1f(x) = mx
this satisfies.
linear eqn. without any constant term.
Well what sort of function can be this which satisfy the function rule as
f(x+y)=f(x)+f(y)?
I mean which function can it be?
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13 Answers
39\large f\left(xy \right)=f\left(x \right)+f\left(y \right) \Rightarrow f\left(x \right)=k\ln x
or even \large f\left(x \right)=0
39\large f\left(xy \right)=f\left(x \right)f\left(y \right)\Rightarrow f\left(x{\right)=x^{n} {n\epsilon R
39\large f\left(x+y \right)=f\left(x \right)f\left(y \right)\Rightarrow f\left(x \right)=a^{kx}
39\large f\left(x+y \right)=f\left(x \right)+f\left(y \right)\Rightarrow f\left(x \right)=kx
66This is one of the Cauchy's functional equation. It can be shown that if we dont's assume continuity, then f(x)=kx for all rational numbers, while if continuity is given, then f(x)=kx for all real x.
62I think for JEE syllabus continuity and differentiability can be safely assumed
for that case, the proof should be simple....
any takers?
39its obvious to see that f(0)=0. and this is an odd function. further f(nx)=n f(x).
but i doubt is this enough to say f(x)= kx
62f(nx)=nf(x)
also f(1)=k let
thus,
(P and q used below are integers)
f(p.1/p)= p f(1/p)
thus f(1/p) = 1/p x f(1) = k. 1/p
also take
f(q/p) = q f(1/p) = q. k. 1/p = k. q/p
thus for all rationals, f(x) = k x
We have not used continuity anywhere..
62If we assume continuity, then there is a direct theorem which proves that f(x)=kx for all x (This is not in syllabus though)
66Proving that the relation f(x)=kx holds also for irrational numbers require continuity. Particularly useful is the definition of continuity as given by Heine (edited after being pointed by TheProphet).
Suppose a function f is defined at at point x=x0 and its neighborhood. This function is said to be continuous at x=x0, if for an arbitrary sequence {xn} of real numbers satifying {xn}→x0, the sequence of the corresponding values of the function, i.e. {f(xn)} satisfies limn→∞{f(xn)}=f(x0).
Here the notation "→" means "tends to as n tends to infinity".
Let us take an arbitrary irrational point, say x0. We need to show f(x0) =kx0.
Note that in the vicinity of any irrational number there are infinitely many rational numbers. So, let us choose a sequence of rational numbers {xn} whose limit is x0. This can always be done owing to the infinite numbers of rational numbers in the vicinity of x0. Owing to the continuity of f at x0, we must have limn→∞{f(xn)}=f(x0).
However, since each xn is a rational number, f(xn)=kxn (it has already been proven for rationals). Therefore, we have
f(x0)=limn→∞ {kxn}=k limn→∞ {xn} = kx0. (since we had already chosen the sequence {xn} whose limit was x0)
Since x0 was arbitrary, it means that for all irrational numbers as well, the relation f(x)=kx holds true. Hence, f(x)=kx for all real x.
341A small correction: the definition of continuity mentioned above is due to Heine. Cauchy's definition uses the well known ε-δ challenge.
