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\title{Homework 2}
\author{Menyoung Lee}
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\subsection*{P1}
\paragraph{1}
$3x+2x^3=1+x^5 \rightarrow x=1+x^5-2x-2x^3$
\paragraph{2}
$3x+2x^3=1+x^5 \rightarrow 3x=1+x^5-2x^3 \rightarrow x=\frac{1+x^5-2x^3}{3}$
\paragraph{3}Newton's Method.
\[g(x) = x-\frac{f(x)}{f'(x)}\]
\[3x+2x^3=1+x^5 \rightarrow 0=1+x^5-3x-2x^3=f(x)\]
\[g(x)=x=x-0=x-\frac{0}{5x^4-3-6x^2} = x-\frac{1+x^5-3x-2x^3}{5x^4-3-6x^2}\]

\subsection*{P2}
Here we use the third function from the above.
\[r_1 = 0.31375559, 
r_2 = 1.68696945, 
r_3 = -1.77101764.\]

\subsection*{P3}
\[f(x) = 1+x^5-3x-2x^3 \rightarrow
f'(x)=5x^4-3-6x^2 \rightarrow f''(x)=20x^3-12x\]
\[g(x) = x-\frac{f(x)}{f'(x)}\]
Applying the quotient rule,
\[g'(r) = 1-\frac{f'(r)f'(r)-f(r)f''(r)}{f'(r)^2}
= -\frac{f(r)f''(r)}{f'(r)^2} = 0\]
by definition that $f(r)=0$.

\subsection*{P4}
Indeed the error ratio does converge to zero,
save pathological weirdness of small numbers at the end.

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