# A review based post

I have a midterm on friday. We are reviewing today in class. I don’t want to review, I want to update. However, that’s not a good compromise. Therefore, I compromise, and choose to update by giving you my review notes. Don’t ask me how I’m going to type out matrix’s if they come up… Maybe I’ll use the table feature.

But until we get to the review, lets talk about something else. Um… I could go off on my google taking over the world theory. But no…

Review Time:

Exam Covers Chap 5, 6, and 7 (up to LUD composition)
Only responsible for material in lecture (in other words all of you at home are screwed)
No Programming
No Calculators
Open Book
1 (8 1/2&quot; x 11&quot;) Crib SheetBasic Trapezoid Rule:

intergral(a,b, f(x)dx = (b-a)((f(a)+f(b))/2)-(f”(E)/12)(b-a)^3)Basic Simpsons Rule:

intergral(a,b, f(x) dx = ((b-a)/6)[f(a)+ 4f((a+b)/2) + f(b)] – ((f””(E)/2880) (b-a)^5))Composite Simpsons Rule:

intergral(a,b, f(x)dx = (h/3)[f(a) + 2 sum(j=1, ((n/2)-1), f(x2j) + 4 sum(j=1, x2j-1) + f(b)] – (b-a) h^4/180 f””(m) WHERE
h=ba/nComposite Trapezoid Rule: Look it up

h = ba/2
s(a,b) = h/3[f(a) + 4f(a+ h) + f(b)]
s2(a,b) = h/6[f(a) + 4f(a+h/2) + f(a + h) + f(a+ h) + 4f(a + 3h/2) + f(b)]
integral(a,b, f(x)dx = s2(a,b) – 1/15[s1(a,b) – s2(a,b)] &lt;- Error estimate.
to get within some e must have following be true:
|s1(a,b) -s2(a,b) &lt; 15e|
if test fails, we set the e/2 and recurse.Romberg Integration:

R(0,0) = (b-a)/2[f(a) + f(b)]
R(n,0) = (1/2)R((n-1),0) + h sum(k=1, 2^(n-1), f[a+ (2k-1) + h] where h = (b-a)/2^n
R(n,m) = R(n,m-1) + (1/(4^m -1))[R(n,m-1) – R(n-1, m-1)
R(0,0)\
R(1,0)\-&gt;R(1,1)\
R(2,0)\- &gt;R(2,1)-&gt;R(2,2)
O(h^2)    O(h^4)    O(h^6)Gaussian quadrature for nodes xi and weights of Ai

intergrate(c,d, f(x)dx = sum(n=0, n, Aif(xi)))
to change the interval from [c,d] to [a,b]:
intergrate(a,b, f(x) dx =
(b – a)/(d-c) sum(i=0, h, Ai(f((b-a)/(d-c))x + (ad-bc)/(d-c))Naive Gaussian Elimination

Ax = b
kth step in forward elimination
mik &lt; aik/bik
No more on this because I know how to do it.
Maybe freshen up though.
Just in case: x = b/a, xi &lt;- [bi – sum(j,n, ajxj)]Gaussian Elimination with scalled partial pivoting:

Look for biggest aij in each row and store in vector
initialize a row order vector l (1,2,3,…)
at the kth step of a forward elimination we look at the largest (alk/ sl) k&lt; i&lt; n to select pivot row, then revise pivot vector.Tridiagonal and pentadiagonal systems:

use forward elimination (efficient version) to create trivial solution for x4, given x4 can solve for x3, x2, and x1.