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yup, that's the expression to which I want to compute dH/dz...
Not from me, I just wanted to know what I was looking at, sorry.
can someone give me a hand with my doubt?
Ask Satellite. He's really smart.
I don't think you can conclude dH/dz is zero. Consider f1 = e^x, f2 = e^y, f3=e^z.
but the integral term should not depend on z since you are integrating with respect to it...
Unless G is a function that maps everything to zero, using function definitions will give a non-zero result. Since a counter-example exists to the statement, it can't be true.
^ using the function definitions I posted earlier, f1 = f2 = f3 = e^whatever
don't get the counterexample. What is inside the integral is integrated and hence, it can't depend on z...
The function parameters for f1, f2, f3 don't matter in the counter-example.
i.e tha H function is only a function of x and not on z (it' sintegrated with respect to it
But if f1=f2=f3=e^x it will be non-zero.
As long as the inner functions are e^anything it will be non-zero.
so you're telling me that a function that does not depend on z changes with z?
Ahh, 1 second, I see Im confusing the conversation
Consider H(e^z)? Then after you evaluate the integral you will get a function of z, specifically e^z. When you differentiate e^z you get e^z which is non-zero.
yup, but the upper limit of the integral does not depend on z by construction...
I mis-read the problem. Sorry for the confusion. It should be zero since after integrating you will have a function of x only.
that's whay I think...but what if you use chain rule?
It shouldn't matter since f1(x) is passed to whatever you get after doing the integration.
Think about what H(x) is....it is an integral evaluated to the bound H(f1(x))
yup..this is what I thought at first. But then I doubted because the function f3 is inside the integral...but I believe you're right. It cannot be the case that f3 changes with z because it is being integrated over all z. Hence, the function H can only vary if x varies...am I correct?
Yes. There is no "z" in H(x) even though f3 = f3(z) b/c the integral is not with respect to z but something else which is how f3 is parameterized.
great! thanks a lot!