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
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).
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
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.
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
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.
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
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!
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
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.
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?
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
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
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.
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
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...
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...
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