How to Inductively Prove a Recursive Sequence Formula?

  • Thread starter Thread starter NuPowerbook
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary

Homework Help Overview

The discussion revolves around a recursive sequence defined by specific initial conditions and recursive relationships. The sequence is characterized by values that depend on the sum of previous terms, with an additional adjustment for odd indices. Participants are exploring how to inductively prove a proposed formula for the sequence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the structure of the recursive sequence and the proposed formulas for odd and even indices. There are attempts to clarify the inductive proof process, including verifying base cases and establishing relationships between terms. Some participants question the validity of the proposed formulas and their alignment with the recursive definitions.

Discussion Status

The discussion is ongoing, with participants providing various approaches to the inductive proof. Some guidance has been offered regarding how to handle the piecewise nature of the function, and there is recognition of potential discrepancies in the proposed formulas. Multiple interpretations of the recursive relationships are being explored.

Contextual Notes

Participants note the challenge of proving the formula due to its dependence on whether the term index is odd or even. There is also mention of the need for clarity in representing the recursive function and its summation form.

NuPowerbook
Messages
8
Reaction score
0
I have the series 2, 2, 6, 10, 22, 42, 86, 170, 342, 682 which follows the following:

M(1)=2
M(2)=2
M(3)=M(1)+M(2)+2
M(4)=M(3)+M(2)+M(1)
M(5)=M(4)+M(3)+M(2)+M(1)+2
M(6)=M(5)+M(4)+M(3)+M(2)+M(1)

Each value for M(N) for N >= 3 is the sum of all previous values, and add 2 if N is odd.

I found that these two functions describe the sum:
[tex]\mbox{If N is odd: }\frac{2^{n+1}+2}{3}[/tex]
[tex]\mbox{If N is even: }\frac{2^{n+1}-2}{3}[/tex]

Can anyone help me inductively prove this?

EDIT: I guess I tried to represent it as a function and failed. If someone can represent this as a summation for me, then I could probably figure out the inductive proof. I guess my whole problem was I tried to represent it as a recursive function, and since the function was wrong, I could not prove it.
 
Last edited:
Physics news on Phys.org
A proof by induction is done as follows. Show the case is true for the first term, n=1. This should be easy, just plug in and verify. Then you need to show that given the expression is true for all terms before the nth term, show that it then must be true for the nth term. In this case, assume the sum expression of the first n-1 terms is accurate, then show that adding the nth term will give the same result as the sum expression evaluated at n.
 
StatusX said:
A proof by induction is done as follows. Show the case is true for the first term, n=1. This should be easy, just plug in and verify. Then you need to show that given the expression is true for all terms before the nth term, show that it then must be true for the nth term. In this case, assume the sum expression of the first n-1 terms is accurate, then show that adding the nth term will give the same result as the sum expression evaluated at n.
I know how to do an inductive proof, but I am just not sure how to prove this. I've never done an inductive proof on a piecewise function where the proof depends on whether the nth term is even or odd.
 
OK, the way I would do it is to use the sum formulas to generate a pair of equations for M(n) if n is even or odd. For example, if n is odd, subtract the even sum expression for n-1 from the odd expression for n. Then prove by induction that these satisfy the generating equation. Remember, if n is odd, n-1 is even and n-2 is odd.

Edit: wait, those expressions are for M(n). Why did you call them the sum? In any case, that just makes your job easier.

second edit: that formula only matches the generating equation for the first few terms (up to 10). 10+6+2 != 22. no wonder you had a hard time proving it.
 
Last edited:
StatusX said:
OK, the way I would do it is to use the sum formulas to generate a pair of equations for M(n) if n is even or odd. For example, if n is odd, subtract the even sum expression for n-1 from the odd expression for n. Then prove by induction that these satisfy the generating equation. Remember, if n is odd, n-1 is even and n-2 is odd.

Edit: wait, those expressions are for M(n). Why did you call them the sum? In any case, that just makes your job easier.

second edit: that formula only matches the generating equation for the first few terms (up to 10). 10+6+2 != 22. no wonder you had a hard time proving it.

Oh bah, I messed up the formula. It should be correct now.
 
Take M(6)-M(5). This should show you a pattern you can use to get two expressions for M(n)-M(n-1), one for even n and one for odd n. Prove this relation holds for your formulas as well.

Also, the equation you're looking for is:

[tex]M(n) = \left\{\begin{array}{cc}\sum_{k=1}^{n-1} M(k), &\mbox{n even}\\ 2+\sum_{k=1}^{n-1} M(k), &\mbox{n odd}\end{array}[/tex]
 
Last edited:

Similar threads

Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
25
Views
2K
Replies
2
Views
1K
Replies
5
Views
1K
Replies
4
Views
3K