Theorem mathematic for relativity

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nulliusinverb
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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 physics 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!
 
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mathman said:
Furthermore the term in parenthesis should be (x-a)i not xi - ai
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.
 
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
 
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!
 
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!