Convergence of the Power Series for $xu''+sin(x)u=0$
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Consider the initial value problem
$$xu''+sin(x)u=0 u(0)=0, u'(0)=2$$
What can be said about the radius of convergence of this series?
I have determined that the first four nonzero terms (about the point $x=0$) of the power series are $$2x,-x^2,frac16x^3,-frac172x^4$$
How can I test for convergence? Intuitively it looks not to converge, but how can I show this?
edit
The solution will have the from $$u(x)=sum_k=0^inftyA_kx^k.$$
Differentiating,
beginalign
sum_k=2^infty k(k-1)A_kx^k-1+sin(x)sum_k=0^inftyA_kx^k&=0 \
sum_k=2^infty k(k-1)A_kx^k-1+left(sum_k=0^infty frac(-1)^k x^2k+1(2k+1)!right)sum_k=0^inftyA_kx^k&=0 \
(2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+left(2x^2+A_2x^3+left(A_3-frac13right)x^4+left(A_4-fracA_26!right)x^5+left(A_5-fracA_33!+frac25!right)x^6+..right)=0
endalign
Equating coefficients, $x^2k=0$ for $kinmathbbZ^+$.
$$6A_3+2=0implies A_3=-frac13$$
$$20A_5+A_3-frac13=0implies A_5=frac130$$
$$42A_7+A_5-fracA_33!+frac25!=0implies A_7=-frac197560$$
Hence $$u=2x-frac13x^3+frac130x^5-frac197560x^7+..$$
differential-equations convergence power-series
asked Aug 19 at 1:56
Bell
741313
 |Â
show 8 more comments
up vote
2
down vote
favorite
Consider the initial value problem
$$xu''+sin(x)u=0 u(0)=0, u'(0)=2$$
What can be said about the radius of convergence of this series?
I have determined that the first four nonzero terms (about the point $x=0$) of the power series are $$2x,-x^2,frac16x^3,-frac172x^4$$
How can I test for convergence? Intuitively it looks not to converge, but how can I show this?
edit
The solution will have the from $$u(x)=sum_k=0^inftyA_kx^k.$$
Differentiating,
beginalign
sum_k=2^infty k(k-1)A_kx^k-1+sin(x)sum_k=0^inftyA_kx^k&=0 \
sum_k=2^infty k(k-1)A_kx^k-1+left(sum_k=0^infty frac(-1)^k x^2k+1(2k+1)!right)sum_k=0^inftyA_kx^k&=0 \
(2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+left(2x^2+A_2x^3+left(A_3-frac13right)x^4+left(A_4-fracA_26!right)x^5+left(A_5-fracA_33!+frac25!right)x^6+..right)=0
endalign
Equating coefficients, $x^2k=0$ for $kinmathbbZ^+$.
$$6A_3+2=0implies A_3=-frac13$$
$$20A_5+A_3-frac13=0implies A_5=frac130$$
$$42A_7+A_5-fracA_33!+frac25!=0implies A_7=-frac197560$$
Hence $$u=2x-frac13x^3+frac130x^5-frac197560x^7+..$$
differential-equations convergence power-series
asked Aug 19 at 1:56
Bell
741313
I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
1
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09
 |Â
show 8 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the initial value problem
$$xu''+sin(x)u=0 u(0)=0, u'(0)=2$$
What can be said about the radius of convergence of this series?
I have determined that the first four nonzero terms (about the point $x=0$) of the power series are $$2x,-x^2,frac16x^3,-frac172x^4$$
How can I test for convergence? Intuitively it looks not to converge, but how can I show this?
edit
The solution will have the from $$u(x)=sum_k=0^inftyA_kx^k.$$
Differentiating,
beginalign
sum_k=2^infty k(k-1)A_kx^k-1+sin(x)sum_k=0^inftyA_kx^k&=0 \
sum_k=2^infty k(k-1)A_kx^k-1+left(sum_k=0^infty frac(-1)^k x^2k+1(2k+1)!right)sum_k=0^inftyA_kx^k&=0 \
(2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+left(2x^2+A_2x^3+left(A_3-frac13right)x^4+left(A_4-fracA_26!right)x^5+left(A_5-fracA_33!+frac25!right)x^6+..right)=0
endalign
Equating coefficients, $x^2k=0$ for $kinmathbbZ^+$.
$$6A_3+2=0implies A_3=-frac13$$
$$20A_5+A_3-frac13=0implies A_5=frac130$$
$$42A_7+A_5-fracA_33!+frac25!=0implies A_7=-frac197560$$
Hence $$u=2x-frac13x^3+frac130x^5-frac197560x^7+..$$
differential-equations convergence power-series
asked Aug 19 at 1:56
Bell
741313
Consider the initial value problem
$$xu''+sin(x)u=0 u(0)=0, u'(0)=2$$
What can be said about the radius of convergence of this series?
I have determined that the first four nonzero terms (about the point $x=0$) of the power series are $$2x,-x^2,frac16x^3,-frac172x^4$$
How can I test for convergence? Intuitively it looks not to converge, but how can I show this?
edit
The solution will have the from $$u(x)=sum_k=0^inftyA_kx^k.$$
Differentiating,
beginalign
sum_k=2^infty k(k-1)A_kx^k-1+sin(x)sum_k=0^inftyA_kx^k&=0 \
sum_k=2^infty k(k-1)A_kx^k-1+left(sum_k=0^infty frac(-1)^k x^2k+1(2k+1)!right)sum_k=0^inftyA_kx^k&=0 \
(2A_2x+6A_3x^2+12A_4x^3+20A_5x^4+30A_6x^5+42A_7x^6+..)+left(2x^2+A_2x^3+left(A_3-frac13right)x^4+left(A_4-fracA_26!right)x^5+left(A_5-fracA_33!+frac25!right)x^6+..right)=0
endalign
Equating coefficients, $x^2k=0$ for $kinmathbbZ^+$.
$$6A_3+2=0implies A_3=-frac13$$
$$20A_5+A_3-frac13=0implies A_5=frac130$$
$$42A_7+A_5-fracA_33!+frac25!=0implies A_7=-frac197560$$
Hence $$u=2x-frac13x^3+frac130x^5-frac197560x^7+..$$
differential-equations convergence power-series
asked Aug 19 at 1:56
Bell
741313
edited Aug 19 at 12:30
asked Aug 19 at 1:56
Bell
741313
asked Aug 19 at 1:56
Bell
741313
asked Aug 19 at 1:56
Bell
741313
741313
I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
1
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09
 |Â
show 8 more comments
I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
1
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09
I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
1
1
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09
 |Â
show 8 more comments
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I have include how I obtained the series. Finding this series was related to my last post. Not sure where I have gone wrong. I took the advice from the last post and thought I had produced the correct solution.
– Bell
Aug 19 at 2:47
If $u=2x+A_2 x^2+A_3 x^3+A_4x^4+dots,$ then $sin(x)u =2x^2+ A_2x^3+(A_3-1/3)x^4+(A_4-A_2/6)x^5+dots.$
– Somos
Aug 19 at 3:14
I'm a bit confused. I was just following the solution provided on this post math.stackexchange.com/a/2886714/557493
– Bell
Aug 19 at 3:25
I think I have found this mistake in that post btw. The answer provided to the OP (me) mistypes the power series expansion for $sin(x)$.
– Bell
Aug 19 at 3:27
1
The problem is with solving for $A_5 = 1/30$. You have $-1/30$ which is a sign error. The next $A_7=-19/7560$ is correct. I have a typo in my 1st comment.
– Somos
Aug 19 at 11:09