9. Exam preparation

Additional homework (only PL): problems, solutions.

Part 1.: Problems, solutions.
Part 2.: Problems, solutions.

Correction test, solutions.

Below I define notion, which are used in the above problems, which have not appeared till now during the classes.

Cyclometric functions

In the same way as the logarithm is the inverse function to the exponential function, we can define the inverse functions for the trigonometric functions, i.e. arcus sinus, arcus cosinus, arcus tangens:

  • \arcsin(x)=y iff \sin(y)=x,
  • \arccos(x)=y iff \cos(y)=x,
  • \text{arctg}(x)=y iff \text{tg}(y)=x,

Derivative of the inverse function

If f is strictly monotone and continuous, then (f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))} if derivative f exists in f^{-1}(y) and is non-zero.

E.g. let f(x)=x^3. Then f^{-1}(y)=\sqrt[3]{y}. Therefore:

    \[(\sqrt[3]{y})'=\frac{1}{f'(\sqrt[3]{y})}=\frac{1}{3(\sqrt[3]{y})^2}=\frac{1}{3} y^{-\frac{2}{3}}.\]

Derivatives and inequalities

Using derivatives one can prove also some inequalities. If f,g are functions, and f(x_0)\leq g(x_0), and f'(x)\leq g'(x) na [x_0,x_1], then f(x)\leq g(x) on (x_0,x_1). Indeed, if at the beginning of the interval f is below g, and increases slower than g, then it is below in the whole interval.

E.g. e^x\geq x+1 on (0,\infty), because e^0=1=0+1, and (e^x)'=e^x\geq 1=(x+1)' on (0,\infty).

Lipschitz’s condition

We shall say that a function satisfies Lipschitz’s condition on the interval X, if there exists L\in\mathbb{R}, such that for all x,y\in X, we have ∣f(x)−f(y)∣\leq L∣x-y∣. We see that Lipschitz condition is stronger than continuity.

Uniform continuity

If we are able to choose \delta for each \varepsilon, in a such way that in any interval of length \delta values of functions do not differ more than \varepsilon, universally regardless of place x, then we say that the function if uniformly continuous. More formally a function f is uniformly continuous, if:

    \[\forall_{\varepsilon>0}\exists_{\delta}\forall_{x_1,x_2} |x_1-x_2|\leq\delta\rightarrow |f(x_1)-f(x_2)|\leq\varepsilon.\]

Function f(x)=x is a very simple example. It is uniformly continuous, because for any \varepsilon set \delta=\varepsilon. Then for all x_1,x_2 such that |x_1-x_2|\leq\delta, then obviously |f(x_1)-f(x_2)|\leq\varepsilon.

On the other hand, function g(x)=\sin\frac{1}{x} on the interval (0,\infty) is continuous but is not uniformly continuous, because if \varepsilon=1, regardless how small \delta we choose, we can find 0<x_1<x_2<\delta, such that \sin\frac{1}{x_1}=-1, \sin\frac{1}{x_2}=-1, and therefore |f(x_1)-f(x_2)|=2>1.