Class 2/17/12 - Reading 4.3.2

 Salut mes amants des maths!

C'est vendredi de nouveau et vous savez ce que ce signifie? (It's Friday again and you know what that means?)  Ouiiii, autre petit examen (Quiz!)

Honestly, the quiz wasn't too bad, but I could have done so much better on it. The first problem was stating two of the seven postulates covered in Chapter 4 and the second was proving, for a specific w1 & w2, Archimedes Law of Lever (Proposition 6) which states: Two magnitudes, whether commensurable or incommensurable, balance at distances inversely proportinal to the magnitudes. 

I talked about this in my last blog post,m but I think the main reason I was confusing the steps of the proof was because I was getting caught up in the fine details when I should have been focusing more on the big picture of what the proof ws trying to accomplish. (mais ma amie intelligente m'a explique tout donc je crois que j'ai compris ce probleme...plus ou moins ^^)

Before the quiz, we had a presentation on Cryptography as well as a lecture during which Dr. Daddel covered topics like using triangles to get the area of a parabola, or the general subject of the area of a parabolic segment by summation of a geometric series. For the latter, the general idea is to construct traingles inside the parabolic segment whose total area differs from that of the segment by less than any given value. 

Archimedes then calculated the total area of the triangles constructed at each stage to be one-fourth of the area of the triangles in the previous stage. The more steps taken, the more closely the sum of the areas approaches the area of the parabolic segment. So, in order to complete the proof, Archimedes essentially needed to find the actual sum of the geometric series  a + (1/4)a + [(1/4)^2]a + ...+ [(1/4 x 4)^n]a + (1/3)[(1/4)^n]a = (4/3)a, which he proved using a double reductio ad absurdum, first assuming that K=(4/3)a is not equal to the area B of the segment. The proof is given as follows: 

If K is less than this area, then triangles can be inscribed so that B - T < B - K, where T is the total area of the inscribed triangles. But then T > K. This is impossible because the summation formula shows that T < (4/3)a = K. On the other hand, if K > B, n is determined so that [(1/4)^n]a < K - B. Because also K - T = (1/3)[(1/4)^n]a < [(1/4)^n]a , it follows that B < T, which is again impossible. Hence K = B. 

Well, that's all for today. Until next time, mes amis!
Bisous

-MB

Class 2/15/12 - Reading 4.2

Salut mes amants des maths! 

j'ai pas beaucoup pour ecrire aujourd'hui (not much too blog today :(  )

Today we had a presentation on Galois and then heard a lecture about some of section 4.2 of the book which covers Archimedes and numerical calculations. Most of what we went over in class were the graphical representations and proofs to Propostions 1, and Lemma 1 & 2 as follows:

Proposition 1: The area of any circle is equal to a rightangled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle.

Observe the following proof.
PROOF:

Let C be the length of the circumference of our circle and r its

radius. Denote by A the area of the circle. The area of the triangle is T = ½rC.

Case I: Assume that A > T. K = A – T is a positive number. By the method of exhaustion, one can find an inscribed regular polygon so that A – (area of the inscribed polygon) < A – T. From this we see that T < area of the inscribed polygon. Since the polygon is inscribed, its perimeter Q is less than C and its apothem a is less than r, so area of the inscribed polygon = ½aQ < ½rC = T, which gives a contradiction.

Case II : A < T . Now T – A is a positive number and we can find a regular polygon circumscribed about the circle so that (area of circumscribed polygon) – A < T – A. or, more simply area of circumscribed polygon < T. Since the polygon is circumscribed, the length of its apothem is r and its perimeter is greater than C so, area of circumscribed polygon = ½rQ > ½rC = T. Another contradiction!

Lemma 1: Suppose OA is the radius of a circle and CA is tangent to the circle at A. Let DO bisect angle(COA) and intersect the tangent at D. Then DA/OA = CA/ (CO+ OA).

Lemma 2: Let AB be the diameter of a circle and ACB a right triangle inscribed in the semicircle. Let AD bisect angle(CAB) and meet the circle at D. Connect DB. The (AB/BD)^2 = 1 + [(AB+Ab)/ BC]^2

I was unable to find illustrations of either the proposition or lemmas, unfortunately. :( But all three are pretty cool. I really hope he puts one of the them on the quiz since I feel pretty comfortable with them.
Wish me luck on the quiz!
Until next time.

Bisous!
-MB

Class 2/13/12 - Readings 4.1-4.3

Salut mes amants des maths!

c'est lundi! voila autre blog, mes amis! (It's Monday and I have another blog for you!)


Today we had a presentation on the Fibonacci & the Fibonacci sequence given by some of mes meilleures amies: Jackie, Kimberly and Michelle!

Leonardo Pisano Bigollo (c. 1170 – c. 1250) also known as Leonardo Fibonacci was an Italian mathematician, considered by some "the most talented western mathematician of the Middle Ages and best known to the modern world for spreading knowledge of the Hidu-Arabic numeral system across Europe through his Book of Calculations, the Liber Abaci. More importantly, he also spread knowledge of the number sequence named after him which was known as the Fibonacci numbers which he did not himself discover but used in the aforementioned book.

An Indian writer, Pigala, of 4th century B.C. discovered and had a general idea of the Fibonacci sequence. In the Fibonacci sequence of numbers, each number is the sum of the previous two numbers, starting with 0 and 1. This sequence begins 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ...                                                                                                                                                                                   ---------------------------------------------------------------------------------- A really cool fact about the Fibonacci numbers is that it can actually be seen in nature! Take Romanesque Broccoli/Cauliflower for example which pretty much looks and tastes like a cross between broccoli and cauliflower. Each floret is peaked and is an identical but smaller version of the whole thing and this makes the spirals easy to see. If you count the number of spirals in each direction as shown in the picture, you can clearly see the pattern that gives the Fibonacci numbers! 

After the presentation, lecture covered some of the topics from chapter 4 some of which include more information about Archimedes and his mathematical contributions related to the concept of Center of Gravity such as  Proposition 6 and properties of parabaloids. 

Proposition 6: Two magnitudes (commensurable) balance at distances inversely proportional to the magnitudes.  This proposition is illustrated here. 

Well that's all for now, mes amis. Until next time. 

Bisous! 

:) 

-MB

Class 2/10/12 - Reading 4.1.1

Salut mes amants des maths!

Today, we had a presentation on A history of pi and answers to the question of what is pi" I thought the presentation was pretty well done overall. We also had a quiz :( 

We started talk about Chapter 4 today, specifically Archimedes and Physics (The Law of the Lever).

  
Archimedes of Syracuse (c. 287-212 BCE)

• Shared his findings liberally

• Known for The Law of the Lever 

• His book was recently auctioned for about $2 million

• Applied his intellect to assist in military defenses

In addition, we went over four of seven of the postulates Archimedese first stated in his treatise Planes in Equilibrium.

 

• 1. Equal weights at equal distances are in equilibrium, and equal weights at unequal distances are not in equilibrium but incline towards the weight that is at the greater distance. 

• 2. If, when weights at certain distances are in equilibrium, something be added to one of the weights, they are not in equilibrium, but incline towards that weight to which the addition was made. 

• 3. Similarly, if anything be taken away from one of the weights, the are not in equilibrium but incline towards the weight from which nothing was taken.
  
• 6. If magnitudes at certain distances be in equilibrium, (other) magnitudes equal to them will also be in equilibrium at the same distances. 

It was also important to note the Archimedes based a lot of his information on the assumption that an "idealistic model" was being considered. A model would agree with the following assumptions/properties:

1. The lever itself was rigid, but weightless
2. The fulcrum and the weghts were mathematical points. 

The postulates previously mentioned served to support his propositions related to equilibrium:

•  Proposition 1: Weights which balance at equal distances are equal

• Proposition 2: Unequal weights at equal distances will not balance but will incline towards the greater weight

• Propositions 6, 7: Two magnitudes, whether commensurable [Proposition 6] or incommensurable [Proposition 7], balance at distances reciprocally proportional to the magnitudes.

He also went over some different properities dealing with centers of gravity. The center of gravity of a:

1. Parallelogram is at the intersection of its diagonals
2. Triangle is at the intersection of two medians 
3. Parabolic segment is at a point on the diameter three-fifths of the distance from the vertex to the base

Well that's all for now, mes amis. Bon weekenda tous!

Bisous!
-MB

Class 2/8/12

Salut mes amants des maths!

J'ai pas beaucoup pour ecrire aujourd'hui. (I don't have much to write today)...mainly because we got our tests back today and I was so disheartened by the score that I couldn't concentrate on the rest of the lecture after the presentation. My test truely was a disaster; I failed miserably.

Today, three students gave a presentation on historically prominent women in mathematics. I found the topic to be quite interesting as well as inspiring considering that fact that even in present day it's not as common to encounter female mathematicians as those who are male. It was really encouraging to learn about women who had accomplished so much even with the obstacles of discrimination and scrutiny they faced from society.

Even with most of the world against them and their great intellect, these women insisted on doing what they loved to do, studying mathematics. They didn't give up their passion simply because either noone believed in their capabilities or society flat out didn't acknowledge them.

How much more should the women involved in the modern world of mathematics endeavor to reach their goals! Fortunately, we as females today are more encouraged to participate and study fields like science and mathematics without much fear of scrutiny from others. I really enjoyed the fact that the students included Professor Abigail Thompson as a prominent woman in mathematics; she's one of my favorite instructors and I really admire her devotion to mathematics, so it was nice to see her mentioned in the presentation.

Well, Dr. Daddel said the quiz would focus mainly on Euclid's propositions that we've been going over recently. He also said we should start reading the first sections of chapter 4 which cover the great mathematician, Archimedes.

Until next time, mes amis...j'ai besoin de pleurer pour les erreurs qui j'ai fait sur l'examen.... :'(

Bisous
-MB

Class 2/6/12 - Reading 3.4

Salut mes amants des maths! 

oh mon dieu.. mon examen a ete horrible :(  (My exam didn't go well)
I'm quite certain I didn't perform well on this exam (I'll have to wait until the grades come back)

The presentation on Parallel Postulates didn't leave much time for lecture today, but Dr. Daddel managed to go over some of section 3.4 which covers circles and the pentagon construction. Specifically, he went over the geometric representations of Propositions III-20, III-32, III-35.


 



Proposition III-20: In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.





 
Proposition III-32: If a straight line touches a circle, and from the point of contact there is drawn across, in the circle, a straight line cutting the circle, then the angles which it makes with the tangent equal the angles in the alternate segments of the circle.

 Proposition III-35: If in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other.




He ended class by asking us to learn the proofs of several other propositions and said we should have our tests back by the next class. 
We're supposed to have a quiz on Friday, something I'm already feeling very discouraged about considering my performance on the midterm.

Until next time, mes amis.

Bisous
-MB

Class 2/1/12 - Reading 3.3

Salut mes amants des maths!

Autre jour, autre blog! (Another day, another blog)

Today we went over the several of the propositions from Book II of Euclid's Elements. He gave us a handout with explanations of some of these to serve as an additional resource to the book. His comments during lecture as well as the handout helped us to understand the modern interpretation of each of the propositions as well as their geometric representations.

Proposition II-1: If there are two straight lines, and one of the them is cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the sum of the rectangles contained by the uncut straight line and each of the segments.

Proposition II-5: If a straight line is cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the striaght line btween the points of secftion is equal to the square on the half.

Modern Interpretation:
(b-x)x + (b/2 - x)^2 = (b/2)^2




Proposition II-6: If a straight line is bisected and a straight line is added to it, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line. 

Modern Interpretation:
(b+x)x + (b/2)^2 = (b/2 + x)^2

Proposition II-11: To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

Modern Interpretation: a (a-x) = x^2  OR x^2+ax = a^2  => x = sqrt((a/2)^2 + a^2 - (a/2))





Dr. Daddel ended class by giving us more information about the material that would be covered on the first exam such as certain propositions in Book 2 in addition to those shown above. I'm so nervous about this test!

Until next time, mes amis!

Bisous!
-MB