Let's calculate the values of the function f ( t , y ) = t − y 2 f\left(t,y\right)=t-y^2 f ( t , y ) = t − y 2 A = { ( t , y ) ∣ − 1 ≤ t ≤ 1 , − 1 ≤ y ≤ 1 } A=\left\{\left(t,y\right)\left|-1\le t\le1,\,\,\,-1\le y\le 1\right.\right\} A = { ( t , y ) ∣ − 1 ≤ t ≤ 1 , − 1 ≤ y ≤ 1 } Δ t = Δ y = 0.5 \Delta t=\Delta y=0.5 Δ t = Δ y = 0.5
t = − 1 : { f ( − 1 , − 1 ) = − 1 − ( − 1 ) 2 = − 2 f ( − 1 , − 0.5 ) = − 1 − ( − 0.5 ) 2 = − 1.25 f ( − 1 , − 0 ) = − 1 − 0 2 = − 1 f ( − 1 , 0.5 ) = − 1 − 0. 5 2 = − 1.25 f ( − 1 , 1 ) = − 1 − 1 2 = − 2 t = − 0.5 : { f ( − 0.5 , − 1 ) = − 0.5 − ( − 1 ) 2 = − 1.5 f ( − 0.5 , − 0.5 ) = − 0.5 − ( − 0.5 ) 2 = − 0.75 f ( − 0.5 , − 0 ) = − 0.5 − 0 2 = − 0.5 f ( − 0.5 , 0.5 ) = − 0.5 − 0. 5 2 = − 0.75 f ( − 0.5 , 1 ) = − 0.5 − 1 2 = − 1.5 t = 0 : { f ( 0 , − 1 ) = 0 − ( − 1 ) 2 = − 1 f ( 0 , − 0.5 ) = 0 − ( − 0.5 ) 2 = − 0.25 f ( 0 , − 0 ) = 0 − 0 2 = − 0 f ( 0 , 0.5 ) = 0 − 0. 5 2 = − 0.25 f ( 0 , 1 ) = 0 − 1 2 = − 1 t = 0.5 : { f ( 0.5 , − 1 ) = 0.5 − ( − 1 ) 2 = − 0.5 f ( 0.5 , − 0.5 ) = 0.5 − ( − 0.5 ) 2 = 0.25 f ( 0.5 , − 0 ) = 0.5 − 0 2 = 0.5 f ( 0.5 , 0.5 ) = 0.5 − 0. 5 2 = 0.25 f ( 0.5 , 1 ) = 0.5 − 1 2 = − 0.5 t = 1 : { f ( 1 , − 1 ) = 1 − ( − 1 ) 2 = 0 f ( 1 , − 0.5 ) = 1 − ( − 0.5 ) 2 = 0.75 f ( 1 , − 0 ) = 1 − 0 2 = 1 f ( 1 , 0.5 ) = 1 − 0. 5 2 = 0.75 f ( 1 , 1 ) = 1 − 1 2 = 0 t=-1 : \left\{\begin{array}{l}
f(-1,-1)=-1-\left(-1\right)^2=-2\\[0.3cm]
f(-1,-0.5)=-1-\left(-0.5\right)^2=-1.25\\[0.3cm]
f(-1,-0)=-1-0^2=-1\\[0.3cm]
f(-1,0.5)=-1-0.5^2=-1.25\\[0.3cm]
f(-1,1)=-1-1^2=-2
\end{array}\right.\\[0.3cm]
t=-0.5 : \left\{\begin{array}{l}
f(-0.5,-1)=-0.5-\left(-1\right)^2=-1.5\\[0.3cm]
f(-0.5,-0.5)=-0.5-\left(-0.5\right)^2=-0.75\\[0.3cm]
f(-0.5,-0)=-0.5-0^2=-0.5\\[0.3cm]
f(-0.5,0.5)=-0.5-0.5^2=-0.75\\[0.3cm]
f(-0.5,1)=-0.5-1^2=-1.5
\end{array}\right.\\[0.3cm]
t=0 : \left\{\begin{array}{l}
f(0,-1)=0-\left(-1\right)^2=-1\\[0.3cm]
f(0,-0.5)=0-\left(-0.5\right)^2=-0.25\\[0.3cm]
f(0,-0)=0-0^2=-0\\[0.3cm]
f(0,0.5)=0-0.5^2=-0.25\\[0.3cm]
f(0,1)=0-1^2=-1
\end{array}\right.\\[0.3cm]
t=0.5 : \left\{\begin{array}{l}
f(0.5,-1)=0.5-\left(-1\right)^2=-0.5\\[0.3cm]
f(0.5,-0.5)=0.5-\left(-0.5\right)^2=0.25\\[0.3cm]
f(0.5,-0)=0.5-0^2=0.5\\[0.3cm]
f(0.5,0.5)=0.5-0.5^2=0.25\\[0.3cm]
f(0.5,1)=0.5-1^2=-0.5
\end{array}\right.\\[0.3cm]
t=1 : \left\{\begin{array}{l}
f(1,-1)=1-\left(-1\right)^2=0\\[0.3cm]
f(1,-0.5)=1-\left(-0.5\right)^2=0.75\\[0.3cm]
f(1,-0)=1-0^2=1\\[0.3cm]
f(1,0.5)=1-0.5^2=0.75\\[0.3cm]
f(1,1)=1-1^2=0
\end{array}\right. t = − 1 : ⎩ ⎨ ⎧ f ( − 1 , − 1 ) = − 1 − ( − 1 ) 2 = − 2 f ( − 1 , − 0.5 ) = − 1 − ( − 0.5 ) 2 = − 1.25 f ( − 1 , − 0 ) = − 1 − 0 2 = − 1 f ( − 1 , 0.5 ) = − 1 − 0. 5 2 = − 1.25 f ( − 1 , 1 ) = − 1 − 1 2 = − 2 t = − 0.5 : ⎩ ⎨ ⎧ f ( − 0.5 , − 1 ) = − 0.5 − ( − 1 ) 2 = − 1.5 f ( − 0.5 , − 0.5 ) = − 0.5 − ( − 0.5 ) 2 = − 0.75 f ( − 0.5 , − 0 ) = − 0.5 − 0 2 = − 0.5 f ( − 0.5 , 0.5 ) = − 0.5 − 0. 5 2 = − 0.75 f ( − 0.5 , 1 ) = − 0.5 − 1 2 = − 1.5 t = 0 : ⎩ ⎨ ⎧ f ( 0 , − 1 ) = 0 − ( − 1 ) 2 = − 1 f ( 0 , − 0.5 ) = 0 − ( − 0.5 ) 2 = − 0.25 f ( 0 , − 0 ) = 0 − 0 2 = − 0 f ( 0 , 0.5 ) = 0 − 0. 5 2 = − 0.25 f ( 0 , 1 ) = 0 − 1 2 = − 1 t = 0.5 : ⎩ ⎨ ⎧ f ( 0.5 , − 1 ) = 0.5 − ( − 1 ) 2 = − 0.5 f ( 0.5 , − 0.5 ) = 0.5 − ( − 0.5 ) 2 = 0.25 f ( 0.5 , − 0 ) = 0.5 − 0 2 = 0.5 f ( 0.5 , 0.5 ) = 0.5 − 0. 5 2 = 0.25 f ( 0.5 , 1 ) = 0.5 − 1 2 = − 0.5 t = 1 : ⎩ ⎨ ⎧ f ( 1 , − 1 ) = 1 − ( − 1 ) 2 = 0 f ( 1 , − 0.5 ) = 1 − ( − 0.5 ) 2 = 0.75 f ( 1 , − 0 ) = 1 − 0 2 = 1 f ( 1 , 0.5 ) = 1 − 0. 5 2 = 0.75 f ( 1 , 1 ) = 1 − 1 2 = 0
In the figure it looks like this
#339914 on Dec 2023
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