Absolute courses - page 79
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Where did you see a "leak"? Are you just wishful thinking?
Uh, that's not fair.
I saw a leak in your profile.
I don't wish anything bad on anyone. I am only stating the facts.
A lot of people have warned you.
Uhh, you shouldn't have done that.
Saw "plum" on your profile.
I don't wish anything bad on anyone. Just stating the facts.
You have been warned by many.
That's unlikely. It's without gloating, the score is still a demo. But a fact. IMHO. You will not.
I have changed the algorithm to version 2. I have not deleted old orders, let them close by themselves or by a stop or reverse signal. We will see.
Solve this by numerical methods))
How about this? There is a suspicion of incorrectness.
Half division method.
The method is based on the assumption that the signs of the first derivative for the resulting range are opposite, with minus signs on the left and plus signs on the right and that the function has a minimum in the range.
This is just the point that it may happen that this minimum is not the inflection point, but only the end of the range - the boundary point is minimal.
Take a growing segment, for example - no inflection point there, but a minimum on the segment.
No, there is always one extremum, but its sign is not necessarily +-+, it can also be -+-.
Did I get it right that if you feed this method with 2 parabolas, one regular and one reversed, then this method will not find minimum parabolas on the segment (the vertex falls in the segment), but exactly the inflection points, i.e. its vertices?
It is very resource-intensive, this method is faster, but is it correct?
Can we do it this way? I have a suspicion that it is not correct.
half division method.
This method is based on the assumption that signs of the first derivative for the resulting range are opposite, with minus signs on the left and plus signs on the right and the function has one minimum in the range.
This is just the point that it may happen that this minimum is not the inflection point, but only the end of the range - the boundary point is minimal.
Take a growing segment, for example - no inflection point there, but a minimum on the segment.
No, there is always one extremum, but its sign is not necessarily +-+, it can also be -+-.
Did I get it right that if you feed this method with 2 parabolas, one regular and one reversed, then this method will not find minimum parabolas on the segment (the vertex falls in the segment), but exactly the inflection points, i.e. its vertices?
It is very resource-intensive and faster, but is it correct?
This is not about solving the equation, but about finding the inflection point of the function itself
Can we do it this way? I have a suspicion that it is not correct.
half division method.
The method is based on the assumption that signs of the first derivative for the resulting range are opposite, with minus signs on the left and plus signs on the right and the function has one minimum in the range.
This is just the point that it may happen that this minimum is not the inflection point, but only the end of the range - the boundary point is minimal.
Take a growing segment, for example - no inflection point there, but a minimum on the segment.
No, there is always one extremum, but its sign is not necessarily +-+, it can also be -+-.
Did I get it right that if you feed this method with 2 parabolas, one regular and one reversed, then this method will not find minimum parabolas on the segment (the vertex falls in the segment), but exactly the inflection points, i.e. its vertices?
It is very resource-intensive and faster, but is it correct?
The inflection point differs from the extremum by the value of the second derivative: in the first case it is 0, in the second it is not.
the inflection point from the extremum differs in the value of the second derivative: in the first case it is 0, in the second it is not.
Yes, I got that wrong, it is the extremum that needs to be found.