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#1
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ᨡ੾ÒСԨ (à¿Ê 3) : Undergraduate and olympiad
¡ÅѺÁÒàÃçÇ¡ÇèÒà´ÔÁ¤ÃѺ ÊÓËÃѺᨡ੾ÒСԨ (à¿Ê 3) ÃѺÇѹÅÍ¡Ãз§¾Í´Õ
à¿Ê¹Õé¨Ð ÁÕ 2 parts ¤ÃѺ â´Â part áá ¨Ðà¹é¹¤ÇÒÁÃÙé·Ò§á¤Å¤ÙÅÑÊ «Ö觤è͹æä»·Ò§ÁËÒÇÔ·ÂÒÅѹԴæ Êèǹ part ·Õè 2 ¨Ðà»ç¹ÊäµÅìâÍÅÔÁ»Ô¡ áµèäÁèâË´¤ÃѺ à¾ÃÒÐ ªèǧ 2 à¿Ê¡è͹˹éÒ ÁÕ¤Ó¶ÒÁẺ¹ÕéàÂÍоÍáÅéÇ ¢Í§·Õè¨Ðᨡ ¡ç´Ù¨Ò¡ËÑǢͧ PART ´éÒ¹ÅèÒ§ä´éàŤÃѺ â´Â PART 1 à»ç¹àÅèÁ·Õè¼ÁªÍºÁÒ¡ áÅéÇ¡ç¡ÇèҨФÇÒ¹ËÒ˹ѧÊ×Íã¹µÓ¹Ò¹àÅèÁ¹Õéä´é¡ç¹Ò¹¾ÍÊÁ¤Çà Êèǹ ã¹ PART 2 à»ç¹¢éÍÊͺÊäµÅìâÍÅÔÁ»Ô¡ ¨Ò¡¡ÒÃá¢è§¢Ñ¹ã¹ÍÍÊàµÃàÅÕ¤ÃѺ ¡µÔ¡Ò¤ÅéÒÂæà´ÔÁ¤ÃѺ ¤×Í (1) ¤¹·Õè¨ÐàÅè¹ µéͧÂѧäÁèÁÕÇØ²Ô â·ËÃ×Í àÍ¡·Ò§ maths áÅÐäÁèãªè moderator ·Õè¹Õè (2) ¤¹·Õèä´é¤Ðá¹¹ÊÙ§ÊØ´ã¹áµèÅÐ part ¨Ðä´é¢Í§·Õè¼ÁÇèÒä» â´Âá¨é§·ÕèÍÂÙèãËé¼Á·Ò§ pm à¾×èͨÐä´éÊ觢ͧãËé·Ò§ä»ÃɳÕÂì (3) ¶éÒÊÁÁµÔã¹¢éÍã´¢éÍ˹Öè§ÁÕ¤¹µÍº¶Ù¡ä»áÅéÇ ¤¹·ÕèµÍº¶Ù¡·ÕËÅѧ㹢é͹Ñé¹ ¤ÇèÐáÊ´§ÇÔ¸Õ·Ó·ÕèµèÒ§¨Ò¡¤¹ááÍÂèÒ§àËç¹ä´éªÑ´ ¨Ö§¨Ðä´é¤Ðá¹¹ (4) »Ô´ÃѺ·Ø¡¤ÓµÍºÇѹÍÒ·ÔµÂì 10 âÁ§àªéÒ àÇÅÒ»ÃÐà·Èä·Â ¤ÃѺ (àÅ×è͹à»ç¹ 5 ·ØèÁ ÈØ¡Ãì·Õè 30 ¾.Â.) áµèã¹ PART 1 ¼Á¢Íà¾ÔèÁ¡µÔ¡ÒÍÕ¡¹Ô´¤×Í moderator ªèÇ HINT ä´é¤ÃѺ ÊÓËÃѺ part ¹Õé PART 1 : "PROBLEM SEMINAR" (Donald J. Newman) (photocopied version) 1. (1 ¤Ðá¹¹) ÊÁÁµÔ $ f, g, f’, g’ $ à»ç¹¿Ñ§¡ìªÑ¹µèÍà¹×èͧº¹ [0,1] áÅÐ $ g(x) \neq 0 \,\, \forall x \in [0,1] $ ¶éÒ $ f(0) = 0 \,\, , g(0)= e \,\, , f(1)=1004 \,\, , g(1)= 1 $ ËÒ¤èÒ $$ \int_0^1 \frac{f(x)g’(x)((g(x))^2-1)+f’(x)g(x)((g(x))^2+1)}{(g(x))^2} \,\, dx $$ 2. (2 ¤Ðá¹¹) Solve initial-valued problem $$ xy' =y \ln(xy) \,\, , y(1)=1 $$ 3. (3 ¤Ðá¹¹) àÅ×Í¡·Ó 1 ¢éͨҡ 3 ¢é͵èÍ仹Õé (3.1) ͹ءÃÁ $$ \sum_{k=2}^{\infty} (-1)^k \ln(1-\frac{1}{k^2}) $$ ÅÙèà¢éÒËÃ×ÍÅÙèÍÍ¡ ͸ԺÒ¾ÍÊѧࢻ áÅжéÒÅÙèà¢éÒ ãËéËÒÇèÒÅÙèà¢éÒÊÙè¤èÒã´ (3.2) partition $ 1,2,3,\cdots $ ÍÍ¡à»ç¹ 2 Êèǹ ¤×ÍÅӴѺ $ a_n$ áÅÐ $ b _n $ â´Â $a_n$ à»ç¹ strictly increasing sequence ·ÕèÁÕ 7 ÃÇÁÍÂÙè´éÇ (¡ÅèÒǤ×Í $ a_1 = 7 \,\, , a_2=17 \,\, , a_3=27 \cdots a_7=70 \,\, , a_8=71 \cdots $ Êèǹ $b_n$ à»ç¹ strictly increasing sequence ¢Í§ àÅ¢·ÕèàËÅ×Í ¡ÅèÒǤ×Í $ b_1 = 1 \,\, , b_2=2 \,\, , b_3=3 \cdots b_7=8 \,\, , b_8=9 \cdots $ ) ãËéà˵ؼÅÇèÒ $ \sum_{n=1}^{\infty} \frac{1}{a_n} $ áÅÐ $ \sum_{n=1}^{\infty} \frac{1}{b_n} $ ÅÙèà¢éÒËÃ×ÍÅÙèÍÍ¡ (3.3) ¡Ó˹´ $ A= 1+1+ \frac{1}{2!}+ \frac{1}{3!}+\cdots $ ¾ÔÊÙ¨¹ìÇèÒ A à»ç¹¨Ó¹Ç¹ÍµÃáÂÐ (â´ÂËéÒÁÍéÒ§ÇèÒ A= e ) 4. (5 ¤Ðá¹¹) ¢éͤÇÒÁµèÒ§æ㹤ӶÒÁ¹Õé µÑ´µÍ¹ÁҨҡ˹ѧÊ×Í letters to a young mathematician ¢Í§ Ian Stewart â´Â 3 ¢éÍÂèÍÂáá ãËéËÒÇèÒ ª×èͤ¹ ËÃ×ͪ×èÍ·ÄɮշÕèËÒÂ令×ÍÍÐäà Êèǹ 2 ¢éÍÂèÍÂÊØ´·éÒ ãËéµÍºÇèÒ ¤Ó·Õè¾ÔÁ¾ìµÑÇË¹Ò Ê×èͶ֧ã¤ÃËÃ×ÍÊÔè§ã´ (4.1) ...ecological models arose in the 1990s , when the Italian mathematician _______________ was trying to understand a serious effect that had been observed by Adriatic fishermen. (4.2) The______________ theorem tells us that if two transfinite numbers are less than or equal to each other , then they must actually be equal. (4.3) In 1998,_____________ announced a computer-assisted proof of the Kepler conjecture that involved hundreds of pages of mathematics….. (4.4) In applied mathematics , he was a genius. But in pure mathematics , he was a god. (4.5) We’re adders, and we can only multiply using logs. 5. (9 ¤Ðá¹¹ ) (5.1) (2 ¤Ðá¹¹ ) ͹ءÃÁ $$ \sum_{n=1}^{\infty} \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}\bigg ( \frac{1}{2^n} \bigg) $$ ÅÙèà¢éÒËÃ×ÍÅÙèÍÍ¡ ͸ԺÒ¾ÍÊѧࢻ áÅжéÒÅÙèà¢éÒ ãËéËÒÇèÒÅÙèà¢éÒÊÙè¤èÒã´ (5.2) (3 ¤Ðá¹¹) ¾ÔÊÙ¨¹ìÇèÒ $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \bigg(1+\frac{1}{2}+\cdots \frac{1}{n} \bigg) $$ ÅÙèà¢éÒáÅÐËÒ´éÇÂÇèÒÅÙèà¢éÒÊÙè¤èÒã´ (5.3) ( 4 ¤Ðá¹¹ ) Evaluate $$ \int_0^ {\infty} \bigg( \ln \bigg( \frac{x^2}{x^2+3x+2} \bigg) \bigg) ^2 \,\, dx $$ (HINT : ºÒ§¢Ñ鹵͹ ãªé¡ÒôѴá»Å§¨Ò¡ (5.2)) ---------------------------------------------------------------------------------- PART 2: Selected years of Australian Maths tournament (photocopied version) 1. $ P $ à»ç¹¨Ø´Ë¹Ö觺¹Ç§ÃÕ áÅÐ $ F_1 , F_2 $ à»ç¹¨Ø´â¿¡ÑʢͧǧÃÕ áÅÐ L à»ç¹àÊé¹ÊÑÁ¼ÑʢͧǧÃÕ·Õè $P$ ÊÁÁµÔ $ X$ à»ç¹¨Ø´ã´æº¹ L ¾ÔÊÙ¨¹ìÇèÒ ÊÒÁàËÅÕèÂÁ $ F_1XF_2$ ÁÕàÊé¹ÃͺÃÙ»¹éÍÂÊØ´ àÁ×èÍ X ·Ñº¡Ñº P ¾Í´Õ ( 1 ¤Ðá¹¹) 2.ÊÓËÃѺ ¨Ó¹Ç¹¹Ñº $ n \geq 2 $ ¾ÔÊÙ¨¹ìÇèÒ ªèǧ $ (2^n+1 ,2^{n+1}-1)$ ºÃèبӹǹ àµçÁ·ÕèÊÒÁÒöà¢Õ¹ä´éã¹ÃÙ»¼ÅºÇ¡¢Í§¨Ó¹Ç¹à©¾ÒÐ n µÑÇ (·Õè«éӡѹä´é) (2 ¤Ðá¹¹) 3.Á´µÑÇ˹Öè§äµè仵ÒÁ ¢Íº¢Í§ÅÙ¡ºÒÈ¡ìáÅШÐàÅÕéÂÇ·ÕèÁØÁ¢Í§ÅÙ¡ºÒÈ¡ìà·èÒ¹Ñé¹ ¶éÒÁ´µÑǹÕé àÅÕéÂÇ·ÕèÁØÁæ˹Öè§ 25 ¤ÃÑé§ à»ç¹ä»ä´éËÃ×ÍäÁèÇèÒ Á´¨ÐàÅÕéÂÇ·ÕèÁØÁÍ×è¹æ¢Í§ÅÙ¡ºÒÈ¡ì ÁØÁÅÐ 20 ¤ÃÑé§¾Í´Õ ( 3 ¤Ðá¹¹) 4.(ÊÒÁÒöãªé·ÄÉ®Õ੾ÒÐÁÒÍéÒ§ä´é) ¾ÔÊÙ¨¹ìÇèÒ ÁÕ¤ÙèÍѹ´Ñº (a, b) ¢Í§¨Ó¹Ç¹¹Ñº à»ç¹¨Ó¹Ç¹Í¹Ñ¹µì ·Õè·ÓãËé (i) $ \frac{a}{b}$ ËÃ×Í $\frac{b}{a} $ à»ç¹¨Ó¹Ç¹¹Ñº (ii) a ¢Ö鹵鹴éÇ 2551 (iii) b ŧ·éÒ´éÇ 001 (iv) a áÅÐ b à¢Õ¹ä´éã¹ÃÙ» $ m^n$ â´Â $m,n$ à»ç¹¨Ó¹Ç¹¹Ñº áÅÐ $ n \geq 2 $ ( 4 ¤Ðá¹¹ )
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à¡ÉÕ³µÑÇàͧ »ÅÒÂÁԶعÒ¹ 2557 áµè¨Ð¡ÅѺÁÒà»ç¹¤ÃÑ駤ÃÒÇ 27 ¾ÄȨԡÒ¹ 2007 22:38 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 4 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ passer-by à˵ؼÅ: edit deadline(again)+ question 4 (part 2) |
#2
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1.(Part1)
$\displaystyle{\int_{0}^{1}\frac{f(x)g'(x)((g(x))^2-1)+f'(x)g(x)((g(x))^2+1)}{(g(x))^2}}dx$ $\displaystyle{=\int_{0}^{1}\left[f(x)g'(x)+f'(x)g(x)\right]dx+\int_{0}^{1}\left[\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\right]dx}$ $\displaystyle{=\left[f(x)g(x)+\frac{f(x)}{g(x)}\right]_{0}^{1}=2008}$ àËç¹â¨·ÂìÅÐà˧×è͵¡àŹФÃѺ ¢é͹ÕéÍÂèÒ§¹ÕéËÃ×Íà»ÅèÒ¤ÃѺ
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$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$ BUT $$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$
25 ¾ÄȨԡÒ¹ 2007 10:59 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Timestopper_STG |
#3
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ÂÒ¡ä»ËÃ×Íà¹Õè àŹÔè§Ê¹Ô·àÅÂ
¤Ø³ Timestopper ¾ÅÒ´µÍ¹¨ºä»¹Ô´à´ÕÂǤÃѺ ÃÕºæÁÒá¡é¹Ð Êèǹ deadline ·Ø¡¤ÓµÍº ¼Á¢ÂÒÂãËé¶Ö§ 10 âÁ§àªéÒ ÇѹàÊÒÃì·Õè 1 ¸Ñ¹ÇÒ¤Á áÅéǡѹ¤ÃѺ áÅéÇ¡ç Hint ¢Í§µÍ¹·Õè 1 ¢ÍãËéäÇéÅÍÂæ Ẻ¹ÕéáÅéǡѹ... ºÒ§¢éÍ ãªé geometric series ÁÒ»Ô´¢Íºà¢µº¹ ºÒ§¢éÍ ÍÒ¨¨Ðµéͧ¾Ô¨ÒÃ³Ò partial sum
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à¡ÉÕ³µÑÇàͧ »ÅÒÂÁԶعÒ¹ 2557 áµè¨Ð¡ÅѺÁÒà»ç¹¤ÃÑ駤ÃÒÇ |
#4
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ÍéÒ§ÍÔ§:
$\therefore a=b=2551...001$ àËç¹ä´éªÑ´ÇèÒàÃÒÊÒÁÒöàµÔÁ$i\in \mathbb{N}$ÃÐËÇèÒ§ $2551$áÅÐ$001$ ä´éà»ç¹¨Ó¹Ç¹Í¹Ñ¹µì ´Ñ§¹Ñé¹ ¤ÙèÍѹ´Ñº (a,b) ÁÕà»ç¹¨Ó¹Ç¹Í¹Ñ¹µì áÅеçµÒÁà§×èÍ¹ä¢ »Å.¾ÔÊÙ¨¹ìẺ¹Õéä´éËÃ×Íà»ÅèÒ¤ÃѺ
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¤ Ç Ò Á ÃÑ º ¼Ô ´ ª Í º $$|I-U|\rightarrow \infty $$ 24 ¾ÄȨԡÒ¹ 2007 19:06 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ kanakon |
#5
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OH! ¼Á¾ÅÒ´¨¹ä´é ·ÕèÁͧ¢éÒÁ ¡Ã³Õ a=b ä»
àÍÒà»ç¹ÇèÒ ¼ÁãËé¤Ðá¹¹¤Ø³ kanakon 2 ¤Ðá¹¹ ¨ÐÇèÒÍÐäüÁÁÑéÂà¹Õè à¾ÃÒе͹áá solution ·ÕèÍÂÙèã¹ã¨Áѹà»ç¹ÍաẺ¹Ö§ àŵÑé§äÇé 4 ¤Ðá¹¹¤ÃѺ
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à¡ÉÕ³µÑÇàͧ »ÅÒÂÁԶعÒ¹ 2557 áµè¨Ð¡ÅѺÁÒà»ç¹¤ÃÑ駤ÃÒÇ |
#6
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ÍéÒ§ÍÔ§:
$$\therefore x\frac{1}{x^2}\left[\,xK'-K \right]=\frac{K}{x}\ln(K)$$ $$K'-\frac{K}{x}=\frac{K}{x}\ln(K)$$ $$K'=\frac{K}{x}(1+\ln(K))$$ $$\frac{dK}{K(1+\ln(K))}=\frac{dx}{x}$$ $$\int\!\frac{dK}{K(1+\ln(K))}=\int\!\frac{dx}{x}$$ $$\ln(\ln(eK))=\ln(x)+C$$ $$y(1)=1\Rightarrow C=0 $$ $$y=\frac{e^{x-1}}{x} $$
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¤ Ç Ò Á ÃÑ º ¼Ô ´ ª Í º $$|I-U|\rightarrow \infty $$ |
#7
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4.4 ¢Íà´ÒÇèÒ Pierre de Fermat
4.5 §Ùº¹àÃ×ͧ͢â¹ÍÒ Source: Math Jokes page
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¤ Ç Ò Á ÃÑ º ¼Ô ´ ª Í º $$|I-U|\rightarrow \infty $$ |
#8
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¢éÍ·Õèà»ç¹ ODE ¡Ñº¢éÍ 4.5 ¶Ù¡áÅéǤÃѺ
à·èҡѺÇèҵ͹¹Õé¤Ø³ kanakon ÁÕ¤Ðṹ㹵͹·Õè 1 ä»áÅéÇ 3 ¤Ðá¹¹ ¾ÃéÍÁ¡Ñº¤Ðá¹¹µÍ¹·Õè 2 ÍÕ¡ 2 ¤Ðá¹¹ Êèǹ¢éÍ 4.4 ÂѧäÁèãªè¤ÃѺ ãºéãËéÍÕ¡¹Ô´ÇèÒ he ã¹·Õè¹Õé à¡ÕèÂÇ¢éͧ¡Ñº telegraph ´éÇÂ
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à¡ÉÕ³µÑÇàͧ »ÅÒÂÁԶعÒ¹ 2557 áµè¨Ð¡ÅѺÁÒà»ç¹¤ÃÑ駤ÃÒÇ |
#9
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4.3 Thomas C. Hales
4.1 Vito Volterra 4.4 ¼ÁÁÕÍÂÙèã¹ã¨ÍÕ¡¤¹áµèáµè¨ÐÅͧµÍº(à»ç¹¤ÃÑé§ÊØ´·éÒÂ)ÇèÒ Carl Friedrich Gauss
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¤ Ç Ò Á ÃÑ º ¼Ô ´ ª Í º $$|I-U|\rightarrow \infty $$ 25 ¾ÄȨԡÒ¹ 2007 08:12 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ kanakon |
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3.1.(Part1)
First, we can easily see that S converges absolutely. $\displaystyle{S=\sum_{k=2}^{\infty}(-1)^{k}\ln\left(1-\frac{1}{k^{2}}\right)=\sum_{k=2}^{\infty}(-1)^{k}\ln\left(\frac{(k-1)(k+1)}{k^{2}}\right)}$ $\displaystyle{S=\sum_{k=1}^{\infty}\ln\left(\frac{(2k-1)(2k+1)}{(2k)^2}\right)-\sum_{k=1}^{\infty}\ln\left(\frac{(2k)(2k+2)}{(2k+1)^2}\right)}$ $\displaystyle{\sum_{k=1}^{\infty}\ln\left(\frac{(2k-1)(2k+1)^3}{(2k+2)(2k)^3}\right)=\ln\prod_{k=1}^{\infty}\left[\frac{(2k-1)(2k+1)^3}{(2k+2)(2k)^3}\right]=\ln P}$ $\displaystyle{P=\lim_{n\rightarrow\infty}\left(\frac{1}{4}\cdot\frac{3}{6}\cdot...\cdot\frac{(2n-1)}{(2n+2)}\right)\left(\frac{3}{2}\cdot\frac{5}{4}\cdot...\cdot\frac{(2n+1)}{(2n)}\right)^3}$ $\displaystyle{P=\lim_{n\rightarrow\infty}2\frac{(2n+1)^3}{(2n+2)}\left(\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{(2n-1)}{(2n)}\right)^4}$ $\displaystyle{P=\lim_{n\rightarrow\infty}2\frac{(2n+1)}{(2n+2)}\left(\frac{1\cdot3\cdot3\cdot5\cdot...\cdot(2n-1)(2n+1)}{2\cdot2\cdot4\cdot4\cdot...\cdot(2n)(2n)}\right)^2=\frac{8}{\pi^{2}}}$ $\displaystyle{\therefore\sum_{k=2}^{\infty}(-1)^{k}\ln\left(1-\frac{1}{k^{2}}\right)=S=\ln P=\ln\frac{8}{\pi^2}}$: Note that : $\dfrac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot...}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot...}=\dfrac{\pi}{2}$ is Wallis's Product
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$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$ BUT $$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$
08 ÁÕ¹Ò¤Á 2008 11:58 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Timestopper_STG à˵ؼÅ: à¾ÔèÁ reference |
#11
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ÍéÒ§ÍÔ§:
ã¹¡Ã³Õ a > b ãËé b=2551001 a=2551001......... â´Â·ÕèµÑÇ¢éÒ§ËÅѧ 2551001 ¢Í§ a à»ç¹ 0 ÍФÃѺ¡ç¨ÐàËç¹ä´éªÑ´ÇèÒÁÕ¤ÙèÍѹ´Ñºà»ç¹¨Ó¹Ç¹Í¹Ñ¹µì àªè¹ a = 25510010 b=2551001 »ÃÐÁÒ³¹Õé
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Rose_joker @Thailand Serendipity 25 ¾ÄȨԡÒ¹ 2007 14:23 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ RoSe-JoKer |
#12
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PART 2: Selected years of Australian Maths tournament (photocopied version)
1. P à»ç¹¨Ø´Ë¹Ö觺¹Ç§ÃÕ áÅÐ F1,F2 à»ç¹¨Ø´â¿¡ÑʢͧǧÃÕ áÅÐ L à»ç¹àÊé¹ÊÑÁ¼ÑʢͧǧÃÕ·Õè P ÊÁÁµÔ X à»ç¹¨Ø´ã´æº¹ L ¾ÔÊÙ¨¹ìÇèÒ ÊÒÁàËÅÕèÂÁ F1XF2 ÁÕàÊé¹ÃͺÃÙ»¹éÍÂÊØ´ àÁ×èÍ X ·Ñº¡Ñº P ¾Í´Õ ( 1 ¤Ðá¹¹) ¨Ò¡¹ÔÂÒÁ¢Í§Ç§ÃÕàÃÒ¨Ðä´éÇèÒ PF1+PF2= 2a ¨Ò¡¡ÒäÇÒÁÂÒǢͧ F1 ä»ËÒ F2 à·èҡѹ ´Ñ§¹Ñé¹µéͧÁÒ¾Ô¨Ò³Ò´éÒ¹·ÕèàËÅ×ͧ͢ÊÒÁàËÅÕèÂÁ XF1F2 ÍÕ¡ 2 ´éÒ¹ ¾Ô¨ÒóÒÊÒÁàËÅÕèÂÁã´æ 2 ÃÙ»·ÕèÁÕ°Ò¹ÃèÇÁ¡Ñ¹¨Ðä´éÇèÒ´éÒ¹·ÕèÃÇÁ¡Ñ¹ 2 ´éÒ¹¢Í§ÊÒÁàËÅÕèÂÁÃٻ㹨йéÍ¡ÇèÒ´éÒ¹·ÕèàËÅ×ÍÊͧ´éÒ¹¢Í§ÊÒÁàËÅÕèÂÁÃÙ»¹Í¡ àÊÁÍ ´Ñ§¹Ñ鹶éҨش X äÁèä´éÍÂÙèÃèÇÁ¡Ñº¨Ø´ P ¨Ðä´éÇèÒÃÙ»ÊÒÁàËÅÕèÂÁÃÙ»¹Ñé¹ÁÕ¤ÇÒÁÂÒÇÃͺÃÙ»à»ç¹ 2a+F1F2+a â´Â·Õè a>0 áµè¶éÒ X ÍÂÙèÃèÇÁ¡Ñº P ¨Ðä´éÇèÒ¤ÇÒÁÂÒÇÃͺÃÙ»à»ç¹ 2a+F1F2 «Öè§ÂÒǹéÍ·ÕèÊØ´
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Rose_joker @Thailand Serendipity |
#13
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¼ÁÇèÒ¢éÍ4(Part2)¾Õèpasser-byµéͧÅ×ÁÍÐäÃÍÕ¡á¹èàŤÃѺ4¤Ðá¹¹·ÓäÁ§èÒ¨ѧ
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$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$ BUT $$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$
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#14
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(1) ¤Ø³ kanakon
µÍ¹¹Õé ¢éÍ 4 µÍ¹·Õè 1 ¡ç¶Ù¡¤Ø³ kanakon ¨Ñ´¡ÒÃä» 4 ¢éÍÂèÍ Ẻ¶Ù¡µéͧáÅéǹФÃѺ àËÅ×Íáµè¢éÍ·Õèà»ç¹ª×èÍ ·ÄÉ®Õ ÊÃØ»ÇèÒ µÍ¹¹Õé ¤Ø³ kanakon ÁÕ 6 ¤Ðá¹¹ ¨Ò¡µÍ¹·Õè 1 áÅÐ 2 ¤Ðá¹¹ ¨Ò¡µÍ¹·Õè 2 (2) ¤Ø³ timestopper ¢éÍ 3.1 ¶Ù¡áÅéÇÅèФÃѺ áµè⨷ÂìãËé͸ԺÒ¾ÍÊѧࢻ´éÇÂÇèÒ ·ÓäÁ converge ·Õè¹éͧ·Óà»ç¹á¤èÇÔ¸Õ·Ó à¾×èÍËÒÇèÒÅÙèà¢éÒÊÙè¤èÒä˹ áµèÂѧäÁèä´é¾ÔÊÙ¨¹ìÊÑé¹æÇèÒ ·ÓäÁÅÙèà¢éÒ ´Ñ§¹Ñ鹵͹¹Õé ¼ÁãËéäÇé 2 ¤Ðá¹¹¡è͹ ÊèǹÍÕ¡ 1 ¤Ðá¹¹ ¨ÐµÒÁÁÒ ¶éÒÁÒ͸ԺÒÂÊÑé¹æä´éÇèÒ ·ÓäÁ convergent Êèǹ¢éÍ 1 µÍ¹·Õè 1 µÍº 2008 ¶Ù¡áÅéǤÃѺ ÊÓËÃѺ¢éÍ 4 µÍ¹·Õè 2 ¼Áà¾ÔèÁà§×è͹ä¢ÍÕ¡¹Ô´¹Ö§ ¨Ðä´éà¢éҡѺ 4 ¤Ðá¹¹ ´Ñ§¹Ñé¹ ¶éÒ¤¹·ÕèµÍºä»áÅéÇ ÍÂÒ¡µÍºãËÁèÊÓËÃѺ¢éÍ 4 version á¡éä¢ à¾×èÍãËéà»ÅÕ蹨ҡ 2 à»ç¹ 4 ¤Ðá¹¹ ¡çµÒÁʺÒ¹ФÃѺ ËÃ×ͨÐÃÑ¡ÉÒÊÔ·¸Ôì 2 ¤Ðá¹¹·Õèä´éäÇéáÅéÇ¡çäÁèÇèÒÍÐäà (3) ¤Ø³ Rose-joker ¢éÍ 1 µÍ¹·Õè 2 ¶éÒÁÒ͸ԺÒºÃ÷Ѵ·ÕèºÍ¡ÇèÒ ¾Ô¨ÒóÒÊÒÁàËÅÕèÂÁã´æ 2 ÃÙ»·ÕèÁÕ°Ò¹ÃèÇÁ¡Ñ¹¨Ðä´éÇèÒ´éÒ¹·ÕèÃÇÁ¡Ñ¹ 2 ´éÒ¹¢Í§ÊÒÁàËÅÕèÂÁÃٻ㹨йéÍ¡ÇèÒ´éÒ¹·ÕèàËÅ×ÍÊͧ´éÒ¹¢Í§ÊÒÁàËÅÕèÂÁÃÙ»¹Í¡ àÊÁÍ ãËé¡ÃШèÒ§ä´é ¡ç¨Ðä´é¤Ðá¹¹àµçÁ¢é͹Õé¤ÃѺ à¾ÃÒÐà»ç¹ËÑÇã¨ÊӤѢͧ¡ÒÃ͸ԺÒ¢é͹Õé Êèǹ¢éÍ 4 (version ¡è͹á¡éä¢) ¼ÁãËé 2 ¤Ðá¹¹äÇé¡è͹áÅéǡѹ¤ÃѺ NOTE: ¢éÍ 3.1 ÂѧÁÕÇÔ¸ÕÍ×è¹ÍÕ¡¹Ð¤ÃѺ áµèÍÒ¨¨ÐÂÒÇ¡ÇèÒ¹Ô´¹Ö§ â´ÂäÁèãªé Wallis product Êèǹ¢éÍ 3.3 ¼Á¤Ô´ÇèÒ ¶éÒã¤Ã·Ó¢é͹Õéä´é ¡ç¹èҨРapply ä»·Ó¢éÍ·Õè ÁÕ¤¹ãËé¾ÔÊÙ¨¹ìÇèÒ cos 1 à»ç¹ irrational number ä´é¹Ð¤ÃѺ
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#15
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3.1.(Part1)
ÊѧࡵÇèÒ $\displaystyle{\sum_{k=2}^{\infty}\ln\left(1-\frac{1}{k^2}\right)}$ ÅÙèà¢éÒÊÙè $-\ln 2$ ´Ñ§¹Ñé¹ $\displaystyle{\sum_{k=2}^{\infty}(-1)^{k}\left(1-\frac{1}{k^2}\right)}$ ÅÙèà¢éÒÍÂèÒ§ÊÑÁºÙóì
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$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$ BUT $$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$
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