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Proving Calculus:Invariant w.r.t. Integral function?

  1. Aug 27, 2011 #1
    http://www.mathhelpforum.com/math-help/attachments/f6/22131d1314389447-proving-question-stucked-question1.gif [Broken]

    Hi.

    How do we solve this? I tried using the hint,but I still couldn't prove it.

    Thank you for your help.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 27, 2011 #2

    Hurkyl

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    Well, can you at least show what you got?
     
  4. Aug 27, 2011 #3
  5. Aug 27, 2011 #4

    Hurkyl

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    You're almost done, do you see it?

    It might help to change your dummy variables back to x, so that what you have more closely resembles the thing you're trying to prove, and maybe to rewrite in full the equation you have proven with your manipulations.
     
  6. Aug 28, 2011 #5
    I got the solution by replacing u with x.

    However,one thing that bothers me is that,can we assume u=pi-x and then later assume u=x?
    Doesn't that make things mathematically incorrect? Since pi-x =/= x?

    Is this what we call invariance w.r.t. integral function? What is invariance w.r.t. integral function anyway?
     
  7. Aug 28, 2011 #6

    Hurkyl

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    I'm not familiar with the phrase. Where did you hear it? Can you use it in a sentence or a paragraph?


    There are two perspectives that lead to the same result.

    The semantics you're taught in elementary calculus, all of the occurrences of x and u in this problem are dummy variables.

    It's not like solving a geometry problem where you have a hypothetical rectangle of and you define L to mean its length and W to mean its width and they have a fixed and relevant meaning throughout the lifetime of the problem.

    Instead, their only meaning is "I am the variable this definite integral is integrating over" and they have no meaning outside of the integral -- its only purpose is to make it easier to write the function inside the integral. The expressions [itex]\int_a^b f(p) \, dp[/itex] and [itex]\int_a^b f(q) \, dq[/itex] literally mean exactly the same thing. Tnd the notation [itex]\int_a^b f[/itex] is sometimes used when feasible, which further emphasizes the irrelevance of dummy variables.




    In the other semantics that's often used (but rarely stated explicitly at your level), the integrals [itex]\int_a^b f(p) \, dp[/itex] and [itex]\int_a^b f(q) \, dq[/itex] might mean different things, but the definition of the integral immediately implies that the two integrals evaluate to the same real number -- you have an identity:
    [tex]\int_a^b f(p) \, dp = \int_a^b f(q) \, dq[/tex]​
    which is just another integral law.
     
  8. Aug 28, 2011 #7
    Hi.
    Thank you for your helpful explanation.
    I understand now.

    The term invariant w.r.t. integral function is basically what you've explained. That is:

    [itex]\int_a^b f(p) \, dp = \int_a^b f(q) \, dq[/itex]

    My tutor said that we can change x to any variable we like( in your words,dummy variables) so long as the other notation/symbols are unchanged. Hence the phrase 'invariant w.r.t to integral function'. Anyway,I have not encountered the term in high school before and I don't think it's very widely used.

    I feel that your explanation is more useful and simpler.

    Thank you.
     
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