Monte Carlo double integral over surface of $|x|+|y| leq 1$
$begingroup$
$iint_{|x|+|y|le1}!x^2,dxdy$
I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.
monte-carlo
edited Dec 9 '18 at 20:01
postmortes
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
$endgroup$
add a comment |
$begingroup$
$iint_{|x|+|y|le1}!x^2,dxdy$
I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.
monte-carlo
edited Dec 9 '18 at 20:01
postmortes
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
$endgroup$
add a comment |
$begingroup$
$iint_{|x|+|y|le1}!x^2,dxdy$
I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.
monte-carlo
edited Dec 9 '18 at 20:01
postmortes
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
$endgroup$
$iint_{|x|+|y|le1}!x^2,dxdy$
I am supposed to calculate this by using Monte Carlo integration. Can anyone give basic hints or directions? I know the idea behind the Monte Carlo integration method but my brain can't seem to be able to grasp any straw at how to solve this.
monte-carlo
monte-carlo
edited Dec 9 '18 at 20:01
postmortes
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
edited Dec 9 '18 at 20:01
postmortes
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
edited Dec 9 '18 at 20:01
postmortes
1,91721117
edited Dec 9 '18 at 20:01
postmortes
1,91721117
edited Dec 9 '18 at 20:01
postmortes
1,91721117
1,91721117
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
asked Dec 9 '18 at 19:40
That guy who is bad at mathThat guy who is bad at math
134
134
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
begin{align}
S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
left[vphantom{Large A}leftvert x_{i}rightvert +
leftvert y_{i}rightvert leq 1right]
\[2mm]
overline{S_{N}} & = N overline{x^{2}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]} =
Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y
\[5mm]
& implies int_{-1}^{1}int_{-1}^{1}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
approx bbox[10px,#ffd,border:1px groove navy]
{4,{S_{N} over N}}
end{align}
The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:
"use strict";
const ITERATIONS = 10000;
let i = 0, theSum = 0, x = null, y = null;
while (i < ITERATIONS) {
x = 2.0*Math.random() - 1.0;
y = 2.0*Math.random() - 1.0;
if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
++i;
}
console.log(4.0*(theSum/ITERATIONS));
A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
$bbox[10px,#ffd,border:1px groove navy]
{displaystyle{1 over 3}}$.
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
begin{align}
S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
left[vphantom{Large A}leftvert x_{i}rightvert +
leftvert y_{i}rightvert leq 1right]
\[2mm]
overline{S_{N}} & = N overline{x^{2}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]} =
Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y
\[5mm]
& implies int_{-1}^{1}int_{-1}^{1}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
approx bbox[10px,#ffd,border:1px groove navy]
{4,{S_{N} over N}}
end{align}
The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:
"use strict";
const ITERATIONS = 10000;
let i = 0, theSum = 0, x = null, y = null;
while (i < ITERATIONS) {
x = 2.0*Math.random() - 1.0;
y = 2.0*Math.random() - 1.0;
if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
++i;
}
console.log(4.0*(theSum/ITERATIONS));
A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
$bbox[10px,#ffd,border:1px groove navy]
{displaystyle{1 over 3}}$.
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
begin{align}
S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
left[vphantom{Large A}leftvert x_{i}rightvert +
leftvert y_{i}rightvert leq 1right]
\[2mm]
overline{S_{N}} & = N overline{x^{2}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]} =
Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y
\[5mm]
& implies int_{-1}^{1}int_{-1}^{1}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
approx bbox[10px,#ffd,border:1px groove navy]
{4,{S_{N} over N}}
end{align}
The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:
"use strict";
const ITERATIONS = 10000;
let i = 0, theSum = 0, x = null, y = null;
while (i < ITERATIONS) {
x = 2.0*Math.random() - 1.0;
y = 2.0*Math.random() - 1.0;
if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
++i;
}
console.log(4.0*(theSum/ITERATIONS));
A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
$bbox[10px,#ffd,border:1px groove navy]
{displaystyle{1 over 3}}$.
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
begin{align}
S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
left[vphantom{Large A}leftvert x_{i}rightvert +
leftvert y_{i}rightvert leq 1right]
\[2mm]
overline{S_{N}} & = N overline{x^{2}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]} =
Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y
\[5mm]
& implies int_{-1}^{1}int_{-1}^{1}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
approx bbox[10px,#ffd,border:1px groove navy]
{4,{S_{N} over N}}
end{align}
The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:
"use strict";
const ITERATIONS = 10000;
let i = 0, theSum = 0, x = null, y = null;
while (i < ITERATIONS) {
x = 2.0*Math.random() - 1.0;
y = 2.0*Math.random() - 1.0;
if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
++i;
}
console.log(4.0*(theSum/ITERATIONS));
A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
$bbox[10px,#ffd,border:1px groove navy]
{displaystyle{1 over 3}}$.
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
$endgroup$
Enclose your integration in $displaystyle{left[-1,1right)^{,2}}$. The Monte-Carlo integration becomes $left(~overline overline{phantom{AAA}}mbox{means average with an uniform distibution over} left[-1,1right)^{,2}~right)$
begin{align}
S_{N} & = sum_{i = 1}^{N}x_{i}^{2}
left[vphantom{Large A}leftvert x_{i}rightvert +
leftvert y_{i}rightvert leq 1right]
\[2mm]
overline{S_{N}} & = N overline{x^{2}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]} =
Nint_{-1}^{1}int_{-1}^{1}{1 over 4}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y
\[5mm]
& implies int_{-1}^{1}int_{-1}^{1}
left[vphantom{Large A}leftvert xrightvert +
leftvert yrightvert leq 1right]x^{2}
,mathrm{d}x,mathrm{d}y = 4,{overline{S_{N}} over N}
approx bbox[10px,#ffd,border:1px groove navy]
{4,{S_{N} over N}}
end{align}
The following code is a $texttt{javascript}$ script which can be run in a terminal with $texttt{node.js}$:
"use strict";
const ITERATIONS = 10000;
let i = 0, theSum = 0, x = null, y = null;
while (i < ITERATIONS) {
x = 2.0*Math.random() - 1.0;
y = 2.0*Math.random() - 1.0;
if (Math.abs(x) + Math.abs(y) <= 1.0) theSum += x*x;
++i;
}
console.log(4.0*(theSum/ITERATIONS));
A typical "run" yields $bbox[10px,#ffd,border:1px groove navy]{displaystyle 0.3302123390009306}$. The exact result is
$bbox[10px,#ffd,border:1px groove navy]
{displaystyle{1 over 3}}$.
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
edited Dec 10 '18 at 18:36
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
answered Dec 9 '18 at 21:56
Felix MarinFelix Marin
67.8k7107142
67.8k7107142
add a comment |
add a comment |
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