# Theorem mathematic for relativity

1. May 15, 2012

### nulliusinverb

hello!:

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

F(x)=F(a) + $\sum^{n}_{i=1}$(x$^{i}$-a$^{i}$)H$_{i}$(x)

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

okk my problem is how to get to the sum $\sum$???

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. May 15, 2012

### mathman

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

3. May 15, 2012

### Fredrik

Staff Emeritus
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.

4. May 16, 2012

### nulliusinverb

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

5. May 16, 2012

### mathman

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!

6. Jun 11, 2012

### nulliusinverb

i apologize for the delay, here is the proof:

F:ℝ$^{n}$$\rightarrow$ℝ

i have:

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

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

finally:

F($\vec{x}$)-F($\vec{a}$)= $\sum$$^{m}_{μ=1}$H$_{μ}$($\vec{x}$)(x$^{μ}$-a$^{μ}$)

qed

thanks you very much to all!!