How people react when they discover you are strudying math?

The main reason I am writing this post is because of an interesting post I recently read. Beans at Me Or My Maths recently wrote about the different reactions he got from people who heard that he studies math. To sum it up, the reactions he gets are of the type "you are crazy". He is not the only blogger from the UK who says he gets this type of reaction, so apparently it is a rather usual thing in that part of the world....

As you probably understood from the last part of the previous paragraph, such reaction is not common in Israel. I remember very clearly how when I had to study math for 4 hours straight in school (with one half hour brake) some other pupils said that they cannot believe that I do this, but there reaction was surprised but not negative. Perhaps they were just glad that they don't have to do this. Also, when asked about this I answered were simply "I enjoy studying math". I guess after this they preferred not to talk to me... Not much of a loss. I am not cynical, it is simply that my interests were very different from theirs.

Unlike Beans I don't do my homework in trains, and I don't speak with random people about my math, but from time to time I share my enthusiasm for math with someone who has no clue of what I am talking about. By sharing enthusiasm I mean that I start talking and I don't really care if I am understood. However, they usually don't faint or run away. They either remain polite or just ignore what I say. Writing this blog helps to control such bursts of enthusiasm, but sometimes I feel the desire to speak with someone....

The above doesn't mean that Israelis love math. However, it might mean that they really don't care. Most of the people here study as little math as possible in school, and then even if they go to college they are likely to never hear about it again. Also, just today I talked with someone who said that he can hardly wait until the end of the semester - unless he will fail in something, he will not have to study any math next year. But even he didn't say anything negative about math, or about those who study it.

Now, to the more mathematical part of the post. I have been thinking about a certain integral in the past few days, but I am still unable to show that it converges (or not). I am probably missing something simple here, this shouldn't be a hard problem:

The limits are zero and plus infinity. If you have the solution you are welcomed to post it, if no solutions will be posted before I will solve this thing, I will write the solution here (unless I will forget about this). Good luck...

The genius of Newton

This is just a short joke I found today:

Archimedes, Pascal, and Newton are playing hide-and-seek.
Archimedes covers his eyes and starts counting.
Pascal looks around and hides behind a bush.
Newton grabs a stick and scrapes a one meter by one meter square in the dirt and stands in it. Otherwise he does not hide at all.
Archimedes opens his eyes and looks around. Of course, he immediately sees Newton and calls "I see Newton" Newton calmly says "But hang on, one Newton in a square meter is a Pascal!"

It is handy when you have physical units named after you......

Formulas for third and forth degree polynomials

In a previous post, the hunt for the roots, I showed how to develop the formula for the second degree polynomial. In this post I want to develop the formulas for the third and forth degree. Unfortunately this means that this whole post will be algebra and nothing else. This formulas were originally developed in the 16 century - you can read more about this in the post linked above.
A little warning - in this post I don't solve numerical examples, so I use letters to denote numbers that would be known in a numerical example, but I freely move them around. This means that if in one line I wrote bx, in the next line I will also write bx even if I should write (b+4)x instead, and the same with the sign of b. To use the formulas you will need to follow the simplification process, and then to aplly the final formula to the result.

Lets start. The idea is to get the general formula for the equation of the form:

Firstly, lets get rid of as many parts as we can. By dividing it by a (it cannot be zero, because then we have a second degree equation) we will get a simpler equation. But it is possible to make it even simpler. To do this lets suppose that x=y-b/3 (b is not the original b, but the original b divided by a). After dividing by a, and writing y-b/3 instead of x we will get an equation of the form:

Writing this step fully requires too much typing, so it is skipped. You can easily verify it yourself. Now we will write this equation in a slightly different way:

Lets suppose that x=u-v (this step is called uglification):

We can choose u,v so that b-3uv=0. Now we get two equations:

,

But those are simple second degree equations, to see this just substitute u^3=t, v^3=s. It is easy to solve and get that:

v^3=

u^3=

Now all we need is to get the solution, x=u-v, is to take the third root. After this we get a rather unpleasing formula, which is however the solution. It might seems surprising that it took so much time to get to this formula (over 1000 years), but lets not forget that it seems relatively simple because of the modern notation. In the ancient world they were without symbols, all they had were words. For us the process is visual, but it wasn't so for them. Note that this formula should be used only after you did the previous steps - that is after you have simplified the equation to the form before the uglification step.

Now to the forth degree. We need to solve an equation of the form:

The first step is to use the same method I used in solving the third degree polynomial, to reduce the problem to:

This reduction can always be done, it is easy to check but requires a lot of typing in latex to show it here, so I am skipping this. Lets complete the equation to a square, and move the rest to the second side (since c,d are unknown numbers, we can ignore the sign):

Now, if only the left side was a square.... Well, it is again time for uglification. Lets look on:

=

==

Now, when this will be a perfect square? The answer is simple. We need the discriminant to be equal zero. This means that:

This is a third degree equation. Now, isn't it great that we have a way to solve a general third degree equation? Without knowing a solution to the third degree polynomial we wouldn't be able to solve this... Now, lets suppose that (t) is a solution. Than we can write the original equation as:

We can now take the root and get the simple equation:

and

This is second degree, so the solution is easy.

Now, this is the part where I apologize for the misleading title - as you can see, while I developed the general way of solving the third and forth degree equations, I didn't write down the formulas. This is because they are nether useful nor nice to write (the formula for the forth degree is huge). I once was told that solving second degree equation using the formula takes 5 minutes for the third degree it takes one day, and for the forth a week. This is not true but it is close. However, all you need to write the final formula is in the post, you just need to write it down. If you are interested in the formulas themselves here are links to them: Third degree, Forth degree.

Multiplication using Vedic mathematics

I was sent this rather interesting video today. It claims to show how to multiply numbers with two and three digits. Probably the method will also work for numbers with four digits.

Since it is from Youtube, I have no idea if this is indeed Vedic mathematics or not. However, the trick used is very simple. What you do is use lines to represent numbers, and crossing lines represent multiplication - this is in fact a step back from the system we use. And it also takes more time that the methods for multiplication we are taught in school. Especially for numbers with a lot of digits....

The Middle Value Theorem

I admit that this is strange, to write about integrals and then to start talking about the Middle value theorem without finishing the previous topic. However, there is a reason for this. Firstly, I intended to write about both, and secondly, to write about integrals means to use latex. Unfortunately, because of the fluctuations in the weather in Jerusalem my head hurts to much for the thought to write in Latex to enter it. So today I will write about the Middle value theorem.

The Middle value theorem:
Let f(x) be a continues function from [a,b] to R. Suppose that f(b)>f(a). Than for any f(b)>y>f(a) there are exists b>x>a such that f(x)=y.

The proof I am going to give is the historical proof, there are others way to get to this result. However, Bolzano who originally proved this theorem used this method. To prove the theorem we firstly need to proof another statement:
Suppose that M is a property which is correct for some, but not all, numbers. Also, there is a number u with the property that M is correct for all numbers strictly less than u. Than there exists a number U with the property that M is correct for all numbers less than U, but there are no numbers large than U with the property that M is correct for all numbers lesser than them.

This is how Bolzano formulated this - in modern notation:

A={u| for all x less than u M is correct}, U=supA

Now to the proof. Balzano proofed this by simple construction. From the problem he new that A is not empty, and that there is a number that is larger than all of the numbers in A (because otherwise M is true for all numbers) we will mark it b. Lets look on the series:

If U>u we will pass it on some step. If this doesn't happen than u=U. If however we passed U on the N step, we can now look on the series:

Again, we repeat the previous step. If we will have to repeat it infinite number of times, we will have a series of numbers (which are all less than U) that is bounded (by U) and increases monotonically - and therefore it has a limit, which is clearly U. This part I am leaving without a proof - it is easy to show using elementary theorems. Bolzano proofed a statement which can be read as "Cauchy series converges" in order to proof this part, but this proof is omitted because it is identical to the proof given in first year Calculus - which means it requires epsilon and thus latex.
It is also interesting to note that as it was given by Bolzano the above proof is incorrect, because without the axiom of completeness he couldn't say that the limit of the Cauchy series is a real number.

Now it is time to use this to prove the Middle value theorem. The proof is very simple - let M be the property: y>f(x). Because this is not true for x=b, we get from the statement that we just proved that there exists U such that for all x less than U, M is true, and for all x greater than U M is false. Therefore f(U)=y.

This proof, unlike the previous one, if perfectly valid. It can be used on an exam today. A question to the reader: Where in the proof the property that f is continues is used?

Definite Integrals

Surprisingly, it is not easy to define what such an integral is. We want it to be equal to the area under the graph of a function - but to get it we need to define a specific algorithm that will produce it. One of such definitions is given by Riemann - it is called the Riemann Integral.
To properly define the Riemann Integral, we need to define a few other things firstly.

Lets suppose that we want a definition that will give us the area under the graph of some function f(x), which goes from the interval [a,b] to the real line (it is not necessary to assume that f is defined in all of [a,b], but it must be bounded in [a,b]). For this interval lets define:

P={a=}.

As you see from the definition P is a group that has n+1 different points and it includes both a and b. Note that according to the definition P is finite. Now for any such group lets define w=max||. Obviously w>0. Now we can write the following sum:

()

In this sum is any number such that .
The next step would be to define a limit "according to w". We will define it in the following way:
We will say that the limit =A if and only if:

|()-A|<

Basically, it says that if for any division of the interval such that the maximum distance between two points in the division is less that delta, the difference between the sum and A is less than epsilon, and such delta exists for any given epsilon (and for any selections of the points ) , than the limit is A.

The limit, if it exists, is the integral according to Riemann. If the limit doesn't exists the function has no integral according to Riemann.

Since we have a definition, lets use it to calculate an integral. For example, lets calculate the integral of f(x)=2 in the interval [0,1]. f()=2 for any part of the interval, so we can get it out of the sum. The rest of the sum is the length of the interval:

lim()=2(0--1)=2

Unsurprisingly the answer is two. Notice that this is indeed the answer only because the calculation here is independent from both P and the points t we selected. But what if the function is slightly more complex, for example f(x)=x^2? It is still possible to calculate the integral using this definition, but it is better to use another definition.

The Riemann definition is a very good definition, but sometimes it is easier to use a different definition. There is more than one other definition, but I want to talk only about one of them - the definition according to Darbo.

In this definition we again make use of P, which is defined in exactly the same way as before. We also need to define two other things:

=inf(f(x)) , =sup(f(x)) ,

Using this we can define the Upper Darbo sum as following:

And also the Lower Darbo sum:

The next step is to define the lower and upper integral. The upper integral is the infimum of the group of the upper sums, and the lower integral is the supremum of the lower Darbo sums. If both of these integrals are equal, we will say that the function has an integral according to Darbo.

In the next posts I will show that these two ways to define the integral are equivalent, and will also discuss ways to calculate the integral.

Bringing out-of-print math books into print

There is a new interesting project on the internet (we have a lot of this, but this one is different). Some mathematicians decided that it is time to do something with the fact that a lot of good math books are now out of print. While it is possible to ask to make a new edition of an old book, those who tried to do so found that they need firstly to convince others that there is a significant number people who want this. Unfortunately, it turned out to be rather difficult.

To solve this, the project "outofprintmath" was created. The idea is to let people suggest what book they think should get back into print, and it also allows to vote on the books suggested. The idea is that this site would show publishers that there is indeed a market for those books. It is a very new project, but there already 53 books suggested, and one of them has 55 votes.
If you have a book you want to see back in print, it is definitely a place to visit.