rearrangement &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/rearrangement/ Feed of posts on WordPress.com tagged "rearrangement" Fri, 04 Dec 2009 15:09:32 +0000 http://en.wordpress.com/tags/ en <![CDATA[The Enterprise Unit]]> http://asifjmir.wordpress.com/2009/09/13/the-enterprise-unit/ Sun, 13 Sep 2009 00:11:32 +0000 Asif Mir http://asifjmir.wordpress.com/2009/09/13/the-enterprise-unit/

Business rearrangements are moving toward the creation of a more horizontally oriented company, one that works faster across its structure than up and down. This form is the next stage in the evolution of the “strategic business unit” concept. This may be called the enterprise unit.

The enterprise unit performs only the activities most vital to its competitiveness, primarily those representing critical and cutting-edge capabilities. Other needed capabilities are purchased in the marketplace or shared with other enterprise units. The enterprise unit relies more on reinforced jobs and composite teams to get things done.

My Consultancy–Asif J. Mir - Management Consultant–transforms organizations where people have the freedom to be creative, a place that brings out the best in everybody–an open, fair place where people have a sense that what they do matters. For details please visit www.asifjmir.com, www.youtube.com/asifjmir, Line of Sight

]]> <![CDATA[Rearranged Room]]> http://44aubleh.wordpress.com/2009/07/04/rearranged-room/ Sat, 04 Jul 2009 02:14:11 +0000 aubleh http://44aubleh.wordpress.com/2009/07/04/rearranged-room/

It was a lot of work, but it looks less 12 year old. More 14 year old now, too bad I’m turning 16 soon. More roomy.

Beds take up a lot of space, It takes up so much less if you push it into a corner, rather than it needing an opening on three sides for walking space.  The weird thing is that my walking space is sorta diaganol, from one corner of the room to the other, and the pathways edges are at 90° angled corners. Turned out pretty weird. Still more practical than before though.

]]> <![CDATA[Dua soal dari IMO 1975]]> http://artofmathematics.wordpress.com/2008/08/17/dua-soal-dari-imo-1975/ Sun, 17 Aug 2008 06:17:53 +0000 Johan http://artofmathematics.wordpress.com/2008/08/17/dua-soal-dari-imo-1975/

Saya akan membahas dua soal pertama hari pertama IMO 1975 di Bulgaria:

1. Misalkan x_1\ge x_2\ge\ldots\ge x_n dan y_1\ge y_2\ge\ldots\ge y_n. Buktikan bahwa

\displaystyle\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2,

di mana z_1,z_2,\ldots,z_n adalah permutasi dari y_1,y_2,\ldots,y_n.

2. Misalkan a_1,a_2,a_3,\ldots adalah barisan tak terbatas bilangan asli yang monoton naik. Buktikan bahwa ada tak terhingga banyaknya m sehingga bisa ditulis a_m=x\cdot a_p+y\cdot a_q dengan x,y bilangan asli dan p\ne q.

Soal yang pertama sangat sederhana. Jika kita uraikan, kita dapat bahwa ketaksamaan itu ekuivalen dengan \sum x_i^2-2\sum x_iy_i+\sum y_i^2\le\sum x_i^2-2\sum x_iz_i+\sum z_i^2 atau \sum x_iz_i\le\sum x_iy_i. Ketaksamaan terakhir ini jelas benar dengan rearrangement.

Sekarang kita lihat soal kedua. Untuk setiap 0\le r\le a_p, nyatakan B_r sebagai subbarisan dari a_1,a_2,\ldots yang kongruen r modulo a_p. Karena ada tak terhingga banyaknya bilangan barisan a_1,a_2,\ldots, pasti ada satu bilangan r sehingga barisan B_r memiliki tak terhingga banyaknya anggota. Misalkan a_p,a_q adalah dua suku terkecil dari B_r. Jadi kita bisa ambil sembarang a_m dari B_r sehingga a_m=xa_p+ya_q, di mana x=\frac{a_m-a_q}{a_q},y=1. Jadi terbukti.

]]> <![CDATA[Ketaksamaan]]> http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-5/ Fri, 20 Jun 2008 13:35:13 +0000 Johan http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-5/

[Problem-Solving Strategies] Jika a,b,c>0, buktikan bahwa abc(a+b+c)\le a^3b+b^3c+c^3a.

Solusi
Karena a,b,c dan ab,ca,ba memiliki urutan yang sama, maka dengan rearrangement

a^2(bc)+b^2(ca)+c^2(ab)\le a^2(ab)+b^2(bc)+c^2(ca),

abc(a+b+c)\le a^3b+b^3c+c^3a.

]]> <![CDATA[Ketaksamaan barisan]]> http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-barisan/ Fri, 20 Jun 2008 13:29:58 +0000 Johan http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-barisan/

[IMO 1978] Misalkan \{a_1,a_2,\ldots,a_n\} adalah barisan bilangan asli yang berbeda. Buktikan bahwa

\displaystyle\sum_{k=1}^n\frac{a_k}{k^2}\ge\sum_{k=1}^n\frac1n.

Solusi
Ruas kanan dapat ditulis menjadi \sum\frac{n}{n^2}. Ruas kiri minimum jika a_1,a_2,\ldots,a_n memiliki nilai 1,2,\ldots,n, dalam suatu urutan. Perhatikan barisan (1,2,\ldots,n), (\frac1{1^2},\frac1{2^2},\ldots,\frac1{n^2}). Barisan ini urutannya terbalik, yang satu naik, yang satu turun. Ruas kanan pada soal adalah perkalian suku-suku dua barisan dengan urutan terbalik, sedangkan ruas kiri belum tentu pada urutan terbalik. Jadi, menurut aturan rearrangement, terbukti.

]]> <![CDATA[Ketaksamaan bilangan dan permutasinya]]> http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-bilangan-dan-permutasinya/ Fri, 20 Jun 2008 13:15:03 +0000 Johan http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-bilangan-dan-permutasinya/

[IMO 1975] Misalkan x_i,y_i adalah bilangan real sehingga

x_1\ge x_2\ge\ldots\ge x_n dan y_1\ge y_2\ge\ldots\ge y_n.

Misalkan z_1,z_2,\ldots,z_n adalah permutasi dari y_1,y_2,\ldots,y_n. Buktikan bahwa

\displaystyle\sum_{i=1}^n(x_i-y_i)^2\le\sum_{i=1}^n(x_i-z_i)^2.

Solusi
Ketaksamaan yang diminta ekuivalen dengan

\displaystyle\sum_{i=1}^n(x_i^2-2x_iy_i+y_i^2)\le\sum_{i=1}^n(x_i^2-2x_iz_i+z_i^2).

Karena \sum y_i^2=\sum z_i^2, maka ketaksamaan tadi ekuivalen dengan

\displaystyle\sum_{i=1}^n(-x_iy_i)\le\sum_{i=1}^n(-x_iz_i),

atau

\displaystyle\sum_{i=1}^n(x_iz_i)\le\sum_{i=1}^n(x_iy_i),

yang terbukti dengan rearrangement.

]]> <![CDATA[Ketaksamaan]]> http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-4/ Fri, 20 Jun 2008 13:07:55 +0000 Johan http://artofmathematics.wordpress.com/2008/06/20/ketaksamaan-4/

[Problem-Solving Strategies] Jika a,b,c>0, buktikan

\displaystyle\frac{a+b+c}{abc}\le\frac1{a^2}+\frac1{b^2}+\frac1{c^2}.

Solusi
WLOG, asumsikan a\ge b\ge c. Dengan rearrangement,

\displaystyle\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge\frac1{ab}+\frac1{bc}+\frac1{ca}

\displaystyle\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge\frac{a+b+c}{abc}.

]]> <![CDATA[New homes ambience for my new mood]]> http://guysarchipelago.wordpress.com/2008/03/17/new-house-ambience-for-my-new-mood/ Mon, 17 Mar 2008 04:12:02 +0000 guysarchipelago http://guysarchipelago.wordpress.com/2008/03/17/new-house-ambience-for-my-new-mood/

Last weekend was my housekeeping days…. thnks to Ardly for my house new arrangement..so we spend whole saturday afternoon searching fresh flowers for my house @ Petaling Street..wow never thought that florist have such a beautiful and cheap fresh flower……we choosed Eucalyptus, Sundal Malam and few ranting’s (forgot arr ranting in english) for our house decoration..heheh….and we have our new house member..very cute and naughty named Lulu Tyra…hehehe…..a female persian type kitten indeed not anak ikan yer..heheheh….why that name?….she got a beautiful fur color likes Tyra Banks from one season of ANTM and Lulu?..ardly gave it..i dunno where he did got an idea for that..just cross over his mind or remind he of Mumu?..whose Mumu?..no comment…heheheheh…..

yesterday we r invited to newly wed Lili’s house warming party…hehehe..never thought we met few YB there from PKR…heheh…yeah Lili’s husband is a keadilan member..hehehe…hidup keadilan!!!…wow wat a very nice party….makan2 minum2..so relaxing….

nie dia rupa Eucalyptus (ada nice smells yer)

ni bunga harum sundal malam (dedicate to pijot as he name it bunga harum sundal siang malam…errrkk…tu u lah pijot)

 

 

 

]]> <![CDATA[Pertidaksamaan tiga bilangan berjumlah 1]]> http://artofmathematics.wordpress.com/2008/01/03/pertidaksamaan-tiga-bilangan-berjumlah-1/ Thu, 03 Jan 2008 14:06:55 +0000 Johan http://artofmathematics.wordpress.com/2008/01/03/pertidaksamaan-tiga-bilangan-berjumlah-1/

[UKMO 1999] p, q dan r adalah bilangan real non negatif yang jumlahnya 1. Buktikan bahwa

7(pq+qr+rp)\le2+9pqr.

Solusi
Pertidaksamaan di atas dapat diubah menjadi

7(pq+qr+rp)(p+q+r)\le2(p+q+r)^3+9pqr.

Kemudian disederhanakan menjadi

p^2q+p^2r+q^2r+q^2p+r^2p+r^2q\le2(p^3+q^3+r^3).

Tanpa mengurangi keumuman, asumsikan p\le q\le r. Dengan aturan menyusun kembali (rearrangement), kita dapat

p^2\cdot p+q^2\cdot q^2\ge p^2q+q^2p, atau p^3+q^3\ge p^2q+q^2p.

p^2\cdot p+r^2\cdot r^2\ge p^2r+r^2p, atau p^3+r^3\ge p^2r+r^2p

q^2\cdot q+r^2\cdot r^2\ge q^2r+r^2q, atau q^3+r^3\ge q^2r+r^2q

Jumlahkan ketiganya, maka

p^2q+p^2r+q^2r+q^2p+r^2p+r^2q\le2(p^3+q^3+r^3).

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