Infinite series manipulation

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I have the series

$$T(x) = sum_n=0^infty t_nx^n$$

Is there anyway to express $sum_n=0^infty nt_nx^n$ in terms of $T(x)$? I tried differentiating $T(x)$ but only got so far as to express the latter as $xT'(x)$.

If that desired form of $T(x)$ is not possible, then is there a way to solve the following for $T(x)$?
$$x^2T'(x) +(3x-1)T(x) +1 = 0$$
My goal is to obtain an algebraic expression for $T(x)$, preferably one that could be then manipulated to obtain an infinite power series.

sequences-and-series differential-equations power-series generating-functions

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edited Jul 22 at 20:00

asked Jul 22 at 19:52

ensbana

279113

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up vote
1
down vote

favorite

I have the series

$$T(x) = sum_n=0^infty t_nx^n$$

Is there anyway to express $sum_n=0^infty nt_nx^n$ in terms of $T(x)$? I tried differentiating $T(x)$ but only got so far as to express the latter as $xT'(x)$.

If that desired form of $T(x)$ is not possible, then is there a way to solve the following for $T(x)$?
$$x^2T'(x) +(3x-1)T(x) +1 = 0$$
My goal is to obtain an algebraic expression for $T(x)$, preferably one that could be then manipulated to obtain an infinite power series.

sequences-and-series differential-equations power-series generating-functions

share|cite|improve this question

edited Jul 22 at 20:00

asked Jul 22 at 19:52

ensbana

279113

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have the series

$$T(x) = sum_n=0^infty t_nx^n$$

Is there anyway to express $sum_n=0^infty nt_nx^n$ in terms of $T(x)$? I tried differentiating $T(x)$ but only got so far as to express the latter as $xT'(x)$.

If that desired form of $T(x)$ is not possible, then is there a way to solve the following for $T(x)$?
$$x^2T'(x) +(3x-1)T(x) +1 = 0$$
My goal is to obtain an algebraic expression for $T(x)$, preferably one that could be then manipulated to obtain an infinite power series.

sequences-and-series differential-equations power-series generating-functions

share|cite|improve this question

edited Jul 22 at 20:00

asked Jul 22 at 19:52

ensbana

279113

I have the series

$$T(x) = sum_n=0^infty t_nx^n$$

Is there anyway to express $sum_n=0^infty nt_nx^n$ in terms of $T(x)$? I tried differentiating $T(x)$ but only got so far as to express the latter as $xT'(x)$.

If that desired form of $T(x)$ is not possible, then is there a way to solve the following for $T(x)$?
$$x^2T'(x) +(3x-1)T(x) +1 = 0$$
My goal is to obtain an algebraic expression for $T(x)$, preferably one that could be then manipulated to obtain an infinite power series.

sequences-and-series differential-equations power-series generating-functions

share|cite|improve this question

edited Jul 22 at 20:00

asked Jul 22 at 19:52

ensbana

279113

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 20:00

edited Jul 22 at 20:00

edited Jul 22 at 20:00

asked Jul 22 at 19:52

ensbana

279113

asked Jul 22 at 19:52

ensbana

279113

asked Jul 22 at 19:52

ensbana

279113

279113

add a comment | 

add a comment | 

1 Answer
1

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oldest

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up vote
1
down vote



accepted

You are correct: $$sum_n=0^infty n t_n x^n = x T'(x)$$.

$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$
is a first-order linear differential equation. Its general solution is
$$
c fracexp(-1/x)x^3 - fracexp(-1/x) Gamma(0,-1/x)2 x^3 - frac12x - frac12x^3
$$
where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.

EDIT: In terms of the $t_n$, your differential equation says
$t_0 = 1$ and $t_n = (n+2) t_n-1$ for $n ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series
$sum_n=0^infty (n+2)!; x^n/2$ diverges for all $x ne 0$.

share|cite|improve this answer

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You are correct: $$sum_n=0^infty n t_n x^n = x T'(x)$$.

$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$
is a first-order linear differential equation. Its general solution is
$$
c fracexp(-1/x)x^3 - fracexp(-1/x) Gamma(0,-1/x)2 x^3 - frac12x - frac12x^3
$$
where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.

EDIT: In terms of the $t_n$, your differential equation says
$t_0 = 1$ and $t_n = (n+2) t_n-1$ for $n ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series
$sum_n=0^infty (n+2)!; x^n/2$ diverges for all $x ne 0$.

share|cite|improve this answer

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

add a comment | 

up vote
1
down vote



accepted

You are correct: $$sum_n=0^infty n t_n x^n = x T'(x)$$.

$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$
is a first-order linear differential equation. Its general solution is
$$
c fracexp(-1/x)x^3 - fracexp(-1/x) Gamma(0,-1/x)2 x^3 - frac12x - frac12x^3
$$
where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.

EDIT: In terms of the $t_n$, your differential equation says
$t_0 = 1$ and $t_n = (n+2) t_n-1$ for $n ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series
$sum_n=0^infty (n+2)!; x^n/2$ diverges for all $x ne 0$.

share|cite|improve this answer

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You are correct: $$sum_n=0^infty n t_n x^n = x T'(x)$$.

$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$
is a first-order linear differential equation. Its general solution is
$$
c fracexp(-1/x)x^3 - fracexp(-1/x) Gamma(0,-1/x)2 x^3 - frac12x - frac12x^3
$$
where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.

EDIT: In terms of the $t_n$, your differential equation says
$t_0 = 1$ and $t_n = (n+2) t_n-1$ for $n ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series
$sum_n=0^infty (n+2)!; x^n/2$ diverges for all $x ne 0$.

share|cite|improve this answer

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

You are correct: $$sum_n=0^infty n t_n x^n = x T'(x)$$.

$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$
is a first-order linear differential equation. Its general solution is
$$
c fracexp(-1/x)x^3 - fracexp(-1/x) Gamma(0,-1/x)2 x^3 - frac12x - frac12x^3
$$
where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.

EDIT: In terms of the $t_n$, your differential equation says
$t_0 = 1$ and $t_n = (n+2) t_n-1$ for $n ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series
$sum_n=0^infty (n+2)!; x^n/2$ diverges for all $x ne 0$.

share|cite|improve this answer

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 20:19

edited Jul 22 at 20:19

edited Jul 22 at 20:19

answered Jul 22 at 20:03

Robert Israel

304k22201441

answered Jul 22 at 20:03

Robert Israel

304k22201441

answered Jul 22 at 20:03

Robert Israel

304k22201441

304k22201441

add a comment | 

add a comment | 

 

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