6 Gauge Freedom of Entropies on q-Gaussian Measures
133
d
n+1
dτ n+1 exp q,a (τ )
=
d
dτ
⎛
⎝ exp q,a (τ )
(n−1)(q−1)+q
(−aτ )
n(1−a)
a
n−1
j=0
b
n
j · (−aτ )
−
j
a
⎞
⎠
=
d
dτ
exp q,a (τ )
(n−1)(q−1)+q
× (−aτ )
n(1−a)
a
n−1
j=0
b
n
j · (−aτ )
−
j
a
+ exp q,a (τ )
(n−1)(q−1)+q
×
d
dτ
⎛
⎝ (−aτ )
n(1−a)
a
n−1
j=0
b
n
j · (−aτ )
−
j
a
⎞
⎠
= {(n − 1)(q − 1) + q} exp q,a (τ )
(n−1)(q−1)+q−1
· exp q,a (τ )
q
(−aτ )
1−a
a
× (−aτ )
n(1−a)
a
n−1
j=0
b
n
j · (−aτ )
−
j
a
+ exp q,a (τ )
(n−1)(q−1)+q
×
⎧
⎨
⎩
−a
n−1
j=0
n(1 − a) − j
a
b
n
j · (−aτ )
n(1−a)− j
a
−1
⎫
⎬
⎭
= exp q,a (τ )
n(q−1)+q
(−aτ )
(n+1)(1−a)
a
×
⎡
⎣ {(n − 1)(q − 1) + q}
n−1
j=0
b
n
j (−aτ )
−
j
a
− exp q,a (τ )
1−q
n−1
j=0
{n(1 − a) − j} b
n
j (−aτ )
−
j+1
a
⎤
⎦ .
We deduce from exp q,a (τ )
1−q
= 1 − (1 − q)(−aτ )
1
a that
exp q,a (τ )
1−q
n−1
j=0
{n(1 − a) − j} b
n
j · (−aτ )
−
j+1
a
=
n−1
j=0
{n(1 − a) − j} b
n
j · (−aτ )
−
j+1
a − (1 − q)
n−1
j=0
{n(1 − a) − j} b
n
j · (−aτ )
−
j
a .
This completes the proof of the lemma.
Remark 3 For q ∈ R and a ∈ R \ {0}, we have that