A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 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
 5 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)?

This Question is Closed

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

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

TuringTest
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0can someone give me a hand with my doubt?

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

anonymous
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0^ using the function definitions I posted earlier, f1 = f2 = f3 = e^whatever

anonymous
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0The function parameters for f1, f2, f3 don't matter in the counterexample.

anonymous
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0But if f1=f2=f3=e^x it will be nonzero.

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

anonymous
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ahh, 1 second, I see Im confusing the conversation

anonymous
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0that's whay I think...but what if you use chain rule?

anonymous
 5 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
 5 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
 5 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
 5 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.