Bisection Newton - Quadratures
The problem states the following:
Find with at least 10 digitis of precisión the roots of the following equation:
$int_x^{x^2} !e^{-t^2},mathrm{d}t = x^5 -3x^2 + 1 $ in the closed Interval [-1,1].
We know that we should use the Gaussian Method of Numerical Integration and the Bisection Method.
However we planned on using the Legendre-Gauss Quadrature, to first approximate the value of the integral, then use the bisection method to find the roots of the equation $int_x^{x^2} !e^{-t^2},mathrm{d}t - x^5 + 3x^2 - 1 = 0 $
Is it an accurate approach? If so, when finding the weights and points to use in the Legendre-Gauss Quadrature we note that the Interval of the integral goes from $x$ to $x^2$. Does the LGQ method work still?
Thank you.
numerical-methods gaussian-integral bisection quadrature
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The problem states the following:
Find with at least 10 digitis of precisión the roots of the following equation:
$int_x^{x^2} !e^{-t^2},mathrm{d}t = x^5 -3x^2 + 1 $ in the closed Interval [-1,1].
We know that we should use the Gaussian Method of Numerical Integration and the Bisection Method.
However we planned on using the Legendre-Gauss Quadrature, to first approximate the value of the integral, then use the bisection method to find the roots of the equation $int_x^{x^2} !e^{-t^2},mathrm{d}t - x^5 + 3x^2 - 1 = 0 $
Is it an accurate approach? If so, when finding the weights and points to use in the Legendre-Gauss Quadrature we note that the Interval of the integral goes from $x$ to $x^2$. Does the LGQ method work still?
Thank you.
numerical-methods gaussian-integral bisection quadrature
add a comment |
The problem states the following:
Find with at least 10 digitis of precisión the roots of the following equation:
$int_x^{x^2} !e^{-t^2},mathrm{d}t = x^5 -3x^2 + 1 $ in the closed Interval [-1,1].
We know that we should use the Gaussian Method of Numerical Integration and the Bisection Method.
However we planned on using the Legendre-Gauss Quadrature, to first approximate the value of the integral, then use the bisection method to find the roots of the equation $int_x^{x^2} !e^{-t^2},mathrm{d}t - x^5 + 3x^2 - 1 = 0 $
Is it an accurate approach? If so, when finding the weights and points to use in the Legendre-Gauss Quadrature we note that the Interval of the integral goes from $x$ to $x^2$. Does the LGQ method work still?
Thank you.
numerical-methods gaussian-integral bisection quadrature
The problem states the following:
Find with at least 10 digitis of precisión the roots of the following equation:
$int_x^{x^2} !e^{-t^2},mathrm{d}t = x^5 -3x^2 + 1 $ in the closed Interval [-1,1].
We know that we should use the Gaussian Method of Numerical Integration and the Bisection Method.
However we planned on using the Legendre-Gauss Quadrature, to first approximate the value of the integral, then use the bisection method to find the roots of the equation $int_x^{x^2} !e^{-t^2},mathrm{d}t - x^5 + 3x^2 - 1 = 0 $
Is it an accurate approach? If so, when finding the weights and points to use in the Legendre-Gauss Quadrature we note that the Interval of the integral goes from $x$ to $x^2$. Does the LGQ method work still?
Thank you.
numerical-methods gaussian-integral bisection quadrature
numerical-methods gaussian-integral bisection quadrature
asked Nov 29 '18 at 20:30
user509468
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1 Answer
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It is still possible to use the quadrature method, call
$$
t = frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)
$$
so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes
$$
int_x^{x^2} e^{-t^2}{rm d}t = frac{x}{2}(x - 1)int_{-1}^1 expleft[- left(frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)right)^2 right]{rm d}u
$$
That being said, it is actually not necessary to calculate the integral at all. Remember that
$$
{rm erf}(x) = frac{2}{sqrt{pi}} int_0^x e^{-t^2}{rm d}t
$$
So that the problem reduces to
$$
frac{sqrt{pi}}{2}[{rm erf}(x^2) - {rm erf}(x)] = x^5 - 3x^2 + 1
$$
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is still possible to use the quadrature method, call
$$
t = frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)
$$
so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes
$$
int_x^{x^2} e^{-t^2}{rm d}t = frac{x}{2}(x - 1)int_{-1}^1 expleft[- left(frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)right)^2 right]{rm d}u
$$
That being said, it is actually not necessary to calculate the integral at all. Remember that
$$
{rm erf}(x) = frac{2}{sqrt{pi}} int_0^x e^{-t^2}{rm d}t
$$
So that the problem reduces to
$$
frac{sqrt{pi}}{2}[{rm erf}(x^2) - {rm erf}(x)] = x^5 - 3x^2 + 1
$$
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
add a comment |
It is still possible to use the quadrature method, call
$$
t = frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)
$$
so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes
$$
int_x^{x^2} e^{-t^2}{rm d}t = frac{x}{2}(x - 1)int_{-1}^1 expleft[- left(frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)right)^2 right]{rm d}u
$$
That being said, it is actually not necessary to calculate the integral at all. Remember that
$$
{rm erf}(x) = frac{2}{sqrt{pi}} int_0^x e^{-t^2}{rm d}t
$$
So that the problem reduces to
$$
frac{sqrt{pi}}{2}[{rm erf}(x^2) - {rm erf}(x)] = x^5 - 3x^2 + 1
$$
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
add a comment |
It is still possible to use the quadrature method, call
$$
t = frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)
$$
so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes
$$
int_x^{x^2} e^{-t^2}{rm d}t = frac{x}{2}(x - 1)int_{-1}^1 expleft[- left(frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)right)^2 right]{rm d}u
$$
That being said, it is actually not necessary to calculate the integral at all. Remember that
$$
{rm erf}(x) = frac{2}{sqrt{pi}} int_0^x e^{-t^2}{rm d}t
$$
So that the problem reduces to
$$
frac{sqrt{pi}}{2}[{rm erf}(x^2) - {rm erf}(x)] = x^5 - 3x^2 + 1
$$
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
It is still possible to use the quadrature method, call
$$
t = frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)
$$
so that $t(u = -1) = x$ and $t(u = +1) = x^2$, the integral then becomes
$$
int_x^{x^2} e^{-t^2}{rm d}t = frac{x}{2}(x - 1)int_{-1}^1 expleft[- left(frac{xu}{2}(x - 1) + frac{x}{2}(x + 1)right)^2 right]{rm d}u
$$
That being said, it is actually not necessary to calculate the integral at all. Remember that
$$
{rm erf}(x) = frac{2}{sqrt{pi}} int_0^x e^{-t^2}{rm d}t
$$
So that the problem reduces to
$$
frac{sqrt{pi}}{2}[{rm erf}(x^2) - {rm erf}(x)] = x^5 - 3x^2 + 1
$$
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
answered Nov 29 '18 at 21:04
caveraccaverac
14k21130
14k21130
add a comment |
add a comment |
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