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Does anyone know of a good expression for this Maclaurin series?
Clash Royale CLAN TAG#URR8PPP
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I'm looking for a usable (something I can program) closed form or recursive formula for the coefficients $a_n$ of this series for an even-symmetry function of a real variable:
$$ sumlimits_n=0^infty a_n x^2n = fracxarctan(x) $$
for $big|xbig| le 1$ .
Ultimately, what I want is an expression like
$$ arctan(x) = fracxsumlimits_n=0^infty a_n x^2n $$
again, for $big|xbig| le 1$ .
I've approached this both from the POV of Taylor/Maclaurin series and tried to get a grip of using the Lagrange inversion theorem, but am not getting to an elegant place.
$$beginalign
sumlimits_n=0^infty a_n x^2n &= fracxarctan(x) \
\
&= fracxsumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1 \
\
&= frac1sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k \
endalign$$
or
$$ left( sumlimits_n=0^infty a_n u^n right) left(sumlimits_k=0^inftyfrac(-1)^k2k+1 u^k right) = 1 $$
for $0 le u le 1$ .
I get this:
$$ sumlimits_n=0^infty a_n sumlimits_k=0^infty frac(-1)^k2k+1 u^k+n = 1 $$
and
$$ sumlimits_n=0^infty (-1)^n a_n sumlimits_k=n^infty frac(-1)^k2k-2n+1 u^k = 1 $$
I can see right away that $a_0 = 1$, but I should be able to yank out a recursion formula out of this. But I forgot how to do that. All of the net coefficients of the higher powers of $u^k$ should be $0$ for $k ge 1$.
Denouement:
It's simpler than I thought.
So we have
$$ sumlimits_n=0^infty sumlimits_k=0^infty a_n frac(-1)^k2k+1 u^k+n = 1 $$
Let's let the exponent of $u$ be fixed to some integer $m ge 0$. So $n+k=m$ and, for a given $n$, the only term in the inner summation that counts for the coefficient of $u^m$ is $k=m-n$.
We know for $m=0$ the summation is $1$.
$$ a_n frac(-1)^k2k+1 u^k+n Big|_n=k=0 = 1 $$
which results in $a_0=1$. For all other $mge 1$, the summation is
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 u^m = 0 qquad forall u: 0 le u le 1$$
or
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 = 0$$
Spinning off the last term for
$$ a_m + sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 = 0 $$
or
$$ a_m = -sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 $$
I confess that I didn't read the reference, I just needed to look at this more clearly. I guess with the additional substitution of $k=m-n$ we get:
$$ a_m = -sumlimits_k=1^m a_m-k frac(-1)^k2k+1 $$
.
power-series taylor-expansion lagrange-inversion
asked Jul 17 at 23:15
robert bristow-johnson
191116
add a comment |Â
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0
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I'm looking for a usable (something I can program) closed form or recursive formula for the coefficients $a_n$ of this series for an even-symmetry function of a real variable:
$$ sumlimits_n=0^infty a_n x^2n = fracxarctan(x) $$
for $big|xbig| le 1$ .
Ultimately, what I want is an expression like
$$ arctan(x) = fracxsumlimits_n=0^infty a_n x^2n $$
again, for $big|xbig| le 1$ .
I've approached this both from the POV of Taylor/Maclaurin series and tried to get a grip of using the Lagrange inversion theorem, but am not getting to an elegant place.
$$beginalign
sumlimits_n=0^infty a_n x^2n &= fracxarctan(x) \
\
&= fracxsumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1 \
\
&= frac1sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k \
endalign$$
or
$$ left( sumlimits_n=0^infty a_n u^n right) left(sumlimits_k=0^inftyfrac(-1)^k2k+1 u^k right) = 1 $$
for $0 le u le 1$ .
I get this:
$$ sumlimits_n=0^infty a_n sumlimits_k=0^infty frac(-1)^k2k+1 u^k+n = 1 $$
and
$$ sumlimits_n=0^infty (-1)^n a_n sumlimits_k=n^infty frac(-1)^k2k-2n+1 u^k = 1 $$
I can see right away that $a_0 = 1$, but I should be able to yank out a recursion formula out of this. But I forgot how to do that. All of the net coefficients of the higher powers of $u^k$ should be $0$ for $k ge 1$.
Denouement:
It's simpler than I thought.
So we have
$$ sumlimits_n=0^infty sumlimits_k=0^infty a_n frac(-1)^k2k+1 u^k+n = 1 $$
Let's let the exponent of $u$ be fixed to some integer $m ge 0$. So $n+k=m$ and, for a given $n$, the only term in the inner summation that counts for the coefficient of $u^m$ is $k=m-n$.
We know for $m=0$ the summation is $1$.
$$ a_n frac(-1)^k2k+1 u^k+n Big|_n=k=0 = 1 $$
which results in $a_0=1$. For all other $mge 1$, the summation is
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 u^m = 0 qquad forall u: 0 le u le 1$$
or
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 = 0$$
Spinning off the last term for
$$ a_m + sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 = 0 $$
or
$$ a_m = -sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 $$
I confess that I didn't read the reference, I just needed to look at this more clearly. I guess with the additional substitution of $k=m-n$ we get:
$$ a_m = -sumlimits_k=1^m a_m-k frac(-1)^k2k+1 $$
.
power-series taylor-expansion lagrange-inversion
asked Jul 17 at 23:15
robert bristow-johnson
191116
You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm looking for a usable (something I can program) closed form or recursive formula for the coefficients $a_n$ of this series for an even-symmetry function of a real variable:
$$ sumlimits_n=0^infty a_n x^2n = fracxarctan(x) $$
for $big|xbig| le 1$ .
Ultimately, what I want is an expression like
$$ arctan(x) = fracxsumlimits_n=0^infty a_n x^2n $$
again, for $big|xbig| le 1$ .
I've approached this both from the POV of Taylor/Maclaurin series and tried to get a grip of using the Lagrange inversion theorem, but am not getting to an elegant place.
$$beginalign
sumlimits_n=0^infty a_n x^2n &= fracxarctan(x) \
\
&= fracxsumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1 \
\
&= frac1sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k \
endalign$$
or
$$ left( sumlimits_n=0^infty a_n u^n right) left(sumlimits_k=0^inftyfrac(-1)^k2k+1 u^k right) = 1 $$
for $0 le u le 1$ .
I get this:
$$ sumlimits_n=0^infty a_n sumlimits_k=0^infty frac(-1)^k2k+1 u^k+n = 1 $$
and
$$ sumlimits_n=0^infty (-1)^n a_n sumlimits_k=n^infty frac(-1)^k2k-2n+1 u^k = 1 $$
I can see right away that $a_0 = 1$, but I should be able to yank out a recursion formula out of this. But I forgot how to do that. All of the net coefficients of the higher powers of $u^k$ should be $0$ for $k ge 1$.
Denouement:
It's simpler than I thought.
So we have
$$ sumlimits_n=0^infty sumlimits_k=0^infty a_n frac(-1)^k2k+1 u^k+n = 1 $$
Let's let the exponent of $u$ be fixed to some integer $m ge 0$. So $n+k=m$ and, for a given $n$, the only term in the inner summation that counts for the coefficient of $u^m$ is $k=m-n$.
We know for $m=0$ the summation is $1$.
$$ a_n frac(-1)^k2k+1 u^k+n Big|_n=k=0 = 1 $$
which results in $a_0=1$. For all other $mge 1$, the summation is
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 u^m = 0 qquad forall u: 0 le u le 1$$
or
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 = 0$$
Spinning off the last term for
$$ a_m + sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 = 0 $$
or
$$ a_m = -sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 $$
I confess that I didn't read the reference, I just needed to look at this more clearly. I guess with the additional substitution of $k=m-n$ we get:
$$ a_m = -sumlimits_k=1^m a_m-k frac(-1)^k2k+1 $$
.
power-series taylor-expansion lagrange-inversion
asked Jul 17 at 23:15
robert bristow-johnson
191116
I'm looking for a usable (something I can program) closed form or recursive formula for the coefficients $a_n$ of this series for an even-symmetry function of a real variable:
$$ sumlimits_n=0^infty a_n x^2n = fracxarctan(x) $$
for $big|xbig| le 1$ .
Ultimately, what I want is an expression like
$$ arctan(x) = fracxsumlimits_n=0^infty a_n x^2n $$
again, for $big|xbig| le 1$ .
I've approached this both from the POV of Taylor/Maclaurin series and tried to get a grip of using the Lagrange inversion theorem, but am not getting to an elegant place.
$$beginalign
sumlimits_n=0^infty a_n x^2n &= fracxarctan(x) \
\
&= fracxsumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1 \
\
&= frac1sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k \
endalign$$
or
$$ left( sumlimits_n=0^infty a_n u^n right) left(sumlimits_k=0^inftyfrac(-1)^k2k+1 u^k right) = 1 $$
for $0 le u le 1$ .
I get this:
$$ sumlimits_n=0^infty a_n sumlimits_k=0^infty frac(-1)^k2k+1 u^k+n = 1 $$
and
$$ sumlimits_n=0^infty (-1)^n a_n sumlimits_k=n^infty frac(-1)^k2k-2n+1 u^k = 1 $$
I can see right away that $a_0 = 1$, but I should be able to yank out a recursion formula out of this. But I forgot how to do that. All of the net coefficients of the higher powers of $u^k$ should be $0$ for $k ge 1$.
Denouement:
It's simpler than I thought.
So we have
$$ sumlimits_n=0^infty sumlimits_k=0^infty a_n frac(-1)^k2k+1 u^k+n = 1 $$
Let's let the exponent of $u$ be fixed to some integer $m ge 0$. So $n+k=m$ and, for a given $n$, the only term in the inner summation that counts for the coefficient of $u^m$ is $k=m-n$.
We know for $m=0$ the summation is $1$.
$$ a_n frac(-1)^k2k+1 u^k+n Big|_n=k=0 = 1 $$
which results in $a_0=1$. For all other $mge 1$, the summation is
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 u^m = 0 qquad forall u: 0 le u le 1$$
or
$$ sumlimits_n=0^m a_n frac(-1)^m-n2(m-n)+1 = 0$$
Spinning off the last term for
$$ a_m + sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 = 0 $$
or
$$ a_m = -sumlimits_n=0^m-1 a_n frac(-1)^m-n2(m-n)+1 $$
I confess that I didn't read the reference, I just needed to look at this more clearly. I guess with the additional substitution of $k=m-n$ we get:
$$ a_m = -sumlimits_k=1^m a_m-k frac(-1)^k2k+1 $$
.
power-series taylor-expansion lagrange-inversion
asked Jul 17 at 23:15
robert bristow-johnson
191116
edited Jul 18 at 9:22
asked Jul 17 at 23:15
robert bristow-johnson
191116
asked Jul 17 at 23:15
robert bristow-johnson
191116
asked Jul 17 at 23:15
robert bristow-johnson
191116
191116
You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06
add a comment |Â
You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06
You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Here's one way of handling this. Note that
beginalign
sumlimits_n=0^infty a_n x^2n arctan(x) &= x implies \
\
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1right) &= x \
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2kright) &= 1 \
endalign
Then you can expand and equate coefficients:
beginalign
a_0cdot 1 &= 1 implies a_0=1\
(a_1-a_0/3)cdot x^2 &= 0 implies a_1=1/3\
vdots
endalign
Another way, from here:
$$a_0 = 1$$
$$a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k$$
Special thanks to Sangchul Lee for changing the indices nicely.
answered Jul 17 at 23:35
Argon
16.1k668122
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
 |Â
show 2 more comments
up vote
0
down vote
$$ tan^ -1 x = x-x^3/3 +x^5/5-x^7/7+.... =x(1-x^2/3 +x^4/5-x^6/7+...)$$
$$frac x tan ^-1 x = frac 11-x^2/3 +x^4/5-x^6/7+...$$
You can get coefficients by long division.
For the first three terms I have $$frac 11-x^2/3 +x^4/5-x^6/7+...=1+x^2/3-4x^4/45-... $$
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here's one way of handling this. Note that
beginalign
sumlimits_n=0^infty a_n x^2n arctan(x) &= x implies \
\
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1right) &= x \
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2kright) &= 1 \
endalign
Then you can expand and equate coefficients:
beginalign
a_0cdot 1 &= 1 implies a_0=1\
(a_1-a_0/3)cdot x^2 &= 0 implies a_1=1/3\
vdots
endalign
Another way, from here:
$$a_0 = 1$$
$$a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k$$
Special thanks to Sangchul Lee for changing the indices nicely.
answered Jul 17 at 23:35
Argon
16.1k668122
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
 |Â
show 2 more comments
up vote
1
down vote
accepted
Here's one way of handling this. Note that
beginalign
sumlimits_n=0^infty a_n x^2n arctan(x) &= x implies \
\
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1right) &= x \
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2kright) &= 1 \
endalign
Then you can expand and equate coefficients:
beginalign
a_0cdot 1 &= 1 implies a_0=1\
(a_1-a_0/3)cdot x^2 &= 0 implies a_1=1/3\
vdots
endalign
Another way, from here:
$$a_0 = 1$$
$$a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k$$
Special thanks to Sangchul Lee for changing the indices nicely.
answered Jul 17 at 23:35
Argon
16.1k668122
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
 |Â
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's one way of handling this. Note that
beginalign
sumlimits_n=0^infty a_n x^2n arctan(x) &= x implies \
\
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1right) &= x \
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2kright) &= 1 \
endalign
Then you can expand and equate coefficients:
beginalign
a_0cdot 1 &= 1 implies a_0=1\
(a_1-a_0/3)cdot x^2 &= 0 implies a_1=1/3\
vdots
endalign
Another way, from here:
$$a_0 = 1$$
$$a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k$$
Special thanks to Sangchul Lee for changing the indices nicely.
answered Jul 17 at 23:35
Argon
16.1k668122
Here's one way of handling this. Note that
beginalign
sumlimits_n=0^infty a_n x^2n arctan(x) &= x implies \
\
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2k+1right) &= x \
left(sumlimits_n=0^infty a_n x^2nright)left( sumlimits_k=0^inftytfrac(-1)^k2k+1 x^2kright) &= 1 \
endalign
Then you can expand and equate coefficients:
beginalign
a_0cdot 1 &= 1 implies a_0=1\
(a_1-a_0/3)cdot x^2 &= 0 implies a_1=1/3\
vdots
endalign
Another way, from here:
$$a_0 = 1$$
$$a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k$$
Special thanks to Sangchul Lee for changing the indices nicely.
answered Jul 17 at 23:35
Argon
16.1k668122
edited Jul 18 at 1:34
answered Jul 17 at 23:35
Argon
16.1k668122
answered Jul 17 at 23:35
Argon
16.1k668122
answered Jul 17 at 23:35
Argon
16.1k668122
16.1k668122
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
 |Â
show 2 more comments
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
is there a general form for the recursion of $a_n$? like how is $a_n$ computed in terms of all of the previous $a_n-1, a_n-2, ...$?
– robert bristow-johnson
Jul 17 at 23:38
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
@robertbristow-johnson mathoverflow.net/a/53402
– Argon
Jul 17 at 23:42
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
well, thanks. i sorta wish, if you would, to see the explicit recursion here.
– robert bristow-johnson
Jul 18 at 0:12
2
2
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
@robertbristow-johnson, You definitely have such a recursion formula: $$ a_0 = 1, qquad a_n = sum_k=1^n frac(-1)^k-12k+1a_n-k, qquad n geq 1. $$
– Sangchul Lee
Jul 18 at 0:21
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
please @SangchulLee make this an answer so i can upvote it and checkmark it.
– robert bristow-johnson
Jul 18 at 0:23
 |Â
show 2 more comments
up vote
0
down vote
$$ tan^ -1 x = x-x^3/3 +x^5/5-x^7/7+.... =x(1-x^2/3 +x^4/5-x^6/7+...)$$
$$frac x tan ^-1 x = frac 11-x^2/3 +x^4/5-x^6/7+...$$
You can get coefficients by long division.
For the first three terms I have $$frac 11-x^2/3 +x^4/5-x^6/7+...=1+x^2/3-4x^4/45-... $$
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
 |Â
show 3 more comments
up vote
0
down vote
$$ tan^ -1 x = x-x^3/3 +x^5/5-x^7/7+.... =x(1-x^2/3 +x^4/5-x^6/7+...)$$
$$frac x tan ^-1 x = frac 11-x^2/3 +x^4/5-x^6/7+...$$
You can get coefficients by long division.
For the first three terms I have $$frac 11-x^2/3 +x^4/5-x^6/7+...=1+x^2/3-4x^4/45-... $$
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
 |Â
show 3 more comments
up vote
0
down vote
up vote
0
down vote
$$ tan^ -1 x = x-x^3/3 +x^5/5-x^7/7+.... =x(1-x^2/3 +x^4/5-x^6/7+...)$$
$$frac x tan ^-1 x = frac 11-x^2/3 +x^4/5-x^6/7+...$$
You can get coefficients by long division.
For the first three terms I have $$frac 11-x^2/3 +x^4/5-x^6/7+...=1+x^2/3-4x^4/45-... $$
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
$$ tan^ -1 x = x-x^3/3 +x^5/5-x^7/7+.... =x(1-x^2/3 +x^4/5-x^6/7+...)$$
$$frac x tan ^-1 x = frac 11-x^2/3 +x^4/5-x^6/7+...$$
You can get coefficients by long division.
For the first three terms I have $$frac 11-x^2/3 +x^4/5-x^6/7+...=1+x^2/3-4x^4/45-... $$
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
edited Jul 18 at 7:08
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
answered Jul 18 at 0:15
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
 |Â
show 3 more comments
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
can you get a recursion relationship that explicitly relates what the coefficient for the $x^2n$ term is as a function of the lower-order coefficients? even if it has a summation in it?
– robert bristow-johnson
Jul 18 at 0:19
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
@robertbristow-johnson I did not go that far, I just divided up to three terms. My guess is that a pattern will develop after few more terms.
– Mohammad Riazi-Kermani
Jul 18 at 0:21
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
i hadn't been seeing the pattern. i don't think the next term is $x^6/12$, is it?
– robert bristow-johnson
Jul 18 at 0:24
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
No, I do not think it was $x^6/12$. I think it was negative. I
– Mohammad Riazi-Kermani
Jul 18 at 1:09
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
The third term is not correct. Using long division with more terms, you should get $$1+fracx^23-frac4 x^445+frac44 x^6945-frac428 x^814175+Oleft(x^10right)$$
– Claude Leibovici
Jul 18 at 5:12
 |Â
show 3 more comments
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You could find the coefficients in sequences $A216272$ and $A195466$ in $OEIS$.
– Claude Leibovici
Jul 18 at 5:14
i must confess that i have absolutely no idea what you mean here, @ClaudeLeibovici.
– robert bristow-johnson
Jul 18 at 8:35
Visit OEIS and type the name of the sequence.
– Claude Leibovici
Jul 18 at 8:59
well, i learn something new every day. had never heard of this On-Line Encyclopedia of Integer Sequences before.
– robert bristow-johnson
Jul 18 at 19:06