• Keine Ergebnisse gefunden

Basismodul Analysis f. MCS-BSc

N/A
N/A
Protected

Academic year: 2022

Aktie "Basismodul Analysis f. MCS-BSc"

Copied!
4
0
0

Wird geladen.... (Jetzt Volltext ansehen)

Volltext

(1)

A TECHNISCHE UNIVERSIT¨ DARMSTADT

Prof. Dr. U. Kohlenbach

AT

Dr. L. Leu¸stean

September 5, 2006 Fachbereich Mathematik

Basismodul Analysis f. MCS-BSc

Please write your na- me on each sheet and number all pages. At the end of the exami- nation, put the sheets with your solutions in the folded examination sheet.

Name: . . . . First name: . . . . Matr.-Nr.: . . . . Studies: . . . . Studienleisteung (Schein)

Pr¨ ufungsleistung (Bachelor)

Important:

• Time for the examination: 240 Minutes. Total amount of points: 100 points.

• Admitted material: 4 handwritten A4 sheets with your signature.

• All steps in the solutions and partial results need sufficient explanation.

• Good luck!

Problem 1 2 3 4 5 6 7 8 9 10 Σ Note

Maximal Points 10 10 10 10 10 10 10 10 10 10 100

Given Points

Analysis I

Problem 1: Logic and Natural Numbers (10 points)

(i) (3 points) Show that the following sentence holds for all nonempty sets X6=∅ and all unary predicates P on X:

(∃x∈X)[P(x)→(∀y∈X)P(y)].

(ii) (3 points) Let f:N→ {0,1}. Show that

(∀n)(∃m > n)(f(m) = 0)∨(∀n)(∃m > n)(f(m) = 1).

(iii) (4 points) Prove by induction that

n

X

i=1

i2=n(n+ 1)(2n+ 1)

6 .

(2)

Problem 2: Functions and Metric Spaces (10 points) (i) (5 points) Let X, Y be nonempty sets and f:X→Y be an arbitrary function. Define x1∼x2:⇐⇒f(x1) =f(x2). Show that ∼ is an equivalence relation on X and that every equivalence relation on X can be obtained in this way (for suitable Y and f).

(ii) (5 points) Let (X, d) be a metric space. Prove that

|d(f, g)−d(u, v)| ≤d(f, u) +d(g, v) for all f, g, u, v∈X.

Problem 3: Sequences and series (10 points)

(i) (5 points) Show that for each sequence (an)n in R

X

k=1

1 2k

|ak|

1 +|ak| converges and that for all sequences (an)n,(bn)n

X

k=1

1 2k

|ak+bk| 1 +|ak+bk|≤

X

k=1

1 2k

|ak| 1 +|ak|+

X

k=1

1 2k

|bk| 1 +|bk| Hint: t7→1+tt is increasing for t >−1.

(ii) (5 points) Let f: [0,1]→[0,1], f(x) = 1−x. For x∈[0,1] define x0:=x, xn+1:=f(xn). For which x∈[0,1] does (xn)n converge?

Problem 4: Continuity (10 points)

(i) (5 points) Determine in which points x∈R the functions f:R→R, f(x) =

(sinx) cos1x

if x6= 0

0 if x= 0

and

g:R→R, g(x) =

(cosx) sin1x

if x6= 0

1 if x= 0

are continuous?

(ii) (5 points) Let f: [0,1]→R be continuous and strictly increasing. Show that f is uniformly strictly increasing in the sense

(∀ε >0)(∃δ >0)(∀x1, x2∈[0,1]) x1+ε≤x2⇒f(x1) +δ≤f(x2) .

Problem 5: Differentiability in R (10 points)

(i) (4 points) Compute the derivatives of the following functions:

(a) f:R→R, f(x) = sin(x5+ 3x); (b) g:]0,∞[→R, g(x) = 2 exp(lnx).

(ii) (3 points) Let f: [a, b]→R be a differentiable function such that f0(x)>0 for all x∈[a, b].

Show that for any c∈R, the equation f(x) =c has at most one solution.

(3)

(iii) (3 points) Determine all differentiable functions f:R→R satisfying (∀x∈R)(f0(x) =f(x)).

Analysis II

Problem 6: Extrema and Taylor series (10 points)

Consider the function f: [0,3]→R, f(x) = 2x3−9x2+ 12x−5.

(i) (4 points) Determine the local minima and maxima of f.

(ii) (3 points) Prove that f has global extrema and determine them.

(iii) (3 points) Prove that for any x∈[0,3], f(x) =f(0) +

3

X

k=1

f(k)(0) k! xk.

Problem 7: Riemann Integral (10 points)

(i) (6 points) Compute the following Riemann integrals:

(a) Z 1

0

x exdx; (b) Z 2

0

|x−1|dx.

(ii) (4 points) Letf: [a, b]→R be a continuous function such that Z b

a

f(x)dx=1

2(b2−a2). Show that there exists x0∈[a, b] such that f(x0) =x0.

Problem 8: Normed spaces (10 points)

(i) (5 points) Suppose that (V,k · kV) and (W,k · kW) are normed spaces over K, where K=R or K=C. Let T :V →W be a linear transformation and define the function k · k:V →R by

kxk:=kxkV +kT(x)kW for all x∈V.

Prove that k · k is a norm on V.

(ii) (5 points) Let V be a normed space over R, A⊆V be closed in V and t∈R, t >0. Define tA:={tx|x∈A}. Prove that tA is closed in V .

Problem 9: Differentiability in Rn (10 points)

(i) (5 points) Let f:Rn→R, f(x1, . . . , xn) =x21+x22+. . .+x2n. Show that f is differentiable on Rn and compute its derivative.

(ii) (5 points) Let g:Rn→R be a differentiable function such that g(tx) =tg(x) for all t∈R and all x∈Rn. Show that g(x) =g0(0)(x) for all x∈Rn.

Problem 10: Inverse Function Theorem (10 points)

Consider the function

f:R2→R2, f(x, y) =

x2−y2 2xy

.

(4)

(i) (4 points) Show that f is continuously differentiable and compute its derivative.

(ii) (3 points) Show that f is locally invertible around every point (x, y)∈R2\ {(0,0)}.

(iii) (3 points) Does f have a global inverse?

Referenzen

ÄHNLICHE DOKUMENTE

At the end of the exami- nation, put the sheets with your solutions in the folded examination

This was partly a consequence of Germany’s strategy of net- worked security (Vernetzte Sicherheit) which sought to prioritize civilian engagement, while keeping Ger- many’s

Bruno Le Maire (france) Minister for Food, Agriculture & Fishing Mark Leonard (united Kingdom) Director, European Council on Foreign Relations. Juan fernando López Aguilar

Sein nächstes Opfer soll die neue Mitschülerin Alyssa (Jessica Barden) sein: Immer wieder stellt er sich vor, wie er das Messer zur Hand nimmt, auf sie einsticht und sie in ihrer

This structure shows the tail por- tion of trigger factor in complex with the large subunit of the ribosome, suggesting the position of intact trigger factor as it might interact

bachelor's or master's thesis, the university shall only accept liability in the event of breaches by the supervisor or examiner. In the event of breaches committed by the

Proof of this is the establishment of the Pan-German League in Germany which was founded as an immediate conse- quence of conflict between the government and nationalists

Session 1 set the scene for the seminar by discussing the economic backdrop that would underpin discussions at the Leaders’ Summit in Brisbane. It was suggested that the