Taylor series expansion of $frac{1}{sqrt{1-beta x(x+1)}}$
$begingroup$
I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x rightarrow frac{1}{sqrt{1-beta x(x+1)}} text{ with } beta in mathbb{R}^{+*}$$
I've tried using the taylor series expansion of $$frac{1}{sqrt{1-X}} = sum_{n=0}^{infty}4^{-n}{2n choose n}X^n text{ }text{ }text{ }text{ with } text{ } X=beta x(x+1)$$
But I can't turn it into a power series because of the $(x+1)^n$...
I've also tried to derive $$frac{1}{n!}cdotfrac{text{d}^n}{text{d}x^n}left(frac{1}{sqrt{1-beta x(x+1)}}right)_{x=0}$$
But no results so far...
Edit : with a more powerful method, I found that, if we call $left(a_nright)_{ninmathbb{N}}$ the coefficients of the taylor series expansion $left(frac{1}{sqrt{1-beta x(x+1)}}=sum_{n=0}^{infty}a_nx^nright)$, the sequence $left(a_nright)_{ninmathbb{N}}$ is then defined by :
$$forall ngeq3, text{} na_n=betaleft(n-frac{1}{2}right)a_{n-1}+betaleft(n-1right)a_{n-2}$$
$$text{with }text{ }a_1 = frac{beta}{2} text{ , } a_2 = frac{3}{8}beta^2+frac{1}{2}beta$$
It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?
real-analysis calculus sequences-and-series power-series taylor-expansion
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
$endgroup$
add a comment |
$begingroup$
I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x rightarrow frac{1}{sqrt{1-beta x(x+1)}} text{ with } beta in mathbb{R}^{+*}$$
I've tried using the taylor series expansion of $$frac{1}{sqrt{1-X}} = sum_{n=0}^{infty}4^{-n}{2n choose n}X^n text{ }text{ }text{ }text{ with } text{ } X=beta x(x+1)$$
But I can't turn it into a power series because of the $(x+1)^n$...
I've also tried to derive $$frac{1}{n!}cdotfrac{text{d}^n}{text{d}x^n}left(frac{1}{sqrt{1-beta x(x+1)}}right)_{x=0}$$
But no results so far...
Edit : with a more powerful method, I found that, if we call $left(a_nright)_{ninmathbb{N}}$ the coefficients of the taylor series expansion $left(frac{1}{sqrt{1-beta x(x+1)}}=sum_{n=0}^{infty}a_nx^nright)$, the sequence $left(a_nright)_{ninmathbb{N}}$ is then defined by :
$$forall ngeq3, text{} na_n=betaleft(n-frac{1}{2}right)a_{n-1}+betaleft(n-1right)a_{n-2}$$
$$text{with }text{ }a_1 = frac{beta}{2} text{ , } a_2 = frac{3}{8}beta^2+frac{1}{2}beta$$
It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?
real-analysis calculus sequences-and-series power-series taylor-expansion
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
$endgroup$
add a comment |
$begingroup$
I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x rightarrow frac{1}{sqrt{1-beta x(x+1)}} text{ with } beta in mathbb{R}^{+*}$$
I've tried using the taylor series expansion of $$frac{1}{sqrt{1-X}} = sum_{n=0}^{infty}4^{-n}{2n choose n}X^n text{ }text{ }text{ }text{ with } text{ } X=beta x(x+1)$$
But I can't turn it into a power series because of the $(x+1)^n$...
I've also tried to derive $$frac{1}{n!}cdotfrac{text{d}^n}{text{d}x^n}left(frac{1}{sqrt{1-beta x(x+1)}}right)_{x=0}$$
But no results so far...
Edit : with a more powerful method, I found that, if we call $left(a_nright)_{ninmathbb{N}}$ the coefficients of the taylor series expansion $left(frac{1}{sqrt{1-beta x(x+1)}}=sum_{n=0}^{infty}a_nx^nright)$, the sequence $left(a_nright)_{ninmathbb{N}}$ is then defined by :
$$forall ngeq3, text{} na_n=betaleft(n-frac{1}{2}right)a_{n-1}+betaleft(n-1right)a_{n-2}$$
$$text{with }text{ }a_1 = frac{beta}{2} text{ , } a_2 = frac{3}{8}beta^2+frac{1}{2}beta$$
It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?
real-analysis calculus sequences-and-series power-series taylor-expansion
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
$endgroup$
I am trying to find the taylor series expansion about $0$ (maclaurin series) of $$x rightarrow frac{1}{sqrt{1-beta x(x+1)}} text{ with } beta in mathbb{R}^{+*}$$
I've tried using the taylor series expansion of $$frac{1}{sqrt{1-X}} = sum_{n=0}^{infty}4^{-n}{2n choose n}X^n text{ }text{ }text{ }text{ with } text{ } X=beta x(x+1)$$
But I can't turn it into a power series because of the $(x+1)^n$...
I've also tried to derive $$frac{1}{n!}cdotfrac{text{d}^n}{text{d}x^n}left(frac{1}{sqrt{1-beta x(x+1)}}right)_{x=0}$$
But no results so far...
Edit : with a more powerful method, I found that, if we call $left(a_nright)_{ninmathbb{N}}$ the coefficients of the taylor series expansion $left(frac{1}{sqrt{1-beta x(x+1)}}=sum_{n=0}^{infty}a_nx^nright)$, the sequence $left(a_nright)_{ninmathbb{N}}$ is then defined by :
$$forall ngeq3, text{} na_n=betaleft(n-frac{1}{2}right)a_{n-1}+betaleft(n-1right)a_{n-2}$$
$$text{with }text{ }a_1 = frac{beta}{2} text{ , } a_2 = frac{3}{8}beta^2+frac{1}{2}beta$$
It's definitely a step forward, but I don't know how to proceed from there. Is there a way to handle sequences that are defined by such a way ?
real-analysis calculus sequences-and-series power-series taylor-expansion
real-analysis calculus sequences-and-series power-series taylor-expansion
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
edited Dec 14 '18 at 0:10
Harmonic Sun
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
asked Dec 13 '18 at 22:27
Harmonic SunHarmonic Sun
71710
71710
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would set $y = beta x(x+1)$. With that:
begin{align}
frac1{sqrt{1 - y}}
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} y^n\
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} beta^n x^n (x+1)^n\
&= sum_{n=0}^{infty}sum_{k=0}^{n} 4^{-n} binom{2n}{n} binom{n}{k} beta^n x^n x^k \
&= sum_{m=0}^{infty} a_m x^m
end{align}
where $m = n + k$, so $a_m = sum_{n=0}^{m} 4^{-n}binom{2n}{n}binom{n}{m-n} beta^n$.
$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, dots, 3$ the $a_m$ are $1,frac{beta }{2},frac{1}{8} beta (3 beta +4),frac{1}{16} beta ^2 (5
beta +12)$.
Mathematica "simplifies" $a_m$ to
$$
a_m = 4^{-m} binom{2 m}{m} beta ^m ,
_2F_1left(frac{1}{2}-frac{m}{2},-frac{m}{2};frac{1}{2}-m;-frac{4}{beta
}right)
$$
where $_2F_1$ denotes the Hypergeometric function.
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
$endgroup$
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I would set $y = beta x(x+1)$. With that:
begin{align}
frac1{sqrt{1 - y}}
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} y^n\
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} beta^n x^n (x+1)^n\
&= sum_{n=0}^{infty}sum_{k=0}^{n} 4^{-n} binom{2n}{n} binom{n}{k} beta^n x^n x^k \
&= sum_{m=0}^{infty} a_m x^m
end{align}
where $m = n + k$, so $a_m = sum_{n=0}^{m} 4^{-n}binom{2n}{n}binom{n}{m-n} beta^n$.
$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, dots, 3$ the $a_m$ are $1,frac{beta }{2},frac{1}{8} beta (3 beta +4),frac{1}{16} beta ^2 (5
beta +12)$.
Mathematica "simplifies" $a_m$ to
$$
a_m = 4^{-m} binom{2 m}{m} beta ^m ,
_2F_1left(frac{1}{2}-frac{m}{2},-frac{m}{2};frac{1}{2}-m;-frac{4}{beta
}right)
$$
where $_2F_1$ denotes the Hypergeometric function.
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
$endgroup$
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
add a comment |
$begingroup$
I would set $y = beta x(x+1)$. With that:
begin{align}
frac1{sqrt{1 - y}}
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} y^n\
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} beta^n x^n (x+1)^n\
&= sum_{n=0}^{infty}sum_{k=0}^{n} 4^{-n} binom{2n}{n} binom{n}{k} beta^n x^n x^k \
&= sum_{m=0}^{infty} a_m x^m
end{align}
where $m = n + k$, so $a_m = sum_{n=0}^{m} 4^{-n}binom{2n}{n}binom{n}{m-n} beta^n$.
$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, dots, 3$ the $a_m$ are $1,frac{beta }{2},frac{1}{8} beta (3 beta +4),frac{1}{16} beta ^2 (5
beta +12)$.
Mathematica "simplifies" $a_m$ to
$$
a_m = 4^{-m} binom{2 m}{m} beta ^m ,
_2F_1left(frac{1}{2}-frac{m}{2},-frac{m}{2};frac{1}{2}-m;-frac{4}{beta
}right)
$$
where $_2F_1$ denotes the Hypergeometric function.
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
$endgroup$
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
add a comment |
$begingroup$
I would set $y = beta x(x+1)$. With that:
begin{align}
frac1{sqrt{1 - y}}
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} y^n\
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} beta^n x^n (x+1)^n\
&= sum_{n=0}^{infty}sum_{k=0}^{n} 4^{-n} binom{2n}{n} binom{n}{k} beta^n x^n x^k \
&= sum_{m=0}^{infty} a_m x^m
end{align}
where $m = n + k$, so $a_m = sum_{n=0}^{m} 4^{-n}binom{2n}{n}binom{n}{m-n} beta^n$.
$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, dots, 3$ the $a_m$ are $1,frac{beta }{2},frac{1}{8} beta (3 beta +4),frac{1}{16} beta ^2 (5
beta +12)$.
Mathematica "simplifies" $a_m$ to
$$
a_m = 4^{-m} binom{2 m}{m} beta ^m ,
_2F_1left(frac{1}{2}-frac{m}{2},-frac{m}{2};frac{1}{2}-m;-frac{4}{beta
}right)
$$
where $_2F_1$ denotes the Hypergeometric function.
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
$endgroup$
I would set $y = beta x(x+1)$. With that:
begin{align}
frac1{sqrt{1 - y}}
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} y^n\
&= sum_{n=0}^{infty} 4^{-n} binom{2n}{n} beta^n x^n (x+1)^n\
&= sum_{n=0}^{infty}sum_{k=0}^{n} 4^{-n} binom{2n}{n} binom{n}{k} beta^n x^n x^k \
&= sum_{m=0}^{infty} a_m x^m
end{align}
where $m = n + k$, so $a_m = sum_{n=0}^{m} 4^{-n}binom{2n}{n}binom{n}{m-n} beta^n$.
$a_m$ is simply an number which you can calculate doing some sums and products. For $m = 0, dots, 3$ the $a_m$ are $1,frac{beta }{2},frac{1}{8} beta (3 beta +4),frac{1}{16} beta ^2 (5
beta +12)$.
Mathematica "simplifies" $a_m$ to
$$
a_m = 4^{-m} binom{2 m}{m} beta ^m ,
_2F_1left(frac{1}{2}-frac{m}{2},-frac{m}{2};frac{1}{2}-m;-frac{4}{beta
}right)
$$
where $_2F_1$ denotes the Hypergeometric function.
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
edited Dec 14 '18 at 0:30
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
answered Dec 14 '18 at 0:03
0x5390x539
1,450518
1,450518
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
add a comment |
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
Well actually, after numerical simulation, your numerical values don't seem to work...
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:17
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
@HarmonicSun Yes, I had forgotten to include $beta$. The values are not as pretty now but you can still easily calculate them.
$endgroup$
– 0x539
Dec 14 '18 at 0:20
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
Ok thanks a lot ! Additionaly, do you think that based on my edit on my initial question, one can derive a prettier closed form for the $a_m$ ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:26
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
I might be picky but, in your formula for $a_m$, ${n choose m-n}$ is not defined for, say, n=0... Or do we take it a 0 conventionaly ?
$endgroup$
– Harmonic Sun
Dec 14 '18 at 0:33
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
$begingroup$
@HarmonicSun Yes, the Binomial coefficient $binom{a}{b}$ is understood to be zero when $a < b$ (and $a, b in mathbb{Z}$).
$endgroup$
– 0x539
Dec 14 '18 at 0:39
add a comment |
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