The Integral Remainder

# The Integral Remainder

We recently saw from Taylor's Theorem and The Lagrange Remainder page that if $f$ is $n + 1$ time differentiable on some interval containing the center of convergence $c$ and $x$ and if $P_n(x)$ is the $n^{\mathrm{th}}$ order Taylor polynomial of $f$ centered at $c$ then for $f(x) + P_n(x) + E_n(x)$, the error remainder $E_n(x)$ can be computed for some $\xi$ between $c$ and $x$ the following formula:

(1)
\begin{align} \quad E_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x - c)^{n+1} \end{align}

This formula is known as the Lagrange Remainder formula for the error $E_n(x)$. We will now look at another form of the remainder error known as the Integral Remainder formula.

 Theorem 1: Suppose that $f$ is $n + 1$ times differentiable on some interval containing the center of convergence $c$ and $x$, and let $P_n(x) = f(c) + \frac{f^{(1)}(c)}{1!}(x - c) + \frac{f^{(2)}(c)}{2!}(x - c)^2 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$ be the $n^{\mathrm{th}}$ order Taylor polynomial of $f$ at $x = c$. Then $f(x) = P_n(x) + E_n(x)$ where $E_n(x)$ is the error term of $P_n(x)$ from $f(x)$ and the error $E_n(x)$ can be computed with the formula $E_n(x) = \frac{1}{n!} \int_c^x (x - t)^n f^{(n+1)}(t) \: dt$.
• Proof: We will use mathematical induction to prove Theorem 1. First consider the case when $n = 1$. Then $P_1(x) = f(c) + f'(c)(x - c)$. We thus get that:
(2)
\begin{align} \quad E_1(x) = f(x) - P_1(x) \\ \quad E_1(x) = f(x) - \left [ f(c) + f'(c)(x - c) \right ] \\ \quad E_1(x) = f(x) - f(c) - f'(c)(x - c) \end{align}
• Now the corresponding integral for $n = 1$ is $\frac{1}{1!} \int_c^x (x - t)^n f''(t) \: dt$. We will evaluate this integral using the technique of integration by parts. Let $u = x - t$ and let $dv = f''(t) \: dt$. Then we have that $du = - \: dt$ and $v = f'(t)$ and so we get that (using $\int u \: dv = uv - \int v \: du$):
(3)
\begin{align} \quad \int_c^x (x - t)f''(t) \: dt = \left [ (x - t)f'(t) \right ]_{t = c}^{t = x} + \int_c^x f'(t) \: dt \\ \quad \int_c^x (x - t)f''(t) \: dt = (x - x)f'(x) - (x - c)f'(c) + f(x) - f(c) \\ \quad \int_c^x (x - t)f''(t) \: dt = f(x) - f(c) - (x - c)f'(c) = E_1(x) \end{align}
• Thus we have that Theorem 1 hold when $n = 1$. Suppose now that Theorem 1 holds for $n =k > 1$, that is the error between the $k^{\mathrm{th}}$ order Taylor polynomial centered at $c$, $P_k$ has error from $f$ given by $E_k(x) = \frac{1}{k!} \int_c^x (x - t)^k f^{(k+1)}(t) \: dt$. We want to then show that the error between the $(k + 1)^{\mathrm{st}}$ order Taylor polynomial centered at $c$, $P_{k+1}$ has error from $f$ given by:
(4)
\begin{align} \quad E_{k+1} = \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: dt \end{align}
• We will use integration by parts once again to prove this. Let $u = (x - t)^{k+1}$ and let $dv = f^{(k+2)}(t)$. Then we have that $du = -(k+1)(x - t)^k \: dt$ and $v = f^{(k+1)}(t)$, and so we have that:
(5)
\begin{align} \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: dt = \left [ \frac{1}{(k+1)!} (x - t)^{k+1}f^{(k+1)}(t) \right ]_{t=c}^{t=x} + \frac{1}{(k+1)!} \int_c^x (k+1)(x - t)^k f^{(k+1)}(t) \: dt \\ \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: dt = - \frac{1}{(k+1)!}(x - c)^{k+1}f^{(k+1)}(c) + \frac{1}{k!} \int_c^x (x - t)^k f^{(k+1)}(t) \: dt \\ \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: dt = R_k(x) - \frac{1}{(k+1)!}(x - c)^{k+1}f^{(k+1)}(c) \\ \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: = f(x) - P_k(x) - \frac{f^{(k+1)}(c)}{(k+1)!}(x - c)^{k+1} \\ \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: = f(x) - P_{k+1}(x) \\ \quad \frac{1}{(k + 1)!} \int_c^x (x - t)^{k+1} f^{(k + 2)}(t) \: = E_{k+1}(x) \\ \end{align}
• Thus we have shown by induction that Theorem 1 is true. $\blacksquare$