Wednesday, August 26, 2009

Why Does 0.999... = 1?

Consider the real number that is represented by a zero and a decimal point, followed by a never-ending string of nines:

0.99999...

It may come as a surprise when you first learn the fact that this real number is actually EQUAL to the integer 1. A common argument that is often given to show this is as follows. If S = 0.999..., then 10*S = 9.999... so by subtracting the first equation from the second, we get

9*S = 9.000...

and therefore S=1. Here's another argument. The number 0.1111... = 1/9, so if we multiply both sides by 9, we obtain 0.9999...=1.
When seeing these arguments, many people feel that there is something shady going on here. Since they do not have a clear idea what a decimal expansion represents, they cannot believe that a number can have two different representations.

We can try to clear that up by explaining what a decimal representation means. Recall that the digit in each place of a decimal expansion is associated with a (positive or negative) power of 10. The k-th place to the left of the decimal corresponds to the power 10^k. The k-th place to the right of the decimal corresponds to the power 10^(-k) or 1/10^k.

If the digits in each place are multiplied by their corresponding power of 10 and then added together, one obtains the real number that is represented by this decimal expansion.

So the decimal expansion 0.9999... actually represents the infinite sum

9/10 + 9/100 + 9/1000 + 10/10000 + ...

which can be summed as a geometric series to get 1. Note that 1 has decimal representation 1.000... which is just 1 + 0/10 + 0/100 + 0/1000 + ... so if one realizes that decimal expansions are just a code for an infinite sum, it may be less surprising that two infinite sums can have the same sum.

Hence 0.999... equals 1.
Here's a cool mathematical magic trick. Write down a three-digit number whose digits are decreasing. Then reverse the digits to create a new number, and subtract this number from the original number. With the resulting number, add it to the reverse of itself. The number you will get is 1089!

For example, if you start with 532 (three digits, decreasing order), then the reverse is 235. Subtract 532-235 to get 297. Now add 297 and its reverse 792, and you will get 1089!
Proof:
If we let a, b, c denote the three digits of the original number, then the three-digit number is 100a+10b+c. The reverse is 100c+10b+a. Subtract: (100a+10b+c)-(100c+10b+a) to get 99(a-c). Since the digits were decreasing, (a-c) is at least 2 and no greater than 8, so the result must be one of 198, 297, 396, 495, 594, 693, 792, or 891. When you add any one of those numbers to the reverse of itself, you get 1089!
How many people do you need in a group to ensure at least a 50 percent probability that 2 people in the group share a birthday?

Let's take a show of hands. How many people think 30 people is enough? 30? 60? 90? 180? 360?

Surprisingly, the answer is only 23 people to have at least a 50 percent chance of a match. This goes up to 70 percent for 30 people, 90 percent for 41 people, 95 percent for 47 people. With 57 people there is better than a 99 percent chance of a birthday match!

Presentation Suggestions:
If you have a large class, it is fun to try to take a poll of birthdays: have people call out their birthdays. But of course, whether or not you have a match proves nothing...

The Math Behind the Fact:
Most people find this result surprising because they are tempted to calculate the probability of a birthday match with one particular person. But the calculation should be done over all pairs of people. Here is a trick that makes the calculation easier.

To calculate the probability of a match, calculate the probability of no match and subtract from 1. But the probability of no match among n people is just
(365/365)(364/365)(363/365)(362/365)...((366-n)/365),
where the k-th term in the product arises from considering the probability that the k-th person in the group doesn't have a birthday match with the (k-1) people before her.

If you want to do this calculation quickly, you can use an approximation: note that for i much smaller than 365, the term (1-i/365) can be approximated by EXP(-i/365). Hence, for n much smaller than 365, the probability of no match is close to

EXP( - SUMi=1 to (n-1) i/365) = EXP( - n(n-1)/(2*365)).

When n=23, this evaluates to 0.499998 for the probability of no match. The probability of at least one match is thus 1 minus this quantity.

For still more fun, if you know some probability: to find the probability that in a given set of n people there are exactly M matches, you can use a Poisson approximation. The Poisson distribution is usually used to model a random variable that counts a number of "rare events", each independent and identically distributed and with average frequency lambda.

Here, the probability of a match in a given pair is 1/365. The matches can be considered to be approximately independent. The frequency lambda is the product of the number of pairs times the probability of a match in a pair: (n choose 2)/365. Then the approximate probability that there are exactly M matches is:
(lambda)^M * EXP(-lambda) / M!
which gives the same formula as above when M=0 and n=-365.

Thursday, August 20, 2009

Finding the N-th digit of Pi

Here is a very interesting formula for pi, discovered by David Bailey, Peter Borwein, and Simon Plouffe in 1995:

Pi = SUMk=0 to infinity 16^-k [ 4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6) ].

The reason this pi formula is so interesting is because it can be used to calculate the N-th digit of Pi (in base 16) without having to calculate all of the previous digits!

Moreover, one can even do the calculation in a time that is essentially linear in N, with memory requirements only logarithmic in N. This is far better than previous algorithms for finding the N-th digit of Pi, which required keeping track of all the previous digits!

Presentation Suggestions:
You might start off by asking students how they might calculate the 100-th digit of pi using one of the other pi formulas they have learned. Then show them this one...

The Math Behind the Fact:
Here's a sketch of how the BBP formula can be used to find the N-th hexadecimal digit of Pi. For simplicity, consider just the first of the sums in the expression, and multiply this by 16N. We are interested in the fractional part of this expression. The numerator of a given term in this sum is 16N-k, and it can be evaluated very easily mod (8k+1) using a binary algorithm for exponentiation. Division by (8k+1) is straightforward via floating point arithmetic. Not many more than N terms of this sum need be evaluated, since the numerator decreases very quickly as k gets large so that terms become negligible. The other sums in the BBP formula are handled similarly. This yields the hexadecimal expansion of Pi starting at the (N+1)-th digit. More details can be found in the Bailey-Borwein-Plouffe reference.

The BBP formula was discovered using the PSLQ Integer Relation Algorithm. However, the Adamchik-Wagon reference shows how similar relations can be discovered in a way that the proof accompanies the discovery, and gives a 3-term formula for a base 4 analogue of the BBP result.

Would you have bet on Charles Coventry improving on Saeed Anwar's 194, the highest individual score in ODIs, by equalling it and remaining unbeaten, when better batsmen with more formidable records - Adam Gilchrist, Sanath Jayasuriya and Sachin Tendulkar, to name only three - had merely got close? Coventry had played only 13 innings and scored 301 runs before his 194 not out against Bangladesh in Bulawayo, and was averaging 23.15 at the time. In this week's column we've looked at batsmen with the lowest averages before they scored a fifty and a century in ODIs and Tests.

The player with the lowest batting average before scoring a Test half-century is New Zealand fast bowler Bob Blair, who is best remembered for coming out to bat on Boxing Day in Johannesburg 1953, hours after learning of his fiancé's death in a train crash at home. Until he scored an unbeaten 64 against England in Wellington in 1963, Blair had reached double figures only once in 25 innings, and averaged 4.23 - two runs fewer than what Glenn McGrath did when he scored his first and only fifty, against New Zealand at the Gabba in 2004.

Sunday, August 16, 2009

Samsung Launches ‘Omnia HD’ and ‘Omnia Pro B7320’ in India

Mobile handset maker Samsung has announced the launch of two new handsets christened Omnia HD and Omnia Pro B7320 in the Indian marke

The first handset OmniaHD, which is equipped with 8 MP camera, is priced at Rs 33,990.

Its main features are 720P high definition (HD) video recording and 9.4 cm Amoled screen. However its other feature includes Bluetooth capabilities, Wi-Fi connectivity, support for DLA, a built-in GPS receiver etc.

The company claims that OmniaHD offer high speed internet access of HSUPA 5.76Mbps and HSDPA 7.2Mbps. In addition, it supports 16GB internal memory, which can be extended to 48GB with the help of memory card.

While, the second handset, Omnia Pro B7320 is priced at Rs 16,500. The model is based on Window Mobile 6.1 and has a rather compact design compared to its peers.

Further, Omnia Pro B7320 is powered with 1Ghz processor, 16:9 screen along with 16M colors, dual stereo speaker, 3 MP camera, Wi-Fi, 3G, EDGE and FM radio.

The handset offers a 70MB internal memory, which can be expandable to 16GB.



Saturday, August 15, 2009

f you can Watch it, you can Record it with Replay Video Capture!

Replay Video Capture is the best way to make production quality videos from hard-to-record sources, including:

* ANY Video Web Site, using ANY streaming protocol
* DVD’s
* Webcam sites
* Powerpoint Presentations
* Skype Video calls
* Video Chat sessions
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With Replay Video Capture, you’ll get amazing video quality, or compact portable files. You can choose between high-quality, high-speed
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Replay Video Capture’s super-fast MPEG-2 codecs don’t bog down your CPU, so you won’t lose video frames or get choppy audio. These are
preferred for high-motion video, or for use on slower PC’s.

For the audio track, Replay Video Capture captures what you hear from your PC’s speakers by default. Or, you can choose to narrate using the
microphone instead, which is great for PowerPoint presentations and software demos.

(OS: Windows XP or Vista)

http://hotfile.com/dl/8841147/6d97675/rEpLayMEdIacAtcHEr302.rar.html

Yahoo Messenger online

Now Yahoo has launched its web based chatting software
http://webmessenger.yahoo.com/

Digsby


digsby is a multiprotocol IM client that lets you chat with all your friends on AIM, MSN, Yahoo, ICQ, Google Talk, and Jabber with one simple to manage buddy list.
Now you don't need to download any of the yahoo messengers and google talks .
this will help you a lot
Download:
http://www.digsby.com/download.php?os=win
Hi this is my new blog, this is my first post.