do derivatives belong to unique functions?


by rhuelu
Tags: belong, derivatives, functions, unique
rhuelu
rhuelu is offline
#1
Sep13-08, 06:26 PM
P: 17
I need to show that f(x)=cexp(-cx) is the only solution to c-[integral from 0 to x](f(x)) = f(x). Is this trivial??? If not how would you suggest I go about showing that functions have unique derivatives (after taking into account constants). Thanks in advance for suggestions!
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
morphism
morphism is offline
#2
Sep13-08, 07:27 PM
Sci Advisor
HW Helper
P: 2,020
Is that equation really

[tex]f(x) = c - \int_0^x f(t) \, dt?[/tex]

Because f(x)=cexp(-cx) isn't a solution to this for all c.
rhuelu
rhuelu is offline
#3
Sep14-08, 11:53 AM
P: 17
the integral should be multiplied by c as well. This is actually a stats problem...the entire question is to show when the hazard/failure rate is constant -> c=f(x) / (1-F(x)) where f=pdf, F=cdf. Its obvious that f(x)=cexp(-cx) is a solution to this. I just don't know how to show its the only solution.

HallsofIvy
HallsofIvy is online now
#4
Sep14-08, 12:12 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

do derivatives belong to unique functions?


The fact that "derivatives belong to unique functions" or, more correctly, that if f(x) and g(x) have the same derivative then f(x) and g(x) differ at most by a constant, is given in most calculus books and follows from the mean value theorem:

Lemma: Suppose f(x) is continuous on [a, b], is differentiable on (a, b), and f'(x)= 0 for all x in (a, b). Then f(x)= C (f(x) is a constant) for all x in (a,b).

Let x be any point in (a, b). By the mean value theorem, (f(x)- f(a))/(x- a)= f '(c) for some c between a and x. Since f'= 0 between a and b, f'(0)= 0 from which it follows that f(x)- f(a)= 0 or f(x)= f(a). That is, f(x) is equal to the number f(a) for all x between a and b and so f(x) is a constant there.

Theorem: If f(x) and g(x) are both continuous on [a, b], differentiable on (a, b), and f'(x)= g'(x) for all x in (a, b), then f(x)= g(x)+ C, a constant.

Let H(x)= f(x)- g(x). Then H(x) is continuous on [a,b] and differentiable on (a, b). Further, for all x in (a, b), H'(x)= f'(x)- g'(x)= 0 because f'(x)= g'(x). By the lemma, H(x)= f(x)+ g(x)= C for some number C. Then f(x)= g(x)+ C.
rhuelu
rhuelu is offline
#5
Sep14-08, 12:55 PM
P: 17
thanks...that makes sense. Do you think this logic still holds if we only know that F(x) is right continuous? This is one of the properties of a cdf.


Register to reply

Related Discussions
derivatives of ln functions Calculus & Beyond Homework 6
derivatives of vector functions Calculus & Beyond Homework 2
Derivatives of Trigonometric Functions Calculus & Beyond Homework 3
Derivatives of Trigonometric Functions Calculus & Beyond Homework 2
Derivatives of Composite Functions Calculus & Beyond Homework 4