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Integration proof

  1. Dec 25, 2007 #1
    Notation: I = integral sign from 0 to 1, D= partial derivative symbol.

    Please help me prove that for any smooth function f:R^n -> R defined on a neighbourhood of a in R^n,

    f(x) = f(a) + I{(D/Dt)f(a+t(x-a))dt}

    Here's my attempt:
    (D/Dt)f(a+t(x-a))dt = d[f(a+t(x-a)] (justification needed?)
    so

    I [(D/Dt)f(a+t(x-a))dt] = I d[f(a+t(x-a)]
    = f(a+1(x-a)) - f(a+0(x-a)) (Fundamental theorem of calculus, right?)
    = f(x)-f(a).

    Am I right, or am I making many unjustified steps here?
     
    Last edited: Dec 25, 2007
  2. jcsd
  3. Dec 25, 2007 #2

    arildno

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    Looks fine to me.

    After all, set [tex]y=a+t(x-a),t=1\to{y}=x,t=0\to{y}=a,\frac{dy}{dt}=(x-a)[/tex]
    Thereby, your integral is readily converted to:
    [tex]I=f(a)+\int_{a}^{x}f'(y)dy=f(x)[/tex]
     
    Last edited: Dec 25, 2007
  4. Dec 25, 2007 #3
    Thanks arildno. But there is one major confusion here. Your dy/dt is supposed to have the partial derivative symbols, right? Because y is a function of x=(x_1,...,x_n) and t, i.e. a function of n+1 variables and not just a function of t alone.

    Also, y= a + t(x-a) is a function that has n components, because x and a are elements of R^n. So what exactly is the meaning of f'(y)dy when your y is not a real number but a variable in R^n???
     
    Last edited: Dec 25, 2007
  5. Dec 25, 2007 #4
    The only thing I can make out of your f'(y) is a Jacobian matrix, in which case the antiderivative cannot be defined.
     
  6. Dec 25, 2007 #5

    arildno

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    You can regard "a" and "x" as arbitrary constants.
     
  7. Dec 26, 2007 #6
    Ok, I think I'm onto something. Please tell me if my explanation below is correct. Remember, I'm using D for the partial derivative symbol.

    Let’s first understand what (D/Dt)f(a+t(x-a)) means: Since x belongs to R^n, then f is a function of the n variables x=(x_1,...,x_n). Now with the introduction of the new independent variable t (which is totally independent of x and vice versa), the expression
    f(a+t(x-a)) is now a function of n+1 variables, and hence the partial derivative symbol D/Dt.

    Having said that, f(a+t(x-a)) also appears as the integrand in I{(D/Dt)f(a+t(x-a))dt} (I'm using I as the integral symbol from 0 to 1), and here we are integrating with respect to t only. Thus we are treating x as a constant within the expression f(a+t(x-a)) during the process of integration, since the integration is with respect to t only (if it was a double integral where we are also integrating with respect to x, then x is certainly no longer treated as a constant). Thus (within the integral only) we can write (D/Dt)f(a+t(x-a) as (d/dt)f(a+t(x-a)) , whereby we get

    I{(D/Dt)f(a+t(x-a))dt} = I{(d/dt)f(a+t(x-a))dt} = I{d[f(a+t(x-a))]},

    which is the formal justification of the first line in my original solution. Am I right?
     
    Last edited: Dec 26, 2007
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