I Like Math

Review of the Math Resourses

2007-04-14T08:47:00.000-07:00

The Big Resources.

Wikipedia mathematics has many interesting articles, but doesn't gives decent explanations in many elementary topics.

MathWorld is a math encyclopedia with lots of articles, but not much of an explanation sometimes.

Cut The Knot has very nice Java applets, is VERY big for a personal site and a good reference! Many discussions and is pretty active. Only complaint is that sometimes its too formal missing many students.

Ask Dr. Math is a gigantic archive of questions/answers, you can ask questions (never did it though), and it sure has lots of nice entries! Excellent reference!

MAA columns are old and new columns of the Mathematica Association of America. Refreshing well written articles.

Mathpages is another big site, has lots of nice stuff, it is more advanced in general, it stopped growing at some point.

MacTutor History of mathematics has many good biographies of mathematicians, and nice articles on math history. It is also active.

Geometry Junkyard has lots of curiosities on geometry, and current research.

Math Atlas and PlanetMath are two less know math encyclopedias.

Smaller sites: Math is Fun is an educational site. Numericana question/answer style with some tricky ones. Inca Geometry has animations of geometry with some tough problems. Baez fun stuff has some nice articles from a known physicist. Art of problem solving is a community site with some different stuff. Words of mathematics is another historical reference.

Found nice proof of the volume of the n-Sphere

2007-03-31T20:24:00.000-07:00

Just find a nice demonstration for volumes and area of the n-spheres here.

First he demonstrates the formula for a general pyramid (by dividing a general cube in pyramids) to get the Volume from the Area. The case for the 3 dimensional sphere he settles by comparing its area with the cylinder, actually he just claims it is possible, pointing to a more famous argument that I already saw before here for example. Then he claims that this kind of argument always works, that I didn't knew!! And derives the general formula from it. It is actually a recurrence formula but we can certainly find the general formula from that.

A bizarre but nice fact of polynomials

2007-03-27T16:48:00.000-07:00

The tangent to the midpoint between two roots in a third degree polynomial(with three roots) is always the third root. Just saw that at Polymath.

Deriving f(x)=K*(x-a)*(x-b)*(x-c) using the product rule we get

f'(x)= K*((x-a)*(x-b)+(x-a)*(x-c)+(x-b)*(x-c))=

K*((x-a)*(x-b)+(x-c)*(2x-a-b))

taking x=v=(a+b)/2 the second term vanishes and we have

f'(v)= K*(v-a)*(v-b)

the crossing point should be

v - f(v)/f'(v)

as in Newton's method and in this case should be:

v - {K*(v-a)*(v-b)*(v-c)}/{K*(v-a)*(v-b)}

= v - (v - c) = c

Comments on proofs of Euler's identity

2007-03-27T15:10:00.000-07:00

You can only proof this identity after you clearly defined what e^(ix) means.

In this sense arguments using calculus and using differential equations are incomplete, without at least a clear definition any argument becomes an heuristic argument.

I know of three ways of defining e^(ix).

(1) From the complex series of e^(z) defined as sum of z^n/n!

(2) as the limit of n to infinity of (1 + ix/n)^n

(3) or directly as cos(x) + i sin(x)

Either way only after defining e^(ix) is that you should show that it has the properties like e^(ia)*e^(ib)=e^(i(a+b)), or (e^ix)'=ie^(ix) that you would expect it to have, and some use them fearlessly.

Number (1) is the most common one, I think number (2) would be a nice thing to use since it is analogous to the real case and can also be interpreted geometrically (as Richard Feynman does for a reference). But using (3) is totally misleading because it doesn't show why it should be true.

That is why I think heuristic arguments are needed to provide "a reason" for us to believe that such a thing should be true. Using circular motion as in my next post seems to me much more simple and much less "out of the blue". While the proof by itself is heuristic it can be made rigorous.

Proof of Euler's formula

2007-03-27T14:06:00.000-07:00

This is a proof of Euler's formula that uses the circular motion

Suppose
P(t)= e^(it)
is a trajectory on the complex plane, using some definition of the exponential (the Taylor series definition for example) it is possible to show that
P'(t)=(e^(it))'= i e^(it) =i P(t)
and
P(0)= e^(i0)= e^(0)=1
In fact any reasonable definition of exponential should have these properties if you want to take these for granted all this becomes an heuristic proof.

then P(t) describes a trajectory in the complex plane that at t = 0 is at 1 and which velocity P(t) is always perpendicular, (90 degrees counter-clockwise) to P(t) and the velocity P(t) also has the same modulus as P(t). Such trajectory can only be the circular movement with radius one, unit velocity, starting at the point 1: cos(t) + i sin(t)!! If you want rigor again use the uniqueness of solution of differential equations the Picard–Lindelöf theorem.

e^(it)= cos(t) + i sin(t).

Map of blog

2007-02-26T16:11:00.000-08:00

Welcome!, strange ideas.

Challenging Sums (arithmetic progressions).
Square and rectangles, averages.
Why 0.99999... = 1 is hard to believe?
Exponents.

Midpoints of a quadrilateral.
Volume of the Pyramid, Sum of squares.
Found nice proof of n-Sphere area and volume.

Complex numbers 1, 2, 3.
tan tan tan identity.
Euler's Formula comments, Euler's Formula proof.

Pythagoras, Pythagoras and physics 1.
Square root of 2 is irrational.

Combinations problem.
A Bizarre but nice fact about polynomials.

Some ideas for the next posts:

-Trigonometry definition as proportions in a right triangle, and relation to the trigonometric circle.
-Trigonometric functions as projections of the circular movement.
-Exponential functions and the definition of the number e.
-The Euler's Formula.
-The circle perimeter, area, and the power of a point.
-Fermat's point a mechanical interpretation.
-The area and volume of a sphere.

More on averages

2007-02-26T14:11:00.000-08:00

This post is a continuation of the post "Squares and rectangles", here I will generalize a bit what we discovered in that post and deduce the geometrical/arithmetical mean inequality, as well as prepare for the next post the many variables case.

In the last post we saw that if we want a bigger area of a rectangle with fixed perimeter we should aim for a square, more specifically, if sides "a" and "b" sum "p" then:

(a+b)/2 =p/2 =M =Mean

and that is the best side possible:

a*b < M*M = (a+b)/2*(a+b)/2 (Ineq. 1)

And if we a set = M-k and b = M+k

a*b = (M-k)*(M+k) = M ² - k ² (Equal. 2)

So for fixed p (and so M) a*b decreases as k increases and vice versa, or the farthest we are from the mean the worst we are.

Taking square roots on both side on Ineq.1 we have:

sqrt(a*b) < (a+b)/2

That is the geometrical mean is less then the arithmetical mean (unless it is equal when a=b).

We could ask the same question for three variables, that is if a+b+c = p is fixed then what is the choice of a, b, and c that gives the largest product?

I guess it is reasonable to say that a= b= c= p/3= (a+b+c)/3 should be the best choice, that is indeed true but it is not easy to prove!

I will prove that in my next post, hint we will use more (Equal. 2).

Combinations problem

2007-02-25T17:50:00.000-08:00

Take a set having let's say 10 elements, if I make a random selection of its elements (each item selected with 1/2 probability) what is more likely, to have an even or an odd number of items selected? Or maybe they have the same probability?

Obs. Allow all or no items selected.

Try this problem with 5 elements instead of 10.

Odd and even selection have the same probability here, why?

Since any selection that has 3 elements leaves 2 unselected items. So you have the same number of ways to select 2 or 3 items and also the same for 1 and 4, or 0 and 5. Making equal probabilities for even or odd selections.

Look closely and see that this logic would also work if instead of 5 we had 7, 9, or 1997 elements. But it wouldn't work if we had 10 elements!! So back to square one!

Any ideas?

As it turns out odd and even selections also have the same probability in that case! And for any quantity of initial elements for that matter. But why anyway?

Maybe you could use the fact that we already now it is true for for 5 elements!

Indeed! Split the 10 elements in two groups of 5, A and B. How can we make an odd selection of 10 terms?

Case 1: Take an odd number of elements from A and even for B, or
Case 2: Take an even number of elements from A and odd from B.

And for an even number of terms?

Case 3: Take an even number of elements from A and even for B, or
Case 4: Take an odd number of elements from A and odd from B.

Here Case 1 and 3 have the same number of possible selections and Case 2 and 4 also.

So even and odd selection have the same probability in that case too, and also in general following similar lines.

Squares and rectangles

2007-02-23T16:59:00.000-08:00

Take a rectangle imagine you have a fixed perimeter, but that you can vary its sides. Of those rectangles which one has greatest area?

Make a "mental movie" of the rectangle with varying sides, from a very big horizontal side to a small one, you can also look at the picture below, making more rectangles with your imagination.

Let's take the perimeter to be 20 and the sides to be "a" and "b".

Then a+b+a+b=20 and a+b=10

Think about big sides (close to 10) what happens to the area? Why?

Big sides makes bad rectangles because the other side will be forced to be small (maybe almost zero) giving a small final area.

Small and big "a" and "b" gives bad choices so maybe the "equilibrium" a=b is the best choice.

Since a+b=10 we get a=b= 5

Indeed the area in this case will be, 5*5= 25, some more cases below:

A=5*5= 25
A=6*4= 24
A=7*3= 21
A=8*2= 16
A=9*1= 9

So a=b=5 definitively looks like the best choice. We could also try

A=5.1*4.9 = 24.99

and many others, making it even more compelling. Can we visualize what is happening here?

In this picture ABCD is our square and AMNP is some rectangle with same perimeter as the square, we want to show that AMNP has a smaller area than ABCD. But think for a moment, why is that?

Here AN is smaller then side AC by the same amount that the side AM is greater than AB (or CN=BM), since they have the same perimeter. In relation to the square the rectangle AMNP gained the area BMEP and lost the area MECD. Those two rectangles have one side in common BM=CN, but BE is less than NE. Thus the rectangle lost more area than it gained from the square and so the square has greater area!
(Idea of argument from Rademacher and Toeplitz, great book by the way)

A more algebraical way uses a similar idea, comparing the sides of a and b with the square with sides 5,5 we get: a= 5-k and b= 5+k and a+b= 10 then:

Area= A= a*b = (5-k)*(5+k)= 25 - k ²

That is always smaller then 25! So the best is the square or k=0.

Actually we got a little more than that, that the farthest the rectangle is from the square (bigger k) the smaller the area it has.

Problem: Use the geometrical reasoning to get to the same conclusion.

More on that in this post.

Some ideas on teaching math

2007-02-16T00:12:00.000-08:00

This is mostly a sketch but I will try to improve it over time. For a introduction to this blog read this post.

Mathematics is about understanding real things. Formulas, methods, and pure logic are just some of the tools mathematics uses for that purpose. One of the purposes of this blog is to stress the importance of learning how to use your intuition, to learn how to think by yourself and understand what those formulas, methods, and pure logic actually do and how they do it.

What could make a better class? Just some general ideas:

Conway's, Arnold's, Moore's, Polya's comments on teaching.(last two are to wikipedia I will try to find better ones later).

1) Simple, intuitive, and natural

Simple ideas are more important than complicated ones. Motivate the ideas. Use natural logic, that is, sometimes reason as if you never saw that topic before. Communicate with the students!! They will tell you the natural logic, try to use those more natural ideas to construct the theory. "Natural" ideas can be wrong but they are always instructive for the teacher. Look for easier ways to explain, maybe organizing the topics differently.

2) Be honest, don't hide anything

Sometimes it is not possible to explain the whole subject with a natural logic. Anyways, managing to be honest an giving most of our ideas in an organized fashion it's a huge achievement. Sometimes I fell the urge to hide some idea to make a cleaner presentation, for me, that is cheating. If you can't find a better way, just be honest, and tell how you thought about it at the first time you saw it, maybe that is the best presentation after all.

3) Makes the student think and participate in creating math.

It is not so impossible, present the theory showing simple examples and problems, give some orientation and let them explore the possibilities too. It is hard to make it work but it gives much more satisfaction to the student and for the teacher.

4) Make math thinking more down to earth.

I am not talking about applications, for example:

In the trigonometric circle notice (make a drawing) that small angles have big cosines and small sines, as the angle grows initially the cosine doesn't change much.... but latter it changes faster...Why? What is happening? ...It is because of the changing slope of the circle! For instance the cosine is 1/2 only at 60 ^o(after 45 ^o=90^o/2, because of the slow beginning) and at that point the sine is big (0.85) or sqrt(3)/2. Why? (leave that to you) Make a imaginary movie of it. There are many things to observe there. Question the students, make them think. That exercise gives a sense of proportions and makes math more real.

My teaching experience is mainly from tutoring individual students, I find it amazing to discover how people think math. Finding more natural thinking is not an easy job. We have to 'unlearn' what we know, understand it in different ways, and to make more connections.

What do you think makes a good class? Give your opinions and ideas!

Pythagoras and physics 1

2007-02-13T15:13:00.000-08:00

Pythagoras appears in many places, the objective here is more than "prove" Pythagoras is to see how it connects to familiar concepts in physics, and what physics has to say about it! For more classical approaches look in this post.

Hypotenuse is "a", leg "b", leg "c", so Pythagoras means that:

a ²= b ² + c ²

Lets try to use physics to show Pythagoras. Since we have something squared we can look for something that is squared in physics...

Energy = 1/2m*v²

Remember that the velocity is a vector (we will use bold letters for vectors) and the formula for energy just uses the length or absolute value of the vector. We will choose m=2 then Energy = v ² to make calculations easier.

Over the sides of the triangle lets put the vectors a, b, and c. Notice that a=b+c!

If E(a) means the energy of an object with velocity "a" so Pythagoras says that:

E(a)=E(b)+E(c)

Why should that be true? Remember that "b" and "c" make 90 ^o. If they don't then it is actually not true!! Because we should replace Pythagoras by the cosine rule.

So why it should be true when "b" and "c" are perpendicular?

The energy of an object is also the work needed (that is force * distance) to make it go from rest to that velocity.

Lets apply a force horizontally to make an object acquire velocity b then its energy is E(b)= b ². Now lets apply a force vertically to make it get an it acquire an upward velocity c. Contributing with an energy E(c)= c ² adding the two vectors "b" and "c" we have the vector "a" that we know should have energy E(a)=a ².

Voila!! E(a)=E(b)+E(c)

Why does this argument works? Notice that the second time we apply the force (now vertically) that force doesn't change velocity b horizontally so all it work (energy) is in the upward direction.

Or more precisely:

Forces perpendicular to the movement don't do any work because they don't change the absolute value of the velocity. But they do change the velocity's direction! Recall the circular movement.

What this blog is about?

2007-02-12T22:53:00.000-08:00

Welcome!

I want to post here some presentations of mostly basic math, that I hope are intuitive, take a look at them, give some comments. I try to explain them in the simpler terms possible. This is a work in progress and suggestions are welcome.

What we can do to make math learning interesting? After all, what is interesting about math? This blog is about trying to answer these questions.

Click here for a map of all the posts.

I also made a list of math resources and blogs take a look at them.

Have Fun!

Technorati Profile

Pythagoras

2007-02-12T21:12:00.000-08:00

We can get so used to the Pythagoras theorem, after all they make us use it again an again in school, but now lets stop and think for a moment. Why the hell it is true??!!

To see what I am mean try to say with a straight face that:

"It absolutely clear and intuitive that:

Hypotenuse ²= (One leg) ² + (Other leg) ²"

Well if you can't say that you have company! Me at least. And if you can please try to convince me!!

There are many different demonstrations for Pythagoras, in this post I will point to some of the more famous ones, some less known, and different/unusual ones.

One thing that bothers me is that squared in the formula, what do squares have to with it any way?

Notice that 3 ² = area of the square of side 3=9, that is why we say "raised to the power two" as "squared"

So we can rephrase Pythagoras as:

The sum of the areas of the two squares over the legs equals the area of the square over the Hypotenuse.

The first demonstrations actually demonstrated that area version of Pythagoras.

Lets begin with the famous ones. Here I will shamelessly point to the collection at Cut the Knot, I will comment some of these.

#4 is an area argument, pretty straightforward. Lots of people like it but doesn't appeal a lot to me.

#6 is the first I ever saw, it uses the similarities with the small triangles formed by the height, you get the same intermediate steps of Euclid's # 1 on the list.

A more honest proof since it uses something we are more used too but kind of tedius to find the right relations to use.

first show a ² m*c
and b ² n*c

and add them together to get a ² + b ² = c *(m+n)=c*c=c ².

#7 Is the second demonstration of Euclid, takes some thinking to understand. But as my friend said you can't hope to get better than that(in some sense).

#9 Another area argument pretty nice (no formulas!!)

#12 is a more 'mechanical' proof (one of my favorites)

#15 is probably the best area argument (uses two different tilings of the plane the two square and the 'diagonal' square) to prove that they have the same area, go here for anothere explanation.

#40 is a "infinitesimals proof" pretty nice if you know calculus.

#41 That is really cheating!!

#43 Is a circle demonstration (prefer that than #11).

Observation 11 before the proofs is a nice generalization of Pythagoras by Edsger W. Dijkstra. Using an idea somewhat similar to demon. #6 with a nice twist.

You can also look at the others demonstrations at Cut the Knot and make your own judgments.

In the next posts I will try to give some different demonstrations, here is a post with a "physical" demonstration.

Midpoints of a quadrilateral

2007-02-01T20:49:00.000-08:00

I was looking at a blog polymathematics and I saw a reference to a nice geometry fact.

If you join the midpoints of a quadrilateral it is always a parallelogram.

When I first saw this result, back in high school, I was totally amazed (couldn't quite believe) specially for weird quadrilaterals.... so I draw a nice big weird quadrilateral in the back of my book, joined the midpoints and saw it!! (I still have the book)

If you don't want spoils don't read below.

The key is to draw the diagonals...

Well, you can find a demonstration here at "Cut the knot" as usual with a nice java application so you can convince yourself.

Just some nice memories...

Why it is so hard to believe that 0.99999 ... = 1?

2007-01-01T22:53:00.000-08:00

This is a hard topic, judging from the amount of discussion about it, you can find more references at: Cut-the-knot Wikipedia Polymathematics

The first time I saw 0.99999 ... = 1 I felt unsettled, seeing again all the tension generated by that equality (see polymath above) I decided to write this post.

That equality might be less troublesome if one remembers that the "real number system" (the mathematical setting where this is demonstrated) is just 'a model' for some quantities and not the absolute truth of the universe. Meaning that only after you assume the axioms of that model you can show that 0.99999...=1. Those axioms imply that arbitrary small numbers exist, the problem is do they?

Let me try to explain myself better.

0.99999...=1 is a proven theorem in mathematics (in the real number system), you can also consider it an annoying characteristic of the "real" numbers, since the equality implies that there are two different decimal representations of the same real number "1".

When I say "real number system" I mean a specific mathematical model generally used to describe quantities. As a mathematical model it has well defined axioms, properties, and proved theorems like the one in the title.

First, the name "real number" doesn't mean those numbers are in the real world, of course the real number system is based on some intuitions on how quantities like length, charge, mass (and many others) should behave, and it is indeed a very successful model, but it is nether less just a model.

In measuring quantities (for example length) in the real world always implies some approximation or error. The "real number system" is a mathematical model which supposes that arbitrary good precision is possible, fixing a maximum precision at some small measure would appear strange and it is actually easier, for the models sake, to suppose "infinite" precision.

But can we achieve arbitrary precision? That is something meaningful?

Consider also that this equality has antecedents in Zeno's paradoxes some simple but also unsettling paradoxes.

The way the real number system solved that paradox, had some "bad" consequences like the one in the title.

Some might say that what I wrote here is obvious but I think part of the reason for so much confusion is that people don't say explicitly that stuff. Now for the more arguable part.

Quantum mechanics says that really small quantities give unusual kinds of problems, and many physical theories put at doubt the arbitrarily small in the usual sense. Maybe we should not worry too much with that equality since many modern physicists don't know how to make sense of it too.

Square root of 2 is Irrational

2006-12-19T22:33:00.000-08:00

I was looking at the in Internet now and found in this site a different proof of this fact:

"Another elegant proof:
In 1952, Robert Gauntt, then a freshman at Purdue University, gave the following proof[2].
a ² = 2b² cannot have a non-zero solution in integers because the last non-zero digit of a square, written in base 3, must be 1, whereas the last non-zero digit of twice a square is 2."

Interesting! note the "last non-zero digit" part.

This proof would be obvious if we were used to base 3.

Here at cut the knot there is a list of different proofs. Proofs number 4 and 3 (in that order) are the ones I like most. The proofs at cut the knot are also reproduced below:

Proof 4

Following is yet another illuminating proof. As in the standard proof, assume p and q are mutually prime (numbers with no common factors). Their squares are still mutually prime for they are built from the same factors. Therefore, the fraction p ² /q² cannot cancel out. In particular, p²/q² cannot cancel down to equal 2. Therefore, p² /q ².

Proof 3

Following is another intuitive one [Lasckovich, p. 4]. This one is based on the assertion that every number is uniquely (up to the order of factors) representable as a product of primes. (A prime is a number divisible only by itself and 1.) The fact which is known as The Fundamental Theorem of Arithmetic. Assuming it as an axiom (or a given fact), let (p/q)² = 2 for some integers p and q. Then p² = 2q². Factor both p and q into a product of primes. p² is factored into a product of the very same primes as p each taken twice. Therefore, p² has an even number of prime factors. So does q². Therefore, 2q² has an odd number of prime factors. Contradiction.

Exponents

2006-12-19T18:07:00.000-08:00

In this post we will try to understand some properties of exponents.

How much is 2³? = 2*2*2 = 8

2⁴ = 2*2*2*2 =16
2⁵ = 2*2*2*2*2 =32
2⁶ = 2*2*2*2*2*2 =64
2² = 2*2 =4

Lets put all that in sequence...

2² 2³ 2⁴ 2⁵ 2⁶
4, 8, 16, 32, 64

Note that when we increase in the exponent we multiply by 2!
So the next terms would be:

128, 256, 512,...

But what the previous terms should be?

When we decrease the exponent... we divide by two!!

2^-1 2⁰ 2¹ 2² 2³ 2⁴ 2⁵ 2⁶
z, y, x, 4, 8, 16, 32, 64

x = 4/2 =2
y = x/2 =2/2=1
z = y/2 =1/2 =1/2

2^-1 2⁰ 2¹ 2² 2³ 2⁴ 2⁵ 2⁶
1/2,1, 2, 4, 8, 16, 32,64

In here notice 2¹ =2 -that does make some sense
and 2⁰ =1!

some more terms:

2^-3 2^-2 2^-1 2⁰ 2¹ 2² 2³
c, b, 1/2, 1, 2, 4, 8

b = (1/2)/2 = 1/4
c = b/2 = 1/8

1/8,1/4,1/2,1,2,4,8

Notice that the numbers (1,2,4,8) 'repeat' 1/8=2 ^-3= 1/(2³)

And that is what happens in general with negative exponents.

2^-n= 1/(2 ⁿ )

Now some more properties

2³*2⁵ = (2*2*2)*(2*2*2*2*2)= 2 "eight times" =2 ⁸

So when we multiply two powers in the same base we add the exponents

For division:

2⁷/2⁴= (2*2*2*2*2*2*2)/(2*2*2*2)= 2 "three times" =2³

There is some cancellation, so when we divide two powers in the same base we subtract the exponents.

(2³)⁴= 2³ 2³ 2³ 2³= 2 "twelve times=3*4" = 2¹²

Kind of strange to put in words (try it!!) but easier to understand.

Sum of the first n squares

2006-12-18T20:48:00.000-08:00

In our last post Volume of the Pyramid we showed that the volume of a truncated solid is the Base*(average of the height at the vertices) for the triangular and square prisms.

Lets now use that idea to calculate the sum of the first n squares.

1² + 2² + 3² + 4² + .... + n²

We can split that in:

1 + 2 + 3 + 4 + 5 + ..... + n
2 + 3 + 4 + 5 + ..... + n
3 + 4 + 5 + ..... + n
4 + 5 + ..... + n
5 + ..... + n
..... + n
. + n
n

If you picture the numbers as the heights of a solid it turns to be a truncated "linear" solid - since the variation is linear, there is also a more formal demonstration of the formula below.

The volume V is:

(B)*(average height at the vertexes)

(n*(n +1)/2)*(1+ n + n)/3

So the sum of the first n squares is

n*(n +1)*(2n +1)/6

You can also put this numbers forming a equilateral triangle rotate by 120^o and 240 ^o and then sum the three versions, getting a nice prism,(it is you can check that). All that is a "3D version" of the classic argument.

1 + 2 +...+(n-1)+ n = S
n +(n-1)+....+2 + 1 = S

n*(n +1) = 2S

S = n*(n +1)/2

We also saw that using averages in the post sums.

Volume of the Pyramid

2006-12-18T16:16:00.000-08:00

For some time I asked myself why the volume of the pyramid is 1/3*B*h.

In this post I will give an unusual demonstration of this fact.

After some reflection that the volume is directly proportional to the Area "B" and the height "h" seems reasonable but the 1/3 puzzled me. I know some different demonstrations for it but I wasn't completely satisfied with them.

If you already saw this idea/demonstration somewhere else please tell me where so I can check it.

For some (important) motivation let's first take a rectangular prism with square base ABCD and cut it with a plane not parallel to the base (such that the cut is between the two bases at the "bottom" and the "top"), considering below such a plane a truncated solid that at each vertex ABCD the solid has heights h_a, h_b, h_c, and h_d.

What is the height of the solid at the center of the square H?

You can show using some geometry (and it should appeal to your intuition) that

H = (h_a + h_c)/2 = (h_b + h_d)/2

That is also equal to our idea of the "average height of the solid" and it is also equal to the average at the points A,B,C ,and D.

H =(h_a + h_b + h_c + h_d)/4 .

The volume of the truncated solid V should be

V = (Area of base B)*(average height) "*"

And what happens if the base is an equilateral triangle ABC? The same idea should apply so its volume would be:

V= B*(h_a + h_b + h_c)/3.

Lets analyze the case when h_a = h_b = 0 and h_c = h... in that case we have a pyramid!!! So it's volume should be:

V= B*(0 + 0 + h)/3 = B*h/3

That is it!

Unfortunately this demonstration is not rigorous. I think it is intuitive thought, and it can be made rigorous using integration, or more simply the idea that sheering doesn't change the volume.

Read on for the integration point of view.

There are three equivalent pyramids we can make:

h_a =0, h_b = 0, and h_c = h
h_a =0, h_b = h, and h_c = 0
h_a =h, h_b = 0, and h_c = 0

These pyramids are all equal so they have the same volume, and each pyramid is determined by a different plane. We can think the volume as the integral of the height(function) over the base ABC of the pyramid. If we add the functions determined by each plane we will also add their volumes so we arrive at a linear function where

h_a = h_b = h_c = h

Such is the constant function, so that the three pyramids "added" together equal the volume of the prism and since each has the same volume and so the volume of the pyramid must be 1/3 of the prism with the same base**.

Why go to all that trouble? There is a very well know way to fit three pyramids in a triangular prism. The problem with that is these pyramid are not easy to "see", and we still have to argue why they should have the same volume***.

The main advantage here is the connection of our idea of average and the 1/3 factor of the pyramid. We will also see in "the sums of squares" that this point of view has nice implications.

For those who want more detail see Observation below:

* Obs. One can also demonstrate this by dividing the solid in prisms, or using "reflection maintain the volume" reasoning.

** Obs. I made the base of the pyramid an equilateral triangle to ease things up, we can use Cavalieri's Principle to get to the other cases, and we have too anyway to get the height off a vertex at the base. We can also use to the same effect the principle that pyramids of equal base area and height have the equal volumes.

*** Obs. The three equal pyramids from the corner of a cube or the six equal pyramids from the center, may be easier to see, but still we have to argue why this is true for more general pyramids.

tan tan tan

2006-12-18T16:13:00.000-08:00

That was a question I saw at Polymathematics raised:

Prove that if A, B, and C are the 3 angles of a triangle,

tan A + tan B + tan C = (tan A)(tan B)(tan C).

There is demonstration for this fact using complex numbers say A + B + C = 180

(cosA + i sinA)*(cosB + i sinB)*(cosC + i sinC)=-1

So this product is real number dividing each factor by the cos we have

(1 + i tanA)*(1 + i tanB)*(1 + i tanC) is real

By foiling and since the imaginary part is 0 we get that identity.

That was the nicest way I could show that identity.

My interest in it was that it can be used to show Hieron's formula, as I was trying to find nicer ways to show it I got the way above.

Sums

2006-12-18T14:28:00.000-08:00

For answers on this post highlight between the xxx's to see

Quickly tell me how much is

95+96+97=

Instead of doing the sum you could just approximate it
Like for example: 3*95

Or even better it could be...

xxx 3*96 xxx

But wait that is exactly 95+96+97

Since xxx 96+96+96 xxx = 95+96+97

As the terms even out.

Now how much is

94+95+96+97+98=

xxx 5*96! xxx

You can guess 93+94+95+96+97+98+99

But wait a second and if you have for example

94+95+96+97=?

There is no central one to average but still

The average should be xxx (95+96)/2 = 95.5 = (94+97)/2 xxx

So the sum is xxx 4*95.5 xxx

It is always the (number of terms)*(average or "middle term")

And now the last sum

37+39+...+79+81

In this case we don't see the middle term but you know what it should be?

xxx (37+81)/2 xxx

That is the (average) is (first+last)/2

So the sum is:

(number of terms)*(average)=(number of terms)*(first+last)/2

That is the sum of an arithmetic progression.

Complex Numbers 3

2006-12-17T13:58:00.001-08:00

So we saw that with the rotation-stretching interpretation multiplying and taking powers of complex numbers makes more sense.

Obs. While this interpretation makes multiplication easier it does not help when making addition. Happily addition is easy in the algebraic form "a + bi".

Now we are going to see three bonus of looking at complex numbers in this way.

First bonus: Our theory is somewhat arbitrary why we defined i to rotate 90^o counterclockwise ? It could be clockwise couldn't it ? The answer is yes! Lets say you defined a different clockwise imaginary number j, because you didn't like mine, then you would have a complex number theory just like mine with j's in the place of i's! Your drawings would be the reflection of my drawings. But the arithmetic would be the same.

But your j does the same job as my -i! So in a way the -i does the same job as i, but clockwise. Changing i's to -i's changes the orientation of my rotations. So if I take a number like 3+4i, then 3+4(-i) = 3 - 4i has opposite rotation but the same stretching factor. The operation changing i's to -i's is called conjugation and this reasoning explain most of it's properties directly.

Second bonus: Let's say our stretching factor is 1 (no stretching at all) in this case our complex number only rotate vectors by angles a and b. So they are:

cos(a) + i sin(b)
cos(b) + i sin(b)

An their product by our interpretation:

cos(a+b) + i sin(a+b)

But we can also multiply
cos(a)+i sin(b))*(cos(b)+i sin(b)) by distribution and that is:

(cos(a)cos(b) - sin(a)sin(b)) + i(cos(a)sin(b) + sin(b)cos(a))

And we have the formulas for cos(a+b) and sin(a+b)!,

Although the geometrical demonstration for those formulas is really nice too (there is at least one nice geom. dem. that I know of).

As an exercise you can find what happens when b = -a.

Third bonus: if you remember the First bonus a+bi and a-bi have same stretching but opposite rotation angles u and -u say. Remembering a+bi forms a right triangle with legs a and b lets call c its hypotenuse, that is also its stretching factor.
Then the product of a+bi and a-bi has u + (-u) = 0 rotation and stretching c*c = c², so it is really c², but algebraically

(a+bi)*(a-bi) = a² - abi + abi - (bi)² = a² -(-b²)=a²+b².

So a² + b² = c²

That is Pythagoras theorem.

Complex Numbers 2

2006-12-16T16:49:00.000-08:00

Last post Complex Numbers 1, we saw that:

Positive numbers like 3 stretches vectors.
Pure positive i numbers like 2i rotate by 90^o and stretches.
Negative numbers like -3 rotate by 180^o and stretches.
Pure negative i numbers like -2i rotate a vector by 270^o and stretches.

What you guess composite numbers like 1+i or 2+3i do?

Returning with our vector v we can phrase the question as what is (1+i)*v ? It is (1+i)*v = v + iv, by distribution.

The vector iv forms 90^o with v, so v + iv forms a square! and the sum is the diagonal of that square. That is v + iv has 45^o, with v and is a little longer by a stretching factor of sqrt(2)=1,41 approximately.

And (1+i)² should rotate by 45^o + 45^o = 90^o and stretch by sqrt(2) * sqrt(2) = 2.

So it should be 2i!! Checking algebraically
(1+i)² = (1+i)*(1+i) = 1 + 2i + i² = 1 + 2i -1 = 2i.

Now what is (3+4i)*v ? It is the diagonal of the rectangle with sides 3v and 4iv, so it is a rotation by the angle of the diagonal with v and a stretching by 5 of v. So what is (3+4i)²? Before doing the multiplication we now that the angle will be the double of 3+4i and the stretching will be of 25!

Take a complex number z₁ that rotates by 30^o and stretches by 7 and another z₂ that rotates by 120^o and stretches by 4 then z₁* z₂ rotates by 30^o+120^o= 150^o and stretches by 4*7= 28.

In the last paragraph I didn't say what those numbers were, with a little trigonometry you find that they are:

7(cos(30^o) + i sin(30^o))
4(cos(120^o) + i sin(120^o))
28(cos(150^o) + i sin(150^o))

And that is the rule of sum the angles and multiply the stretching factors.

This way to write complex numbers is called the polar form it is more useful when we want to multiply or divide complex numbers, and it also makes powering much more easier.

Like if we want to calculate

(2(cos(30^o) + i sin(30^o)))¹⁵

We get 2¹⁵ of stretching and an angle of 15*30^o= 450^o= 90^o since 360^o is a complete turn.

More on the next post.

Complex Numbers 1

2006-12-15T19:54:00.000-08:00

In this post we will try to find some light in those strange complex numbers. There is a nice way to present them as rotations and stretchings of vectors, where complex numbers concepts and properties are all very natural, look for that later in this post.

Let's begin by telling my first experiences with complex numbers.

I think my first impressions were based on those mysterious names, after all we would learn about "complex numbers", about the number " i " that was the "square root of -1", that the teacher said before it didn't existed, and also about the "imaginary" numbers in opposition to the more "real" numbers learned before.

The names "complex numbers" and "imaginary numbers" were not a good choice of words from the pedagogical point of view. Well, "i" is the number that squared is -1, what is the logic in that? Lets make some experiments.

For example: 3² = 9 and 5² = 25 so positive numbers squared are positive. (-3)² = (-3)x(-3) = 9, because negative times negative is positive, so negative numbers squared are also positive!

Concluding that all numbers squared are positive! So no number squared is negative! And in particular no number squared is -1! at worst it is 0.

So they actually invented a number " i " not a "normal number" that squared would be -1, that is what my teacher said to me.

After some algebra facts they let you know that you can plot a complex number like " 3+ 4i " in an x-y axis were the point (3,4) is, unfortunately that was not so helpful. Later on that class my teacher showed me lots of even more bizarre properties of complex numbers.

All that together was a dazzling sequence of "out of the blue" concepts and relations. Actually, complex numbers have a pretty decent history, real flesh and bone people invented it you know, and that history is not "out of the blue" I may tell something of that story latter.

Now I will present complex numbers in a different and I hope better way, first lets explore vectors.

A vector can represent something as velocity or force on a body and lots of other concepts. Lets think of it as a force on an object (which I am not drawing) to fix ideas.

When we apply two forces on an object what happens?

We have a resulting force! That depends on the strength of the original ones and on their direction. There is a way to add them, say u = AB v = AC move the vector u (maintaining direction) to the endpoint of the other C making a vector CD the sum u + v is AD that is the parallelogram law.

If we add v + v we have a vector twice the size in the same direction or v + v = 2v, also, v + v + v = 3v is the vector in the same direction three times longer. 1.5v is the vector with 1.5 the size, -v = -1v is the vector with opposite direction and same size (v + -v = 0), and -5v has opposite direction and is 5 times the size of v.

So when we multiply by a positive number we "stretch" the vector, or maybe shrink it if we multiply by a number smaller than 1, and when we multiply by a negative number the vector inverts its direction and is stretched appropriately.

If we multiply by 3 and after that multiply by 2 we get a vector six times larger, or just 2(3v) = 6v.

Imagine what we get with -3(4v), 5(-1v) and -3(-2v)? Think geometrically!!

-12v, -5v and 6v respectively

Obs. A negative number times a negative number should be positive since we invert the vector two times! - a good justification for that rule/theorem in my opinion.

If we put another vector w somewhere else multiplying by real numbers would also work exactly in the same way.

We can also think that multiplying by -1 rotates vectors by 180^o degrees, in that spirit we could rotate vectors by other angles like 90^o or 120^o. Let's say the number "j" rotates vectors by 90^o degrees counterclockwise.

If we multiply our vector by j two times we rotate it by:
90^o+ 90^o = 180^o that is -v, or

j*jv = -v = -1v

So j*j = j² should be -1, and we have a number that squared is -1! j is really our imaginary number i, time for more experiments.

2i rotates 90^o and multiplies or stretches by 2
2i*2i rotates by 90^o+ 90^o=180^o and multiplies by 2x2=4 or 2ix2i = -4.

-i rotates by 90^o and inverts the direction so it rotates by 270^o.

i³ = i*i*i rotates by 90^o + 90^o + 90^o= 270^o also, so i³ = -i.

i⁴ = i*i*i*i rotates 4*90^o = 360^o so i⁴= 1 = i² * i².

You can guess i⁵, i⁶, and i³⁷, note that the algebra coincides with our geometrical ideas.

All properties of complex numbers can be understood in this geometric "rotation-stretching" way, more on that on next post.