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Please explain result - DrD - 03-29-2018 12:19 PM

After a restart, entering: int(sin(x),x,0,∞), repeatedly, produces the results shown in the image. What is meant by "bounded_function(n)+1", where n increments by 2, after each entry?

[attachment=5792]


RE: Please explain result - Arno K - 03-29-2018 05:24 PM

Hm, "undef" would be a reasonable answer, as the integral does not converge. Bounded-function is clear, so the number that is increased by 2 is interesting.
Arno


RE: Please explain result - Arno K - 03-29-2018 05:40 PM

Found bounded-function in an xcas document:
For regular series expansion, order_size is a bounded function, but for non
regular series expansion, it might tend slowly to infinity, for example like a power of ln(x).
So its left over by the tried taylor expansion of sin(x) to get the integral done, then simply returned as the result, but the argument isn't explained.
Arno


RE: Please explain result - DrD - 03-29-2018 06:19 PM

It's a divergent integral, limit does not exist, but I didn't grasp the idea of "boundary_function(n)+1", where the n is always even. What does the "n" tell us, and is the +1, just there to represent a constant? Sometimes the way these things are taught varies by country, but I haven't encountered this kind of result anywhere.


RE: Please explain result - Arno K - 03-29-2018 06:57 PM

I think the one simply is -(-cos(0)), so the antiderivative is first evaluated at its lower bound, n may be an exponent used by taylor, using cosine instead of sine makes it increase by 6, but here Bernard may have an answer.
Arno


RE: Please explain result - parisse - 03-29-2018 07:32 PM

I introduced bounded functions in order to solve limits like
limit(sin(x+1/x)-sin(x),x,inf)


RE: Please explain result - DrD - 03-29-2018 07:54 PM

What does the function value represent?

bounded_function(2)+1


RE: Please explain result - parisse - 03-30-2018 06:14 AM

Exactly that: a bounded function + 1 (the +1 comes from evaluation of cos at 0).


RE: Please explain result - Nigel (UK) - 03-30-2018 08:31 AM

(03-30-2018 06:14 AM)parisse Wrote:  Exactly that: a bounded function + 1 (the +1 comes from evaluation of cos at 0).

Does the “2” just enumerate bounded functions?

Nigel (UKj


RE: Please explain result - DrD - 03-30-2018 10:57 AM

Interesting ... I shut down the system last night, and this morning when I turned it back on and ran the emulator, all the pre-existing "bounded_function(n)" entries had changed to "undef+1"

Why does the number increment by 2 each time similar integrals are entered?

Further divergent entries like:

int(cos(x),x,0,∞);
int(sin(x),x,0,∞);

still result in bounded_function(2*n) (I am using n to represent the number of divergent integrals entered). Can programmers use that information for some suitable purpose?


RE: Please explain result - parisse - 03-30-2018 11:07 AM

No. I'm thinking of replacing bounded_function by undef in the limit output.


RE: Please explain result - DrD - 03-30-2018 11:44 AM

It seems like undef would be more meaningful, Parisse, and thank you, very much, for your explanations, and guidance!

-Dale-


RE: Please explain result - parisse - 03-30-2018 11:50 AM

Change has been committed.