Getting the equation for the line of intersection using Mathematica - wolfram-mathematica

I have a nasty expression that I am playing around with on Mathematica.
(-X + (2 X - X^2)/(
2 (-1 + X)^2 ((1 + 2 (-1 + p) X - (-1 + p) X^2)/(-1 + X)^2)^(3/2)))/X
I graphed it along with the plane z = 0 where X and p are both restricted from 0 to 1:
Plot3D[{nasty equation is here, 0}, {p , 0, 1}, {X, 0, 1}]
I decided it would be interesting to obtain the equation for the intersection of the plane generated from the nasty equation and z = 0. So I used solve:
Solve[{that nasty equation == 0}, {p, X}, Reals]
and the output was even nastier with some results having the # symbol in it ( I have no idea what it is, and I am new to Mathematica). Is there a way to get an equation for the nice line of intersection between the nasty equation and z = 0 where p and X are restricted from 0 to 1? In the graph generated from Plot3D I see that the line of intersection appears to be some nice looking half parabola looking thing. I would like the equation for that if possible. Thank you!

For complicated nasty equations Reduce is often more powerful and less likely to give you something that you will later find has hidden assumptions inside the result. Notice I include your constraint about the range of p and X to give Reduce the maximum amount of
information that I can to help it produce the simplest possible solution for you.
In[1]:= Reduce[(-X + (2 X-X^2)/(2 (-1 + X)^2 ((1 + 2 (-1 + p) X - (-1 + p) X^2)/
(-1 + X)^2)^(3/2)))/X == 0 && 0 < X < 1 && 0 < p < 1, {X, p}]
Out[1]= 0<X<1 && p == Root[12 - 47*X + 74*X^2 - 59*X^3 + 24*X^4 - 4*X^5 + (-24 +
108*X - 192*X^2 + 168*X^3 - 72*X^4 + 12*X^5)*#1 + (-48*X + 144*X^2 - 156*X^3 +
72*X^4 - 12*X^5)*#1^2 + (-32*X^2 + 48*X^3 - 24*X^4 + 4*X^5)*#1^3 & , 1]
Root is a Mathematica function representing a root of a usually complicated polynomial
that would often be much larger if the actual root were written out in algebra, but we
can see whether the result is understandable enough to be useful by using ToRadicals.
Often Reduce will return several different alternatives using && (and) and || (or) to
let you see the details you must understand to correctly use the result. See how I
copy the entire Root[...] and put that inside ToRadicals. Notice how Reduce returns
answers that include information about the ranges of variables. And see how I give Simplify the domain information about X to allow it to provide the greatest possible simplification.
In[2]:= Simplify[ToRadicals[Root[12 - 47 X + 74 X^2 - 59 X^3 + 24 X^4 - 4 X^5 +
(-24 + 108 X - 192 X^2 + 168 X^3 - 72 X^4 + 12 X^5) #1 + (-48 X + 144 X^2 -
156 X^3 + 72 X^4 - 12 X^5) #1^2 + (-32 X^2 + 48 X^3 - 24 X^4+ 4 X^5)#1^3&,1]],
0 < X < 1]
Out[2]= (8*X - 24*X^2 + 26*X^3 - 12*X^4 + 2*X^5 + 2^(1/3)*(-((-2 + X)^8*(-1 +
X)^2*X^3))^(1/3))/(2*(-2 + X)^3*X^2)
So your desired answer of where z= 0 will be where X is not zero, to avoid 0/0 in
your original equation, and where 0 < X < 1, 0 < p < 1 and where p is a root of that
final complicated expression in X. That result is a fraction and to be a root you
might take a look at where the numerator is zero to see if you can get any more
information about what you are looking for.
Sometimes you can learn something by plotting an expression. If you try to plot that final result you may end up with axes, but no plot. Perhaps the denominator is causing problems. You can try plotting just the numerator. You may again get an empty plot. Perhaps it is your cube root giving complex values. So you can put your numerator inside Re[] and plot that, then repeat that but using Im[]. Those will let you plot just the real and imaginary parts. You are doing this to try to understand where the roots might be. You should be cautious with plots because sometimes, particularly for complicated nasty expressions, the plot can make mistakes or hide desired information from you, but when used with care you can often learn something from this.
And, as always, test this and everything else very carefully to try to make sure that no mistakes have been made. It is too easy to "type some stuff into Mathematica, get some stuff out", think you have the answer and have no idea that there are significant errors hidden.

Related

15 digit floating variable calculation in microcontroller

I want to calculate an equation within a controller(Arduino)
y = -0.0000000104529251928664x^3 + 0.0000928316793270531x^2 - 0.282333029643959x + 297.661280719026
Now the decimal values of the coefficients are important because "x" varies in thousands so cube term cannot be ignored. I have tried manipulating the equation in excel to reduce the coefficients but R^2 is lost in the process and I would like to avoid that.
Max variable size available in Arduino is 4byte. And on google search, I was not able to find an appropriate solution.
Thank you for your time.
Since
-0.0000000104529251928664 ^ (1/3) = - 0.0021864822
0.0000928316793270531 ^ (1/2) = 0.00963491978
The formula
y = -0.0000000104529251928664x^3 + 0.0000928316793270531x^2 - 0.282333029643959x + 297.661280719026
Can be rewritten:
y = -(0.0021864822 * x)^3 + (0.00963491978 * x)^2 - 0.282333029643959 * x + 297.661280719026
Rounding all coefficients to 10 decimal places, we get:
y = -(0.0021864822 * x)^3 + (0.00963491978 * x)^2 - 0.2823330296 * x + 297.6612807
But I don't know Arduino, I'm not sure what the correct number of decimal places is, nor do I know what the compiler will accept or refuse.

What's an algorithm to get a number closest to a constant that can evenly (within a margin) divide into two other constants?

So let't say I have numbers A=1483 and B = 635. My X=100.0
Let's say my allowed MARGIN is 10.0
What's the best way to get the closest number to X (can be floating point) that can divide into A and B with a remainder that is less that MARGIN?
For an answer K. A % K <= MARGIN, B % K <= MARGIN, with K being as close to X as possible, for example |K - X| < 100
Let's try and write the problem with mathematical notations.
What you have is Euclidean divisions:
A = Q1*X + R1
B = Q2*X + R2
You want to find the minimal |x| such that
A = Q1'*(X+x) + R1' , |R1'| <= M
B = Q2'*(X+x) + R2' , |R2'| <= M
To help you finding such x, you have relations like:
A = Q1*(X+x) + R1-Q1*x
B = Q2*(X+x) + R2-Q2*x
From here, you should first concentrate on how to solve the example you gave, then try and generalize.
1483 = 14*100 + 83 = 15*100 - 17
635 = 6*100 + 35 = 7*100 - 65
Should you can take x > 0 in order to reduce R2 (35) down to 10, or x < 0 to increase R1 (-17) up to -10?
In the first case, x should be in interval [25/6 , 45/6] to bring |R2'| <= M, but at the same time it must be in interval [73/14 , 93/14] to bring |R1'| <= M.
Do these intervals overlap?
if yes you have a solution.
if no, then you have to try further (decrement quotients Q1' and/or Q2')
Just check with any decent interpreter (Squeak/Pharo Smalltalk here)
{25/6 . 45/6. 73/14 . 93/14} sorted
= {(25/6) . (73/14) . (93/14) . (15/2)}
So they overlap, starting at x=73/14.
But maybe you would get a closer x in the other direction?
I have not given an algorithm, just a clue, up to you to continue. But you see that increment does not have to be random (like 0.001).
For now the best way I have found is a brute force method by finding the GCD of A and B and decrease by a small interval (0.001) and find the smallest c(K) where K >= X and c(x) = A % x + B % x
If I had found a way to differentiate c(x) correctly, I would've liked to find its gradient and use gradient descent to find the most optimal value without brute force.

Mapping function for two integers

SO,
The problem
I have two integers, which are in first case, positive, and in second case - just any integers. I need to create a map function F from them to some another integer value, which will be:
Result should be integer value. For first case (x>0, y>0), positive integer value
Symmetric. That means F(x, y) = F(y, x)
Unique. That means F(x0, y0) = F(x1, y1) <=> (x0 = x1 ^ y0 = y1) V (y0 = x1 ^ x0 = y1)
My approach
At first glance, for positive integers we could use expression like F(x, y) = x2 + y2, but that will fail - for example, 892 + 232 = 132 + 912 As for second (common) case - that's even more complicated.
Use-case
That may be useful when dealing with some things, which supposed to be order-independent and need to be unique. For example, if we want to find cartesian product of many arrays and we want result to be unique independent of order, i.e. <x,z,y> is equal to <x,y,z>. It may be done with:
function decartProductPair($one, $two, $unique=false)
{
$result = [];
for($i=0; $i<count($one); $i++)
{
for($j=0; $j<count($two); $j++)
{
if($unique)
{
if($i!=$j)
{
$result[$i*$i+$j*$j]=array_merge((array)$one[$i],(array)$two[$j]);
// ^
// |
// +----//this is the place where F(i,j) is needed
}
}
else
{
$result[]=array_merge((array)$one[$i], (array)$two[$j]);
}
}
}
return array_values($result);
}
Another use-case is to properly group sender and receiver in some SQL table, so that different senders/receivers will be differed while they should stay symmetric. Something like:
SELECT
COUNT(1) AS message_count,
sender,
receiver
FROM
test
GROUP BY
-- this is the place where F(sender, receiver) is needed:
sender*sender + receiver*receiver
(By posting samples I wanted to show that issue is certainly related to programming)
The question
As mentioned, the question is - what can be used as F? I want as simple F as it's possible. Keep in mind two cases:
Integer x>0, y>0. F(x,y) > 0
Any integer x, y and so any integer F(x,y) as a result
May be F isn't just an expression - but some algorithm to find desired result for any x,y (so tagging with algorithm too). However, expression is better because it's more like that it will be able to use that expression in SQL or PHP or whatever. Feel free to edit tagging because I'm not sure if two tags here is enough
Most simple solution: f(x,y) = x^5 + y^5
No positive integer is known which can be written as the sum of two fifth powers in more than one way.
As for now, this is unsolved math problem.
You need a MAX_INTEGER constant, and the result will need to hold MAX_INTEGER**2 (say: be a long, if both are int's). In that case, one such function is:
f(x,y) = min(x,y)*MAX_INTEGER + max(x,y)
But I propose a different solution: use a hash function (say md5) of the string resulting from the concatenation of str(min(x,y)), a separator (say ".") and str(max(x,y)). That is:
f(x,y) = md5(str(min(x,y)) + "." + str(max(x,y)))
It is not unique, but collisions are very rare, and probably OK for most use cases. If still worried about collisions, save the actualy {x,y} along with f(x,y), and check if collisions happened.
Sort input numbers and interleave their bits:
x = 5
y = 3
Step 1. Sorting: 3, 5
Step 2. Mixing bits: 11, 101 -> 1_1_, 1_0_1 -> 11011 = 27
So, F(3, 5) = 27
A compact representation is x*(x+3)/2 + y*(x+1) + (y*(y-1))/2, which comes from an arrangement like this:
x->
y 0 1 3 6 10 15
| 2 4 7 11 16
v 5 8 12 17
9 13 18
14 19
20
According to [Stackoverflow:mapping-two-integers-to-one-in-a-unique-and-deterministic-way][1], if we symmetrize the formula we would have the following:
(x + y) * (x + y + 1) / 2 + min(x, y)
This might just work. For
(x + y) * (x + y + 1) / 2 + x
is unique, then the first formula is also unique.
[1]: Mapping two integers to one, in a unique and deterministic way

How can I get rid of irritating prefactor of one in Mathematica?

I am trying to refine algebraic expressions into a convenient form using Mathematica to make sure I don't drop a sign or make some other trivial slip. After a lot of swearing, I have come to accept that this is not a deterministic process and I will have to do it step by step and also that the algebraic manipulation palette is my friend. However, there is still one thing that's driving me nuts. Sometimes mathematica spits out expressions with these seemingly extraneous leading ones. For example, right now I'm looking at:
0.5*(1.*Log[-1.*a^2 + 1.*bigr^2] - 1.*Log[1.*a^2 - 2.*a*bigr + 1.*bigr^2])
when I would much rather be looking at:
0.5*(Log[-a^2 + bigr^2] - Log[a^2 - 2.*a*bigr + bigr^2])
It's more than just a cosmetic problem because it confuses Factor[] when I try to apply it to some of the obvious quadratic factorizations in the above expression. Any clean fixes?
your = 0.5*(1.*Log[-1.*a^2 + 1.*bigr^2] - 1.*Log[1.*a^2 - 2.*a*bigr + 1.*bigr^2])
your /. {1. -> 1, -1. -> -1}
(* -> 0.5 (Log[-a^2 + bigr^2] - Log[a^2 - 2. a bigr + bigr^2]) *)
The dot after the 1 tells mathematica to treat the number as a non-exact quantity. So for example
1. * 1/2
(* -> .5 *)
but
1 * 1/2
(* -> 1/2 *)
Use exact quantities in your calculations (1, 2, 1/2) instead of decimal numbers (1., 2. ,0.5) if you need exact results
expr = 0.5*(1.*Log[-1.*a^2 + 1.*bigr^2] - 1.*Log[1.*a^2 - 2.*a*bigr + 1.*bigr^2]);
Convert all approximate numbers to precise:
expr // Rationalize
1/2 (Log[-a^2 + bigr^2] - Log[a^2 - 2 a bigr + bigr^2])
Selectively:
expr /. x_ /; x == 1 -> 1
0.5 (Log[-1. a^2 + bigr^2] - 1. Log[a^2 - 2. a bigr + bigr^2])

How can I use Piecewise[] with a variable number of graphs/intervals

I'm writing a program to do cubic spline interpolation. Basically the program will piece together cubic polynomials over certain intervals. I would like to graph this result if all possible with piecewise[] or another similar function.
In my code I have my equations in an array that outputs like this (for example):
{2+3/4 (-1+X$6836)+1/4 (-1+X$6836)^3,3+3/2 (-2+X$6836)+3/4 (-2+X$6836)^2-1/4 (-2+X$6836)^3}
I also have another array that stores the specific intervals to graph over for each equation above, respectively:
{{1<=X$6836<=2},{2<=X$6836<=3}}
The number of equations in both arrays can be variable so I need to be able to account for this in piecewise[].
Just to make sure I understand you, you mean something like this?
eq = {2 + 3/4 (-1 + x) + 1/4 (-1 + x)^3,
3 + 3/2 (-2 + x) + 3/4 (-2 + x)^2 - 1/4 (-2 + x)^3};
cond = {{1 <= x <= 2}, {2 <= x <= 3}};
p = Piecewise[Thread[{eq, cond}]]

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