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Hello Everyone.
From :
A Reason for a missing-point ?
Thanks in advance.
Regards.
Loi.
Solved! Go to Solution.
Again and again and again and again ....
You always run into numerical inaccuracies! You cannot expect a 100% exact result when using the numerics with its IEEE format and its limited about 15 digits precision.
The numerics in your case thinks that the first value is non-real and so it is not plotted:
In the meantime you should already know what to look for and be able to find out things like that yourself.
Why don't you just use
I guess I don't fully understand above that, but a number with 30-decimal-digit It did show the point. And again : " I guess I don't fully understand above that " :
Best Regards.
Oh... so far I guess I somewhat understand about that, now the point to be shown but now I need help with a explanation.
Best Regards.
As you can see - the roots you have in mind are roots of DIFFERENT MEMBERS of that family.
Mathcad can't really deal with that kind of family of curves and "arbitrary" choses one, but not the same for every value of x you put in.
And again and again, so far the MATHCAD is the best one to me to use. And I have only a bit of knowledge and only about the MATHCAD. And I have not had any other Math software, but MATHCAD. And to me, the MATHCAD make MATH very simplest. As my demostration, The g(x) = cot(x) + e^(x)*{ e^-pi - e^(-3*pi) } and my function is NOT equivalent.
Best Regards.
Loi.
Now, anyway so far I still need help with a brief explanation about original question.
Best Regards.
Loi.
You don't understand that "your" function is not just ONE single function but its an expression which splits into an infinite number of different functions.
You have no control over which of the many possible functions Mathcad uses when you let it calculate g(-2) or g(4) but chances are that Mathcad uses different ones.
Hope the following will make it clear to you, otherwise I can't help
For now , all words I have to say is I greatly appreciate your time and help.
Best Best Regards.
Loi.
And : The g(k,x) = cot(x) + e^x*{ e^-pi - e^(2k-1)*pi } and g_(,x) = cot(x) + { {e^x - 1/{ [e^(x*1i)]^1i} }*e^-pi is NOT equivalent , EITHER :
Best Regards.
Now it's my turn to use :
and :
That's not root-point I wish to solve it.
and I guess : "The world is not the world".
Best Regards.
Loi.