I have this equation that I can't insert in Mathematica Abs[x-1]+Abs[x+2]=3. I tried solve [Abs[x-1]+Abs[x+2]==3] but it gives no answer.
I tried NSolve and DSolve and Evaluate but nothing worked. I can get the answer by hand -2 <= x <= 1 but I need to solve in Mathematica without Wolfram Alpha.
Reduce[Abs[x - 1] + Abs[x + 2] == 3, x, Reals]
-2 <= x <= 1
Related
I watched and tried to understand bunch of sites and videos about this, and I came into a weird conclusions and some questions. I need some help to explain which one of the method is right, or even both of them are right (but I got different result from each methods).
I'm sorry that I'm bad at explaining things, the first method is solve the equations normally. But, here the link for the video I tried to learn from
https://www.youtube.com/watch?v=o7CfCDkRwfY
Second method is to do cross product for the direction and find the point by set one of the variables as 0. https://www.youtube.com/watch?v=jozabh0lFmo
I tried for this example
x+2y+z−1=0
2x+3y−2z+2=0
and turned out for different answers. Is both of the method are correct, or which one? Thank you.
You have two equations with three unknowns.
You can eliminate one variable and solve for a relationship between the remaining two. Let's eliminate z.
Multiply the first equation by 2:
2x + 4y + 2z = 2
Add this to the second equation:
4x + 7y = 0
You can solve for y as a function of x:
y = -4x/7
Substitute this back into the first equation:
x - 8x/7 + z = 1
Simplify by combining the first and second terms:
-x/7 + z = 1
Solve for z:
z = 1 + x/7
Now you have an equation for the line in 3D space.
-inf <= x <= +inf
y = -4x/7
z = 1 + x/7
Both your equations are satisfied by these two points. Since two points are enough to define a line in Euclidean space I'd say I've got the correct answer.
This line goes through the point (0, 0, 1). It also goes through (7, -4, 2)
Here's a parametric representation of that line for -inf <= t <= +inf:
(x, y, z) = (0, 0, 1) + t*(7, -4, 1)
You can see that when t = 0 (x, y, z) = (0, 0, 1) (first point above) and when t = 1 (x, y, z) = (7, -4, 2) (second point above).
I didn't look at either of your videos. This is how I'd solve it.
This is high school algebra.
I'm trying to solve the following equation in R for x:
(z + x)/y = remainder of zero
In other words, I'm trying to find what value should be added to my number "z" so that it divided by "y" gives a remainder of zero. I couldn't find anything about it so any help would be appreciated!
Since you are referring to a "remainder", I assume you are only dealing with integers.
z <- 8
y <- 7
(x <- ceiling(z / y) * y - z)
#[1] 6
(z + x) %% y
#[1] 0
It's not the most optimised way but you can do this :
X,Y in R*
X*Y = S
S-X = Z
With this exept mistake from me your equation will alway give a remainder of zero
I am trying to help a friend in finance research. I wish to solve, for x, the equation that looks like :
g,h,c,p,a,b are all constants.
I guess the first step would be to find it's derivative. This I did, using an online derivative calculator at http://www.derivative-calculator.net/. I got this :
Further, I am trying to solve for x, assuming this is equal to zero. None of the online tools for 'solve for x' are able to do it. I have tried, Wolfram Alpha's online tool for 'solve-for-x', QuickMath, CynMath etc. All of them, 'cannot be solved'. I am looking for a solution like : x = blah-blah-blah. I have also tried the online Mathlab/Octave tools at CompileOnline/TutorialPoint. What can I do to solve for x, (preferably not having to install MathLab etc). Is there anything about these equations that render them incapable to be solved by Wolfram Alpha or such online tools for 'solve for x' ?
You could just write a function in R to solve your equation for a given value of x:
solvex <- function(x) {
g = 1
h = 1
c = 1
p = 1
b = 1
a = 1
g * (1 - exp(h*x + c)) + p * (1 - exp(b*x + a))
}
Then, to plot the solutions over some range, do something like:
x <- seq(-100,100,1)
yseq <- lapply(x, solvex)
plot(x, yseq, type = "l")
Finding a solution using Mathematica:
I'm trying to force Mathematica to implicitly differentiate an ellipse equation of the form:
x^2/a^2+y^2/b^2 == 100
with a = 8 and b = 6.
The command I'm using looks like this:
D[x^2/a^2 + y^2/b^2 == 100/. y -> 3/4*Sqrt[6400-x^2], x]
where, y->3/4*Sqrt[6400-x^2] comes from solving y in terms of x.
I got this far by following the advice found here: http://www.hostsrv.com/webmaa/app1/MSP/webm1010/implicit
Input for this script is the conventional way that an implicit
relationship beween x and y is expressed in calculus textbooks. In
Mathematica you need to make this relationship explicit by using y[x]
in place of y. This is done automatically in the script by replacing
all occurances of y with y[x].
But the solution Mathematica gives does not have y' or dy/dx in it (like when I solved it by hand). So I don't think it's been solved correctly. Any idea on what command would get the program to solve an implicit differential? Thanks.
The conceptually easiest option (as you mentioned) is to make y a function of x and use the partial derivative operator D[]
In[1]:= D[x^2/a^2 + y[x]^2/b^2 == 100, x]
Solve[%, y'[x]]
Out[1]= (2 x)/a^2 + (2 y[x] y'[x])/b^2 == 0
Out[2]= {{y'[x] -> -((b^2 x)/(a^2 y[x]))}}
But for more complicated relations, it's best to use the total derivative operator Dt[]
In[3]:= SetOptions[Dt, Constants -> {a, b}];
In[4]:= Dt[x^2/a^2 + y^2/b^2 == 100, x]
Solve[%, Dt[y, x]]
Out[4]= (2 x)/a^2 + (2 y Dt[y, x, Constants -> {a, b}])/b^2 == 0
Out[5]= {{Dt[y, x, Constants -> {a, b}] -> -((b^2 x)/(a^2 y))}}
Note that it might be neater to use SetAttributes[{a, b}, Constant] instead of the SetOptions[Dt, Constants -> {a, b}] command... Then the Dt doesn't carry around all that extra junk.
The final option (that you also mentioned) is to solve the original equation for y[x], although this is not always possible...
In[6]:= rep = Solve[x^2/a^2 + y^2/b^2 == 100, y]
Out[6]= {{y -> -((b Sqrt[100 a^2 - x^2])/a)}, {y -> (b Sqrt[100 a^2 - x^2])/a}}
And you can check that it satisfies the differential equation we derived above for both solutions
In[7]:= D[y /. rep[[1]], x] == -((b^2 x)/(a^2 y)) /. rep[[1]]
Out[7]= True
You can substitute your values a = 8 and b = 6 anytime with replacement rule {a->8, b->6}.
If you actually solve your differential equation y'[x] == -((b^2 x)/(a^2 y[x]) using DSolve with the correct initial condition (derived from the original ellipse equation) then you'll recover the solution for y in terms of x given above.
I know this isn't exactly programming related per se, but programmers are the most
probable of all people who will recognize this maybe.
I have the following (X and Y are arrays, both with 3 elements), and I cannot recognize (although it reminds me of a few things, but none quite!) what is being done here. Does it ring any bells for anyone else ?
I gather you can disregard the lower part; the upper should probably give it away ... but I still cannot see it.
At first it reminded me of linear interpolation in 3d space ...
SUBROUTINE TRII(X,Y,XR,YR)
DIMENSION X(3),Y(3)
D=X(1)*(X(2)**2-X(3)**2)+
> X(2)*(X(3)**2-X(1)**2)+
> X(3)*(X(1)**2-X(2)**2)
D1=Y(1)*(X(2)*X(3)**2-X(3)*X(2)**2)+
> Y(2)*(X(3)*X(1)**2-X(1)*X(3)**2)+
> Y(3)*(X(1)*X(2)**2-X(2)*X(1)**2)
D2=Y(1)*(X(2)**2-X(3)**2)+
> Y(2)*(X(3)**2-X(1)**2)+
> Y(3)*(X(1)**2-X(2)**2)
D3=X(2)*(Y(3)-Y(1))+
> X(1)*(Y(2)-Y(3))+
> X(3)*(Y(1)-Y(2))
A=D1/D
B=D2/D
C=D3/D
YR=A+B*XR+C*XR**2
RETURN
END
SUBROUTINE TRIM(X,Y,XR,YR,XM,YM)
DIMENSION X(3),Y(3)
D=X(1)*(X(2)**2-X(3)**2)+
> X(2)*(X(3)**2-X(1)**2)+
> X(3)*(X(1)**2-X(2)**2)
D1=Y(1)*(X(2)*X(3)**2-X(3)*X(2)**2)+
> Y(2)*(X(3)*X(1)**2-X(1)*X(3)**2)+
> Y(3)*(X(1)*X(2)**2-X(2)*X(1)**2)
D2=Y(1)*(X(2)**2-X(3)**2)+
> Y(2)*(X(3)**2-X(1)**2)+
> Y(3)*(X(1)**2-X(2)**2)
D3=X(2)*(Y(3)-Y(1))+
> X(1)*(Y(2)-Y(3))+
> X(3)*(Y(1)-Y(2))
A=D1/D
B=D2/D
C=D3/D
XR=-B/(2.*C)
YR=A+B*XR+C*XR**2
XM=XR
IF(XR.GT.X(1).OR.XR.LT.X(3))XM=X(1)
YM=A+B*XM+C*XM**2
IF(YM.LT.Y(1))XM=X(1)
IF(YM.LT.Y(1))YM=Y(1)
RETURN
END
">" is a continuation sign.
The code run as follows
Routine TRII takes as input the coordinates of three points (x,y) and interpolates a parabola using Lagrange interpolation. Also takes as input the coordinate XR. Returns in YR the value at XR for the interpolating parabola.
I guess the name of the routine comes from "TRI" (Croatian for "three" (points)) and "I" for Interpolation.
Routine TRIM also calculates the same parabola, and returns the minimun value of the function in the interval {X(1),X(3)}.The name comes from "TRI" and "M" (minimum)
(I "really" executed the program) >)
Note that this is FORTRAN code and the parameters are passed by reference, so the results are returned back in the same parameters (very odd!)
Edit
Just for fun, let's run TRII
TRII[X_, Y_, XR_] :=
Module[{D0, D1, D2, D3, A, B, C},
D0 = X[[1]]*(X[[2]]^2 - X[[3]]^2) +
X[[2]]*(X[[3]]^2 - X[[1]]^2) +
X[[3]]*(X[[1]]^2 - X[[2]]^2);
D1 = Y[[1]]*(X[[2]]*X[[3]]^2 - X[[3]]*X[[2]]^2) +
Y[[2]]*(X[[3]]*X[[1]]^2 - X[[1]]*X[[3]]^2) +
Y[[3]]*(X[[1]]*X[[2]]^2 - X[[2]]*X[[1]]^2);
D2 = Y[[1]]*(X[[2]]^2 - X[[3]]^2) +
Y[[2]]*(X[[3]]^2 - X[[1]]^2) +
Y[[3]]*(X[[1]]^2 - X[[2]]^2);
D3 = X[[2]]*(Y[[3]] - Y[[1]]) +
X[[1]]*(Y[[2]] - Y[[3]]) +
X[[3]]*(Y[[1]] - Y[[2]]);
A = D1/D0;
B = D2/D0;
C = D3/D0;
Return[A + B*XR + C*XR^2];];
X = RandomReal[1, 3];
Y = RandomReal[1, 3];
Show[Plot[TRII[X, Y, x], {x, 0, 1}],
ListPlot[Transpose[{X, Y}], PlotMarkers -> Automatic]]
D is the determinant of the matrix:
| x(1) x(1)² 1 |
D = det | x(2) x(2)² 1 |
| x(3) x(3)² 1 |
In D1, the rightmost column has been replaced with Y:
| x(1) x(1)² Y(1) |
D1 = det | x(2) x(2)² Y(2) |
| x(3) x(3)² Y(3) |
In D2, and D3 it's the first and second columns, respectively. Is it easier to recognize now? Looks a lot like using Cramer's rule to solve a linear equation to me.
Edit: To be more precise: (A, B, C) is the solution to the system:
A + x(1)*B + x(1)²*C = Y(1)
A + x(2)*B + x(2)²*C = Y(2)
A + x(3)*B + x(3)²*C = Y(3)
YR is the square of the solution to the quadratic equation (nb, different x!):
C*x² + B*x + A = 0
I feel like this should be obvious now, but I can't quite grasp it...
This code represents a kind of interpolation/quadratic curve fitting on three 2d points together with a way to compute the minimum or maximum value of such a fitted quadratic within the interval itself. I guess that TRII stands for triple (point)-interpolation and TRIM stands for triple (point) minimum or maximum.
To be more precised TRII solves the problem :- find a quadratic curve that passes through the points (x1,y1),(x2,y2) and (x3,y3) in the form Y=A+BX+CX^2 and compute the Y value of the quadratic at the point XR and return as YR. This is basically a way to interpolate smoothly between three 2d points. It is often used to find a better approximation for the max or min value of a set of discrete data points.
All the D, D1, D2, D3 stuff is to solve the matrix equation:
(1 X1 X1^2) *(A) = (Y1)
(1 X2 X2^2) *(B) = (Y2)
(1 X3 X3^2) *(C) = (Y3)
using Cramers rule as mentioned in one of the other comments, D is the matrix determinant and D1, D2, D3 are co-factors.
TRIM again computes the quadratic Y=A+BX+CX^2 and then finds a max/min of this quadratic (XM, YM). This is done by initially finding the point where the quadratic has a turning point: if F(X)=A+BX+CX^2, F'(XR)=B+2*C*XR=0, or XR=-B/2*C, YR=A+BXR+CXR^2. There is then some logic to force the returned XM, YM min or max values to lie within certain bounds.
The code:
XM=XR
.
.
.
IF(YM.LT.Y(1))YM=Y(1)
Is a little weird since if we assume that GT and LT mean greater than and less than respectively then we need to assume that X3'<'X1 otherwise the condition (XR.GT.X(1).OR.XR.LT.X(3)) is trivial and XM,YM are set to X1, Y1.
So X3'<'X1 and the condition says that if the quadratics max/min value is outside the interval (X1,X3) then set (XM,YM) to (X1, Y1) as before. If not then if Y1 is above the min/max value in Y then again set (XM,YM) to (X1, Y1).
It is hard to understand what this means and I suspect the code may be wrong! Any thoughts?
Ivan
I'm not sure what language this is, but it's clear that this is some sort of solver for quadratic equations. The XR and YR expressions are a dead giveaway:
XR = -B / (2.*C)
YR = A + B*XR + C*XR**2
Without knowing what the X(1..3) and Y(1..3) expressions are, however, it's not going to be possible to infer too much more about what the A/B/C coefficients represent, however. Lots of things use quadratic equations -- area of a circle given the radius, intensity of light at a given distance, et cetera. More contextual data is required.
Update: The OP indicated that he can't be too much more specific for secrecy reasons. Here are some hints, though:
What does the subroutine return? How are those results used later on? That may lead to better insights.
It appears that Y(1) is some sort of magic lower bound for the result of this computation. Notice that if YM is less than Y(1), then both XM and YM are set to X(1) and Y(1), respectively.
The "D" expressions look like this, in more natural syntax:
d = x1 * [x2^2 - x3^2] + x2 * [x3^2 - x1^2] + x3 * [x1^1 - x2^2]
d1 = y1 * [x2*x3^2 - x3*x2^2] + y2 * [x3*x1^2 - x1*x3^2] + y3 * [x1*x2^2 - x1*x2^2]
d2 = y1 * [x2^2 - x3^2] + y2 * [x3^2 - x1^2] + y3 * [x1^2 - x2^2]
d3 = x2 * [y3 - y1] + x1 * [y2 - y3] * x3 * [y1 - y2]
This looks very much like some sort of matrix operation; D is almost certainly for "determinant". But there are other things that have the same mathematical relationship.
This is a way to solve linear equation systems, specifically cramers rule. Also have a look at the rule of sarrus. After that, you seem to construct a quadratic equation out of it.