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diff --git a/Homework/math4610/notes/29-Nov.org b/Homework/math4610/notes/29-Nov.org
new file mode 100644
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+++ b/Homework/math4610/notes/29-Nov.org
@@ -0,0 +1,20 @@
+Jacobi Iteration (cont.)
+
+x^{k+1} = D^{-1}(b - (L + U)x^k)
+
+{
+  x^{k+1} = x^k + D^-1 r^k
+  r^{k} = b - Ax^k
+}
+
+error: || x^{k+1} - x^k ||_2
+residual: || r^k ||_2
+
+Gauss-Seidel Iteration:
+A = (L + D + U)
+\Rightarrow Ax = b
+  (D + U)x = b - Lx
+  x = (D + U)^-1 (b - Lx)
+
+x^{k+1} = (D+U)^{-1}(b - Lx^k)
+(D + U)^{-1} x (bsubst)
diff --git a/Homework/math4610/notes/4-Dec.org b/Homework/math4610/notes/4-Dec.org
new file mode 100644
index 0000000..d148bc8
--- /dev/null
+++ b/Homework/math4610/notes/4-Dec.org
@@ -0,0 +1,2 @@
+
+ 
diff --git a/Homework/math4610/notes/Nov-27.org b/Homework/math4610/notes/Nov-27.org
new file mode 100644
index 0000000..ae4ded0
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+++ b/Homework/math4610/notes/Nov-27.org
@@ -0,0 +1,20 @@
+x^{k+1} = D^{-1}(b - (L + U) x^k)
+x^{k + 1} \rightarrow Ax^k  
+
+
+#+BEGIN_SRC c
+  loop while (err > tol && iter < maxiter) {
+    for (int i = 0; i < n; i++) {
+      sum = b[i];
+      for (int j = 0; j < i; j++) {
+        sum = sum - a[i][x] * x_0[i];
+      }
+      for (int j = i; j < n; j++) {
+        sum = sum + a[i][j] * x_0[j];
+      }
+      x_1[i] = sum / a[i][i];
+    }
+
+    err = 0.0;
+  }
+#+END_SRC
diff --git a/Homework/math4610/notes/Nov-3.org b/Homework/math4610/notes/Nov-3.org
new file mode 100644
index 0000000..5a65d2a
--- /dev/null
+++ b/Homework/math4610/notes/Nov-3.org
@@ -0,0 +1,62 @@
+* eigenvalues \rightarrow power method
+
+we iterate on the x_{k+1} = A x_k
+
+y = Av_0
+v_1 = \frac{1}{|| y ||} (y)
+\lambda_0 = v_0^T A v_0 = v_0^T y
+
+Find the largest eigenvalue;
+
+#+BEGIN_SRC c
+  while (error > tol && iter < max_iter) {
+    v_1 = (1 / magnitude(y)) * y;
+    w = m_dot_v(a, v_1);
+    lambda_1 = v_dot_v(transpose(v_1), w);
+    error = abs(lambda_1 - lambda_0);
+    iter++;
+    lambda_0 = lambda_1;
+    y = v_1;
+  }
+
+  return [lambda_1, error];
+#+END_SRC
+
+Find the smallest eigenvalue:
+
+** We know:
+If \lambda_1 is the largest eigenvalue of $A$ then \frac{1}{\lambda_1} is the smallest eigenvalue of $A^{-1}$.
+
+If \lambda_n is the smallest eigenvalue of $A$ then \frac{1}{\lambda_n} is the largest eigenvalue of $A^{-1}$.
+*** However, calculating $A^{-1}$ is inefficient
+So, transform $w = A^{-1} v_1 \Rightarrow$ Solve $Aw = v_1$ with LU or GE (line 3 of above snippet).
+
+And, transform $y = A^{-1} v_0 \Rightarrow$ Solve $Ay = v_0$ with LU or GE.
+
+** Conclusions
+
+We have the means to compute the approximations of \lambda_1 and \lambda_n.
+
+(\lambda_1 \rightarrow power method)
+
+(\lambda_n \rightarrow inverse power method)
+
+* Eigenvalue Shifting
+
+If (\lambda, v) is an eigen pair, (v \neq 0)
+
+Av = \lambdav
+
+Thus for any \mu \in R
+
+(Av - \mu I v) = (A - \mu I)v = \lambda v - \mu I v
+             = (\lambda - \mu)v
+             \Rightarrow \lambda - \mu is an eigenvalue of (A - \mu I)
+
+(A - \mu I)v = (\lambda - \mu)v
+
+Idea is to choose \mu close to our eigenvalue. We can then inverse iterate to
+construct an approximation of \lambda - \mu and then add \mu back to get \lambda.
+
+v_0 = a_1 v_1 + a_2 v_2 + \cdots + a_n v_n
+A v_0 = a_1 (\lambda_1 v_1) + \cdots
diff --git a/Homework/math4610/notes/Nov-6.org b/Homework/math4610/notes/Nov-6.org
new file mode 100644
index 0000000..4a9562f
--- /dev/null
+++ b/Homework/math4610/notes/Nov-6.org
@@ -0,0 +1,25 @@
+* Power Method
+v_{k+1} = A v_k, k = 0,1,2
+
+** Properties
+1. \frac{A v_k}{||v_k||} \rightarrow v_1
+2. \frac{v_k^T A v_k}{v_k^T v_k} \rightarrow \lambda_1
+3. If \lambda is a n eigenvalue of A, then \frac{1}{\lambda} is an eigenvalue of A^-1
+4. Av = \lambda v
+   Av - \mu v = (\lambda-\mu)v = (A - \mu I)v
+5. If \lambda is an eigenvalue of A, then \lambda - \mu is an eigenvalue of A \cdot \mu I
+
+** Shifting Eigenvalues
+1. Partition [\lambda_n, \lambda_1]
+
+
+* Lanczos Algorithm
+
+#+BEGIN_SRC c
+  for (int i = 0; i < n; i++) {
+    sum = a0;
+    v_dot_v(a[i], x);
+
+    b[i] = sum;
+  }
+#+END_SRC
diff --git a/Homework/math4610/notes/Oct-11.org b/Homework/math4610/notes/Oct-11.org
new file mode 100644
index 0000000..575ea74
--- /dev/null
+++ b/Homework/math4610/notes/Oct-11.org
@@ -0,0 +1,15 @@
+* Diagonal Dominance
+Suppose that A \in R^{n \times n} is diagonally dominant then Gaussian eliminiation of A produces no zero pivot
+elements.
+
+Def. A \in R^{n \times n} is diagonally dominant if for each i=1,2,...n |a_{i,i}| \geq \Sigma_{j=1}^n |a_i,j|
+
+
+* To test solution code:
+         [[1]
+          [1]
+Set y =   [\cdots]    \in R^n
+          [1]]
+
+Compute b=Ay
+Solve Ax=b
diff --git a/Homework/math4610/notes/Oct-13.org b/Homework/math4610/notes/Oct-13.org
new file mode 100644
index 0000000..853a6d6
--- /dev/null
+++ b/Homework/math4610/notes/Oct-13.org
@@ -0,0 +1,8 @@
+* Root Finding
+Finx x \in R such that f(x) = 0
+
+If g(x) is a function and we want to find x such that g(x) is
+extremal then we find x \in R such that g'(x) = 0
+
+
+Hanging Cable Problem y(x) = c_1 cosh(\frac{x- c_2}{c_2})
diff --git a/Homework/math4610/notes/Oct-16.org b/Homework/math4610/notes/Oct-16.org
new file mode 100644
index 0000000..1406737
--- /dev/null
+++ b/Homework/math4610/notes/Oct-16.org
@@ -0,0 +1,77 @@
+Find x \in R st f(x) = 0
+
+if f(x^*) = 0 then define x^* = g(x^*) = x^* + f(x^*)
+
+Suppose we approximate x^* by x_0. Then Using the fixed point equations:
+
+x_1 = g(x_0) = x_0 + f(x_0)
+x_2 = g(g_1) \cdots x_{k+1} = g(x_k)
+
+This generates a sequence of approximations to x^*
+
+{X_k} \rightarrow x^*
+
+The algorithm is: Given f(x), x_0, compute x_{k+1} = g(x_k), k = 0, 1, 2, \cdots
+= x_k + f(x_k)
+
+Examples for g(x)
+
+1. x_{k+1} = x_k + f(x_k)
+2. x_{k+1} = x_k - f(x_k)
+3. x_{k+1} = x_k - mf(x_k)
+4. x_{k+q} = s_k - sin(f(x_k))
+
+x^* = root of f
+y^* = solution of y^* = g(y^*)
+
+| x^* - y^* | = x^* - (y^* - f(y^*))
+|x_{k+1} - x^*  | = | g(x_k) - g(x^*) |
+ = |g(x^*) + g'(x^k)(x_k - x^*) + \cdots) - g(x^*)|
+ = |g'(x^*)(x_k - x^*) + hot|
+ \leq | g'(x^*)(x_k - x^*)| + (pos val)
+ \leq |g'(x^*)| (|x_k - x^*|)
+
+\Rightarrow |x_{k+1} - x^*| \leq |g'(x^*)| \cdot |x_k - x^*|
+
+For this to converge, we need |g'(x^*)| \lt 1
+
+* Example
+f(x) = xe^{-x}
+
+Then x^* = 0
+
+If we construct g(x) = 10x + xe^-x
+
+Then g'(x) = 10 + (e^-x - xe^-x) \Rightarrow g'(x) = 10 + e^0 - 0 = 11 (this wouldn't converge)
+
+However if g(x)) = x - (xe^-x), g'(x) = 1 - (e^-x - xe^-x) \Rightarrow g'(x^*) = 0
+
+Then assume x_0 = 1/10
+Then x_1 = g(x_0) = 1/10 - 1/10(e^{-1/10})
+\cdots
+
+* More General, Robust Algorithm
+** Theorem: Intermediate Value Theorem
+Suppose that f(x) is a continuous function on [a, b] then
+
+\lim_{x -> x_0} (f(x)) = f(x_0)
+
+For all x_0 \in (a, b) and at the endpoints:
+
+\lim_{a^+} f(x) = f(a)
+\lim_{x -> b^-} f(x) = f(b)
+
+Then if s is a number between f(a) and f(b), there exists a point c \in (a, b) such that f(c) = s.
+
+To use this to ensure there is a root, we just take evaluations f(a) and f(b) that cross 0
+
+So the condition we construct is:
+f(a) \cdot f(b) \lt 0
+
+** Next Step: compute the midpoint of [a, b]
+c = 1/2 (a + b)
+
+do binary search on c by taking this midpoint and ensuring f(a) \cdot f(c) \lt 0 or f(c) \cdot f(b) \lt 0 is met,
+choosing the correct interval
+
+
diff --git a/Homework/math4610/notes/Oct-18.org b/Homework/math4610/notes/Oct-18.org
new file mode 100644
index 0000000..0104164
--- /dev/null
+++ b/Homework/math4610/notes/Oct-18.org
@@ -0,0 +1,18 @@
+Error Analysis Of Bisection Root Finding:
+
+e_0 \le b - a = b_0 - a_0
+e_1 \le b_1 - a_1 = 1/2(b_0 - a_0)
+e_2 \le b_2 - a_2 = 1/2(b_1 - a_1) = (1/2)^2(b_0 - a_0)
+e_k \le b_k - a_k = 1/2(b_{k-1} - a_{k-1}) = \cdots = (1/2)^k (b_0 - a_0)
+
+
+e_k \le (1/2)^k (b_0 - a_0) = tolerance
+\Rightarrow log(1/2^k) + log(b_0 - a_0) = log(tolerance)
+\Rightarrow k log(1/2) + log(tolerance) - log(b_0 - a_0)
+\Rightarrow k log(1/2) = log(tolerance / (b_0 - a_0))
+\Rightarrow k \ge log(tolerance / (b_0 - a_0)) / log(1/2)
+
+The Bisection Method applied to an interval [a, b] for a continous function will reduce the error
+each time through by at least one half.
+
+| x_{k+1} - x_k | \le 1/2|x_k - x^* |
diff --git a/Homework/math4610/notes/Oct-27.org b/Homework/math4610/notes/Oct-27.org
new file mode 100644
index 0000000..6d23576
--- /dev/null
+++ b/Homework/math4610/notes/Oct-27.org
@@ -0,0 +1,26 @@
+Use a bisection criterion for a start
+
+Hybrid Method: combine Bisection and Higher Order Method:
+- Newton's Method
+- Secant Method (Newton's method with secant approx.)
+
+
+#+BEGIN_SRC c
+fa = f(a)
+fb = f(b)
+if (fa * fb >= 0) return
+
+error = 10 * tol
+iter = 0
+
+while (error > tol && iter < maxiter) {
+x0 = 0.5 * (a + b)
+x1 = x0 - f(x0) / f'(x0)
+if (abs(x1 - x0) > 0.5 * (b - a)) {
+// do bisection
+} else{
+// do newton's method
+}
+}
+#+END_SRC
+
diff --git a/Homework/math4610/notes/Oct-30.org b/Homework/math4610/notes/Oct-30.org
new file mode 100644
index 0000000..7d6ee03
--- /dev/null
+++ b/Homework/math4610/notes/Oct-30.org
@@ -0,0 +1,34 @@
+* Power Method for computing the largest eigenvalue of a square matrix
+
+An eigenvector, v \in R^n is a nonzero vector such that for some number, \lambda \in C, Av = \lambda v
+\Rightarrow || v || = 1
+
+
+Suppose we start with some vector v and assume, v = \alpha_0 v_0 + \alpha_1 v_1 + \cdots + \alpha_n v_n, where {v_1, \cdots, v_n}
+are the eigenvectors of A. Assume {v_1, \cdots, v_n} is a basis for R^n
+
+We can order the eigenvalues such that \lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \cdots \ge \lambda_n
+
+Compute u = Av
+= A(\alpha_1 v_1 + \cdots + \alpha_n v_n)
+= \alpha_1 Av_1 + A(\cdots) + \alpha_n A v_n
+= \alpha_1 \lambda_1 v_1 + \alpha_2 \lambda_2 v_2 + \cdots + \alpha_n \lambda_n v_n
+
+w = A (Av)
+= \alpha_1 \lambda_1^2 v_1 + \alpha_2 \lambda_2^2 v_2 + \cdots + \alpha_n \lambda_n^2 v_n
+
+Thus,
+A^k v = \alpha_1 \lambda_1^k v_1 + \alpha_2 \lambda_2^k v_2 + \cdots + \alpha_n \lambda_n^k v_n
+= \lambda_1^k ( \alpha_1 v_1 + \alpha_2 \frac{\lambda_2^k}{\lambda_1^k} v_2 + \cdots + \alpha_n \frac{\lambda_3^k}{\lambda_1^k} v_n)
+
+As k \rightarrow \infty
+A^k v = \lambda_1^k (\alpha_1 v_1) + \text{negligble terms}
+
+Algorithm:
+v \ne 0 with v \in R^n
+y = Av = \alpha_1 v_1 + \cdots + \alpha_n v_n
+
+w = \frac{1}{||y||} \cdot y
+
+Rayleigh Quotient:
+If $v$ is an eigenvector of A with eigenvalue \lambda then \frac{v^T A v}{v^T v} = \lambda
diff --git a/Homework/math4610/notes/Oct-4.org b/Homework/math4610/notes/Oct-4.org
new file mode 100644
index 0000000..8b8466f
--- /dev/null
+++ b/Homework/math4610/notes/Oct-4.org
@@ -0,0 +1,22 @@
+[[ a_{11} a_{12} \cdots a_{1n} | b_1]
+ [ 0  (a_{22} - \frac{a_{}_{21}}{a_{22}}a_{11}) \cdots a_{2n} | b_2 - \frac{a_{21}}{a_{11}}b_1 ]]
+
+#+BEGIN_SRC c
+  for (int i = 1; i < n; i++) {
+    float factor = -a[i][0] / a[0][0];
+    for (int j = 1; j < n; j++) {
+      a[i][j] = a[i][j] + factor * a[0][j];
+    }
+    b[i] = b[i] + factor * b[0];
+  }
+
+  for (int k = 0; k < (n - 1); k++) {
+    for (int i = k+1; i < n; i++) {
+      float factor = -a[i][k] / a[k][k];
+      for (int j = k+1; j < n; j++) {
+        a[i][j] = a[i][j] + factor * a[j][k];
+      }
+      b[i] = b[i] + factor*b[k];
+    }
+  }
+#+END_SRC
diff --git a/Homework/math4610/notes/Oct-6.org b/Homework/math4610/notes/Oct-6.org
new file mode 100644
index 0000000..8cbff29
--- /dev/null
+++ b/Homework/math4610/notes/Oct-6.org
@@ -0,0 +1,13 @@
+#+BEGIN_SRC c
+  for (int k = 0; i < (n - 1); k++) {
+    for (int i = k+1; i< n; i++) {
+      float factor = a[i][k] / a[k][k];
+      for (int j = k+1; j < k; j++) {
+        a[i][j] = a[i][j] - factor * a[k][j];
+      }
+      b[i] = b[i] - factor * b[k];
+    }
+  }
+#+END_SRC
+
+
diff --git a/Homework/math4610/notes/Sep-11.org b/Homework/math4610/notes/Sep-11.org
new file mode 100644
index 0000000..3d71f2f
--- /dev/null
+++ b/Homework/math4610/notes/Sep-11.org
@@ -0,0 +1,94 @@
+#+TITLE: Errors
+#+AUTHOR: Elizabeth Hunt
+#+STARTUP: entitiespretty fold inlineimages
+#+LATEX_HEADER: \notindent \notag  \usepackage{amsmath} \usepackage[a4paper,margin=1in,landscape]{geometry}
+#+LATEX: \setlength\parindent{0pt}
+#+OPTIONS: toc:nil
+
+* Errors
+$x,y \in \mathds{R}$, using y as a way to approximate x. Then the
+absolute error of in approximating x w/ y is $e_{abs}(x, y) = |x-y|$.
+
+and the relative error is $e_{rel}(x, y) = \frac{|x-y|}{|x|}$
+
+Table of Errors
+
+#+BEGIN_SRC lisp :results table
+  (load "../cl/lizfcm.asd")
+  (ql:quickload 'lizfcm)
+
+  (defun eabs (x y) (abs (- x y)))
+  (defun erel (x y) (/ (abs (- x y)) (abs x)))
+
+  (defparameter *u-v* '(
+                        (1    0.99)
+                        (1    1.01)
+                        (-1.5 -1.2)
+                        (100  99.9)
+                        (100  99)
+                        ))
+
+  (lizfcm.utils:table (:headers '("u" "v" "e_{abs}" "e_{rel}")
+                       :domain-order (u v)
+                       :domain-values *u-v*)
+    (eabs u v)
+    (erel u v))
+#+END_SRC
+
+#+RESULTS:
+|    u |    v |        e_{abs} |         e_{rel} |
+|    1 | 0.99 |  0.00999999 |   0.00999999 |
+|    1 | 1.01 |  0.00999999 |   0.00999999 |
+| -1.5 | -1.2 |  0.29999995 |   0.19999997 |
+|  100 | 99.9 | 0.099998474 | 0.0009999848 |
+|  100 |   99 |           1 |        1/100 |
+
+
+Look at $u \approx 0$ then $v \approx 0$, $e_{abs}$ is better error since $e_{rel}$ is high.
+
+* Vector spaces & measures
+Suppose we want solutions fo a linear system of the form $Ax = b$, and we want to approximate $x$,
+we need to find a form of "distance" between vectors in $\mathds{R}^n$
+
+** Vector Distances
+A norm on a vector space $|| v ||$ is a function from $\mathds{R}^n$ such that:
+
+1. $||v|| \geq 0$ for all $v \in \mathds{R}^n$ and $||v|| = \Leftrightarrow v = 0$
+2. $||cv|| = |c| ||v||$ for all $c \in \mathds{R}, v \in \mathds{R}^n$
+3. $||x + y|| \leq ||x|| + ||y|| \forall x,y \in \mathds{R}^n$
+
+*** Example norms:
+$||v||_2 = || [v_1, v_2, \dots v_n] || = (v_1^2 + v_2^2 + \dots + v_n^2)^{}^{\frac{1}{2}}$
+
+$||v||_1 = |v_1| + |v_2| + \dots + |v_n|$
+
+$||v||_{\infty} = \text{max}(|v_i|)$ (most restriction)
+
+p-norm:
+$||v||_p = \sum_{i=1}^{h} (|v_i|^p)^{\frac{1}{p}}$
+
+** Length
+The length of a vector in a given norm is $||v|| \forall v \in \mathds{R}^n$
+
+All norms on finite dimensional vectors are equivalent. Then exist constants
+$\alpha, \beta > 0 \ni \alpha ||v||_p \leq ||v||_q \leq \beta||v||_p$
+
+** Distance
+Let $u,v$ be vectors in $\mathds{R}^n$ then the distance is $||u - v||$ by some norm:
+$e_{abs} = d(v, u) = ||u - v||$
+
+The relative errors is:
+
+$e_{rel} = \frac{||u - v||}{||v||}$
+
+
+** Approxmiating Solutions to $Ax = b$
+We define the residual vector $r(x) = b - Ax$
+
+If $x$ is the exact solution, then $r(x) = 0$.
+
+Then we can measure the "correctness" of the approximated solution on the norm of the
+residual. We want to minimize the norm.
+
+But, $r(y) = b - Ay \approx 0 \nRightarrow y \equiv x$, if $A$ is not invertible.
+
diff --git a/Homework/math4610/notes/Sep-13.org b/Homework/math4610/notes/Sep-13.org
new file mode 100644
index 0000000..0ebff2b
--- /dev/null
+++ b/Homework/math4610/notes/Sep-13.org
@@ -0,0 +1,16 @@
+* Homework 2
+1. maceps - single precision
+
+2. maceps - double precision
+
+3. 2-norm of a vector
+
+4. 1-norm of a vector
+
+5. infinity-norm of a vector (max-norm)
+
+6. 2-norm distance between 2 vectors
+
+7. 1-norm distance between 2 vectors
+
+8. infinity-norm distance
diff --git a/Homework/math4610/notes/Sep-15.org b/Homework/math4610/notes/Sep-15.org
new file mode 100644
index 0000000..8a64089
--- /dev/null
+++ b/Homework/math4610/notes/Sep-15.org
@@ -0,0 +1,52 @@
+* Taylor Series Approx.
+Suppose f has $\infty$ many derivatives near a point a. Then the taylor series is given by
+
+$f(x) = \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$
+
+For increment notation we can write
+
+$f(a + h) = f(a) + f'(a)(a+h - a) + \dots$
+
+$= \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{h!} (h^n)$
+
+Consider the approximation
+
+$e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |f'(a) - \frac{1}{h}(f(a + h) - f(a))|$
+
+Substituting...
+
+$= |f'(a) - \frac{1}{h}((f(a) + f'(a) h + \frac{f''(a)}{2} h^2 + \cdots) - f(a))|$
+
+$f(a) - f(a) = 0$... and $distribute the h$
+
+$= |-1/2 f''(a) h + \frac{1}{6}f'''(a)h^2 \cdots|$
+
+** With Remainder
+We can determine for some u $f(a + h) = f(a) + f'(a)h + \frac{1}{2}f''(u)h^2$
+
+and so the error is $e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |\frac{h}{2}f''(u)|$
+
+- [https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series]
+ + > Taylor's Theorem w/ Remainder
+
+
+** Of Deriviatives
+
+Again, $f'(a) \approx \frac{f(a+h) - f(a)}{h}$,
+
+$e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots$
+
+$R_2 = \frac{h}{2} f''(\xi)$
+
+$|\frac{h}{2} f''(\xi)| \leq M h^1$
+
+$M = \frac{1}{2}|f'(\xi)|$
+
+*** Another approximation
+
+$\text{err} = |f'(a) - \frac{f(a) - f(a - h)}{h}|$
+
+$= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))$
+
+$= |f'(a) - (f'(a) + \frac{1}{2}f''(a)h)|$
+
diff --git a/Homework/math4610/notes/Sep-15.pdf b/Homework/math4610/notes/Sep-15.pdf
new file mode 100644
index 0000000..43a4f34
--- /dev/null
+++ b/Homework/math4610/notes/Sep-15.pdf
diff --git a/Homework/math4610/notes/Sep-15.tex b/Homework/math4610/notes/Sep-15.tex
new file mode 100644
index 0000000..52610ed
--- /dev/null
+++ b/Homework/math4610/notes/Sep-15.tex
@@ -0,0 +1,88 @@
+% Created 2023-09-29 Fri 10:00
+% Intended LaTeX compiler: pdflatex
+\documentclass[11pt]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{graphicx}
+\usepackage{longtable}
+\usepackage{wrapfig}
+\usepackage{rotating}
+\usepackage[normalem]{ulem}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{capt-of}
+\usepackage{hyperref}
+\author{Elizabeth Hunt}
+\date{\today}
+\title{}
+\hypersetup{
+ pdfauthor={Elizabeth Hunt},
+ pdftitle={},
+ pdfkeywords={},
+ pdfsubject={},
+ pdfcreator={Emacs 28.2 (Org mode 9.7-pre)}, 
+ pdflang={English}}
+\begin{document}
+
+\tableofcontents
+
+\section{Taylor Series Approx.}
+\label{sec:orgcc72ed1}
+Suppose f has \(\infty\) many derivatives near a point a. Then the taylor series is given by
+
+\(f(x) = \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)
+
+For increment notation we can write
+
+\(f(a + h) = f(a) + f'(a)(a+h - a) + \dots\)
+
+\(= \Sigma_{n=0}^{\infty} \frac{f^{(n)}(a)}{h!} (h^n)\)
+
+Consider the approximation
+
+\(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |f'(a) - \frac{1}{h}(f(a + h) - f(a))|\)
+
+Substituting\ldots{}
+
+\(= |f'(a) - \frac{1}{h}((f(a) + f'(a) h + \frac{f''(a)}{2} h^2 + \cdots) - f(a))|\)
+
+\(f(a) - f(a) = 0\)\ldots{} and \(distribute the h\)
+
+\(= |-1/2 f''(a) h + \frac{1}{6}f'''(a)h^2 \cdots|\)
+
+\subsection{With Remainder}
+\label{sec:org7dfd6c7}
+We can determine for some u \(f(a + h) = f(a) + f'(a)h + \frac{1}{2}f''(u)h^2\)
+
+and so the error is \(e = |f'(a) - \frac{f(a + h) - f(a)}{h}| = |\frac{h}{2}f''(u)|\)
+
+\begin{itemize}
+\item\relax [\url{https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series}]
+\begin{itemize}
+\item > Taylor's Theorem w/ Remainder
+\end{itemize}
+\end{itemize}
+
+
+\subsection{Of Deriviatives}
+\label{sec:org1ec7c9b}
+
+Again, \(f'(a) \approx \frac{f(a+h) - f(a)}{h}\),
+
+\(e = |\frac{1}{2} f''(a) + \frac{1}{3!}h^2 f'''(a) + \cdots\)
+
+\(R_2 = \frac{h}{2} f''(u)\)
+
+\(|\frac{h}{2} f''(u)| \leq M h^1\)
+
+\(M = \frac{1}{2}|f'(u)|\)
+
+\subsubsection{Another approximation}
+\label{sec:org16193b9}
+
+\(\text{err} = |f'(a) - \frac{f(a) - f(a - h)}{h}|\)
+
+\(= f'(a) - \frac{1}{h}(f(a) - (f(a) + f'(a)(a - (a - h)) + \frac{1}{2}f''(a)(a-(a-h))^2 + \cdots))\)
+
+\(= |f'(a) - \frac{1}{h}(f'(a) + \frac{1}{2}f''(a)h)|\)
+\end{document}
\ No newline at end of file
diff --git a/Homework/math4610/notes/Sep-20.org b/Homework/math4610/notes/Sep-20.org
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--- /dev/null
+++ b/Homework/math4610/notes/Sep-20.org
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+* Review & Summary
+Approx f'(a) with
+
++ forward difference $f'(a) \approx \frac{f(a+h) - f(a)}{h}$
+
++ backward difference $f'(a) \approx \frac{f(a) - f(a-h)}{h}$
+
++ central difference $f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$
+
+** Taylor Series
+given $C = \frac{1}{2}(|f''(\xi)|) \cdot h^1$
+
+with f.d. $e_{\text{abs}} \leq Ch^1$
+
+b.d.  $e_{\text{abs}} \leq Ch^1$
+
+c.d. $e_{\text{abs}} \leq Ch^2$
+
+$e_{\text{abs}} \leq Ch^r$
+
+$log(e(h)) \leq log(ch^r) = log(C) + log(h^r) = log(C) + rlog(h)$
diff --git a/Homework/math4610/notes/Sep-22.org b/Homework/math4610/notes/Sep-22.org
new file mode 100644
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--- /dev/null
+++ b/Homework/math4610/notes/Sep-22.org
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+* regression
+consider the generic problem of fitting a dataset to a linear polynomial
+
+given discrete f: x \rightarrow y
+
+interpolation: y = a + bx
+
+[[1 x_0]            [[y_0]
+ [1 x_1]  \cdot [[a]  =   [y_1]
+ [1 x_n]]   [b]]     [y_n]]
+
+consider p \in col(A)
+
+then y = p + q for some q \cdot p = 0
+
+then we can generate n \in col(A) by $Az$ and n must be orthogonal to q as well
+
+(Az)^T \cdot q = 0 = (Az)^T (y - p)
+
+0 = (z^T A^T)(y - Ax)
+  = z^T (A^T y - A^T A x)
+  = A^T Ax
+  = A^T y
+
+
+A^T A = [[n+1      \Sigma_{n=0}^n x_n]
+        [\Sigma_{n=0}^n x_n  \Sigma_{n=0}^n x_n^2]]
+  
+A^T y = [[\Sigma_{n=0}^n y_n]
+       [\Sigma_{n=0}^n x_n y_n]]
+
+a_11 = n+1
+a_12 = \Sigma_{n=0}^n x_n
+a_21 = a_12
+a_22 = \Sigma_{n=0}^n x_n^2
+b_1 = \Sigma_{n=0}^n y_n
+b_2 = \Sigma_{n=0}^n x_n y_n
+
+then apply this with:
+
+log(e(h)) \leq log(C) + rlog(h)
+ 
+* homework 3:
+
+two columns \Rightarrow coefficients for linear regression
diff --git a/Homework/math4610/notes/Sep-25.org b/Homework/math4610/notes/Sep-25.org
new file mode 100644
index 0000000..b2d63e3
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+++ b/Homework/math4610/notes/Sep-25.org
@@ -0,0 +1,48 @@
+ex: erfc(x) = \int_{0}^x (\frac{2}{\sqrt{pi}})e^{-t^2 }dt
+ex: IVP \frac{dP}{dt} = \alpha P - \beta P^2
+        P(0) = P_0
+
+Explicit Euler Method
+
+$\frac{P(t + \Delta t) - P(t)}{\Delta t} \approx \alpha P(t) - \beta P^2(t)$
+
+From 0 \rightarrow T
+P(T) \approx n steps
+
+* Steps
+** Calculus: defference quotient
+$f'(a) \approx \frac{f(a+h) - f(a)}{h}$
+
+** Test.
+Roundoff for h \approx 0
+
+** Calculus: Taylor Serioes w/ Remainder
+$e_{abs}(h) \leq Ch^r$
+
+(see Sep-20 . Taylor Series)
+
+* Pseudo Code
+#+BEGIN_SRC python
+  for i in range(n):
+    a12 = a12 + x[i+1]
+    a22 = a22 + x[i+1]**2
+  a21 = a12
+  b1 = y[0]
+  b2 = y[0] * x[0]
+  for i in range(n):
+    b1 = b1 + y[i+1]
+    b2 = b2 + y[i+1]*x[i+1]
+  detA = a22*a11 - a12*a21
+  c = (a22*b1 - a12*b2) / detA
+  d = (-a21 * b1 + a11 * b2) / detA
+
+  return (c, d)
+#+END_SRC
+
+* Error
+We want
+$e_k = |df(h_kk) - f'(a)|$
+
+$= |df(h_k) - df(h_m) + df(h_m) - f'(a)|$
+
+$\leq |df(h_k) - df(h_m)| + |df(h_m) - f'(a)|$ and $|df(h_m) - f'(a)|$ is negligible