Mixing
Solve $(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$ via power series method.
$begingroup$
I have trouble finding a closed expression for the following problem:
$$(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$$ where also $f'(0)=0$ and $f(0)=1$
Solve this via a power series method, so suppose $f(x)= sum_{n=0} ^{infty}a_n x^n$. Find its radius of convergence and a closed expression.
I tried solving this, but I get a really nasty expression. We first plug in our Ansatz into the differential eqution, with the usual derivatives:
$$ (1+x^2)sum_{n=0} ^{infty}n (n-1)a_n x^{n-2}+ 4xsum_{n=0} ^{infty}n a_n x^{n-1} + 2 sum_{n=0} ^{infty}a_n x^{n}=0$$
We now multiply out the factors:
$$ sum_{n=0} ^{infty}n (n-1)a_n x^{n-2} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We apply a shift to the first sum:
$$ sum_{n=0} ^{infty}(n+2) (n+1)a_{n+2} x^{n} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We now use the identity theorem for power series to compare coefficients and get for $n geq 2$:
$$ (n+2)(n+1)a_{n+2} + n(n-1)a_n + 4n a_n +2a_n=0$$
After rewriting, we get:
$$ (n+2)(n+1)a_{n+2} + (n+2)(n+1)a_{n}=0$$
$$ a_{n+2} = - a_n$$
Which is not a nice expression, since $a_0=1$ and $a_1=0$ this tells me that $a_2=-1$, $a_4 =1$, $a_6=-1$ and all odd terms are $0$.
What am I missing here?
real-analysis ordinary-differential-equations power-series
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
$endgroup$
|
show 1 more comment
$begingroup$
I have trouble finding a closed expression for the following problem:
$$(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$$ where also $f'(0)=0$ and $f(0)=1$
Solve this via a power series method, so suppose $f(x)= sum_{n=0} ^{infty}a_n x^n$. Find its radius of convergence and a closed expression.
I tried solving this, but I get a really nasty expression. We first plug in our Ansatz into the differential eqution, with the usual derivatives:
$$ (1+x^2)sum_{n=0} ^{infty}n (n-1)a_n x^{n-2}+ 4xsum_{n=0} ^{infty}n a_n x^{n-1} + 2 sum_{n=0} ^{infty}a_n x^{n}=0$$
We now multiply out the factors:
$$ sum_{n=0} ^{infty}n (n-1)a_n x^{n-2} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We apply a shift to the first sum:
$$ sum_{n=0} ^{infty}(n+2) (n+1)a_{n+2} x^{n} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We now use the identity theorem for power series to compare coefficients and get for $n geq 2$:
$$ (n+2)(n+1)a_{n+2} + n(n-1)a_n + 4n a_n +2a_n=0$$
After rewriting, we get:
$$ (n+2)(n+1)a_{n+2} + (n+2)(n+1)a_{n}=0$$
$$ a_{n+2} = - a_n$$
Which is not a nice expression, since $a_0=1$ and $a_1=0$ this tells me that $a_2=-1$, $a_4 =1$, $a_6=-1$ and all odd terms are $0$.
What am I missing here?
real-analysis ordinary-differential-equations power-series
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
$endgroup$
$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
3
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
1
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07
|
show 1 more comment
$begingroup$
I have trouble finding a closed expression for the following problem:
$$(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$$ where also $f'(0)=0$ and $f(0)=1$
Solve this via a power series method, so suppose $f(x)= sum_{n=0} ^{infty}a_n x^n$. Find its radius of convergence and a closed expression.
I tried solving this, but I get a really nasty expression. We first plug in our Ansatz into the differential eqution, with the usual derivatives:
$$ (1+x^2)sum_{n=0} ^{infty}n (n-1)a_n x^{n-2}+ 4xsum_{n=0} ^{infty}n a_n x^{n-1} + 2 sum_{n=0} ^{infty}a_n x^{n}=0$$
We now multiply out the factors:
$$ sum_{n=0} ^{infty}n (n-1)a_n x^{n-2} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We apply a shift to the first sum:
$$ sum_{n=0} ^{infty}(n+2) (n+1)a_{n+2} x^{n} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We now use the identity theorem for power series to compare coefficients and get for $n geq 2$:
$$ (n+2)(n+1)a_{n+2} + n(n-1)a_n + 4n a_n +2a_n=0$$
After rewriting, we get:
$$ (n+2)(n+1)a_{n+2} + (n+2)(n+1)a_{n}=0$$
$$ a_{n+2} = - a_n$$
Which is not a nice expression, since $a_0=1$ and $a_1=0$ this tells me that $a_2=-1$, $a_4 =1$, $a_6=-1$ and all odd terms are $0$.
What am I missing here?
real-analysis ordinary-differential-equations power-series
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
$endgroup$
I have trouble finding a closed expression for the following problem:
$$(1+x^2) f''(x) + 4x f'(x) + 2 f(x)=0$$ where also $f'(0)=0$ and $f(0)=1$
Solve this via a power series method, so suppose $f(x)= sum_{n=0} ^{infty}a_n x^n$. Find its radius of convergence and a closed expression.
I tried solving this, but I get a really nasty expression. We first plug in our Ansatz into the differential eqution, with the usual derivatives:
$$ (1+x^2)sum_{n=0} ^{infty}n (n-1)a_n x^{n-2}+ 4xsum_{n=0} ^{infty}n a_n x^{n-1} + 2 sum_{n=0} ^{infty}a_n x^{n}=0$$
We now multiply out the factors:
$$ sum_{n=0} ^{infty}n (n-1)a_n x^{n-2} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We apply a shift to the first sum:
$$ sum_{n=0} ^{infty}(n+2) (n+1)a_{n+2} x^{n} +sum_{n=0} ^{infty}n (n-1)a_n x^{n} + sum_{n=0} ^{infty}4n a_n x^{n} + sum_{n=0} ^{infty}2a_n x^{n}=0$$
We now use the identity theorem for power series to compare coefficients and get for $n geq 2$:
$$ (n+2)(n+1)a_{n+2} + n(n-1)a_n + 4n a_n +2a_n=0$$
After rewriting, we get:
$$ (n+2)(n+1)a_{n+2} + (n+2)(n+1)a_{n}=0$$
$$ a_{n+2} = - a_n$$
Which is not a nice expression, since $a_0=1$ and $a_1=0$ this tells me that $a_2=-1$, $a_4 =1$, $a_6=-1$ and all odd terms are $0$.
What am I missing here?
real-analysis ordinary-differential-equations power-series
real-analysis ordinary-differential-equations power-series
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
edited Feb 1 at 16:12
Wesley Strik
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
asked Feb 1 at 13:44
Wesley StrikWesley Strik
2,189424
2,189424
$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
3
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
1
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07
|
show 1 more comment
$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
3
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
1
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07
$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
3
3
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
1
1
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8dots$$
Notice that:
$$ frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 dots $$
So our closed expression is the geometric series variation $$frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
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$begingroup$
We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8dots$$
Notice that:
$$ frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 dots $$
So our closed expression is the geometric series variation $$frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
$begingroup$
We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8dots$$
Notice that:
$$ frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 dots $$
So our closed expression is the geometric series variation $$frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
$begingroup$
We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8dots$$
Notice that:
$$ frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 dots $$
So our closed expression is the geometric series variation $$frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
$endgroup$
We recognise this to be:
$$ 1 - x^2 + x^4 - x^6 +x^8dots$$
Notice that:
$$ frac{1}{1 - (-x^2)}= 1- x^2 + x^4 - x^6 +x^8 dots $$
So our closed expression is the geometric series variation $$frac{1}{1+x^2}$$
Its radius of convergence follows from the geometric series to be $|x^2|<1$ so $|x|<1$. This means the series has radius of convergence $1$.
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
answered Feb 1 at 16:15
Wesley StrikWesley Strik
2,189424
2,189424
add a comment |
add a comment |
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$begingroup$
Why are all odd terms $0$?
$endgroup$
– Bernard
Feb 1 at 13:55
$begingroup$
@Bernard $f'(0)=0$.
$endgroup$
– Floris Claassens
Feb 1 at 13:58
$begingroup$
But you wrote $a_1=1$!
$endgroup$
– Bernard
Feb 1 at 14:01
3
$begingroup$
Why do you think that is not nice? The geometric series can easily be identified as $frac1{1+x^2}$.
$endgroup$
– LutzL
Feb 1 at 14:02
1
$begingroup$
general solution is $frac{c_1x+c_2}{1+x^2}$
$endgroup$
– Aleksas Domarkas
Feb 1 at 14:07