Is it possible to get the value of 'A' in a table format of 11x4 dimension

function main
format long
Pr=0.72;
xa=0;xb=2;
solinit=bvpinit(linspace(xa,xb,11),[0 0 0 1 0]);
sol=bvp4c(@ode,@bc,solinit);
x=linspace(xa,xb,11);
sxint=deval(sol,x);
f0 = deval(sol,0);
P=f0(3);,Q=f0(5);
f=x.^2*(P/2)+x.^5*(P^2)/120-x.^3*(1/6)-x.^4*(Q/24)-x.^8*(P^3)*(13/40320)+x.^7*(P*Q)/1008+x.^7*(Pr*P*Q)/1680+...
    x.^11*(P^4)*(229/13305600)-x.^10*(P^2)*Q*Pr^2*(1/40320)-x.^10*(P^2)*Q*Pr*(37/1209600)+...
    x.^9*(P^2)/20160-x.^10*(P^2)*Q/22400-x.^9*(Pr*Q^2)/120960-x.^8*Q*Pr/13440+x.^8*Q/8064+x.^9*Q^2/72576;
g=1+x.*Q-x.^4*Pr*P*Q/8+x.^7*P^2*Q/56-x.^10*P^3*Q*Pr*(13/1209600)-x.^9*P*Pr^2*Q^2*(19/13440)-...
    x.^8*P*Q*Pr^2/128- x.^9*P*Pr*Q^2/24192;
function res=bc(ya,yb)
res=[ya(1); ya(2); yb(2); ya(4)-1;  yb(4)];  
end
 function dydx=ode(x,y)    
 dydx=[y(2);  y(3);  -3*y(1)*y(3)+2*y(2)^2-y(4);  y(5);  -3*Pr*y(1)*y(5)];              
 end
A = [x; sxint(4,:);g;E2]';
end

%%%   Is it possible to get the value of 'A' in a table format of 11x4 dimension

[Walter Roberson]

E2 is not defined.

[Walter Roberson]

 array2table(A, 'VariableNames', {'x', 'sxint', 'g', 'E2'})

[MINATI]

Dear Walter

if we open 'A' ,this table is coming BUT what I need is a table with BORDER. Is it possible?

[Walter Roberson]

table() objects are objects for organizing data and computing with it. They are not for display purposes.

For display purposes you can use uitable(), or you can use fprintf() to format something for the command window.

[MINATI]

how to find f '(0) from the code given above from f

[Walter Roberson]

E2 is undefined.

By examination we can see that f is the sum of x to various powers, with no x^1 term -- a polynomial with 0*x . Take the derivative with respect to x and you get a polynomial with constant term 0 . Evaluate at 0 and the end result is going to be that constant term, 0.

[MINATI]

@Walter but if the function is a nonvanishing one, what is the code to find f ' (0)

[Walter Roberson]

The equation is

(229*P^4)/13305600*x^11 + (-1/40320*P^2*Q*Pr^2 - 37/1209600*P^2*Q*Pr - 1/22400*P^2*Q)*x^10 + (1/20160*P^2 - 1/120960*Pr*Q^2 + 1/72576*Q^2)*x^9 + (-13/40320*P^3 - 1/13440*Q*Pr + 1/8064*Q)*x^8 + (1/1008*P*Q + 1/1680*Pr*P*Q)*x^7 + P^2/120*x^5 - Q/24*x^4 - 1/6*x^3 + P/2*x^2

The first derivative of this will have an x^1 term and a 0 x^0 (constant) term. Substitute in x = 0 and since that is 0 to various powers multiplied by things, you get 0 for all of them and there is no constant term to add, so you will get 0 for f'(0)

The second derivative of this will have an x^0 (constant) term of P. Substitute in x = 0 and all the other terms would become 0 and you would get P for f''(0)

The function is not a non-vanishing one.

If you want to ask about other hypothetical functions, then (A) Do you have the symbolic toolbox; and (B) what are the known properties of the function, if you are not permitted to read the properties out of the defining expression (such as if you were given a function handle.) Is it real-valued? Is the derivative known to be real-valued at 0? What is the region in which we can assume that the function is continuous C1 (continuous in the first derivative) ?

[MINATI]

(A) Do you have the symbolic toolbox;

NO

Is it real-valued?

YES

What is the region in which we can assume that the function is continuous C1 (continuous in the first derivative) ?

HERE we can take [0, 2].

[Walter Roberson]

Provided that the function is known to be a polynomial:

Let MD be the maximum degree that the polynomial could be.

X = linspace(0, 2, MD+1);
Y = f(X);
p = polyfit(X, Y, MD);

then f'(0) = p(end-1)

This will not work if f is not a polynomial.

If the actual degree is 1 more than MD, then it might work. If the actual degree is 2 or more than MD, then it will probably not work.

[MINATI]

oook

thanks Walter

[Walter Roberson]

Coefficient may be approximate. You can estimate the degree by looking at the number of leading coefficients that are approximately 0. You can also estimate by starting with a high degree fit and extracting the coefficient mentioned and then reducing the degree and examining again. When the coefficient changes drastically then you have gone one step too far.