 Malay  124069

144 # FORM 4 高级数学 (Additional Mathematics) NOTES 分享 发表于 9-10-2008 10:46 PM | 显示全部楼层
 2x(2x+1)^4 + 8x^2(2x+1)^3 = 2x(2x+1)^3×(2x+1) + 2x(2x+1)^3×4x = 2x(2x+1)^3 [(2x+1) + 4x] 这是根据乘法对加法的分配律。 发表于 10-10-2008 11:34 PM | 显示全部楼层

## 回复 101# mathlim 的帖子

 o。。。明白了。。。原来是酱factorize出来的。。。 发表于 11-10-2008 09:48 PM | 显示全部楼层     additional mathematics 是高级数学咩？ 字面解释addition是附加， mathematics是数学，additional mathematics应该叫附数（附加数学）吧？ 高数（高级数学）的程度应该没有酱低啦…… 注:高级数学的英文是advanced mathematics…… 发表于 11-10-2008 09:53 PM | 显示全部楼层
 顺便问一题 the sum of first nth terms in an  arithmetic progression is said to be   Sn = 6n^2-5n.Find the value of a6 S(6)=6*6^2-5*6        =186 S(5)=6*5^2-5*5        =125 a6 = S(6)-S(5)      =186-125      =61 这样做对吗？会不会被扣分？ 发表于 12-10-2008 01:59 AM | 显示全部楼层
 如果我是评阅老师的话，我会给满分。   发表于 12-10-2008 01:29 PM | 显示全部楼层

## 回复 105# mathlim 的帖子

 谢啦      发表于 17-1-2009 03:56 PM | 显示全部楼层
 1.given that function f(x)= 2x+1 , find f^n (x),where n is a positive integer. 2.given that 2^a=5^b=10^c ,express a in term of b and c . 3.Solve the equation 4^x -7 .  2^x - 8 = 0 4.Solve the equation  x - 4√x - 5 = 0 5.Solve the equation x^4 = 7x^2 +18. 帮帮忙.谢谢.  发表于 17-1-2009 11:45 PM | 显示全部楼层
 1. Given that function f(x) = 2x+1, find f^n (x), where n is a positive integer. f^2(x) = f[f(x)] = f(2x+1) = 2(2x+1)+1 = 4x+3 f^3(x) = f[f^2(x)] = f(4x+3) = 2(4x+3)+1 = 8x+7 f^4(x) = f[f^3(x)] = f(8x+7) = 2(8x+7)+1 = 16x+15 … … f^n(x) = (2^n)x+(2^n)-1 2. Given that 2^a = 5^b = 10^c, express a in term of b and c. 2^a = 10^c 2^ab = 10^bc —— ⑴ 5^b = 10^c 5^ab = 10^ca —— ⑵ ⑴×⑵ 10^ab = 10^(bc+ca) ab = bc + ca ab – ca = bc ∴ a = bc/(b-c) 3. Solve the equation 4^x - 7·2^x - 8 = 0. 4^x - 7·2^x - 8 = 0 (2^x)^2 - 7·2^x - 8 = 0 (2^x + 1)(2^x - 8) = 0 2^x = -1 或 2^x = 8   不合       x = 3 ∴ x = 3 4. Solve the equation x - 4√x - 5 = 0. x - 4√x - 5 = 0 (√x + 1)(√x - 5) = 0 √x = -1 或 √x = 5   不合       x = 25 ∴ x = 25 5. Solve the equation x^4 = 7x^2 + 18. x^4 = 7x^2 + 18 (x^2)^2 - 7x^2 - 18 = 0 (x^2 + 2)(x^2 - 9) = 0 x^2 = -2 或 x^2 = 9   不合       x = ±3 ∴ x = ±3 发表于 17-1-2009 11:51 PM | 显示全部楼层
 1. f^1(x)=2x+1     f^2(x)=2(2x+1)+1=4x+3     f^3(x)=4(2x+1)+1=8x+7    ..... induction  f^n(x)=(2^n)x+(2^n)-1 f^(n+1)=(2^n)(2x+1)+(2^n)-1=(2^(n+1))x+(2^(n+1))-1 so f^n(x)=(2^n)x+(2^n)-1 2. 2^a=5^b=10^c (2^a)^2=5^b*10^c 2^(2a)=5^b*10^c (2a)ln2=(b)ln5+(c)ln10 a=.......... 3. tak tahu 4. x - 4 √x- 5 = 0 x-5=4√x (x-5)^2=16x ............. warning:x>0 5. x^4 = 7x^2 +18. let y=x^2 y^2=7y+18 y=............ x=√y=........... warning y>=0 发表于 17-1-2009 11:55 PM | 显示全部楼层
 靠！第三题原来是7*2^2,我当作是7.2^2，冤枉咯！ 第二题原来是这样算的，我这种解法只是投机取巧而已！ 发表于 18-1-2009 06:52 AM | 显示全部楼层
 2^a = 10^c 2^ab = 10^bc —— ⑴ 5^b = 10^c 5^ab = 10^ca —— ⑵ ⑴×⑵ 我不明白这2个equation怎样来. 完全不明白,为什么2^ab = 10^bc? 不可以2^ac = 10^ba ? (2^x)^2 - 7·2^x - 8 = 0 (2^x + 1)(2^x - 8) = 0 (√x + 1)(√x - 5) = 0 (x^2 + 2)(x^2 - 9) = 0 原来数学有这种事发生,我只学过 (ax+c)(bx+d)=0 的quadratic equation而已,从来没有想到可以这样...原来数学是那么的广大 你们的数学真厉害,怎样练出来的? 原帖由 puangenlun 于 17-1-2009 11:55 PM 发表 靠！第三题原来是7*2^2,我当作是7.2^2，冤枉咯！ 第二题原来是这样算的，我这种解法只是投机取巧而已！ 什么是In?我没有学过,但是可以在calculator 看到. form 5也多多指教哦 去上学了 [ 本帖最后由 白羊座aries 于 18-1-2009 06:54 AM 编辑 ] 发表于 18-1-2009 11:51 AM | 显示全部楼层
 2^a = 10^c 两边取b次方 (2^a)^b = (10^c)^b 2^(ab) = 10^(cb) 发表于 18-1-2009 03:53 PM | 显示全部楼层
 原帖由 mathlim 于 18-1-2009 11:51 AM 发表 2^a = 10^c 两边取b次方 (2^a)^b = (10^c)^b 2^(ab) = 10^(cb) 原来如此,明白了  发表于 19-1-2009 12:45 AM | 显示全部楼层
 ln 跟 log 类似 只不过一个取2            一个取e 我那题完全不能和mathlim的解答作比较 献丑了！ 发表于 19-1-2009 09:32 PM | 显示全部楼层
 原帖由 puangenlun 于 19-1-2009 12:45 AM 发表 ln 跟 log 类似 只不过一个取2            一个取e 我那题完全不能和mathlim的解答作比较 献丑了！ in calculator, log is base on 10 not 2 ln is log which base on e, e = ??? i forget le 发表于 19-1-2009 11:26 PM | 显示全部楼层
 请参考： 请问ln和log有什么差别? 发表于 20-1-2009 09:46 AM | 显示全部楼层
 in calculator, log a=log10 a the base of log is 10 or you can change ln a = loge a e = 2.7.... ln is natural log i forget how to get e le.... 发表于 20-1-2009 04:51 PM | 显示全部楼层
 原帖由 TKCboy 于 20-1-2009 09:46 AM 发表 in calculator, log a=log10 a the base of log is 10 or you can change ln a = loge a e = 2.7.... ln is natural log i forget how to get e le.... 请用中文回复 e 的defination是从intergration那边拿来的 int a^x=a^x+c a=e=2.718281828459... 发表于 20-1-2009 05:58 PM | 显示全部楼层
 来源， 第一次提到常數e，是約翰·納皮爾於1618年出版的對數著作附錄中的一張表。但它沒有記錄這常數，只有由它為底計算出的一張自然對數列表，通常認為是由威廉·奧特雷德(William Oughtred)製作。第一次把e看為常數的是雅各·伯努利(Jacob Bernoulli) The compound-interest problem Jacob Bernoulli discovered this constant by studying a question about compound interest. One simple example is an account that starts with \$1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is \$2.00; but if the interest is computed and added twice in the year, the \$1 is multiplied by 1.5 twice, yielding \$1.00×1.5² = \$2.25. Compounding quarterly yields \$1.00×1.254 = \$2.4414…, and compounding monthly yields \$1.00×(1.0833…)12 = \$2.613035…. Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields \$2.692597…, while compounding daily yields \$2.714567…, just two cents more. Using n as the number of compounding intervals, with interest of 1/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach \$2.7182818…. More generally, an account that starts at \$1, and yields (1+R) dollars at simple interest, will yield eR dollars with continuous compounding. 来自 wikipedia.. 发表于 20-1-2009 08:14 PM | 显示全部楼层
 原帖由 笨蛋一个 于 20-1-2009 04:51 PM 发表 请用中文回复 e 的defination是从intergration那边拿来的 int a^x=a^x+c a=e=2.718281828459... 原来我现在学的intergration竟然是那么地简单 ,我只学 过  integrate ax^n dx = ax^n+1/n+1 +c . 虽然我不知道那个a 和x 代表什么,但是我想问 什么时候学到In 这sign

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