how can i solve this?

I am trying to solve this really simple code problems, but I wan not able:

Omega=zeros(1,800);

Y=cos(km2rad(20));

for n=1:800

          Omega(n)=(1/2)*[(legendreP(n-1,Y))-(legendreP(n+1,Y))];

end 

Omegazero=(1/2)*(1-Y);

OmegaT=[Omegazero Omega];

********************************************************************

The real formulation is:

Omega(n)=1/2[Pn-1(Y)-Pn+1(Y)],

OmegaZero=1/2(1-Y)

Y=cos(20km) that should be in radian.

I know the answer but I can not reach to real answer. I got something really different.

Thanks

Answer 1

[Walter Roberson]

I could see how

Bn=2n-1/n+1*Bn-1*cos(psi)-(n-2/n+1)*Bn-2

could translate to

B(n) = (2*n-1)/(n+1) * B(n-1) * cos(psi) - (n-2)/(n+1)*B(n-2);

but where are your "3" coming from in your line

B(n)=((((2*n)-3)/(n))*B(n-1)*cos(psi))-(((n-3)/(n))*B(n-2));

?

[Mahdiye]

Thank you Walter, I changed my question, could you please answer this??? Thanks

[Walter Roberson]

I have no idea what it means to express km in radians.

LP(1) = legendreP(1,Y);
LP(2) = legendreP(2,Y);

for n = 1:800
  LP(n+2) = legendreP(n+2,Y);
  Omega(n) =(1/2)*(LP(n)-LP(n+2));
end

You might need to define

legendreP = @(n,x) double( feval(symengine, 'legendreP', n, x) );

[Mahdiye]

Thanks Walter, some points:

LP(1) = legendreP(0,Y);
LP(2) = legendreP(1,Y);

LP(n+2) = legendreP(n+1,Y);

??? Am I right???

and:

what about cos(Y)???

When I use just Y=km2rad(20) instead Y=cos(km2rad(20)), it seems near to have wisely results.

It should be stated that, this is a disk on a sphere that has radius 20 km which we need angular radius in radian. So, we use km2rad for converting distance to radians.

And why we need to use:

 legendreP = @(n,x) double( feval(symengine, 'legendreP', n, x) );

??

Thanks

[Walter Roberson]

Yes you are right that LP(n+2) = legendreP(n+1,Y)

I had never seen km2rad before.

You have defined the size of the sphere (20 km) but you have not defined the angle to be subtended. km2rad(Distance) by default means

fraction = Distance / (2 * pi * radius_of_Earth_in_km)
radians = fraction * (2 * pi)

which works out as

radians = Distance / radius_of_Earth_in_km

If you want to use a sphere with a radius different than the radius of the Earth then you would need to use km2rad with two parameters. km2rad(Distance, radius_in_km). For a sphere of radius 20 km you would use km2rad(Distance, 20) which would come out as

radians = Distance / 20

But notice that you need to define the Distance.

Radians is 2 * Pi * fraction around the circle that you have traveled. And that information about how far around the circle you have traveled is missing at this time: you have only defined how big the circle is.

[Mahdiye]

Thank you Walter; For the Earth, we have 111 km on surface for 1 degree and as my disk is on the Earth with radius 20 km: angular radius: 20/111 in degree and then *pi/180 for radian. Am I right???

And still another question: Why we need to use:

 legendreP = @(n,x) double( feval(symengine, 'legendreP', n, x) );
??

Thanks

[Walter Roberson]

That calculation looks okay.

legendreP is part of the symbolic toolbox, so it is only invoked if at least one of the arguments to it is symbolic, but you are wanting to invoke it with arguments that are completely numeric. legendreP(numeric,numeric) would not be found -- without a symbolic parameter MATLAB does not have reason to look in the symbolic toolbox for a match.

legendreP = @(n,x) double( feval(symengine, 'legendreP', n, x) );

locally defines legendreP as an anonymous function with two arguments that invokes the symbolic engine (MuPAD) directly and tells the symbolic engine to execute the MuPAD function legendreP with the two given arguments. With this direct call, MATLAB will be able to find the correct function and execute it, returning a symbolic number as a solution; double() then converts the symbolic number to a floating point number that MATLAB can use.

[Walter Roberson]

Mahdiye comments "Wonderful answer"