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No possible solution Mathcad 15

jlindrup
1-Newbie

No possible solution Mathcad 15

I have MathCad V.15, and theres one equation I can't solve. I've tried a simplified solution which works, but I can't figure whats wrong. Can you please help me?

See attachments.

Thank you.

1 ACCEPTED SOLUTION

Accepted Solutions
StuartBruff
23-Emerald II
(To:jlindrup)

Jens Lindrup wrote:

I have MathCad V.15, and theres one equation I can't solve. I've tried a simplified solution which works, but I can't figure whats wrong. Can you please help me?

See attachments.

Thank you.

Stuart

(PS. Please don't post the same question multiple times.   Those who are likely to answer will see it no matter what category you post it under, and it can lead to people wasting their time and effort answering a question that's been already been answered elsewhere).

View solution in original post

13 REPLIES 13
StuartBruff
23-Emerald II
(To:jlindrup)

Jens Lindrup wrote:

I have MathCad V.15, and theres one equation I can't solve. I've tried a simplified solution which works, but I can't figure whats wrong. Can you please help me?

See attachments.

Thank you.

Stuart

(PS. Please don't post the same question multiple times.   Those who are likely to answer will see it no matter what category you post it under, and it can lead to people wasting their time and effort answering a question that's been already been answered elsewhere).

Could do it symbolically this way:

integralsolve.PNG

Removing the "assume b = real" gives the other, complex, solutions as well.

Alan

RichardJ
19-Tanzanite
(To:jlindrup)

Another example of just how poorly the symbolic engine in Mathcad handles definite integrals. It can find the indefinite integral, so you can do the substitution for the limits by hand. In this case even that step is not really necessary though, because the lower limit is 0 and there are no terms that do not contain x.

StuartBruff
23-Emerald II
(To:RichardJ)

Richard Jackson wrote:

Another example of just how poorly the symbolic engine in Mathcad handles definite integrals. It can find the indefinite integral, so you can do the substitution for the limits by hand. In this case even that step is not really necessary though, because the lower limit is 0 and there are no terms that do not contain x.

I was wondering how Find managed to symbolically do what it's name suggests if it couldn't evaluate a complex integral?  I now wonder if it's possible that it solves the indefinite integral and then tries the find method?

Stuart

RichardJ
19-Tanzanite
(To:StuartBruff)

I was wondering how Find managed to symbolically do what it's name suggests if it couldn't evaluate a complex integral?  I now wonder if it's possible that it solves the indefinite integral and then tries the find method?

Stuart

It's not clear to me what the symbolic processor does when given a solve block with find. Certainly not the same thing it does with the solve keyword.

StuartBruff
23-Emerald II
(To:RichardJ)

Richard Jackson

I was wondering how Find managed to symbolically do what it's name suggests if it couldn't evaluate a complex integral?  I now wonder if it's possible that it solves the indefinite integral and then tries the find method?

Stuart

It's not clear to me what the symbolic processor does when given a solve block with find. Certainly not the same thing it does with the solve keyword.

No, I don't know what's going on.  I've had a quick browse through some of the mpl, lisp and mupad files, but, although I know that MuPad passes the equations to a function called solve, I haven't yet dug further into the MuPad library to find out whether there is any more information on what solve does.

Stuart

Hi Jens!

It should work the way you did it but it doesn't because of a bug obviously introduced with MC14.

I ran into the very same bug a short while ago -> Bug in Mathcad 15 M030 (symbolics)

You have already seen some workarounds and I may add another one. As the problem in the symbolics are the floating point decimal numbers, we can rewrite r(x) using fractions of integers. And - Voilà! Mathcads symbolics is capable to solve for all solutions, even though it switches to float display itself:

If you manually force it in float mode (even after solve), it fails again - clearly a bug!

Regards, Werner

StuartBruff
23-Emerald II
(To:Werner_E)

Werner Exinger wrote:

Hi Jens!

It should work the way you did it but it doesn't because of a bug obviously introduced with MC14.

I ran into the very same bug a short while ago -> Bug in Mathcad 15 M030 (symbolics)

You have already seen some workarounds and I may add another one. As the problem in the symbolics are the floating point decimal numbers, we can rewrite r(x) using fractions of integers. And - Voilà! Mathcads symbolics is capable to solve for all solutions, even though it switches to float display itself:

If you manually force it in float mode (even after solve), it fails again - clearly a bug!

Thanks,Werner - you've literally just saved me from doing the same thing!   I was idly browsing through the (non-compressed) mpl, lisp and mupad files and wondered whether the symbolic processor might handle a rational solution any differently.

zsort is a simple sort function I put together to sort complex numbers by their Real values and sort any values with equal Reals by their Imaginary values, simply so I could look at the differences between the solutions..

Stuart

RichardJ
19-Tanzanite
(To:Werner_E)

Very interesting! I wonder if that explains some of the other examples of horrible handling of definite integrals. I think not all of them, but perhaps some.

As you say, clearly a bug.

MikeArmstrong
5-Regular Member
(To:jlindrup)

Sorry to sound really stupid, but could someone please explain why r(x) is raised is changed to (r(x))^2 in both the examples below.

Thanks,

Mike

Mike Armstrong wrote:

Sorry to sound really stupid, but could someone please explain why r(x) is raised is changed to (r(x))^2 in both the examples below.

Thanks,

Mike

What do you mean, Mike?  It only seems to have the extra pair of parentheses in the integral that you show?  (The reason it's in my answer is simply that it was that way in the original post; it doesn't affect the results).

Stuart

MikeArmstrong
5-Regular Member
(To:StuartBruff)

I believe it may be down to my lack of knowledge.

Mike

Mike Armstrong wrote:

I believe it may be down to my lack of knowledge.

Mike

OK. 

Stuart

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