What is analysis and why do analysis — part 2 of 2 | For the love of Maths: the Queen of all Sciences

We had discussed this on Nov 17 2015 blog. We finish the article with more examples from the work of Prof. Terence Tao. (If you like it, please send a thanks to him :-))

Example 1. (Interchanging limits and integrals).

For any real number y, we have

$\int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\arctan (x-y)\mid_{x=-\infty}^{\infty}$ which equals

$(\pi/2)-(-\pi/2)=\pi$ .

Taking limits as $y \rightarrow \infty$ , we should obtain

$\int_{-\infty}^{\infty}\lim_{y \rightarrow \infty}\frac {dx}{1+(x-y)^{2}}=\lim_{y \rightarrow \infty} \int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\pi$

But, for every x, have $\lim_{y \rightarrow \infty} \frac {1}{1+(x-y)^{2}}=0$ . So, we seem to have concluded that $0=\pi$ . What was the problem with the above argument? Should one abandon the (very useful) technique of interchanging limits and integrals?

Example 2. Interchanging limits and derivatives.

Observe that if $\in > 0$ , then

$\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}=\frac {3x^{2}(\in^{2}+x^{2})+x^{2}-2x^{4}}{(\in^{2}+x^{2})^{2}}$ ,

and in particular that

$\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}\mid_{x=0}=0$ .

Taking limits as $\in \rightarrow 0$ , one might then expect that

$\frac {d}{dx}\frac {x^{3}}{0+x^{2}}\mid_{x=0}=0$ .

But, the right hand side is $\frac {dx}{dx}=1$ . Does this mean that it is always illegitimate to interchange limits and derivatives?

Example 3. Interchanging derivatives.

$Let^{1}$ $f(x,y)$ be the function $f(x,y)=\frac {xy^{3}}{x^{2}+y^{2}}$ . A common manoeuvre in analysis is to interchange two partial derivatives, thus one expects

$\frac {\partial^{2}f(0,0)}{\partial x \partial y}=\frac {\partial^{2}f(0,0)}{\partial y \partial x}$ .

But, from the quotient rule, we have

$\frac {\partial f(x,y)}{\partial y}=\frac {3xy^{2}}{x^{2}+y^{2}}=\frac {2xy^{4}}{(x^{2}+y^{2})^{2}}$

and in particular,

$\frac {\partial f(x,0)}{\partial y}=\frac {0}{x^{2}}-\frac{0}{x^{4}}=0$ .

Thus, $\frac {\partial^{2}f(0,0)}{\partial x \partial y}=0$ .

On the other hand, from the quotient rule again, we have

$\frac {\partial f(x,y)}{\partial x}=\frac {y^{3}}{x^{2}+y^{2}} - \frac {2x^{2}y^{3}}{(x^{2}+y^{2})^{2}}$ and hence,

$\frac {\partial f(0,y)}{\partial x}=\frac {y^{3}}{y^{2}}-\frac {0}{y^{4}}=y$ .

Thus, $\frac {\partial^{2}f(0,0)}{\partial y \partial x}=1$ .

Since $1 \neq 0$ , we thus seem to have shown that interchange of two derivatives is untrustworthy. But, are there any other circumstances in which the interchange of derivatives is legitimate?

Example 4. $L^{'} H\hat {o}pital's Rule$

We are familiar with the beautifully simple $L^{'}H \hat{0}pital's$ rule

$\lim_{ x \rightarrow x_{0}} \frac {f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac {f^{'}(x)}{g^{'}(x)}$

but one can still get led to incorrect conclusions if one applies it incorrectly. For instance, applying it to $f(x)=x$ , $g(x)=1+x$ and $x_{0}=0$ we would obtain

$\lim_{x \rightarrow 0}\frac {x}{1+x}=\lim_{x \rightarrow 0} \frac {1}{1}=1$ .

But this is an incorrect answer since $\lim_{x \rightarrow 0}\frac {x}{1+x}=\frac {0}{1+0}=0$ .

Of course, all that is going on here is that $L^{'}H \hat{o}pital's rule$ is only applicable when both $f(x), g(x)$ go to zero as $x \rightarrow x_{0}$ , a condition which was violated in the previous example. But, even when $f(x)$ and $g(x)$ do go to zero as $x \rightarrow x_{0}$ , there is still a possibility for an incorrect conclusion. For instance, consider the limit

$\lim_{x \rightarrow 0} \frac {x^{2} \sin (x^{-4})}{x}$ .

Both numerator and denominator go to zero as $x \rightarrow 0$ , so it seems pretty safe to apply the rule, to obtain

$\lim_{x \rightarrow 0} \frac {x^{2}\sin (x^{-4})}{x}=\lim_{x \rightarrow 0} \frac {2x \sin (x^{-4})-4x^{-3}\cos (x^{-4})}{1}$ which equals

$\lim_{x \rightarrow 0}2x \sin (x^{-4})-\lim_{x \rightarrow 0}4x^{-3}\cos (x^{-4})$ .

The first limit converges to zero by the Sandwich theorem (since the function $2xsin(x^{-4})$ is bounded above by $2|x|$ and below by $-2|x|$ , both of which go to zero at 0). But the second limit is divergent (because $x^{-3}$ goes to infinity as $x \rightarrow 0$ , and $\cos (x^{-4})$ does not go to zero.) So the limit $\lim_{x \rightarrow 0} \frac {2x \sin(x^{-4})-4x^{-2}\cos (x^{-4})}{1}$ diverges. One might then conclude using $L^{'}H\hat{o}pital's Rule$ that $\lim_{x \rightarrow 0}\frac {x^{2}\sin (x^{-4})}{x}$ also diverges; however, we can clearly rewrite this limit as $\lim_{x \rightarrow 0}x\sin(x^{-4})$ , which goes to zero when $x \rightarrow 0$ by the Sandwich Theorem again. This does not show that $L^{"}H\hat opital's Rule$ is untrustworthy. Indeed, it is quite rigorous, but it still requires some care when applied.

That is all, once again, if you like this, please send a thanks note to Prof. Terence Tao.

More later,

Nalin Pithwa

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