Warm-up-exercises Let
x
∈
R
{\displaystyle {}x\in \mathbb {R} }
f
:
R
+
⟶
R
,
t
⟼
f
(
t
)
=
t
x
e
−
t
.
{\displaystyle f\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} ,t\longmapsto f(t)=t^{x}e^{-t}.}
Determine the extrema of this function.
Prove that for the factorial function the relationship
Fac
(
2
k
−
1
2
)
=
∏
i
=
1
k
(
2
i
−
1
)
2
k
⋅
π
{\displaystyle {}\operatorname {Fac} \,{\left({\frac {2k-1}{2}}\right)}={\frac {\prod _{i=1}^{k}(2i-1)}{2^{k}}}\cdot {\sqrt {\pi }}\,}
holds.
a) Prove that for
x
≥
1
{\displaystyle {}x\geq 1}
∫
1
∞
t
x
e
−
t
d
t
≤
1
{\displaystyle {}\int _{1}^{\infty }t^{x}e^{-t}dt\leq 1\,}
holds.
b)
Prove that the function
H
(
x
)
{\displaystyle {}H(x)}
H
(
x
)
=
∫
1
∞
t
x
e
−
t
d
t
{\displaystyle {}H(x)=\int _{1}^{\infty }t^{x}e^{-t}dt\,}
for
x
≥
1
{\displaystyle {}x\geq 1}
c) Prove that
10
!
≥
e
11
+
1
{\displaystyle {}10!\geq e^{11}+1}
d) Prove that for the factorial function for
x
≥
10
{\displaystyle {}x\geq 10}
Fac
x
≥
e
x
{\displaystyle {}\operatorname {Fac} \,x\geq e^{x}\,}
holds.
Solve the initial value problem
y
′
=
sin
t
with
y
(
π
)
=
7
{\displaystyle y'=\sin t{\text{ with }}y(\pi )=7}
Solve the initial value problem
y
′
=
3
t
2
−
4
t
+
7
with
y
(
2
)
=
5
{\displaystyle y'=3t^{2}-4t+7{\text{ with }}y(2)=5}
Find all the solutions for the ordinary differential equation
y
′
=
y
.
{\displaystyle {}y'=y\,.}
Convince yourself that in a location-independent differential equation (i.e.
f
(
t
,
y
)
{\displaystyle {}f(t,y)}
y
{\displaystyle {}y}
y
1
{\displaystyle {}y_{1}}
y
2
{\displaystyle {}y_{2}}
y
1
(
t
)
−
y
2
(
t
)
{\displaystyle {}y_{1}(t)-y_{2}(t)}
Hand-in-exercises Prove that for the factorial function the relationship
Fac
x
=
∫
0
1
(
−
ln
t
)
x
d
t
{\displaystyle {}\operatorname {Fac} \,x=\int _{0}^{1}(-\ln t)^{x}dt\,}
holds.
Solve the initial value problem
y
′
=
3
t
3
−
2
t
+
5
with
y
(
3
)
=
4
{\displaystyle y'=3t^{3}-2t+5{\text{ with }}y(3)=4}
Find a solution for the ordinary differential equation
y
′
=
t
+
y
.
{\displaystyle {}y'=t+y\,.}
Solve the initial value problem
y
′
=
t
3
t
2
+
1
with
y
(
1
)
=
2
{\displaystyle y'={\frac {t^{3}}{t^{2}+1}}{\text{ with }}y(1)=2}
