I've been trying to solve a system of linear equations , but I didn't get this done. I hope getting an answer if it's solvable .
Say we have an angle A and this system :
y - x = sin(A)
x - z = sin(A+100)
z - y = sin(A+200)
I'm looking for `x and y and z , the value of the angle isn't constant, but it's known !
If there is a numerical way to solve it I'll be intressed .
PS: this NOT a homework. and everything I've tried is written on papers that 's the reason I didn't put here.
If you add all equations you get
sin(A) + sin(A + 100) + sin(A + 200) = 0
So your system isn't solvable if the above sum is not zero. If it's zero then you have infinite solutions:
x = x, y = x + sin(A), z = x - sin(A + 100)
where x can have any value.
The system can be rearranged to
0 = sin(A) - sin(A+100) - sin(A+200)
x - z = sin(A+100)
- y + z = sin(A+200)
which means that there is no solution if sin(A) - sin(A+100) - sin(A+200) != 0 holds. Otherwise, chose y at will and setting x=sin(A+100)+y and z=sin(A)-y yields a solution.
Related
I want to calculate the sum of the series as below
Lim
X->1 (2/3 - x/3 -(x^2)/3 +(x^3)*2/3 -..). I am not sure whether we have a formula for finding the sum of this kind of series. Tried a lot but couldn't find any. Any help is appreciated.
This seems to be more maths than computing.
It factorises as (1 + x^3 + x^6 + ...)(2 - x - x^2)/3
If x = 1-d (where d is small), then to first order in d, the (2 - x - x^2) term becomes (2 - (1-d) - (1-2d)) = 3d
And the (1 + x^3 + x^6 + ...) term is a geometric progression, with sum 1/(1-x^3), or here 1/(1-(1-d)^3), and the denominator to first order in d is (1 - (1-3d)) = 3d
Hence the whole thing is (1/3d) (3d) / 3 = 1/3
But we can also verify computationally with a value close to 1 (Python code here):
x = 0.999999
s = 0
f = (2 - x - x*x) / 3.
x3 = x ** 3
s_prev = None
while s != s_prev:
s_prev = s
s += f
f *= x3
print(s)
gives:
0.33333355556918565
First of all, this looks like a fair amount of Calculus, so I predict that it would get forwarded to Cross-Validated by someone who thinks that this is TL;DR. But I think this is a programming question, so here me out.
Imagine that I have the following functions in terms of x: f(x), g(x), h(x) ...
f(x) = 2x^2 + 4x - 30
g(x) = x^2 - x + 12
h(x) = f(x) - g(x) = (2x^2 + 4x - 30) - (x^2 - x + 12) = x^2 + 5*x - 42
Note: If I were to compute g(x) - f(x) here I would get a different polynomial, but I would get the same roots so it doesn't really matter because if I took the coefficients from g(x) - f(x), then polyroot() would return the same x-intercept intersection points as f(x) = g(x).
I am able to resolve h(x) = (2x^2 + 4x - 30) - (x^2 - x + 12), but I can't resolve it to x^2 + 5*x - 42 which is just a more simplified version of the same function of h(x). But I need it in this form to compute the intersections of these functions where I need the coefficients of the difference function. Then I would use the points of intersection to compute the difference integral over the greater function minus the smaller function over the range where the functions intersect, and this difference integral is simply the area between the functions.
So my goal is to compute the area between two intersecting functions.
My problem is that I want to automate the whole process, and I want to simply the h(x) difference function to 1*x^2 + 5*(x) - 42, where the coefficients of this polynomial function in increasing order are -42, 5, 1 in that order.
So let's just write the code:
fx <- function(x){2*x^2 + 4*x - 30}
gx <- function(x){1*x^2 - 1*x + 12}
hx <- function(x){fx - gx} # doesn't work because I can't pass it to curve(hx)
hx <- function(x){(2*x^2 + 4*x - 30) - (1*x^2 - 1*x + 12)} # works
but it is not in the form that I want.
> hx
function(x){(2*x^2 + 4*x - 30) - (1*x^2 - 1*x + 12)}
<bytecode: 0x000000001c0bfc10>
Errors:
> curve(hx)
Error in expression(fx) - expression(gx) :
non-numeric argument to binary operator
See this is why I need the coefficients.
> z <- polyroot(c(-42, 5, 1)) # polyroot functions give you the x-intercepts of a polynomial function.
> z
[1] 4.446222-0i -9.446222+0i
Of course I could just compute "x^2 + 5*x - 42" on pen and paper, but they say that programmers always want to find the most efficient algorithmic process with the least amount of work.
Now I need to see which function is greater than the other, over the given range. Two ways visually or incrementally. (This is for the Calculus II part.)
x = seq(from = -9.4, to = 4.4, by = 0.2)
fx_range = 2*x^2 + 4*x - 30
> table(fx_range >= gx_range)
FALSE
70
> table(gx_range >= fx_range)
TRUE
70
It looks like the g(x) function is greater than or equal to the f(x) function over the range of the intersection points. So should evalulate the integral of g(x) - f(x) according to calculus. I was just doing f(x) - g(x) earlier for the polyroot function.
Areabetween curves = (from -9.446222 to 4.446222) ∫[g(x) - f(x)]dx
= (from -9.446222 to 4.446222) ∫[(x^2 - x + 12) - (2*x^2 + 4*x - 30)]
gx_minus_fx = function(x){(x^2 - x + 12) - (2*x^2 + 4*x - 30)}
Area = integrate(gx_minus_fx, lower = -9.446222, upper = 4.446222)
Area
446.8736 with absolute error < 5e-12 # This is exactly what I wanted to compute!
Now let's graphically check if I was supposed to subtract g(x) - f(x):
> curve(fx, main = "Functions with their Intersection Points", xlab = "x", ylab = "Functions of x", from = -9.446222, to = 4.446222)
> curve(gx, col = "red", add = TRUE)
> legend("topright", c("f(x) = 2x^2 + 4x - 30", "g(x) = x^2 - x + 12"), fill = c("black", "red"))
Yeah, I did it right!
So again, what I would like help with is figuring out how I could simplify
h(x) = f(x) - g(x) to x^2 + 5*x - 42.
This appears to be an algebraic problem. I showed that I could do high-level Calculus 2 in R, and I would just like to know if there is a way that I can automate this whole process for the h(x) function.
Thank you!!!
This is a fairly simple question. I need need an equation to determine whether two 2 dimensional lines collide with each other. If they do I also need to know the X and Y position of the collision.
Put them both in general form. If A and B are the same then they're parallel. Otherwise, create two simultaneous equations and solve for x and y.
Let A and B represented by this parametric form : y = mx + b
Where m is the slope of the line
Now in the case of parallelism of A and B their slope should be equal
Else they will collide with each other at point T(x,y)
For finding the coordinates of point T you have to solve an easy equation:
A: y = mx + b
B: y = Mx + B
y(A) = y(B) means : mx + b = Mx + B which yields to x = (B - b)/(m - M) and by putting
the x to the line A we find y = ((m*(B - b))/(m - M)) + b
so : T : ((B - b)/(m - M) , ((m*(B - b))/(m - M)) + b)
I have a square bitmap of a circle and I want to compute the normals of all the pixels in that circle as if it were a sphere of radius 1:
The sphere/circle is centered in the bitmap.
What is the equation for this?
Don't know much about how people program 3D stuff, so I'll just give the pure math and hope it's useful.
Sphere of radius 1, centered on origin, is the set of points satisfying:
x2 + y2 + z2 = 1
We want the 3D coordinates of a point on the sphere where x and y are known. So, just solve for z:
z = ±sqrt(1 - x2 - y2).
Now, let us consider a unit vector pointing outward from the sphere. It's a unit sphere, so we can just use the vector from the origin to (x, y, z), which is, of course, <x, y, z>.
Now we want the equation of a plane tangent to the sphere at (x, y, z), but this will be using its own x, y, and z variables, so instead I'll make it tangent to the sphere at (x0, y0, z0). This is simply:
x0x + y0y + z0z = 1
Hope this helps.
(OP):
you mean something like:
const int R = 31, SZ = power_of_two(R*2);
std::vector<vec4_t> p;
for(int y=0; y<SZ; y++) {
for(int x=0; x<SZ; x++) {
const float rx = (float)(x-R)/R, ry = (float)(y-R)/R;
if(rx*rx+ry*ry > 1) { // outside sphere
p.push_back(vec4_t(0,0,0,0));
} else {
vec3_t normal(rx,sqrt(1.-rx*rx-ry*ry),ry);
p.push_back(vec4_t(normal,1));
}
}
}
It does make a nice spherical shading-like shading if I treat the normals as colours and blit it; is it right?
(TZ)
Sorry, I'm not familiar with those aspects of C++. Haven't used the language very much, nor recently.
This formula is often used for "fake-envmapping" effect.
double x = 2.0 * pixel_x / bitmap_size - 1.0;
double y = 2.0 * pixel_y / bitmap_size - 1.0;
double r2 = x*x + y*y;
if (r2 < 1)
{
// Inside the circle
double z = sqrt(1 - r2);
.. here the normal is (x, y, z) ...
}
Obviously you're limited to assuming all the points are on one half of the sphere or similar, because of the missing dimension. Past that, it's pretty simple.
The middle of the circle has a normal facing precisely in or out, perpendicular to the plane the circle is drawn on.
Each point on the edge of the circle is facing away from the middle, and thus you can calculate the normal for that.
For any point between the middle and the edge, you use the distance from the middle, and some simple trig (which eludes me at the moment). A lerp is roughly accurate at some points, but not quite what you need, since it's a curve. Simple curve though, and you know the beginning and end values, so figuring them out should only take a simple equation.
I think I get what you're trying to do: generate a grid of depth data for an image. Sort of like ray-tracing a sphere.
In that case, you want a Ray-Sphere Intersection test:
http://www.siggraph.org/education/materials/HyperGraph/raytrace/rtinter1.htm
Your rays will be simple perpendicular rays, based off your U/V coordinates (times two, since your sphere has a diameter of 2). This will give you the front-facing points on the sphere.
From there, calculate normals as below (point - origin, the radius is already 1 unit).
Ripped off from the link above:
You have to combine two equations:
Ray: R(t) = R0 + t * Rd , t > 0 with R0 = [X0, Y0, Z0] and Rd = [Xd, Yd, Zd]
Sphere: S = the set of points[xs, ys, zs], where (xs - xc)2 + (ys - yc)2 + (zs - zc)2 = Sr2
To do this, calculate your ray (x * pixel / width, y * pixel / width, z: 1), then:
A = Xd^2 + Yd^2 + Zd^2
B = 2 * (Xd * (X0 - Xc) + Yd * (Y0 - Yc) + Zd * (Z0 - Zc))
C = (X0 - Xc)^2 + (Y0 - Yc)^2 + (Z0 - Zc)^2 - Sr^2
Plug into quadratic equation:
t0, t1 = (- B + (B^2 - 4*C)^1/2) / 2
Check discriminant (B^2 - 4*C), and if real root, the intersection is:
Ri = [xi, yi, zi] = [x0 + xd * ti , y0 + yd * ti, z0 + zd * ti]
And the surface normal is:
SN = [(xi - xc)/Sr, (yi - yc)/Sr, (zi - zc)/Sr]
Boiling it all down:
So, since we're talking unit values, and rays that point straight at Z (no x or y component), we can boil down these equations greatly:
Ray:
X0 = 2 * pixelX / width
Y0 = 2 * pixelY / height
Z0 = 0
Xd = 0
Yd = 0
Zd = 1
Sphere:
Xc = 1
Yc = 1
Zc = 1
Factors:
A = 1 (unit ray)
B
= 2 * (0 + 0 + (0 - 1))
= -2 (no x/y component)
C
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2 + (0 - 1) ^ 2 - 1
= (X0 - 1) ^ 2 + (Y0 - 1) ^ 2
Discriminant
= (-2) ^ 2 - 4 * 1 * C
= 4 - 4 * C
From here:
If discriminant < 0:
Z = ?, Normal = ?
Else:
t = (2 + (discriminant) ^ 1 / 2) / 2
If t < 0 (hopefully never or always the case)
t = -t
Then:
Z: t
Nx: Xi - 1
Ny: Yi - 1
Nz: t - 1
Boiled farther still:
Intuitively it looks like C (X^2 + Y^2) and the square-root are the most prominent figures here. If I had a better recollection of my math (in particular, transformations on exponents of sums), then I'd bet I could derive this down to what Tom Zych gave you. Since I can't, I'll just leave it as above.
how can i calculate the polynomial that has the tangent lines (1) y = x where x = 1, and (2) y = 1 where x = 365
I realize this may not be the proper forum but I figured somebody here could answer this in jiffy.
Also, I am not looking for an algorithm to answer this. I'd just like like to see the process.
Thanks.
I guess I should have mentioned that i'm writing an algorithm for scaling the y-axis of flotr graph
The specification of the curve can be expressed as four constraints:
y(1) = 1, y'(1) = 1 => tangent is (y=x) when x=1
y(365) = 1, y'(365) = 0 => tangent is (y=1) when x=365
We therefore need a family of curves with at least four degrees of freedom to match these constraints; the simplest type of polynomial is a cubic,
y = a*x^3 + b*x^2 + c*x + d
y' = 3*a*x^2 + 2*b*x + c
and the constraints give the following equations for the parameters:
a + b + c + d = 1
3*a + 2*b + c = 1
48627125*a + 133225*b + 365*c + d = 1
399675*a + 730*b + c = 0
I'm too old and too lazy to solve these myself, so I googled a linear equation solver to give the answer:
a = 1/132496, b = -731/132496, c = 133955/132496, d = -729/132496
I will post this type of question in mathoverflow.net next time. thanks
my solution in javascript was to adapt the equation of a circle:
var radius = Math.pow((2*Math.pow(365, 2)), 1/2);
var t = 365; //offset
this.tMax = (Math.pow(Math.pow(r, 2) - Math.pow(x, 2), 1/2) - t) * (t / (r - t)) + 1;
the above equation has the above specified asymptotes. it is part of a step polynomial for scaling an axis for a flotr graph.
well, you are missing data (you need another point to determine the polynomial)
a*(x-1)^2+b*(x-1)+c=y-1
a*(x-365)^2+b*(x-365)+c=y-1
you can solve the exact answer for b
but A depends on C (or vv)
and your question is off topic anyways, and you need to revise your algebra