2010 ws analysis 1 philip klausur B 10 PDF

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Σ

n ∈ N0

n!

5! ·

Y

∞ X (k − 1)−i .

k∈{−1,3} i=0

∀

n∈N0

∀

x,y∈R

(y − x)

n X i=0

xi y n−i = y n+1 − xn+1.

f : R −→ R,

f (x) := arctan ln(x2 + 1). ln

arctan

∅= 6 M ⊆R f

g : [−1, 1] −→ R,

f : M −→ R ξ∈M

   |x| sin(cot x) g(x) := 0   42 10

x∈ / {0, 1/42}, x = 0, x = 1/42,

ξ = 0 sin cot

sin : R −→ R f : R × R −→ R

x, y ∈ R

x 6= y

f (x, y) 6= f (y, x)

R limn→∞ M ⊆R

P∞

n=0

an an ∈ R

α∈R

α = sup M

limn→∞ ean = 1 M...