How to solve a recurrence relation with generating functions?












2












$begingroup$


I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n geq 2$



This is what I was thinking,
begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\
g(x) &= a_0 + a_1x^1 + sum^{infty}_{n=2}a_nx^n\
g(x) &= 1 + 2x + sum^{infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\
g(x) &= 1 + 2x + sum^{infty}_{n=2}a_{n-1}x^n + sum^{infty}_{n=2}(2n-2)x^n\
g(x) &= 1 + 2x + xsum^{infty}_{n=1}a_{n-1}x^{n-1} + sum^{infty}_{n=2}2n x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + xsum^{infty}_{m=1}a_{m}x^{m} + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + x(g(x) - a_0) + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
end{align*}






combinatorics recurrence-relations relations generating-functions







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$endgroup$












  • $begingroup$
    A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:09










  • $begingroup$
    Fixed it -- thanks!
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:22
















2












$begingroup$


I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n geq 2$



This is what I was thinking,
begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\
g(x) &= a_0 + a_1x^1 + sum^{infty}_{n=2}a_nx^n\
g(x) &= 1 + 2x + sum^{infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\
g(x) &= 1 + 2x + sum^{infty}_{n=2}a_{n-1}x^n + sum^{infty}_{n=2}(2n-2)x^n\
g(x) &= 1 + 2x + xsum^{infty}_{n=1}a_{n-1}x^{n-1} + sum^{infty}_{n=2}2n x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + xsum^{infty}_{m=1}a_{m}x^{m} + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + x(g(x) - a_0) + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
end{align*}






combinatorics recurrence-relations relations generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:09










  • $begingroup$
    Fixed it -- thanks!
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:22














2












2








2





$begingroup$


I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n geq 2$



This is what I was thinking,
begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\
g(x) &= a_0 + a_1x^1 + sum^{infty}_{n=2}a_nx^n\
g(x) &= 1 + 2x + sum^{infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\
g(x) &= 1 + 2x + sum^{infty}_{n=2}a_{n-1}x^n + sum^{infty}_{n=2}(2n-2)x^n\
g(x) &= 1 + 2x + xsum^{infty}_{n=1}a_{n-1}x^{n-1} + sum^{infty}_{n=2}2n x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + xsum^{infty}_{m=1}a_{m}x^{m} + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + x(g(x) - a_0) + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
end{align*}






combinatorics recurrence-relations relations generating-functions







share|cite|improve this question











$endgroup$




I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n geq 2$



This is what I was thinking,
begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\
g(x) &= a_0 + a_1x^1 + sum^{infty}_{n=2}a_nx^n\
g(x) &= 1 + 2x + sum^{infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\
g(x) &= 1 + 2x + sum^{infty}_{n=2}a_{n-1}x^n + sum^{infty}_{n=2}(2n-2)x^n\
g(x) &= 1 + 2x + xsum^{infty}_{n=1}a_{n-1}x^{n-1} + sum^{infty}_{n=2}2n x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + xsum^{infty}_{m=1}a_{m}x^{m} + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
g(x) &= 1 + 2x + x(g(x) - a_0) + sum^{infty}_{n=2}2 binom{n}{1}x^n - sum^{infty}_{n=2}2x^n\
end{align*}






combinatorics recurrence-relations relations generating-functions




combinatorics recurrence-relations relations generating-functions






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share|cite|improve this question













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edited Dec 12 '18 at 3:21







Math Newbie

















asked Dec 12 '18 at 2:55









Math NewbieMath Newbie

428




428












  • $begingroup$
    A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:09










  • $begingroup$
    Fixed it -- thanks!
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:22


















  • $begingroup$
    A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:09










  • $begingroup$
    Fixed it -- thanks!
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:22
















$begingroup$
A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:09




$begingroup$
A linear recurrence relation and two initial conditions? That's strange, but easily fixable.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:09












$begingroup$
Fixed it -- thanks!
$endgroup$
– Math Newbie
Dec 12 '18 at 3:22




$begingroup$
Fixed it -- thanks!
$endgroup$
– Math Newbie
Dec 12 '18 at 3:22










1 Answer
1






active

oldest

votes


















1












$begingroup$

You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,ldots,N$ to get that $a_n$ depends on $sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $ngeq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=sum_{ngeq 0} a_n x^n = 2+sum_{ngeq 1}a_n x^n$$
we have
$$ xcdot f(x) = sum_{ngeq 0} a_n x^{n+1} = sum_{ngeq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2sum_{ngeq 1}(n-1)x^n = 2+frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ frac{1}{(1-x)^{m+1}}=sum_{ngeq 0}binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = frac{2}{1-x}+frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:15










  • $begingroup$
    @MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:18










  • $begingroup$
    makes sense - thank you.
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:21










  • $begingroup$
    $a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:29










  • $begingroup$
    Sorry, understand that now. Thanks for the clarification
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:29











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,ldots,N$ to get that $a_n$ depends on $sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $ngeq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=sum_{ngeq 0} a_n x^n = 2+sum_{ngeq 1}a_n x^n$$
we have
$$ xcdot f(x) = sum_{ngeq 0} a_n x^{n+1} = sum_{ngeq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2sum_{ngeq 1}(n-1)x^n = 2+frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ frac{1}{(1-x)^{m+1}}=sum_{ngeq 0}binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = frac{2}{1-x}+frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:15










  • $begingroup$
    @MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:18










  • $begingroup$
    makes sense - thank you.
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:21










  • $begingroup$
    $a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:29










  • $begingroup$
    Sorry, understand that now. Thanks for the clarification
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:29
















1












$begingroup$

You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,ldots,N$ to get that $a_n$ depends on $sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $ngeq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=sum_{ngeq 0} a_n x^n = 2+sum_{ngeq 1}a_n x^n$$
we have
$$ xcdot f(x) = sum_{ngeq 0} a_n x^{n+1} = sum_{ngeq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2sum_{ngeq 1}(n-1)x^n = 2+frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ frac{1}{(1-x)^{m+1}}=sum_{ngeq 0}binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = frac{2}{1-x}+frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:15










  • $begingroup$
    @MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:18










  • $begingroup$
    makes sense - thank you.
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:21










  • $begingroup$
    $a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:29










  • $begingroup$
    Sorry, understand that now. Thanks for the clarification
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:29














1












1








1





$begingroup$

You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,ldots,N$ to get that $a_n$ depends on $sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $ngeq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=sum_{ngeq 0} a_n x^n = 2+sum_{ngeq 1}a_n x^n$$
we have
$$ xcdot f(x) = sum_{ngeq 0} a_n x^{n+1} = sum_{ngeq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2sum_{ngeq 1}(n-1)x^n = 2+frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ frac{1}{(1-x)^{m+1}}=sum_{ngeq 0}binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = frac{2}{1-x}+frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$






share|cite|improve this answer









$endgroup$



You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,ldots,N$ to get that $a_n$ depends on $sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $ngeq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=sum_{ngeq 0} a_n x^n = 2+sum_{ngeq 1}a_n x^n$$
we have
$$ xcdot f(x) = sum_{ngeq 0} a_n x^{n+1} = sum_{ngeq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2sum_{ngeq 1}(n-1)x^n = 2+frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ frac{1}{(1-x)^{m+1}}=sum_{ngeq 0}binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = frac{2}{1-x}+frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 3:08









Jack D'AurizioJack D'Aurizio

289k33281661




289k33281661












  • $begingroup$
    I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:15










  • $begingroup$
    @MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:18










  • $begingroup$
    makes sense - thank you.
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:21










  • $begingroup$
    $a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:29










  • $begingroup$
    Sorry, understand that now. Thanks for the clarification
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:29


















  • $begingroup$
    I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:15










  • $begingroup$
    @MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:18










  • $begingroup$
    makes sense - thank you.
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:21










  • $begingroup$
    $a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
    $endgroup$
    – Jack D'Aurizio
    Dec 12 '18 at 3:29










  • $begingroup$
    Sorry, understand that now. Thanks for the clarification
    $endgroup$
    – Math Newbie
    Dec 12 '18 at 3:29
















$begingroup$
I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
$endgroup$
– Math Newbie
Dec 12 '18 at 3:15




$begingroup$
I was solving the question to "Find and solve the recurrence relation for the number of regions created by n mutually intersecting circles on a piece of paper (no three circles have a common intersection point)." so I incorrectly "set" the initial values. sorry! Do you have any recommendations for what the initial conditions should be?
$endgroup$
– Math Newbie
Dec 12 '18 at 3:15












$begingroup$
@MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:18




$begingroup$
@MathNewbie: you may just start with $a_1=2$, the linear recurrence relation and the generating function defined as a sum over $ngeq 1$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:18












$begingroup$
makes sense - thank you.
$endgroup$
– Math Newbie
Dec 12 '18 at 3:21




$begingroup$
makes sense - thank you.
$endgroup$
– Math Newbie
Dec 12 '18 at 3:21












$begingroup$
$a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:29




$begingroup$
$a_1=2$ makes sense: a circle (simple closed curve) divides the plane into two connected components (Jordan's theorem). What is debatable is to assign a value to $a_0$.
$endgroup$
– Jack D'Aurizio
Dec 12 '18 at 3:29












$begingroup$
Sorry, understand that now. Thanks for the clarification
$endgroup$
– Math Newbie
Dec 12 '18 at 3:29




$begingroup$
Sorry, understand that now. Thanks for the clarification
$endgroup$
– Math Newbie
Dec 12 '18 at 3:29


















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