We have the following equation, that we want to implement into our code with arrays/matrices
h(x, y, z) = ax + by + cz
pseudeo code:
X = [a, b, c]
A = [x, y, z]
Often I see the equation being implemented like this:
h = transpose(A) * T
Is there any difference of just using the dot product?
h = dotproduct(A, X)
Is there a specific reason why the transpose is used over the dotproduct?
Mathematically, there's no difference. In implementation- dotproduct(...) may actually be faster, since transpose(...) may eagerly move the elements around prior to the matrix multiplication, doing "unnecessary" work.
Related
How do you write a F# recursive function 'something' that takes a single point (x, y) and index i as
an arguments and returns an ๐๐กโ element of the infinite list corresponding to S(x, y).
ex: let something (x,y) i =
F# function F(๐ฅ,๐ฆ) should be defined as:
F(x,y)(u, v) = (u2 โ v2 + x, 2uv + y)
F# function 'something' ๐(x, y) of a point (x, y) should be an infinite list of items:
S(x, y) = {(0, 0), F(x,y)(0, 0), F(๐ฅ,y)(F(x,y)(0, 0)), F(x,y)(F(x,y)(F(x,y)(0, 0))), โฆ}
This looks like an assignment to me. I don't mind helping out but at the same time I don't want to give the full solution.
The exercise itself looks like generating the series:
Z' = Z*Z + C
For a complex number Z and C. This is commonly done when generating the mandelbrot or julia set.
The function F can be written almost like the definition in F#:
let f (x, y) (u, v) = (u*u - v*v + x, 2.*u*v + y)
The infinite set S is generated from a starting point (0,0) and applying the output of f on itself over and over again.
An infinite set in F# can be represented using seq and you can create them using seq comprehesions
Once you have an infinite seq with the right values you can pick the ith value by using Seq.item
I need to store 3D normal vectors, that is vectors (x, y, z) such that x^2 + y^2 + z^2 = 1. But due to space constraints I can only use 2 floats to store it.
So by storing only x and y, the third component can be computed as sqrt(1 - x^2 - y^2), i.e. one square root, two products and two subtractions.
What would be the most efficient way to store the vectors, so that reading them is as fast as possible, and if possible without bias towards one spatial direction?
Edit
Now using the values (a, b) with a = x - y and b = x + y.
You could satisfy your space constraint by storing the vectors via spherical coordinates. As is well known, each point on the unit sphere, i.e., each unit vector, has at least one pair of spherical coordinates characterizing it.
Or if you want something convoluted: The complex square function maps the unit disk to a double cover of it. So you could use the left half-disk for the upper half-sphere and the right half-disk for the lower half-sphere.
SphereFromDisk(a,b)
a2=a*a; b2=b*b; r2=a2+b2; // assert r2 <= 1
x = a2 - b2;
y = 2*a*b
z = sqrt(1-r2*r2)
if(a<0 or (a=0 and b<0) z=-z
return (x,y,z)
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.
Does anyone know how to minimize a function containing an integral in MATLAB? The function looks like this:
L = Int(t=0,t=T)[(AR-x)dt], A is a system parameter and R and x are related through:
dR/dt = axRY - bR, where a and b are constants.
dY/dt = -xRY
I read somewhere that I can use fminbnd and quad in combination but I am not able to make it work. Any suggestions?
Perhaps you could give more details of your integral, e.g. where is the missing bracket in [AR-x)dt]? Is there any dependence of x on t, or can we integrate dR/dt = axR - bR to give R=C*exp((a*x-b)*t)? In any case, to answer your question on fminbnd and quad, you could set A,C,T,a,b,xmin and xmax (the last two are the range you want to look for the min over) and use:
[x fval] = fminbnd(#(x) quad(#(t)A*C*exp((a*x-b)*t)-x,0,T),xmin,xmax)
This finds x that minimizes the integral.
If i didn't get it wrong you are trying to minimize respect to t:
\int_0^t{(AR-x) dt}
well then you just need to find the zeros of:
AR-x
This is just math, not matlab ;)
Here's some manipulation of your equations that might help.
Combining the second and third equations you gave gives
dR/dt = -a*(dY/dt)-bR
Now if we solve for R on the righthand side and plug it into the first equation you gave we get
L = Int(t=0,t=T)[(-A/b*(dR/dt + a*dY/dt) - x)dt]
Now we can integrate the first term to get:
L = -A/b*[R(T) - R(0) + Y(T) - Y(0)] - Int(t=0,t=T)[(x)dt]
So now all that matters with regards to R and Y are the endpoints. In fact, you may as well define a new function, Z which equals Y + R. Then you get
L = -A/b*[Z(T) - Z(0)] - Int(t=0,t=T)[(x)dt]
This next part I'm not as confident in. The integral of x with respect to t will give some function which is evaluated at t = 0 and t = T. This function we will call X to give:
L = -A/b*[Z(T) - Z(0)] - X(T) + X(0)
This equation holds true for all T, so we can set T to t if we want to.
L = -A/b*[Z(t) - Z(0)] - X(t) + X(0)
Also, we can group a lot of the constants together and call them C to give
X(t) = -A/b*Z(t) + C
where
C = A/b*Z(0) + X(0) - L
So I'm not sure what else to do with this, but I've shown that the integral of x(t) is linearly related to Z(t) = R(t) + Y(t). It seems to me that there are many equations that solve this. Anyone else see where to go from here? Any problems with my math?
I know the coordinates of A, B and C.. I also know of a vector V originating from C..
I know that the vector intersects A and B, I just don't know how to find i.
Can anyone explain the steps involved in solving this problem?
Thanks alot.
http://img34.imageshack.us/img34/941/triangleprob.png
If you know A and B, you know equation for the line AB, and you said you know V, so you can form the equation for Line V.... Well i is only point that satisfies both those equations.
Equation for Line AB:
(bx-ax)(Y-ay) = (by-ay)(X-ax)
If you knpow the direction (or slope = m) of the vector, and any point that lies on the vector, then the equation of the line for vector V is
Y = mX = b
where m is the slope or direction of the line, and b is the y coordinate where it crosses thevertical y=axis (where X = 0)
if you know a point on the line (i.e., C = (s, t) then you solve for b by:
t = ms + b ==> b = t - ms,
so equation becomes
Y = mX + t-ms
i = C+kV
Lets call N the normal to the line A,B so N = [-(B-A).y, (B-A).x]
Also, for any point on the line:
(P-A)*N = 0 -- substitute from line 1 above:
(C+kV-A)*N = 0
(kV+C-A)*N = 0
kV*N + (C-A)*N = 0
kV*N = (A-C)*N
k = [(A-C)*N]/V*N
Now that we have k, plug it into line 1 above to get i.
Here I'm using * to represent dot product so expanding to regular multiplication:
k = ((A.x-C.x)*-(B.y-A.y) + (A.y-C.y)*(B.x-A.x)) / (V.x*-(B.y-A.y) + V.x*(B.x-A.x))
I.x = C.x + k*V.x
I.y = C.y + k*V.y
Unless I screwed something up....
Simple algebra. The hard part is often just writing down the basic equations, but once written down, the rest is easy.
Can you define a line that emanates from the point C = [c_x,c_y], and points along the vector V = [v_x,v_y]? A nice way to represent such a line is to use a parametric representation. Thus,
V(t) = C + t*V
In terms of the vector elements, we have it as
V(t) = [c_x + t*v_x, c_y + t*v_y]
Look at how this works. When t = 0, we get the point C back, but for any other value of t, we get some other point on the line.
How about the line segment that passes through A and B? One way to solve this problem would be to define a second line parametrically in the same fashion. Then solve for a system of two equations in two unknowns to find the intersection.
An easier approach is to look at the normal vector to the line segment AB. That vector is given as
N = [b_y - a_y , a_x - b_x]/sqrt((b_x - a_x)^2 + (b_y - a_y)^2)
Note that N is defined here to have a unit norm.
So now, when do we know if a point happens to lie along the line that connects A and B? This is easy now. That will happen when the dot product defined below is exactly zero.
dot(N,V(t) - A) = 0
Expand this, and solve for the parameter t. We can write it down using dot products.
t = dot(N,A-C)/dot(N,V)
Or, if you prefer,
t = (N_x*(a_x - c_x) + N_y*(a_y - c_y)) / (N_x*v_x + N_y*v_y))
And once we have t, substitute into the expression above for V(t). Lets see all of this work in practice. I'll pick some points A,B,C and a vector V.
A = [7, 3]
B = [2, 5]
C = [1, 0]
V = [1, 1]
Our normal vector N, after normalization, will look something like
N = [0.371390676354104, 0.928476690885259]
The line parameter, t, is then
t = 3.85714285714286
And we find the point of intersection as
C + t*V = [4.85714285714286, 3.85714285714286]
If you plot the points on a piece of paper it should all fit together, and all in only a few simple expressions.