FORM 4 高级数学 (Additional Mathematics) NOTES 分享

mathlim

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。

play

o。。。明白了。。。原来是酱factorize出来的。。。

MaoMao92

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……

MaoMao92

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？

mathlim

如果我是评阅老师的话，我会给满分。

MaoMao92

谢啦

白羊座aries

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.

mathlim

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

puangenlun

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

puangenlun

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！

白羊座aries

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 ]

mathlim

2^a = 10^c
两边取b次方
(2^a)^b = (10^c)^b
2^(ab) = 10^(cb)

白羊座aries

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

puangenlun

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！

TKCboy

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

mathlim

请参考：
请问ln和log有什么差别?

TKCboy

in calculator,
log a=log₁₀ a
the base of log is 10 or you can change
ln a = log_e a
e = 2.7....
ln is natural log
i forget how to get e le....

笨蛋一个

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...

TKCboy

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..

白羊座aries

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