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Let $n\in\mathbb{Z}^+$. Let $A$ be an open subset of $\mathbb{R}^n$. Let $f : A\to\mathbb{R}$ be a map. Let $K = [a_1,b_1]\times\cdots\times[a_n,b_n]$ be an $n$-cell contained in $A$ $(a_i<b_i \hspace{1mm}\forall i\in\{1,\ldots,n\})$. To be concise, suppose that for each $k\in\{1,\ldots,n\},$ the partial derivative $$ \frac{\partial^k f(c)}{\partial x_k \partial x_{k-1} \cdots \partial x_1}(c) $$ exists at every $c \in K$.


In POMA (3e), Rudin shows (Thm 9.40) that in the case $n=2$, there exists a $(c_1,c_2) \in K$ such that $$ \frac{\partial^2f(c_1,c_2)}{\partial x_2 \partial x_1} (b_2-a_2)(b_1-a_1) \hspace{1mm}=\hspace{1mm} f(b_1,b_2) - f(b_1,a_2) - \big(f(a_1,b_2) - f(a_1,a_2)\big). $$

I believe I understand the proof well, which essentially consists of applying the ordinary MVT to the function $[a_1,b_1]\to\mathbb{R} : t \mapsto f(t,b_2) - f(t,a_2)$. Then, $$ f(b_1,b_2) - f(b_1,a_2) - \big(f(a_1,b_2) - f(a_1,a_2)\big) \hspace{1mm}=\hspace{1mm} (b_1-a_1)\left(\frac{\partial f(c_1,b_2)}{\partial x_1} - \frac{\partial f(c_1,a_2)}{\partial x_1}\right) \hspace{1mm}=\hspace{1mm} (b_1-a_1)(b_2-a_2) \frac{\partial^2f(c_1,c_2)}{\partial x_2 \partial x_1} $$ where the second equality follows from applying the MVT to the function $t \mapsto \frac{\partial f(c_1,t)}{\partial x_2}$. My question is other.


Preceding the theorem, he writes (pg. 235)

For simplicity (and without loss of generality) we state our next two theorems for real functions of two variables.

Man, that wouldn't have bothered me, if not for his unjustified claim that the omission of cases $n>2$ is WLOG.

I'm currently stumped trying to derive an analogous formula for ${n=3}$ (I keep getting partial derivatives that don't cancel). My quesiton is: does someone happen to know the general formula? I'm really tempted to try to derive it, but realistically I just don't know if I can spare the time. If I were going to, a hint might not be a bad thing. My first several google searches have turned up nothing.

Also, Rudin's choice of words suggests to me that we may yet get analogous results for functions which are not only real-valued. What would such results be?

Thanks for your time.

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    $\begingroup$ Did you mean $$\frac{\partial^2f(c_1,c_2)}{\partial x_2 \partial x_1} (b_2-a_2)(b_1-a_1)= f(a_2,b_2)-f(a_1,b_2) - \left(f(a_2,b_1) - f(a_1,b_1)\right)?$$ $\endgroup$ – Angina Seng Jan 6 at 4:19
  • $\begingroup$ probably, let me double check what I wrote real quick $\endgroup$ – Thomas Winckelman Jan 6 at 4:22
  • $\begingroup$ No, I think actually something different. Thank you for pointing out the typo (hopefully it is the only one) $\endgroup$ – Thomas Winckelman Jan 6 at 4:29
  • $\begingroup$ The 2-cell is $[a_1,b_1]\times[a_2,b_2]$. In retrospect, I admit that is confusing, since one sometimes writes $f(a,b)$ (in fact, I think Rudin writes this). I'm sorry about the mix up. Does that check out now? $\endgroup$ – Thomas Winckelman Jan 6 at 4:30
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The "obvious" formula to induct (assuming your formula in 2D is right) to me would be that: there exists $c=(c_1\dots c_n)$ such that $$\frac{\partial^nf}{\partial x_1\dots \partial x_n}(c) \ (b_n-a_n)\dots (b_1-a_1) = \delta_1(\dots (\delta_n f)\dots))$$ where $\delta_k g := g(b_k)-g(a_k)$ only changes the inputs in the $k$th variable. Writing $\partial_i$ for $\partial/\partial x_i$ and $V_n(a,b):=(b_n-a_n)\dots (b_1-a_1) $ we have \begin{align} \partial_1f(c) V_1(a,b) &= \delta_1 f=f(b_1)-f(a_1)\\ \partial_1\partial_2f(c) V_2(a,b) &= \delta_1 \delta_2 f =\delta_1(f(\bullet,b_2)-f(\bullet,a_2)) \\&= f(b_1,b_2)-f(b_1,a_2) - [f(a_1,b_2)-f(a_1,a_2)]\end{align} matching the two earlier formulas, and for your convenience here's the third one:

\begin{align} \partial_1\partial_2\partial_3f(c) V_3(a,b)&= \delta_1\delta_2\delta_3 f=\delta_1(f(\bullet , b_2,b_3)-f(\bullet , b_2,a_3) - [f(\bullet , a_2,b_3)-f(\bullet , a_2,a_3)])\\ &= (f(b_1 , b_2,b_3)-f(b_1, b_2,a_3) - [f(b_1 , a_2,b_3)-f(b_1, a_2,a_3)]) \\ &\quad - (f(a_1, b_2,b_3)-f(a_1 , b_2,a_3) - [f(a_1 , a_2,b_3)-f(a_1 , a_2,a_3)]) \end{align}

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    $\begingroup$ Yes! That works as far as I can tell. You apply MVT to $$[a_1,b_1]\to\mathbb{R} \hspace{1mm}:\hspace{1mm} t \mapsto f(t,b_2,b_3) - f(t,b_2,a_3) - \big( f(t,a_2,b_3) - f(t,a_2,a_3) \big),$$ were $c_1 \in (a_1,b_1)$ is that which is given by MVT, then to $$[a_2,b_2]\to\mathbb{R} \hspace{1mm}:\hspace{1mm} t \mapsto \partial_1 f(c_1,t,b_3) - \partial_1 f(c_1,t,a_3)$$ and, finally, to $$[a_3,b_3]\to\mathbb{R} \hspace{1mm}:\hspace{1mm} t \mapsto \partial_2 \partial_2 f(c_1,c_2,t).$$ $\endgroup$ – Thomas Winckelman Jan 6 at 15:41
  • $\begingroup$ Also, I think my formula for 2D is right. Is it not? Also also, was that simply an intuitive educated guess based on experience? Or, was there a specific approach you used to come up with that formula?? $\endgroup$ – Thomas Winckelman Jan 6 at 15:44
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    $\begingroup$ @ThomasWinckelman yes, its true, but I didn't check at the time of writing (sorry!) Maybe it was experience? The approach was, "finite differences make MVT (and other things) look extra clean, you're just using MVT twice, so maybe write it like that and see if its easier to generalise..." Also I got the order of derivatives maybe wrong but $\delta_i\delta_j f = \delta_j \delta_i f$ so it doesn't matter oh and last thing, "the same" should hold for functions $f:\mathbb R^n \to \mathbb R^m$ by just using this once for each of the $m$ components $f_1,\dots, f_m$ $\endgroup$ – Calvin Khor Jan 7 at 5:46

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