MAT3, Mat-Øk 3, Mat-Tek3 -- Efterår 2014

Analyse 1

Onsdag den 8. oktober 2014,  kl. 08:15
Sted: Rum NJV14, 4-117.

Dagens program

08:15-10:15

Repetition og forelæsning: kontinuerte og differentiable reelle funktioner.

10:30--12:00

Exercises: page 260 in [Lay], exercises 3, 4, 5 and 6.
Hint for exercise 3:
c). Note that $f(1)=1$. Let $x_n$ be any sequence which converges to $1$ and $x_n<1$. Then $(f(x_n)-f(1))/(x_n-1)=3$ for all $n$, thus the left derivative exists and equals the value $3$. Now let $x_n$ be any sequence which converges to $1$ and $x_n > 1$. Then $(f(x_n)-f(1))/(x_n-1)=(x_n^2-1)(x_n-1)=x_n+1$ for all $n$, thus the right derivative exists and equals the value $2$. Hence $f$ is not differentiable at $x=1$.

Hint for exercise 4:
e). Let $c>0$ and write $(f(x)-f(c))/(x-c)=-1/(\sqrt{x}+\sqrt{c}) \times 1/\sqrt{x} \times 1/\sqrt{c}$. Conclude that $f'(c)=-1/2 c^{-3/2}$.

Hint for exercise 5:
a). Use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
b). Since $(f(x)-f(0))(x-0)=1/x^{2/3}$ for $x\neq 0$, we see that if $x_n=1/n$ then $1/(x_n)^{2/3}=n^{2/3}$, which defines a divergent sequence.

Hint for exercise 6:
a). Show that $f'(x)=2x \sin(1/x) -\cos(1/x)$ if $x\neq 0$.
b). We have $(f(x)-f(0))/(x-0)=x\sin(1/x)$ for all $x\neq 0$. The right hand side converges to $0$ when $x$ goes to zero, hence $f'(0)=0$.
c). If $f'$ were continuous at $0$, then from the formula of $f'(x)$ at $x\neq 0$ we could conclude that $\lim_{x\to 0}\cos(1/x)=0$. But this limit does not exist.

Literatur: Vi skal gå gennem afsnit 1 og 2 fra kapitlen 'Differentiation' (starter på side 251) fra [Lay]. De vigtigste sætninger er sætning 1.7 på side 255 (regneregler med differentiable funktioner), sætning 1.10 på side 257 (kædereglen) og sætning 2.2 på side 263 (Rolles sætning).