- #1
sphyrch
- 35
- 9
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
$$F(x)=F(a)+\sum_{i=1}^n(x^i-a^i)H_i(x)$$
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
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