Decision Boundary Line (Linear/Non-Linear)

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Homework Statement



Given a non-linear decision boundary line: (1 + X1)^2 + (2 − X2)^2 = 4

Argue that while the decision boundary is not linear in terms of X1 and X2, it is linear in terms of X1,X1^2 , X2, and X2^2 .

The Attempt at a Solution



I'm honestly not sure. I realize the curve is a circle, but I don't understand how it could be turned linear by having it terms of X1,X1^2 , X2, and X2^2
 
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Is it because we are extending the feature space by including quadratic terms that can address this non-linearity?
 
It's pretty basic algebra that (1+ X1)^2+ (2- X2)^2= X1^2- 2X1+ 1+ X2^2- 4X2+ 4= 4
so X1^2- 2X1+ X2^2- 4X2+ 1= 0.

If you let Y1= X1^2 and Y2= X2^2, then you have Y1- 2X1+ Y2- 4Y1+ 1= 0 which is 'linear in X1, X2, Y1, and Y2".
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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