Hard Calculus

(a)
Assume that f'(a)<0 then since f'(x) is strictly decreasing
This implies that f(b)<f(a) which is a contradiction
So f'(a)>0
(b)
x1=a-f(a)/f'(a)
Since f(a)<0,f'(a)>0 x1=a+(positive thing) >a
Consider
b-x1
=b- [a-f(a)/f'(a)]
=(b-a)+f(a)/f'(a)
=[(b-a)f'(a)+f(a)]/f'(a)
>[f(b)-f(a)+f(a)]/f'(a)
>0
Apple mean value theorem on (a,x1)
f(x1)
=f(a)+f'(d)(x1-a)
<f(a)+f'(a)(x1-a)
=0
Apple mean value theorem on (a,x1)
f'(e)
=[f(b)-f(x1)]/(b-x1)
>0
Also
f'(x1)
>f'(e)
>0
(c)
The result is proved by MI
when n=1, the result is true by (b)
Assume that when n=k is true
i.e a<xk<b f(xk<0) and f'(xk>0)
when n=k+1
xk+1=f(xk)-f(xk)/f'(xk)
>xk
>a
b-xk+1
=b-xk+f(xk)/f'(xk)
=[(b-xk)f'(xk)+f(xk)]/f'(xk)
>[((b)-f(xk)+f(xk)]/f'(xk)
>0
Apple mean value theorem on (xk,xk+1)
f(xk+1)
=f(xk)+f'(d)(xk+1-xk)
<f(xk)+f'(xk)(xk+1-xk)
=0
Apple mean value theorem on (xk+1,b)
f'(e)
=[f(b)-f(xk+1)]/(b-xk+1)
>0
Also
f'(xk+1)
>f'(e)
>0
By MI, a<xn<b and f(xn)<0 f'(xn)>0
(d)
Since a<x1<x2<...<xn<b
xn is monotonic increasing and is bounded from above by (b)
Hence lim xn exist, and lim(xn-xn+1)=0
Also as xk>a and f'(x) is strictly decreasing f'(xk)<f'(a)
lim(xn-xn+1)f'(xn)=0
f(xn)=(xn-xn+1)f'(xn) and so lim f(xn)=0
Since f is differentiable, f is continuous
Hence lim f(xn)=0 => f [lim xn]=0
That is lim xn is the zero of f

2009-01-23 17:57:21 補充：
maximal_ideal_space,

我估不到你會答﹐因為這些a_level標準題(根本上就是93年的問題)你好像興趣不大。我本來覺得煩惱即是菩提﹐EMK或者冷凝液答的可能性大些。等STEVIE-G™君自己揀最佳啦

不過朋友你答得這麼快﹐當年一定拿A啦。

另外條INTEGRATION做不到。

2009-01-23 17:58:56 補充：
Apple mean value theorem on (a,x1) 應是(x1,b)

2009-01-23 18:23:12 補充：
你再 + EMK + 菩提 = 天下無敵 Maths group

無可否認菩提計工程數學是好快﹐組合數學都有一手﹐不過代數或者數論﹐imo好像及不上 maximal_ideal_space 您。

應該再加上貓朋的dynamical system+ computer graphics

可惜無概率﹐統計加精算類﹐我是不行的﹐可以找及時而出或者找你張遼囉。

至於模糊數學﹐分形﹐劇變論那些10年都無人問一條就算啦

2009-01-24 02:37:15 補充：
有既﹐那幾條培X數學邀請賽幾何題目非閣下莫屬

多謝咁多位知識+的勁人黎討論呢題^^
令我困擾主要係b part的個3舊野...雖然都係用mvt去做...但要代數的卻有點兒混亂...
同埋第c part個個...主要諗唔明點用番b part去計...

岩 login 至睇到呢題當年第一次做煩足我成粒鐘的題目, 記憶猶新.
maximal_ideal_space, why 你最近少了來交流 ? 當年你答 D Univ math 答得好精彩, 呢度不能失去你的, 請回歸吧.
你再 + EMK + 菩提 = 天下無敵 Maths group

2009-01-23 18:06:54 補充：
既然整左, 唔爭在呢度發表埋, 好助 STEVIE-G™ 考 AL:
http://i117.photobucket.com/albums/o61/billy_hywung/Jan09/CrazyPP1.jpg

2009-01-23 18:34:37 補充：
另一題 hard Calculus 都係等落班至再諗啦, 食完飯又要開工, 差館忙死人.
希望今晚回家後可見高手解決o個題啦.

(a) By mean value theorem, there is c such that a<c<b, and
f(b)-f(a) = f'(c) (b-a).
Note that f(b)-f(a)>0, and b-a>0. So f'(c)>0.
Since f'(x) is strictly decreasing, so f'(a) > f'(c) > 0.

(b) x_1 = a - f(a)/f'(a).
Since f(a)<0, f'(a)>0, we have x_1 > a.
On the other hand,
-f(a)/(b-a) < (f(b)-f(a))/(b-a) = f'(c) < f'(a)
where c is defined in part (a).
Since f'(a) > 0, b-a>0, we have
-f(a)/f'(a) < b-a.
Therefore, x_1 = a - f(a)/f'(a) < b.

f(x_1) = f(a) + ∫(t=a to x_1) f'(t) dt
< f(a) + ∫(t=a to x_1) f'(a) dt (because f'(x) is strictly decreasing)
= f(a) + f'(a)(x_1 - a)
= f(a) + f'(a)(-f(a)/f'(a))
= 0.

Lastly, to show f'(x_1)>0, do exactly as in part (a).

(c) Use M.I. The case n=1 is done in part (b).
The induction step is done exactly as in part (b), with a replaced by x_(n-1).

(d) Note that
x_n = x_(n-1) - f(x_(n-1))/f'(x_(n-1)) > x_(n-1).
So {x_n} is an increasing sequence, bounded above by b.
So lim x_n exists. Let the limit be L.

Since f'(x) is decreasing, we have
f(a) < f'(t) < f(b) for all t between a and b.
Therefore, there is a constant K such that
|f(x_n)| < K for all n.
We have
f(x_n) = f'(x_n) (x_n - x_(n+1)),
so

0 < |f(x_n)| = |f'(x_n)| |x_n - x_(n+1)| < K(x_(n+1) - x_n).
Let n tends to infinity, we have
0 =< lim |f(x_n)| =< K(L-L) = 0.
So lim |f(x_n)|=0.
Therefore, lim f(x_n) = 0.
Since f is diffenentiable, f is continuous.
So lim f(x_n) = f(lim x_n).
Therefore, lim x_n is a zero of f(x).


2009-01-23 17:46:08 補充：
This is a special case for Newton's method.
This is a standard exam question for AL/AS applied maths, at least at the time when I took the exam.

2009-01-23 17:49:13 補充：
myisland8132,
我都估到你會答，如果唔係就應該係張遼會答。
不過今日有Ｄ時間，一時手痕，就答左先喇，估唔到快你少少。

2009-01-24 00:59:22 補充：
我都係忙，唔係好多時間答問題，剛好今天比較有時間，先上黎答下問題。而且現在高手眾多，都無乜要我出手的題目啦。大家都答唔到的，我也不一定答到。點都好啦，見到有值得答的問題，我都會盡力而為。