Friday, May 16, 2014
Partial fractions video for funnnnnnn
Vectors in space



For a general vector space, the scalars are members of a field
, in which case
is called a vector space over
.
Euclidean
-space
is called a real vector space, and
is called a complex vector space.
In order for
to be a vector space, the following conditions must hold for all elements
and any scalars
:
1. Commutativity:
(1) |
2. Associativity of vector addition:
(2) |
3. Additive identity: For all
,
(3) |
4. Existence of additive inverse: For any
, there exists a
such that
(4) |
5. Associativity of scalar multiplication:
(5) |
6. Distributivity of scalar sums:
(6) |
7. Distributivity of vector sums:
10.5 Parametric Equations
Definition of a plane curve
If f and g are continuous functions of t on an interval l, the set of ordered pairs (f(t),g(t)) is a plane curve C. The equations:
x=f(t) and y=g(t)
are parametric equations for C, and t is the parameter.
Eliminating the Parameter
1. Parametric equations
2. solve for t in one equation
3. substitute into second equation
4.rectangular equation (only 2 variable)
Example:
step 1: subtract both side of x=t+3 by three:
x-3=t
then, take x-3 into another equation
get y= - (x-3)^2
Tuesday, May 6, 2014
Aristotle and logic : creating something from nothing
Mathematical Induction
This is the section of my presentation!!
1. Definition:
mathematical induction is a method for proving that statements involving natural numbers are true for all natural numbers.
2. Theorem:
suppose that 他the following two conditions are satisfied with regard to a statement bout natural numbers:
CONDITIONI:The statement is true for natural number 1
Condition II: if the statement is true for some natural number k, it is also true for k+1
Thus the statement is true for all numbers
3. sums of powers of integers
Examples:
n=1 1 = 1(2)/2 which is clearly true.
Here we must prove the following assertion: "If there is a k such that P(k) is true, then (for this same k) P(k+1) is true."
assume there is a k such that 1 + 2 + ... + k = k (k+1)/2.
We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2.
1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2.
a0=1 an=a(n-1)+2
a0=1
a1=3
a2=5
a3=7


