simultaneous sine functions

Posted on 2006-10-28 20:46 By sungyg (

47)

Categories:

Hello!

I try to solve the eq1, eq2, eq3, and eq4.
I simplified the 4 equations to rs1 and rs2
to get the solution t3 and t5.
Then, I check the solution by subsituting Eq(14)
into Eq(10). The result is right. The rs1 equation
was found by numerical tests.

However, it is not correct for the eq1, eq2, eq3,
or eq4. I do not understand why it is not correct.

Please tell me the reason and how I can solve and obtain
correct solution. The worksheet can be found in the
File Manager

Thank you for your time and help.

Y-G Sung

sungyg's blog
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Comments

Straightforward

Comment on 2006-10-29 01:53 from alec ( Maple Rating 4

1607)

It seems as if your system can be solved without tricks,

t[1],t[2],t[4],t[6]:=0,delta,t[3]+delta,t[5]+delta:
u1:=(t[2]-t[1])+(t[4]-t[3])+(t[6]-t[5])=U/2:
delta:=solve(u1,delta):
eq1:=expand(add((-1)^(i+1)*cos(omega*t[i]),i=1..6)=0):
eq2:=expand(add((-1)^(i+1)*sin(omega*t[i]),i=1..6)=0):
eq3:=expand(add((-1)^(i+1)*cos(r*omega*t[i]),i=1..6)=0):
eq4:=expand(add((-1)^(i+1)*sin(r*omega*t[i]),i=1..6)=0):
_EnvAllSolutions:=true:
_EnvExplicit:=true:
s:=solve({eq1,eq2,eq3,eq4},{t[3],t[5],r});

            12 Pi _Z1~         2/3 Pi + 2 Pi _Z4~
  s := {r = ----------, t[5] = ------------------,
             omega U                 omega

               -2/3 Pi + 2 Pi _Z6~        12 Pi _Z1~
        t[3] = -------------------}, {r = ----------,
                      omega                omega U

               -2/3 Pi + 2 Pi _Z4~         2/3 Pi + 2 Pi _Z6~
        t[5] = -------------------, t[3] = ------------------}
                      omega                      omega

Note, that both solutions satisfy condition r*ω*U/12/Pi = an integer,

about(_Z1,_Z4,_Z6);

Originally _Z1, renamed _Z1~:
  is assumed to be: integer

Originally _Z4, renamed _Z4~:
  is assumed to be: integer

Originally _Z6, renamed _Z6~:
  is assumed to be: integer

Substituting integer values for _Zs, we obtain 2 series of solutions, for example,

s1:=eval(s,[_Z1=1,_Z4=1,_Z6=1]);

              12 Pi           8 Pi            4 Pi
  s1 := {r = -------, t[5] = -------, t[3] = -------},
             omega U         3 omega         3 omega

              12 Pi           4 Pi            8 Pi
        {r = -------, t[5] = -------, t[3] = -------}
             omega U         3 omega         3 omega

eval([eq1,eq2,eq3,eq4],s1[1]);

                     [0 = 0, 0 = 0, 0 = 0, 0 = 0]

__________
Alec Mihailovs
http://mihailovs.com/Alec/

Thanks, but

Comment on 2006-10-29 03:34 from sungyg (

47)

In SOLVE command,
Why do you include the variable r?
Is there any way to use the following
command instead of your command?

s1: = solve({eq1,eq2,eq3,eq4},{t[3],t[5]});

Could you explain why you use "r" variable
which is positive real?

Thank you in Smoking Section.

Y-G Sung

Well

Comment on 2006-10-29 03:43 from alec ( Maple Rating 4

1607)

Well, one of r, U, or ω has to be added to t[3] and t[5] -because of the restriction that I mentioned - solutions exist only if rωU/(12π) is an integer. Otherwise, solve won't give an answer.

_________
Alec Mihailovs
http://mihailovs.com/Alec/

Thanks a lot, Dr.

Comment on 2006-11-01 00:38 from sungyg (

47)

Thanks a lot, Dr. Alec.
However, I try to get the same answer as yours
by setting up the equations. Here is my code.
I think that s2 should be the same as s1.
Please tell me why it does not give me the same results.

Thank you

Sung

restart:
t[1] := 0:t[2] := delta:t[4] := t[3]+delta:t[6] := t[5]+delta:
u1 := t[2]-t[1]+t[4]-t[3]+t[6]-t[5] = 1/2*U:
delta := solve(u1, delta):
eq1 := expand(sum((-1)^(i+1)*cos(omega*t[i]), i = 1 .. 6) = 0):
eq2 := expand(sum((-1)^(i+1)*sin(omega*t[i]), i = 1 .. 6) = 0):
eq3 := expand(sum((-1)^(i+1)*cos(r*omega*t[i]), i = 1 .. 6) = 0):
eq4 := expand(sum((-1)^(i+1)*sin(r*omega*t[i]), i = 1 .. 6) = 0):
_EnvAllSolutions := true:
_EnvExplicit := true:
s1:=solve({eq1, eq2, eq3, eq4}, {r, t[3], t[5]});
trs1 := simplify(eq1*sin(omega*delta)+eq2*(1-cos(omega*delta))):
trs2 := simplify(eq3*sin(r*omega*delta)+eq4*(1-cos(r*omega*delta))):
rs1 := factor(trs1):
rs2 := factor(trs2):
s2:=solve({rs2, rs1}, {r, t[3], t[5]});

The systems are not equivalent

Comment on 2006-11-01 04:13 from alec ( Maple Rating 4

1607)

The system {rs1,rs2} in addition to solutions of the system {eq1,eq2,eq3,eq4} has other solutions. For example, when the lhs of the first term of trs1 equals minus lhs of the second term, and similarly for trs2, and these lhs' are not necessarily zeros, or even when they are zeros, it could be because of cos(ω*δ)=cos(r*ω*δ)=1 independently of values of eq1, eq2, eq3, and eq4.

__________
Alec Mihailovs
http://mihailovs.com/Alec/