Surviving…

I have been up seen four the last three days so I am running low on energy, but before class starts today I want to place my POW #7 post up…

1) Getting help from my math major suitemate, Liz Mosher!

2) I think my favorite Theorem I learned thus far would have to be the Fundmental Theorem of Calculus. The reason why I like this theorem is mainly due to the fact that Calculus has been my favorite math class to date (of course FOM will probably replace it after this semester…wink, wink)

3) I have actually been working hard in this class. I think the hardest part of  FOM is keeping up with the reading, only because it is hard to follow where we are in class. Most of the times I end up reading too much of the chapter. I have done all the homework and POW as well as been keeping up with my blog weekly. To conclude I think I have been working hard.

I almost forgot?!?

Ahhh this week has been long it almost slipped my mind to post, but no need to worry I will make this post extra long!

I figured to make up for lost time I might as well go over the worksheet for this week…

#1 a) How many functions are there from (x){1,2,3} to (y){1,2,3,4,5}?

–personally speaking it is easier for me to look at functions vertically (like elementary school), it gives me the ability to draw lines between the two functions.

To find how many functions it is the number of elements in the second function powered to the number of elements in the first function, therefore the answer is 5^3.

b) How many of them are onto?

–this is where drawing the functions vertically really helps

none of them are onto because every element in y is not an output for f

c) How many of them are one-to-one

3 functions are one-to-one because the three elements in x go to a different output

#2 a) How many functions are there from (x){1,2,3,4,5} to (y){1,2,3}?
similar to before y^x thus, 5^3

b) How many of them are onto?

there are 69 functions that are onto…getting 69 takes alot of addition

c) How many of them are one-to-one?

none because the five elements in x can’t go to different outputs in y

#3

a) If f & g are surjective, then g0f  is surjective–True the best explanation my group could think of was that g is the co-domain of f therefore if both are surjective their composite with also be surjective

b) If g0f is surjective, then f is surjective–false

f(x)=sqrt (X), g(x)= y^2

#4 (because I have trouble using latex I am just going to write our explanations of each statement)

a) the statement states that the functions are not onto

b) If 3/(n^2), then 3/n (are group couldn’t figure out a simpler way to put it)

c) f is a decreaseing function

#5  Due to my inability to latex and the mere fact that two of the problems are due for ths weeks homework I will pass this question!

Phewwwwwww!!! If you have any questions or comments feel free to respond 🙂

GOOD LUCK ON THE MIDTERM TOO!!!

This blew my mind today in class and I thought I might as well blog about it.  I thought it was really cool how negating something to the millionth time would actually just be  the first negation. Therefore this means that we could figure out how to negate things to the trillionth. This is possible because negation runs in cycles similar to imginary numbers.  Mind blown.

On another note I was interested in Cyrus question today about  Fibonacci and golden numbers so i check out this website (the one highlighted in blue) to get a better explanaion. Anywho its is actually really cool and I recommend checking it out, in fact I recommend checking out the entire website it is about Fibonacci and Nature.

More Functions

To continue from last post….functions!!

A function f: A–>B

Surjective=onto=EPIC (can be used interchangably)

A surjective function means every element in B is an output for f (the function).

Example of surjective function:

y=x^3

Injective=one-to-one=monic

A injective function means for every input there is a single output (aka: the horizontal line test)

Example of  injective function:

y=e^x

Bijecture

A bijecture function means a function that is both epic and one-to-one

Example of bijecture function:

y=3x+5

Prisoners of FOM

FUNCTIONS:

Yesterday in class Casey talked about functions, I thought that it would be beneficial to reenforce some of the concepts.

The first concept that I thought is important is that for every input there is an output. This means that for a function to exist it must pass the vertical line test. Thus for every input there must be a single output.

Gentle Definition:
A function from a set A to a set B is a rule that associates some $b/in B$ to elements $a/in A$ (latex fail)

The second conceept that I thought to reenforce would be the hardcore definition that Casey explained in class. The reason why “math people” like the hardcore definition is mainly because it is easier to place in a proof. It has teo elements in the defintion that must satisfied before the function is consider a true function (aka pass the vertical line test)

To conclude, don’t forget to do this week’s homework and worksheet on functions!!!

POW #5

Helping my roomie with vector!!

POW #4

10 coins in a bag, each having a different value, consecutively:

1,2,3,4,5,6,7,8,9, and 10 dollar coins

We know that a minimum \$10 of coins will be donated…

(1,9) (2,8) (3,7) (4,6) (10)

to at least 5 charities. But If we only donated minimum of \$10 to each charitiy we would not be using all the coins, in fact the (5) coin is missing. Supicious, we try donating \$11 to each charitiy by using ALL the coins.

TRY \$11:

(1,10) (2,9) (3,8) (4,7) (5,6)

We still serve 5 charities, but would be giving each charity more money and use all the coins!

So does a PATTERN emerge?

9 coins—-> Each charity gets \$9 each
10 coins—> Each charity gets \$11 each
11 coins—> Each charity gets \$11 each
12 coins—> Each charity gets \$13 each

In terms of n:

Odd number of coins n is an element of 2Z+1, (n-1)/2 charities would recieve \$n each

Even number of coins n is an element of 2Z, (n)/2 charities would recieve \$n+1 each

“Die Hard” doing FOM

The other day in class Casey told us to look up the Die Hard: Jug Problem, so what else was I suppose to do except look it up. In short, the jug problem has to do with basic addition and subtraction. The two men each have a container, one container holds 5 gallons while the other one holds 3 gallons. In order for them to get 4 gallons to disarm the bomb they do the following mathematical operations:

5-3=2 (Poured the full 5 gallon jug into the three gallon jug, leaving only two gallons in the 5 gallon jug)

3-2=1 (Poured te 2 gallons from the 5 gallon jug and empty the 3 gallons into the 5 gallon jug)

5-1=4 (Fill up the 5 gallon jug, then fill up the 3 gallon jug (containing two gallons of water) from the five gallon jug, leaving 4 gallons in the five gallon jug)

At first I thought it was confusing, but once you go through it slowly it make complete sense! This trick it actually really cool and useful if I ever need to disarm a bomb with exactly 4 gallons of water. Thanks to Bruce Willis for saving the day!

Music to my ears

I just wanted to mention that last semester I attended Professor Kung’s presentation at St. Mary’s Hall about music and math’s relationship. I thought the presentation was by far the most interesting lecture I have been to thus far. The musicians playing at the presentations were very talented and had great stage presence. Professor Kung’s presentation was basically about the wavelengths of musical instruments and how he could capture the sound than analyze it significance using advance technology. He uses props to show how the scientific finding which allows for less advance math students to follow. To conclude I recommend seeing his presentation it was AMAZING!!!!

It’s a miracle

Following up on my last post, yestrday inclass we learned that the conjecture if |s|=n then |P(s)|=2^n. Ultimately this means that the answer to number seven on the in-class worksheet is 65,536.
Example: 2^2=4
2^4=16
2^16= 65,536