# Is it generally difficult to memorize 'multivariable calculus' theorems?

There are many weak forms of "Fubini's theorem" with strong hypotheses in elementary calculus texts. However, these strong hypotheses are very unnatural and are thus hard to memorize. Compared to that, the full Fubini's theorem (in measure space) is actually relatively easy to memorize in comparison.

Furthermore: I am wondering if this is also the case for multivariable calculus.

I'm currently studying "implicit function theorem", and I find the hypotheses for the theorem to be quite hard to memorize. Are theorems in mutivariable calculus easier to memorize in the context of manifolds?

• Are you seeking to understand said theorems, or memorize them? – Display Name Mar 23 '14 at 9:18
• @Display To memorize. It is usual that if once understood, then it becomes easier to memorize, but this theorem is hard to memorize even though i understamd this theorem. – John. p Mar 23 '14 at 9:20
• @Display Specifically Rudin uses the operator norm to tackle those multivariable calculus theorems, but i think this way is quite unnatural and makes one hard to imagine why and how those theorems must be true – John. p Mar 23 '14 at 9:23
• Well, I never can. Except when I'm gonna sit an exam tomorrow : / – Vim Jul 9 '15 at 14:11

It really does depend. I try to understand what the broad meaning of a particular theorem is, and attempt to gain a geometric understanding of it (if I can). For example, take the implicit function theorem. It says that we can convert various relations to functions of several variables. More precisely, take a point in the locus $L=(x_0,y_0)$. The implicit function theorem states that we can write a function $F(x,y)=0$ in the form $y=f(x)$ with $x$ chosen from the interval $x_0-h<x<x_0+h,~h\in\mathbb{R}^+$.
Another way of stating it is that if there exists a some rectangle with side lengths $|x-x_0|<h,~|y-y_0|<k,~k\in\mathbb{R}^+$, then the part of the locus $L$ of $F(x,y)=0$ inside this rectangle lies on an arc of the form $y=f(x)$ as long as $f$ is differentiable.