- #1
Hariraumurthy
- 14
- 0
1. The problem statement, all variables and given/known
Given a function f satisfying the following two conditions for all x and y;
(a)f(x+y) = f(x)*f(y)
(b)f(x) = 1 + x*g(x) where the limit as x tends to 0 of g(x) =1
prove
(a) the derivative f'(x) exists
(b) f'(x)=f(x)
2. Homework Equations
this book uses the starting point ln(x)= integral from 1 to x of 1/x.
3. The Attempt at a Solution
NOTE* DISCLAIMER:the below solution is very roundabout and I have given the entire solution but what I want is a much more elegant solution. This problem was given before the theory of logarithimic and exponential functions. In fact it was given in the section on continuity and differentiability so that is all you need. To answer this question there is no need to read my highly roundabout solution below.
from this definition it is obvious that the derivative of ln(x) exists when x>0,( for x less than or equal to 0, the integral is obviously divergent) and is equal to 1/x.
and it is easy to show that for rational x from the power rule since
if y=ln(x^n)
dy/dx=(1/x^n)*nx^(n-1)=n/x=d/dx(n*ln(x)
therefore ln(x^n)=n*ln(x) for rational x(constant of integration is 0 since ln(1)=0 from the definition)
therefore ln(e^x)=xlne=x (since by definition e is the number whose logarithm is 1)
this implies e^x=the ln inverse of x for rational x
and we define e^x for x irrational = ln inverse of x
therefore y=e^x\Leftrightarrow x=ln(y)(since dx/dy is 1/y it is always positive and therefore maps one to one on e^x)
dy/dx=1/(dx/dy)=1/(1/y)=y=e^x
which shows e^x is differentiable as long as y\neq0 which will never happen.
the only thing left to show is that e^x is the only such function satisfying f(x+y)=f(x)*f(y) and f(x) = 1 + x*g(x) where the limit as x tends to 0 of g(x) =1.
differentiating with respect to x and y in turn,
f'(x+y)=f'(x)f(y) and f'(x+y)=f(x)f'(y)
hence f'(y)/f(y)=f'(x)/f(x) and therefore each is constant. Thus if x=f(y), dx/dy=ax where a is a constant so that x= Ke^ay so that the function won't be fundamentally different from e^y, and since f'(y)=f(y), the only function is x=e^y
this is the entire solution but it is very round about. This problem was given before the theory of exponential functions and logarithmic functions were given. It was given on the section of continuity and differentiability. So there must be a much less roundabout way to do this.
Given a function f satisfying the following two conditions for all x and y;
(a)f(x+y) = f(x)*f(y)
(b)f(x) = 1 + x*g(x) where the limit as x tends to 0 of g(x) =1
prove
(a) the derivative f'(x) exists
(b) f'(x)=f(x)
2. Homework Equations
this book uses the starting point ln(x)= integral from 1 to x of 1/x.
3. The Attempt at a Solution
NOTE* DISCLAIMER:the below solution is very roundabout and I have given the entire solution but what I want is a much more elegant solution. This problem was given before the theory of logarithimic and exponential functions. In fact it was given in the section on continuity and differentiability so that is all you need. To answer this question there is no need to read my highly roundabout solution below.
from this definition it is obvious that the derivative of ln(x) exists when x>0,( for x less than or equal to 0, the integral is obviously divergent) and is equal to 1/x.
and it is easy to show that for rational x from the power rule since
if y=ln(x^n)
dy/dx=(1/x^n)*nx^(n-1)=n/x=d/dx(n*ln(x)
therefore ln(x^n)=n*ln(x) for rational x(constant of integration is 0 since ln(1)=0 from the definition)
therefore ln(e^x)=xlne=x (since by definition e is the number whose logarithm is 1)
this implies e^x=the ln inverse of x for rational x
and we define e^x for x irrational = ln inverse of x
therefore y=e^x\Leftrightarrow x=ln(y)(since dx/dy is 1/y it is always positive and therefore maps one to one on e^x)
dy/dx=1/(dx/dy)=1/(1/y)=y=e^x
which shows e^x is differentiable as long as y\neq0 which will never happen.
the only thing left to show is that e^x is the only such function satisfying f(x+y)=f(x)*f(y) and f(x) = 1 + x*g(x) where the limit as x tends to 0 of g(x) =1.
differentiating with respect to x and y in turn,
f'(x+y)=f'(x)f(y) and f'(x+y)=f(x)f'(y)
hence f'(y)/f(y)=f'(x)/f(x) and therefore each is constant. Thus if x=f(y), dx/dy=ax where a is a constant so that x= Ke^ay so that the function won't be fundamentally different from e^y, and since f'(y)=f(y), the only function is x=e^y
this is the entire solution but it is very round about. This problem was given before the theory of exponential functions and logarithmic functions were given. It was given on the section of continuity and differentiability. So there must be a much less roundabout way to do this.
