Dear Saad,
It happens that I want to find more than one root numerically, then I use
brute force as in code below. It is not possible to
have any general heuristics for arbitrary function about where its roots
are, so if one knows more about the function in some special case
then the interval
So what is likely the difference between how Sage solves this problem and
how Mathematica solves this problem that makes Mathematica show more
solutions?
On Friday, March 16, 2018 at 4:31:33 AM UTC-5, Dima Pasechnik wrote:
>
> This makes sense. It most probably has a pre-set minimal interval le
This makes sense. It most probably has a pre-set minimal interval length, so it
stops splitting at certain moment.
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OK, that was too easy, as there was an exact symbolic solution.
Is this a solution Sage could have given symbolically as well?
Here are the results for sin(1/x)==x on [-1,1] :
It apparently computed many solutions and originally asked me whether I
wanted to show all the found solutions or ju
On Thursday, March 15, 2018 at 10:23:44 PM UTC, saad khalid wrote:
>
> I have to apologize, I gave a slightly incorrect Mathematica code earlier,
> the actual code was:
>
> Solve[ Exp[-2*a*x]-1+4*a*x==0,x]//N
>
> The earlier code gave the wrong answer. Anyways:
>
> it could be a different functi
I have to apologize, I gave a slightly incorrect Mathematica code earlier,
the actual code was:
Solve[ Exp[-2*a*x]-1+4*a*x==0,x]//N
The earlier code gave the wrong answer. Anyways:
it could be a different function, which potentially would run much longer,
> by repetitive splitting of the inter
On 2018-03-15, Dima Pasechnik wrote:
> it could be a different function, which potentially would run much longer,
> by repetitive splitting of the interval
> (I guess that's what Mathematica is doing)
I have toyed with the idea of repurposing whatever adaptive splitting
code is in the plottin
On Thursday, March 15, 2018 at 2:25:00 AM UTC, saad khalid wrote:
>
> That is a good point, however I feel an even better solution in that case
> would be giving some of the roots and then giving some indication that
> there are an infinite number of roots. Disregarding the case where there
>
That is a good point, however I feel an even better solution in that case
would be giving some of the roots and then giving some indication that
there are an infinite number of roots. Disregarding the case where there
are infinitely many roots, I don't see why it wouldn't be preferable to
have
On 05/03/2018 20:01, saad khalid wrote:
Hello, and thank you for your response. While I agree that the behaviour of
the function certainly complies with the specifications listed in its
description, I think everyone would agree that it would be better if it did
give all of the roots in a given in
Hello, and thank you for your response. While I agree that the behaviour of
the function certainly complies with the specifications listed in its
description, I think everyone would agree that it would be better if it did
give all of the roots in a given interval. Would you happen to know
anyth
Hi Khalid,
On 2018-03-02, saad khalid wrote:
> I'm running this code:
> find_root(e^(-2*x*1)-(1 - 4*x),-2,2)
>
> It returns
>
> 2.4011774461136836e-13
>
> which is approximately 0. However, there should be another root around x =
> -0.628. Why isn't it finding this root?
> Is there any way I c
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