## 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)?

1. TuringTest

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

2. anonymous

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

3. anonymous

any help?

4. TuringTest

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

5. anonymous

can someone give me a hand with my doubt?

6. Mertsj

7. anonymous

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

8. anonymous

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

9. anonymous

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

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

Ahh, 1 second, I see Im confusing the conversation

18. anonymous

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

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

20. anonymous

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

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

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

great! thanks a lot!