A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 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)?

  • This Question is Closed
  1. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[H=\int_{0}^{f1(x)}[G(f2(ξ))G(f3(ξ))dξ\]?

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup, that's the expression to which I want to compute dH/dz...

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any help?

  4. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not from me, I just wanted to know what I was looking at, sorry.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can someone give me a hand with my doubt?

  6. Mertsj
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ask Satellite. He's really smart.

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't think you can conclude dH/dz is zero. Consider f1 = e^x, f2 = e^y, f3=e^z.

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but the integral term should not depend on z since you are integrating with respect to it...

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^ using the function definitions I posted earlier, f1 = f2 = f3 = e^whatever

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    don't get the counterexample. What is inside the integral is integrated and hence, it can't depend on z...

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The function parameters for f1, f2, f3 don't matter in the counter-example.

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i.e tha H function is only a function of x and not on z (it' sintegrated with respect to it

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    But if f1=f2=f3=e^x it will be non-zero.

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    As long as the inner functions are e^anything it will be non-zero.

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you're telling me that a function that does not depend on z changes with z?

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahh, 1 second, I see Im confusing the conversation

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup, but the upper limit of the integral does not depend on z by construction...

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mis-read the problem. Sorry for the confusion. It should be zero since after integrating you will have a function of x only.

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's whay I think...but what if you use chain rule?

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It shouldn't matter since f1(x) is passed to whatever you get after doing the integration.

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Think about what H(x) is....it is an integral evaluated to the bound H(f1(x))

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    great! thanks a lot!

  27. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.