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Theorem mathematic for relativity

  1. May 15, 2012 #1
    hello!:

    my problem is about of a theorem mathematic,as I prove the following theorem?

    F(x)=F(a) + [itex]\sum^{n}_{i=1}[/itex](x[itex]^{i}[/itex]-a[itex]^{i}[/itex])H[itex]_{i}[/itex](x)

    good first start with the fundamental theorem of calculus: (for proof):

    F(x) - F(a) = [itex]\int^{x}_{a}[/itex]F'(s)ds sustitution: s=t(x - a) + a [itex]\Rightarrow[/itex] [a,x] to [0,1] then:
    ds=dt(x - a) later:
    f(x) - F(a)= (x - a)[itex]\int^{1}_{0}[/itex]F'(t(x - a) +a)dt

    okk my problem is how to get to the sum [itex]\sum[/itex]???

    is physic relativistic forum, because of this theorem I can get to the change of coordinates in the Einstein equations and find bases for the manifolds of space-time. thanks!!!
     
  2. jcsd
  3. May 15, 2012 #2

    mathman

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    http://en.wikipedia.org/wiki/Taylor_series

    Your representation is a modification of the standard Taylor series representation, where Hi(a) (not Hi(x)) is related to the ith derivative of F.

    Furthermore the term in parenthesis should be (x-a)i not xi - ai
     
  4. May 15, 2012 #3

    Fredrik

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    I know it's not clear from what he said, but he has a different theorem in mind. Wald's statement of the theorem goes like this:
    If ##F:\mathbb R^n\to\mathbb R## is ##C^\infty##, then for each ##a=(a^1,\dots,a^n)\in\mathbb R^n## there exist ##C^\infty## functions ##H_\mu## such that for all ##x\in\mathbb R^n## we have
    $$F(x)=F(a)+\sum_{\mu=1}^n(x^\mu-a^\mu)H_\mu(x).$$ Furthermore, we have
    $$H_\mu(a)=\frac{\partial F}{\partial x^\mu}\bigg|_{x=a}.$$​
    A similar theorem is stated and proved in Isham's book on differential geometry, page 82. So nulliusinverb, I suggest you take a look at that.
     
  5. May 16, 2012 #4
    Fredrick thank you very much, the book is recommended to study this issue, although the theorem is raised from other values ​​is the same. thank!!

    ps: "Modern Differential Geometry for Physicists" autor: Isham
     
  6. May 16, 2012 #5

    mathman

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    Notation is a major problem here. xi usually means the ith power of x. For this theorem it means the ith component of a vector x ε Rn.

    I suggest that, when anyone asks a question, make sure the notation is clear!
     
  7. Jun 11, 2012 #6
    i apologize for the delay, here is the proof:

    F:ℝ[itex]^{n}[/itex][itex]\rightarrow[/itex]ℝ

    i have:

    F([itex]\vec{x}[/itex])-F([itex]\vec{a}[/itex])= [itex]\sum[/itex][itex]^{m}_{μ=1}[/itex]F(t(x[itex]^{μ}[/itex]-a[itex]^{μ}[/itex])+a[itex]^{μ}[/itex],0...,0)[itex]^{t=1}_{t=0}[/itex]
    then:
    =[itex]\sum[/itex][itex]^{m}_{μ=1}[/itex](x[itex]^{μ}[/itex]-a[itex]^{μ}[/itex])[itex]\int[/itex][itex]^{1}_{0}[/itex][itex]\frac{\partial F}{\partial u^{μ}}[/itex]((t(x[itex]^{μ}[/itex]-a[itex]^{μ}[/itex])+a[itex]^{μ}[/itex],0...,0)dt
    where:

    H[itex]_{μ}[/itex]([itex]\vec{x}[/itex])=[itex]\int[/itex][itex]^{1}_{0}[/itex][itex]\frac{\partial F(\vec{x})}{\partial u^{μ}}[/itex]dt

    finally:

    F([itex]\vec{x}[/itex])-F([itex]\vec{a}[/itex])= [itex]\sum[/itex][itex]^{m}_{μ=1}[/itex]H[itex]_{μ}[/itex]([itex]\vec{x}[/itex])(x[itex]^{μ}[/itex]-a[itex]^{μ}[/itex])

    qed

    thanks you very much to all!!
     
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