Binomial series expansion of a trinomial? The Next CEO of Stack OverflowSolving an equation arising from method of image chargesElectrostatic Potential Energy integral in spherical coordinatesBinomial Expansion without inifinty seriesOver an integral arising from Kepler's problem [also: generally useful integral, NOT DUPLICATE!]Integral of solid angle of closed surface from the exteriorBinomial Series Expansionlimit of integration is a vector, but I can rewrite the integrand as a scalar. What do I do?Need help understanding a particular proof regarding integration to find the net electric fieldExpansion question involving Taylor or Binomial series?Analytical solutions to integrals
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Binomial series expansion of a trinomial?
The Next CEO of Stack OverflowSolving an equation arising from method of image chargesElectrostatic Potential Energy integral in spherical coordinatesBinomial Expansion without inifinty seriesOver an integral arising from Kepler's problem [also: generally useful integral, NOT DUPLICATE!]Integral of solid angle of closed surface from the exteriorBinomial Series Expansionlimit of integration is a vector, but I can rewrite the integrand as a scalar. What do I do?Need help understanding a particular proof regarding integration to find the net electric fieldExpansion question involving Taylor or Binomial series?Analytical solutions to integrals
$begingroup$
In electrostatics, the potential of a charge $q$ placed on the $z$-axis at $z=a$ is
beginequation
phi=frac14pi epsilon_0fracqr_1
endequation
where $r_1$ is the distance from the observation point to the charge.
Using law of cosines we get
beginequation
phi=fracq4pi epsilon_0 (r^2+a^2-2artext textcostheta)^-frac12
endequation
That will be Eq.(12.2).
Arkfen's Mathematical Methods for Physicist$^1$ states:
Consider the case of $r > a$ or, more precisely, $r^2 > |a^2 − 2artext cos θ|$. The radical in
Eq. (12.2) may be expanded in a binomial series and then rearranged in powers of ($a/r$).
The expression is a trinomial: $((fracar)^2+1-frac2artextcostheta)^-frac12$.
How can it be expanded through a binomial series? I can't find any source making this expansion explicitly, instead jumping directly into Legendre polynomials.
[1] George Arkfen and Hans Weber. Mathematical Methods for Physicist (6th ed. pp. 742)
taylor-expansion binomial-coefficients physics binomial-theorem electromagnetism
edited Mar 20 at 2:09
Andrews
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
$endgroup$
add a comment |
$begingroup$
In electrostatics, the potential of a charge $q$ placed on the $z$-axis at $z=a$ is
beginequation
phi=frac14pi epsilon_0fracqr_1
endequation
where $r_1$ is the distance from the observation point to the charge.
Using law of cosines we get
beginequation
phi=fracq4pi epsilon_0 (r^2+a^2-2artext textcostheta)^-frac12
endequation
That will be Eq.(12.2).
Arkfen's Mathematical Methods for Physicist$^1$ states:
Consider the case of $r > a$ or, more precisely, $r^2 > |a^2 − 2artext cos θ|$. The radical in
Eq. (12.2) may be expanded in a binomial series and then rearranged in powers of ($a/r$).
The expression is a trinomial: $((fracar)^2+1-frac2artextcostheta)^-frac12$.
How can it be expanded through a binomial series? I can't find any source making this expansion explicitly, instead jumping directly into Legendre polynomials.
[1] George Arkfen and Hans Weber. Mathematical Methods for Physicist (6th ed. pp. 742)
taylor-expansion binomial-coefficients physics binomial-theorem electromagnetism
edited Mar 20 at 2:09
Andrews
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
$endgroup$
1
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51
add a comment |
$begingroup$
In electrostatics, the potential of a charge $q$ placed on the $z$-axis at $z=a$ is
beginequation
phi=frac14pi epsilon_0fracqr_1
endequation
where $r_1$ is the distance from the observation point to the charge.
Using law of cosines we get
beginequation
phi=fracq4pi epsilon_0 (r^2+a^2-2artext textcostheta)^-frac12
endequation
That will be Eq.(12.2).
Arkfen's Mathematical Methods for Physicist$^1$ states:
Consider the case of $r > a$ or, more precisely, $r^2 > |a^2 − 2artext cos θ|$. The radical in
Eq. (12.2) may be expanded in a binomial series and then rearranged in powers of ($a/r$).
The expression is a trinomial: $((fracar)^2+1-frac2artextcostheta)^-frac12$.
How can it be expanded through a binomial series? I can't find any source making this expansion explicitly, instead jumping directly into Legendre polynomials.
[1] George Arkfen and Hans Weber. Mathematical Methods for Physicist (6th ed. pp. 742)
taylor-expansion binomial-coefficients physics binomial-theorem electromagnetism
edited Mar 20 at 2:09
Andrews
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
$endgroup$
In electrostatics, the potential of a charge $q$ placed on the $z$-axis at $z=a$ is
beginequation
phi=frac14pi epsilon_0fracqr_1
endequation
where $r_1$ is the distance from the observation point to the charge.
Using law of cosines we get
beginequation
phi=fracq4pi epsilon_0 (r^2+a^2-2artext textcostheta)^-frac12
endequation
That will be Eq.(12.2).
Arkfen's Mathematical Methods for Physicist$^1$ states:
Consider the case of $r > a$ or, more precisely, $r^2 > |a^2 − 2artext cos θ|$. The radical in
Eq. (12.2) may be expanded in a binomial series and then rearranged in powers of ($a/r$).
The expression is a trinomial: $((fracar)^2+1-frac2artextcostheta)^-frac12$.
How can it be expanded through a binomial series? I can't find any source making this expansion explicitly, instead jumping directly into Legendre polynomials.
[1] George Arkfen and Hans Weber. Mathematical Methods for Physicist (6th ed. pp. 742)
taylor-expansion binomial-coefficients physics binomial-theorem electromagnetism
taylor-expansion binomial-coefficients physics binomial-theorem electromagnetism
edited Mar 20 at 2:09
Andrews
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
edited Mar 20 at 2:09
Andrews
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
edited Mar 20 at 2:09
Andrews
1,2812422
edited Mar 20 at 2:09
Andrews
1,2812422
edited Mar 20 at 2:09
Andrews
1,2812422
1,2812422
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
asked Mar 18 at 19:41
IchVerlorenIchVerloren
21310
21310
1
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51
add a comment |
1
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51
1
1
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The multinomial theorem, for the case $n=3$, says that
$$(x_1 + x_2 + x_3)^n =
sum_k_1+k_2+k_3 = n fracn!k_1!k_2!k_3! x_1^k_1 x_2^k_2 x_3^k_3
$$
This generalizes to possibly non-integer real exponents $r$ by writing $r! = Gamma(r+1)$ and
$$
sum_k_1,k_2,k_3 in Bbb Z fracr!k_1!k_2!k_3!(r-k_1-k_2-k_3)! x_1^k_1 x_2^k_2 x_3^k_3
$$
(Notice that in the case of positive integer $r$ this series ends because a $(-1)!$ starts appearing in the denominator, and that is infinite leading to the terms being zero.)
In this case, $(-frac12)! = Gamma(frac12)$ is $sqrtpi$ and the factors of $sqrtpi$ will cancel.
For the expression at hand, the lowest order terms in an expansion for $a << r$ are:
$$
left[ 1+ frac2arcostheta + left(fracar right)^2 right]^-frac12=
1 - fraca2r + left( frac38-costhetaright)fraca^2r^2 -
left( frac516-frac32costhetaright)fraca^3r^3+ cdots
$$
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
$endgroup$
add a comment |
$begingroup$
$$frac1sqrt1+x sim 1-frac12x, x<<1.$$
So,
$$frac1sqrt1+(a/r)(a/r+2cos(theta)) sim 1-frac12fracar(fracar+2cos(theta)) = 1-cos(theta)fracar-frac12(fracar)^2.$$
answered Mar 18 at 19:55
D.B.D.B.
1,34019
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The multinomial theorem, for the case $n=3$, says that
$$(x_1 + x_2 + x_3)^n =
sum_k_1+k_2+k_3 = n fracn!k_1!k_2!k_3! x_1^k_1 x_2^k_2 x_3^k_3
$$
This generalizes to possibly non-integer real exponents $r$ by writing $r! = Gamma(r+1)$ and
$$
sum_k_1,k_2,k_3 in Bbb Z fracr!k_1!k_2!k_3!(r-k_1-k_2-k_3)! x_1^k_1 x_2^k_2 x_3^k_3
$$
(Notice that in the case of positive integer $r$ this series ends because a $(-1)!$ starts appearing in the denominator, and that is infinite leading to the terms being zero.)
In this case, $(-frac12)! = Gamma(frac12)$ is $sqrtpi$ and the factors of $sqrtpi$ will cancel.
For the expression at hand, the lowest order terms in an expansion for $a << r$ are:
$$
left[ 1+ frac2arcostheta + left(fracar right)^2 right]^-frac12=
1 - fraca2r + left( frac38-costhetaright)fraca^2r^2 -
left( frac516-frac32costhetaright)fraca^3r^3+ cdots
$$
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
$endgroup$
add a comment |
$begingroup$
The multinomial theorem, for the case $n=3$, says that
$$(x_1 + x_2 + x_3)^n =
sum_k_1+k_2+k_3 = n fracn!k_1!k_2!k_3! x_1^k_1 x_2^k_2 x_3^k_3
$$
This generalizes to possibly non-integer real exponents $r$ by writing $r! = Gamma(r+1)$ and
$$
sum_k_1,k_2,k_3 in Bbb Z fracr!k_1!k_2!k_3!(r-k_1-k_2-k_3)! x_1^k_1 x_2^k_2 x_3^k_3
$$
(Notice that in the case of positive integer $r$ this series ends because a $(-1)!$ starts appearing in the denominator, and that is infinite leading to the terms being zero.)
In this case, $(-frac12)! = Gamma(frac12)$ is $sqrtpi$ and the factors of $sqrtpi$ will cancel.
For the expression at hand, the lowest order terms in an expansion for $a << r$ are:
$$
left[ 1+ frac2arcostheta + left(fracar right)^2 right]^-frac12=
1 - fraca2r + left( frac38-costhetaright)fraca^2r^2 -
left( frac516-frac32costhetaright)fraca^3r^3+ cdots
$$
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
$endgroup$
add a comment |
$begingroup$
The multinomial theorem, for the case $n=3$, says that
$$(x_1 + x_2 + x_3)^n =
sum_k_1+k_2+k_3 = n fracn!k_1!k_2!k_3! x_1^k_1 x_2^k_2 x_3^k_3
$$
This generalizes to possibly non-integer real exponents $r$ by writing $r! = Gamma(r+1)$ and
$$
sum_k_1,k_2,k_3 in Bbb Z fracr!k_1!k_2!k_3!(r-k_1-k_2-k_3)! x_1^k_1 x_2^k_2 x_3^k_3
$$
(Notice that in the case of positive integer $r$ this series ends because a $(-1)!$ starts appearing in the denominator, and that is infinite leading to the terms being zero.)
In this case, $(-frac12)! = Gamma(frac12)$ is $sqrtpi$ and the factors of $sqrtpi$ will cancel.
For the expression at hand, the lowest order terms in an expansion for $a << r$ are:
$$
left[ 1+ frac2arcostheta + left(fracar right)^2 right]^-frac12=
1 - fraca2r + left( frac38-costhetaright)fraca^2r^2 -
left( frac516-frac32costhetaright)fraca^3r^3+ cdots
$$
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
$endgroup$
The multinomial theorem, for the case $n=3$, says that
$$(x_1 + x_2 + x_3)^n =
sum_k_1+k_2+k_3 = n fracn!k_1!k_2!k_3! x_1^k_1 x_2^k_2 x_3^k_3
$$
This generalizes to possibly non-integer real exponents $r$ by writing $r! = Gamma(r+1)$ and
$$
sum_k_1,k_2,k_3 in Bbb Z fracr!k_1!k_2!k_3!(r-k_1-k_2-k_3)! x_1^k_1 x_2^k_2 x_3^k_3
$$
(Notice that in the case of positive integer $r$ this series ends because a $(-1)!$ starts appearing in the denominator, and that is infinite leading to the terms being zero.)
In this case, $(-frac12)! = Gamma(frac12)$ is $sqrtpi$ and the factors of $sqrtpi$ will cancel.
For the expression at hand, the lowest order terms in an expansion for $a << r$ are:
$$
left[ 1+ frac2arcostheta + left(fracar right)^2 right]^-frac12=
1 - fraca2r + left( frac38-costhetaright)fraca^2r^2 -
left( frac516-frac32costhetaright)fraca^3r^3+ cdots
$$
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
edited Mar 18 at 20:25
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
answered Mar 18 at 19:56
Mark FischlerMark Fischler
33.8k12552
33.8k12552
add a comment |
add a comment |
$begingroup$
$$frac1sqrt1+x sim 1-frac12x, x<<1.$$
So,
$$frac1sqrt1+(a/r)(a/r+2cos(theta)) sim 1-frac12fracar(fracar+2cos(theta)) = 1-cos(theta)fracar-frac12(fracar)^2.$$
answered Mar 18 at 19:55
D.B.D.B.
1,34019
$endgroup$
add a comment |
$begingroup$
$$frac1sqrt1+x sim 1-frac12x, x<<1.$$
So,
$$frac1sqrt1+(a/r)(a/r+2cos(theta)) sim 1-frac12fracar(fracar+2cos(theta)) = 1-cos(theta)fracar-frac12(fracar)^2.$$
answered Mar 18 at 19:55
D.B.D.B.
1,34019
$endgroup$
add a comment |
$begingroup$
$$frac1sqrt1+x sim 1-frac12x, x<<1.$$
So,
$$frac1sqrt1+(a/r)(a/r+2cos(theta)) sim 1-frac12fracar(fracar+2cos(theta)) = 1-cos(theta)fracar-frac12(fracar)^2.$$
answered Mar 18 at 19:55
D.B.D.B.
1,34019
$endgroup$
$$frac1sqrt1+x sim 1-frac12x, x<<1.$$
So,
$$frac1sqrt1+(a/r)(a/r+2cos(theta)) sim 1-frac12fracar(fracar+2cos(theta)) = 1-cos(theta)fracar-frac12(fracar)^2.$$
answered Mar 18 at 19:55
D.B.D.B.
1,34019
edited Mar 18 at 20:01
answered Mar 18 at 19:55
D.B.D.B.
1,34019
answered Mar 18 at 19:55
D.B.D.B.
1,34019
answered Mar 18 at 19:55
D.B.D.B.
1,34019
1,34019
add a comment |
add a comment |
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1
$begingroup$
Treat the $1$ and the $(a/r)^2+2(a/r)cos(theta)$ as the two terms in the binomial. You can always expand out the polynomial you get.
$endgroup$
– D.B.
Mar 18 at 19:51