what the value of array in decimal

F(1)=0;
F(2)=1;
F(3)=-5;
 for k=1:10
     A=0;
       for m=1:k
          A=A-2*a1*Re*F(m)*(k-m+1)*F(k-m+2);
       end
       F(k+3)=(A/((k)*(k+1)*(k+2)))-((4*a1^2*k*F(k+1))/((k)*(k+1)*(k+2)));
 end
 % disp(F)
  for k=1:1:12
      series1(x)=series1(x)+F(k)*(power(x,k-1));
  end
  series1
for x=1:11
      e=(x-1)/10
      U1(x)=series1(e)

end

I want the value of U1(x) in decimal form .I use the command disp( U1) ,it is no doubt displaying the value but in the fraction form and Iwant in decimal form.In addition to that I want to plot the graph for different value of e.

Answer 1

[Walter Roberson]

a1 and Re are undefined.

series1 and x are undefined at the point where you first use them.

I suspect that you initialized

syms x

which is not the proper thing to do in this case: if x is symbolic then series1(x) would be a symbolic function, but you cannot define a symbolic function in terms of itself (the series(x) on the right hand side.)

Please read https://www.mathworks.com/matlabcentral/answers/395816-why-do-i-get-array-indices-must-be-positive-integers-or-logical-values-error-when-using-angle-fu#comment_882605

[YOGESHWARI PATEL]

syms a 
syms Re
syms a1
syms x
series1(x)=sym(zeros(1));
F=zeros(1,2,'sym','sym','sym');
%n=input('enter the value of n')
F(1)=1;
F(2)=0;
F(3)= -0.000514754075397 ;
 a1=input('enter the value of alpha=')
  Re=input('enter the value of reynold number R=')
 for k=1:10
     A=0;
       for m=1:k
          A=A-2*a1*Re*F(m)*(k-m+1)*F(k-m+2);
       end
       F(k+3)=(A/((k)*(k+1)*(k+2)))-((4*a1^2*k*F(k+1))/((k)*(k+1)*(k+2)));
 end
 % disp(F)
  for k=1:1:10
      series1(x)=series1(x)+F(k)*(power(x,k-1));
  end
  series1
for x=1:11
      e=(x-1)/10
      U1(x)=series1(e)
 
end

This is the modification I have done and now I want the value of U1 t x=0,0.1,0.2,0.3,.......in fraction form and want to plat the graph for x=0,0.1,0.2,......1.

[Walter Roberson]

vpa() the appropriate variable.

[Walter Roberson]

syms a 
syms Re
syms a1
syms x
series1(x)=sym(zeros(1));
F=zeros(1,2,'sym','sym','sym');
%n=input('enter the value of n')
F(1)=1;
F(2)=0;
F(3)= -0.000514754075397 ;

if ispc()
 a1=input('enter the value of alpha=')
   Re=input('enter the value of reynold number R=')
else
    a1 = 1/sqrt(2);
    Re = pi/2;
end

 for k=1:10
     A=0;
       for m=1:k
          A=A-2*a1*Re*F(m)*(k-m+1)*F(k-m+2);
       end
       F(k+3)=(A/((k)*(k+1)*(k+2)))-((4*a1^2*k*F(k+1))/((k)*(k+1)*(k+2)));
 end
 % disp(F)
  for k=1:1:10
      series1(x)=series1(x)+F(k)*(power(x,k-1));
  end
  series1

series1(x) = 

  vpa(series1, 5)

ans(x) = 

  
for x=1:11
      e=(x-1)/10;
      U1(x)=series1(e);
end

vpa(U1, 5)

ans = 

plot(0:0.1:1, U1)

[YOGESHWARI PATEL]

Thank you for solving the problem.

I want to know that why you have included this part and what it means

if ispc()

a1=input('enter the value of alpha=')

Re=input('enter the value of reynold number R=')

else

a1 = 1/sqrt(2);

Re = pi/2;

end

[Walter Roberson]

It means that my demonstration is running on a system that does not support input(), so I arranged the code to detect that it is not running on Windows and to use constant values for alpha and Re. If the same code were to be run on Windows it would prompt for inputs.

[YOGESHWARI PATEL]

Thank you .I got the solution.