User:Dylanwhs
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About
Dylan likes to read science fiction and fantasy novels. He also speaks Hakka, English and some Cantonese. He has been known to have written a couple of websites (1 (http://www.sungwh.freeserve.co.uk), 2 (http://www.dylanwhs.ukgateway.net)), and is omnivorous.
Since beginning to edit and create articles on Monday 16th February 2004, I find that today Saturday 21st February 2004, almost 2500 new articles have been written. Wow! Even I've lost count of what I've added to, apart from the following articles, if it wasn't for "user contributions" on the left here.
He has added to or edited parts of
Hakka Hakka (linguistics) Oracle bones Empress Wu Zetian of China Shang Dynasty Guangdong Taiping Rebellion Cantonese_(linguistics) Tonal language Kangxi Dictionary Demographics of Hong Kong Fermented egg Tam Dalyell She GB Simplified Chinese character Shanghai Shaoxing Chinese spoken language
New Pages I've begun
Shuowen Jiezi Old Chinese Seal Script Qieyun Guangyun Rime tables Kangxi Dictionary Sha Tau Kok Sexagesimal cycle Yin Oracle bone Kangxi Emperor of China Fanqie Hakka Hill Songs Tone Name Tone Contour Anyang Pan Geng Jiyun Rime book Maya numerals Haab Tzolkin Calendar Round Meixian Mayan Long Count Calendar Li Si Xiaozhuan Xu Shen Duan Yucai Historical Chinese Phonology Yue You tiao Rice congee Siyi Bronzeware Script Wu dialect Shanghainese GB 18030 Suzhouhua Kuk Po Fung Hang Nam Chung San Tsuen Luk Keng Chan Uk Buwei Yang Chao
Personal Page
<math>c = \pi r^2\,\!<math>
<math>e^{i\theta} = \cos\theta + i \sin\theta\,\!<math>
The numbers which are subtracted from the Root form a series <math>
S_n = 1 + 7 + 19 + 37 + ..... <math>
If one looks at the sum, the number of terms relates to the cube root of the sum itself.
1:~ 1 2:~ 1 + 7 3:~ 1 + 7 + 19 4:~ 1 + 7 + 19 + 37 5:~ 1 + 7 + 19 + 37 + 61 . . n:~ 1 + 7 + 19 + 37 + 61 + ....... +
Notice also that these terms can be written as
n:~ 1 + (1+6) + (1+18) + (1+36) + (1+60) + .... +
= 1 + (1 + 6) + (1 +6 + 12) + (1+6+12+18) + (1+6+12+18+24) + .... = 1 + (1 + A) + (1 + B) + (1 + C) + (1 + D) +
Where
A = 6 B = 6 + 12 C = 6 + 12 + 18 D = 6 + 12 + 18 + 24
1 occurs n times 6 = 6.1 occurs n-1 times
12 = 6.2 occurs n-2 times 18 = 6.3 occurs n-3 times 24 = 6.4 occurs n-4 times
The term which occurs 1 times only will be 6.(n-1)
Thus the sum becomes
<math> S_n = 1 + n + 6[ 1.(n-1) + 2.(n-2) + 3(n-3) + ..... + (n-1)(n - (n-1))] <math>
<math> = \sum{r=1}^n 1 + 6[ 1n-1^2 + 2n - 2^2 + 3n - 3^2 + ...... + n(n-1) - (n-1)^2 ] <math>
<math> = n + 6[ n( 1 + 2 + 3 + .... + n-1 ) ] - 6[ 1^2 + 2^2 + 3^3 + .... + (n-1)^2] <math> <math>\sun_n = n + 6n\sum{r=1}^n-1 r - 6\sum{r=1}^n-1 r^2<math>
<math>\sun_n = n + 6n(\sum{r=1}^n r - n ) - 6(\sum{r=1}^n r^2 - n^2)<math>
<math>\sun_n = n + 6\sum{r=1}^n r - 6n^2 - 6\sum{r=1}^{n} r^2 + 6n^2<math>
<math>\sun_n = n + 6\sum{r=1}^n r - 6\sum{r=1}^n r^2<math>
<math>\sun_n = n + 6 \frac{1}{2} n(n+1) - 6 \frac{1}{6}n(n+1)(2n+1)<math>
<math>\sun_n = n + 3n^2(n+1) - n(n+1)(2n+1)<math>
<math>\sun_n = n + 3n^3 + 3n^2 - [2n^3 + 3n^2 + n]<math>
<math>\sun_n = (3-2)n^3 + (3-3)n^2 + (1-1)n = n^3 + 0 + 0<math>
<math>\sun_n = n^3<math>
Therefore:
<math>S_n = n^3<math>
</math>
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This user supports the return of the Old Wikipedia, before it obtained this new "blog" look, and would like it changed back. |