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Solution of $y''+xy=0$
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The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= sum_{n=0}^{infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= sum_{n=1}^{infty} n a_n x^{n-1}= sum_{n=0}^{infty} (n+1) a_{n+1} x^n$$
$$y''(x)= sum_{n=1}^{infty} (n+1) n a_{n+1} x^{n-1}= sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n+ x sum_{n=0}^{infty} a_n x^n=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=0}^{infty} a_n x^{n+1}=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=1}^{infty} a_{n-1} x^n=0 \ Rightarrow 2a_2+sum_{n=1}^{infty} left[ (n+2) (n+1) a_{n+2}+ a_{n-1}right] x^n=0$$
So it has to hold:
$$a_2=0 \ (n+2) (n+1) a_{n+2}+a_{n-1}=0, forall n=1,2,3, dots$$
For $n=1$: $3 cdot 2 cdot a_3+ a_0=0 Rightarrow a_3=-frac{a_0}{6}$
For $n=2$: $4 cdot 3 cdot a_4+a_1=0 Rightarrow a_4=-frac{a_1}{12}$
For $n=3$: $5 cdot 4 cdot a_5+a_2=0 Rightarrow a_5=0$
For $n=4$: $6 cdot 5 cdot a_6+a_3=0 Rightarrow 30 a_6-frac{a_0}{6}=0 Rightarrow a_6=frac{a_0}{6 cdot 30}=frac{a_0}{180}$
For $n=5$: $7 cdot 6 cdot a_7+ a_4=0 Rightarrow 7 cdot 6 cdot a_7-frac{a_1}{12}=0 Rightarrow a_7=frac{a_1}{12 cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
differential-equations power-series
asked Mar 19 '15 at 21:06
evinda
4,01031749
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up vote
9
down vote
favorite
The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= sum_{n=0}^{infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= sum_{n=1}^{infty} n a_n x^{n-1}= sum_{n=0}^{infty} (n+1) a_{n+1} x^n$$
$$y''(x)= sum_{n=1}^{infty} (n+1) n a_{n+1} x^{n-1}= sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n+ x sum_{n=0}^{infty} a_n x^n=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=0}^{infty} a_n x^{n+1}=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=1}^{infty} a_{n-1} x^n=0 \ Rightarrow 2a_2+sum_{n=1}^{infty} left[ (n+2) (n+1) a_{n+2}+ a_{n-1}right] x^n=0$$
So it has to hold:
$$a_2=0 \ (n+2) (n+1) a_{n+2}+a_{n-1}=0, forall n=1,2,3, dots$$
For $n=1$: $3 cdot 2 cdot a_3+ a_0=0 Rightarrow a_3=-frac{a_0}{6}$
For $n=2$: $4 cdot 3 cdot a_4+a_1=0 Rightarrow a_4=-frac{a_1}{12}$
For $n=3$: $5 cdot 4 cdot a_5+a_2=0 Rightarrow a_5=0$
For $n=4$: $6 cdot 5 cdot a_6+a_3=0 Rightarrow 30 a_6-frac{a_0}{6}=0 Rightarrow a_6=frac{a_0}{6 cdot 30}=frac{a_0}{180}$
For $n=5$: $7 cdot 6 cdot a_7+ a_4=0 Rightarrow 7 cdot 6 cdot a_7-frac{a_1}{12}=0 Rightarrow a_7=frac{a_1}{12 cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
differential-equations power-series
asked Mar 19 '15 at 21:06
evinda
4,01031749
The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
1
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27
|
show 2 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= sum_{n=0}^{infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= sum_{n=1}^{infty} n a_n x^{n-1}= sum_{n=0}^{infty} (n+1) a_{n+1} x^n$$
$$y''(x)= sum_{n=1}^{infty} (n+1) n a_{n+1} x^{n-1}= sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n+ x sum_{n=0}^{infty} a_n x^n=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=0}^{infty} a_n x^{n+1}=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=1}^{infty} a_{n-1} x^n=0 \ Rightarrow 2a_2+sum_{n=1}^{infty} left[ (n+2) (n+1) a_{n+2}+ a_{n-1}right] x^n=0$$
So it has to hold:
$$a_2=0 \ (n+2) (n+1) a_{n+2}+a_{n-1}=0, forall n=1,2,3, dots$$
For $n=1$: $3 cdot 2 cdot a_3+ a_0=0 Rightarrow a_3=-frac{a_0}{6}$
For $n=2$: $4 cdot 3 cdot a_4+a_1=0 Rightarrow a_4=-frac{a_1}{12}$
For $n=3$: $5 cdot 4 cdot a_5+a_2=0 Rightarrow a_5=0$
For $n=4$: $6 cdot 5 cdot a_6+a_3=0 Rightarrow 30 a_6-frac{a_0}{6}=0 Rightarrow a_6=frac{a_0}{6 cdot 30}=frac{a_0}{180}$
For $n=5$: $7 cdot 6 cdot a_7+ a_4=0 Rightarrow 7 cdot 6 cdot a_7-frac{a_1}{12}=0 Rightarrow a_7=frac{a_1}{12 cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
differential-equations power-series
asked Mar 19 '15 at 21:06
evinda
4,01031749
The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= sum_{n=0}^{infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= sum_{n=1}^{infty} n a_n x^{n-1}= sum_{n=0}^{infty} (n+1) a_{n+1} x^n$$
$$y''(x)= sum_{n=1}^{infty} (n+1) n a_{n+1} x^{n-1}= sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$sum_{n=0}^{infty} (n+2) (n+1) a_{n+2} x^n+ x sum_{n=0}^{infty} a_n x^n=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=0}^{infty} a_n x^{n+1}=0 \ Rightarrow sum_{n=0}^{infty} (n+2)(n+1) a_{n+2} x^n+ sum_{n=1}^{infty} a_{n-1} x^n=0 \ Rightarrow 2a_2+sum_{n=1}^{infty} left[ (n+2) (n+1) a_{n+2}+ a_{n-1}right] x^n=0$$
So it has to hold:
$$a_2=0 \ (n+2) (n+1) a_{n+2}+a_{n-1}=0, forall n=1,2,3, dots$$
For $n=1$: $3 cdot 2 cdot a_3+ a_0=0 Rightarrow a_3=-frac{a_0}{6}$
For $n=2$: $4 cdot 3 cdot a_4+a_1=0 Rightarrow a_4=-frac{a_1}{12}$
For $n=3$: $5 cdot 4 cdot a_5+a_2=0 Rightarrow a_5=0$
For $n=4$: $6 cdot 5 cdot a_6+a_3=0 Rightarrow 30 a_6-frac{a_0}{6}=0 Rightarrow a_6=frac{a_0}{6 cdot 30}=frac{a_0}{180}$
For $n=5$: $7 cdot 6 cdot a_7+ a_4=0 Rightarrow 7 cdot 6 cdot a_7-frac{a_1}{12}=0 Rightarrow a_7=frac{a_1}{12 cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
differential-equations power-series
differential-equations power-series
asked Mar 19 '15 at 21:06
evinda
4,01031749
asked Mar 19 '15 at 21:06
evinda
4,01031749
edited May 28 '15 at 17:43
asked Mar 19 '15 at 21:06
evinda
4,01031749
asked Mar 19 '15 at 21:06
evinda
4,01031749
asked Mar 19 '15 at 21:06
evinda
4,01031749
4,01031749
The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
1
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27
|
show 2 more comments
The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
1
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27
The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
1
1
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27
|
show 2 more comments
3 Answers
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$$y''+xy=0$$
$$y=sum_{n=0}^{infty} c_n x^n, y'=sum_{n=1}^{infty} nc_n x^{n-1}, y''=sum_{n=2}^{infty} n(n-1)c_n x^{n-2}$$
$$therefore y''+xy=0=underbrace{sum_{n=2}^{infty} n(n-1)c_n x^{n-2}}_{k=n-2Rightarrow n=k+2}+underbrace{sum_{n=0}^{infty} c_n x^{n+1}}_{k=n+1Rightarrow n=k-1}=sum_{k=0}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k$$
$$=2c_2+sum_{k=1}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k=2c_2+sum_{k=1}^{infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0Rightarrow c_{k+2}=-frac{c_{k-1}}{(k+2)(k+1)}forall k=1,2,3,ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-frac{1}{6}x^3+frac{1}{180}x^6+ldots$$
$$y_2=x-frac{1}{12}x^4+frac{1}{504}x^7+dots$$
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
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Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -frac{a_0}{3cdot2}$$
$$a_6 = -frac{a_3}{6cdot5} = frac{a_0}{6cdot5cdot3cdot2} $$
This can be extrapolated to
$$ a_{n} = pmfrac{a_0}{n(n-1)(n-3)(n-4)cdotscdot6cdot5cdot3cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^kfrac{a_0}{(3k)!}prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^kfrac{a_1}{(3k+1)!}prod_{i=0}^{k-1} (3i+2)$$
edited yesterday
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answered Mar 20 '15 at 2:24
Dylan
11.6k31026
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
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Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know): 
The viewport is $,0 < x < 40,$ and $,-1.5 < y < +1.5,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):
{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx)
------------------------------ + x.y(x) = 0 ==>
dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0
y(1.dx) = y(0.dx) + v.dx
y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx)
y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx)
..............................................
y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx)
}
procedure bereken;
const
N : integer = 40000;
v : double = 1;
var
x,dx,y0,y1,y2 : double;
k : integer;
begin
x := 0; dx := 0.001;
y0 := 0; y1 := y0+v*dx;
Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0));
for k := 0 to N-1 do
begin
x := x + dx;
y2 := 2*y1 - y0 - dx*dx*x*y1;
Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2));
y0 := y1; y1 := y2;
end;
end;
Note that the solution becomes very oscillatory (i.e. singular) for $xtoinfty$ . However, the frequency only varies with the square root of the distance: $omega=sqrt{x}$ , therefore it doesn't happen immediately.
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
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3 Answers
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3 Answers
3
active
oldest
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active
oldest
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active
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up vote
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$$y''+xy=0$$
$$y=sum_{n=0}^{infty} c_n x^n, y'=sum_{n=1}^{infty} nc_n x^{n-1}, y''=sum_{n=2}^{infty} n(n-1)c_n x^{n-2}$$
$$therefore y''+xy=0=underbrace{sum_{n=2}^{infty} n(n-1)c_n x^{n-2}}_{k=n-2Rightarrow n=k+2}+underbrace{sum_{n=0}^{infty} c_n x^{n+1}}_{k=n+1Rightarrow n=k-1}=sum_{k=0}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k$$
$$=2c_2+sum_{k=1}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k=2c_2+sum_{k=1}^{infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0Rightarrow c_{k+2}=-frac{c_{k-1}}{(k+2)(k+1)}forall k=1,2,3,ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-frac{1}{6}x^3+frac{1}{180}x^6+ldots$$
$$y_2=x-frac{1}{12}x^4+frac{1}{504}x^7+dots$$
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
add a comment |
up vote
3
down vote
$$y''+xy=0$$
$$y=sum_{n=0}^{infty} c_n x^n, y'=sum_{n=1}^{infty} nc_n x^{n-1}, y''=sum_{n=2}^{infty} n(n-1)c_n x^{n-2}$$
$$therefore y''+xy=0=underbrace{sum_{n=2}^{infty} n(n-1)c_n x^{n-2}}_{k=n-2Rightarrow n=k+2}+underbrace{sum_{n=0}^{infty} c_n x^{n+1}}_{k=n+1Rightarrow n=k-1}=sum_{k=0}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k$$
$$=2c_2+sum_{k=1}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k=2c_2+sum_{k=1}^{infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0Rightarrow c_{k+2}=-frac{c_{k-1}}{(k+2)(k+1)}forall k=1,2,3,ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-frac{1}{6}x^3+frac{1}{180}x^6+ldots$$
$$y_2=x-frac{1}{12}x^4+frac{1}{504}x^7+dots$$
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
add a comment |
up vote
3
down vote
up vote
3
down vote
$$y''+xy=0$$
$$y=sum_{n=0}^{infty} c_n x^n, y'=sum_{n=1}^{infty} nc_n x^{n-1}, y''=sum_{n=2}^{infty} n(n-1)c_n x^{n-2}$$
$$therefore y''+xy=0=underbrace{sum_{n=2}^{infty} n(n-1)c_n x^{n-2}}_{k=n-2Rightarrow n=k+2}+underbrace{sum_{n=0}^{infty} c_n x^{n+1}}_{k=n+1Rightarrow n=k-1}=sum_{k=0}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k$$
$$=2c_2+sum_{k=1}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k=2c_2+sum_{k=1}^{infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0Rightarrow c_{k+2}=-frac{c_{k-1}}{(k+2)(k+1)}forall k=1,2,3,ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-frac{1}{6}x^3+frac{1}{180}x^6+ldots$$
$$y_2=x-frac{1}{12}x^4+frac{1}{504}x^7+dots$$
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
$$y''+xy=0$$
$$y=sum_{n=0}^{infty} c_n x^n, y'=sum_{n=1}^{infty} nc_n x^{n-1}, y''=sum_{n=2}^{infty} n(n-1)c_n x^{n-2}$$
$$therefore y''+xy=0=underbrace{sum_{n=2}^{infty} n(n-1)c_n x^{n-2}}_{k=n-2Rightarrow n=k+2}+underbrace{sum_{n=0}^{infty} c_n x^{n+1}}_{k=n+1Rightarrow n=k-1}=sum_{k=0}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k$$
$$=2c_2+sum_{k=1}^{infty} (k+2)(k+1)c_{k+2} x^{k}+sum_{k=1}^{infty} c_{k-1} x^k=2c_2+sum_{k=1}^{infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0Rightarrow c_{k+2}=-frac{c_{k-1}}{(k+2)(k+1)}forall k=1,2,3,ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-frac{1}{6}x^3+frac{1}{180}x^6+ldots$$
$$y_2=x-frac{1}{12}x^4+frac{1}{504}x^7+dots$$
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
answered May 29 '15 at 21:11
Yagna Patel
6,39912345
6,39912345
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
add a comment |
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
@evinda: Does this solve your problem?
– Yagna Patel
May 31 '15 at 16:07
add a comment |
up vote
3
down vote
Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -frac{a_0}{3cdot2}$$
$$a_6 = -frac{a_3}{6cdot5} = frac{a_0}{6cdot5cdot3cdot2} $$
This can be extrapolated to
$$ a_{n} = pmfrac{a_0}{n(n-1)(n-3)(n-4)cdotscdot6cdot5cdot3cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^kfrac{a_0}{(3k)!}prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^kfrac{a_1}{(3k+1)!}prod_{i=0}^{k-1} (3i+2)$$
edited yesterday
Community♦
1
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
|
show 1 more comment
up vote
3
down vote
Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -frac{a_0}{3cdot2}$$
$$a_6 = -frac{a_3}{6cdot5} = frac{a_0}{6cdot5cdot3cdot2} $$
This can be extrapolated to
$$ a_{n} = pmfrac{a_0}{n(n-1)(n-3)(n-4)cdotscdot6cdot5cdot3cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^kfrac{a_0}{(3k)!}prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^kfrac{a_1}{(3k+1)!}prod_{i=0}^{k-1} (3i+2)$$
edited yesterday
Community♦
1
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
|
show 1 more comment
up vote
3
down vote
up vote
3
down vote
Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -frac{a_0}{3cdot2}$$
$$a_6 = -frac{a_3}{6cdot5} = frac{a_0}{6cdot5cdot3cdot2} $$
This can be extrapolated to
$$ a_{n} = pmfrac{a_0}{n(n-1)(n-3)(n-4)cdotscdot6cdot5cdot3cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^kfrac{a_0}{(3k)!}prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^kfrac{a_1}{(3k+1)!}prod_{i=0}^{k-1} (3i+2)$$
edited yesterday
Community♦
1
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
Follow-up from my comment, every $a_{n+3}$ can be expressed in terms of $a_n$.
$$a_{n+3} = -frac{a_{n}}{(n+3)(n+2)}$$
Since $a_2 = 0$, all coefficients with $n = 3k + 2$ will be $0$
The rest will have to based on the initial conditions of $a_0 = y(0)$ and $a_1 = y'(0)$
For $n = 3k$
$$a_3 = -frac{a_0}{3cdot2}$$
$$a_6 = -frac{a_3}{6cdot5} = frac{a_0}{6cdot5cdot3cdot2} $$
This can be extrapolated to
$$ a_{n} = pmfrac{a_0}{n(n-1)(n-3)(n-4)cdotscdot6cdot5cdot3cdot2} $$
Basically, the denominator is a product of all integers from $1$ to $n$, skipping every 3rd number and alternating signs. An alternate notation is
$$a_{3k} = (-1)^kfrac{a_0}{(3k)!}prod_{i=0}^{k-1} (3i+1) $$
The same can be done for $n = 3k+1$
$$a_{3k+1} = (-1)^kfrac{a_1}{(3k+1)!}prod_{i=0}^{k-1} (3i+2)$$
edited yesterday
Community♦
1
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
edited yesterday
Community♦
1
edited yesterday
Community♦
1
edited yesterday
Community♦
1
1
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
answered Mar 20 '15 at 2:24
Dylan
11.6k31026
11.6k31026
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
|
show 1 more comment
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
Isn't it as follows, Dylan? $$$$ $$a_{n+3}=-frac{a_n}{(n+3)(n+1)}$$ $$a_3=-frac{a_0}{3 cdot 1}$$ $$a_6=frac{a_0}{3 cdot 4 cdot 6}$$ $$a_9=frac{-a_0}{3 cdot 4 cdot 6 cdot 7 cdot 9}$$ So: $$a_{3k}=(-1)^k frac{a_0}{n(n-2)(n-3) cdots 3}$$ $$a_4=-frac{a_1}{2 cdot 4}$$ $$a_7=frac{a_1}{2 cdot 4 cdot 5 cdot 7}$$ $$a_{3k+1}=(-1)^k frac{a_1}{n(n-2)(n-3) cdots 4 cdot 2}$$ Don't we also have to check the case $n=3k+2$? $$$$ Also what can we deduce for the radius of convergence?
– evinda
Mar 22 '15 at 23:06
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I fixed the calculation. Also, as I said above, $a_{3k+2} = 0$ since $a_2 = 0$
– Dylan
Mar 23 '15 at 23:49
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
I think that it should be as follows: $$$$ For $n=3k$: $$a_{n}=(-1)^k frac{a_0}{(3k)!} prod_{i=0}^{k-1} (3i+1)$$ and for $n=3k+1$: $$a_n=(-1)^k frac{a_1}{(3k+1)!} prod_{i=0}^{k-1} (3i+2)$$ Or am I wrong?
– evinda
Mar 24 '15 at 22:39
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Also taking the limit of the ratios $frac{a_{3(k+1)}}{a_{3k}}$ and $frac{a_{3(k+1)+1}}{a_{3k+1}}$, we will get $0$ and so the radii of convergence of $sum_k a_{3k}x^{3k}$ and $sum_k a_{3k+1} x^{3k+1}$ will be $+infty$, so the series converge . $$$$ So the solution of the differential equation will be: $$sum_k a_{3k} x^{3k}+ sum_k a_{3k+1} x^{3k+1}$$ where $a_{3k}, a_{3k+1}$ have the above formulas and the radius of convergence of the solution will be $+infty$. Right?
– evinda
Mar 24 '15 at 22:56
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
Looks right to me :)
– Dylan
Mar 25 '15 at 23:18
|
show 1 more comment
up vote
1
down vote
Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know):
The viewport is $,0 < x < 40,$ and $,-1.5 < y < +1.5,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):
{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx)
------------------------------ + x.y(x) = 0 ==>
dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0
y(1.dx) = y(0.dx) + v.dx
y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx)
y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx)
..............................................
y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx)
}
procedure bereken;
const
N : integer = 40000;
v : double = 1;
var
x,dx,y0,y1,y2 : double;
k : integer;
begin
x := 0; dx := 0.001;
y0 := 0; y1 := y0+v*dx;
Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0));
for k := 0 to N-1 do
begin
x := x + dx;
y2 := 2*y1 - y0 - dx*dx*x*y1;
Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2));
y0 := y1; y1 := y2;
end;
end;
Note that the solution becomes very oscillatory (i.e. singular) for $xtoinfty$ . However, the frequency only varies with the square root of the distance: $omega=sqrt{x}$ , therefore it doesn't happen immediately.
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
add a comment |
up vote
1
down vote
Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know):
The viewport is $,0 < x < 40,$ and $,-1.5 < y < +1.5,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):
{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx)
------------------------------ + x.y(x) = 0 ==>
dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0
y(1.dx) = y(0.dx) + v.dx
y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx)
y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx)
..............................................
y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx)
}
procedure bereken;
const
N : integer = 40000;
v : double = 1;
var
x,dx,y0,y1,y2 : double;
k : integer;
begin
x := 0; dx := 0.001;
y0 := 0; y1 := y0+v*dx;
Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0));
for k := 0 to N-1 do
begin
x := x + dx;
y2 := 2*y1 - y0 - dx*dx*x*y1;
Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2));
y0 := y1; y1 := y2;
end;
end;
Note that the solution becomes very oscillatory (i.e. singular) for $xtoinfty$ . However, the frequency only varies with the square root of the distance: $omega=sqrt{x}$ , therefore it doesn't happen immediately.
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
add a comment |
up vote
1
down vote
up vote
1
down vote
Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know):
The viewport is $,0 < x < 40,$ and $,-1.5 < y < +1.5,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):
{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx)
------------------------------ + x.y(x) = 0 ==>
dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0
y(1.dx) = y(0.dx) + v.dx
y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx)
y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx)
..............................................
y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx)
}
procedure bereken;
const
N : integer = 40000;
v : double = 1;
var
x,dx,y0,y1,y2 : double;
k : integer;
begin
x := 0; dx := 0.001;
y0 := 0; y1 := y0+v*dx;
Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0));
for k := 0 to N-1 do
begin
x := x + dx;
y2 := 2*y1 - y0 - dx*dx*x*y1;
Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2));
y0 := y1; y1 := y2;
end;
end;
Note that the solution becomes very oscillatory (i.e. singular) for $xtoinfty$ . However, the frequency only varies with the square root of the distance: $omega=sqrt{x}$ , therefore it doesn't happen immediately.
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
Disclaimer. This is not an answer, but rather a too long comment with some graphics in it.
The equation looks like the common ODE for a harmonic oscillator: $y'' + omega^2 y = 0$ with the square of the frequency varying proportional to "time" $x$ . Numerical simulation with the initial conditions $y(0)=0$ and $y'(0)=1$ reveals that the solution indeed looks like that (I hate solutions without a picture, you know):
The viewport is $,0 < x < 40,$ and $,-1.5 < y < +1.5,$.
Program (Delphi Pascal) snippet for doing the calculations and the drawing (not optimized at all):
{
The equations of motion are solved numerically as follows.
Start with: y(0) = 0 ; x = 0 ;
(y(dx) - y(0))/dx = v ==> y(dx) = y(0) + v.dx
y(x + dx) - 2.y(x) + y(x - dx)
------------------------------ + x.y(x) = 0 ==>
dx^2
y(x + dx) = 2.y(x) - y(x - dx) - dx^2.x.y(x) ==>
y(0.dx) = 0
y(1.dx) = y(0.dx) + v.dx
y(2.dx) = 2.y(1.dx) - y(0.dx) - dx^2.x.y(1.dx)
y(3.dx) = 2.y(2.dx) - y(1.dx) - dx^2.x.y(2.dx)
..............................................
y(k+1).dx) = 2.y(k.dx) - y((k-1).dx) - dx^2.x.y(k.dx)
}
procedure bereken;
const
N : integer = 40000;
v : double = 1;
var
x,dx,y0,y1,y2 : double;
k : integer;
begin
x := 0; dx := 0.001;
y0 := 0; y1 := y0+v*dx;
Form1.Image1.Canvas.MoveTo(x2i(0),y2j(0));
for k := 0 to N-1 do
begin
x := x + dx;
y2 := 2*y1 - y0 - dx*dx*x*y1;
Form1.Image1.Canvas.LineTo(x2i(x),y2j(y2));
y0 := y1; y1 := y2;
end;
end;
Note that the solution becomes very oscillatory (i.e. singular) for $xtoinfty$ . However, the frequency only varies with the square root of the distance: $omega=sqrt{x}$ , therefore it doesn't happen immediately.
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
edited Jun 4 '15 at 20:19
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
answered Jun 4 '15 at 11:03
Han de Bruijn
12.1k22361
12.1k22361
add a comment |
add a comment |
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The coefficients aren't too nice. Check out the Wolfram Alpha result here and go to the documentation on the Airy functions here. It'll help you figure out what the general form of the coefficients is. Particularly, see equations $(9)$ and $(10)$ on the Wolfram MathWorld page.
– Cameron Williams
Mar 19 '15 at 21:09
You can find them through recursion by rewriting it as $$(n+2)(n+1)a_{n+3} + a_{n} = 0 $$ $$ a_{n+3} = -frac{a_{n}}{(n+2)(n+1)}$$ Since $a_2 = 0$, you have $a_{2+3k} = 0$ for all integer $k$
– Dylan
Mar 19 '15 at 21:20
@Dylan Shouldn't it be $(n+3)(n+2) a_{n+3}+a_n=0$ ?
– evinda
Mar 19 '15 at 22:43
@evinda You're right, I screwed up. But you get my point, right?
– Dylan
Mar 20 '15 at 2:09
1
What are the initial conditions for $y(0)$ and $y'(0)$ ?
– Han de Bruijn
Jun 4 '15 at 9:27