Mixing
Taylor series of $1/(e^x-a)$ about 0?
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I know what the Taylor series for $frac{1}{e^x-1}$ about 0 is, and it's a function of Bernoulli numbers.
However, $frac{1}{e^x-a}$ seems to be much more convoluted. Does anybody know the general term of this series?
If it exists, finding it on Google is challenging, due to the mathematical notation.
taylor-expansion closed-form
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
$endgroup$
add a comment |
$begingroup$
I know what the Taylor series for $frac{1}{e^x-1}$ about 0 is, and it's a function of Bernoulli numbers.
However, $frac{1}{e^x-a}$ seems to be much more convoluted. Does anybody know the general term of this series?
If it exists, finding it on Google is challenging, due to the mathematical notation.
taylor-expansion closed-form
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
$endgroup$
$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
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@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
$endgroup$
– JR Sousa
Jan 26 at 7:52
add a comment |
$begingroup$
I know what the Taylor series for $frac{1}{e^x-1}$ about 0 is, and it's a function of Bernoulli numbers.
However, $frac{1}{e^x-a}$ seems to be much more convoluted. Does anybody know the general term of this series?
If it exists, finding it on Google is challenging, due to the mathematical notation.
taylor-expansion closed-form
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
$endgroup$
I know what the Taylor series for $frac{1}{e^x-1}$ about 0 is, and it's a function of Bernoulli numbers.
However, $frac{1}{e^x-a}$ seems to be much more convoluted. Does anybody know the general term of this series?
If it exists, finding it on Google is challenging, due to the mathematical notation.
taylor-expansion closed-form
taylor-expansion closed-form
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
asked Jan 26 at 7:35
JR SousaJR Sousa
1009
1009
$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
$begingroup$
@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
$endgroup$
– JR Sousa
Jan 26 at 7:52
add a comment |
$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
$begingroup$
@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
$endgroup$
– JR Sousa
Jan 26 at 7:52
$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
$begingroup$
@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
$endgroup$
– JR Sousa
Jan 26 at 7:52
$begingroup$
@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
$endgroup$
– JR Sousa
Jan 26 at 7:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So, maybe the Binomial Theorem is needed.
See https://en.wikipedia.org/wiki/Binomial_theorem.
Actually, we can obtain that
begin{align}
(e^x-a)^{-1} &= -a^{-1}sum_{k=0}^{infty} left(frac{e^x}{a}right)^k\
&=-frac{1}{a} sum_{k=0}^{infty} frac{1}{a^k} sum_{n=0}^{infty} frac{(kx)^n}{n!}\
&=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n.
end{align}
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
$endgroup$
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
add a comment |
$begingroup$
Starting from @Aaron Jia's answer
$$frac{1}{e^x-a}=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n$$ let
$$c_n=sum_{k=0}^{infty} frac{k^n}{a^{k+1}}=frac{1}{a},Phi left(frac{1}{a},-n,0right) $$ where appears the Hurwitz-Lerch transcendent function.
So,
$$frac{1}{e^x-a}=sum_{n=0}^{infty} -frac{Phi left(frac{1}{a},-n,0right)}{a, n!} x^n$$ and the coefficients are then
$$left(
begin{array}{cc}
0 & frac{1}{1-a} \
1 & -frac{1}{(a-1)^2} \
2 & frac{-a-1}{2 (a-1)^3} \
3 & frac{-a^2-4 a-1}{6 (a-1)^4} \
4 & frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \
5 & frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6}
end{array}
right)$$ to be compared to the expansion
$$frac{1}{1-a}-frac{x}{(a-1)^2}+frac{(-a-1) x^2}{2 (a-1)^3}+frac{left(-a^2-4
a-1right) x^3}{6 (a-1)^4}+frac{left(-a^3-11 a^2-11 a-1right) x^4}{24
(a-1)^5}+frac{left(-a^4-26 a^3-66 a^2-26 a-1right) x^5}{120
(a-1)^6}+Oleft(x^6right)$$
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
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@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So, maybe the Binomial Theorem is needed.
See https://en.wikipedia.org/wiki/Binomial_theorem.
Actually, we can obtain that
begin{align}
(e^x-a)^{-1} &= -a^{-1}sum_{k=0}^{infty} left(frac{e^x}{a}right)^k\
&=-frac{1}{a} sum_{k=0}^{infty} frac{1}{a^k} sum_{n=0}^{infty} frac{(kx)^n}{n!}\
&=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n.
end{align}
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
$endgroup$
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
add a comment |
$begingroup$
So, maybe the Binomial Theorem is needed.
See https://en.wikipedia.org/wiki/Binomial_theorem.
Actually, we can obtain that
begin{align}
(e^x-a)^{-1} &= -a^{-1}sum_{k=0}^{infty} left(frac{e^x}{a}right)^k\
&=-frac{1}{a} sum_{k=0}^{infty} frac{1}{a^k} sum_{n=0}^{infty} frac{(kx)^n}{n!}\
&=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n.
end{align}
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
$endgroup$
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
add a comment |
$begingroup$
So, maybe the Binomial Theorem is needed.
See https://en.wikipedia.org/wiki/Binomial_theorem.
Actually, we can obtain that
begin{align}
(e^x-a)^{-1} &= -a^{-1}sum_{k=0}^{infty} left(frac{e^x}{a}right)^k\
&=-frac{1}{a} sum_{k=0}^{infty} frac{1}{a^k} sum_{n=0}^{infty} frac{(kx)^n}{n!}\
&=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n.
end{align}
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
$endgroup$
So, maybe the Binomial Theorem is needed.
See https://en.wikipedia.org/wiki/Binomial_theorem.
Actually, we can obtain that
begin{align}
(e^x-a)^{-1} &= -a^{-1}sum_{k=0}^{infty} left(frac{e^x}{a}right)^k\
&=-frac{1}{a} sum_{k=0}^{infty} frac{1}{a^k} sum_{n=0}^{infty} frac{(kx)^n}{n!}\
&=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n.
end{align}
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
answered Jan 26 at 8:13
Aaron JiaAaron Jia
383
383
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
add a comment |
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
$begingroup$
That is not it yet because the second sum is infinite, and I want it to be a closed-form, right?
$endgroup$
– JR Sousa
Jan 26 at 8:21
add a comment |
$begingroup$
Starting from @Aaron Jia's answer
$$frac{1}{e^x-a}=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n$$ let
$$c_n=sum_{k=0}^{infty} frac{k^n}{a^{k+1}}=frac{1}{a},Phi left(frac{1}{a},-n,0right) $$ where appears the Hurwitz-Lerch transcendent function.
So,
$$frac{1}{e^x-a}=sum_{n=0}^{infty} -frac{Phi left(frac{1}{a},-n,0right)}{a, n!} x^n$$ and the coefficients are then
$$left(
begin{array}{cc}
0 & frac{1}{1-a} \
1 & -frac{1}{(a-1)^2} \
2 & frac{-a-1}{2 (a-1)^3} \
3 & frac{-a^2-4 a-1}{6 (a-1)^4} \
4 & frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \
5 & frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6}
end{array}
right)$$ to be compared to the expansion
$$frac{1}{1-a}-frac{x}{(a-1)^2}+frac{(-a-1) x^2}{2 (a-1)^3}+frac{left(-a^2-4
a-1right) x^3}{6 (a-1)^4}+frac{left(-a^3-11 a^2-11 a-1right) x^4}{24
(a-1)^5}+frac{left(-a^4-26 a^3-66 a^2-26 a-1right) x^5}{120
(a-1)^6}+Oleft(x^6right)$$
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
add a comment |
$begingroup$
Starting from @Aaron Jia's answer
$$frac{1}{e^x-a}=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n$$ let
$$c_n=sum_{k=0}^{infty} frac{k^n}{a^{k+1}}=frac{1}{a},Phi left(frac{1}{a},-n,0right) $$ where appears the Hurwitz-Lerch transcendent function.
So,
$$frac{1}{e^x-a}=sum_{n=0}^{infty} -frac{Phi left(frac{1}{a},-n,0right)}{a, n!} x^n$$ and the coefficients are then
$$left(
begin{array}{cc}
0 & frac{1}{1-a} \
1 & -frac{1}{(a-1)^2} \
2 & frac{-a-1}{2 (a-1)^3} \
3 & frac{-a^2-4 a-1}{6 (a-1)^4} \
4 & frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \
5 & frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6}
end{array}
right)$$ to be compared to the expansion
$$frac{1}{1-a}-frac{x}{(a-1)^2}+frac{(-a-1) x^2}{2 (a-1)^3}+frac{left(-a^2-4
a-1right) x^3}{6 (a-1)^4}+frac{left(-a^3-11 a^2-11 a-1right) x^4}{24
(a-1)^5}+frac{left(-a^4-26 a^3-66 a^2-26 a-1right) x^5}{120
(a-1)^6}+Oleft(x^6right)$$
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
add a comment |
$begingroup$
Starting from @Aaron Jia's answer
$$frac{1}{e^x-a}=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n$$ let
$$c_n=sum_{k=0}^{infty} frac{k^n}{a^{k+1}}=frac{1}{a},Phi left(frac{1}{a},-n,0right) $$ where appears the Hurwitz-Lerch transcendent function.
So,
$$frac{1}{e^x-a}=sum_{n=0}^{infty} -frac{Phi left(frac{1}{a},-n,0right)}{a, n!} x^n$$ and the coefficients are then
$$left(
begin{array}{cc}
0 & frac{1}{1-a} \
1 & -frac{1}{(a-1)^2} \
2 & frac{-a-1}{2 (a-1)^3} \
3 & frac{-a^2-4 a-1}{6 (a-1)^4} \
4 & frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \
5 & frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6}
end{array}
right)$$ to be compared to the expansion
$$frac{1}{1-a}-frac{x}{(a-1)^2}+frac{(-a-1) x^2}{2 (a-1)^3}+frac{left(-a^2-4
a-1right) x^3}{6 (a-1)^4}+frac{left(-a^3-11 a^2-11 a-1right) x^4}{24
(a-1)^5}+frac{left(-a^4-26 a^3-66 a^2-26 a-1right) x^5}{120
(a-1)^6}+Oleft(x^6right)$$
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Starting from @Aaron Jia's answer
$$frac{1}{e^x-a}=-sum_{n=0}^{infty} left(frac{1}{n!} sum_{k=0}^{infty} frac{k^n}{a^{k+1}} right)x^n$$ let
$$c_n=sum_{k=0}^{infty} frac{k^n}{a^{k+1}}=frac{1}{a},Phi left(frac{1}{a},-n,0right) $$ where appears the Hurwitz-Lerch transcendent function.
So,
$$frac{1}{e^x-a}=sum_{n=0}^{infty} -frac{Phi left(frac{1}{a},-n,0right)}{a, n!} x^n$$ and the coefficients are then
$$left(
begin{array}{cc}
0 & frac{1}{1-a} \
1 & -frac{1}{(a-1)^2} \
2 & frac{-a-1}{2 (a-1)^3} \
3 & frac{-a^2-4 a-1}{6 (a-1)^4} \
4 & frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \
5 & frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6}
end{array}
right)$$ to be compared to the expansion
$$frac{1}{1-a}-frac{x}{(a-1)^2}+frac{(-a-1) x^2}{2 (a-1)^3}+frac{left(-a^2-4
a-1right) x^3}{6 (a-1)^4}+frac{left(-a^3-11 a^2-11 a-1right) x^4}{24
(a-1)^5}+frac{left(-a^4-26 a^3-66 a^2-26 a-1right) x^5}{120
(a-1)^6}+Oleft(x^6right)$$
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
edited Jan 27 at 1:20
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 26 at 9:39
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
add a comment |
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
The first term, n=0, is 0, so the above needs a little correction.
$endgroup$
– JR Sousa
Jan 26 at 20:45
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
@JRS. I do not agree : it is not $0$ so it does not need any little correction. Cheers.
$endgroup$
– Claude Leibovici
Jan 27 at 1:18
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
$begingroup$
You didn't understand what I said, I checked it on Mathematica, your $c_0$ gives the coefficient of $x$, instead of the independent term, your $c_1$ gives the coefficient of $x^2$, and so on. Should be $1/a*phi(1/a,-n-1,0)$. But, whatever, if you wanna stay ignorant, be my guest.
$endgroup$
– JR Sousa
Jan 28 at 16:22
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$begingroup$
If you have trouble Googling it, consider using a calculator. I used Symbolab and found Maclaurin series of $dfrac{1}{e^x-a}$ as:$$dfrac{1}{1-a}-dfrac{1}{1-a^2}x+dfrac{a+1}{2(1-a)^3}x^2-+cdots$$
$endgroup$
– Paras Khosla
Jan 26 at 7:41
$begingroup$
@ParasKhosla What is general term? I can easily produce the series with Mathematica, but I need to know the general term, not just the first few terms.
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– JR Sousa
Jan 26 at 7:52