Übungsblatt 10

Übungsblatt 10
Analysis 1, HS14
Ausgabe Donnerstag, 13. November.
Abgabe Donnerstag, 20. November. Bitte Lösungen bis spätestens 17 Uhr im Briefkasten des jeweiligen Übungsleiters am J- oder K-Geschoss von Bau Y27 einreichen.
Übung 1.
i) Von Korollar 8.4.I.ii weiss man dass cos : [0, π] → [−1, 1] bijektiv ist, so dass es ein
Inverses hat welches wir arcus cosinus (arccos) nennen. Bestimme dessen Urbild
und Bild.
ii) Zeige auf gleicher Weise wie für 8.4.I.ii dass der Sinus bijektiv ist wenn er auf geeigneter Weise eingeschränkt wird (wie?) und zeige damit die Existenz der inversen
Funktion arcus sinus (arcsin).
sin(x)
iii) Zeige schliesslich die Existenz des Inversen arctan von tan(x) = cos(x)
.
Hinweis: Benütze die Definition von lim f (x) = ±∞ um den Zwischenwertsatz
x→x0
auf ähnlicher Weise wie im Beweis der Surjektivität vom Cosinus anwenden zu
können.
Lösung 1.
i) We know that cos is bijective as a function from [0, π] to [−1, 1]. Thus arccos is a
function from [−1, 1] to [0, π].
ii) Let us first look at the range [0, π2 ].
pThere cos is strictly monotonically decreasing
from 1 to 0. Thus using sin(x) = 1 − cos2 (x) as in the proof of (8.4.I.ii) we see
that the sine is strictly monotonically increasing from 0 to 1. Now we know from
(8.4.E.ii) that sin is an odd function. Thus since
x > y ⇒ sin(x) > sin(y)
for x, y ∈ [0, π2 ] it follows that
−x < −y ⇒ sin(−x) = − sin(x) < − sin(y) = sin(−y)
and sin is also strictly monotonically increasing on [− π2 , 0] from −1 to 0. Thus sin
is injective.
1
We have from the proof of (8.4.I.ii) that the sine is continuous, sin( π2 ) = 1 and
sin(− π2 ) = −1. From the intermediate value theorem it follows that sin takes
every value between −1 and 1 (it also follows via sin2 (x) + cos2 (x) = 1 from the
surjectivity of cos) and so is surjective. In conclusion sin : [− π2 , π2 ] → [−1, 1] is
bijective and its inverse is a function from [−1, 1] to [− π2 , π2 ].
iii) We first show that tan is strictly monotonically increasing on [0, π2 [.
x > y ⇒ sin(x) > sin(y), cos(x) < cos(y)
sin(x)
sin(y)
⇒
>
cos(x)
cos(y)
sin(−x)
sin(x)
Since tan is also odd: tan(−x) = cos(−x)
= −cos(x)
= − tan(x)
it follows by the same argument than in ii) that tan is strictly monotonically
increasing and thus injective on ] − π2 , π2 [. Furthermore we suspect that at the
boundaries it grows unbounded to −∞ respectively to +∞. We thus next need
to show that tan takes all values in R. Let A ∈ R. We consider x > 0, thus
A
1
∃m, 0 < m < 1 : sin(x) > m. Let M = m
and such that < M
. Then from the
continuity of cos follows according to (7.5.7) that for all sequences (xn )n∈N ∈ [0, π2 [
which converge to π2 :
lim cos(x) = 0 ⇔ ∃N ∈ N : cos(xn ) < x→ π2
1
1
> >M
cos(xn )
sin(xn )
> mM = A ∀n > N.
cos(xn )
sin(0)
= 0 it follow from the continuity of tan and the intermediate
Since tan(0) = cos(0)
value theorem that on [0, xN +1 ] tan(x) takes all values between 0 and some number
bigger than A, including A. The same steps can be performed for x < 0. Since this
holds for all A ∈ R it follows that tan is surjective and thus bijective. The inverse
is a function from R to ] − π2 , π2 [.
Übung 2.
i) Benütze cos(x) = 21 (eix + e−ix ) und den Binomialsatz um einen geschlossenen Ausdruck für cosn (x), x ∈ R, n ∈ N in Abhängigkeit von cos(kx), k ∈ N zu finden.
Bemerkung: Natürlich existiert auch ein entsprechender Ausdruck für sinn (x).
ii) Analog zu sin(x) und cos(x) werden sinus hyperbolicus und cosinus hyperbolicus
definiert durch:
2
ex − e−x
2
ex + e−x
cosh(x) =
.
2
sinh(x) =
Zeige für x, y ∈ R:
cosh2 (x) − sinh2 (x) = 1
cos(x + iy) = cos(x) cosh(y) − i sin(x) sinh(y)
sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y)
wobei Sinus und Cosinus einer komplexen Zahl z ∈ C definiert sind durch
eiz − e−iz
2i
iz
e + e−iz
cos(z) =
.
2
sin(z) =
Lösung 2.
i) We have
2n cosn (x) = (eix + e−ix )n
=
=
=
n
X
k=0
n
X
k=0
n
X
k=0
!
n
(eix )k (e−ix )n−k
k
!
n ix(2k−n)
e
k
!
n cos((2k − n)x) + i sin((2k − n)x)
k
Where 2k − n ranges from −n to n. Now since sin(x) is odd (see 4i) ) we have
i
n
X
k=0
bnc
!
!
!
2 X
n
n
n
sin((2k − n)x) = i
sin((2k − n)x) +
sin(−(2k − n)x)
k
k
n−k
k=0
=i
bn
c
2 X
k=0
!
For n odd, where for n even for k =
term. We remain with
2n cosn (x) =
!
n
n
sin((2k − n)x) −
sin((2k − n)x) = 0
k
k
n
2
n
X
⇒ 2k − n = 0 there is an additional zero
!
n
cos((2k − n)x)
k
k=0
n
n
1 X
⇒ cos (x) = n
cos((2k − n)x).
2 k=0 k
!
n
3
ii) We have
1 2x
e + 2e0 + e−2x − e2x + 2e0 − e−2x
4
1
= (4) = 1
4
1 ix
cos(x) cosh(x) − i sin(x) sinh(x) =
(e + e−ix )(ey + e−y ) − (eix − e−ix )(ey − e−y )
4
1 ix+y
e
+ eix−y + e−ix+y + e−ix−y
=
4
cosh2 (x) − sinh2 (x) =
− eix+y + eix−y + e−ix+y − e−ix−y
1 ix−y
2e
+ 2e−ix+y
4
1 i(x+iy)
=
e
+ e−i(x+iy) = cos(x + iy).
2
=
i
− (eix − e−ix )(ey + e−y ) + (eix + e−ix )(ey − e−y )
4
1
=
− eix+y − eix−y + e−ix+y + e−ix−y
4
sin(x) cosh(x) + i cos(x) sinh(x) =
+ eix+y − eix−y + e−ix+y − e−ix−y
1 ix−y
e
− e−ix+y
2i
1 i(x+iy)
e
− e−i(x+iy) = sin(x + iy).
=
2i
=
Übung 3.
i) Bestimme den Differenzenquotient lim
y→x
f (x)−f (y)
x−y
oder lim
h→0
f (x+h)−f (x)
h
der folgen-
den Funktionen oder zeige dass sie nicht differenzierbar sind:
1) f1 : R → R , x 7→ ax , a ∈ R+
2) f2 : R → R+ ∪ {0} , x 7→ |x|
3) f3 : R+ ∪ {0} → R+ ∪ {0} , x 7→
√
x
ii) Es können nun alle zur Verfügung stehenden Sätze benützt werden um die Ableitung (dort wo sie existiert) folgender Funktionen zu bestimmen:
1) f4 : R → R , x 7→
x2 −1
3x
2) f5 : R → [−1, 1] , x 7→ cos(sin(x))
3) f6 : R → R , x 7→ xx
4
Lösung 3.
i1)
ax+h − ax
ax (ah − 1)
=
h
h
ax+h − ax
lim
= log(a)ax
h→0
h
using (8.3.8.v)
i2)
|x + h| − |x|
1 if x > 0
=
−1 if x < 0
h→0
h
independently of the sign of h, since in the limit x + h and x come arbitrarily close
to each other, and thus have the same sign for a finite x. So f2 is differentiable at
x 6= 0 and f20 (x) = 1 if x > 0, f20 (x) = −1 if x < 0. However,
lim
|0 + h| − |0|
=
h
1 if h > 0
−1 if h < 0
Since the limits of the above function as h ↑ 0 and h ↓ 0 are different, the limit as
as h → 0 does not exist. Thus |x| is not differentiable at x = 0.
i3)
√
x+h−
h
√
x
h
√
√
h( x + h + x)
1
=√
√
x+h+ x
√
√
x+h− x
1
lim
= √ , x 6= 0
h→0
h
2 x
√
and the limit does not exist at x = 0, so x is not differentiable at x = 0
=
ii1)
x 2 − 1 0
3x
(x2 − 1)0 (3x) − (x2 − 1)(3x)0
(3x)2
(2x)(3x) − (x2 − 1)(3)
=
9x2
2
x −1
=
.
3x2
=
(quotient rule)
ii2)
cos(sin(x))0 = cos(y)0 y=sin(x) sin(x)0
= − sin(y)y=sin(x) cos(x)
= − sin(sin(x)) cos(x).
5
(chain rule)
ii3)
(xx )0 = ex log(x)
0
0 = ey y=x log(x)
(x log(x))0
(chain rule)
= ey y=x log(x) (log(x) + 1)
= ex log(x) (log(x) + 1)
= xx (log(x) + 1).
Übung 4.
i) Zeige, dass der Sinus Nullstellen hat bei x = kπ , k ∈ Z, und eins ist bei x =
π
2 + 2kπ , k ∈ Z.
ii) Sei f : R → R definiert als:
(
f (x) =
0
|x|a sin( x1 )
, falls x = 0
, falls x 6= 0.
Zeige an der Stelle x = 0 dass für a = 0 die Funktion nicht stetig ist, für a =
1 die Funktion stetig aber nicht differenzierbar ist und für a = 2 die Funktion
differenzierbar ist.
Lösung 4.
i) From cos2 + sin2 = 1 we know that the sine is zero whenever the cosine is one or
minus one, that is, for all 2πk, k ∈ Z (cos = 1) and all π + 2πk, k ∈ Z (cos = −1).
Thus the sine is zero for all x = kπ , k ∈ Z. Furthermore the sine is ∈ {1, −1}
when cos is zero, that is for π2 + kπ, k ∈ Z. We know from the proof of 8.4.I.ii that
sin( π2 ) = 1. Since sin(−x) = − sin(x) we have sin(− π2 ) = −1. From Periodicity
it follows then that sin(x) = 1, x = π2 + 2πk, k ∈ Z and sin(x) = −1, x =
3π
2 + 2πk, k ∈ Z.
ii)
a = 0: We find 2 sequences, such that sin( x1n ) = 1 and sin( y1n ) = −1 ∀n ∈ N.
1
1
<
,
2πn
+ 2πn
1
1
yn = 3π
<
.
2πn
2 + 2πn
xn =
π
2
Since xn , yn > 0 with the theorem of gend’armes we see that lim xn = 0 =
n→∞
lim yn . However lim sin( x1n ) = 1 6= −1 = lim sin( y1n ). Thus we see that
n→∞
n→∞
n→∞
lim sin( x1 ) does not exist and thus sin( x1 ) is not continuous at x = 0.
x→0
6
a = 1 Since the sine is bounded: −1 ≤ sin( x1 ) ≤ 1 we have −x ≤ x sin( x1 ) ≤ x.
Therefore, taking arbitrary and choosing δ = we have for each x satisfying
|x − 0| = |x| < δ that
|f (x) − f (0)| = |f (x)| ≤ |x| < δ = .
showing that x sin( x1 ) is continuous at x = 0. However evaluating the difference quotient
x sin( x1 ) − 0
x sin( x1 )
1
=
= sin( )
x−0
x
x
we get back to sin( x1 ) from which we know from the a = 0 case that it doesn’t
converge for x going to zero. Thus x sin( x1 ) is not differentiable.
a = 2 We build the difference quotient:
x2 sin( x1 ) − 0
x2 sin( x1 )
1
=
= x sin( )
x−0
x
x
In a similar way as for the a = 1 case, we notice that −x ≤ x sin( x1 ) ≤ x.
Thus for any sequence converging to zero:
lim xn = 0 ⇔ ∀ > 0 ∃N ∈ N : |xn | < ∀n > N
n→∞
1
)| ≤ |xn | < xn
1
⇒ lim x sin( ) = 0.
x→0
x
|xn sin(
Thus x2 sin( x1 ) is differentiable at x = 0 and its derivative there is zero.
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