I know what a Taylor series expansion is and I know how to find the Lagrange remainder but what does it mean intuitively? I need an explanation of what the Lagrange remainder represents in terms of the Taylor series expansion without proofs, strictly intuition.

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    $\begingroup$ Any difference between remainder and Lagrange remainder? I think it's the difference between the function f and the Taylor polynomial that approximates it. So it's like an error term. It approaches zero as we add more and more terms to the Taylor polynomial to approximate f $\endgroup$
    – BCLC
    Dec 19 '15 at 16:42
  • $\begingroup$ I guess any remainder because the Lagrange remainder is just one way of calculating the same remainder correct? $\endgroup$
    – idknuttin
    Dec 19 '15 at 16:46

Taking your comment about “any remainder” in mind, BCLC’s comment is effectively the answer you seem to be looking for, i.e. the remainder $R_n(x)$ is simply the difference between the $n^{th}$ order Taylor polynomial $T_n(x)$ and the function it is approximating. That is, given $T_n(x) = \sum_{i=0}^{n} \frac{f^{(i)}( c)}{i!}(x-c)^i$, the remainder is:

$$ R_n(x) = f(x) - T_n(x) $$

There are various forms for this remainder, one of which is the Lagrange remainder you mentioned. Wikipedia groups the various forms under “Taylor’s theorem,” and provides a more detailed discussion about the technical details.

Broadly speaking, for applications, having an expression for the remainder allows one to, e.g.:

  1. Determine the error on a given approximation (i.e. for a given number of terms used in the approximation, determine the error);
  2. Determine the number of terms required to get an approximation that will be within some desired error (i.e. to obtain a value of the function accurate to a given number of decimal places, determine the number of terms $n$);
  3. Determine the largest interval over which the error in a given approximation will be below a desired level (i.e. fix the number of terms and the error, determine the interval).

One can also use the remainder to prove that the limit (i.e. as $n \to \infty$) of the sequence of partial sums of $T_n(x)$ converges to $f(x)$, by showing that $R_n(x) \to 0$, in the same limit, i.e.:

$$ \begin{aligned} f(x) & = T_n(x) + R_n(x) \\[5pt] % \Leftrightarrow \qquad \lim_{n \to \infty} f(x) & = \lim_{n \to \infty} T_n(x) + \lim_{n \to \infty} R_n(x) \\[5pt] % = \qquad f(x) & = \lim_{n \to \infty} T_n(x) + 0 \end{aligned} $$

Although, in practice, it is usually impractical to work with $R_n(x)$ directly (even using Taylor’s inequality) and other methods may be simpler (e.g. showing that the series satisfies the D.E. and initial condition which define the function; for example $\frac{df}{dx} = f$ and $f(0) = 1$ for $f(x) = e^x$, or $(1+x)f = rf$ and $f(0) = 1$ for $f(x) = (1+x)^r$, etc.).

Hope this gives you a non-technical sense for the remainder.


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