Finding the explicit formula for the recursion $a_n = 2a_{n-1} + a_{n-2}$ with $a_0 = 1$ and $a_1 = 3$?












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I'm trying to find the explicit formula for the recursion $a_n = 2a_{n-1} + a_{n-2}$ with $a_0 = 1$ and $a_1 = 3$. I already found the generating function: $f(x) = frac{1+x}{1-2x-x^2}$, and the characteristic polynomial: $x^2 - 2x + 1 = 0$, which has the roots $1+√2$ and $1-√2$. What is the explicit formula for this recursion, and how do I find it?






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  • 1




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    You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:35










  • $begingroup$
    @SauhardSharma Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 14 at 7:36










  • $begingroup$
    @Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:38










  • $begingroup$
    @SauhardSharma Thank you so much!!!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38






  • 1




    $begingroup$
    @Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:42
















1












$begingroup$


I'm trying to find the explicit formula for the recursion $a_n = 2a_{n-1} + a_{n-2}$ with $a_0 = 1$ and $a_1 = 3$. I already found the generating function: $f(x) = frac{1+x}{1-2x-x^2}$, and the characteristic polynomial: $x^2 - 2x + 1 = 0$, which has the roots $1+√2$ and $1-√2$. What is the explicit formula for this recursion, and how do I find it?






generating-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:35










  • $begingroup$
    @SauhardSharma Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 14 at 7:36










  • $begingroup$
    @Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:38










  • $begingroup$
    @SauhardSharma Thank you so much!!!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38






  • 1




    $begingroup$
    @Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:42














1












1








1





$begingroup$


I'm trying to find the explicit formula for the recursion $a_n = 2a_{n-1} + a_{n-2}$ with $a_0 = 1$ and $a_1 = 3$. I already found the generating function: $f(x) = frac{1+x}{1-2x-x^2}$, and the characteristic polynomial: $x^2 - 2x + 1 = 0$, which has the roots $1+√2$ and $1-√2$. What is the explicit formula for this recursion, and how do I find it?






generating-functions







share|cite|improve this question









$endgroup$




I'm trying to find the explicit formula for the recursion $a_n = 2a_{n-1} + a_{n-2}$ with $a_0 = 1$ and $a_1 = 3$. I already found the generating function: $f(x) = frac{1+x}{1-2x-x^2}$, and the characteristic polynomial: $x^2 - 2x + 1 = 0$, which has the roots $1+√2$ and $1-√2$. What is the explicit formula for this recursion, and how do I find it?






generating-functions




generating-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 7:31









Drew WeissermanDrew Weisserman

205




205








  • 1




    $begingroup$
    You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:35










  • $begingroup$
    @SauhardSharma Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 14 at 7:36










  • $begingroup$
    @Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:38










  • $begingroup$
    @SauhardSharma Thank you so much!!!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38






  • 1




    $begingroup$
    @Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:42














  • 1




    $begingroup$
    You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:35










  • $begingroup$
    @SauhardSharma Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 14 at 7:36










  • $begingroup$
    @Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:38










  • $begingroup$
    @SauhardSharma Thank you so much!!!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38






  • 1




    $begingroup$
    @Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 7:42








1




1




$begingroup$
You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
$endgroup$
– Sauhard Sharma
Jan 14 at 7:35




$begingroup$
You have done almost everything. Roots of the characteristic polynomial are solutions of the recursion. Your solution is of the form $$a_n = c_1(1+sqrt{2})^n + c_2(1-sqrt{2})^n$$ Plug in the initial values of $a_0$ and $a_1$ to get the values of the constants $c_1$ and $c_2$
$endgroup$
– Sauhard Sharma
Jan 14 at 7:35












$begingroup$
@SauhardSharma Why are you answering in a comment?
$endgroup$
– Arthur
Jan 14 at 7:36




$begingroup$
@SauhardSharma Why are you answering in a comment?
$endgroup$
– Arthur
Jan 14 at 7:36












$begingroup$
@Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
$endgroup$
– Sauhard Sharma
Jan 14 at 7:38




$begingroup$
@Arthur Because I thought that he had done most of the work and required only a small nudge to get to the solution.
$endgroup$
– Sauhard Sharma
Jan 14 at 7:38












$begingroup$
@SauhardSharma Thank you so much!!!
$endgroup$
– Drew Weisserman
Jan 14 at 7:38




$begingroup$
@SauhardSharma Thank you so much!!!
$endgroup$
– Drew Weisserman
Jan 14 at 7:38




1




1




$begingroup$
@Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
$endgroup$
– Sauhard Sharma
Jan 14 at 7:42




$begingroup$
@Arthur Mea Culpa then. I won't do this from now on. Thanks for the advice !!!
$endgroup$
– Sauhard Sharma
Jan 14 at 7:42










1 Answer
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$begingroup$

Since the characteristic function has roots $1+sqrt{2}$ and $1-sqrt{2}$, the formula for the general term is $$a_n = A(1+sqrt{2})^n+B(1-sqrt{2})^n$$ for some constants $A$ and $B$.



Using the initial conditions $a_0 = 1$ and $a_1 = 3$, you get $$1 = a_0 = A+B$$ $$3 = a_1 = (1+sqrt{2})A+(1-sqrt{2})B$$ You simply need to solve this system of equations for $A$ and $B$.



In general, if the characteristic polynomial has distinct roots $r_1, r_2, cdots, r_k$ each with multiplicity $1$, then the general term will be $a_n = C_1r_1^n+C_2r_2^n+cdots+C_kr_k^n$.



If the roots $r_1,ldots,r_n$ have multiplicities $m_1,ldots,m_k$, then the formula is a bit more complicated: $a_n = p_1(n)r_1^n+cdots+p_k(n)r_k^n$ where $p_i(n)$ is a polynomial with degree $le m_i-1$.






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$endgroup$













  • $begingroup$
    Thank you so much!!! I'll accept the answer as soon as the timer allows it!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38











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$begingroup$

Since the characteristic function has roots $1+sqrt{2}$ and $1-sqrt{2}$, the formula for the general term is $$a_n = A(1+sqrt{2})^n+B(1-sqrt{2})^n$$ for some constants $A$ and $B$.



Using the initial conditions $a_0 = 1$ and $a_1 = 3$, you get $$1 = a_0 = A+B$$ $$3 = a_1 = (1+sqrt{2})A+(1-sqrt{2})B$$ You simply need to solve this system of equations for $A$ and $B$.



In general, if the characteristic polynomial has distinct roots $r_1, r_2, cdots, r_k$ each with multiplicity $1$, then the general term will be $a_n = C_1r_1^n+C_2r_2^n+cdots+C_kr_k^n$.



If the roots $r_1,ldots,r_n$ have multiplicities $m_1,ldots,m_k$, then the formula is a bit more complicated: $a_n = p_1(n)r_1^n+cdots+p_k(n)r_k^n$ where $p_i(n)$ is a polynomial with degree $le m_i-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much!!! I'll accept the answer as soon as the timer allows it!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38
















2












$begingroup$

Since the characteristic function has roots $1+sqrt{2}$ and $1-sqrt{2}$, the formula for the general term is $$a_n = A(1+sqrt{2})^n+B(1-sqrt{2})^n$$ for some constants $A$ and $B$.



Using the initial conditions $a_0 = 1$ and $a_1 = 3$, you get $$1 = a_0 = A+B$$ $$3 = a_1 = (1+sqrt{2})A+(1-sqrt{2})B$$ You simply need to solve this system of equations for $A$ and $B$.



In general, if the characteristic polynomial has distinct roots $r_1, r_2, cdots, r_k$ each with multiplicity $1$, then the general term will be $a_n = C_1r_1^n+C_2r_2^n+cdots+C_kr_k^n$.



If the roots $r_1,ldots,r_n$ have multiplicities $m_1,ldots,m_k$, then the formula is a bit more complicated: $a_n = p_1(n)r_1^n+cdots+p_k(n)r_k^n$ where $p_i(n)$ is a polynomial with degree $le m_i-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much!!! I'll accept the answer as soon as the timer allows it!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38














2












2








2





$begingroup$

Since the characteristic function has roots $1+sqrt{2}$ and $1-sqrt{2}$, the formula for the general term is $$a_n = A(1+sqrt{2})^n+B(1-sqrt{2})^n$$ for some constants $A$ and $B$.



Using the initial conditions $a_0 = 1$ and $a_1 = 3$, you get $$1 = a_0 = A+B$$ $$3 = a_1 = (1+sqrt{2})A+(1-sqrt{2})B$$ You simply need to solve this system of equations for $A$ and $B$.



In general, if the characteristic polynomial has distinct roots $r_1, r_2, cdots, r_k$ each with multiplicity $1$, then the general term will be $a_n = C_1r_1^n+C_2r_2^n+cdots+C_kr_k^n$.



If the roots $r_1,ldots,r_n$ have multiplicities $m_1,ldots,m_k$, then the formula is a bit more complicated: $a_n = p_1(n)r_1^n+cdots+p_k(n)r_k^n$ where $p_i(n)$ is a polynomial with degree $le m_i-1$.






share|cite|improve this answer











$endgroup$



Since the characteristic function has roots $1+sqrt{2}$ and $1-sqrt{2}$, the formula for the general term is $$a_n = A(1+sqrt{2})^n+B(1-sqrt{2})^n$$ for some constants $A$ and $B$.



Using the initial conditions $a_0 = 1$ and $a_1 = 3$, you get $$1 = a_0 = A+B$$ $$3 = a_1 = (1+sqrt{2})A+(1-sqrt{2})B$$ You simply need to solve this system of equations for $A$ and $B$.



In general, if the characteristic polynomial has distinct roots $r_1, r_2, cdots, r_k$ each with multiplicity $1$, then the general term will be $a_n = C_1r_1^n+C_2r_2^n+cdots+C_kr_k^n$.



If the roots $r_1,ldots,r_n$ have multiplicities $m_1,ldots,m_k$, then the formula is a bit more complicated: $a_n = p_1(n)r_1^n+cdots+p_k(n)r_k^n$ where $p_i(n)$ is a polynomial with degree $le m_i-1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 7:43

























answered Jan 14 at 7:36









JimmyK4542JimmyK4542

41.3k245107




41.3k245107












  • $begingroup$
    Thank you so much!!! I'll accept the answer as soon as the timer allows it!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38


















  • $begingroup$
    Thank you so much!!! I'll accept the answer as soon as the timer allows it!
    $endgroup$
    – Drew Weisserman
    Jan 14 at 7:38
















$begingroup$
Thank you so much!!! I'll accept the answer as soon as the timer allows it!
$endgroup$
– Drew Weisserman
Jan 14 at 7:38




$begingroup$
Thank you so much!!! I'll accept the answer as soon as the timer allows it!
$endgroup$
– Drew Weisserman
Jan 14 at 7:38


















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