Review for test 3

Review questions for test 3. Some edits made 12/13. Test 3 will now only cover chapters 6-8

Ch 6: Polynomial Interpolation

The main goal: if we have points $(x_0, y_0), \dots, (x_n, y_n)$ where $y_i = f(x_i)$, can we approximate the function $f(x)$ with a polynomial $p(x)$?

Thm: if the $x_i$'s are distinct, then there is a unique polynomial of degree $n$ with $p(x_i) = y_i$.

There are two different ways to write this:

$$~ l_i(x) = \prod_{j=0, j \neq i}^n \frac{x - x_j}{x_i - x_j}. ~$$

The Newton form represents $p$ in terms of polynomials of degree $i$ whereas the Lagrange form has terms all of degree $n$.


$$~ f(x) - p_n(x) = \frac{1}{(n+1)!} f^{(n+1)}(\xi) \prod_{i=0}^n (x - x_i) ~$$

This gives a bound on the error.


Even if it is true that for each $x$, $f(x) - p_n(x) \rightarrow 0$ it need not be the case that $\|f - p_n\|_\infty = max_{a \leq x \leq b}|f(x) - p_n(x)| \rightarrow 0$. As we saw on the computer there can be wild oscillations near the edges in $p_n$. In fact: if the interpolation points are evenly spaced the interpolating polynomials become unbounded.

HW Problems

6.1: 1, 2, 12

6.2: 4, 12, 13 (n=1 only)

Ch7: Numeric differentiation and Integration

Again, we have a table of function values $(x_0, y_0), \dots, (x_n, y_n)$ where $y_i = f(x_i)$, What can we say about the derivate of $f$? The integral of $f$?

We had a basic scheme:


For differentiation it was mentioned that if we applied the above to points $x$, $x-h$ and $x_h$, we would get the central difference formula:

$$~ f'(x) \approx \frac{f(x+h) - f(x-h)}{2h}. ~$$

The forward difference equation is $(f(x+h) - f(x))/h$.

The error in the approximation – when done on the computer – has two sources of error:

For the forward difference, the truncation error is roughly $\mathcal{O}(h)$, whereas the floating point error is $\mathcal{O}(\delta/h)$. So there needs to be a balance if using: take $h$ small enough so the truncation error isn't large but not too small so that the floating point error isn't large.

The central difference has similar floating point error, but truncation error like $\mathcal{O}(h^2)$. (It is basically $f'''(\xi)/6h^2$.)

We mentioned automatic differentiation, but this won't be on the test.


The process of approximating $f$ by $p$ and integrating $p$ leads to 3 familar concepts:

Rather than globally approximate $[a,b]$ with just a few points, what is done if that interval is partitioned and on each subpartition the approximation is used. With this we saw errors, $\int_a^b f(x) dx - \int_a^b p_n(x) dx$, given by


The expression $\int_a^b p_n(x)dx$ can be written differently taking LaGrange's form for polynomial interpolation:

$$~ \int_a^b f(x) dx \approx \int_a^b p_n(x) dx - \int_a^b \sum_{k=0}^n f(x_k) l_k(x) dx = \sum_{k=0}^n f(x_k) \int_a^b l_k(x)dx = \sum_{k=0}^n f(x_k) A_k. ~$$

The points $x_k$ are called the nodes and the terms $A_k$ the weights. Both can be precomputed, as they do not depend on $f$.

The main use here are quadrature formulas:

(p493) Let $w$ be a positive weight function (like $w(x) = 1$) and let $q$ be a non-zero polynomial of degree $n+1$ that is $w$-orthogonal to the space of polynomials of degree $n$ or less. Then If $x_0, x_1, \dots, x_n$ are the zeros of $q$, the quadrature formula derived by using these zeros as the nodes and used in the weight computation will be exact for any polynomial of degree $2n+1$ or less.

We saw one family of orthogonal polynomials, those when $w=1$. These were the Legendre polynomials satisfying the recursion:

$$~ P_0(x) = 1, \quad P_1(x) =x, \quad (n+1)P_{n+1}(x) = (2n + 1) x P_n(x) - n P_{n-1}(x). ~$$


If we defined

$$~ \int_a^b f(x) w(x) dx = \sum_{i=0}^n A_i f(x_i) + E ~$$

Then for $f$ in $C^{2(n+1)}([a,b])$ the error can be written as:

$$~ E = \frac{1}{(2n)!} f^{(2(n+1))}(\xi) \int_a^b (\prod(x-x_i))^2 w(x) dx ~$$


7.1 6 (for $f'(x)$ only),

7.2: 1, 2, 27 (using 26 as given)

7.3: 3, 6

nodes = [-sqrt(3 / 5), 0.0, sqrt(3 / 5)]
weights = [5 // 9, 8 // 9, 5 // 9]
3-element Array{Rational{Int64},1}:

Use these to estimate $\int_{-1}^1 \sin(x) dx$.

Use the error term and the fact that $|f^{(2n)}(\xi) |<1$ to estimate the error.

Chapter 8. Differential Equations

The IVP (initial value problem) is a specification about a function $x(t)$ through a relation involving its derivative at time $t$:

$$~ x'(t) = f(t, x), \quad x(t_0) = x_0 ~$$

In 8.1, we see that an IVP may not have an answer for all $t$; it may not have an answer at all; and if it does have an answer, it may not be unique.

There are theorems which will vouch for uniqueness and existence:

Thm 1. Existence theorem

If $f$ is continuous on a rectangle centered at $(t_0, x_0)$, say:

$$~ R = \{ (t,x) : |t - t_0| \leq \alpha, |x - x_0| \leq \beta \}. ~$$

Then the inital value problem has a solution $x(t)$ for $|t - t_0| \leq min(\alpha, \beta/M)$ where $M$ maximizes $|f|$ in $R$.

Thm 2 (p526). If both $f$ and $\partial f/\partial x$ are continuous in R, then there is a unique solution for $|t - t_0| \leq min(\alpha, \beta/M)$

Thm 3 If f is Lipshitz then the intial value problem will have a unique solution in some interval.

Precisely, if $f$ is continuous on the strip $a \leq t\leq b$ and $x \in (-\infty, \infty)$ and satisfies the inequality for a fixed $L$:

$$~ | f(t,x_1) - f(t, x_2) | \leq L | x_1 - x_2| ~$$

then the solution exists on the interval $[a,b]$.

The last two give conditions on $f$ that guarantee an answer exists and is unique, at least for some values of $t$.

Euler's method

The granddaddy of all methods to numerically approximate the solution to an IVP is Euler's method:

From a sequence of time steps $t_i = t_0 + ih$, where $h$ is the small time step, Euler's method defines a sequence of $x_i$ values through

$$~ x_{i+1} = x_{i} + f(t_i, x_i) ~$$

The approximate local truncation error at each step is like $\mathcal{O}(h^2)$, so if things are nice, the global truncation error will be like $\mathcal{O}(h)$, when the number of steps is basically $1/h$.

Taylor methods

The Euler method can be derived by starting with a Taylor series:

$$~ x(t + h) = x(t) + x'(t)h + x''(t)/2 \cdot h^2 + \cdots + x^{(n)}(t)/n! \cdot h^n + \mathcal{O}(h^{n+1}) ~$$

Then truncating at the first order. As the IVP defines $x'(t)$ in terms of $f$, Euler's method is nothing more than the tangent line approximation.

Using more terms can be done by differentiating, though that gets tricky.

Runge-Kutte methods

The Runge - Kutte methods generalize Euler's method by adding various combinations of the approximations for $f$. The two discussed in class were:

$$~ \begin{align} x_{n+1} &= x_n + \frac{1}{2} F_1 + \frac{1}{2} F_2, \text{ where}\\ F_1(x) &= h\cdot f(t, x)\\ F_2(x) &= h\cdot f(t+h, F_1) \end{align} ~$$

This has local truncation errors of $\mathcal{O}(h^3)$

$$~ \begin{align} x_{n+1} &= x_n + \frac{1}{6}( F_1 + 4F_2 + 4F_3 + F_4), \text{ where}\\ F_1 &= h\cdot f(t_n, x_n)\\ F_2 &= h\cdot f(t_n+h, F_1/2)\\ F_3 &= h\cdot f(t_n+h, F_2/2)\\ F_2 &= h\cdot f(t_n+h, F_3)\\ \end{align} ~$$

This has local truncation errors of $\mathcal{O}(h^5)$.

Multistep methods

The general multistep model might look like this (from p557)

$$~ a_k x_n + a_{k-1}x_{n-1} + \cdots + a_0 x_{n-k} = h( b_k f_n + f_{k-1} f_{n-1} + \cdots + b_0 f_{n-k}). ~$$

Some special cases:


$$~ |f(t,x_1) - f(t,x_2)| \leq L |x_1 - x_2| ~$$ $$~ x'(t) = 10 - 3x \cdot t\quad x(0) = 1 ~$$ $$~ x'(t) = 1 + x^2,\quad x(0)=0 ~$$ $$~ x_{n+1} = x_n + h[(3/2) f_n - (1/2) f_{n-1}] ~$$

Take $h=1/4$. Using Euler's method to find $x_1$, find $x_4$ when

$$~ x'(t) = t + x,\quad x'(0) = 1 ~$$ $$~ x_{n+1} = x_n + h f(t_{n+1}, x_{n+1}). ~$$

For the IVP

$$~ x'(t) = -10x, \quad x(0) =1 ~$$

The implicitness can be solved directly, as $f$ is linear. Using $h=1/3$, find $x_3$.

Show the second-order Adams-Bashworth method

$$~ x_{n+1} = x_n + h[(3/2) f_n - (1/2) f_{n-1}] ~$$

is convergent.

Evals  order
2  2
3  3
4  4
5  4
6  5
7  6
8  6
9  7