Root of an quadratic equation
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I have the following quadratic equation :
$m^2 + m(p-1/l) - (Omega_x^2 + Omega_y^2)=0$
I would like to get the solution in terms of $Omega_x, Omega_y$ with some approximations i.e. neglecting $(p-1/l)$ term.
Is it possible to express $mapproxOmega_x + Omega_y$ ? or any other form. Since I do not want roots in my approximated solution.
Thanks
roots quadratics approximation
asked Nov 26 at 10:34
newstudent
226
add a comment |
up vote
0
down vote
favorite
I have the following quadratic equation :
$m^2 + m(p-1/l) - (Omega_x^2 + Omega_y^2)=0$
I would like to get the solution in terms of $Omega_x, Omega_y$ with some approximations i.e. neglecting $(p-1/l)$ term.
Is it possible to express $mapproxOmega_x + Omega_y$ ? or any other form. Since I do not want roots in my approximated solution.
Thanks
roots quadratics approximation
asked Nov 26 at 10:34
newstudent
226
Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following quadratic equation :
$m^2 + m(p-1/l) - (Omega_x^2 + Omega_y^2)=0$
I would like to get the solution in terms of $Omega_x, Omega_y$ with some approximations i.e. neglecting $(p-1/l)$ term.
Is it possible to express $mapproxOmega_x + Omega_y$ ? or any other form. Since I do not want roots in my approximated solution.
Thanks
roots quadratics approximation
asked Nov 26 at 10:34
newstudent
226
I have the following quadratic equation :
$m^2 + m(p-1/l) - (Omega_x^2 + Omega_y^2)=0$
I would like to get the solution in terms of $Omega_x, Omega_y$ with some approximations i.e. neglecting $(p-1/l)$ term.
Is it possible to express $mapproxOmega_x + Omega_y$ ? or any other form. Since I do not want roots in my approximated solution.
Thanks
roots quadratics approximation
roots quadratics approximation
asked Nov 26 at 10:34
newstudent
226
asked Nov 26 at 10:34
newstudent
226
edited Nov 26 at 10:53
asked Nov 26 at 10:34
newstudent
226
asked Nov 26 at 10:34
newstudent
226
asked Nov 26 at 10:34
newstudent
226
226
Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51
add a comment |
Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51
Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51
add a comment |
1 Answer
1
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oldest
votes
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1
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accepted
You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l approx 0$ and $Omega_x > Omega_y$ then
$$
mapprox (Omega_x^2 + Omega_y^2)^{1/2} = Omega_xleft[1 + left(frac{Omega_y}{Omega_x}right)^2 right]^{1/2} approx Omega_x left[1 + frac{1}{2}left(frac{Omega_y}{Omega_x}right)^2 - frac{1}{8}left(frac{Omega_y}{Omega_x}right)^4 + cdots right]
$$
answered Nov 26 at 10:51
caverac
12.4k21027
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l approx 0$ and $Omega_x > Omega_y$ then
$$
mapprox (Omega_x^2 + Omega_y^2)^{1/2} = Omega_xleft[1 + left(frac{Omega_y}{Omega_x}right)^2 right]^{1/2} approx Omega_x left[1 + frac{1}{2}left(frac{Omega_y}{Omega_x}right)^2 - frac{1}{8}left(frac{Omega_y}{Omega_x}right)^4 + cdots right]
$$
answered Nov 26 at 10:51
caverac
12.4k21027
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
add a comment |
up vote
1
down vote
accepted
You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l approx 0$ and $Omega_x > Omega_y$ then
$$
mapprox (Omega_x^2 + Omega_y^2)^{1/2} = Omega_xleft[1 + left(frac{Omega_y}{Omega_x}right)^2 right]^{1/2} approx Omega_x left[1 + frac{1}{2}left(frac{Omega_y}{Omega_x}right)^2 - frac{1}{8}left(frac{Omega_y}{Omega_x}right)^4 + cdots right]
$$
answered Nov 26 at 10:51
caverac
12.4k21027
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l approx 0$ and $Omega_x > Omega_y$ then
$$
mapprox (Omega_x^2 + Omega_y^2)^{1/2} = Omega_xleft[1 + left(frac{Omega_y}{Omega_x}right)^2 right]^{1/2} approx Omega_x left[1 + frac{1}{2}left(frac{Omega_y}{Omega_x}right)^2 - frac{1}{8}left(frac{Omega_y}{Omega_x}right)^4 + cdots right]
$$
answered Nov 26 at 10:51
caverac
12.4k21027
You can use a binomial approximation for the square root, otherwise there's not way around it. If $p- 1 / l approx 0$ and $Omega_x > Omega_y$ then
$$
mapprox (Omega_x^2 + Omega_y^2)^{1/2} = Omega_xleft[1 + left(frac{Omega_y}{Omega_x}right)^2 right]^{1/2} approx Omega_x left[1 + frac{1}{2}left(frac{Omega_y}{Omega_x}right)^2 - frac{1}{8}left(frac{Omega_y}{Omega_x}right)^4 + cdots right]
$$
answered Nov 26 at 10:51
caverac
12.4k21027
answered Nov 26 at 10:51
caverac
12.4k21027
answered Nov 26 at 10:51
caverac
12.4k21027
answered Nov 26 at 10:51
caverac
12.4k21027
12.4k21027
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
add a comment |
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
Thanks. I think this is what I was looking for.
– newstudent
Nov 26 at 10:54
@newstudent Happy to help
– caverac
Nov 26 at 10:55
@newstudent Happy to help
– caverac
Nov 26 at 10:55
add a comment |
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Why you don't just use the Quadratic formula? en.wikipedia.org/wiki/Quadratic_formula
– Stockfish
Nov 26 at 10:41
if you neglect the $m(p−1/l)$ term then your equation becomes $m^2 -(Omega_x^2 + Omega_y^2)=0$. Thus $$m = pmsqrt(Omega_x^2 + Omega_y^2)$$
– TheD0ubleT
Nov 26 at 10:43
@TheD0ubleT. Exactly, but I would like to avoid square root.
– newstudent
Nov 26 at 10:44
@newstudent In that case you can use a Taylor expansion if $Omega_y << Omega_x$ (or the other way around) $$m = Omega_xsqrt(1 + (frac{Omega_y}{Omega_x})^2)approxOmega_x(1+(frac{Omega_y}{2Omega_x}))$$
– TheD0ubleT
Nov 26 at 10:51