سلام!نمیدونم برای خوندن مطالب مشکل خواهین داشت یا نه.این طوری خوبه!میدونین چندین قرن غربیا علوم مختلف مارو ترجمه کردن به اسم خودشون به دنیا عرضه کردند حالا نوبت ماست که زبونای غربی رو یاد بگیریم وتا چند سال آینده به امید خدا و با اعتماد به نفس همگام با علم روز دنیا پیشرفت کنیم.به امید روزی که کشور عزیزمون ایران طلایه دار رشته های مختلف علوم پایه و مهندسی(مخصوصا علم شیرین ریاضی!!) تو دنیا باشه.

                                                                                        شپره 

Persian or Arabic Mathematics!

 Before I start criticizing to the great works of two Scottish mathematicians who have worked on the history of mathematics, I congratulate for their works and I express that I have tried to seek for reality and there in no place for nationalism in my critics 

The question is that why most of the Iranian (Persian) scholars in the history are considered as Arabs. For example while Khayyam is considered as a Persian poet, he is introduced as an Arab mathematician!

 

Iran is a big country that different nations live beside each other and some of them are Arab too, but they live in Kuzestan, a southern province of Iran. Also we should not forget that in the past Iran (Persian Empire) was bigger and after different separations, Iran has become smaller.

 

This is right that many of those scholars have written their works in Arabic (the international scientific language among people of that time), but this should not cause us to think that they have been Arab. Today most of the scholars write in English, then should we consider them American for instance?

 

Though it is wonderful that the two esteemed mathematicians have started publishing some great articles about the contribution of Iranian mathematicians in the history of mathematics in Internet, but it is surprisingly strange that why they have categorized it as Arabic, the mathematics that has been nurtured and flourished by these great mathematicians, most of them non-Arab mathematicians!

 

The title of this great article is “Arabic mathematics: forgotten brilliance?” and then they start their article by this passage that “recent research paints a new picture of the debt that we owe to Arabic/Islamic mathematics. Certainly many of the ideas which were previously thought to have been brilliant new conceptions due to European mathematicians of the sixteenth, seventeenth and eighteenth centuries are now known to have been developed by Arabic/Islamic mathematicians around four centuries earlier. In many respects the mathematics studied today is far closer in style to that of the Arabic/Islamic contribution than to that of the Greeks.”

 

Again the question is why Arabic/Islamic contribution?

And this passage becomes more interesting when they add that “there is a widely held view that, after a brilliant period for mathematics when the Greeks laid the foundations for modern mathematics, there was a period of stagnation before the Europeans took over where the Greeks left off at the beginning of the sixteenth century. The common perception of the period of 1000 years or so between the ancient Greeks and the European Renaissance is that little happened in the world of mathematics except that some Arabic translations of Greek texts were made which preserved the Greek learning so that it was available to the Europeans at the beginning of the sixteenth century.

That such views should be generally held is of no surprise. Many leading historians of mathematics have contributed to the perception by either omitting any mention of Arabic/Islamic mathematics in the historical development of the subject or with statements such as that made by Duhem in [3]:-

... Arabic science only reproduced the teachings received from Greek science.”

And at the end when they want to describe the period that they want to discuss they write:

“Before we proceed it is worth trying to define the period that this article covers and give an overall description to cover the mathematicians who contributed. The period we cover is easy to describe: it stretches from the end of the eighth century to about the middle of the fifteenth century. Giving a description to cover the mathematicians who contributed, however, is much harder. The works [6] and [17] are on "Islamic mathematics", similar to [1] which uses the title the "Muslim contribution to mathematics". Other authors try the description "Arabic mathematics", see for example [10] and [11]. However, certainly not all the mathematicians we wish to include were Muslims; some were Jews, some Christians, some of other faiths. Nor were all these mathematicians Arabs, but for convenience we will call our topic "Arab mathematics".”

The inconvenience shows itself when they vividly express that “the regions from which the "Arab mathematicians" came was centred on Iran/Iraq but varied with military conquest during the period. At its greatest extent it stretched to the west through Turkey and North Africa to include most of Spain, and to the east as far as the borders of China.”

And when one refers to the biographies of these mathematicians understands that most of them have been from Persia (now Iran), so why those mathematicians must be considered as Arabs. And since Persians has had a very great and glorious culture and civilization in pre-Islamic ages, then why there is no mention to the probable influence of pre-Islamic Persian mathematics on “Persian mathematics from the end of the eighth century to about the middle of the fifteenth century”!

And at the end, despite of this critic to their work, I consider their works very useful for the history of mathematics, since thanks to the great efforts of these specialists of the history of mathematics, today we know that what they did was not just a reproduction of the works of Greeks!

 

 

 

By Peyman Nasehpour

References:

Arabic Mathematics: forgotten brilliance

List of Iranian (Persian) Mathematicians

Birthday Paradox

I'm sure that most of you readers have either heard of, or will hear of the birthday paradox. Here's one way of posing it: A statistics professor tells his class of 30 that he will bet that at least two people in the class share the same birthday. The students thought they had a good chance of winning, seeing as how there are 365 days in the year, and took the bet — the professor won. In fact, assuming a uniform distribution of birthdays, he was more than 70% likely to be correct. How many students are required for the professor to be correct with probability 50%? 90%?

Let us examine some possible solutions or approximations. This first problem is most easily approached by considering the probability that no two people share the same birthday. We thus pick any random person and fix the birthday. The chance that the next person has a different birthday is 1 – 1/365 (we typically assume the year has 365 days for simplicity — sorry Feb 29ers), since he or she can have any birthday except for the one we fixed initially. Likewise, the probability the third person has a different birthday from either of the first two people is 1 – 2/365, as we have two taboo birthdays. We continue in this trend, then multiply all the probabilities (we assume independence) to arrive at: the probability that n people have distinct birthdays is 365!/(365ⁿ(365 – n)!), so the probability that at least two people share the same birthday is simply 1 – 365!/(365ⁿ(365 – n)!). I don't know about you, but my calculator can't compute 365!, so this formula isn't terribly useful. Well ok, my Maple calculator can, and it turns out to be something around 2.51e778, and completing the calculation tells us that 23 people will give a 50.7% of at least two people sharing a birthday, and 41 people will give a 90.3% chance.

Working with numbers in the 1e778 range is somewhat annoying, so perhaps we can find a better way of finding a solution, or at least an approximate solution. Notice that for the n^th additional person we required to have a differing birthday, we were simply multiplying our current probability by 1 – (n – 1)/365, so we could just use recursion to arrive at our answer. But that's fairly tedious as well, so we make a sacrifice in precision for the sake of our sanity. Recall the linear approximation of exp(x) is given by 1 + x (we simply truncated the Taylor series expansion). Now this is nice for us, since at each stage we were multiplying by 1 – x for some x, which by the linear approximation is nearly exp(–x). Hence the probability that no two people share the same birthday out of n people is

Include graphic InterestingMath0.eps here

.

We subtract this from 1 to get the probability we want. This is so very much easier to compute, and we quickly get that 23 people will give a 50.0% of at least two people sharing a birthday, and 42 people will give a 90.5% chance — which is fairly close to what we had before. You may notice that the disagreement between the actual probability and the approximation gets worse as n increases; this is simply because as n increases, so does x, and the linear approximation we used works best near 0 while getting worse farther out. There are a few other ways to get an approximation; I leave it as an exercise to the reader to find some.

As pure mathematicians always do, I like to generalize results. In this case, we have only considered n people, out of which at least two people share a birthday, assuming a uniform distribution of birthdays in a 365-day year. We could throw back in February 29, but the problem is nearly the same, as you might discover. We could look at alternate distributions, since a uniform distribution might be unrealistic — in fact, statistically (in North America) September to November has more birthdays than other months, perhaps because it is about 9 months from New Year's Eve/Day. On the flip side, birthdays in April or May are fairly scarce, which happen to be about 9 months from the hottest days of the year.

Something someone said recently had me thinking about a different sort of generalization, however. What is the probability that at least 3, or 4, or k people have the same birthday? After some brute force calculation, one can arrive at the fact that to ensure a 50% probability of having at least 1, 2, 3, 4, 5, 6, ... coincident birthdays, the minimum number of people required is: 1, 23, 88, 187, 313, 460, ... respectively. That's fairly restrictive, so I would like to consider another kind of generalization: what if we simply wanted "nearly" coincident birthdays? This certainly makes the problem more difficult, and requires the inclusion-exclusion principle. I leave it as an exercise to prove it, but if you're interested: to ensure a 50% probability of having at least 2 people with birthdays at most 1, 2, 3, 4, 5, 6, ... days apart, the minimum number of people required is: 23, 14, 11, 9, 8, 8, ... respectively. So if you want an interesting twist on an old classic, bet groups of at least 9 people that there will be at least two birthdays within a week (l = 3.5), and you will have more than a 50% success rate. The results may surprise people!

We can exhibit an explicit formula for situations similar to this. Let Q_i(n,d) denote the probability that exactly i people out of n people share a birthday, out of d possible birthdays (uniformly distributed). Let P_k(n,d) be the probability that at least k people out of n people share a birthday, out of d possible birthdays (uniformly distributed). Then we get the relation

Include graphic InterestingMath1.eps here

We can calculate Q_k(n,d) via the recurrence relation

Include graphic InterestingMath2.eps here

where

Include graphic InterestingMath3.eps here

denotes the floor of x. Clearly, this is not a fun thing to calculate, and the time it takes to calculate grows exponentially with k. A reasonable approximation can be used however; Diaconis and Mosteller (1989) produce a relationship that approximates the number of people n such that p = P_k(n,d):

Include graphic InterestingMath4.eps here

For example, suppose we wanted to determine the probability that out of 30 people, 4 people had a birthday within a 2 week span. Using the assumption that birthdays are uniformly distributed, this is nearly equivalent to the problem of asking for the probability of 4 people having the same "birthday" if there were only d = 26 possible birthdays (52 weeks/2 weeks = 26). By the formula above, the probability p is about 54.5%. Thus, if you're planning to have a combined birthday party for people out of a group of 30 for birthdays within 2 weeks of a specific date, you're more than likely to have at least 4 "birthday people". Try this out with your friends and see if this is close!

Vince's problem of the issue

Suppose there is a contest: contestants will enter a room one by one. The winner is the first person to have the same birthday as someone already inside the room. The catch: you, being special, have the option of entering whenever you wish. Now, there are 367 contestants (including yourself), so there is a guaranteed winner by pigeonhole. It is clear that the first person to enter will have a 0% chance of winning, so you definitely don't want to go first. On the other hand, going last will almost certainly be a bad idea, since the only way you win is if every possible birthday occurs before you. Here's the question: What position should you place yourself in in order to maximize the chance of winning?

Vince Chan