I need to find the coordinates of point M. Has anyone got any idea of how to find them?
Thanks in advance
MATLAB: Find coordinates point M
coordinates
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You should use more than 3 points to get better noise/error-resistance. Also, if your transformation is a rotation/translation, you need to constrain M to be of this form. This can be done with ABSOR ( Download ),
reg=absor(P,T)There is no consistent numeric value you can obtain when taking the ration inf/inf.
Should it be 1? Clearly x/x is always 1, for any non-zero NUMBER x. But inf is not really a number. And since 2*inf is also inf, then logically, we could conclude that if inf/inf was 1, then
2 = 2*1 = 2*(inf/inf) = (2*inf)/inf = inf/inf = 1Just as easily, we could construct any number of other nonsense results, that is, if inf/inf were defined to take on any value.
The point is, once your computations have reduced to inf/inf, you are already in a deep hole, one from which you will not escape.
Instead, you need to do some thinking. See if you can avoid computing those infs in the first place.
syms x ra=1; b=0; dx=5; t=40;K=10;f(x) = a*x+b;df(x)=diff(f(x));r(1)=inf;Now, lets see what happens when you try to start that loop, trying to create r(2).
i = 1;K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))ans =NaNThat expression has a few terms in it.
K*int((1+df(x)).^(1/2),x,i,i+dx)ans =50*2^(1/2)(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))ans =NaN(r(i)*(K*f(i*dx)+t))ans =Inf(r(i)+(K*f(i*dx)+t))ans =InfSo, you have inf/inf, which as I showed before, is an expression that has no reasonable value that can be assigned.
You may have "done it by hand" but if you did so, you made up a result that you decided you like. So, lets go more deeply into that expression for r(2).
You have what is essentially
50*2^(1/2) + inf/infand now, you want the inf/inf to magically be 90?
Hmm. I think you are now making up a result. As we saw before, inf/inf is anything you wish it to be. Personally, I think 17 or 42 are better choices for results there. But what have you done? And why do you think the result of
(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))ans =NaNshould be 90?
(K*f(i*dx)+t)ans =90Hmm, so now you have
(inf*90)/(inf+90)And now you think that
inf + 90 = infso 90*inf/inf should be 90, since inf/inf should be 1.
SHEESH!
It seems that you are playing fast and loose with infs here.
Lets see what happens if r(1) is just some large number.
r(1) = 1/eps;for i=1:10 r(i+1)=K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t)); enddouble(r)ans = Columns 1 through 7 4.5035996273705e+15 160.710678118653 145.531748914597 153.120263450657 164.190629789399 175.546135998982 186.482443382689 Columns 8 through 11 196.869165257471 206.723594803462 216.097689853589 225.046227726715Thus, by making r(1) a finite value, while still large in context of double precision arithmetic, it is not infinite. Now we get a value that is at least somewhat more reasonable.
So, can we do this using mathematics? A limit seems like what you really wanted to do.
syms u realr(1) = u;for i=1:10 r(i+1)=K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t));endr = limit(r,u,inf);r(1)ans =Infr(2)ans =50*2^(1/2) + 90r(3)ans =50*2^(1/2) + (7000*2^(1/2) + 12600)/(50*2^(1/2) + 230)double(r(11))ans = 225.046227726715I imagine that is what you wanted to see.
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