Solving Exponential Equations



#10

Hi folks,
Please consider the very simple exponential equation 3^x-9=0.

The obvious solution for x is 2, but there is an infinite series of complex solutions as well, as alluded to by the fact that when solving this equation symbolically on my 50G , I get as a solution:
(2*e^(ipi/2)*n1*pi + 2*ln3)/ln3. Indeed, this solution satisfies the original equation when any integer is substituted for n1. Quite an impressive feat of number crunching by the 50G, I must say.

When I solved this equation on my TI89 emulator, and noted that only the principle solution of 2 was returned, with no mention of the complex solutions. I was just wondering if there is a reason for this seemingly incomplete solution from the TI (perhaps the complex series is considered superfluous?), and is there a way to get it to deliver the complex solutions?


Any insights appreciated.
Best regards, Hal


#11

The TI-89 and TI-NSpire CAS use cSolve() to solve for complex (and real) roots.


#12

Here is what the Nspire CAS Mac/PC Software returns:

The second line was created by just selecting >Rect from the menu, Ans has been replaced by the calculator in the history view.

BTW, the HP-50g returns essentially the same solution:

I used the SOLVEVX softkey (F5) after a DUP to have both expressions on the screen. I just couldn't find a simple way to split the solution.

I couldn't get it to work on my HP-40gs: SOLVEVX(3X=9) returns an error: Non unary operator.


Edited: 12 Mar 2009, 6:48 a.m.

#13

What's odd is that the 50G doesn't make it a little simpler: x = (n 2 pi i)/ln(3) + 2 (for any integer n).

Edited: 11 Mar 2009, 7:40 p.m.


#14

I think you're asking too much of it. It would take a pretty sophisticated simplification algorithm to figure out that:

e^(i pi/2) = i

#15

Chuck wrote:


Quote:
What's odd is that the 50G doesn't make it a little simpler: x = (n 2 pi i)/ln(3) + 2 (for any integer n).

If you delete the "X=" leaving


(2*LN(3)+2*i*n1*pi)/LN(3)



and then execute PARTFRAC, you will get

2+(2*i*n1*pi)/LN(3)


#16

Quote:
If you delete the "X=" leaving

(2*LN(3)+2*i*n1*pi)/LN(3)

and then execute PARTFRAC, you will get


2+(2*i*n1*pi)/LN(3)


Hmm, this doesn't work on my HP-49G -- I'm still seeing

2*n1*pi    [   pi]
-------*EXP|i*---| + 2
LN(3) [ 2 ]

[sarcasm]Maybe flag -279 isn't set set correctly...[/sarcasm]

Seriously, the question everyone ought to answer for themselves is... "Can I solve this on pen and paper?"

-- KS


#17

May be du to the 'general solution' versus 'principal value' mode setting.
On the HP-28C/S, this is drive by the flag 34:

34 SF '3^X=9' X ISOL  returns                        2.
which is the ‘principal value’ or obvious value.
34 CF '3^X=9' X ISOL  returns   (2.1972+2*pi*n1)/1.0986
which is the ‘general solution’ solution.

The ‘principal value domain’ is definite for each analytical function of your calculator and is give on any good user manual by a graph.
I am not sure that the ‘principal value domain’ definitions are exactly the same on any calculators, variations may exist between HP and TI, or between models from the same brand (due to technology evolution and mathematical choices).

As it can be seen there, the HP28S/C don’t give a pure analytical solution with ISOL command such as HP49 or HP50. To get it the same form as HP-50g, a HP28C/S user have to follow the following method which is close to the ‘by hand’ resolution.


Principal(real) solution
To avoid resolution of an unknown x at power of a number a, one may use the logarithm to transform the equation from ax = b to ln(ax) = ln(b) assuming ax is strictly positive (principal or obvious solution) and NOT complex (general solution).

Remember the logarithm property of transforming products into additions: ln(ax) is transformed into x.ln(a).

Thus, the equation is now writen as:

x.ln(a) = ln(b).

As far as a is not unity, ln(a) is not zero, so we get:

x = ln(b)/ln(a).

N.A.: ‘3^X=9’ give principal solution ‘X=2’ since ln 9 = 2.ln 3.
We also have to verify that 32 > 0 and 3=/=1, which is true, so ‘X=2’ is a valid ‘principal’ solution.


The general (complex) solution
As the principal one, the general solution is obtained by using logarithm, but considering X as a complex.


If x is complex, we know that it can be expressed as a function of its argument tand module (absolute value) r: x = r.ei.t
The equation is now :
ar.ei.t = b to

which may be transform into by using logarithm :

[italic]ln(ar.ei.t) = ln(b)


r.ei.t. ln(a) = ln(b)

r.ei.t = ln(b)/ln(a)

To found the general solution of 3x=9, we have to found all the possible value of (r,t) which satisfy the equation:
r.ei.t = 2

Two complex are equal when their modules (absolute value) and arguments are both simultaneously equal.


Since 2 = 2+0.i = 2.e0i, we have to solve the system:
  1. r = 2
  2. ei.t = 1


By definition, ei.t is cos(t)+i.sin(t):
  1. r = 2
  2. cos t + i.sin t = 1

Equation 2 is true for any t = ... -6pi, -4pi, -2pi, 0, 2pi, 4pi, 6pi, ...


We can express that t have to be zero modulo 2pi or t = 2.pi.n
Thus x = [ ln(b) + 2pi.n.i ]/ln(a)
x = [ 2.ln(3) + 2pi.n.i ]/ln(3)
x = 2 +2pi.n.i/ln(3)
x = (2, 5.7192n)


Conclusion, ‘general solutions’ are all complex number of module (absolute value) 2 and argument 0 modulo (2pi).
This can be expresse as 3x=9 for any x = 2 + 2pi*i*n/ln(3) where n is any integer.

The ‘principal value’ solution is found at n = 0 where x =2


#18

"C.Ret" --

Thank you for showing about flag 34 on the HP-28C/S, and its effect on ISOL.

Here is an old thread that suggests increasingly better performance of ISOL on the HP-28, HP-48, and HP-49 for a fairly straightforward problem:

http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv014.cgi?read=70012

Thanks also for writing your approach for obtaining the solution manually. The solution seems fairly obvious in retrospect, after it has been methodically derived by this or another method.

-- KS

Edited: 16 Mar 2009, 12:09 a.m.


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