Pure Maths continuous function

1(a)

if x>0,then f(lx|)=f(x)
if x<0, then f(|x|=f(-x)
Consider f(√(x^2+y^2))=f(x)f(y)
Substitute y=0 f(√(x^2))=f(x)f(0)=f(x)
=>f(-x)=f(x)
So f(x)=f(lxl) for all real x
(b)
Using M.I.
when n=1
the statement is obvious true
Assume that when n=k, the statement is true
i.e. f(√k(x))=[f(x)]^k for all positive integers n and for all real x
when n=k+1
[f(x)]^(k+1)
=[f(x)]^kf(x)
= f(√k(x))f(x)
= f(√(kx^2+x^2))
= f(√(k+1)(x))
when n=k+1, the statemett is true
By M.I. f(√n(x))=[f(x)]^n for all positive integers n and for all real x
(c)
[f(1)]^k
=[f(1)]^(r^2)
=f(√(r^2)*1) by (b)
=f(r) (no matter r is positive or negative given by (a))
So, f(r)=[f(1)]^k for any rational number r, where k= r^2
(d)
Since we can construct rational number r_1,r_2,...r_n such that lim n-> infinity r_n = x
On the other hand, we can also construct k_n=r_n^2 such that
lim n-> infinity k_n = lim n-> infinity (r_n)^2 = x^2
From f(r_n)=[f(1)]^k_n
lim n-> infinity f(r_n)= lim n-> infinity [f(1)]^k_n
Since f is a non-negative continous function
f {lim n-> infinity (r_n)}=[f(1)]^ (lim n-> infinity k_n)
f(x)=[f(1)]^m for all real x, where m = x^2





2009-01-31 18:51:17 補充：
Actually, I miss the important point in part (c) f(√(1/q)(x))=[f(x)]^(1/q) since I don't know how to prove it ( I believe that it is true). However, after see the answer of 及時而出, I think that this statement may be wrong!!

2009-01-31 22:20:46 補充：
及時而出的part (c)正確, I would like to write it more clear
Let r=p/q then p = rq.
f(p)
= f( q*r )
= f(r)^(q^2) ... by (b)
On the other hand,
f(p)
= f(p*1)
= f(1)^(p^2) ... by (b)
Hence,
f(r)^(q^2) = f(1)^(p^2)
or
f(r) = f(1)^( (p/q)^2 ) = f(1)^( r^2) = f(1)^k
where k=r^2

2009-01-31 22:21:36 補充：
我原本個諗法這題用不到

2009-01-31 22:23:53 補充：
part (d) 是經驗。基本上係「吹水」﹐你這樣寫就full marks

lim n-> infinity f(r_n)= f (lim n-> infinity r_n)

即函數和極限可交換次序﹐基本上f是continuous就用得架啦。

2009-01-31 22:24:34 補充：
好似二項式sigma和積分可交換0米又係「老吹」﹐要証明是好難的

2009-01-31 22:27:14 補充：
因為題目好明顯想由(c)証(d)。條條sequence都是這種模式。就是想考你識用這招

a.) Let p be a non-negative number.
By sub x=p, y=-p into (2), we have f(p)=f(p)f(-p), which gives f(p)=0 or f(-p)=f(p);
By sub x=y=-p into (2), we have f(-p)f(-p)=f(p)f(p), which gives f(-p)=f(p) or f(-p)=-f(p);
combining, we have f(-p)=f(p) for all non-negative p,
i.e. f(x)=f(|x|) for all real x
.
Remarks: this question wants us to show that "the functioned value of any negative number is nothing but the

functioned value of its magnitude", and in fact, it has nothing to do with the functioned value at zero.
..
Next, we try to prove in the sequence of
N -> Z -> Q -> R
..
N)) Let x be any real number.
The statement holds when n=1 trivially. Assume it holds when n=k.
[f(x)]^k+1=f(x)f(√k(x))=f(√(x^2+k*x^2))=f(√(k+1)(x))
so, by induction, hold for all n belongs to N.
.
Z)) Let p be any non-negative number.
f(p)=f(√p^2*(1))=[f(1)]^(p^2),
f(-p)=f(√p^2*(-1))=[f(-1)]^(p^2)=[f(1)]^(p^2)=[f(1)]^((-p)^2)
Thus, f(a)=[f(1)]^(a^2) for all integer a.
..
Remarks: AL answer can be very precise(精簡), and actually you don't need to bother with the k=r^2, that's just due to limitations(不能上標再上標) of old word processing program.
..
Q)) Let p,q be any integer.
[f(1)]^(p^2)=f(p)=f(p/q*q)=f(√(q^2)(p/q))=[f(p/q)]^(q^2)
Therefore, f(p/q)=[f(1)]^(p^2/q^2),
i.e. f(r)=[f(1)]^(r^2) for all rational number r.
.
R)) Let x be any irrational number,
then there must exists an infinite sequence of rational number a1,a2,.... such that lim[n->inf]{ an } = x
Let k be any natural number, As ak is rational, f(ak)=[f(1)]^(ak^2)
Taking limit on both sides, lim[k->inf]{ f(ak) } = lim[k->inf]{ [f(1)]^(ak^2) }
which implies f( lim[k->inf]{ ak } ) = [f(1)]^( ( lim[k->inf]{ ak } ) ^2 ) as both [f(1)]^(x^2) and f(x) are
continuous for all real number x.
Thus, f(x) = [f(1)]^(x^2) for all real number x.

2009-02-01 15:03:54 補充：
For part d, the fact used are
(i) property of continuous function lim[x->a]{ f(x) } = f(a) iff f is cont at x=a.

2009-02-01 15:05:17 補充：
(ii)for every irrational number, we can approximate it by taking an infinite sequence of rational numbers.
See [ http://hk.myblog.yahoo.com/yahoo-knowledger/article?mid=8 ] for such sequence.

2009-02-04 11:09:28 補充：
及時而出 朋友：
(i) "Hence, f(x)=0 for all x if f(0)=0"
呢度真係唔駛分case咁累贅le....
後面的proof會自動包括埋呢個情況!
(ii) "f(1)>=0"
諗得好詳細，用 f ( x / root(2) )squared 證亦好精彩~
不過題目話: 「1. f is a non-negative continous function」.....

From (2),
f(√(0^2+0^2))=f(0)f(0)
implies f(0)=0 or f(0)=1
If f(0)=0, for any x>0,
f(x) = f(√(x^2+0^2))=f(x)f(0)=0 and
for any x<0, f(x)f(x) = f(√(x^2+x^2)) = f( |√2 x|) =0
Hence, f(x)=0 for all x if f(0)=0
a)
If f(0)=0, then f(x)=0=f(|x|)
If f(0)=1,
f(x)=f(x)f(0) = f(√(x^2+0^2))=f(|x|)
b)
f(√n(x))
= f(√n|x|) ... by (a)
= f(√{ nx^2 } )
= f(√{ (n-1)x^2 + x^2} )
= f(√{n-1}*x ) f(x)
...
= f(√{n-2}*x ) f(x) f(x)
...
= f(x) f(x) ... f(x) ... n terms
= {f(x) }^n
c)
By (a), it is sufficient to consider r>0.
There exists positive integers m and n such that r*n = m.
f(m)
= f( n*r )
= f(r)^(n^2) ... by (b)
On the other hand,
f(m)
= f(m*1)
= f(1)^(m^2) ... by (b)
Hence,
f(r)^(n^2) = f(1)^(m^2)
or
f(r) = f(1)^( (m/n)^2 ) = f(1)^( r^2) = f(1)^k
d)
By (a), it is sufficient to consider x >0.
Let {r(j) } be a positive rational sequence such that
r(j1) < r(j2) if j1<j2 and lim r(j) = x
Now, let s(j)^2 = x^2 - r(j)^2
f(x)
= f(√(r(j)^2+ s(j)^2))
f( r(j) ) f( s(j) )
Hence,
f(x)
= lim f(x)
= lim f( r(j) ) f( s(j) )
= lim f( r(j) ) lim f( s(j) )
= lim {f(1)^( r(j)^2) } lim f( s(j) ) .... by (c)
= {lim f(1)^( r(j)^2) } f( lim s(j) ) .... since f is continuous
= { f(1)^(x^2) } f(0)
= 0 if f(0)=0
= f(1)^(x^2) if f(0)=1
Since, f(1)=0 if f(0)=0, we have f(1)^(x^2) =0 if f(0)=0
Hence, we always have
f(x) = f(1)^(x^2) for all x

2009-01-31 17:26:14 補充：
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2009-01-31 17:29:22 補充：
影相輸馬鼻，但(c)要研訊。

2009-01-31 23:53:24 補充：
(d)這招背後的理論是任何實數都可以用一有理數列的極限去表示，上了大學數學系一定會學。點解係 AL 題目有的咁既，聽的老鬼講話哩的係以前 AL set theory裡面既number system教，所以會係舊題目裡見到。但係 set theory已經 out syl. 好耐，老鬼可能恃老吹水，信不信由你。

2009-02-01 20:32:50 補充：
我犯了嚴重錯誤!
在(a)前的東西要重寫，否則(c)和(d)會全盤爛掉。

2009-02-01 20:32:59 補充：
From (2),
f(√(0^2+0^2))=f(0)f(0)
implies f(0)=0 or f(0)=1
If f(0)=0, for any x>0,
f(x) = f(√(x^2+0^2))=f(x)f(0)=0
and hence
for any x<0, f(x)f(x) = f(√(x^2+x^2)) = f( |√2 x|) =0
Hence, f(x)=0 for all x if f(0)=0

If f(0)=1, for x > 0 ,
f(x) = f(√{(x^2+x^2)/2} ) = f(x/√2)f(x/√2) > = 0

2009-02-01 20:42:01 補充：
Especially, f(1) > = 0.
( Otherwise, (c) and (d) will be undefined.)

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現在三個一手答案都有錯，平頭起步。希望後來的答案會更準，更清晰或有更特別的敘述，否則就太令人失望了。

2009-02-05 09:28:57 補充：
呵呵，我已經唔點名，只係發下牢騷。既然有人沙圈叫陣，咁就唔可以唔出場，讓投了票的觀眾失望。
閣下唔講"f(0)=0"問題，我都睇唔到你"x in Z"裡面有問題。但係哩個唔係我睇到閣下証明裡的錯誤，我有意見的是(a)。

2009-02-05 09:29:21 補充：
我話"三個一手答案都有錯"只係客氣說話，比機會所有人有位執生，亦唔想比人有"就勁，錯就要收皮"哩種錯覺。就算我話樓上"(c)要研訊"，都只係因為我輸馬鼻，討些口頭便宜。但如果看得香港賽馬多的人，就知道"研訊不是要分對錯定輸贏，而是要令比賽結果毫無爭議地被肯定"。
雖然我計數係好隨意兼成日計錯數，但我至今仍未看到我的一手答案有任何其他問題，如有任何發現，歡迎各位指教。

2009-02-05 09:30:28 補充：
反而閣下的表現就一再令人失望。
http://hk.knowledge.yahoo.com/question/question?qid=7009013101844
大聲喊錯無問題，只是閣下對數學比其他人都熱情。追答有錯都無問題，及時更正就可以。只是計錯數”sorry”都無聲，只係”oops”一聲算數，閣下有咁大聲話人，是不是應該被期望有相應的力度話自己？

2009-02-05 09:31:45 補充：
無論閣下會如何再回應，我也無所謂。因為我最近狀態好，本來要再擺上網問的問題已解決，亦完成有問有答的義務，所以會提早離場復課。拜拜。


呀! 特別送上一個沒有"研訊"的例子給大家。
http://hk.news.yahoo.com/article/090204/4/aiv2.html
http://hk.news.yahoo.com/article/090204/4/ais6.html
這就是沒有"研訊"的例子。

這些應該不是ALE的題目，只係普通練習書的題目而已。

2009-01-31 17:34:36 補充：
myisland8132: 你點知 f(0)=1？

這是ALE的題目嗎? 寫下來很長吔!