Mixing
$u_{i,j}=frac{u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}}{4}$
$begingroup$
Let your living room be the unit square. We denote the coordinates in the square by x and y. The dots are at locations $(x_i,y_j)$ (see the image here: https://imgur.com/a/IPbyuzp). We denote the temperature in $(x_i ,y_j)$ by $u_{i,j}$. The temperature is the average of the temperatures in the four neighboring points:
$$u_{i,j}=frac{u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}}{4} (1).$$
Now assume that $u_{i,j}=u(x_i,y_j)$ with $u(x,y)=x^2-y^2$. Show that $(1)$ holds.
My attempt: The average of the temperatures in the four neighboring points is
$$frac{x_{i-1}^2-y_j^2+x_{i+1}^2-y_j^2+x_i^2-y_{j-1}^2+x_i^2-y_{j+1}^2}{4}.$$ We note that $frac{x_{i-1}+x_{i+1}}{2}=x_i, frac{y_{j-1}+y_{j+1}}{2}=y_j$. Then consider x coordinates first: $x_{i-1}^2+x_{i+1}^2+2x_i^2=(x_{i-1}+x_{i+1})^2-2x_{i-1}x_{i+1}+2x_i^2=4x_i^2-2x_{i-1}x_{i+1}+2x_i^2$. So we get $4x_i^2$, which is what we need, but I don't know how to deal with the rest...
numerical-linear-algebra
asked Jan 27 at 16:51
dxdydzdxdydz
42110
$endgroup$
add a comment |
$begingroup$
Let your living room be the unit square. We denote the coordinates in the square by x and y. The dots are at locations $(x_i,y_j)$ (see the image here: https://imgur.com/a/IPbyuzp). We denote the temperature in $(x_i ,y_j)$ by $u_{i,j}$. The temperature is the average of the temperatures in the four neighboring points:
$$u_{i,j}=frac{u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}}{4} (1).$$
Now assume that $u_{i,j}=u(x_i,y_j)$ with $u(x,y)=x^2-y^2$. Show that $(1)$ holds.
My attempt: The average of the temperatures in the four neighboring points is
$$frac{x_{i-1}^2-y_j^2+x_{i+1}^2-y_j^2+x_i^2-y_{j-1}^2+x_i^2-y_{j+1}^2}{4}.$$ We note that $frac{x_{i-1}+x_{i+1}}{2}=x_i, frac{y_{j-1}+y_{j+1}}{2}=y_j$. Then consider x coordinates first: $x_{i-1}^2+x_{i+1}^2+2x_i^2=(x_{i-1}+x_{i+1})^2-2x_{i-1}x_{i+1}+2x_i^2=4x_i^2-2x_{i-1}x_{i+1}+2x_i^2$. So we get $4x_i^2$, which is what we need, but I don't know how to deal with the rest...
numerical-linear-algebra
asked Jan 27 at 16:51
dxdydzdxdydz
42110
$endgroup$
1
$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12
add a comment |
$begingroup$
Let your living room be the unit square. We denote the coordinates in the square by x and y. The dots are at locations $(x_i,y_j)$ (see the image here: https://imgur.com/a/IPbyuzp). We denote the temperature in $(x_i ,y_j)$ by $u_{i,j}$. The temperature is the average of the temperatures in the four neighboring points:
$$u_{i,j}=frac{u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}}{4} (1).$$
Now assume that $u_{i,j}=u(x_i,y_j)$ with $u(x,y)=x^2-y^2$. Show that $(1)$ holds.
My attempt: The average of the temperatures in the four neighboring points is
$$frac{x_{i-1}^2-y_j^2+x_{i+1}^2-y_j^2+x_i^2-y_{j-1}^2+x_i^2-y_{j+1}^2}{4}.$$ We note that $frac{x_{i-1}+x_{i+1}}{2}=x_i, frac{y_{j-1}+y_{j+1}}{2}=y_j$. Then consider x coordinates first: $x_{i-1}^2+x_{i+1}^2+2x_i^2=(x_{i-1}+x_{i+1})^2-2x_{i-1}x_{i+1}+2x_i^2=4x_i^2-2x_{i-1}x_{i+1}+2x_i^2$. So we get $4x_i^2$, which is what we need, but I don't know how to deal with the rest...
numerical-linear-algebra
asked Jan 27 at 16:51
dxdydzdxdydz
42110
$endgroup$
Let your living room be the unit square. We denote the coordinates in the square by x and y. The dots are at locations $(x_i,y_j)$ (see the image here: https://imgur.com/a/IPbyuzp). We denote the temperature in $(x_i ,y_j)$ by $u_{i,j}$. The temperature is the average of the temperatures in the four neighboring points:
$$u_{i,j}=frac{u_{i-1,j}+u_{i+1,j}+u_{i,j-1}+u_{i,j+1}}{4} (1).$$
Now assume that $u_{i,j}=u(x_i,y_j)$ with $u(x,y)=x^2-y^2$. Show that $(1)$ holds.
My attempt: The average of the temperatures in the four neighboring points is
$$frac{x_{i-1}^2-y_j^2+x_{i+1}^2-y_j^2+x_i^2-y_{j-1}^2+x_i^2-y_{j+1}^2}{4}.$$ We note that $frac{x_{i-1}+x_{i+1}}{2}=x_i, frac{y_{j-1}+y_{j+1}}{2}=y_j$. Then consider x coordinates first: $x_{i-1}^2+x_{i+1}^2+2x_i^2=(x_{i-1}+x_{i+1})^2-2x_{i-1}x_{i+1}+2x_i^2=4x_i^2-2x_{i-1}x_{i+1}+2x_i^2$. So we get $4x_i^2$, which is what we need, but I don't know how to deal with the rest...
numerical-linear-algebra
numerical-linear-algebra
asked Jan 27 at 16:51
dxdydzdxdydz
42110
asked Jan 27 at 16:51
dxdydzdxdydz
42110
edited Jan 28 at 0:13
dxdydz
asked Jan 27 at 16:51
dxdydzdxdydz
42110
asked Jan 27 at 16:51
dxdydzdxdydz
42110
asked Jan 27 at 16:51
dxdydzdxdydz
42110
42110
1
$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12
add a comment |
1
$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12
1
1
$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12
$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12
add a comment |
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$begingroup$
Hint: Get rid of redundant variables. If $d$ is the unit length of the grid, and $left(a, bright)$ is the lower left corner of the room, then $x_i = a + id$ for all $i$, and $y_j = b + jd$ for all $j$. Now substitute these into the left and right hand sides of the equation.
$endgroup$
– darij grinberg
Jan 28 at 1:12