prove that if the function is infinitely differentiable on an interval (r,s) containing a, then for any x in (r,s) f(x) can be expand :
f(a) + (x-a) f'(a) + (x-a)^2*f''(a)/2! +....+ (x-a)^n * f(n')(a) + Rn(f,a,x)
Rn is the remainder
prove that if the function is infinitely differentiable on an interval (r,s) containing a, then for any x in (r,s) f(x) can be expand :
f(a) + (x-a) f'(a) + (x-a)^2*f''(a)/2! +....+ (x-a)^n * f(n')(a) + Rn(f,a,x)
Rn is the remainder
Of course a function can be expanded in this way. Just expand a function as a polynomial. Then add the remainder term to make it equal to the function. I assume you are not asking what you want to ask. You want to bound the remainder, and therefore say that the approximation is a good approximation.
Here is another proof of Lagrange's estimate using repeated application of Rolle's theorem.
Let and let be the solution to the equation:
.The argument will be complete if we can show for some between and .
Now define the function,
.Notice that .
Since by definition of we know by Rolle's theorem there is between and so that . But since by Rolle's theorem again there is an between and so that . This continues until we get an between and so that . But then by definition of it follows that , choose and the proof is complete.