This is an example of procedural programming borrowed from Paul Wellin's book. It's a simple implementation of the Newton-Rhapson method.

I overloaded the findRoot function to accommodate different kinds of arguments. It works fine in the following first case:

f[x_]:= x^2 - 2; findRoot[fun_Symbol, {var_, init_}, \[Epsilon]_] := Module[{xi = init}, While[Abs[fun[xi]] > \[Epsilon],  xi = N[xi - fun[xi]/fun'[xi]]]; {var -> xi}] findRoot[f,{x,2},.0001] {x-> 1.41422} 

But when I overload it to accommodate expressions, like the following for example, it doesn't evaluate.

findRoot[expr_== val_, {var_, init_}, \[Epsilon]_] := Module[{xi = init, fun = Function[fvar,expr - val]}, While[Abs[fun[xi]] > \[Epsilon],  xi = N[xi - fun[xi]/fun'[xi]]]; {var -> xi}] findRoot[x^2-2==0,{x, 2.0}, 0.0001] {x->2.} 

How do I make Mathematica evaluate it numerically all the way through?

This is an example of procedural programming borrowed from Paul Wellin's book. It's a simple implementation of the Newton-Rhapson method.

I overloaded the findRoot function to accommodate different kinds of arguments. It works fine in the following first case:

findRoot[fun_Symbol, {var_, init_}, ϵ_] :=  Module[{xi = init},  While[Abs[fun[xi]] > ϵ, xi = N[xi - fun[xi]/fun'[xi]]];  {var -> xi}] f[x_] := x^2 - 2; findRoot[f,{x,2},.0001] 

{x-> 1.41422} 

But when I overload it to accommodate expressions, like the following for example, it doesn't evaluate.

findRoot[expr_== val_, {var_, init_}, ϵ_] :=  Module[{xi = init, fun = Function[fvar,expr - val]},  While[Abs[fun[xi]] > ϵ, xi = N[xi - fun[xi]/fun'[xi]]];  {var -> xi}] findRoot[x^2 - 2 == 0, {x, 2.0}, 0.0001] 

{x - >2.} 

How do I make Mathematica evaluate it numerically all the way through?