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Maclaurin series for $f(x)=ln(1+tanx)$
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How to find the Maclaurin series for $f(x)=ln(1+tan x)$. I tried it by successively differentiating it up to four times but it is very time consuming.
calculus power-series taylor-expansion
edited Jul 14 at 17:30
Bernard
110k635103
asked Jul 14 at 16:51
jiren
284
add a comment |Â
up vote
0
down vote
favorite
How to find the Maclaurin series for $f(x)=ln(1+tan x)$. I tried it by successively differentiating it up to four times but it is very time consuming.
calculus power-series taylor-expansion
edited Jul 14 at 17:30
Bernard
110k635103
asked Jul 14 at 16:51
jiren
284
It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to find the Maclaurin series for $f(x)=ln(1+tan x)$. I tried it by successively differentiating it up to four times but it is very time consuming.
calculus power-series taylor-expansion
edited Jul 14 at 17:30
Bernard
110k635103
asked Jul 14 at 16:51
jiren
284
How to find the Maclaurin series for $f(x)=ln(1+tan x)$. I tried it by successively differentiating it up to four times but it is very time consuming.
calculus power-series taylor-expansion
edited Jul 14 at 17:30
Bernard
110k635103
asked Jul 14 at 16:51
jiren
284
edited Jul 14 at 17:30
Bernard
110k635103
edited Jul 14 at 17:30
Bernard
110k635103
edited Jul 14 at 17:30
Bernard
110k635103
110k635103
asked Jul 14 at 16:51
jiren
284
asked Jul 14 at 16:51
jiren
284
asked Jul 14 at 16:51
jiren
284
284
It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58
add a comment |Â
It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58
It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58
It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
According to OEIS, the series can be written as $$log (1+tan x) = sum_n=0^infty fraca_nn! x^n,$$ where $$a_n = sum_m=0^(n-1)/2 sum_j=0^2m binomj+n-2m-1n-2m-1 frac(j+n-2m)!n-2m , 2^2m-j (-1)^n-m+j-1 leftn atop j+n-2mright$$ where $a atop b$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
S[u_] := (Range[u + 1] - 1)! CoefficientList[Normal[Series[Log[1 + Tan[x]],
x, 0, u]], x] - Table[Sum[Binomial[j + n - 2 m - 1, n - 2 m - 1]
(j + n - 2 m)! 2^(2 m - j) (-1)^(n - m + j - 1) StirlingS2[n,
j + n - 2 m]/(n - 2 m), m, 0, (n - 1)/2, j, 0, 2 m], n, 0, u]
Then S[50] for example gives a list of zeros, indicating equality.
answered Jul 15 at 8:08
heropup
59.8k65895
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ usingSeriesand then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.
– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
add a comment |Â
up vote
3
down vote
For this kind of work, I think that composition of series is a simple way
$$tan(x)=x+fracx^33+frac2 x^515+frac17 x^7315+Oleft(x^9right)$$
Let $t$ to be the above development and use
$$log(1+t)=t-fract^22+fract^33-fract^44+cdots$$ and use the binomial expansion (stopping when you find powers of $x$ greater than $7$ to get
$$ln(1+tan(x))=x-fracx^22+frac2 x^33-frac7 x^412+frac2 x^53-frac31
x^645+frac244 x^7315+Oleft(x^8right)$$
Notice that using this process, we could continue for ever; for example
$$sin (log (tan (x)+1))=x-fracx^22+fracx^32-fracx^43+frac13 x^560-frac23
x^6360-frac271 x^72520+Oleft(x^8right)$$
$$tan (sin (log (tan (x)+1)))=x-fracx^22+frac5 x^36-frac5 x^46+frac11 x^510-frac229
x^6180+frac323 x^7210+Oleft(x^8right)$$ and so on.
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
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
According to OEIS, the series can be written as $$log (1+tan x) = sum_n=0^infty fraca_nn! x^n,$$ where $$a_n = sum_m=0^(n-1)/2 sum_j=0^2m binomj+n-2m-1n-2m-1 frac(j+n-2m)!n-2m , 2^2m-j (-1)^n-m+j-1 leftn atop j+n-2mright$$ where $a atop b$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
S[u_] := (Range[u + 1] - 1)! CoefficientList[Normal[Series[Log[1 + Tan[x]],
x, 0, u]], x] - Table[Sum[Binomial[j + n - 2 m - 1, n - 2 m - 1]
(j + n - 2 m)! 2^(2 m - j) (-1)^(n - m + j - 1) StirlingS2[n,
j + n - 2 m]/(n - 2 m), m, 0, (n - 1)/2, j, 0, 2 m], n, 0, u]
Then S[50] for example gives a list of zeros, indicating equality.
answered Jul 15 at 8:08
heropup
59.8k65895
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ usingSeriesand then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.
– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
add a comment |Â
up vote
3
down vote
According to OEIS, the series can be written as $$log (1+tan x) = sum_n=0^infty fraca_nn! x^n,$$ where $$a_n = sum_m=0^(n-1)/2 sum_j=0^2m binomj+n-2m-1n-2m-1 frac(j+n-2m)!n-2m , 2^2m-j (-1)^n-m+j-1 leftn atop j+n-2mright$$ where $a atop b$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
S[u_] := (Range[u + 1] - 1)! CoefficientList[Normal[Series[Log[1 + Tan[x]],
x, 0, u]], x] - Table[Sum[Binomial[j + n - 2 m - 1, n - 2 m - 1]
(j + n - 2 m)! 2^(2 m - j) (-1)^(n - m + j - 1) StirlingS2[n,
j + n - 2 m]/(n - 2 m), m, 0, (n - 1)/2, j, 0, 2 m], n, 0, u]
Then S[50] for example gives a list of zeros, indicating equality.
answered Jul 15 at 8:08
heropup
59.8k65895
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ usingSeriesand then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.
– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
add a comment |Â
up vote
3
down vote
up vote
3
down vote
According to OEIS, the series can be written as $$log (1+tan x) = sum_n=0^infty fraca_nn! x^n,$$ where $$a_n = sum_m=0^(n-1)/2 sum_j=0^2m binomj+n-2m-1n-2m-1 frac(j+n-2m)!n-2m , 2^2m-j (-1)^n-m+j-1 leftn atop j+n-2mright$$ where $a atop b$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
S[u_] := (Range[u + 1] - 1)! CoefficientList[Normal[Series[Log[1 + Tan[x]],
x, 0, u]], x] - Table[Sum[Binomial[j + n - 2 m - 1, n - 2 m - 1]
(j + n - 2 m)! 2^(2 m - j) (-1)^(n - m + j - 1) StirlingS2[n,
j + n - 2 m]/(n - 2 m), m, 0, (n - 1)/2, j, 0, 2 m], n, 0, u]
Then S[50] for example gives a list of zeros, indicating equality.
answered Jul 15 at 8:08
heropup
59.8k65895
According to OEIS, the series can be written as $$log (1+tan x) = sum_n=0^infty fraca_nn! x^n,$$ where $$a_n = sum_m=0^(n-1)/2 sum_j=0^2m binomj+n-2m-1n-2m-1 frac(j+n-2m)!n-2m , 2^2m-j (-1)^n-m+j-1 leftn atop j+n-2mright$$ where $a atop b$ is a Stirling number of the second kind. Numeric calculation with Mathematica supports this; e.g.,
S[u_] := (Range[u + 1] - 1)! CoefficientList[Normal[Series[Log[1 + Tan[x]],
x, 0, u]], x] - Table[Sum[Binomial[j + n - 2 m - 1, n - 2 m - 1]
(j + n - 2 m)! 2^(2 m - j) (-1)^(n - m + j - 1) StirlingS2[n,
j + n - 2 m]/(n - 2 m), m, 0, (n - 1)/2, j, 0, 2 m], n, 0, u]
Then S[50] for example gives a list of zeros, indicating equality.
answered Jul 15 at 8:08
heropup
59.8k65895
answered Jul 15 at 8:08
heropup
59.8k65895
answered Jul 15 at 8:08
heropup
59.8k65895
answered Jul 15 at 8:08
heropup
59.8k65895
59.8k65895
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ usingSeriesand then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.
– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
add a comment |Â
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ usingSeriesand then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.
– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
May I ask ? How did you find it ? What is amazing is the asymptotics given in the EOIS page. Cheers.
– Claude Leibovici
Jul 15 at 8:49
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ using
Series and then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.– heropup
Jul 15 at 19:19
@ClaudeLeibovici I used Mathematica to numerically compute $a_n$ for $n le 10$ using
Series and then used OEIS to search for the sequence, which brought me to that page and the rest of the information. If you read the linked preprint by Kruchinin, it tells you a lot about how the formula was obtained. As for the asymptotics, I have no clue.– heropup
Jul 15 at 19:19
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
@ClaudeLeibovici Possibly related.
– Simply Beautiful Art
Aug 7 at 19:05
add a comment |Â
up vote
3
down vote
For this kind of work, I think that composition of series is a simple way
$$tan(x)=x+fracx^33+frac2 x^515+frac17 x^7315+Oleft(x^9right)$$
Let $t$ to be the above development and use
$$log(1+t)=t-fract^22+fract^33-fract^44+cdots$$ and use the binomial expansion (stopping when you find powers of $x$ greater than $7$ to get
$$ln(1+tan(x))=x-fracx^22+frac2 x^33-frac7 x^412+frac2 x^53-frac31
x^645+frac244 x^7315+Oleft(x^8right)$$
Notice that using this process, we could continue for ever; for example
$$sin (log (tan (x)+1))=x-fracx^22+fracx^32-fracx^43+frac13 x^560-frac23
x^6360-frac271 x^72520+Oleft(x^8right)$$
$$tan (sin (log (tan (x)+1)))=x-fracx^22+frac5 x^36-frac5 x^46+frac11 x^510-frac229
x^6180+frac323 x^7210+Oleft(x^8right)$$ and so on.
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
add a comment |Â
up vote
3
down vote
For this kind of work, I think that composition of series is a simple way
$$tan(x)=x+fracx^33+frac2 x^515+frac17 x^7315+Oleft(x^9right)$$
Let $t$ to be the above development and use
$$log(1+t)=t-fract^22+fract^33-fract^44+cdots$$ and use the binomial expansion (stopping when you find powers of $x$ greater than $7$ to get
$$ln(1+tan(x))=x-fracx^22+frac2 x^33-frac7 x^412+frac2 x^53-frac31
x^645+frac244 x^7315+Oleft(x^8right)$$
Notice that using this process, we could continue for ever; for example
$$sin (log (tan (x)+1))=x-fracx^22+fracx^32-fracx^43+frac13 x^560-frac23
x^6360-frac271 x^72520+Oleft(x^8right)$$
$$tan (sin (log (tan (x)+1)))=x-fracx^22+frac5 x^36-frac5 x^46+frac11 x^510-frac229
x^6180+frac323 x^7210+Oleft(x^8right)$$ and so on.
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For this kind of work, I think that composition of series is a simple way
$$tan(x)=x+fracx^33+frac2 x^515+frac17 x^7315+Oleft(x^9right)$$
Let $t$ to be the above development and use
$$log(1+t)=t-fract^22+fract^33-fract^44+cdots$$ and use the binomial expansion (stopping when you find powers of $x$ greater than $7$ to get
$$ln(1+tan(x))=x-fracx^22+frac2 x^33-frac7 x^412+frac2 x^53-frac31
x^645+frac244 x^7315+Oleft(x^8right)$$
Notice that using this process, we could continue for ever; for example
$$sin (log (tan (x)+1))=x-fracx^22+fracx^32-fracx^43+frac13 x^560-frac23
x^6360-frac271 x^72520+Oleft(x^8right)$$
$$tan (sin (log (tan (x)+1)))=x-fracx^22+frac5 x^36-frac5 x^46+frac11 x^510-frac229
x^6180+frac323 x^7210+Oleft(x^8right)$$ and so on.
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
For this kind of work, I think that composition of series is a simple way
$$tan(x)=x+fracx^33+frac2 x^515+frac17 x^7315+Oleft(x^9right)$$
Let $t$ to be the above development and use
$$log(1+t)=t-fract^22+fract^33-fract^44+cdots$$ and use the binomial expansion (stopping when you find powers of $x$ greater than $7$ to get
$$ln(1+tan(x))=x-fracx^22+frac2 x^33-frac7 x^412+frac2 x^53-frac31
x^645+frac244 x^7315+Oleft(x^8right)$$
Notice that using this process, we could continue for ever; for example
$$sin (log (tan (x)+1))=x-fracx^22+fracx^32-fracx^43+frac13 x^560-frac23
x^6360-frac271 x^72520+Oleft(x^8right)$$
$$tan (sin (log (tan (x)+1)))=x-fracx^22+frac5 x^36-frac5 x^46+frac11 x^510-frac229
x^6180+frac323 x^7210+Oleft(x^8right)$$ and so on.
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
edited Jul 16 at 3:13
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
answered Jul 15 at 6:39
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
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It is not a "nice" expansion as you can see by just checking WolframAlpha (link)
– Benjamin Dickman
Jul 14 at 16:58