Proving Recurrence Relation by Induction

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SUMMARY

The discussion focuses on proving a recurrence relation using mathematical induction. The relation is defined as \( x_{n} = 5x_{n-1} - 6x_{n-2} \) for \( n \geq 2 \) with initial conditions \( x_{1} = 1 \) and \( x_{0} = 0 \). The proof involves showing that \( \begin{bmatrix} x_{n} \\ x_{n-1} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix}^{n-1} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) holds true for \( n = k + 1 \) given that it is true for \( n = k \). The key step is to demonstrate the relationship \( \begin{bmatrix} x_{n+1} \\ x_{n} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_{n} \\ x_{n-1} \end{bmatrix} \).

PREREQUISITES
  • Understanding of mathematical induction
  • Familiarity with matrix multiplication
  • Knowledge of recurrence relations
  • Basic linear algebra concepts
NEXT STEPS
  • Study mathematical induction proofs in detail
  • Learn about matrix exponentiation techniques
  • Explore advanced recurrence relations and their applications
  • Investigate linear transformations and their properties
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Students and educators in mathematics, particularly those focusing on discrete mathematics, linear algebra, and mathematical proofs. This discussion is also beneficial for anyone looking to strengthen their understanding of recurrence relations and induction methods.

dragonkiller1
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x_{xn-1}= 5_{xn-1} - 6_{xn-2};for \ n≥2 \\ x_{1}=1 \\ x_{0}=0

prove by induction that:
\begin{bmatrix}<br /> x_{n}\\ x_{n-1} <br /> \end{bmatrix} = \begin{bmatrix}<br /> 5 &amp;-6 \\ <br /> 1 &amp; 0<br /> \end{bmatrix}^{n-1}\begin{bmatrix}<br /> 1\\0 <br /> \end{bmatrix}
 
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welcome to pf!

hi dragonkiller1! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
I've showed the base step for n=1
For the inductive step:
Suppose P(k) is true for some k
\begin{bmatrix}<br /> x_{k}\\ x_{k-1} <br /> \end{bmatrix} = \begin{bmatrix}<br /> 5 &amp;-6 \\ <br /> 1 &amp; 0<br /> \end{bmatrix}^{k-1}\begin{bmatrix}<br /> 1\\0 <br /> \end{bmatrix}
Then for n=k+1
\begin{bmatrix}<br /> x_{k+1}\\ x_{k} <br /> \end{bmatrix} = \begin{bmatrix}<br /> 5 &amp;-6 \\ <br /> 1 &amp; 0<br /> \end{bmatrix}^{k}\begin{bmatrix}<br /> 1\\0 <br /> \end{bmatrix}
?
I'm stuck on the inductive step. >.<
 
hi dragonkiller1! :smile:

you obviously need to prove

\begin{bmatrix}<br /> x_{n+1}\\ x_{n} <br /> \end{bmatrix} = \begin{bmatrix}<br /> 5 &amp;-6 \\ <br /> 1 &amp; 0<br /> \end{bmatrix}\begin{bmatrix}<br /> x_{n}\\ x_{n-1} <br /> \end{bmatrix}

which is easier …

the top line or the bottom line? :wink:
 
pre multiply 5,-6,1,0 matrix on both the sides(to the k equation) and you will have your proof
 
Oh got it. Thanks :)
 

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