computation of symbolic terms and error analysis

tic
syms x y  m  %alpha beta b lambda eta phi  
phi=sqrt(0.5)

phi = 0.7071

b=10

b = 10

eta=1

eta = 1

alpha=2;
beta=1;
T=zeros(1,2,'sym');
G=zeros(1,2,'sym');
A=zeros(1,'sym');
B=zeros(1,'sym');
C=zeros(1,'sym');
D=zeros(1,'sym');
E=zeros(1,'sym');
F=zeros(1,'sym');
G_val=zeros(1,'sym');
series(x)=sym(zeros(1,1));
series1(x)=sym(zeros(1,1));
% Transform Initial  condition 
T(1)=m; 
T(2)=0;
% Initial transform value of exp function

for i=1:4
    G(i)=y;
    if i==1
        E(1)=exp((eta*T(1))/(eta+T(1)));
    else
        A(1)=0;
        for j=1:i-1
            E(i)=A(1)+(j/(i-1))*G(j)*E(i-j);
        end
        B(1)=0;
        for j=1:i
            B(1)=B(1)+T(j)*G(i-j+1);
        end
        eq1=G(i)-T(i)+(1/eta)*B(1);
        G_val=simplify(solve(eq1,y));
        G(i)=simplify(G_val);
        E_val=subs(E(i),y,G_val);
        E(i)=E_val
    end
    C(1)=0
    for r=1:i
        C(1)=C(1)+T(r)*E(i-r+1);
    end
    T(i + alpha*beta) = (phi^2*C(1)-b*phi^2*E(i))*(gamma(1+(sym(i-1)/beta))/gamma(1+alpha+(sym(i-1)/beta)));
    %T(i+2)=simplify((1/(i*(i+1)))*(phi^2*C(1)-b*phi^2*E(i)));
end  

C = 

0

E = 

C = 

0

E = 

C = 

0

E = 

C = 

0

for k=1:3
    series(x)=simplify(series(x)+T(k)*(power(x,((k-1)*beta))));
end
series

series(x) = 

e1=subs(series,x,1);
format long 
accuracy=1e-4;%input('enter the accuracy')
f=e1(x)

f = 

g=inline(f)

g =

     Inline function:
     g(m) = m-exp(m./(m+1.0)).*(5.0./2.0)+(m.*exp(m./(m+1.0)))./4.0

a=1;%input('enter the ist approximation=')
b=5;%input('enter the 2nd approximation=')
fa=feval(g,a)

fa = 
  -2.709622859075289

fb=feval(g,b)

fb = 
   2.123780136383969

while fa*fb>0
    a=input('enter the ist approximation=')
    b=input('enter the 2nd approximation=')
    fa=feval(g,a)
    fb=feval(g,b)   
end
for i=1:50
    c=(a+b)/2;
    fc=feval(g,c);
    disp([i   a      fa     b       fb        c       fc     abs(b-a)])
    if fc==accuracy
        fprintf('the root of the equation is %f',c)
        break;
    elseif abs(b-a)<=accuracy
        fprintf('the root of the equation is %f',c) 
        break;
    elseif fa*fc<=0
        b=c;
        fb=fc;
    else  
        a=c;
        fa=fc;
    end
end

   1.000000000000000   1.000000000000000  -2.709622859075289   5.000000000000000   2.123780136383969   3.000000000000000  -0.704750029072181   4.000000000000000

   2.000000000000000   3.000000000000000  -0.704750029072181   5.000000000000000   2.123780136383969   4.000000000000000   0.661688607261298   2.000000000000000

   3.000000000000000   3.000000000000000  -0.704750029072181   4.000000000000000   0.661688607261298   3.500000000000000  -0.037023639038904   1.000000000000000

   4.000000000000000   3.500000000000000  -0.037023639038904   4.000000000000000   0.661688607261298   3.750000000000000   0.309004610855695   0.500000000000000

   5.000000000000000   3.500000000000000  -0.037023639038904   3.750000000000000   0.309004610855695   3.625000000000000   0.135098495155879   0.250000000000000

   6.000000000000000   3.500000000000000  -0.037023639038904   3.625000000000000   0.135098495155879   3.562500000000000   0.048806291793267   0.125000000000000

   7.000000000000000   3.500000000000000  -0.037023639038904   3.562500000000000   0.048806291793267   3.531250000000000   0.005832481645888   0.062500000000000

   8.000000000000000   3.500000000000000  -0.037023639038904   3.531250000000000   0.005832481645888   3.515625000000000  -0.015610425294633   0.031250000000000

   9.000000000000000   3.515625000000000  -0.015610425294633   3.531250000000000   0.005832481645888   3.523437500000000  -0.004892666418066   0.015625000000000

  10.000000000000000   3.523437500000000  -0.004892666418066   3.531250000000000   0.005832481645888   3.527343750000000   0.000468986086110   0.007812500000000

  11.000000000000000   3.523437500000000  -0.004892666418066   3.527343750000000   0.000468986086110   3.525390625000000  -0.002212070812502   0.003906250000000

  12.000000000000000   3.525390625000000  -0.002212070812502   3.527343750000000   0.000468986086110   3.526367187500000  -0.000871599991724   0.001953125000000

  13.000000000000000   3.526367187500000  -0.000871599991724   3.527343750000000   0.000468986086110   3.526855468750000  -0.000201321355804   0.000976562500000

  14.000000000000000   3.526855468750000  -0.000201321355804   3.527343750000000   0.000468986086110   3.527099609375000   0.000133828764920   0.000488281250000

  15.000000000000000   3.526855468750000  -0.000201321355804   3.527099609375000   0.000133828764920   3.526977539062500  -0.000033747195564   0.000244140625000

  16.000000000000000   3.526977539062500  -0.000033747195564   3.527099609375000   0.000133828764920   3.527038574218750   0.000050040559656   0.000122070312500

  17.000000000000000   3.526977539062500  -0.000033747195564   3.527038574218750   0.000050040559656   3.527008056640625   0.000008146625789   0.000061035156250

the root of the equation is 3.527008

fprintf('the value of c=%f', c);

the value of c=3.527008

series1(x)=subs(series,m,c)

series1(x) = 

residualError=sym(zeros(1,1));
residual=sym(zeros(1,1));
residualError(x) = abs(diff(series1,x,2)+ B*phi^2*(1-(series1)/B)*exp((eta*series1)/(eta+series1)))

residualError(x) = 

residual_error=double(subs(residualError,x,0.2))

Error using mupadengine/evalin2double
Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for variables.

Error in mupadengine/feval2double

Error in sym/double (line 756)
X = feval2double(symengine, "symobj::doubleDirect", S);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

var =double(residual_error);
x=0:0.1:1
error=zeros(1)
row=0;
for i=1:length(x)
    row=row+1;
    error(row)=(residualErrorx(i))
    max_error=max(error)
end
fprintf('The residual error is %f\n', max_error);
%-----------------------------------------------------------------

the error appear as Error using mupadengine/evalin2double

Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for variables.

Error in mupadengine/feval2double

Error in sym/double (line 756)

X = feval2double(symengine, "symobj::doubleDirect", S);

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Related documentation

now i substutite the value of m using sysntax series1(x)=subs(series,m,c) but when i later on call series 1 in line residualError(x) = abs(diff(series1,x,2)+ B*phi^2*(1-(series1)/B)*exp((eta*series1)/(eta+series1))) the final expression is having m and y . That is symbolic and craet an error

[yogeshwari patel]

I find the solution . Thanks

Answer 1

[Steven Lord]

You indicated you've already found the solution (the expression you're trying to convert to double still has symbolic variables m and y inside it) but I wanted to call out one change you should make. It is with this line of code (left in a block comment so that I can run later code in this answer):

%{
g=inline(f)
%}

Don't use the inline function. It is not recommended and has been "not recommended" for years if not decades. To convert a symbolic expression into something you can evaluate directly, use matlabFunction instead. Or use subs to substitute values into a symbolic expression.

syms x
f = x^2+2*x+3

f = 

fh = matlabFunction(f)

fh = function_handle with value:
    @(x)x.*2.0+x.^2+3.0

Let's check by using subs to substitute values into the symbolic expression and calling the function handle.

check1 = subs(f, x, 1:10)

check1 = 

check2 = fh(1:10)

check2 = 1×10

6 11 18 27 38 51 66 83 102 123

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

double(check1) == check2

ans = 1×10 logical array
   1   1   1   1   1   1   1   1   1   1