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Finding the fundamental set of solutions and expressing the functions in terms of power series
Finding coefficients of a differential equation represented by power seriesFind the power series for $d/dx(arcsin x)$Finding a recurrence relation, first few terms of power series solution to differential equationThe number of solutions for infinite power series equations $sum_k=0^inftyA_k z^k=0$?Finding power series and ROC of complex functionDifficulty in finding interval of convergence with power seriesFinding linearly independent power series solutions of a differential equationHow can I express the natural logarithm of a power series in terms of another power series?Power Series and RecursionFinding the first non-zero terms of a power series
$begingroup$
Find the fundamental set of solutions to the equation
$$xy'' + y' + y = 0$$
Express the functions in terms of power series.
I have found the general solution to be
$$a_n+1 = frac-a_n (n + 1)^2$$
However, I am having trouble finding the fundamental set of solutions.
Thank you.
calculus power-series
edited Mar 16 at 21:18
NoChance
3,76721221
asked Mar 16 at 21:13
benrobenro
1
$endgroup$
add a comment |
$begingroup$
Find the fundamental set of solutions to the equation
$$xy'' + y' + y = 0$$
Express the functions in terms of power series.
I have found the general solution to be
$$a_n+1 = frac-a_n (n + 1)^2$$
However, I am having trouble finding the fundamental set of solutions.
Thank you.
calculus power-series
edited Mar 16 at 21:18
NoChance
3,76721221
asked Mar 16 at 21:13
benrobenro
1
$endgroup$
$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
1
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27
add a comment |
$begingroup$
Find the fundamental set of solutions to the equation
$$xy'' + y' + y = 0$$
Express the functions in terms of power series.
I have found the general solution to be
$$a_n+1 = frac-a_n (n + 1)^2$$
However, I am having trouble finding the fundamental set of solutions.
Thank you.
calculus power-series
edited Mar 16 at 21:18
NoChance
3,76721221
asked Mar 16 at 21:13
benrobenro
1
$endgroup$
Find the fundamental set of solutions to the equation
$$xy'' + y' + y = 0$$
Express the functions in terms of power series.
I have found the general solution to be
$$a_n+1 = frac-a_n (n + 1)^2$$
However, I am having trouble finding the fundamental set of solutions.
Thank you.
calculus power-series
calculus power-series
edited Mar 16 at 21:18
NoChance
3,76721221
asked Mar 16 at 21:13
benrobenro
1
edited Mar 16 at 21:18
NoChance
3,76721221
asked Mar 16 at 21:13
benrobenro
1
edited Mar 16 at 21:18
NoChance
3,76721221
edited Mar 16 at 21:18
NoChance
3,76721221
edited Mar 16 at 21:18
NoChance
3,76721221
3,76721221
asked Mar 16 at 21:13
benrobenro
1
asked Mar 16 at 21:13
benrobenro
1
asked Mar 16 at 21:13
benrobenro
1
1
$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
1
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27
add a comment |
$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
1
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27
$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
1
1
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$ fraca_na_0=prod_k=0^n-1fraca_k+1a_k=(-1)^nprod_k=0^n-1frac1(k+1)^2=(-1)^nfrac1(n!)^2$$
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
$endgroup$
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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$begingroup$
$$ fraca_na_0=prod_k=0^n-1fraca_k+1a_k=(-1)^nprod_k=0^n-1frac1(k+1)^2=(-1)^nfrac1(n!)^2$$
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
$endgroup$
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
add a comment |
$begingroup$
$$ fraca_na_0=prod_k=0^n-1fraca_k+1a_k=(-1)^nprod_k=0^n-1frac1(k+1)^2=(-1)^nfrac1(n!)^2$$
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
$endgroup$
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
add a comment |
$begingroup$
$$ fraca_na_0=prod_k=0^n-1fraca_k+1a_k=(-1)^nprod_k=0^n-1frac1(k+1)^2=(-1)^nfrac1(n!)^2$$
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
$endgroup$
$$ fraca_na_0=prod_k=0^n-1fraca_k+1a_k=(-1)^nprod_k=0^n-1frac1(k+1)^2=(-1)^nfrac1(n!)^2$$
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
edited Mar 16 at 22:07
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
answered Mar 16 at 21:28
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,478211
1,478211
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
add a comment |
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
I think the minus one factor has to be retained inside the product, or else pulled out as a power of minus one; please check. In any case the Question itself is not explicit in identifying terms $a_n$ as the power series coefficients, so it would be helpful to include that identification in your Answer. Finally a "fundamental set of solutions" should (for this singular second order homogeneous ODE) consist of a linearly independent pair of solutions.
$endgroup$
– hardmath
Mar 16 at 22:04
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
Yes you’re right
$endgroup$
– HAMIDINE SOUMARE
Mar 16 at 22:07
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
$begingroup$
How does this answer satisfy the differential equation at the top?
$endgroup$
– NoChance
Mar 17 at 5:25
add a comment |
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$begingroup$
Please ensure that the LaTex edit is correct.
$endgroup$
– NoChance
Mar 16 at 21:18
1
$begingroup$
You have a regular singular point at $x=0$ so you should use the Method of Frobenius.
$endgroup$
– John Wayland Bales
Mar 16 at 21:27