Politics

The only good thing in politics, in my opinion, is that most of the time it doesn't influence you directly. Unfortunately, this is not the case for me now. The rest of the post is my complains, so feel free not to read it.
First of all, there are elections to the mayor office at Jerusalem at the middle of next month. As of now, I don't even intend to vote because I don't have the time nor the desire to check the information available on the candidates. Also, I will probably be too busy studying anyways. What annoys my most is that for some reason I got a call asking me to help with the election process - to take attendance notes. I have no idea why they called me with this..

Secondly, next week I suppose to start a new semester. However the university already sent letters telling all the students that the semester will not open because the government doesn't give them the money they need. They even included a long list of emails in the letter to where we can send complains. I really doubt that anything will happen to the semester, but who knows...

Thirdly, due to government policies and economical crisis the apartments in Jerusalem are rather expensive now, and my family and I are currently looking for a new apartment. Horrible timing... I certainly hope that we will find something soon.

On a somewhat brighter note, here are link to some political jokes I found:

Two political candidates were having a hot debate. Finally one of
them jumped up and yelled at the other: "What about the powerful
interests that control you?"

The other guy screamed back, "You leave my wife out of this."

Found this one here, follow the link for more.
Also check the Circle vs. Square webcomic, its author did an excellent job making fun of politics without insulting anybody.

Free Books

I found today a fairly large collection of books in pdf format. The books cover a lot of subject, but there are also a few books about programing and math. If you are interested follow the link.

Series Part 4

Unlike the previous posts on this subject, in this post instead of talking about the subject in general I will just fully solve one rather nontrivial problem. The problem is to find the formula for the n-th term of the Fibonacci progression.

The Fibonacci progression is a progression defined by the following relation:

a(1)=a(2)=1, a(n)=a(n-1)+a(n-2)

It is of course possible to find any term in the progression by simple calculations, but buy using this relation we are required to calculate all the terms which come before the one we want to know. It turn out that for high values of n this task is not practical even for a computer.

To find the formula we will need to use linear algebra. Firstly, lets define a group of progressions which we will call generalized Fibonacci progressions. Those are all the progressions that follow the recursive rule, but the first terms are allowed to be any numbers. Now, lets look on the vector space of all infinite progressions. Obviously, the generalized Fibonacci progressions belong to this space. Moreover they form a subspace of dimension two in the vector space.
The last sentence requires a proof - all we need to show is that the group of the generalized Fibonacci progressions is closed under multiplication and addition. We will take two such series, a(n) and b(n) and a scalar r:

a(n)=a(n-1)+a(n-2)
g(n)=ra(n)=ra(n-1)+ra(n-2)=g(n-1)+g(n-2)

Thus, g(n) is also a generalized Fibonacci progression, and we proved that the group is closed to multiplication.

c(n)=a(n)+b(n)=[a(n-1)+a(n-2)]+[b(n-1)+b(n-2)]
c(n)=a(n-1)+b(n-1)+a(n-2)+b(n-2)=c(n-1)+c(n-2)

The only thing we need to prove is that the dimension is two. To do this it is enough to show that there is a basis for this subspace that has two vectors in it. This is simple, the basis is the two progressions a(n)=1,0,... , b(n)=0,1,....
To prove that this is a basis we need to show that the vectors are independent (left as an exercise) and to show that any other vector is a linear combination of these two. Let c(n) be a vector in this space. All the generalized Fibonacci progressions are uniquely defined be their two first terms, so lets suppose that the two first terms are c(1)=d and c(2)=r. Then, f(n)=da(n)+rb(n)=d,r,....
Since we are in a vector space the resutl is another progression, and because of the unique definiton by the first two terms, we get that c(n)=f(n).

Now, is there a genarilized Fibonacci progression which is also a geometrical progression? We get the following conditions:

a(n)=aq^(n-1), a(n)=a(n-1)+a(n-2)

aq^(n-1)=aq^(n-2)+aq^(n-3)
q^(n-1)=q^(n-2)+q^(n-3)
q^2=q+1
q^2-q-1=0

There is obviously a solution to this equation. Moreover, there are obviuosly two different solutions, and therefore two different progressions. Lets suppose that q(1) and q(2) are the solutions. We can also select a=1 because it doesn't influence the solution. We get then two progressions: a(n) and b(n). The only important thing left to do is to show that they are a basis to our vector space. Because I already prooved that the dimension is two, we just need to show that they are lineary independant. It ois enough to show that the solution of the following equations is a=b=0:

a+b=0, aq(1)+bq(2)=0

Since q(1) is not equal to q(2) this is indeed so and therefore we have a basis.

Now we know that any progression in our space can be written as c(n)=da(n)+fb(n) for some d,f. We also know that a(n)=q(1)^(n-1), b(n)=q(2)^(n-1). Deriving the final formula from this is easy, but since I still cannot use latex in blogger, completing the last step is left to the reader.

Series Part 3

As promised, in this post I will shortly discuss infinite series. I don't want to say a lot about this because this would require developing precise definition, so I will just describe the main ideas and give some classical examples.

The main question we can ask about an infinite series is if it has a limit, or in a case of an infinite sum, the result of the summation.
Firstly, lets define what a limit of a series is. Simply put, a limit of a series is a single number to which the series is close from some point. For example, the limit of the series:

1,1,1,1,1,.....

Is 1 (we assume that the series continues to infinity following the same pattern). The same is true is we change the first number - the limit will remain the same. In a counter example, the series: 0,1,0,1,01.... Doesn't have a limit at all, because it is wrong to say that there is single number to which the series is getting close to.
It is important to notice that the definition I gave is not exact. Moreover, if you will write such a definition on a exam you will get null point for it. It doesn't mean that it is wrong - but it is not useful. It is simple to fix, all that is needed is to define what does it mean "close to" and "from some point". Unfortuantely, the script I used to write latex in posts no longer works, so I cannot write a definition (it requires writing symbols). Therefore completing the definition is left as an exercise for the reader :).

Since we have the "definition" lets try to use it. Lets look on the series:

a(n)=2^(-n)

It is very easy to see to what number this series is getting close to. All you need to do is to imagine that n is very big (or use a calculator). Either way you will see immediately that the result is very close to zero. However, according to the definition we need to show that there is only one number that the series is getting close to. To do this lets suppose that r is another number that the series is close to. Since the series is positive, if r is negative than the series is always closer to zero than to r, so r is not the limit. Otherwise, r is positive. So lets find n such that 2^(-n)<0.25r. Because the series is monotonically decreasing, we get again that from some point the series is closer to zero than to r.

Lets look on the series: a(n)=4+n. According to our definition it doesn't have a limit because the limit must be a number and it is clear that this series grows "to infinity". (It is of course possible to define the limit differently). From this two examples we get a simple result - all arithmetic series don't have a limit and all the geometric series for which |q|<1 have a limit. Also, for all such geometric series the limit is non other that zero, because we get: a(n)=aq^n. If q is less that one, this number approaches zero.

Lets now look on infinite sums. We will say that a sum converges if the sum is a real number. (Again, the definition is not precise.) This brings a question - how is it possible at all? After all it doesn't matter how small the numbers we sum, if we sum an infinite number of them the result "should" also be infinite. At least that is what our logic tries to say. But, this is wrong. Lets see an example:

1, 05, 0.25, ....

This is a simple geometric series, in which q=0.5 and a=1. From the previous post the sum is:

S(n)=(q^n-1)/(q-1)

We can think about this as a new series - a series of partial sums of the original infinite series. Does it has a limit? Of course - it is obvious that the limit is simply 1/(1-q). Now, if we will define the infinite sum as the limit of the series of the partial sums, we will get the immediate result - the sum of the infinite series. In the case of p=0.5, the sum is 2.

In the next (and final) post I want to discuss an example of a series that is not an arithmetical or a geometrical progression.

What is persistence?

The following story is from Clientopia. I do not know if it is real or not, but after reading it you will surely understand what does it means to completely concentrate on your goal leaving behind any doubts...
---------------------------------------------------------------


I have had my company mobile for nearly 2 years, out of the 50 odd calls I've had, only 3 were legitimate, the rest were wrong numbers. This is a transcript of the one man who has called me more than 35 times.

Me: Hello, XXX IT, XXXX speaking, how can I help?

Random: Is that Brian?

Me: No, this is XXXX

Random: Are you sure?

Me: I should think I am. I've never been called Brian, and there isn't a Brian involved with this mobile

Random: Oh, can you put me through to him?

Me: No, as there isn't a Brian here, or in XXX IT

Random: Now I know you're lying, this is the number he gave me, now put him on the phone!

Me: I'm sorry, you've got the wrong number

Random: NO I HAVE NOT! BRIAN GAVE ME THIS NUMBER

Me: Then Brian got the wrong number

Random: I THINK HE KNOWS HIS OWN MOBILE NUMBER

Me: On the current evidence, I would say he doesn't, I cannot help, goodbye. *click*

----

He rings back, I don't answer.

----

12 (!) calls later he gives up. Then calls back the next day:

Me: I'm sorry you've got the wrong number, you rang yesterday several times.

Random: THIS NUMBER WAS GIVEN TO ME SO PUT ME THROUGH

Me: *click*

He has since called back several more times.

Ubuntu Intrepid

I am installing the latest version of Ubuntu now. It is still a beta version, the stable release is due on the 30 of this month. Hopefully I will not break my system.... If everything will go fine, I will update this post with a short review of this beta release.

Update: Well, the installation itself went just fine. However I am unable to connect to the internet. To be precise, I a,m able to connect to my ISP and to ping sites, but I am unable to view any web pages. I am also unable to download files using wget or to install new packages. However amule works fine. From what I found on the net this is a rather common problem but there is still no solution.
Right now I am using a liveCD to connect to the internet. I will probably reinstall Ubuntu Hardy tomorrow (unless I will find a solution).

As promised - a short review of the new release. I noticed three main changes. The first one is tabbed browsing in Nautilus. It works well, but for some reason if you start a second session it is not opened automatically as a tab in the first one. It is possible that this behavior can be configured somewhere, but in my opinion it is a required addition.
The second change is that the shutdown button is now changed into a log-out button. I cannot say that a like or dislike this change, but it seems a good idea.
The third one is a change is the network manager. For me it was a disaster - I am no longer able to connect to the internet because of this. It also looks much more complicated than the previous one.
There is also new artwork is this release. In this area there is a clear improvement.

Update2: I did a clean install of Ubuntu Hardy today and everything is working fine again. This is the first time I was forced to reinstall Ubuntu...

Series Part 2

Continuing from the previous post on this subject, lets find the formulas for the sum of the two series I mentioned in the previous post, the arithmetic and the geometric progressions:

Arithmetic progression:
Firstly, lets look on a very simple case - the progression 1+2+3+...+n=S. If we just look on it, it is not very clear what the sum is. However, if we will write it in a slightly different manner the answer will be obvious:

1,2,3,4,5,6,7,8,9,...n
n...,9,8,7,6,5,4,3,2,1

Obviously the sum of all the numbers is 2S. We also know that there are n columns and the sum of any column is n+1. Therefore 2S=n(n+1). From this we get the formula by division.

Now lets look on the general case. The general arithmetic series is:

a+(a+d)+(a+2d)+...+(a+(n-1)d)=S
na+d+2d+3d+...+(n-1)d=S
na+d(1+2+...+(n-1))=S
na+0.5d(n-1)n=S
0.5[2a+(n-1)d]n=S

By the way, this formula requires to know the first number in the progression. However, if you know the last one instead of the first you can also use this formula - after changing d to (-d).


Geometric progression:
Firstly, lets again consider a specific case. Lets look on the sum of the following progression:

1+q+q^2+q^3+..+q^(n-1)=S
q+q^2+q^3+..+q^(n-1)+q^n=Sq
q^n-1=Sq-S
S=(q^n-1)/(q-1)

Getting from this to the general case is extremely easy. The only thing we need to do is to remember that a general geometric progression is just a multiplication of the case we took care of by a constant number a. Therefore, the formula for the sum is:

a+aq+aq^2+....+aq^(n-1)=S
S=a(q^n-1)/(q-1)

In the next post I will talk about infinite series and their sums.

University vs the Government

I received an email from the university today, as expected it was an official statement that unless the government will provide the necessary funds the upcoming semester will be canceled. The previous year the situation was the same, and in the last moment the government decided to give the necessary funds. I wonder what will happen this year...
The letter also contained a call for students to join the fight with the government. While I am not going to participate in such activities, I am sure there will be enough people who will.
I just hope that the semester will open in time and in an orderly fashion..

Series - Part 1

In math a series is a function from the natural number to some set B. The set B can be the real numbers but it can also be a vector space or some other set.
The most basic question we can ask about a series is finding the generating formula. Such formula can be extremely difficult to find, and since a series of random number is by all means a series such a formula may not even exist. Lets start with a few simple series:

Arithmetic progression:
This is probably the simplest example. An arithmetic progression is obtained by adding a fixed number to the previous number in the progression. For example, if we choose to start with the number 5 and to add seven, we will get:

5, 13, 20, 27, 34.....

In the general case we get:

a1=a, a2=a+d, a3=a+2d, a4=a+3d,...

It is very simple to find the generating formula for this progression. To do this we need to look on the difference:

a2-a1=d
a3-a2=d
.
.
a(n)-a(n-1)=d

If we add all of the equations we will get the generating formula:

a(n)-a1=(n-1)d

This method of looking on the difference can be used for many series. For it to work, we need to know how to sum the right side of the equation. In the example it was very simple to do, but usually it is much more difficult. Also, since the differences are in fact a series of its own, we can try to find the formula for the difference, and from it get the formula to the original series.

Geometric progression:
A geometric progression is a progression in which the ratio of the successive terms is fixed, that is a(n+1)/a(n)=const for all n. For example:

1,2,4,8,16....

And the general progression:

a,aq,aq^2,aq^3,aq^4...

In this case, finding the generating formula is trivial. Just from looking on the progression we get that:

a(n)=aq^(n-1)

However, if we would try to find this formula by looking on the difference we would just get the original progression back. Therefore to get the formula this way we would need to know the sum of the geometrical progression.
In the next post I will develop the formulas for the sum of both arithmetical and geometrical progressions.

I am back

I didn't post anything for a lot of time, but I am still here. Sorry for suddenly stopping posting, I was busy with exams so I had to stop posting for some time. I already got all the exam results, I did good in all except two (but this was expected from the start, so this is not much of a problem).
Originally I planned to stop posting only for a few days, maximum a week, but for some reason not posting anything was so relaxing that I just couldn't force myself to write anything. I even didn't read my rss :).

The next semester will start only next month, so I still have a lot of time to prepare for it. I already have some books for next year, and I intend to buy more. Hopefully I will be able to prepare well - learning seven courses in one semester is not a nice thing...