Multivariable fundamental calculus theorem in Wald

[sphyrch]

i want to prove that if $F:\mathbb{R}^n\to\mathbb{R}$ is a differentiable function, then
$$F(x)=F(a)+\sum_{i=1}^n(x^i-a^i)H_i(x)$$
where $H_i(a)=\frac{\partial F}{\partial x^i}\bigg|_{x=a}$. the hint is that with the 1-dimensional case, convert the integral into one with limits from $0$ to $1$ and then we'll get the 1-dimensional version of what we're trying to prove. then we have to extend it to $n$-dimensional case. my try is like this -
$$F(x)=F(a)+\int_a^xF'(s)ds$$
if I substitute $s=(x-a)t+a$, then the above becomes
$$F(x)=F(a)+(x-a)\int_0^1F'((x-a)t+a)dt$$
so the rhs integral should be my $H(x)$ so that $H(a)=\frac{dF}{dx}\bigg|_{x=a}=F'(a)$

but first problem: if i evaluate the integral, i get $H(x)=\frac{F((x-a)t+a)}{x-a}\big|_{t=1}-\frac{F((x-a)t+a)}{x-a}\big|_{t=0}=\frac{F(x)-F(a)}{x-a}$ but i don't see how $H(a)=\frac{dF}{dx}\bigg|_{x=a}=F'(a)$

second problem is, how should I extend to the $n$-dimensional case? the most I can think of is that $n$-dimensional $F$ will have several component functions $F_1,\ldots,F_n$ - to each of which we can apply the 1-D result, but how does that get us to the final result? Please help

 

[fresh_42]

Look up "Taylor n-dimensional series proof pdf". I found
https://www.math.cuhk.edu.hk/course_builder/1516/math1010c/Taylor.pdf
https://www.rose-hulman.edu/~bryan/lottamath/mtaylor.pdf

 

[wrobel]

$$x_1,x_2\in\mathbb{R}^m,\quad\int_0^1\frac{d}{ds}f(sx_1+(1-s)x_2)ds=f(x_1)-f(x_2);$$
$$\int_0^1\frac{d}{ds}f(sx_1+(1-s)x_2)ds=\int_0^1\frac{\partial f}{\partial x}(sx_1+(1-s)x_2)ds(x_1-x_2);$$
$$H_i=\int_0^1\frac{\partial f}{\partial x^i}(sx_1+(1-s)x_2)ds$$
ok?