1-1 Sketch the curve \((1 + X_1)^2 + (2 - X_2)^2 = 4\).

由式子可以直接看出圓心在 (-1,2) ，半徑是 2 的圓

library(plotrix)
plot(-3:2,0:5,type="n",xlab="X1",ylab="X2",asp=1)
draw.circle(-1,2,2,border="red",lty=1,lwd=1)
points(-1,2,pch=20)
text(-1,2,labels = "(-1,2)",pos = 3)

1-2 On your sketch, indicate the set of points for which \((1 + X_1)^2 + (2 - X_2)^2 > 4\), as well as the set of points for which \((1 + X_1)^2 + (2 - X_2)^2 \leq 4\).

中間畫線的地方包含圓本身是 \((1 + X_1)^2 + (2 - X_2)^2 \leq 4\) 的範圍，相反的其他留白為 \((1 + X_1)^2 + (2 - X_2)^2 > 4\) 的範圍

plot(-3:2,0:5,type="n",xlab="X1",ylab="X2",asp=1)
draw.circle(-1,2,2,border="red",col="yellow",lty=1,density=5,angle=30,lwd=3)
points(-1,2,pch=20)

1-3 Suppose that a classifier assigns an observation to the blue class if \((1 + X_1)^2 + (2 - X_2)^2 > 4\) and to the red class otherwise. To what class is the observation (0, 0) classified? (-1, 1)? (2, 2)? (3, 8)?

由下圖不難看出只有 (-1,1) 會被分在圓內屬於紅色的範圍，而其他則分為藍色

plot(-3:5,0:8,type="n",xlab="X1",ylab="X2",asp=1)
draw.circle(-1,2,2,border="red",lty=1,lwd=1)
points(0,0,pch=20)
text(0,0,labels = "(0,0)",pos = 3)
points(-1,1,pch=20)
text(-1,1,labels = "(-1,1)",pos = 3)
points(2,2,pch=20)
text(2,2,labels = "(2,2)",pos = 3)
points(3,8,pch=20)
text(3,8,labels = "(3,8)",pos = 1)

1-4 Argue that while the decision boundary in (1-3) is not linear in terms of \(X_1\) and \(X_2\), it is linear in terms of \(X_1\), \(X^2_1\) , \(X_2\), and \(X_2^2\).

boundary 的數學式為 \((1 + X_1)^2 + (2 - X_2)^2 = 4\), 展開可得 \(2X_1+X^2_1-4X_2+X_2^2=-1\) 是 \(X_1\), \(X^2_1\) , \(X_2\), \(X_2^2\) 的線性組合