Sunday, December 8, 2013

SP #6: Unit K Concept 10: Writing Repeating Decimals as Rational Numbers

 What is this problem about?
Sometimes in life, or in math analysis, you will come across a repeating decimal. Now we want to find its geometric infinite sum without a calculator! The sum we find will be the number that the term converges on but will never reach. 

What must the viewer pay special attention to in order to understand?
Viewer, please, please, please pay careful attention to your decimal places especially when solving for the ratio.
Also be sure to multiply both sides of the fraction by the denominator's reciprocal. Save yourself the hassle and be sure to plug in the correct terms into the formula. At the end do not forget to add your whole number to the decimal part or else your answer will be incomplete. Once you have done that, always check to see if the fraction can simplify!

Tuesday, November 26, 2013

Fibonacci Beauty Ratio Reflective Essay

**word count: 991**

Golden Ratio in Human Body
The Fibonacci sequence terms are the sum of the preceding two numbers. But when you divide one number in the series by the number before it, you get numbers very close to one another. This number becomes fixed after the thirteenth term. This fixed number (1.618) is known as the Golden Ratio. In this video we learn that we can actually measure what humans find as "aesthetically pleasing" with the human body. Artists, sculptors and designers like Leonardo di Vinci and Le Corbusier use those proportions in their work. The Golden Ratio was also used architecture such as in the Great Pyramids in Egypt, the Parthenon in Greece. The Golden Ratio can be found in a myriad of surprising ways in the human body: distance from the fingertips to the elbow by the wrist to the elbow, the distance between the navel and the top of the head by the distance of the shoulder line to the top of the head, the total width of the front two teeth over their height, the length of face over the width of face, the length of mouth over the width of nose, the width of nose over the distance between nostrils(weird, but cool!) In 1985-1987 study, American physicists were able to prove the existence of the Golden Ratio in our lungs. The asymmetrical structure of the bronchi (one branch is shorter than the other) allows for a Golden Ratio to occur.  

The Beauty of the Gold Ratio
As mentioned before, the Golden Ratio was present in both Egyptian and Greek cultures but it can also be founded in our modern culture such as United Nations building, the width of the building compared with the height of every ten floors is a Golden Ratio, which is also known as Divine Ratio, the Golden Mean, the Golden Number, and the Golden Section. The Fibonacci Sequence is an infinite sequence, it will go on forever and ever. The ratio of the consecutive terms gets closer to the Golden Ratio of 1.618. But the ratio of Fibonacci number to preceding number will never reach Nth Fibonacci term, it is always approaching the one particular constant but will never get there, so sad L This Golden Ratio has shown up all over history and has had a great deal of influence on what humans find aesthetically pleasing. But here are some cold hard facts:  Phi was first understood and used by the ancient mathematicians in Egypt due to its frequent appearance in geometry. It was later given its name Phidias (500BC-432 BC), a Greek sculptor and mathematician, who studied Phi and used phi in his work for example, the statue of the goddess Athena in Athena, and the state of god Zeus in Olympiad. The symbol came from mathematician in 1900’s named Mark Barr, who represented the Golden Ratio by using a Greek symbol Φ.

Nature’s Number: 1.618033988…
Fibonacci was a man known as Leonardo of Pisa (like the Leaning Tower of Pisa). Phi, which comes from his Fibonacci sequence, is completely underrated in the world with everybody’s fascination with pi, infinity, etc. The Golden Rectangle which is broken up into two parts; one part is a square with a length and width of one and the other part is a rectangle, length is 1 and width is 0.618 which gives us phi (golden ratio) 1.618. This is present in the Greek’s Parthenon, Mona Lisa, and various other paintings. Leonardo da Vinci was convinced the Golden Ratio (1.618) defined perfect proportions within the human body. Within the Golden Rectangle, you can get the Spiral of Archimedes. This spiral is found in snails, in crustaceans and so on!

Golden Ratio in Art and Architecture
Le Corbusier at first wasn’t very…fond of the Golden Ratio. He was completely adamant to it. Before 1927, it is shown in his work that he never used the Golden Ratio. His fascination with aesthetics and such came from his curiosity but it also came from his search for a standardized proportion. From this came the Modulor, a new proportional system based on the Fibonacci series (1,1,2,3,5,8,13), providing, “a harmonic measure to the human scale, universally applicable to architecture and mechanics.” The Modulor was developed between 1943 and 1955, where mathematics was significant in showing people universal truths. It was also a time where there was an increasing need for a mass production, the Modulor was to be that standard model. Fun fact: the entire building "Unite d'" was based on the Modulor proportioning system. Also the musical scales are based on Fibonacci numbers; the piano keyboard of C to C above of 13 keys has 8 white keys and 5 black keys, split into groups of 3 and 2. Musical instruments design such as the violin are also based on phi!

Personal Reflection
From my watching and my reading of the Fibonacci Sequence and the Golden Ratio, I have just learned so much more about it. The last time I read about phi was in Dan Brown’s The Da Vinci Code. It was a book I enjoyed thoroughly and after reading it, I was more intrigued by phi but never really researched it until now. I’m still awestruck by how the Golden Ratio can appear everywhere and anywhere, even in snails! Those little suckers that just crawl by unappreciated have the Spiral of Archimedes on their protection gear and we aren’t even aware of it half the time! I loved learning more about how the Golden Ratio is present in architecture because again it’s another thing that goes by unnoticed. I had studied the Great Pyramids of Giza and the Parthenon in my AP Art History class but the Golden Ratio wasn’t mentioned in our study of the two pieces. After learning all of this, I know I’m going to catch myself glancing at people’s noses and nostrils and wonder if they have the Golden Ratio going on there.

Sunday, November 24, 2013

Fibonacci Beauty Ratio

Beauty Ratio (phi) is 1.618

And the most mathematically beautiful person based on the beauty ratio out of these five is .... *drumroll please* Mr. Lee!!!

It was a very close tie between Mr. Lee and Victoria but by a thousandth Mr. Lee won. He was the closest to 1.618. Third place would go to Jorge, with Sandibel and Julian following behind in fourth and fifth.**

The Beauty Ratio is pretty swell since Leonardo da Vinci's Vitruvian man demonstrates body mathematical perfection. But to have that mathematical perfection is hard to actually have since we are human and humans are complicated creatures faaar from perfection. Even Mrs. Kirch said she hasn't seen a person be exactly 1.618! The results of the experiment aren't all that accurate when you consider all the dependent factors in it, like shoes, some people's measurements could be off if their shoe had a slight heel that increased their overall height. Also, as the class period neared the end, measurements were hastily taken (Victoria and Mr. Lee in my case) However if this experiment was to happen again with more time and exact calculations, the results would have more validity.


Saturday, November 23, 2013

Fibonacci Haiku: The Best of Friends.

Surprisingly smart.
The true homie.
Orange soda and strawberry ice cream.
She listens to my frantic worries: chin cancer.
2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, an exciting journey. 
 There is no plan of the journey ever ending as school will come to an end but our lives will carry on. 

Sunday, November 17, 2013

SP #5: Unit J Concept 6: Partial Fraction Decomposition with repeated factors

What is this problem about?

This problem is very similar to SP#5 with partial fraction decomposition. There is one important difference however: this problem has repeated factors. It is still a very similar way of solving it however, you just need to pay more attention to factoring as seeing that one little slip-up can give you a wrong answer. This one is much more difficult than others because it has four values to solve for but you can do it with a little perserverance. 

What must the viewer pay special attention to in order to understand? 

Please be sure to properly count up to the powers and don't confuse the number of the power for a term in the binomial. Like I said before with the foiling, be careful! You do not want to drop a negative somewhere. Be sure to distribute the letter of the value as that will be very important to get your system. Also be careful of combining like terms. Other than that, you should get through this problem just fine!

Thursday, November 14, 2013

SP #4: Unit J Concept 5: Partial Fraction Decomposition with distinct factors


What is this problem about?

This problem is about breaking down partial fractions, ooh my favorite! Ha...but we use our knowlegdge of systems that we have learned in the last few concepts here. This algebraic skill will be very useful to us in process of integration in calculus next year woo hoo. So we're basically breaking fractions apart here.

What must the viewer pay special attention to in order to understand?

You, as the viewer, must pay attention to your foiling, I can gurantee you will get this wrong if you do not properly foil (I say this because I originally got my own problem wrong the first time around, so trust me because I speak from experience.) But other than that, make sure you don't drop a negative and be sure to properly distribute and combine like terms. Also do not forget simple little rules like dropping the x^2 and x's near the end to set up your system. Other than this, you should be good!

**little tip: if the writing is hard to make out, just click on the image to zoom (in case you didn't know)**

Monday, November 11, 2013

SV #5: Unit J Concept 3-4: Solving Systems of Equations

What is this problem about?

This problem is about solving a matrix! Wooo so much FUN. This problem covers and explains all the steps needed to solve a system of equations.

What must the viewer pay special attention to in order to understand?

Please pay special attention to all the steps and remember to solve for things in row-echelon form. Do not forget that y=a and that this system is consistent dependent and not inconsistent because one of the rows gives you 0 0 0  0 and not a false statement like 0 0 0 9. It is also not consistent independent because it is not a certain solution. It has an infinite amount of solution under the constraints of the solution we got. 

Sunday, October 27, 2013

SV #4: Unit I Concept 2: Graphing Logarithmic Equations

What is this problem about?

This problem goes over how to graph logarithmic equations and how to find key points such as x and y-intercepts. We also identify the asymptote, domain, and range. This type of graph is very different than exponential ones.

What does the viewer need to pay special attention to in order to understand?

Make sure you are correctly able to find your h because it will be crucial to finding your asymptote. In these types of graph we only care about h and k. Our asymptote equation will be x = h. P.S. h will be the opposite of what you see in the equation. Be sure to use the change of base formula when solving for your y-intercept and to correctly exponentiate for your x-intercept.

Happy solving!

Thursday, October 24, 2013

SP #3: Unit I Concept 1: Finding Parts & Graphing Exponential Functions

Let's start off by breaking off the steps to solve this problem!
So you first find the parts of the equation such as a, h, b, and k as in the first box. You then find your asymptote because your equation for this is y=k (remember YAK). You then proceed to find x and y intercepts (hint, you won't be able to find one of these *nudge nudge*) After this you can plot in the equation into your graphing calculator as is but remember to put parantheses around (x-1). Your domain and range should be especially easy to find, domain anyway. Your range depends on the location of the asymptote.

Ok so now:

What is this problem about?

This problem covers our favorite exponential functions and how to solve for their key parts. We use horizontal asymptotes for exponential graphs. So this is like intertwining two concepts (the graphing functions and exponential stuff)

What do you need to pay special attention to in order to understand? 

The exponentual YaK died! Don't forget that there will be no x-intercept because a is positive and it must go above the asymptote of y = 2. Range will depend on the location of the asymptote and direction the graph goes off in. Other than that don't forget to pay attention to what you input in your calculator! And do not make simple mistakes as I do and think h will be -1 instead of +1. OH and do not forget that when (1/2) is raised to -1, it will be become because you flip the fraction to get its reciprocal.

Tuesday, October 15, 2013

SV #3: Unit H Concept 7: Finding logs when given approximations

What is this problem about?

This problem is about finding logs when given approximations of course. The clues are given and the equation we are aiming to solve for. We need to break down the problem and expand our log in order to use the clues given to us. You then substitute in your clues. These problems demonstrate our ability to solve these without a calculator (for the most part because sometimes you do need help to break down the terms)

What does the viewer need to pay special attention to in order to understand?

Please be sure to remember log base 8 of 8 will equal to one. Do not forget how the quotient and power property will play a role in expansion and solving our log. Please please please take into careful consideration what these properties will do when you write out the expanded form. Also be sure to use the correct clues when you end up with your final form with your approximations!

Monday, October 7, 2013

SV #2: Unit G Concepts 1-7: Finding all parts and graphing a rational function

What is this problem about?

This video covers rational functions. Horizontal asymptotes, slant asymptotes, vertical asymptotes, holes, domain, interval notation are included. These factors of the problem significantly contribute to our ability to graph the function. This problem however specifically does not deal with horizontal asymptotes. We focus on slant/vertical asymptotes and especially holes.

What does the viewer need to pay special attention to in order to understand? 

In order to better understand this problem, you must pay attention to every single detail. In order to find the equation for slant asympote, you need not go further than y = mx + b. Once you have the needed information, you can forget about the remainder since it is not necessary to us at the moment. Also remember even if it seems like the graph is touching the asympotote, it never, ever does. 

Sunday, September 29, 2013

SV #1: Unit F Concept 10: Finding all real and imaginary zeroes of a polynomial

What is this problem about?
This problem utilizes rational roots theorem and Descartes' Rule of Signs. It is similar to Concept 6 but this time we find all real and complex zeroes. This makes it a bit more difficult to solve since we are now dealing with a possibility of imaginary zeroes and harder quadratics formulas to solve once that step is reached.

What does the viewer need to pay special attention to in order to understand?
The viewer needs to pay special attention to every tiny detail because it is very possible to make a mistake in Descartes' Rule of Signs or some other step. They should also pay special attention to imaginary numbers and always remember they come in conjugates. Be sure to carefully factor the x's that come out of quadratic equations.

**also try to watch the video on full screen or on youtube for best viewing
OH and thank you for watching!

Sunday, September 15, 2013

SP #2: Unit E Concept 7: Graphing a polynomial and identifying all key parts

1. What is this problem about?
This problem allows us to see how polynomials behave at extremes, in the middle, where their highest and lowest points are, where their intercepts, under what intervals they are increasing or decreasing. But most of all the multiplicities of the zeroes show us how to act around the x-axis.

2. What do you need to pay special attention to in order to understand?
Be sure to pay attention to the leading coefficient and its exponent to make sure what direction(s) your end behavior will go off in. Also look at the multiplicities of your zero, these have everything to do with the graph. They determine whether the line will go through, bounce off, or curve through the x-axis. Make sure to find your y-intercept as that will also help develop the line.

Monday, September 9, 2013

WPP #3: Unit E Concept 2: Path of a Record

SP#1: Unit E Concept 1: Graphing a quadratic and identifying all key parts

1. What is this problem about? 

This problem is a quadratic one, where our sketches will be more accurate and detailed. Our equation starts off in standard form: f(x) = ax^2 + bx + c. But to make it easier to graph we complete the square to put in parent function form: f(x) = a(x - h)^2 + k. This form allows us to find the vertex(max/min), axis, y-intercept, and two x-intercepts much easier and more efficiently.

2. What do you need to pay special attention to in order to understand?

Be sure to correctly write out your steps along the process. To find your parent function form, you begin to complete the square by subtracting 1 from both sides. Continue completing the square by using (b/2)^2 to get 2(x + 2)^2 = 7. To get the parent function form you subtract 7 and your equation is 2(x + 2)^2 -7. You then find your y-intercept by plugging 0 in for x to get y=1. Vertex is (h,k) so that would be (-2, -7). Please note that the x-coordinate is -2 not 2 because h is the opposite of the equation when graphed. Your axis of symmetry is also x=-2 which divides the parabola perfectly in half and in a perfectly symmetrical fashion. You find your x-intercepts by solving 2(x + 2)^2 = 7 and you will get x= -2 +/- square root of 7/2 which will help the symmetry with the plotting of extra points to make the graph even more accurate.