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anonymous
 4 years ago
Suppose you have three functions, f1(x),f2(y),f3(z). Consider the following expression: H=\int_{0}^{f1(x)}[G(f2(ξ))G(f3(ξ))dξ, where G is some continuously differentiable function. What is dH/dz? is it zero? or should I apply the chain rule dH/dz=(dH/df3)(df3/dz), and conclude that dH/dz is not zero because dH/df3 and df2/dz are both nonzero (known data)?
anonymous
 4 years ago
Suppose you have three functions, f1(x),f2(y),f3(z). Consider the following expression: H=\int_{0}^{f1(x)}[G(f2(ξ))G(f3(ξ))dξ, where G is some continuously differentiable function. What is dH/dz? is it zero? or should I apply the chain rule dH/dz=(dH/df3)(df3/dz), and conclude that dH/dz is not zero because dH/df3 and df2/dz are both nonzero (known data)?

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0\[H=\int_{0}^{f1(x)}[G(f2(ξ))G(f3(ξ))dξ\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup, that's the expression to which I want to compute dH/dz...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0Not from me, I just wanted to know what I was looking at, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can someone give me a hand with my doubt?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Ask Satellite. He's really smart.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think you can conclude dH/dz is zero. Consider f1 = e^x, f2 = e^y, f3=e^z.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the integral term should not depend on z since you are integrating with respect to it...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Unless G is a function that maps everything to zero, using function definitions will give a nonzero result. Since a counterexample exists to the statement, it can't be true.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^ using the function definitions I posted earlier, f1 = f2 = f3 = e^whatever

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't get the counterexample. What is inside the integral is integrated and hence, it can't depend on z...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The function parameters for f1, f2, f3 don't matter in the counterexample.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i.e tha H function is only a function of x and not on z (it' sintegrated with respect to it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But if f1=f2=f3=e^x it will be nonzero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As long as the inner functions are e^anything it will be nonzero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you're telling me that a function that does not depend on z changes with z?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh, 1 second, I see Im confusing the conversation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Consider 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 nonzero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup, but the upper limit of the integral does not depend on z by construction...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I misread the problem. Sorry for the confusion. It should be zero since after integrating you will have a function of x only.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's whay I think...but what if you use chain rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It shouldn't matter since f1(x) is passed to whatever you get after doing the integration.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Think about what H(x) is....it is an integral evaluated to the bound H(f1(x))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup..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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. 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.
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