differentiations
susugoodboy2022-10-04 11:39:541条回答
differentiations
suppose a function f is defined on the real numbers that for each x and y, f(x+y)=f(x)+f(y)
show that
a) if f is continuous at 0 then f is continuous at any x
b) f(n)=n*f(1); f(-n)=-f(n); f(1/n)=(1/n)*f(1), f(m/n)=(m/n)*f(1) for each positive integers m and n
c) if f is continuous for each x, then f(x)=x*f(1)
suppose a function f is defined on the real numbers that for each x and y, f(x+y)=f(x)+f(y)
show that
a) if f is continuous at 0 then f is continuous at any x
b) f(n)=n*f(1); f(-n)=-f(n); f(1/n)=(1/n)*f(1), f(m/n)=(m/n)*f(1) for each positive integers m and n
c) if f is continuous for each x, then f(x)=x*f(1)
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逃课的水手 共回答了24个问题 |采纳率87.5%- 来来来回答你：
a)
first,f(0)=0,obvious.
because f is continuous at 0,limit of f(X)=0 when X->0.
Now,for any x,f(x)=f(x-x')+f(x').let x'->x,we see that x-x'->0.
As we mentioned,when X->0,f(X)=0.
So,f(x')=f(x),for any x'->x.
As a result,f is continuous at any x.
b)
use induction.obvious.you can prove it.
c)
we just prf that for any rational number q,f(q)=qf(1).
now,for any real number x,we have two sequences of rational number pn and qn,s.t.
pn - 1年前
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