Solving a nonhomogeneous recurrence relation?











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1
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I was asked to find a first order linear recurrence relation for



$$
a_n=3n^2-2n+1
$$



Here is what I did
begin{align}label{1}
a_{n-1} &= 3(n-1)^2-2(n-1)+1\
&=3(n^2-2n+1)-2n+2+1\
&=underbrace{3n^2-2n+1}_{a_n}-6n+5\
&=a_n-6n+5
end{align}
Thus,
begin{align}
a_n-a_{n-1}=6n-5tag{1}label{2}
end{align}
with $a_0=1$ is a first order recurrence relation for the given sequence.



But I was unable to retrieve the given sequence from this recurrence relation. Clearly, the recurrence is nonhomogeneous. So, its solution is of the form
$$
a_n=a_n^h+a_n^p
$$
Now,
$$
a_n^h=c, mbox{any constant}
$$
Since $f(n)=-6n+5$o to find a particular solution for the non-homogeneous part, we set
$a_n^p=A_1n+A_0$, where $A_1,A_0$ are constant. Substituting this into (ref{2}) yield
begin{align*}
A_1n+A_0-[A_1(n-1)+A_0]&=6n-5\
A_1 &= 6n-5
end{align*}
I have tried this again and again but I couldn't tell what is happening? What is wrong with me?




Edited: Let me put it in this way, solve
begin{align}
a_n-a_{n-1}=6n-5, a_0=1.
end{align}







recurrence-relations







share|cite|improve this question




















  • 1




    Try $$a_n^p=A_1n^2+A_0n.$$
    – bof
    Nov 22 at 13:07










  • @bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
    – marya
    Nov 22 at 13:15








  • 1




    In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
    – bof
    Nov 22 at 21:56










  • It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
    – bof
    Nov 23 at 4:18










  • Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
    – bof
    Nov 23 at 4:20















up vote
1
down vote

favorite












I was asked to find a first order linear recurrence relation for



$$
a_n=3n^2-2n+1
$$



Here is what I did
begin{align}label{1}
a_{n-1} &= 3(n-1)^2-2(n-1)+1\
&=3(n^2-2n+1)-2n+2+1\
&=underbrace{3n^2-2n+1}_{a_n}-6n+5\
&=a_n-6n+5
end{align}
Thus,
begin{align}
a_n-a_{n-1}=6n-5tag{1}label{2}
end{align}
with $a_0=1$ is a first order recurrence relation for the given sequence.



But I was unable to retrieve the given sequence from this recurrence relation. Clearly, the recurrence is nonhomogeneous. So, its solution is of the form
$$
a_n=a_n^h+a_n^p
$$
Now,
$$
a_n^h=c, mbox{any constant}
$$
Since $f(n)=-6n+5$o to find a particular solution for the non-homogeneous part, we set
$a_n^p=A_1n+A_0$, where $A_1,A_0$ are constant. Substituting this into (ref{2}) yield
begin{align*}
A_1n+A_0-[A_1(n-1)+A_0]&=6n-5\
A_1 &= 6n-5
end{align*}
I have tried this again and again but I couldn't tell what is happening? What is wrong with me?




Edited: Let me put it in this way, solve
begin{align}
a_n-a_{n-1}=6n-5, a_0=1.
end{align}







recurrence-relations







share|cite|improve this question




















  • 1




    Try $$a_n^p=A_1n^2+A_0n.$$
    – bof
    Nov 22 at 13:07










  • @bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
    – marya
    Nov 22 at 13:15








  • 1




    In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
    – bof
    Nov 22 at 21:56










  • It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
    – bof
    Nov 23 at 4:18










  • Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
    – bof
    Nov 23 at 4:20













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was asked to find a first order linear recurrence relation for



$$
a_n=3n^2-2n+1
$$



Here is what I did
begin{align}label{1}
a_{n-1} &= 3(n-1)^2-2(n-1)+1\
&=3(n^2-2n+1)-2n+2+1\
&=underbrace{3n^2-2n+1}_{a_n}-6n+5\
&=a_n-6n+5
end{align}
Thus,
begin{align}
a_n-a_{n-1}=6n-5tag{1}label{2}
end{align}
with $a_0=1$ is a first order recurrence relation for the given sequence.



But I was unable to retrieve the given sequence from this recurrence relation. Clearly, the recurrence is nonhomogeneous. So, its solution is of the form
$$
a_n=a_n^h+a_n^p
$$
Now,
$$
a_n^h=c, mbox{any constant}
$$
Since $f(n)=-6n+5$o to find a particular solution for the non-homogeneous part, we set
$a_n^p=A_1n+A_0$, where $A_1,A_0$ are constant. Substituting this into (ref{2}) yield
begin{align*}
A_1n+A_0-[A_1(n-1)+A_0]&=6n-5\
A_1 &= 6n-5
end{align*}
I have tried this again and again but I couldn't tell what is happening? What is wrong with me?




Edited: Let me put it in this way, solve
begin{align}
a_n-a_{n-1}=6n-5, a_0=1.
end{align}







recurrence-relations







share|cite|improve this question















I was asked to find a first order linear recurrence relation for



$$
a_n=3n^2-2n+1
$$



Here is what I did
begin{align}label{1}
a_{n-1} &= 3(n-1)^2-2(n-1)+1\
&=3(n^2-2n+1)-2n+2+1\
&=underbrace{3n^2-2n+1}_{a_n}-6n+5\
&=a_n-6n+5
end{align}
Thus,
begin{align}
a_n-a_{n-1}=6n-5tag{1}label{2}
end{align}
with $a_0=1$ is a first order recurrence relation for the given sequence.



But I was unable to retrieve the given sequence from this recurrence relation. Clearly, the recurrence is nonhomogeneous. So, its solution is of the form
$$
a_n=a_n^h+a_n^p
$$
Now,
$$
a_n^h=c, mbox{any constant}
$$
Since $f(n)=-6n+5$o to find a particular solution for the non-homogeneous part, we set
$a_n^p=A_1n+A_0$, where $A_1,A_0$ are constant. Substituting this into (ref{2}) yield
begin{align*}
A_1n+A_0-[A_1(n-1)+A_0]&=6n-5\
A_1 &= 6n-5
end{align*}
I have tried this again and again but I couldn't tell what is happening? What is wrong with me?




Edited: Let me put it in this way, solve
begin{align}
a_n-a_{n-1}=6n-5, a_0=1.
end{align}







recurrence-relations




recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 13:03

























asked Nov 22 at 12:23









marya

373217




373217








  • 1




    Try $$a_n^p=A_1n^2+A_0n.$$
    – bof
    Nov 22 at 13:07










  • @bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
    – marya
    Nov 22 at 13:15








  • 1




    In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
    – bof
    Nov 22 at 21:56










  • It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
    – bof
    Nov 23 at 4:18










  • Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
    – bof
    Nov 23 at 4:20














  • 1




    Try $$a_n^p=A_1n^2+A_0n.$$
    – bof
    Nov 22 at 13:07










  • @bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
    – marya
    Nov 22 at 13:15








  • 1




    In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
    – bof
    Nov 22 at 21:56










  • It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
    – bof
    Nov 23 at 4:18










  • Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
    – bof
    Nov 23 at 4:20








1




1




Try $$a_n^p=A_1n^2+A_0n.$$
– bof
Nov 22 at 13:07




Try $$a_n^p=A_1n^2+A_0n.$$
– bof
Nov 22 at 13:07












@bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
– marya
Nov 22 at 13:15






@bof, Why? $f(n)=6n-5$, a polynomial of degree one, shouldn't our particular solution take the same form?
– marya
Nov 22 at 13:15






1




1




In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
– bof
Nov 22 at 21:56




In your $a_n^p=A_1n+A_0$ the $A_0$ term is useless, because $a_n=A_0$ is a solution of the homogeneous recurrence $a_n-a_{n-1}=0$.
– bof
Nov 22 at 21:56












It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
– bof
Nov 23 at 4:18




It's basically the same reason why, in solving the nonhomogeneous DIFFERENTIAL equation $y'-y=(6x+1)e^x$, you would look for a particular solution of the form $y_p=(Ax^2+Bx)e^x$ instead of $y_p=(Ax+B)e^x$.
– bof
Nov 23 at 4:18












Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
– bof
Nov 23 at 4:20




Actually, you could turn your recurrence into a differential equation by considering the exponential generating function $$y(x)=sum_{n=0}^inftyfrac{a_n}{n!}x^n.$$
– bof
Nov 23 at 4:20










1 Answer
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$implies$



$$a_n-a_{n-1}=6n-5iff6n=?$$
$$ a_{n+1}-a_n=6(n+1)-5iff6n=?$$



Compare the two values of $6n?$






share|cite|improve this answer





















  • @marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
    – lab bhattacharjee
    Nov 22 at 12:49











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1 Answer
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active

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up vote
1
down vote













$implies$



$$a_n-a_{n-1}=6n-5iff6n=?$$
$$ a_{n+1}-a_n=6(n+1)-5iff6n=?$$



Compare the two values of $6n?$






share|cite|improve this answer





















  • @marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
    – lab bhattacharjee
    Nov 22 at 12:49















up vote
1
down vote













$implies$



$$a_n-a_{n-1}=6n-5iff6n=?$$
$$ a_{n+1}-a_n=6(n+1)-5iff6n=?$$



Compare the two values of $6n?$






share|cite|improve this answer





















  • @marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
    – lab bhattacharjee
    Nov 22 at 12:49













up vote
1
down vote










up vote
1
down vote









$implies$



$$a_n-a_{n-1}=6n-5iff6n=?$$
$$ a_{n+1}-a_n=6(n+1)-5iff6n=?$$



Compare the two values of $6n?$






share|cite|improve this answer












$implies$



$$a_n-a_{n-1}=6n-5iff6n=?$$
$$ a_{n+1}-a_n=6(n+1)-5iff6n=?$$



Compare the two values of $6n?$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 12:28









lab bhattacharjee

222k15155273




222k15155273












  • @marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
    – lab bhattacharjee
    Nov 22 at 12:49


















  • @marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
    – lab bhattacharjee
    Nov 22 at 12:49
















@marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
– lab bhattacharjee
Nov 22 at 12:49




@marya, $$3{n^2-(n-1)^2}-2{n-(n-1)}=3(2n-1)-2=6n-5$$ is correct
– lab bhattacharjee
Nov 22 at 12:49


















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