![](https://pdfstore-manualsonline.prod.a.ki/pdfasset/c/3a/c3a0bcf5-a285-451a-ad11-c09255eee3e1/c3a0bcf5-a285-451a-ad11-c09255eee3e1-bg38.png)
3-26
k Σ Calculations [OPTN]-[CALC]-[Σ (]
To perform Σ calculations, first display the function analysis menu, and then input the values
using the syntax below.
K4(CALC)6(g)3(Σ ( )
ak , k ,
α
,
β
, n !/( ) )
(
n: distance between partitions)
Example To calculate the following:
Use
n = 1 as the distance between partitions.
AK4(CALC)6(g)3(Σ ( )
a3(K-O)1(K)x-d1(K)
+f,1(K),c,g,b
!/( ) )w
Σ Calculation Precautions
• The value of the specified variable changes during a Σ calculation. Be sure to keep separate
written records of the specified variable values you might need later before you perform the
calculation.
• You can use only one variable in the function for input sequence
ak.
• Input integers only for the initial term (
α
) of sequence ak and last term (
β
) of sequence ak.
• Input of
n and the closing parentheses can be omitted. If you omit n, the calculator
automatically uses n = 1.
• Make sure that the value used as the final term
β
is greater than the value used as the initial
term
α
. Otherwise, an error will occur.
• To interrupt an ongoing Σ calculation (indicated when the cursor is not on the display), press
the A key.
• You cannot use a differential, quadratic differential, integration, Σ , maximum/minimum value,
Solve, RndFix or log
a
b calculation expression inside of a Σ calculation term.
k Maximum/Minimum Value Calculations [OPTN]-[CALC]-[FMin]/[FMax]
After displaying the function analysis menu, you can input maximum/minimum calculations
using the formats below, and solve for the maximum and minimum of a function within interval
a < x < b.
u Minimum Value
K4(CALC)6(g)1(FMin) f (x) , a , b , n !/( ) )
(
a: start point of interval, b: end point of interval, n: precision ( n = 1 to 9))
β
Σ
(
a
k
,
k
,
α
,
β
,
n
)
=
Σ
a
k
=
a
α
+
a
α
+1
+........+
a
β
k =
α
β
Σ
(
a
k
,
k
,
α
,
β
,
n
)
=
Σ
a
k
=
a
α
+
a
α
+1
+........+
a
β
k =
α
6
Σ
(
k
2
–3
k
+5)
k = 2
6
Σ
(
k
2
–3
k
+5)
k = 2