CSS322 - Scientific Computing: Proofs Cheat Sheet

Final Exam Reference (Lectures 8-17)

Lecture 8: Numerical Integration

Proof: Method of Undetermined Coefficients = Integrating the Interpolant

Claim: If we use the same set of nodes, the method of undetermined coefficients gives the same weights as integrating the interpolant.

Proof:

• Let $w_1, \ldots, w_n$ denote the weights obtained by the method of undetermined coefficients
• Let $Q_n(f) = \sum_{i=1}^n w_i f(x_i)$ denote the resulting quadrature rule
• Let $p_{n-1}(x) = c_1 + c_2x + \ldots + c_nx^{n-1}$ denote the polynomial of degree $n-1$ that interpolates $f$ at $x_1, \ldots, x_n$

$\int_a^b p_{n-1}(x)dx = \int_a^b (c_1 + c_2x + \cdots + c_nx^{n-1})dx$

This equals:
$c_1\int_a^b dx + c_2\int_a^b x dx + \cdots + c_n\int_a^b x^{n-1} dx$

By the method of undetermined coefficients, these integrals equal:
$c_1(w_1 + \cdots + w_n) + c_2(w_1x_1 + \cdots + w_nx_n) + \cdots + c_n(w_1x_1^{n-1} + \cdots + w_nx_n^{n-1})$

Rearranging:
$w_1(c_1 + c_2x_1 + \cdots + c_nx_1^{n-1}) + w_2(c_1 + c_2x_2 + \cdots + c_nx_2^{n-1}) + \cdots$

This simplifies to:
$w_1p_{n-1}(x_1) + w_2p_{n-1}(x_2) + \cdots + w_np_{n-1}(x_n) = w_1f(x_1) + w_2f(x_2) + \cdots + w_nf(x_n) = Q_n(f)$

The last line follows because $p_{n-1}$ interpolates $f$ at $x_1, \ldots, x_n$, so $p_{n-1}(x_i) = f(x_i)$.

This shows that the quadrature rule obtained by the method of undetermined coefficients is exactly the integral of the interpolant. ∎

Proof: Midpoint Rule is Degree 1

Claim: The midpoint rule $M(f) = (b-a)f\left(\frac{a+b}{2}\right)$ is of degree 1.

Proof:

• By construction, midpoint rule is exact for degree 0 (constants)
• For degree 1, any polynomial has form $p(x) = cx + e$

Exact integral:
$\int_a^b (cx + e)dx = c\left[\frac{x^2}{2}\right]_a^b + e[x]_a^b = \frac{c}{2}(b^2 - a^2) + e(b - a)$

Midpoint rule gives:
$(b - a)p\left(\frac{a + b}{2}\right) = (b - a)\left(c\frac{a + b}{2} + e\right)$
$= c(b-a)\frac{a+b}{2} + e(b-a) = \frac{c}{2}(b^2 - a^2) + e(b - a)$

These match! ✓

Counterexample for degree 2:
$\int_0^2 3x^2 dx = [x^3]_0^2 = 8$
But:
$M(f) = 2(3)(1)^2 = 6 \neq 8$

Therefore, midpoint rule is NOT exact for all degree 2 polynomials. ∎

Proof: Midpoint is Twice as Accurate as Trapezoid

Setup: Let $m = \frac{a + b}{2}$ (midpoint)

Taylor expansion of $f(x)$ about $m$:
$f(x) = f(m) + f'(m)(x - m) + \frac{f''(m)}{2}(x - m)^2 + \frac{f^{(3)}(m)}{6}(x - m)^3 + \frac{f^{(4)}(m)}{24}(x - m)^4 + \ldots$

For Midpoint Rule:

Integrate both sides from $a$ to $b$. Odd powers integrate to zero due to symmetry about $m$:

$\int_a^b f(x)dx = f(m)(b - a) + 0 + \frac{f''(m)}{24}(b - a)^3 + 0 + \frac{f^{(4)}(m)}{1920}(b - a)^5 + \ldots$

Therefore:
$I(f) = M(f) + E(f) + F(f) + \ldots$

where:
$\boxed{E(f) = \frac{f''(m)}{24}(b - a)^3}$
$\boxed{F(f) = \frac{f^{(4)}(m)}{1920}(b - a)^5}$

For Trapezoid Rule:

Starting from Taylor expansions at $a$ and $b$, then adding them and multiplying by $(b-a)/2$:

$T(f) = M(f) + 3E(f) + 5F(f) + \ldots$

Rearranging:
$I(f) = T(f) - 2E(f) - 4F(f) + \ldots$

Comparison:

• Midpoint error: $E(f) + F(f) + \ldots$
• Trapezoid error: $2E(f) + 4F(f) + \ldots$

The trapezoid rule has twice the error of the midpoint rule (when $(b - a)$ is not too large and $f^{(4)}$ is well-behaved). ∎

Error Estimation Formula:

From $T(f) = M(f) + 3E(f) + 5F(f) + \ldots$:

$\boxed{E(f) \approx \frac{T(f) - M(f)}{3}}$

Derivation: Simpson's Rule from Midpoint and Trapezoid

Setup: From the expansions:

• $I(f) = M(f) + E(f) + F(f) + \ldots$
• $I(f) = T(f) - 2E(f) - 4F(f) + \ldots$

Multiply midpoint by 2/3:
$\frac{2}{3}I(f) = \frac{2}{3}M(f) + \frac{2}{3}E(f) + \frac{2}{3}F(f) + \ldots$

Multiply trapezoid by 1/3:
$\frac{1}{3}I(f) = \frac{1}{3}T(f) - \frac{2}{3}E(f) - \frac{4}{3}F(f) + \ldots$

Add them:
$I(f) = \left(\frac{2}{3}M(f) + \frac{1}{3}T(f)\right) - \frac{2}{3}F(f) + \ldots$

But:
$\frac{2}{3}M(f) + \frac{1}{3}T(f) = \frac{2}{3}(b - a)f(m) + \frac{1}{3}\frac{b - a}{2}(f(a) + f(b))$
$= \frac{b - a}{6}\left(f(a) + 4f\left(\frac{a + b}{2}\right) + f(b)\right) = S(f)$

Therefore:
$\boxed{I(f) = S(f) - \frac{2}{3}F(f) + \ldots}$

This shows the dominant error term of Simpson's rule is $O((b-a)^5)$, making it much more accurate! ∎

Lecture 10: Linear Least Squares

Lemma 1: $A^T A$ is Symmetric Positive Definite

Theorem: Suppose $A \in \mathbb{R}^{m \times n}$ has rank $n$. Then $A^T A$ is symmetric positive definite.

Proof:

Part 1 - Symmetry:
$(A^T A)^T = A^T (A^T)^T = A^T A$ ✓

Part 2 - Positive Definiteness:

Suppose $x \in \mathbb{R}^n$ is nonzero. Then:
$x^T A^T Ax = (Ax)^T Ax = \|Ax\|_2^2 \geq 0$

But $Ax$ is a nontrivial linear combination of columns of $A$ (since $x \neq 0$).

Since columns of $A$ are linearly independent (because rank of $A$ is $n$), we have $Ax \neq 0$.

Therefore: $x^T A^T Ax = \|Ax\|_2^2 > 0$ ✓

This proves $A^T A$ is symmetric positive definite. ∎

Lemma 2: Minimizer of Quadratic Forms

Theorem: Suppose $f(x) = x^T Cx - 2h^T x + d$, where $C$ is symmetric positive definite. Then $f$ has a unique minimizer at $\boxed{x^* = C^{-1}h}$.

Proof:

Step 1: Evaluate $f(x^*)$

$f(x^*) = f(C^{-1}h) = (C^{-1}h)^T CC^{-1}h - 2h^T C^{-1}h + d$
$= h^T C^{-T}h - 2h^T C^{-1}h + d$

Since $C$ is symmetric: $C^{-T} = (C^T)^{-1} = C^{-1}$

Therefore:
$f(x^*) = h^T C^{-1}h - 2h^T C^{-1}h + d = -h^T C^{-1}h + d$

Step 2: Evaluate $f(x^* + y)$ for any nonzero vector $y$

Let $y \in \mathbb{R}^n$ be any arbitrary nonzero vector:

$f(x^* + y) = (C^{-1}h + y)^T C(C^{-1}h + y) - 2h^T(C^{-1}h + y) + d$
$= (h^T C^{-1} + y^T)(h + Cy) - 2h^T C^{-1}h - 2h^T y + d$
$= h^T C^{-1}h + h^T y + y^T h + y^T Cy - 2h^T C^{-1}h - 2h^T y + d$

Since $h^T y = y^T h$ (scalars are symmetric):
$f(x^* + y) = -h^T C^{-1}h + d + y^T Cy = f(x^*) + y^T Cy$

Step 3: Show $x^*$ is the unique minimizer

But $C$ is symmetric positive definite, so: $y^T Cy > 0$ for all nonzero $y$

Therefore: $f(x^* + y) = f(x^*) + y^T Cy > f(x^*)$ for all nonzero $y$

So $x^*$ is the unique minimizer. ∎

Main Theorem: Normal Equations Solution

Theorem:
$\boxed{x = (A^T A)^{-1}A^T b}$
is the unique solution to the linear least squares problem when $A$ has rank $n$.

Proof: This follows directly from Lemma 1 and Lemma 2 by setting:

• $f(x) = \|Ax - b\|_2^2 = x^T A^T Ax - 2b^T Ax + b^T b$
• $C = A^T A$ (symmetric positive definite by Lemma 1)
• $h = A^T b$
• $d = b^T b$

By Lemma 2: $x^* = C^{-1}h = (A^T A)^{-1}A^T b$ ∎

QR Factorization Theorems

Theorem 1: Any $A \in \mathbb{R}^{m \times n}$ ($m \geq n$) can be factored $A = QR$ where $Q$ is an $m \times m$ orthogonal matrix and $R$ is an $m \times n$ upper triangular matrix.

Theorem 2: If $A \in \mathbb{R}^{m \times n}$, $m \geq n$, has rank $n$ and $A = QR$ is the QR factorization of $A$, then $R$ has rank $n$ and $R_1$ is nonsingular.

Main Result: Let $A = QR$ be the QR factorization of $A$. Assume $A \in \mathbb{R}^{m \times n}$, $m \geq n$, $\text{rank}(A) = n$.

Let $c = Q^T b$. Let $c_1$ be the top $n$ entries of $c$. Then $x$ minimizes $\|Ax - b\|_2$ if:
$\boxed{R_1 x = c_1}$

Proof:
$\|Ax - b\|_2 = \|QRx - b\|_2 = \|Q(Rx - Q^T b)\|_2 = \|Rx - c\|_2$
$= \left\|\begin{bmatrix} R_1 \\ 0 \end{bmatrix} x - \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}\right\|_2 = \left\|\begin{bmatrix} R_1 x - c_1 \\ -c_2 \end{bmatrix}\right\|_2$
$= \sqrt{\|R_1 x - c_1\|_2^2 + \|c_2\|_2^2}$

Since $c_2$ is constant, minimizing this is equivalent to minimizing $\|R_1 x - c_1\|_2$.

If $R_1$ is nonsingular (which follows from Theorem 2), then $R_1 x = c_1$ has a unique solution that minimizes the norm. ∎

Householder Transformation Properties

Theorem: Let $v \in \mathbb{R}^m$ be a nonzero vector. Then $H = I - 2\frac{vv^T}{v^T v}$ is symmetric and orthogonal.

Proof:

Symmetry:
$H^T = \left(I - 2\frac{vv^T}{v^T v}\right)^T = I^T - 2\frac{(vv^T)^T}{v^T v} = I - 2\frac{vv^T}{v^T v} = H$ ✓

Orthogonality:
$HH^T = HH = \left(I - 2\frac{vv^T}{v^T v}\right)\left(I - 2\frac{vv^T}{v^T v}\right)$
$= I - 4\frac{vv^T}{v^T v} + 4\frac{vv^T vv^T}{(v^T v)^2}$
$= I - 4\frac{vv^T}{v^T v} + 4\frac{v(v^T v)v^T}{(v^T v)^2}$
$= I - 4\frac{vv^T}{v^T v} + 4\frac{vv^T}{v^T v} = I$ ✓ ∎

Lecture 14: Singular Value Decomposition

Lemma 3: Norm of Diagonal Matrix

Lemma: Let $D \in \mathbb{R}^{m \times n}$ be diagonal. Then $\|D\|_p = \max |d_{ii}|$ for any $p$.

Proof: [Exercise for student]

Key steps:

• For any vector $x$, $Dx$ has entries $d_{ii}x_i$
• $\|Dx\|_p \leq (\max |d_{ii}|) \|x\|_p$
• Equality achieved when $x$ is the unit vector in direction of largest $|d_{ii}|$
• Therefore $\|D\|_p = \max |d_{ii}|$ ∎

Theorem: Matrix 2-Norm from SVD

Theorem: If $A = U\Sigma V^T$ (SVD of $A$), then $\|A\|_2 = \sigma_1$ (largest singular value).

Proof:

From Lemma 1 and Lemma 2 (from lecture notes):

• $\|QA\|_2 = \|A\|_2$ for orthogonal $Q$
• $\|AQ\|_2 = \|A\|_2$ for orthogonal $Q$

Therefore:
$\|A\|_2 = \|U\Sigma V^T\|_2 = \|\Sigma V^T\|_2 = \|\Sigma\|_2$

By Lemma 3: $\|\Sigma\|_2 = \sigma_1$ (largest diagonal entry). ∎

Theorem: Condition Number from SVD

Theorem: If $\sigma_1, \ldots, \sigma_n$ are singular values of $A \in \mathbb{R}^{n \times n}$ and $A$ is invertible, then:
$\boxed{\|A^{-1}\|_2 = \frac{1}{\sigma_n}}$

Proof:

From SVD $A = U\Sigma V^T$:
$A^{-1} = (U\Sigma V^T)^{-1} = V^{-T} \Sigma^{-1} U^{-1} = V\Sigma^{-1} U^T$

Note that:
$\Sigma^{-1} = \begin{bmatrix} \frac{1}{\sigma_1} & 0 & \cdots & 0 \\ 0 & \frac{1}{\sigma_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{1}{\sigma_n} \end{bmatrix}$

However, this is not an SVD of $A^{-1}$ because diagonal entries are not in non-increasing order.

Consider permutation matrix $P = \begin{bmatrix} & & 1 \\ & \ddots & \\ 1 & & \end{bmatrix}$

Then:
$P\Sigma^{-1} P = \begin{bmatrix} \frac{1}{\sigma_n} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \frac{1}{\sigma_1} \end{bmatrix}$

has diagonal entries in non-increasing order.

Since $PP = I$, $VP$ is orthogonal, and $PU^T$ is orthogonal:
$A^{-1} = V\Sigma^{-1} U^T = (VP)(P\Sigma^{-1} P)(PU^T)$

is the SVD of $A^{-1}$.

Thus: $\|A^{-1}\|_2 = \frac{1}{\sigma_n}$ (largest singular value of $A^{-1}$). ∎

Corollary:
$\boxed{\text{cond}_2(A) = \|A\|_2 \|A^{-1}\|_2 = \frac{\sigma_1}{\sigma_n}}$

Theorem: Condition Number of $A^T A$

Theorem: Suppose $A \in \mathbb{R}^{m \times n}$ with $\text{rank}(A) = n$. Then:
$\boxed{\text{cond}_2(A^T A) = \text{cond}_2(A)^2}$

Proof: [Exercise for student]

Key steps:

• If $A = U\Sigma V^T$, then $A^T A = V\Sigma^T U^T U\Sigma V^T = V\Sigma^T\Sigma V^T$
• $\Sigma^T\Sigma$ has diagonal entries $\sigma_i^2$
• $\text{cond}_2(A^T A) = \frac{\sigma_1^2}{\sigma_n^2} = \left(\frac{\sigma_1}{\sigma_n}\right)^2 = \text{cond}_2(A)^2$ ∎

Theorem: Rank Determination

Theorem: If $A = U\Sigma V^T$, then:
$\boxed{\text{rank}(A) = \text{number of nonzero singular values}}$

Proof: [Standard linear algebra result - relies on orthogonal transformations preserving rank]

Lecture 15: Eigenvalues and Eigenvectors

Theorem: Eigenvalues and Eigenvectors of Inverse Matrix

Theorem: If $A \in \mathbb{R}^{n \times n}$ is invertible, then $A$ and $A^{-1}$ have the same eigenvectors, and if $\lambda$ is an eigenvalue of $A$, then $1/\lambda$ is an eigenvalue of $A^{-1}$.

Proof:

Suppose $(\lambda, x)$ is an eigenpair of $A$:
$Ax = \lambda x$

Multiply both sides by $A^{-1}$:
$A^{-1}(Ax) = A^{-1}(\lambda x)$
$x = \lambda A^{-1}x$

Divide both sides by $\lambda$ (assuming $\lambda \neq 0$):
$\left(\frac{1}{\lambda}\right)x = A^{-1}x$

Therefore, $(1/\lambda, x)$ is an eigenpair of $A^{-1}$. ∎

Theorem: Shifted Matrix Properties

Theorem: Suppose $(A - \sigma I)$ is invertible. Then:

1. $A$ and $(A - \sigma I)^{-1}$ have the same eigenvectors
2. $\lambda$ is an eigenvalue of $A$ if and only if $\frac{1}{\lambda - \sigma}$ is an eigenvalue of $(A - \sigma I)^{-1}$

Proof:

Suppose $(\lambda, x)$ is an eigenpair of $A$:
$Ax = \lambda x$

Subtract $\sigma x$ from both sides:
$(A - \sigma I)x = (\lambda - \sigma)x$

Multiply both sides by $(A - \sigma I)^{-1}$:
$x = (\lambda - \sigma)(A - \sigma I)^{-1}x$

Divide by $(\lambda - \sigma)$ (assuming $\lambda \neq \sigma$):
$\left(\frac{1}{\lambda - \sigma}\right)x = (A - \sigma I)^{-1}x$

Therefore, $(1/(\lambda - \sigma), x)$ is an eigenpair of $(A - \sigma I)^{-1}$. ∎

Lecture 13: Initial Value Problems for ODEs

Theorem: Symmetry of Mixed Partials

Theorem: If both $\frac{\partial^2 f}{\partial x_j \partial x_i}$ and $\frac{\partial^2 f}{\partial x_i \partial x_j}$ are well-defined and continuous, then:
$\boxed{\frac{\partial^2 f}{\partial x_j \partial x_i} = \frac{\partial^2 f}{\partial x_i \partial x_j}}$

Proof: [Standard multivariable calculus result - Schwarz's theorem/Clairaut's theorem]

Lecture 16: Boundary Value Problems

Integration by Parts Formula

Formula:
$\boxed{\int_a^b v''(t) \phi(t) \, dt = v'(t)\phi(t)\Big|_a^b - \int_a^b v'(t) \phi'(t) \, dt}$

When $\phi(a) = \phi(b) = 0$:
$\boxed{\int_a^b v''(t) \phi(t) \, dt = -\int_a^b v'(t) \phi'(t) \, dt}$

Proof:

Standard integration by parts with $u = \phi(t)$, $dv = v''(t)dt$:
$\int_a^b v''(t) \phi(t) \, dt = v'(t)\phi(t)\Big|_a^b - \int_a^b v'(t) \phi'(t) \, dt$

If $\phi(a) = \phi(b) = 0$:
$v'(b)\phi(b) - v'(a)\phi(a) = 0$

Therefore:
$\int_a^b v''(t) \phi(t) \, dt = -\int_a^b v'(t) \phi'(t) \, dt$ ∎

Important Relationships and Formulas

Finite Difference Approximations

First derivative (forward): $O(h)$
$\boxed{u'(t) \approx \frac{u(t+h) - u(t)}{h}}$

First derivative (backward): $O(h)$
$\boxed{u'(t) \approx \frac{u(t) - u(t-h)}{h}}$

First derivative (centered): $O(h^2)$
$\boxed{u'(t) \approx \frac{u(t+h) - u(t-h)}{2h}}$

Second derivative (centered): $O(h^2)$
$\boxed{u''(t) \approx \frac{u(t+h) - 2u(t) + u(t-h)}{h^2}}$

Richardson Extrapolation

Formula:
$\boxed{a_0 = F(h) + \frac{F(h) - F(h/q)}{q^{-p} - 1}}$

where:

• $F(h)$ is the approximation with step size $h$
• $F(h/q)$ is the approximation with step size $h/q$
• $p$ is the order of the method
• The extrapolated value has error $O(h^r)$ where $r > p$

Power Method Convergence

For $A$ with eigenvalues $\lambda_1, \ldots, \lambda_n$ where $|\lambda_1| > |\lambda_j|$ for all $j > 1$:

If $x^{(0)} = \sum_{j=1}^{n} \alpha_j v_j$ (linear combination of eigenvectors), then:

$\boxed{x^{(k)} = \sum_{j=1}^{n} \lambda_j^k \alpha_j v_j = \lambda_1^k \left(\alpha_1 v_1 + \sum_{j=2}^{n} \left(\frac{\lambda_j}{\lambda_1}\right)^k \alpha_j v_j\right)}$

As $k \to \infty$:
$\boxed{x^{(k)} \to \lambda_1^k \alpha_1 v_1}$ (multiple of dominant eigenvector)

Rayleigh Quotient

$\boxed{\lambda = \frac{(x^{(k)})^T Ax^{(k)}}{(x^{(k)})^T x^{(k)}}}$

(Denominator may be omitted if $x^{(k)}$ is normalized)

Error Analysis for ODEs

Local Truncation Error: $l_k = y_k - y(t_k)$ assuming $y_i = y(t_i)$ for $i < k$

Global Truncation Error: $e_k = y_k - y(t_k)$ (accumulated error)

Relationship: For a wide class of ODEs and methods:

• If local truncation error is $O(h^{p+1})$
• Then global truncation error is $O(h^p)$

Stability Conditions

Explicit Euler for Heat Equation:
$\boxed{\Delta t \leq \frac{(\Delta x)^2}{2c}}$

General Form for Parabolic PDEs: Explicit methods require $\Delta t = O((\Delta x)^2)$

Implicit Methods: Typically unconditionally stable (no restriction on $\Delta t$)

Quick Reference Tables

Numerical Integration Methods

ODE Methods

PDE Methods (Heat Equation)

Key Theorems for Exam

1. Normal Equations

• Solution exists and is unique when $\text{rank}(A) = n$
• Solution: $x = (A^T A)^{-1}A^T b$
• $A^T A$ is symmetric positive definite

2. QR Factorization

• Every matrix has QR factorization
• If $\text{rank}(A) = n$, then $R_1$ is nonsingular
• Least squares solution: solve $R_1 x = c_1$ where $c_1$ are first $n$ entries of $Q^T b$

3. SVD Properties

• $\|A\|_2 = \sigma_1$ (largest singular value)
• $\text{cond}_2(A) = \frac{\sigma_1}{\sigma_n}$
• $\text{rank}(A) =$ number of nonzero singular values
• Pseudoinverse: $A^+ = V\Sigma^+ U^T$

4. Eigenvalue Methods

• Power method finds largest $|\lambda|$
• Inverse power method finds smallest $|\lambda|$
• Shifted inverse power method finds $\lambda$ closest to shift $\sigma$
• All require unique target eigenvalue for convergence

5. Stability vs Accuracy

• Stability: errors don't grow unboundedly
• Accuracy: truncation error order
• Backward stable: computes exact solution to nearby problem
• Conditionally stable: requires relationship between parameters

Common Mistakes to Avoid

1. Forgetting to normalize in power methods
2. Using wrong order formulas together (e.g., $O(h)$ with $O(h^2)$)
3. Not checking rank conditions before using normal equations
4. Confusing local and global truncation error
5. Forgetting boundary conditions in PDE problems
6. Not checking stability conditions for explicit methods
7. Computing matrix inverses explicitly (use factorizations instead)
8. Mixing up conditioning and stability

Final Exam Tips

1. Proofs: Focus on understanding the logic flow, not memorizing word-for-word
2. Formulas: Know when each method applies and its limitations
3. Order of accuracy: Always check consistency of approximations
4. Stability: Know which methods are stable and under what conditions
5. Relationships: Understand how different concepts connect (e.g., SVD → condition number → stability)
6. Special cases: Remember homogeneous vs non-homogeneous boundary conditions
7. Computational cost: Know the big-O complexity of major algorithms

Last-Minute Review Checklist

• Can derive normal equations from minimization
• Understand QR vs Cholesky for least squares
• Know SVD properties and applications
• Understand power method variants
• Know stability conditions for PDEs
• Can set up finite difference schemes
• Understand explicit vs implicit methods
• Know integration by parts for Galerkin method
• Understand relationship between local and global error
• Can identify method type and order

Good luck on your final exam! 🎓