NumericalMethods/lect_root_finding.html at gh-pages · toperkin/NumericalMethods · GitHub

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		<title>Numerical Methods</title>

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				<section>
					<h1># Numerical Methods with Python</h1>
					<h3>Root finding</h3>
					<p>
						<small>Created by <a href="http://toperkin.github.io/nm.html">Tony Perkins</a></small>
					</p>
				</section>

<section>
	<h2>Three algorithms to implement.</h2>
	<ol>
		<li>Fixed point iteration</li>
		<li>Bisection method</li>
		<li>Newton's method</li>
	</ol>
</section>

<section>
	<h2>Fixed points.
	</h2>
	<p>A number $p$ is a <i>fixed point</i> for a given function $g$ if
		\[g(p)=p.\]
	</p>
</section>

<section>
	<h2>Fixed points and root finding.
	</h2>
	<p>Given a root-finding problem $f(z)=0$, there are many $g$ with fixed points at $z$:
		\[ g(x) = x - f(x) \]
		\[ g(x) = x + 3f(x) \]
		\[ g(x) = x + cf(x) \]
	</p>
</section>

<section>
	<h2>Existence of fixed points.
	</h2>
	<p>If $g\in C[a, b]$ and $g(x)\in [a, b]$ for all $x\in [a, b]$, then $g$ has a fixed point in $[a,b]$
	</p>
</section>

<section>
<h3>proof.</h3><p>Consider $f(x)=g(x)-x$.  As $g$ is continuous, so must $f$ be.
</p>
<p>
What happens at the end points?  If $g(a)=a$, we're done.  $g(a)\nless a$ otherwise $g(a)\not\in [a,b]$.  So \[g(a)>a \Leftrightarrow f(a)>0\]
</p>
<p>
If $g(b)=b$, we're done.  $g(b)\ngtr b$ otherwise $g(b)\not\in [a,b]$.  So \[g(b)< b \Leftrightarrow f(b)< 0\]
</p>
<p>
By the <i>intermediate value theorem</i> $f$ must have a zero, and consequentally $g$ must have a fixed point.
</p>
</section>

<section>
	<h2>Uniqueness of fixed points.
	</h2>
	<p>Suppose $g\in C[a, b]$ and $g(x)\in [a, b]$ for all $x\in [a, b]$.</p>
	<p>If, in addition, $g'(x)$ exists on $(a, b)$, and a positive constant $k < 1$ exists with
		\[|g'(x)| < k \text{ for all } x\in (a, b)\]
then the fixed point in $[a, b]$ is unique.
	</p>
</section>

<section>
	<h3>proof.</h3><p>Suppose $p$ and $q$ are both fixed points in $[a, b]$. Then the Mean Value Theorem implies there exists a $\zeta$ between $p$ and $q$ with
		\[g'(\zeta) = \frac{g(p)-g(q)}{p-q} = \frac{p-q}{p-q} = 1\]
a contradiction.
	</p>
</section>

<section>
	<p>There are a lot of fixed point theorems.  They are awesome.  Here are some of my favorites:</p>

	<ul>
	<li>Brouwer fixed-point theorem</li>
	<li>Banach fixed-point theorem</li>
	<li>Schauder fixed-point theorem</li>
	<li>Sperner's lemma</li>
	</ul>
</section>

<section>
	<h2>Fixed-Point Iteration</h2>
	<p>For initial $p_0$, generate the sequence $\{p_n\}_{n=0}^\infty$ by $p_n=g(p_{n-1})$.
	If the sequence converges to $p$, then

\[p = \lim_{n\rightarrow \infty}p_n = \lim_{n\rightarrow \infty} g(p_n) = g\left(\lim_{n\rightarrow \infty} p_n\right) = g(p) \]
	</p>
</section>


<section>
	<h2>Fixed-Point Theorem.
	</h2>
	<p>Suppose $g\in C[a, b]$ and $g(x)\in [a, b]$ for all $x\in [a, b]$.</p>
	<p>If, in addition, $g'(x)$ exists on $(a, b)$, and a positive constant $k < 1$ exists with
		\[|g'(x)| < k \text{ for all } x\in (a, b).\]
Then, for any number $p_0$ in $[a,b]$ the sequence defined by $p_n=g(p_{n-1})$ converges to the unique fixed point $p$ in $[a, b]$.
	</p>
</section>


<section>
	<h2>Corollary.
	</h2>
	<p>Suppose $g$ satisfies the hypothesis above, then bounds for the error are given by</p>
	<p>
		\[|p_n-p| \le k^n \max\{p_0-a, b-p_0\}\]
	</p>
		<p>
		\[|p_n-p| \le \frac{k^n}{1-k} |p_1-p_0|\]
	</p>
</section>

<section>
	<h2>Follow from MVT.
	</h2>
	<p>\[g'(\zeta) = \frac{g(p)-g(q)}{p-q}\]</p>

	<p>
		\[|g(p)-g(q)| = |g'(\zeta)|\cdot |p-q| \le k\cdot |p-q| \]
	</p>
</section>

<section>
	<h2>Bisection Method
	</h2>
	<p>
		<ul>
			<li>Suppose $f$ is continuos on $[a, b]$, and $f(a)$, $f(b)$ have opposite signs.</li>

			<li>By the IVT, there exists an $x_0$ in $(a, b)$ with $f(x_0)=0$</li>

			<li>Divide the interval $[a, b]$ by computing the midpoint
			\[p=(a+b)/2\]</li>

			<li>If $f(p)$ has the same sign as $f(a)$, consider new interval $[p, b]$</li>

			<li>If $f(p)$ has the same sign as $f(b)$, consider new interval $[a, p]$</li>

			<li>Repeat until the interval is small enough to approximate $x_0$ well.</li>
		</ul>
	</p>
</section>

<section>
	<h2>Termination criteria</h2>
	<p>There are many ways to decide when to stop</p>
	\[ |p_n - p_{n-1}| < \varepsilon \]
	\[ \frac{|p_n-p_{n-1}|}{|p_{n-1}|} < \varepsilon \]
	\[ |f(p_n)| < \varepsilon \]
	<p>None is perfect.  In real software, people use a combination.</p>
</section>

<section>
	<h2>Convergence criteria</h2>
	<p>Suppose that $f\in C[a, b]$ and $f(a)\cdot f(b) < 0$.  The Bisection method generates a sequence $\{p_n\}_{n=1}^\infty$ approximating a zero $p$ of $f$
	\[|p_n-p|\le \frac{b-a}{2^n}, \text{ when } n\ge 1.\]</p>
</section>

<section>
	<h2>Newton's Method</h2>
	<h3>Taylor Polynomial Derivation</h3>
	<p>Suppose $f \in C^2[a, b]$ and $p_0\in [a, b]$ with $f'(p_0)\neq 0.$ Expand $f(x)$ about $p_0$:
	\[f(p) = f(p_0) + (p-p_0)f'(p_0) + \frac{(p-p_0)^2}{2}f''( \zeta(p) ) \]
Set $f(p)=0$, assume $(p-p_0)^2$ neglible:
\[p\approx p_1=p_0-\frac{f(p_0)}{f'(p_0)}\]
This gives the sequence $\{p_n\}_{n=1}^\infty$
\[p_n = p_{n-1} - \frac{ f(p_{n-1}) }{ f'(p_{n-1}) }\]
	</p>
</section>

<section>
<img src="http://toperkin.github.io/NumericalMethods/images-lect_root_finding/newmet.png" alt="">
</section>

<section>
<h3>Newton's Method - Fixed point formulation</h3>
<p>Newton's method is fixed point iteration $p_n=g(p_{n-1})$ where
\[g(x) = x-\frac{f(x)}{f'(x)}\]</p>
</section>

<section>
<p><b>Theorem</b> Let $f\in C^2[a, b]$.  If $p\in [a, b]$ is such that $f(p)=0$ and $f'(p)\neq 0$, then there exists a $\delta>0$ such that Newton's method generates a sequence $\{p_n\}_{n=1}^\infty$ converging to $p$ for any initial approximation $p_0\in [p-\delta, p+\delta]$</p>
</section>

<section>
<h1>Bonus topic!</h1>
</section>

<section>
<p>Consider a polynomial of degree $n$</p>
\[p(x) = \sum_{i=0}^na_ix^i = a_0 + a_1x+a_2x^2+\cdots a_nx^n\]
</section>

<section>
<p>How many basic operations (add, subtract, multiply and divide) does it take to evaluate the polynomial at a point $x_0$?</p>
</section>

<section>
<h2>Horner's Method</h2>

\[p(x) = a_0 + x(a_1 + x (a_2 + \cdots+ x (a_{n-1} + a_nx)\cdots ))\]
</section>

<section>
	<h2>Moral</h2>
</section>

<section>
	<h2>Advanced cleverness is everywhere</h2>
	<p>Just because it is the way you where taught to evaluate something, does not mean that it is the best way to evaluate it.</p>
</section>

<section>
<p>Now suppose you have two degree $n$ polynomials $f$ and $g$.</p>
</section>

<section><p>How many basic operations will it take to find the coefficients of their product?</p></section>

<section><p>It can be done in $\mathcal{O}(n^3)$ using the fast Fourier transform.</p></section>

<section><h2>Moral 2.</h2></section>

<section><p>Sometimes the advanced cleverness to do a problem optimally requires mathematics well beyond the 'obvious' method.</p>
<p>The is a paper listed in the project topics list that will explain how to do Fast Fourier Transforms.</p>
</section>

<section>
<h2>That's all folks (for now)!</h2>
<p>These slides were prepared by <a href="http://toperkin.github.io/nm.html">Tony Perkins</a></p>
</section>

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