Sarah.: ....folio addmaths...

aper nih? ya ini la pi..or simbol dia macam kat atas tuh..
yeah..
kene wat folio addmaths..
gna pc..

task1 ialah sy pnya
so sy kener cari..
tp kalau anda usha link nih..
mau xberpusing kpala anda...

tgk taw..
kalo x tgk,..
xder feel

tp xper la..
sy try bg hint kpusingan dia.

Numerical value

See also: Numerical approximations of π

The numerical value of π truncated to 50 decimal places is:^[12]

3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510

See the links below and those at sequence A000796 in OEIS for more digits.

While the value of π has been computed to more than a trillion (10¹²) digits,^[13] elementary applications, such as calculating the circumference of a circle, will rarely require more than a dozen decimal places. For example, a value truncated to 11 decimal places is accurate enough to calculate the circumference of a circle the size of the earth with a precision of a millimeter, and one truncated to 39 decimal places is sufficient to compute the circumference of any circle that fits in the observable universe to a precision comparable to the size of a hydrogen atom.^[14]^[15]

Because π is an irrational number, its decimal expansion never ends and does not repeat. This infinite sequence of digits has fascinated mathematicians and laymen alike, and much effort over the last few centuries has been put into computing more digits and investigating the number's properties.^[16] Despite much analytical work, and supercomputer calculations that have determined over 1 trillion digits of π, no simple base-10 pattern in the digits has ever been found.^[17] Digits of π are available on many web pages, and there is software for calculating π to billions of digits on any personal computer.

....
ni baru skit taw..

ader lagi

In 2006, Simon Plouffe found a series of beautiful formulas.^[37] Let q = e^π, then

$\frac{\pi}{24} = 3\sum_{n=1}^\infty \frac{1}{n(q^n-1)} -4\sum_{n=1}^\infty \frac{1}{n(q^{2n}-1)} + \sum_{n=1}^\infty \frac{1}{n(q^{4n}-1)}$

$\frac{\pi^3}{180} = 4\sum_{n=1}^\infty \frac{1}{n^3(q^n-1)} -5\sum_{n=1}^\infty \frac{1}{n^3(q^{2n}-1)} + \sum_{n=1}^\infty \frac{1}{n^3(q^{4n}-1)}$

and others of form,

$\pi^k = a\sum_{n=1}^\infty \frac{1}{n^k(q^n-1)} +b\sum_{n=1}^\infty \frac{1}{n^k(q^{2n}-1)} + c\sum_{n=1}^\infty \frac{1}{n^k(q^{4n}-1)}$

where q = e^π, k is an odd number, and a,b,c are rational numbers. If k is of the form 4m+3, then the formula has the particularly simple form,

$p\pi^k = 2^{k-1}\sum_{n=1}^\infty \frac{1}{n^k(q^n-1)} -(2^{k-1}+1)\sum_{n=1}^\infty \frac{1}{n^k(q^{2n}-1)} + \sum_{n=1}^\infty \frac{1}{n^k(q^{4n}-1)}$

ok..
mmgla kener tapis maklumat dia
spya equivalent ngan otak bdak form 5

tp............

entahlah satu perasaan yg skar ditafsirkan.

n sy da tanyer ckgu samat (gna sms tuh)
dia kater
cuba jerk....
try gna pc nak wat solution addmaths tuh....

adoiyai...

Sarah.

Sunday, April 19, 2009

....folio addmaths...

Numerical value

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