Mathematical Concept
In mathematics , the indefinite product operator is the inverse operator of
Q
(
f
(
x
)
)
=
f
(
x
+
1
)
f
(
x
)
{\textstyle Q(f(x))={\frac {f(x+1)}{f(x)}}}
. It is a discrete version of the geometric integral of geometric calculus, one of the non-Newtonian calculi.
Thus
Q
(
∏
x
f
(
x
)
)
=
f
(
x
)
.
{\displaystyle Q\left(\prod _{x}f(x)\right)=f(x)\,.}
More explicitly, if
∏
x
f
(
x
)
=
F
(
x
)
{\textstyle \prod _{x}f(x)=F(x)}
, then
F
(
x
+
1
)
F
(
x
)
=
f
(
x
)
.
{\displaystyle {\frac {F(x+1)}{F(x)}}=f(x)\,.}
If F (x ) is a solution of this functional equation for a given f (x ), then so is CF (x ) for any constant C . Therefore, each indefinite product actually represents a family of functions, differing by a multiplicative constant.
Period rule
If
T
{\displaystyle T}
is a period of function
f
(
x
)
{\displaystyle f(x)}
then
∏
x
f
(
T
x
)
=
C
f
(
T
x
)
x
−
1
{\displaystyle \prod _{x}f(Tx)=Cf(Tx)^{x-1}}
Connection to indefinite sum
Indefinite product can be expressed in terms of indefinite sum :
∏
x
f
(
x
)
=
exp
(
∑
x
ln
f
(
x
)
)
{\displaystyle \prod _{x}f(x)=\exp \left(\sum _{x}\ln f(x)\right)}
Alternative usage
Some authors use the phrase "indefinite product" in a slightly different but related way to describe a product in which the numerical value of the upper limit is not given. e.g.
∏
k
=
1
n
f
(
k
)
{\displaystyle \prod _{k=1}^{n}f(k)}
.
Rules
∏
x
f
(
x
)
g
(
x
)
=
∏
x
f
(
x
)
∏
x
g
(
x
)
{\displaystyle \prod _{x}f(x)g(x)=\prod _{x}f(x)\prod _{x}g(x)}
∏
x
f
(
x
)
a
=
(
∏
x
f
(
x
)
)
a
{\displaystyle \prod _{x}f(x)^{a}=\left(\prod _{x}f(x)\right)^{a}}
∏
x
a
f
(
x
)
=
a
∑
x
f
(
x
)
{\displaystyle \prod _{x}a^{f(x)}=a^{\sum _{x}f(x)}}
List of indefinite products
This is a list of indefinite products
∏
x
f
(
x
)
{\textstyle \prod _{x}f(x)}
. Not all functions have an indefinite product which can be expressed in elementary functions.
∏
x
a
=
C
a
x
{\displaystyle \prod _{x}a=Ca^{x}}
∏
x
x
=
C
Γ
(
x
)
{\displaystyle \prod _{x}x=C\,\Gamma (x)}
∏
x
x
+
1
x
=
C
x
{\displaystyle \prod _{x}{\frac {x+1}{x}}=Cx}
∏
x
x
+
a
x
=
C
Γ
(
x
+
a
)
Γ
(
x
)
{\displaystyle \prod _{x}{\frac {x+a}{x}}={\frac {C\,\Gamma (x+a)}{\Gamma (x)}}}
∏
x
x
a
=
C
Γ
(
x
)
a
{\displaystyle \prod _{x}x^{a}=C\,\Gamma (x)^{a}}
∏
x
a
x
=
C
a
x
Γ
(
x
)
{\displaystyle \prod _{x}ax=Ca^{x}\Gamma (x)}
∏
x
a
x
=
C
a
x
2
(
x
−
1
)
{\displaystyle \prod _{x}a^{x}=Ca^{{\frac {x}{2}}(x-1)}}
∏
x
a
1
x
=
C
a
Γ
′
(
x
)
Γ
(
x
)
{\displaystyle \prod _{x}a^{\frac {1}{x}}=Ca^{\frac {\Gamma '(x)}{\Gamma (x)}}}
∏
x
x
x
=
C
e
ζ
′
(
−
1
,
x
)
−
ζ
′
(
−
1
)
=
C
e
ψ
(
−
2
)
(
z
)
+
z
2
−
z
2
−
z
2
ln
(
2
π
)
=
C
K
(
x
)
{\displaystyle \prod _{x}x^{x}=C\,e^{\zeta ^{\prime }(-1,x)-\zeta ^{\prime }(-1)}=C\,e^{\psi ^{(-2)}(z)+{\frac {z^{2}-z}{2}}-{\frac {z}{2}}\ln(2\pi )}=C\,\operatorname {K} (x)}
(see K-function )
∏
x
Γ
(
x
)
=
C
Γ
(
x
)
x
−
1
K
(
x
)
=
C
Γ
(
x
)
x
−
1
e
z
2
ln
(
2
π
)
−
z
2
−
z
2
−
ψ
(
−
2
)
(
z
)
=
C
G
(
x
)
{\displaystyle \prod _{x}\Gamma (x)={\frac {C\,\Gamma (x)^{x-1}}{\operatorname {K} (x)}}=C\,\Gamma (x)^{x-1}e^{{\frac {z}{2}}\ln(2\pi )-{\frac {z^{2}-z}{2}}-\psi ^{(-2)}(z)}=C\,\operatorname {G} (x)}
(see Barnes G-function )
∏
x
sexp
a
(
x
)
=
C
(
sexp
a
(
x
)
)
′
sexp
a
(
x
)
(
ln
a
)
x
{\displaystyle \prod _{x}\operatorname {sexp} _{a}(x)={\frac {C\,(\operatorname {sexp} _{a}(x))'}{\operatorname {sexp} _{a}(x)(\ln a)^{x}}}}
(see super-exponential function )
∏
x
x
+
a
=
C
Γ
(
x
+
a
)
{\displaystyle \prod _{x}x+a=C\,\Gamma (x+a)}
∏
x
a
x
+
b
=
C
a
x
Γ
(
x
+
b
a
)
{\displaystyle \prod _{x}ax+b=C\,a^{x}\Gamma \left(x+{\frac {b}{a}}\right)}
∏
x
a
x
2
+
b
x
=
C
a
x
Γ
(
x
)
Γ
(
x
+
b
a
)
{\displaystyle \prod _{x}ax^{2}+bx=C\,a^{x}\Gamma (x)\Gamma \left(x+{\frac {b}{a}}\right)}
∏
x
x
2
+
1
=
C
Γ
(
x
−
i
)
Γ
(
x
+
i
)
{\displaystyle \prod _{x}x^{2}+1=C\,\Gamma (x-i)\Gamma (x+i)}
∏
x
x
+
1
x
=
C
Γ
(
x
−
i
)
Γ
(
x
+
i
)
Γ
(
x
)
{\displaystyle \prod _{x}x+{\frac {1}{x}}={\frac {C\,\Gamma (x-i)\Gamma (x+i)}{\Gamma (x)}}}
∏
x
csc
x
sin
(
x
+
1
)
=
C
sin
x
{\displaystyle \prod _{x}\csc x\sin(x+1)=C\sin x}
∏
x
sec
x
cos
(
x
+
1
)
=
C
cos
x
{\displaystyle \prod _{x}\sec x\cos(x+1)=C\cos x}
∏
x
cot
x
tan
(
x
+
1
)
=
C
tan
x
{\displaystyle \prod _{x}\cot x\tan(x+1)=C\tan x}
∏
x
tan
x
cot
(
x
+
1
)
=
C
cot
x
{\displaystyle \prod _{x}\tan x\cot(x+1)=C\cot x}
See also
References
Algorithms for Nonlinear Higher Order Difference Equations , Manuel Kauers
Further reading
External links
Categories :
Text is available under the Creative Commons Attribution-ShareAlike License. Additional terms may apply.
**DISCLAIMER** We are not affiliated with Wikipedia, and Cloudflare.
The information presented on this site is for general informational purposes only and does not constitute medical advice.
You should always have a personal consultation with a healthcare professional before making changes to your diet, medication, or exercise routine.
AI helps with the correspondence in our chat.
We participate in an affiliate program. If you buy something through a link, we may earn a commission 💕
↑