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Math Help - check this proof(is there any fallacy)

  1. #1
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    check this proof(is there any fallacy)

    ( f(x) is a polynomial function and when f(x+y)=f(x)+f(y) given-------------(1) f(
    solution
    f(0)=0 (proved)
    now put y=h where h approaches to zero in (1)
    we get f(x+h)=f(x)+f(h)
    implies f(x+h)-f(x)=f(h)
    implies[ f(x+h)-f(x)]/h=f(h)/h
    implies f'(x)=f'(h) { as f(0)=0 }

    implies f'(x)=k (as f'(0) is constant) here k is a constant
    implies f(x)=kx hence proved
    is there any fallacy in this proof.
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  2. #2
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    I don't understand what is given.

    By the way, there is exist a real function not of the form k*x, which satisfies for all real a,b:

    f(a+b)=f(a)+f(b)
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  3. #3
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    What is given is that f is a polynomial and that, for all x and y, f(x+ y)= f(x)+ f(y).

    What is needed for the proof given here is simply that f is differentiable so "f is a polynomial" is sufficient but not necessary.

    Yes, ayushdadhwal, that is a valid proof. You have shown that the derivative is a constant and so f(x)= cx.

    It is, by the way, possible to prove this without assuming differentiabllity, only continuity. However, there exist non-continuous functions satisfying f(x+ y)= f(x)+ f(y). They, of course, are nothing like "f(x)= cx".
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    Quote Originally Posted by HallsofIvy View Post
    What is given is that f is a polynomial and that, for all x and y, f(x+ y)= f(x)+ f(y).

    What is needed for the proof given here is simply that f is differentiable so "f is a polynomial" is sufficient but not necessary.

    Yes, ayushdadhwal, that is a valid proof. You have shown that the derivative is a constant and so f(x)= cx.

    It is, by the way, possible to prove this without assuming differentiabllity, only continuity. However, there exist non-continuous functions satisfying f(x+ y)= f(x)+ f(y). They, of course, are nothing like "f(x)= cx".
    can you give me some other examples
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  5. #5
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    What do you mean by "example"? An example of what?

    If you are referring to my statement that you really only need continuity, what I did was look at the specific kinds of numbers.

    The simplest numbers are the "counting numbers" or positive integers. The defining property of the counting numbers is- counting! Or, in other terms, induction. So I prove by induction that if f(x+ y)= f(x)+ f(y), then f(nx)= nf(x) for any x and n any positive integer.

    f(1x)= f(x)= 1f(x). Assume that, for some k, f(kx)= kf(x). Then f((k+1)x)= f(kx+ x)= f(kx)+ f(x)= kf(x)+ f(x)= (k+1)f(x).

    The next set of numbers are the "whole numbers" which are just the positive integers together with the number 0. And 0 is, of course, the additive identity: x+ 0= x. f(x)= f(x+ 0)= f(x)+ f(0) so that, subtracting f(x) from each side, f(0)= 0. (Which also says, of course, f(0x)= 0f(x).

    The set of all integers includes negatives. If n is any positive integer then f((n+ (-n)x)= f(nx)+ f(-nx). but f(n+ (-n))= f(0)= 0 so we have f(nx)+ f(-nx)= 0. From that, f(-nx)= -f(nx)= -(nf(x))= (-n)f(x). So we have shown that, for all integers n and x any real number, f(nx)= nf(x).

    The rational numbers are ratios of integers- the difference is that denominator. If n is any non-zero integer, then n(1/n)= 0 so that f(x)= f(n(1/n)x) and that, from just above, is equal to nf((1/n)x). 'That is, f(x)= nf((1/n)x so that f((1/n)x)= (1/n)f(x).

    Of course, then, f((m/n)x)= f(m((1/n)x))= mf((1/n)x)= m((1/n)f(x))= (m/n)f(x).

    So far, we have shown that if r is any rational number and x is any real number, f(rx)= rf(x) and, taking x= 1, f(r)= rf(1). Writing c= f(1), that gives f(r)= cr.

    The reason we need continuity is that the the set of all real numbers cannot be defined "algebraically"- they must be defined "analytically" using some form of a limit process. The simplest way is this- if x is any real number then there exist a sequence of rational numbers converging to x (This is exactly what we mean by saying that we can write a real number in decimal form. If, say, x= 3.141592... then the sequence 3, 3.1, 3.14, 3.131, 3.1415, 3.14159, 3.141592, ... is a sequence of rational numbers (because they are terminating decimals) that converges to x.)

    So for any real number, x, there exist {r_n} converging to x: f(x)= f(\lim r_n) and, with f continuous, that is \lim f(r_n)= \lim cr= c\lim r= cx.

    And, unlike "differentiable", "continuous is a necessary condition. If we remove that condition, there exist functions satisfying f(x+y)= f(x)+ f(y) that do not satisfy f(x)= cx. ]

    To do that, observe that f(x+y)= f(x)+ f(y) is one of the two conditions for a "linear transformation" on a vector space. The other is that f(cx)= cf(x) for c any number. Even without continuity, we were able to prove that f(cx)= cf(x) for c any rational number. So think of the real numbers as a vector space over the field of ratioal numbers. We can define a linear transformation by defining it on a basis for the vector space. The real numbers, as a vector space over the rational numbers has infinite dimension (in fact, uncountable) but we can do the following- since the rational numbers are the underlying field, we can take the number "1" as one basis "vector" and, say \sqrt{2} for another, then whatever (independent) numbers you wish after that. Define f(1)= 1, f(\sqrt{2})= 2. Every rational number, r, can be written as "r(1)" so that f(r)= rf(1)= r. But f(\sqrt{2})= 2\ne \sqrt{2}f(1)=\sqrt{2}.

    While all continuous functions that satisfy f(x+y)= f(x)+ f(y) are of the form cx, non-continuous functions are nasty! In fact, it can be shown the the graph of such a function, far from being a straight line, is dense in the plane. That is, given any point in the plane, there exist a point on the graph arbitrarily close to the given point. Think how awful that would be to graph!
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