HS17 (Imamoglu)

1.

c)

Use the squeeze theorem:

• $0\le |x^4\cdot \sin (1/x)|\le x^4$.
• As both $0\to 0$ and $x^4\to 0$, $a_n$ must converge to 0 as well.

2

Since $f$ is strictly monotone decreasing, it is injective.
Dadurch ist er auf dem Bild von $f((0,+\infty ))$ bijektiv.

Umkehrsatz

Da $f:(0,\infty )\to \mathbb{R}$ stetig und strikt monoton fallend ist, gilt für $J:=\text{Im}(f)$

• $f$ ist eine Bijektion auf $I\to J$.
• es existiert $f^{-1}:J\to (0,\infty )$ stetig und strikt fallend

3

Keep the following in mind!

Simplify $z,\bar{z}\in C$ Solutions of DiffEq

If $C_1{e}^{ax+ibx}+C_2{e}^{ax-ibx}$ is a solution of a differential equation, we can simplify it to $e^{ax}(D_1\cos (bx)+D_2\sin (bx))$.

Proof: Expand using Euler formula and then group terms (note we group the $i$ with the $\sin$ constants, giving us a complex constant).

This makes solving DiffEqs involving cosines and sines much easier!

4

a)

For a factor $(ax+b{)}^k$, there are $k$ unknowns, over the powers $1,\dots ,k$ → keep this in mind for partial fractions.

Then don’t make a stupid mistake integrating!

b)

Don’t fuck up IBP → the $[\dots {]}_a^b$ term has one factor being integrated. Then the integral has the integrated factor * the other derived!

c)

Use double substitution.
First we factor out the $e^2$ and get that out to have a clean $\frac{1}{\sqrt{e^{x-2}-1}}$.

• first sub: $u=x-2$ → makes sense, since $a=2$, we get a perfect $a=0$.
• second sub $t=\sqrt{e^u-1}$ for $t$
  • leaves us with $\int \frac{1}{t^2+1}$ = arctan!