Is it possible to have a function such that:
Examples would be welcome. But I am mostly interested in whether it is possible or not.
Let me try that again.
Does there exist a function such that:
In words-- does a function exist such that the definite integral over some interval a,b with respect to some dummy variable "t" of the derivative of the function equal the original function divided by two?
Ah, and I'm not sure whether to call f(x) a function or a number. In the problem I am trying to solve it is the arithmetic-geometric mean of two numbers.
Oh wait a second, this is really messed up.... let me get back to you
I'll try just one more time:
Does there exist some number and some function such that:
In words-- does there exist some number x and some function f(x)=x such that the definite integral of the derivative of the function over some interval a,b with respect to some dummy variable "t" equals x divided by two?
One way I can makes sense of it is if the "x" on the left side, of each equation, is intended to be a constant- a number. If that is the case, then f(x) is just some other number. In that case you are asking either, "given a function g(x), does there exist a function g, and a number, G, such that and .
Of course, by the "Fundamental Theorem of Calculus" so that is asking if there exist a function g such that . That, in turn, is asking whether there is a function, g, whose anti-derivative is g/2- . Differentiating both sides of that, we are asking whether there is a function g(x) such that g'(x)= g(x)/2.
The answer to that, of course, is "Yes, there is". And the function, g, is any solution to the differential equation y'= y/2. The general solution to that is for any constant C. In that case, the two equations become and which is correct.
Another way to make sense of it is to assume you really did not mean the limits of integration but are looking at f being some anti-derivative of g.
That is, you want and . Differentiating those equatkons gives and [tex]f'(x)/2= g'(x)[tex].
Again, it is not f that is important but g. From the second equation, f'(x)= 2g'(x) so we have f'(x)= 2g(x)= g(x) as before. Now this is true for and, since the integral of is , which is still a constant time , f(x) is also .
In the meantime I think I've come up with a solution that satisfies and :
In this solution and
I might just be chasing my tail around in circles--does this amount to a pointless statement?