Mixing
Power series representation for f''(x)-xf(x)=x
$begingroup$
Given is $sumlimits_{k=0}^infty a_kx^k$ and $f''(x)-xf(x)=x$, $f'(0)=0$, $f(0)=0$.
newline
I need to find a series representation of the upper sum, through determining an expression for $a_k$ using the given relations. I know how to solve such an expression when $f''(x)-xf(x)=x$ is instead $f''(x)-xf(x)=0$, but if I remove the x here I get $frac{f''(x)}{x}-f(x)=1$ and I am not sure how to solve that, as I do not get a logical answer when I tried. So how can I solve this? I got:
$sumlimits_{k=1}^infty (k+2)(k+3)a_{k+3}x^{k}-sumlimits_{k=3}^infty a_{k}x^{k}=1$
But how do I need to work with that last 1 in order to be able to work with the identity theorem and thus obtain a reursive formula?`
real-analysis convergence power-series recursion
asked Jan 22 at 11:03
C. EliasC. Elias
223
$endgroup$
add a comment |
$begingroup$
Given is $sumlimits_{k=0}^infty a_kx^k$ and $f''(x)-xf(x)=x$, $f'(0)=0$, $f(0)=0$.
newline
I need to find a series representation of the upper sum, through determining an expression for $a_k$ using the given relations. I know how to solve such an expression when $f''(x)-xf(x)=x$ is instead $f''(x)-xf(x)=0$, but if I remove the x here I get $frac{f''(x)}{x}-f(x)=1$ and I am not sure how to solve that, as I do not get a logical answer when I tried. So how can I solve this? I got:
$sumlimits_{k=1}^infty (k+2)(k+3)a_{k+3}x^{k}-sumlimits_{k=3}^infty a_{k}x^{k}=1$
But how do I need to work with that last 1 in order to be able to work with the identity theorem and thus obtain a reursive formula?`
real-analysis convergence power-series recursion
asked Jan 22 at 11:03
C. EliasC. Elias
223
$endgroup$
add a comment |
$begingroup$
Given is $sumlimits_{k=0}^infty a_kx^k$ and $f''(x)-xf(x)=x$, $f'(0)=0$, $f(0)=0$.
newline
I need to find a series representation of the upper sum, through determining an expression for $a_k$ using the given relations. I know how to solve such an expression when $f''(x)-xf(x)=x$ is instead $f''(x)-xf(x)=0$, but if I remove the x here I get $frac{f''(x)}{x}-f(x)=1$ and I am not sure how to solve that, as I do not get a logical answer when I tried. So how can I solve this? I got:
$sumlimits_{k=1}^infty (k+2)(k+3)a_{k+3}x^{k}-sumlimits_{k=3}^infty a_{k}x^{k}=1$
But how do I need to work with that last 1 in order to be able to work with the identity theorem and thus obtain a reursive formula?`
real-analysis convergence power-series recursion
asked Jan 22 at 11:03
C. EliasC. Elias
223
$endgroup$
Given is $sumlimits_{k=0}^infty a_kx^k$ and $f''(x)-xf(x)=x$, $f'(0)=0$, $f(0)=0$.
newline
I need to find a series representation of the upper sum, through determining an expression for $a_k$ using the given relations. I know how to solve such an expression when $f''(x)-xf(x)=x$ is instead $f''(x)-xf(x)=0$, but if I remove the x here I get $frac{f''(x)}{x}-f(x)=1$ and I am not sure how to solve that, as I do not get a logical answer when I tried. So how can I solve this? I got:
$sumlimits_{k=1}^infty (k+2)(k+3)a_{k+3}x^{k}-sumlimits_{k=3}^infty a_{k}x^{k}=1$
But how do I need to work with that last 1 in order to be able to work with the identity theorem and thus obtain a reursive formula?`
real-analysis convergence power-series recursion
real-analysis convergence power-series recursion
asked Jan 22 at 11:03
C. EliasC. Elias
223
asked Jan 22 at 11:03
C. EliasC. Elias
223
edited Jan 22 at 20:40
C. Elias
asked Jan 22 at 11:03
C. EliasC. Elias
223
asked Jan 22 at 11:03
C. EliasC. Elias
223
asked Jan 22 at 11:03
C. EliasC. Elias
223
223
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
add a comment |
$begingroup$
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
add a comment |
$begingroup$
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
$endgroup$
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
answered Jan 22 at 13:09
Robert IsraelRobert Israel
326k23215469
326k23215469
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
add a comment |
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
How did you get f(x)=-1?
$endgroup$
– C. Elias
Jan 22 at 20:46
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
$begingroup$
I looked for a constant solution.
$endgroup$
– Robert Israel
Jan 23 at 0:06
add a comment |
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