Learned techniques

Analysis I FS24 f)

Let sequence $x_n\to 0$ and $x_n{y}_n\to L$, then

$$\underset{n\to \infty }{\lim }(1+x_n{)}^{y_n}=e^L$$

$e$.$g$. ${\lim }_{n\to \infty }{\left(1-\frac{1}{n}\right)}^n=e^{-1}$ because ${\lim }_{n\to \infty }-\frac{1}{n}\cdot n=-1$

Analysis I FS24 o)

Counterexample

A Smooth Function That Is Not Equal to Its Taylor Series

Consider the function

$$f(x)=\{\begin{array}{ll}{e}^{-1/x^2},& x\ne 0,\\ 0,& x=0.\end{array}$$

This function is infinitely differentiable (smooth) on $\mathbb{R}$.

Smoothness at $x=0$

Although the expression $e^{-1/x^2}$ looks singular at $x=0$, the exponential term decays to zero faster than any power of $x$ as $x\to 0$.

In fact, for every $n\in \mathbb{N}$,

$$\underset{x\to 0}{\lim }\frac{e^{-1/x^2}}{|x{|}^n}=0.$$

From this one can show that

$$f^{(k)}(0)=0\text{for all }k\in \mathbb{N}.$$

So all derivatives of $f$ at $0$ vanish. The Taylor series of $f$ at $0$ is therefore $0$

However, For every $x\ne 0$, $f(x)=e^{-1/x^2}>0.$ Hence, in any neighborhood of $0$, the function is not identically zero. Therefore,

$$f(x)\ne \sum _{k=0}^{\infty }\frac{f^{(k)}(0)}{k!}{x}^k\text{for }x\ne 0.$$

Such a function is called smooth but not analytic at $x=0$.

Analysis I FS24 t)

Counterexample

Convergent Improper Integral Does Not Imply $f(x)\to 0$ A standard counterexample is

$$f(x)=\sin (x^2).$$

The improper integral

$${\int }_0^{\infty }\sin (x^2)dx$$

converges. This is a Fresnel-type integral. However, ${\lim }_{x\to \infty }\sin (x^2)$ does not exist.

The only necessary conclusion is weaker:

$$\forall \epsilon >0,\forall N\in \mathbb{N},\exists x\in [N,\infty )\text{such that}|f(x)|<\epsilon .$$

In words: no matter how far out we go, the function must still get arbitrarily close to $0$ somewhere. But it does not have to stay close to $0$.

Analysis I FS24 q)

We define $f,g:[0,1]\to \mathbb{R}$ with

$$d(f,g)={\int }_0^1|f(x)-g(x)|dx.$$

Observation:

$$d(f\cdot g,0)\le \sqrt{d(f^2,0)d(g^2,0)}$$

the inequality:

$${\int }_0^1|f(x)g(x)|dx\le \sqrt{{\int }_0^{1}f(x{)}^{2}dx}\sqrt{{\int }_0^{1}g(x{)}^{2}dx}.$$

This is the Cauchy-Schwarz inequality.

Interpretation: functions behave like infinite-dimensional vectors.

$$\left|\sum _{i=1}^n{u}_i{v}_i\right|\le \sqrt{\sum _{i=1}^n{u}_i^2}\sqrt{\sum _{i=1}^n{v}_i^2}$$

and we defined the inner product as

$$⟨f,g⟩={\int }_0^{1}f(x)g(x)dx$$