lines at a unique point.
    • These intersection points divide the nn-th line into nn segments.
    • Each of these segments creates a new region in the plane.

    Hence, the nn-th line adds nn new regions to the plane.

  4. Recurrence Relation:

    • The total number of regions after adding the nn-th line is therefore: Rn=Rn1+n.R_n = R_{n-1} + n. This is because Rn1R_{n-1} accounts for all regions created by the previous n1n-1 lines, and nn new regions are added by the nn-th line.
  5. Initial Condition:

    • As stated earlier, when n=0n = 0, the plane has only one region: R0=1.R_0 = 1.

Final Recurrence Relation

Thus, the recurrence relation is: Rn=Rn1+n,R0=1.R_n = R_{n-1} + n, \quad R_0 = 1.

Verification

To confirm, let’s compute the first few terms:

  • R0=1R_0 = 1.
  • R1=R0+1=1+1=2R_1 = R_0 + 1 = 1 + 1 = 2.
  • R2=R1+2=2+2=4R_2 = R_1 + 2 = 2 + 2 = 4.
  • R3=R2+3=4+3=7R_3 = R_2 + 3 = 4 + 3 = 7.
  • R4=R3+4=7+4=11R_4 = R_3 + 4 = 7 + 4 = 11.

These match the expected number of regions for n=0,1,2,3,n = 0, 1, 2, 3, and 44.


Key Insight:

The recurrence captures the incremental increase in regions caused by the addition of each new line. This systematic addition arises from the geometric constraints of the problem.


Would you like me to explain further details or help with a related concept? Here are some questions to explore further:

  1. How can you derive a closed-form expression for RnR_n instead of a recurrence relation?
  2. Can we generalize this problem to kk-dimensional space (e.g., planes in 3D space)?
  3. What if the lines are parallel? How does the recurrence change?
  4. How does the number of intersection points relate to the number of regions?
  5. How would the recurrence change if three lines were allowed to meet at a point?

Tip: Drawing the regions for small values of nn (e.g., n=1,2,3n = 1, 2, 3) can help build intuition for recurrence relations in geometry problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Plane Geometry
Combinatorics

Formulas

Rn = Rn−1 + n
R0 = 1

Theorems

Geometric Region Division Theorem

Suitable Grade Level

Grades 10-12