Solve equation with First and Second type of Elliptical Integral?

Questions
1

Hi,

I am trying to solve elliptical equation that contain First and Second kind of elliptical. I did not get any result. I tried below steps

k = Simplify[[[4*R*r]/[[f - Y]^2 + [r + R]^2]]^1/2, Assumptions -> {r = 0.05, R = 0.3, f = 0.1, Y = 0.8}]
A = EllipticK[k]
B = EllipticE[k]
N = [R*r]^1/2/[2*Pi*k][[2 - k^2]*A - [2*B]] // FullSimplify

Thanks

2

Several syntax errors. N is a reserved symbol for numerical value. A and B should be defined as functions A[k_]:= , etc. [ ] cannot be used as grouping brackets; use ( ). [ ] reserved for function definitions. Here is corrected sample of part of your code:

In[16]:= A[k_] := EllipticK[k];
B[k_] := EllipticE[k];
(R*r)^1/2/(2*Pi*k) ((2 - k^2)*A[k] - 2*B[k]) // FullSimplify

Out[18]= -((r R (2 EllipticE[k] + (-2 + k^2) EllipticK[k]))/(
 4 k \[Pi]))
3

Hello,

Thank you so much for your reply. Please I have the value of r,R,f, and Y. and I define k as

k = Simplify[((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2, Assumptions -> {r = 0.05, R = 0.3, f = 0.1, Y = 0.8}]

I want to get out of K(k) and E(k) in final result.

Thanks

4

Generally, rules are used with Mathematica to substitute numerical values into expressions containing symbols. Something like the following should work with the expression that you tried.

((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2 /. {r -> 0.05, R -> 0.3, 
  f -> 0.1, Y -> 0.8}
5

Hello,

Thank you so much for your reply. Please I still got the result as function of E(K). How I can get it without E(k). In addition, In original equation I have k^2. How I can assign the result from first equation to be k for second, third, and forth equations?

((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2 /. {r -> 0.05, R -> 0.3, f -> 0.1, Y -> 0.8}
A[k_] := EllipticK[k];
B[k_] := EllipticE[k];
(R*r)^1/2/(2*Pi*k) ((2 - k^2)*A[k] - 2*B[k]) // FullSimplify

Output

(r R (2 EllipticE[5] + 23 EllipticE[(4 r R)/((r + R)^2 + (-2 + Y)^2)][5]))/(20 \[Pi])
6

Rules are only applied using RelpaceAll[] for each individual expression, so the substitutions are not done globally. Also, you did not assign the result of the first expression to k. I think this is what you are trying to do.

k = ((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2 /. {r -> 0.05, R -> 0.3, 
    f -> 0.1, Y -> 0.8};
A[k_] := EllipticK[k];
B[k_] := EllipticE[k];
(R*r)^1/2/(2*Pi*k) ((2 - k^2)*A[k] - 2*B[k]) /. {r -> 0.05, R -> 0.3}
7

Hello,

Really I appreciate that. It is work!

 k = ((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2 /. {r -> 0.05, R -> 0.3,  f -> 0.1, Y -> 0.8};
 A[k_] := EllipticK[k];
 B[k_] := EllipticE[k];
 J = (R*r)^1/2/(2*Pi*k) ((2 - k^2)*A[k] - 2*B[k]) /. {r -> 0.05, R -> 0.3, Y -> 0.8} // Simplify

Output

 -0.0756162 + 0.0486827 1.55561[0.0489796]
8

For some reason, Y was not evaluated. You could check to see where the problem arises by evaluating each line in its own cell. Often just clearing the symbols, and re-running the code will solve the problem. However, I just copied your code and pasted into Mathematica, and the result was 0.0156334.

A simpler alternative in this case might be to globally assign your symbols to their input numerical values

r = 0.5; R = 0.3; f = 0.1; Y = 0.8;
k = ((4*R*r)/((f - Y)^2 + (r + R)^2))^1/2;
A[k_] := EllipticK[k];
B[k_] := EllipticE[k];
(R*r)^1/2/(2*Pi*k) ((2 - k^2)*A[k] - 2*B[k])

This also gives the output as 0.0156334.

9

Hello,

Thank you so much for all your help!