11-Garnet

Solved

Math equation solve issue

Forum|Forum|1 year ago
June 13, 2024
4 replies
6801 views

i Mathcad experts,

I have some questions about Mathcad function. I don’t have any useful idea to solve my issues.

I have one equation, and I can draw the curve of this equation.

And then I try to solve the peak value of curve (fig.1) by using the method in EX2.

Below are my questions.

1.Any method I can solve this equation to get the fn_Y2, please see EX1?

2.How do I use solve function to get the fn_Y2, if I have different inputs Q2 at the same time (please see EX3). I can do this when I only have one input, please see EX2.

3.If the issue 2 can be solved, and then I would try to input the fn_Y2 to get the peak value . Also I would like to draw the relationship of Q and peak value, please see the fig.2 (this fig is from TI’s datasheet), could you please give me some suggestions.

1_Math-issue.zip

Best answer by Werner_E

Prime is not able to solve your equation symbolically.

But as you had shown yourself it can provide solution(s) if you provide numeric values for the other two variables

You may also use the "assume" modifier to limit the solution to the only positive, real one

But you are just looking for numeric results (depending on Q and L(ambda), so I would suggest that you use Mathcads numeric methods. Either a solve block with "find" or one of the two ways to use the "root" function.

Here is the latter, assuming the solution is within the range from 10^-10 to 10:

You can turn this calculation into a function of lambda and Q

So it can be used to get the result for any of these values

From here its not difficult to plot either the My function for various Q-values and their peaks

or the plot you seem to be looking for

Mathcad 15 sheet attached

C

Cornel

19-Tanzanite

I am not sure why do you want to differentiate with respect to fn_Y2, and not fn_Y.

Its too complicated to solve for fn_Y...also having these 2 as parameters

But you can do something like this:

as you already did in fact here:

Its based on a manual way to do this what you want as you did...

Here...someone will need to find a way on how to choose maximum value as we can see for example at Q4 index is 3 and at Q6 index is 5, so index is changing:

And then how to iterate Q parameter from 0.1, 0.2,...1, then for each Q value to find the maximum, and then to plot, these in a more automatical way...Maybe one could find how to do it...

1_Math-issue.zip

W

Werner_E

25-Diamond I

@Cornel wrote:

I am not sure why do you want to differentiate with respect to fn_Y2, and not fn_Y.

Above that expression he already had defined a value for f.n_Y and so Mathcads symbolics would refuse to solve for this variable. So he had chosen a different name to be used for the variable Mathcad should solve for.

W

Werner_E

Answer

25-Diamond I

Prime is not able to solve your equation symbolically.

But as you had shown yourself it can provide solution(s) if you provide numeric values for the other two variables

You may also use the "assume" modifier to limit the solution to the only positive, real one

But you are just looking for numeric results (depending on Q and L(ambda), so I would suggest that you use Mathcads numeric methods. Either a solve block with "find" or one of the two ways to use the "root" function.

Here is the latter, assuming the solution is within the range from 10^-10 to 10:

You can turn this calculation into a function of lambda and Q

So it can be used to get the result for any of these values

From here its not difficult to plot either the My function for various Q-values and their peaks

or the plot you seem to be looking for

Mathcad 15 sheet attached

2_WE_20240613_Math-issue.zip

J

J_power

Author

11-Garnet

Thanks for your response.

May I know why you use Vector in M function and F function?

J

J_power

Author

11-Garnet

it is more clear to me, thanks.

W

Werner_E

25-Diamond I

I played around and added a 3D plot with the L (lambda), Q an max My values, showing the curve corresponding to L=6.3 in red

I also added a way to use the symbolics to get the necessary fnY-values, but I won't suggest using it:

1_WE_20240613(2)_Math-issue.zip

J

J_power

Author

11-Garnet

What's the function of this equation?

W

Werner_E

25-Diamond I

@J_power wrote:

What's the function of this equation?

This expression does the very same (or something similar) as the function f.positive in my last reply (posting a worksheet with Lucs approach).

F is the result of calling the function "temp" which uses the symbolics to return a vector with six elements.

In case of your example just one of these six values is real and positive and this is the one to pick. I won't bet that its always the first or always the second one (even this may be the case here).

Furthermore sometimes because of round-off errors even the real solutions are returned with a very tiny imaginary part.

1) |Im(F)|<^10^.10 returns a 6-element vector with a 1 where F is a real value (Imaginary part 0 or very small) and a zero if F is non.real.

2) F is multiplied b the vector created in the first step, so all non-real values are set to zero

3) this new vector is sorted using the "sort" function. That way the largest (positive ?) value comes last in the vector returned

4) "reverse" changes the order of the vector elements so that that positive value is the first element

5) This first element is now chosen using the matrix index ORIGIN

And, no, this sure is not the most elegant or intuitive way to achieve this task 😉

After all I think its best to avoid the symbolics altogether and just use Mathcads numeric facilities.

L

LucMeekes