Equation Solver (plain DB2)

Lenny77 · #1 (**permalink**) 08-07-09, 15:38

Using the DB2 power it is a very easy to create the simple Equation Solver.

For example you have to find the root of Equation 2 * SIN(X) = X.

Transform this equation to f(X) = 2 * SIN(X) - X = 0.

One root could be find between 1 and 2 because of
f(1) * f(2) = (2 *Sin(1) - 1) * (2 * Sin(2) - 2) = -1.23889187945779E-001 < zero.

So Xleft = 1 and Xright = 2....

When we find the boundaries we will iterate by finding the middle of the interval:

So the first iteration of the X will be (1 + 2) / 2 = 1.5.

Then we find where f(X) * f(Xright), or f(X) * f(Xleft) is less then 0.

If f(X) * f(Xright) > 0 we change Xright on (Xleft + Xright) / 2,
If f(X) * f(Xleft) > 0 we change Xleft on (Xleft + Xright) / 2.

We will finish the process when abs (Xleft - Xright) <= eps

Quote:

with control_tbl (Equation, Xleft, Xright, eps) as
(select '2 * sin(X) = X ==> 2 * sin(X) - X = 0', double(1), double(2),
double(1.e-12)
from sysibm.sysdummy1
)
,
eq_solution (Equation, Xleft, Xright, eps, X) as
(select ct.*, (Xleft + Xright) / 2
from control_tbl ct
union all
select Equation,
case
when (2 * sin ((Xleft + Xright) / 2) - (Xleft + Xright) / 2) *
(2 * sin (Xleft) - Xleft) > 0
then (Xleft + Xright) / 2
else Xleft
end,
case
when (2 * sin ((Xleft + Xright) / 2) - (Xleft + Xright) / 2) *
(2 * sin (Xright) - Xright) > 0
then (Xleft + Xright) / 2
else Xright
end,
eps,
(Xleft + Xright) / 2
from eq_solution
where abs(Xleft - Xright) > eps
)
select Equation, abs(Xleft - Xright) "Real Accurancy", X "Equation Solution"
from eq_solution
where abs(Xleft - Xright) <= eps

Lenny K.

Lenny77 · #2 (**permalink**) 08-10-09, 10:50

We can make better and compacter view with the small transform of the SQL statement:

Quote:

with control_tbl (Equation, Xleft, Xright, eps) as
(select '2 * sin(X) = X ==> 2 * sin(X) - X = 0',
double(-2), double(-1), double(1.e-12)
from sysibm.sysdummy1
)
,
eq_solution (Equation, Xleft, Xright, eps, X) as
(select ct.*, (Xleft + Xright) / 2
from control_tbl ct
union all
select Equation,
case
when (md.E_Function * lf.E_Function ) > 0
then (Xleft + Xright) / 2
else Xleft
end,
case
when (md.E_Function * rt.E_Function ) > 0
then (Xleft + Xright) / 2
else Xright
end,
eps,
(Xleft + Xright) / 2
from eq_solution eq,
table(select 2 * sin (X) - X as E_Function
from (select Xleft as X from sysibm.sysdummy1 ) ii) lf,
table(select 2 * sin (X) - X as E_Function
from (select Xright as X from sysibm.sysdummy1 ) ii) rt,
table(select 2 * sin (X) - X as E_Function
from (select (Xleft + Xright) / 2 as X from sysibm.sysdummy1 ) ii) md

where abs(Xleft - Xright) > eps
)
select Equation, abs(Xleft - Xright) "Real Accurancy", X "Equaty Solution"
from eq_solution
where abs(Xleft - Xright) <= eps

Result: X = -1.89549426703434

Lenny

DB2Plus · #3 (**permalink**) 10-03-09, 22:52

Not bad, I have checked.

But what we have to do with equation
SIN(X) = 0 on interval [p/2, p/2 + 2pi] having the same sign on the ends:

SIN(pi/2) = +1 and SIN(p/2 + 2pi) = +1 ?

Same time this equation has 2 roots on this interval:
SIN(pi) = 0 and SIN(2pi) = 0 are solutions.

Kara S.

Lenny77 · #4 (**permalink**) 10-05-09, 16:08

Quote:

Originally Posted by DB2Plus

Not bad, I have checked.

But what we have to do with equation
SIN(X) = 0 on interval [p/2, p/2 + 2pi] having the same sign on the ends:

SIN(pi/2) = +1 and SIN(p/2 + 2pi) = +1 ?

Same time this equation has 2 roots on this interval:
SIN(pi) = 0 and SIN(2pi) = 0 are solutions.

Kara S.

This change is not too complicated I think.

Problem is in definition of step length to locate the root.

Lenny.

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