pure maths function

2009-02-17 4:14 am
Q1 f is not identically equal to zero, show f(0)>0.
Assume for all x belongs to real such that f(x0)=0.


Q2 show f(x)=2x is injective but not surjective.

Q3 show g(x)=x/2 if x is even g(x)=(x-1)/2 if x is odd is surjective but not injective.

Q4 Suppose f:R-->R is a function satisfying f(a+x)=f(a-x) and f(b+x)=f(b-x) for all x, where a,b are constants and a>b. Let w=2(a-b). Show that w is a period of f, i.e. f(x+w)=f(x) for all x belongs to R.


Q5 Suppose g:R-->R is a periodic function with period T>0 satisfying g(x)=f(-x) for all x. Show that there exists c with 0<c<T such that g(c+x)=f(c-x) for all x.
更新1:

sorry i missed part of the question. Q1 Let f: R --->R be a real function such that f(x+y)=f(x)f(y) for all x, y belongs to R. if f(x) is not identically equal to zero, show f(0)>0. Hint: Assume for all x belongs to real such that f(x0)=0.

更新2:

Q2 why not surjective?

更新3:

Q5 why g(x-c) =f(-(x-c)) =f(c-x) ??

回答 (1)

2009-02-17 4:44 am
✔ 最佳答案
你第一條寫得不清楚
Q2
if x is defined on Z
Let 2x1=2x2=>x1=x2 So f(x)=2x is injective
There is no x such that f(x)=1
Q3
For each y, we can find an even x=2y such that g(x)=y.
Since for x=2y+1, g(x)=y this shows that there are two differents x such that g(x)=y. So g(x)=x/2 if x is even g(x)=(x-1)/2 if x is odd is surjective but not injective.
Q4 Suppose f:R-->R is a function satisfying f(a+x)=f(a-x) and f(b+x)=f(b-x) for all x, where a,b are constants and a>b. Let w=2(a-b). Show that w is a period of f, i.e. f(x+w)=f(x) for all x belongs to R.
f(x+w)
=f(x-b-b+a+a)
=f(a-x+b+b-a)
=f(b+b-x)
=f(b-b+x)
=f(x)
w is a period of f
Q5 g(x)=f(-x)=g(x+T)
Let T=2c=>c=T/2
Then
g(c+x)
=g(x+T/2)
=g(x-T/2+T)
=g(x-T/2)
=g(x-c)
=f(-(x-c))
=f(c-x)



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