Estimation with Taylor Polynomials and Error Bounds

# Estimation with Taylor Polynomials and Error Bounds

Recall that if $f$ is a function that is $n$ times differentiable on an open interval containing $c$ then the Taylor polynomial of $f$ at $c$ is:

(1)
\begin{align} \quad P_n(x) = f(c) + \frac{f'(c)}{1!} (x - c) + \frac{f''(c)}{2!} (x - c)^2 + \frac{f'''(c)}{3!} (x - c)^3 + ... + \frac{f^{(n)}(c)}{n!} (x - c)^n \end{align}

Furthermore, if $c = 0$ then the polynomial above is sometimes called the Maclaurin polynomial of $f$.

We recently saw on the Taylor's Theorem and The Lagrange Remainder page that the Lagrange Remainder $E_n(x) = f(x) - P_n(x)$ can be obtained for some $\xi$ between $c$ and $x$ with the following formula:

(2)
\begin{align} \quad E_n(x) = \frac{f^{(n+1)}(\xi)}{(n + 1)!} (x - c)^{n+1} \end{align}

We will now look at some examples of estimating sums with Taylor polynomials and bounding the error in these approximations.

## Example 1

Use the second order Taylor polynomial $P_2(x)$ for $f(x) = \sqrt{x}$ centered about $64$ to approximate the value of $\sqrt{61}$. Bound the error of this approximation.

We have that $f'(x) = \frac{1}{2\sqrt{x}}$ and $f''(x) = \frac{-1}{4(\sqrt{x})^3}$. Therefore the second order Taylor polynomial $P_2(x)$ for $f(x) = \sqrt(x)$ centered about $64$ is:

(3)
\begin{align} \quad P_2(x) = \sqrt{64} + \frac{1}{2\sqrt{64}}(x - 64) + \frac{1}{2!}\frac{-1}{4(\sqrt{64})^3} (x - 64)^2 \end{align}

Plugging in $x = 61$ gives us:

(4)
\begin{align} \quad P_2(61) = 8 + \frac{1}{16} (61 - 64) - \frac{1}{2!}\frac{1}{4(512)}(61 - 64)^2 \\ \quad P_2(61) \approx 7.810303... \end{align}

Now let's bound the error. For some $\xi$ between $61$ and $64$ we have that:

(5)
\begin{align} \quad E_n(61) = \frac{f^{(3)}(\xi)}{3!} (61 - 64)^3 \end{align}

We compute the third derivative of $\sqrt{x}$ as $f^{(3)}(x) = \frac{3}{8x^{5/2}}$. Note that $\max_{[61, 64]} \mid f^{(3)} (x) \mid = \frac{3}{8(\sqrt{61})^5}$

Therefore we have that:

(6)
\begin{align} \quad \mid E_n(61) \mid = \biggr \rvert \frac{f^{(3)}(\xi)}{3!} (-3)^3 \biggr \rvert = \frac{27}{3!} \mid f^{(3)} (\xi) \mid ≤ \frac{27}{3!} \frac{3}{8(\sqrt{61})^5} \end{align}