19-Tanzanite

Solved

Why root function does not find the second imaginary solution?

Forum|Forum|1 year ago
October 4, 2024
4 replies
4317 views

Hi,

Why is it not found also this solution by the root function?

1_Untitled234.zip

Mathcad Usage

Best answer by Werner_E

I was a bit surprised to see that root would even return non-real results with real guesses.

Here are my 2 Cents, using random non-real guesses.

S

StuartBruff

23-Emerald V

y runs from -50. Indices have to start from ORIGIN.

2024 10 04 F.png

Stuart

C

Cornel

Author

19-Tanzanite

Ok, but still one solution is missing:

The second imaginary solution:

I don't know why this solution it is not catched/found it...

S

StuartBruff

23-Emerald V

Try polyroots. root's algorithm (without a search range) doesn't cover all complex results).

Stuart

C

Cornel

Author

19-Tanzanite

I understand. Also, I can use solve command and received all 3 solutions.

But I want to understand why in the way as I shown above (with x as a range, and with root function), this way does not find the second imaginary solution as I should be the case.

S

StuartBruff

23-Emerald V

I enjoy conversations where messages cross each other—so much more fun than a strictly linear thread. 😊

You'd have to dig into the code to understand why Ridder's (or Brent's) algorithm returns one complex root and not the other. It would help if one could specify a complex range. Perhaps somebody could write a wrapper function around f(x) that allows root to stray further into the complex plane. Alternatively, somebody could write a new root function. I've got vague memories of rolling my own Levenberg–Marquardt solver to better understand what it did.

Stuart

S

StuartBruff

23-Emerald V

As to why, ...

Mathcad Help:

The root of a function is the value at which the function equals zero.

•polyroots(v)—Returns a vector containing the roots of the polynomial whose coefficients are in v.

•By default, polyroots uses the LaGuerre method which is iterative and searches for solutions in the complex plane. If you want to use the Companion Matrix method instead, right-click on the polyroots function and select the method from the menu. The Companion Matrix method converts the equations into an eigenvalue problem. Learn more about choosing a solving algorithm for the polyroots function here.

•root(f(var1, var2, ...), var1, [a, b])—Returns the value of var1 to make the function f equal to zero. If a and b are specified, root finds var1 on the interval [a, b]. Otherwise, var1 must be defined with a guess value before root is called. When a guess value is used, root uses the secant or Mueller’s method; in the case where root bracketing is used, root uses Ridder’s or Brent’s method.

Stuart

I have a strange feeling of deja vu, as if this subject has been discussed in the not too distant past.

W

Werner_E

Answer

25-Diamond I

I was a bit surprised to see that root would even return non-real results with real guesses.

Here are my 2 Cents, using random non-real guesses.

3_WE_20241005_Untitled234-1.zip

S

StuartBruff

23-Emerald V

Now that's interesting. When I tried that, Mathcad squawked like a parrot on a sugar high. "No, take that nasty complexity out right now! Honestly, some people. Pieces of Eight. Who's a pretty CAS. The horror, The horror ...".

The local definition I used above arose from such an attempt. However, with freshly typed expressions this time, the parrot was rather quiet.

2024 10 05 A.png