1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral equality paradox?

  1. Jul 10, 2007 #1
    if we have that:

    [tex] \int_{a}^{b}dxf(x) - \int_{a}^{b}dxg(x)= \int_{a}^{b}dx(f(x)-g(x)) [/tex]

    where the integral over (a,b) of f(x) and g(x) exist separately then my question is if

    [tex] \int_{a}^{b}dx(f(x)-g(x)) =0 [/tex] then

    does this imply necessarily that [tex] f(x)=g(x)+h'(x) [/tex]

    where h(a)=h(b)=0 and its derivative is 0 almost everywhere on the interval (a,b)
     
  2. jcsd
  3. Jul 10, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    We have that [tex]\int^b_a f(x) dx = \int^b_a g(x) dx[/tex] and from that condition you ask if that implies [itex]f(x) = g(x) \in [b,a][/itex](Neglecting small value of h'(x) you proposed, i know don't why you did, though it doesn't end up making a difference). As a simple counter example, f(x) and g(x) may be very different Odd functions, and the integral of these two between b and -b will be 0.
     
  4. Jul 10, 2007 #3

    VietDao29

    User Avatar
    Homework Helper

    No, this is not correct, you just need that h(a) = h(b), since:

    [tex]\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = \int_a ^ b \left( f(x) - g(x) \right) dx = \int_a ^ b \left( g(x) + h'(x) - g(x) \right) dx[/tex]

    [tex]= \int_a ^ b h'(x) dx = \left. h(x) \right|_a ^ b = h(b) - h(a)[/tex]

    Now, if h(a) = h(b), then we'll have: [tex]\int_a ^ b f(x) dx - \int_a ^ b g(x) dx = 0[/tex]

    No, why should it's derivative is 0? =.="
     
  5. Jul 10, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    Are you sure that worked out right VietDao29 >.<? He asks if [tex] \int^b_a f(x) dx = \int^b_a g(x) dx[/tex] then does that imply f(x) = g(x) + h'(x), where h(a)=h(b)=0. It seems from your post you assumed f(x)=g(x) + h'(x), to get that all we need is h(a)=h(b), but that is somewhat the converse of what he is asking I think >.<
     
  6. Jul 11, 2007 #5
    He also said h'=o a.e.
     
  7. Jul 11, 2007 #6

    VietDao29

    User Avatar
    Homework Helper

    I was pointing out that his claim that h(a) = h(b) = 0 is wrong, and it's even wronger, when he said that h' = 0.

    You can find h'(x) simply by subtracting f(x) from g(x).

    Say, we have:
    [tex]\int_0 ^ \frac{\pi}{2} \sin x = \int_0 ^ \frac{\pi}{2} \cos x = 1[/tex]

    We also have:
    [tex]\sin x = \cos x + (\sin x - \cos x)[/tex]

    So h'(x) = sin(x) - cos(x)
    Integrate it with respect to x, we have:
    h(x) = -cos(x) - sin(x) + C

    [tex]h(0) = -1 + C[/tex]

    [tex]h \left( \frac{\pi}{2} \right) = -1 + C[/tex]

    [tex]\Rightarrow h(0) = h \left( \frac{\pi}{2} \right)[/tex] (but not equal to 0, unless C = 1)

    However, h(0) = h(pi/ 2) does not imply that h'(x) = 0 on the interval (0; pi/2).

    So, in conclusion, both of his assertions (?, can I use plural here) are wrong. =.="
     
    Last edited: Jul 11, 2007
  8. Jul 11, 2007 #7

    Gib Z

    User Avatar
    Homework Helper

    Ok we'll good we agree on that :) Pretty pointless though, "Klaus" never comes back to his threads when he's proven wrong, and matt grime reckons its some other guy named Jose whose not good >.<
     
  9. Jul 11, 2007 #8
    ok,.. thank you i fooled myself once again.
     
  10. Jul 11, 2007 #9

    VietDao29

    User Avatar
    Homework Helper

    Well, he did come back. :rolleyes: o:) ^.^
     
  11. Jul 11, 2007 #10
    If he is Jose he's calmed down a lot. Which is annoying actually, because jose was crazy enough to be funny, where as this person is simply obnoxious.
     
  12. Jul 11, 2007 #11

    VietDao29

    User Avatar
    Homework Helper

    Well, it must be some of my missing days here, at PF, but who's the well-known Jose that you guys are talking about? I don't think I know anyone named Jose. :cry: And, btw, what has he done to make him this famous?
     
  13. Jul 11, 2007 #12
    Well his username was eljose. You can search for posts by him. Let's just say he was a genius on the level of gauss, and solved multiple millennium prize problems but the mathematical community kept holding him down because they didn't appreciate proper mathematics.
     
  14. Jul 12, 2007 #13

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The snobbish mathematical community, I think you'll find, who only let famous people publish in journals.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integral equality paradox?
  1. Integral equality (Replies: 2)

  2. An integral paradox ? (Replies: 3)

Loading...