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
Determinants of a matrix
Wednesday, April 30, 2014
Limits
Sometimes you can't work something out directly ... but you can see what it should be as you get closer and closer! We call that the limit of a function.
![]() | For example, what is the value of (x2-1)/(x-1) when x=1? (Notice in the graph there is actually a "hole" at that point). Let us substitute "1" for "x" and see the result: (12-1)/(1-1) = (1-1)/(1-1) = 0/0 |
But 0/0 is "indeterminate", meaning we can't determine its value, so instead of trying to work it out for x=1 let's try approaching it closer and closer:
| x | (x2-1)/(x-1) |
|---|---|
| 0.5 | 1.50000 |
| 0.9 | 1.90000 |
| 0.99 | 1.99000 |
| 0.999 | 1.99900 |
| 0.9999 | 1.99990 |
| 0.99999 | 1.99999 |
| ... | ... |
Now we can see that as x gets close to 1, then (x2-1)/(x-1) gets close to 2, so we say:
The limit of (x2-1)/(x-1) as x approaches 1 is 2
And it is written in symbols as:
![]()
We don't know the value at x=1 (it is indeterminate), but the limit has a value of 2
More Formal
But you can't just say a limit equals some value because it looked like it was going to. We need a more formal definition.
So let's start with the general idea
From English to Mathematics
Let's say it in English first:
"f(x) gets close to some limit as x gets close to some value"
If we call the Limit "L", and the value that x gets close to "a" we can say
"f(x) gets close to L as x gets close to a"
![]()
Calculating "Close"
Now, what is a mathematical way of saying "close" ... could we subtract one value from the other?
Example 1: 4.01 - 4 = 0.01 ![]()
Example 2: 3.8 - 4 = -0.2 ![]()
Hmmm ... negatively close? That doesn't work ... we really need to say "I don't care about positive or negative, I just want to know how far" The solution is to use the absolute value.
"How Close" = |a-b|
Example 1: |4.01-4| = 0.01 ![]()
Example 2: |3.8-4| = 0.2 ![]()
And if |a-b| is small we know we are close, so we write:
"|f(x)-L| is small when |x-a| is small"
| And this animation shows you what happens with the function f(x) = (x2 - 1) / (x-1)
So
|
Delta and Epsilon
But "small" is still English and not "Mathematical-ish".
Let's choose two values to be smaller than:
| that |x-a| must be smaller than | |
| that |f(x)-L| must be smaller than |
And we have:
"|f(x)-L|< |
That actually says it! So if you understand that you understand limits ...
... but to be absolutely precise we need to add these conditions:
| 1) | 2) | 3) |
| it is true for any | x not equal to a means0<|x-a| |
And this is what we get:
"for any |
That is the formal definition. It actually looks pretty scary, doesn't it!
But in essence it still says something simple: when x gets close to a then f(x) gets close to L.
How to Use it in a Proof
To use this definition in a proof, we want to go
| From: | To: | |
| 0<|x-a|< | |f(x)-L|< |
This usually means finding a formula for
(in terms of
) that works.
How do we find such a formula?
Guess and Test!
That's right, you can:
- Play around till you find a formula that might work
- Test to see if that formula works.
Example: Let's try to show that
![]()
Using the letters we talked about above:
- The value that x approaches, "a", is 3
- The Limit "L" is 10
So we want to know:
| How do we go from: | 0<|x-3|< | to | |(2x+4)-10|< |
Step 1: Play around till you find a formula that might work
| Start with: | |(2x+4)-10|< |
| Simplify: | |2x-6|< |
| Move 2 outside: | 2|x-3|< |
| Move 2 across: | |x-3|< |
So we can now guess that
=
/2 might work
Step 2: Test to see if that formula works.
So, can we get from 0<|x-3|<
to |(2x+4)-10|<
... ?
Let's see ...
| Start with: | 0<|x-3|< |
| Replace | 0<|x-3|< |
| Move 2 across: | 0<2|x-3|< |
| Move 2 inside: | 0<|2x-6|< |
| Replace "-6" with "+4-10" | 0<|(2x+4)-10|< |
Yes! We can go from 0<|x-3|<
to |(2x+4)-10|<
by choosing
=
/2
DONE!
We have seen then that if we choose a
we can find a
, so it is true that:
"for any
, there is a
so that |f(x)-L|<
when 0<|x-a|<
"
And we have proved that




