Equation with log LN

[Jan 11]

I have another equation that I have a problem with. This is the following equation: x ^ 2-100 = LN (x ^ 2-1). The equation has 4 solutions. The calculator gives only two solutions. Why.

   

[Eddie W. Shore]

Does it have complex numbers for solutions? Try cSolve.

[Jan 11]

The cSolve command does not solve the problem. There is probably some problem in CAS. The author of CAS, can speak as an authority.
I await his opinion.

[jfdHP]

In fact, the calculator is right, there are only two real solutions, as can be shown by plotting the function. Why do you believe there are four solutions?

[Jan 11]

There are four solutions to this system of equations.
I solved it graphically in the Advance Graphing application.
See attached screenshots. I have no doubts. CAS does not provide all solutions. This is subject to improvement by the CAS author.

               

[Mark Hardman]

Remove the "Y=" from the equations in the Function Symbolic View. You will see that there is a discontinuity of LN(X^2-1) between -1 and 1. The resulting plot only shows two solutions.

Wolfram Alpha agrees:

https://www.wolframalpha.com/input/?i=x%...x%5E2-1%29

[Jan 11]

I still think there are 4 solutions. The calculator only calculates two. I made a similar equation x ^ 2-100 = LN (x). In this case, the calculator correctly solves the equation and gives 2 solutions. One solution x = 10.115 and x = 3.720E-44. It also solves unevenness well (see calculator screenshots).
Solution x = 3.720E-44 is a number almost equal to 0, but it is not zero. Similarly, in the equation x ^ 2-100 = LN (x), one solution is the number x> 1 and the other is the number x <-1. The calculator, however, does not calculate these values. He only gives x = -10.229 and x = 10.229.
But let this issue be decided by authority, the author of CAS.

       

[ijabbott]

(03-24-2020 07:47 PM)Jan 11 Wrote:  I have another equation that I have a problem with. This is the following equation: x ^ 2-100 = LN (x ^ 2-1). The equation has 4 solutions. The calculator gives only two solutions. Why.

On the bottom two solve()'s, it looks like you are trying to zoom in on the other two solutions, but you altered the equation from \(x^2-100=\ln(x^2-1)\) to \(x^2-100=\ln(x^2)\).

[ijabbott]

The nSolve on TI Voyage 200 finds a solution at x=1.00000000001 with a "Questionable accuracy" warning.

[Jan 11]

This is an attempt to solve this equation by other methods. But it was ineffective. Please do not pay attention to this. Today's post and screenshots are current.

[ijabbott]

(03-26-2020 07:33 AM)Jan 11 Wrote:  This is an attempt to solve this equation by other methods. But it was ineffective. Please do not pay attention to this. Today's post and screenshots are current.

Nevertheless, you are trying to find numerical solutions at the limits of the numerical precision of the calculator, very close to the discontinuities.

[Jan 11]

Ijabbott, thank you very much. I had the same thoughts as you. That's why I downloaded OS 2.09 from the network (for my VOYAGE 200). Screenshots with the solution attached. However, there are 4 solutions, not two.


  SCREEN01.BMP (Size: 90.05 KB / Downloads: 14)
  SCREEN02.BMP (Size: 90.05 KB / Downloads: 10)
  SCREEN03.BMP (Size: 90.05 KB / Downloads: 8)

[CyberAngel]

(03-26-2020 07:58 AM)Jan 11 Wrote:  Ijabbott, thank you very much. I had the same thoughts as you. That's why I downloaded OS 2.09 from the network (for my VOYAGE 200). Screenshots with the solution attached. However, there are 4 solutions, not two.

I have a small fever.
Have to inserted all these roots into the original equation in order to check them?

[parisse]

Yes, it's not possible to solve this equation numerically, because the singularity of the logarithm that will raise the crossing will not appear numerically. The solution near 1 (or -1) verifies x^2-100 is almost -99, therefore ln(x^2-1) is almost -99, and x is about sqrt(1+exp(-99)) i.e. about 1+exp(-99)/2, the difference with 1 is 5e-44. And with floats, 1+5e-44==1.

[rombva]

How can I solve the equation x^2-100=log(x^2-1) in Xcas with Digits:=50?

[parisse]

You don't need multi-precision floats, but you have to desingularize the equation if you want to improve the accuracy of 1+exp(-99)/2. The next command will also work on the Prime:
f:=x^2-100-log(x^2-1); x:=1+h; h:=exp(-99)/2*(1+k); [kk]:=fsolve(normal(f),k=-0.1..0.1);
Therefore x-1 is approximatively exp(-99)/2*(1+kk)
kk being approx 4.7e-13, in fact 1+exp(-99)/2 is already a numeric approximation of x with 50 digits.

[rombva]

My question concerns exactly Xcas. When I change the Digits the number of digits after decimal point in the roots of some equation isn't changed. How can some equations with multi-precision floats be solved?

Does the Digits parametr refer only to the evalf command or does it influence the number of digits returned by the solve command?

[CyberAngel]

(03-26-2020 01:00 PM)rombva Wrote:  My question concerns exactly Xcas. When I change the Digits the number of digits after decimal point in the roots of some equation isn't changed. How can some equations with multi-precision floats be solved?

Does the Digits parametr refer only to the evalf command or does it influence the number of digits returned by the solve command?

HP Prime Digits := 50 ; // or 1000
More χCas features?
– –
VPN
PS: more sick, still flu, Ibuprofen gives ½ brains back

[parisse]

(03-26-2020 01:00 PM)rombva Wrote:  My question concerns exactly Xcas. When I change the Digits the number of digits after decimal point in the roots of some equation isn't changed. How can some equations with multi-precision floats be solved?

Does the Digits parametr refer only to the evalf command or does it influence the number of digits returned by the solve command?

There is indeed something preventing newton to work with multi-precision floats, it will be fixed in the next version of Xcas.

[rombva]

Another issue I found with rand().

Commandline evaluation with rand() command causes the unexpected close of the Xcas (x64 Win 10, win32 Xcas 1.5.0-37).

Thanks!