Es soll die Funktion
f
:
R
2
→
R
,
x
↦
exp
(
x
1
−
x
2
)
⋅
log
(
1
−
x
2
)
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ,~x\mapsto \exp(x_{1}-x_{2})\cdot \log(1-x_{2})}
mit
x
=
(
x
1
,
x
2
)
∈
R
2
{\displaystyle x=(x_{1},x_{2})\in \mathbb {R} ^{2}}
a
=
(
a
1
,
a
2
)
=
(
1
,
0
)
∈
R
2
{\displaystyle a=(a_{1},a_{2})=(1,0)\in \mathbb {R} ^{2}}
In diesem Beispiel soll die Funktion bis zum zweiten Grad entwickelt werden. Es gilt also
k
=
2
{\displaystyle k=2}
|
j
|
≤
k
{\displaystyle |j|\leq k}
Multiindexschreibweise die Tupel
(
0
,
0
)
{\displaystyle (0,0)}
(
1
,
0
)
{\displaystyle (1,0)}
(
0
,
1
)
{\displaystyle (0,1)}
(
2
,
0
)
{\displaystyle (2,0)}
(
1
,
1
)
{\displaystyle (1,1)}
(
0
,
2
)
{\displaystyle (0,2)}
Die partiellen Ableitungen der Funktion lauten:
∂
f
∂
x
1
(
a
)
=
[
exp
(
x
1
−
x
2
)
⋅
log
(
1
−
x
2
)
]
x
=
(
1
,
0
)
=
0
{\displaystyle {\frac {\partial f}{\partial x_{1}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0}
∂
f
∂
x
2
(
a
)
=
[
−
exp
(
x
1
−
x
2
)
⋅
(
log
(
1
−
x
2
)
+
1
1
−
x
2
)
]
x
=
(
1
,
0
)
=
−
e
{\displaystyle {\frac {\partial f}{\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e}
∂
2
f
∂
x
1
2
(
a
)
=
[
exp
(
x
1
−
x
2
)
⋅
log
(
1
−
x
2
)
]
x
=
(
1
,
0
)
=
0
{\displaystyle {\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0}
∂
2
f
∂
x
1
∂
x
2
(
a
)
=
[
−
exp
(
x
1
−
x
2
)
⋅
(
log
(
1
−
x
2
)
+
1
1
−
x
2
)
]
x
=
(
1
,
0
)
=
−
e
{\displaystyle {\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e}
∂
2
f
∂
x
2
(
a
)
=
[
exp
(
x
1
−
x
2
)
(
log
(
1
−
x
2
)
+
2
1
−
x
2
−
1
(
1
−
x
2
)
2
)
]
x
=
(
1
,
0
)
=
e
{\displaystyle {\frac {\partial ^{2}f}{\partial x_{2}}}(a)=\left[\exp(x_{1}-x_{2})\left(\log(1-x_{2})+{\frac {2}{1-x_{2}}}-{\frac {1}{(1-x_{2})^{2}}}\right)\right]_{x=(1,0)}=e}
Es folgt mit der mehrdimensionalen Taylor-Formel:
f
(
x
)
≈
f
(
a
)
+
1
1
!
∂
f
∂
x
1
(
a
)
(
x
1
−
a
1
)
+
1
1
!
∂
f
∂
x
2
(
a
)
(
x
2
−
a
2
)
+
1
2
!
∂
2
f
∂
x
1
2
(
a
)
(
x
1
−
a
1
)
2
+
1
1
!
1
!
∂
2
f
∂
x
1
∂
x
2
(
a
)
(
x
1
−
a
1
)
(
x
2
−
a
2
)
+
1
2
!
∂
2
f
∂
x
2
2
(
a
)
(
x
2
−
a
2
)
2
=
0
+
0
−
e
(
x
2
−
0
)
+
0
−
e
(
x
1
−
1
)
(
x
2
−
0
)
+
1
2
e
(
x
2
−
0
)
2
=
−
x
1
x
2
e
+
1
2
x
2
2
e
{\displaystyle {\begin{aligned}f(x)\approx &f(a)+{\frac {1}{1!}}{\frac {\partial f}{\partial x_{1}}}(a)~(x_{1}-a_{1})+{\frac {1}{1!}}{\frac {\partial f}{\partial x_{2}}}(a)~(x_{2}-a_{2})\\&+{\frac {1}{2!}}{\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)~(x_{1}-a_{1})^{2}+{\frac {1}{1!1!}}{\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)~(x_{1}-a_{1})(x_{2}-a_{2})\\&+{\frac {1}{2!}}{\frac {\partial ^{2}f}{\partial x_{2}^{2}}}(a)~(x_{2}-a_{2})^{2}\\&=0+0-e(x_{2}-0)+0-e(x_{1}-1)(x_{2}-0)+{\frac {1}{2}}e(x_{2}-0)^{2}\\&=-x_{1}x_{2}e+{\frac {1}{2}}x_{2}^{2}e\end{aligned}}}
Eine alternative Darstellung der Taylorentwicklung basiert nicht auf der Multiindexdarstellung, sondern auf einer Matizendarstellung der Taylor-Formel.
So gilt auch:
f
(
x
)
≈
f
(
a
)
+
J
f
(
a
)
(
x
−
a
)
+
1
2
(
x
−
a
)
T
H
f
(
a
)
(
x
−
a
)
=
f
(
a
)
+
(
∂
f
∂
x
1
(
a
)
∂
f
∂
x
2
(
a
)
)
(
x
1
−
a
1
x
2
−
a
2
)
+
1
2
(
x
1
−
a
1
x
2
−
a
2
)
(
∂
2
f
∂
x
1
2
(
a
)
∂
2
f
∂
x
2
∂
x
1
(
a
)
∂
2
f
∂
x
1
∂
x
2
(
a
)
∂
2
f
∂
x
2
2
(
a
)
)
(
x
1
−
a
1
x
2
−
a
2
)
=
0
+
(
0
−
e
)
(
x
1
−
1
x
2
)
+
1
2
(
x
1
−
1
x
2
)
(
0
−
e
−
e
e
)
(
x
1
−
1
x
2
)
=
−
x
1
x
2
e
+
1
2
x
2
2
e
{\displaystyle {\begin{aligned}f(x)&\approx f(a)+J_{f}(a)(x-a)+{\frac {1}{2}}(x-a)^{T}H_{f}(a)(x-a)\\&=f(a)+{\begin{pmatrix}{\frac {\partial f}{\partial x_{1}}}(a)&{\frac {\partial f}{\partial x_{2}}}(a)\end{pmatrix}}{\begin{pmatrix}x_{1}-a_{1}\\x_{2}-a_{2}\end{pmatrix}}\\&\qquad +{\frac {1}{2}}{\begin{pmatrix}x_{1}-a_{1}&x_{2}-a_{2}\end{pmatrix}}{\begin{pmatrix}{\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)&{\frac {\partial ^{2}f}{\partial x_{2}\partial x_{1}}}(a)\\{\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)&{\frac {\partial ^{2}f}{\partial x_{2}^{2}}}(a)\end{pmatrix}}{\begin{pmatrix}x_{1}-a_{1}\\x_{2}-a_{2}\end{pmatrix}}\\&=0+{\begin{pmatrix}0&-e\end{pmatrix}}{\begin{pmatrix}x_{1}-1\\x_{2}\end{pmatrix}}+{\frac {1}{2}}{\begin{pmatrix}x_{1}-1&x_{2}\end{pmatrix}}{\begin{pmatrix}0&-e\\-e&e\end{pmatrix}}{\begin{pmatrix}x_{1}-1\\x_{2}\end{pmatrix}}\\&=-x_{1}x_{2}e+{\frac {1}{2}}x_{2}^{2}e\end{aligned}}}
mit der Jacobi-Matrix
J
f
{\displaystyle J_{f}}
Hesse-Matrix
H
f
{\displaystyle H_{f}}
![{\displaystyle {\frac {\partial f}{\partial x_{1}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80a34287647244879f4bda63a4390809aa833d72)
![{\displaystyle {\frac {\partial f}{\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55df47c8d0210eccaca74dfbca03dc43fca5b193)
![{\displaystyle {\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5198d1f21234848dbe1c5d6be6e181b9a802a337)
![{\displaystyle {\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca8d1453d75beb6f7ea1452ba92794e96eee3ff3)
![{\displaystyle {\frac {\partial ^{2}f}{\partial x_{2}}}(a)=\left[\exp(x_{1}-x_{2})\left(\log(1-x_{2})+{\frac {2}{1-x_{2}}}-{\frac {1}{(1-x_{2})^{2}}}\right)\right]_{x=(1,0)}=e}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bb4f30e6b1493bfd8f4f428a37b141a7bb3d5ce)
