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Series expansion of $tan^2$ and $tanh^2$
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Are there known closed formula expressions for their power series expansion at the origin? I couldn't find anything online.
Edit: (to clarify)
Of course we could simply take the series expansion of
$$tanh(x) = sum_n=1^infty frac2^2n(2^2n-1)B_2n(2n)!x^2n-1 $$
And square it with a Cauchy product:
$$tanh(x)^2 = sum_n=2^infty Big[sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)!Big] x^2n-2 $$
The question is whether or not one can simplify the coefficient term
$$ sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)! $$
in any meaningful way.
I tried to do the following:
$$ tanh^2 =fracsinh^2cosh^2 iff tanh^2(1+sinh^2) = sinh^2$$
Now letting $tanh(x)^2 = sum_n=0^infty a_2n x^2n$, and using the known formula $sinh(x)^2 = sum_n=1^infty frac2^2n-1(2n)!x^2n$ yields a Volterra type difference equation:
$$a_2n = frac2^2n(2n)! - sum_k=0^n a_2kfrac2^2n-2k(2n-2k)! qquadtextfor nge1 $$
which I had no luck solving so far (it's a hairy business!)
taylor-expansion hyperbolic-functions
asked Aug 2 at 16:25
Hyperplane
3,1831624
add a comment |Â
up vote
2
down vote
favorite
Are there known closed formula expressions for their power series expansion at the origin? I couldn't find anything online.
Edit: (to clarify)
Of course we could simply take the series expansion of
$$tanh(x) = sum_n=1^infty frac2^2n(2^2n-1)B_2n(2n)!x^2n-1 $$
And square it with a Cauchy product:
$$tanh(x)^2 = sum_n=2^infty Big[sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)!Big] x^2n-2 $$
The question is whether or not one can simplify the coefficient term
$$ sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)! $$
in any meaningful way.
I tried to do the following:
$$ tanh^2 =fracsinh^2cosh^2 iff tanh^2(1+sinh^2) = sinh^2$$
Now letting $tanh(x)^2 = sum_n=0^infty a_2n x^2n$, and using the known formula $sinh(x)^2 = sum_n=1^infty frac2^2n-1(2n)!x^2n$ yields a Volterra type difference equation:
$$a_2n = frac2^2n(2n)! - sum_k=0^n a_2kfrac2^2n-2k(2n-2k)! qquadtextfor nge1 $$
which I had no luck solving so far (it's a hairy business!)
taylor-expansion hyperbolic-functions
asked Aug 2 at 16:25
Hyperplane
3,1831624
Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
2
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Are there known closed formula expressions for their power series expansion at the origin? I couldn't find anything online.
Edit: (to clarify)
Of course we could simply take the series expansion of
$$tanh(x) = sum_n=1^infty frac2^2n(2^2n-1)B_2n(2n)!x^2n-1 $$
And square it with a Cauchy product:
$$tanh(x)^2 = sum_n=2^infty Big[sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)!Big] x^2n-2 $$
The question is whether or not one can simplify the coefficient term
$$ sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)! $$
in any meaningful way.
I tried to do the following:
$$ tanh^2 =fracsinh^2cosh^2 iff tanh^2(1+sinh^2) = sinh^2$$
Now letting $tanh(x)^2 = sum_n=0^infty a_2n x^2n$, and using the known formula $sinh(x)^2 = sum_n=1^infty frac2^2n-1(2n)!x^2n$ yields a Volterra type difference equation:
$$a_2n = frac2^2n(2n)! - sum_k=0^n a_2kfrac2^2n-2k(2n-2k)! qquadtextfor nge1 $$
which I had no luck solving so far (it's a hairy business!)
taylor-expansion hyperbolic-functions
asked Aug 2 at 16:25
Hyperplane
3,1831624
Are there known closed formula expressions for their power series expansion at the origin? I couldn't find anything online.
Edit: (to clarify)
Of course we could simply take the series expansion of
$$tanh(x) = sum_n=1^infty frac2^2n(2^2n-1)B_2n(2n)!x^2n-1 $$
And square it with a Cauchy product:
$$tanh(x)^2 = sum_n=2^infty Big[sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)!Big] x^2n-2 $$
The question is whether or not one can simplify the coefficient term
$$ sum_i+j=nfrac2^2i(2^2i-1)B_2i(2i)!frac2^2j(2^2j-1)B_2j(2j)! $$
in any meaningful way.
I tried to do the following:
$$ tanh^2 =fracsinh^2cosh^2 iff tanh^2(1+sinh^2) = sinh^2$$
Now letting $tanh(x)^2 = sum_n=0^infty a_2n x^2n$, and using the known formula $sinh(x)^2 = sum_n=1^infty frac2^2n-1(2n)!x^2n$ yields a Volterra type difference equation:
$$a_2n = frac2^2n(2n)! - sum_k=0^n a_2kfrac2^2n-2k(2n-2k)! qquadtextfor nge1 $$
which I had no luck solving so far (it's a hairy business!)
taylor-expansion hyperbolic-functions
asked Aug 2 at 16:25
Hyperplane
3,1831624
edited Aug 2 at 16:38
asked Aug 2 at 16:25
Hyperplane
3,1831624
asked Aug 2 at 16:25
Hyperplane
3,1831624
asked Aug 2 at 16:25
Hyperplane
3,1831624
3,1831624
Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
2
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31
add a comment |Â
Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
2
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31
Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
2
2
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31
add a comment |Â
2 Answers
2
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oldest
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up vote
3
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accepted
The series expansions of $tan^2 x$ and $tanh^2 x$ can be obtained by differentiating the series expansions of $tan x$ and $tanh x$ and utilising $sec^2 x = 1+tan^2 x$ and $DeclareMathOperatorsechsech sech^2 x = 1-tanh^2 x$ respectively
As $fracd ; tan xdx=sec^2x$ and $ fracd ; tanh xdx=sech^2x$
$$tan^2 x=fracd ; tan xdx-1 tag1$$
and
$$tanh^2 x=1-fracd ; tanh xdx tag2$$
I am not sure this is what you meant though as you mentioned closed form expressions in your question.
[Corrected hyperbolic identity]
answered Aug 2 at 17:38
James Arathoon
1,155420
add a comment |Â
up vote
0
down vote
This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $ tan(x). $ Because of the identity
$$ fracddx tan (x) = sec^2(x) = 1 + tan^2(x) =
1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +dots $$
the function $ 1+tan^2(x) $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $ x. $ A very similar relationship holds for the hyperbolic tangent function since
$$ fracddx tanh (x) = sech^2(x) = 1 - tanh^2(x) =
1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +dots . $$
As a consequence of this, we get the recursion
$ a_n = sum_k=1^n a_k-1a_n-k 2n choose 2k-1 $ for $ n>0. $
answered Aug 2 at 19:55
Somos
10.9k1831
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The series expansions of $tan^2 x$ and $tanh^2 x$ can be obtained by differentiating the series expansions of $tan x$ and $tanh x$ and utilising $sec^2 x = 1+tan^2 x$ and $DeclareMathOperatorsechsech sech^2 x = 1-tanh^2 x$ respectively
As $fracd ; tan xdx=sec^2x$ and $ fracd ; tanh xdx=sech^2x$
$$tan^2 x=fracd ; tan xdx-1 tag1$$
and
$$tanh^2 x=1-fracd ; tanh xdx tag2$$
I am not sure this is what you meant though as you mentioned closed form expressions in your question.
[Corrected hyperbolic identity]
answered Aug 2 at 17:38
James Arathoon
1,155420
add a comment |Â
up vote
3
down vote
accepted
The series expansions of $tan^2 x$ and $tanh^2 x$ can be obtained by differentiating the series expansions of $tan x$ and $tanh x$ and utilising $sec^2 x = 1+tan^2 x$ and $DeclareMathOperatorsechsech sech^2 x = 1-tanh^2 x$ respectively
As $fracd ; tan xdx=sec^2x$ and $ fracd ; tanh xdx=sech^2x$
$$tan^2 x=fracd ; tan xdx-1 tag1$$
and
$$tanh^2 x=1-fracd ; tanh xdx tag2$$
I am not sure this is what you meant though as you mentioned closed form expressions in your question.
[Corrected hyperbolic identity]
answered Aug 2 at 17:38
James Arathoon
1,155420
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The series expansions of $tan^2 x$ and $tanh^2 x$ can be obtained by differentiating the series expansions of $tan x$ and $tanh x$ and utilising $sec^2 x = 1+tan^2 x$ and $DeclareMathOperatorsechsech sech^2 x = 1-tanh^2 x$ respectively
As $fracd ; tan xdx=sec^2x$ and $ fracd ; tanh xdx=sech^2x$
$$tan^2 x=fracd ; tan xdx-1 tag1$$
and
$$tanh^2 x=1-fracd ; tanh xdx tag2$$
I am not sure this is what you meant though as you mentioned closed form expressions in your question.
[Corrected hyperbolic identity]
answered Aug 2 at 17:38
James Arathoon
1,155420
The series expansions of $tan^2 x$ and $tanh^2 x$ can be obtained by differentiating the series expansions of $tan x$ and $tanh x$ and utilising $sec^2 x = 1+tan^2 x$ and $DeclareMathOperatorsechsech sech^2 x = 1-tanh^2 x$ respectively
As $fracd ; tan xdx=sec^2x$ and $ fracd ; tanh xdx=sech^2x$
$$tan^2 x=fracd ; tan xdx-1 tag1$$
and
$$tanh^2 x=1-fracd ; tanh xdx tag2$$
I am not sure this is what you meant though as you mentioned closed form expressions in your question.
[Corrected hyperbolic identity]
answered Aug 2 at 17:38
James Arathoon
1,155420
edited Aug 4 at 14:14
answered Aug 2 at 17:38
James Arathoon
1,155420
answered Aug 2 at 17:38
James Arathoon
1,155420
answered Aug 2 at 17:38
James Arathoon
1,155420
1,155420
add a comment |Â
add a comment |Â
up vote
0
down vote
This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $ tan(x). $ Because of the identity
$$ fracddx tan (x) = sec^2(x) = 1 + tan^2(x) =
1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +dots $$
the function $ 1+tan^2(x) $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $ x. $ A very similar relationship holds for the hyperbolic tangent function since
$$ fracddx tanh (x) = sech^2(x) = 1 - tanh^2(x) =
1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +dots . $$
As a consequence of this, we get the recursion
$ a_n = sum_k=1^n a_k-1a_n-k 2n choose 2k-1 $ for $ n>0. $
answered Aug 2 at 19:55
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $ tan(x). $ Because of the identity
$$ fracddx tan (x) = sec^2(x) = 1 + tan^2(x) =
1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +dots $$
the function $ 1+tan^2(x) $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $ x. $ A very similar relationship holds for the hyperbolic tangent function since
$$ fracddx tanh (x) = sech^2(x) = 1 - tanh^2(x) =
1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +dots . $$
As a consequence of this, we get the recursion
$ a_n = sum_k=1^n a_k-1a_n-k 2n choose 2k-1 $ for $ n>0. $
answered Aug 2 at 19:55
Somos
10.9k1831
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $ tan(x). $ Because of the identity
$$ fracddx tan (x) = sec^2(x) = 1 + tan^2(x) =
1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +dots $$
the function $ 1+tan^2(x) $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $ x. $ A very similar relationship holds for the hyperbolic tangent function since
$$ fracddx tanh (x) = sech^2(x) = 1 - tanh^2(x) =
1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +dots . $$
As a consequence of this, we get the recursion
$ a_n = sum_k=1^n a_k-1a_n-k 2n choose 2k-1 $ for $ n>0. $
answered Aug 2 at 19:55
Somos
10.9k1831
This is a nice question. The OEIS sequence A000182 is known as the Tangent numbers because the exponential generating function (e.g.f.) (with interpolated zeros) is $ tan(x). $ Because of the identity
$$ fracddx tan (x) = sec^2(x) = 1 + tan^2(x) =
1 + 2x^2/2! + 16 x^4/4! + 272 x^6/6! +dots $$
the function $ 1+tan^2(x) $ is also the e.g.f. of A000182 except the coefficients are for the even powers of $ x. $ A very similar relationship holds for the hyperbolic tangent function since
$$ fracddx tanh (x) = sech^2(x) = 1 - tanh^2(x) =
1 - 2x^2/2! + 16 x^4/4! - 272 x^6/6! +dots . $$
As a consequence of this, we get the recursion
$ a_n = sum_k=1^n a_k-1a_n-k 2n choose 2k-1 $ for $ n>0. $
answered Aug 2 at 19:55
Somos
10.9k1831
edited yesterday
answered Aug 2 at 19:55
Somos
10.9k1831
answered Aug 2 at 19:55
Somos
10.9k1831
answered Aug 2 at 19:55
Somos
10.9k1831
10.9k1831
add a comment |Â
add a comment |Â
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Do you know the Maclaurin expansions for $tan$ and $tanh$?
– Simply Beautiful Art
Aug 2 at 16:28
@SimplyBeautifulArt Yes of course. But if you use Cauchy product to square them you get some really ugly sums which I want to get rid of if possible....
– Hyperplane
Aug 2 at 16:30
2
I don't see how you could.
– Simply Beautiful Art
Aug 2 at 16:31