With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!
Showing posts with label pi. Show all posts
Showing posts with label pi. Show all posts
Saturday, 1 June 2013
100-digit Pi
One fine day when we reach above 80 years old, if the doctor accuses us of having dementia, then prove the doctor wrong by shocking him with 100-digit Pi memory :)
With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!
With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!
Saturday, 20 April 2013
Transcendental Number
Transcendental numbers: e, Π, L...
What about $Latex e^{e}, \pi^{\pi} ,\pi^{e} $ ?
Aleksander (Alexis) Osipovich Gelfond (1906-68):
Gelfond-Schneider Theorem
$Latex a^ b $ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number
Examples:
$latex \sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $latex 2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}
Is log 2 transcendental ?
[log = logarithm Base 10]
Proof:
$latex 10^{log 2} = 2$
1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$latex 10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$latex 2^{q} = 10^{p} = (2.5)^{p}$
$latex 2^{q} = 2^{p}.5^{p}$
Case 1: p>q
$latex 1 = 2^{p-q}.5^{p}$
=> False
Case 2: q>p
$latex 2^{q-p}= 5^{p}$
Left is even : $latex 2^{m} \text { = even} $
Right is odd: $latex 5^{n} \text {= ....5} $
=> False
Therefore p,q do not exist,
=> log 2 irrational
Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html
What about $Latex e^{e}, \pi^{\pi} ,\pi^{e} $ ?
Aleksander (Alexis) Osipovich Gelfond (1906-68):
Gelfond-Schneider Theorem
$Latex a^ b $ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number
Examples:
$latex \sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $latex 2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}
Is log 2 transcendental ?
[log = logarithm Base 10]
Proof:
$latex 10^{log 2} = 2$
1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$latex 10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$latex 2^{q} = 10^{p} = (2.5)^{p}$
$latex 2^{q} = 2^{p}.5^{p}$
Case 1: p>q
$latex 1 = 2^{p-q}.5^{p}$
=> False
Case 2: q>p
$latex 2^{q-p}= 5^{p}$
Left is even : $latex 2^{m} \text { = even} $
Right is odd: $latex 5^{n} \text {= ....5} $
=> False
Therefore p,q do not exist,
=> log 2 irrational
Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html
Tuesday, 16 April 2013
π in 1King 7:23
King Solomon - The Temple's Furnishing:
"He made the Sea of cast metal, circular in shape,
measuring 10 cubits from rim to rim and 5 cubits
high. It took a line of 30 cubits to measure around
it." - 1King 7: 23
[10 cubits = 15 feet, 30 cubits = 45 feet]
So mathematically by today's primary school geometry:
Circumference = 30 = 2 *π *r
Diameter = 10 = 2 * r
π = 30/10 = 3
Note: Solomon's period is equivalent to China's Zhou 周 dynasty of the famous Prince "Zhou Gong" 周公
"He made the Sea of cast metal, circular in shape,
measuring 10 cubits from rim to rim and 5 cubits
high. It took a line of 30 cubits to measure around
it." - 1King 7: 23
[10 cubits = 15 feet, 30 cubits = 45 feet]
So mathematically by today's primary school geometry:
Circumference = 30 = 2 *π *r
Diameter = 10 = 2 * r
π = 30/10 = 3
Note: Solomon's period is equivalent to China's Zhou 周 dynasty of the famous Prince "Zhou Gong" 周公
Monday, 1 April 2013
Google Math
Google = Googol means '1 followed by 100 zeroes'
Google raised Capital at IPO = $2.718281828 Billion = e Billion
(e= 2.718281828...)
Google’s 3 building names: e, pi, ø (Golden ratio)
Google raised Capital at IPO = $2.718281828 Billion = e Billion
(e= 2.718281828...)
Google’s 3 building names: e, pi, ø (Golden ratio)
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