what im doing wrong

 g=@(tu((i),j+1)) tu((i),j+1)-(1/2*(xu((i),j)-xu((i-1),j)+tu((i),j)+tu((i-1),j))+h/8*(r*(tu((i),j)-(1/f)*(tu((i),j))^3)+2*(r*(tu((i),j+1)-(1/f)*(tu((i),j+1))^3))+r*(tu((i-1),j)-(1/f)*(tu((i-1),j))^3)));

         p=fzero(@(tu((i),j+1)) g,0);

 but i got this answer
         Unbalanced or unexpected parenthesis or bracket.

Answer 1

[Walter Roberson]

When you construct an anonymous function, the part directly after the @ must be pure variable names and not expressions or indexed variables.

Something like

g = @(tu, i, j) tu((i),j+1)-(1/2*(xu((i),j)-xu((i-1),j)+tu((i),j)+tu((i-1),j))+h/8*(r*(tu((i),j)-(1/f)*(tu((i),j))^3)+2*(r*(tu((i),j+1)-(1/f)*(tu((i),j+1))^3))+r*(tu((i-1),j)-(1/f)*(tu((i-1),j))^3)));

p = fzero(@(i) g(tu, i, j), 0)

[Walter Roberson]

The syntax you are using in

fzero(@(tu((i),j+1)) g,0)

is wrong. What goes in the () after the @ can only be variable names. With what you used, MATLAB is confused when it sees the "g" after the ")" .

[Walter Roberson]

If you have an expression of the form

f(x) = g(x)

where g(x) is a function of x and includes the term f(x) somewhere inside, then rearrange the expression to

g(x) - f(x) = 0

and then you can solve for the x that makes it zero.

For example if

 tu(i) = tu(i) * exp(-i^2) - cos(i)

then you can rearrange that to

tu(i) * exp(-i^2) - cos(i) - tu(i) = 0

and then you

fzero(@(i) tu(i) * exp(-i^2) - cos(i) - tu(i), InitialValue)

Answer 2

[Gentian Zavalani]

 As you see this equation is implicit ,i wanna to solve this equation and to find  tu((i),j+1) for each iteration so i don't know if you could suggest me any way,how to proceed to  for example tu((i),j+1)=s i dont know something like that 
pleas if you could give me a answer
regards