Question

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)?

Answer 1

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

Answer 2

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

Answer 3

any help?

Answer 4

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

Answer 5

can someone give me a hand with my doubt?

Answer 6

Ask Satellite. He's really smart.

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

Ahh, 1 second, I see Im confusing the conversation

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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?

Answer 25

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.

Answer 26

great! thanks a lot!