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20-Turquoise
Dec 14, 2025 10:21 PM UTC
Solved

How to prove f(x) and its symbolic evaluate ----> 0 ?

Hello Everyone.
From :

Z1.PNGZ2.PNG

To :
How to prove f.1(phi) symbolic evaluate ----> 0 ?

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Z3.PNG
Thanks in advance.
Regards.

Werner_E
Werner_EAnswer
25-Diamond I
Dec 15, 2025 4:52 AM UTC

1) Before you symbolically evaluate you have to write phi:=phi to make phi unknown to the symbolics. Otherwise it tries to use the defined range variable phi (and therefore the result is too large ...).

 

2) The graph is not that unusual if you look at the scale of the y-axis. What you see are just minor numerical round-off errors in the range of +-10-15. This is to be expected with numerical evaluation - after all that's the usual precision of values in IEEE format which is used by Mathcad and most other numerical number cruncher.

 

Werner_E_0-1765773848457.png

 

Here is what you see when you manually scale the y-axis from, let's say -0.1 to +0.1

Werner_E_0-1765774801358.png

 

11 replies

Werner_E
25-Diamond I
Dec 14, 2025 11:10 PM UTC

As you are squaring anyway you can do without the absolute value operator

Werner_E_0-1765753825812.png

 

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lvl10720-TurquoiseAuthor
20-Turquoise
Dec 14, 2025 11:17 PM UTC

Thanks, Werner. How about f.2(phi) ? Too many (green) points, See the graph above. And we should compare the above with the following :

Z4.PNG

 

Since :

Z7.PNG

 

Regards.

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Werner_E
25-Diamond I
Dec 15, 2025 12:10 AM UTC

Not sure what you mean and I am too lazy to retype 😉

Why you think that you should get zero for any angle phi?

And why do you need Mathcad to confirm?

 

And what about f4 and f5? They obviously are not equal.

They are equal only when the sine is positive which is the case for -pi/0 <= phi <= pi/3 and you may add or subtract integer multiples of  8 pi /9 to the limits.

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