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sihaozi61

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wurongjun

ר¼Ò¹ËÎÊ (Ö°Òµ×÷¼Ò)

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ÒýÓûØÌû:
3Â¥: Originally posted by wurongjun at 2013-09-29 08:04:57
0<xn
=(1-1/2)(1-1/4)...(1-1/2^n)
=1/2*3/4*¡­¡­*£¨2^n-1)/2^n
<2/3*4/5*¡­¡­*2^n/(2^n+1),¼ÇΪyn.
xn^2<xn*yn=1/(2^n+1)¡ú0£¨n¡ú¡Þ£©,
¡àxn¡ú0£¨n¡ú¡Þ£©.

ûÓÐ×Ðϸ¿´!¸ã´íÁË!
Ö±½ÓÉÏͼ!
È¡x=1/2¼´¿É!
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x1c2.JPG

ÉÆ¶ñµ½Í·ÖÕÓб¨,È˼äÕýµÀÊDz×É£.
12Â¥2013-09-29 20:51:22
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

wurongjun

ר¼Ò¹ËÎÊ (Ö°Òµ×÷¼Ò)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

¸Ðл²ÎÓ룬ӦÖúÖ¸Êý +1
0<xn
=(1-1/2)(1-1/4)...(1-1/2^n)
=1/2*3/4*¡­¡­*£¨2^n-1)/2^n
<2/3*4/5*¡­¡­*2^n/(2^n+1),¼ÇΪyn.
xn^2<xn*yn=1/(2^n+1)¡ú0£¨n¡ú¡Þ£©,
¡àxn¡ú0£¨n¡ú¡Þ£©.
ÉÆ¶ñµ½Í·ÖÕÓб¨,È˼äÕýµÀÊDz×É£.
3Â¥2013-09-29 08:04:57
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

ÒýÓûØÌû:
3Â¥: Originally posted by wurongjun at 2013-09-29 08:04:57
0<xn
=(1-1/2)(1-1/4)...(1-1/2^n)
=1/2*3/4*¡­¡­*£¨2^n-1)/2^n
<2/3*4/5*¡­¡­*2^n/(2^n+1),¼ÇΪyn.
xn^2<xn*yn=1/(2^n+1)¡ú0£¨n¡ú¡Þ£©,
¡àxn¡ú0£¨n¡ú¡Þ£©.

ÎÊÌâ³öÔÚ
xn*yn=1/(2^n+1)¡ú0  Ô¼¼òÉÏ¡£
4Â¥µÄ´ð¸´Âí»¢ÁË¡£
7Â¥2013-09-29 15:27:53
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

hank612

ÖÁ×ðľ³æ (ÖøÃûдÊÖ)

ÒýÓûØÌû:
11Â¥: Originally posted by sihaozi61 at 2013-09-29 18:30:13
лл  Í¬Ñùллhank612~\(¨R¨Œ¨Q)/~
Ï£Íû¸ü¶àµÄÈËÄܶÔÕâµÀÌâÄ¿Ìá³öÒâ¼û
...

Â¥Ö÷£¬ ²»ºÃÒâ˼£¬ ÎÒÊÇÀ´´ò»÷ÄãÏë»ñµÃ¾«È·½âµÃÐÅÐĵġ£

°ÑÄãµÄÁ¬³ËÈ¡Log, µÃµ½ (ËùÓÐÇóºÍ£¬¶Ôk, ¶Ôn, ¶¼ÊÇ´Ó1µ½ÎÞÇî)
\Sum_{n}  Log(1-2^{-n})
=- \Sum_{k} \Sum_{n} 1/(k*2^{nk})
=- \Sum_{k} 1/(k*(2^k-1))

È»ºóÎÒÈ¥MathematicaÍøÕ¾ÊÔÁËһϣ¬http://www.wolframalpha.com/inpu ... 2%88%9E%29+1%2F%28k*%282%5Ek+-+1%29%29+

È»ºó¾ÍûÓÐÈ»ºóÁË£¬ÒòΪ´ð°¸¾ÍÊÇÂ¥Ö÷µÄ±í´ïʽ£¬ËüÊÇq-Pochhammer symbol ÀïÃæµÄ¡¡(1/2: 1/2)_{\infty}. ËùÒÔÄãÒªÇóµÄ¾ÍÊÇ¡¡(1/2: 1/2)_{\infty}£»¡¡Î޽⡣

http://en.wikipedia.org/wiki/Q-Pochhammer_symbol
We_must_know. We_will_know.
13Â¥2013-09-30 02:46:36
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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mengbiao

гæ (³õÈëÎÄ̳)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

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Ç°Ãæ³Ë¶þ·ÖÖ®Ò»ÔÙ³ý¶þ·ÖÖ®Ò»

[ ·¢×ÔÊÖ»ú°æ http://muchong.com/3g ]
2Â¥2013-09-29 07:17:46
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

ÒýÓûØÌû:
3Â¥: Originally posted by wurongjun at 2013-09-29 08:04:57
0<xn
=(1-1/2)(1-1/4)...(1-1/2^n)
=1/2*3/4*¡­¡­*£¨2^n-1)/2^n
<2/3*4/5*¡­¡­*2^n/(2^n+1),¼ÇΪyn.
xn^2<xn*yn=1/(2^n+1)¡ú0£¨n¡ú¡Þ£©,
¡àxn¡ú0£¨n¡ú¡Þ£©.

ÕâµÀ°ÂÊýÌâ×öµÃºÃ¡£
4Â¥2013-09-29 08:27:15
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

sihaozi61

½û³æ (³õÈëÎÄ̳)

±¾ÌûÄÚÈݱ»ÆÁ±Î

5Â¥2013-09-29 13:04:09
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

hank612

ÖÁ×ðľ³æ (ÖøÃûдÊÖ)

ÒýÓûØÌû:
5Â¥: Originally posted by sihaozi61 at 2013-09-29 13:04:09
ллÌṩ°ïÖú£¡
...

N=5000; a=zeros([1,N]);
a(1)=1/2;
for i=2:N
    a(i)=a(i-1)*(1-1/2^i);
end
b=a(N)

Õâ¶ÎMatlab³ÌÐòÅÜÍêºó£¬ b=0.2888. ËùÒÔ¼«ÏÞÒ»¶¨ÊÇ´æÔڵģ¬ ¾ÍÊDz»ÖªµÀ¾«È·½âÊÇʲô¡£
We_must_know. We_will_know.
6Â¥2013-09-29 13:57:04
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

ÒýÓûØÌû:
6Â¥: Originally posted by hank612 at 2013-09-29 13:57:04
N=5000; a=zeros();
a(1)=1/2;
for i=2:N
    a(i)=a(i-1)*(1-1/2^i);
end
b=a(N)

Õâ¶ÎMatlab³ÌÐòÅÜÍêºó£¬ b=0.2888. ËùÒÔ¼«ÏÞÒ»¶¨ÊÇ´æÔڵģ¬ ¾ÍÊDz»ÖªµÀ¾«È·½âÊÇʲô¡£...

Ñ­»·n¶à´Î£¬½á¹û±£³ÖÔÚ0.2887881ÉÏ.
8Â¥2013-09-29 15:30:22
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

sihaozi61

½û³æ (³õÈëÎÄ̳)

±¾ÌûÄÚÈݱ»ÆÁ±Î

9Â¥2013-09-29 17:32:53
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

ÒýÓûØÌû:
9Â¥: Originally posted by sihaozi61 at 2013-09-29 17:32:53
¸Õµ½¼Ò  È·Êµ¸Ðµ½ÓÐЩÒÉ»ó
ÕâµÀÌâÄ¿»¹ÊÇûÓÐÕæÕý½â¾ö
...

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ÁíÍâ£¬ÌæÄãΪhank612ËÍ»¨Ò»Êø¡£
10Â¥2013-09-29 17:52:18
ÒÑÔÄ   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
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