Lagrange Error Bound

What is the maximum error when finding a Maclaurin polynomial for \(\sin x\) on the interval \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]\)? What can you conclude about the Maclaurin series for \(\sin x\)?

Solution:

First, remember that a Maclaurin polynomial is just a Taylor polynomial centered at \(x=0\). Looking at some of the derivatives of \(f(x)=\sin x\) along with their function values at \(x=0\) you get:

\[ \begin{array}{ccc} &f(x) = \sin x & \quad \quad & f(0) = 0\\ &f'(x) = \cos x & \quad \quad & f'(0)= 1 \\ &f''(x) = -\sin x & \quad \quad & f''(0)=0 \\ &f'''(x) = -\cos x & \quad \quad & f'''(0)= -1 \\ &f^{(4)}(x) = \sin x & \quad \quad & f^{(4)}(0) = 0. \end{array} \]

As you can see it cycles back around to the start of the list when you get to the \(4^{\text{th}}\) derivative. So the Maclaurin polynomial of order \(n\) for \(\sin x\) is

\[\begin{align} T_n(x) &= 0 + \frac{1}{1!}x + 0 + \frac{-1}{3!}x^3 + 0 + \dots \\ & \quad + \begin{cases} 0 & \text{ if } n \text{ is even} \\ \dfrac{f^{(n)}(0)}{n!}x^n & \text{ if } n \text{ is odd} \end{cases} \end{align}\]

and the Lagrange error will have a different formula depending on if \(n\) is odd or even as well.

However you want to find the maximum error, and that certainly isn't going to happen when the error term is zero! This polynomial is centered at \(x=0\), and the interval is

\[\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right].\]

That means \(R = \frac{\pi}{2}\). Because all of the derivatives involve sine and cosine, you also know that

\[|f^{(n+1)}(c) |<1\]

for any \(c\) in the interval \(I\). Therefore

\[\begin{align} |R_n(x) | &\le M\frac{R^{n+1}}{(n+1)!} \\ &= 1\cdot \dfrac{\left(\dfrac{\pi}{2}\right)^{n+1} }{(n+1)!} \\ &= \left(\dfrac{\pi}{2}\right)^{n+1} \frac{1}{(n+1)!}, \end{align}\]

and that is the maximum error.

You would like to draw a conclusion about the Maclaurin series for \(\sin x\). To do that you need to look at

\[\lim\limits_{n\to \infty} |R_n(x) | = \lim\limits_{n\to \infty} \left(\dfrac{\pi}{2}\right)^{n+1} \frac{1}{(n+1)!} .\]

Since this sequence converges to \(0\) as \(n \to \infty\), you can conclude that the Maclaurin series does converge. In fact the Maclaurin series is equal to the function on the entire interval \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]\).

When you are estimating

\[\sin \left(\dfrac{\pi}{16}\right)\]

using the Maclaurin polynomial, what is the smallest degree of the polynomial that guarantees the error will be less than \(\dfrac{1}{100}\)?

Solution:

From the previous example you know that the error on the interval \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]\) has the property that

\[|R_n(x) | \le \left(\dfrac{\pi}{2}\right)^{n+1} \frac{1}{(n+1)!} \]

You want that error to be less than \(\dfrac{1}{100}\), or in other words that

\[ \left(\dfrac{\pi}{2}\right)^{n+1} \frac{1}{(n+1)!} < \frac{1}{100}.\]

Unfortunately solving for \(n\) is quite challenging! So the only thing you can do is try out values of \(n\) and see which one makes the Lagrange error bound sufficiently small.

But what if you don't have a calculator handy? The problem is really that the interval is too big, which makes \(\dfrac{\pi}{2} >1\). Can you change the interval so that \(\dfrac{\pi}{16} \) is inside the interval, but the bound is smaller? Sure thing!

The maximum error when finding a Maclaurin polynomial for \(\sin x\) on the interval \( \left[ -\dfrac{\pi}{4}, \dfrac{\pi}{4} \right]\) has the property that

\[|R_n(x) | \le \left(\dfrac{\pi}{4}\right)^{n+1} \frac{1}{(n+1)!} ,\]

where you have used the same technique as in the previous example. Then

\[ \dfrac{\pi}{16} \in \left[ -\dfrac{\pi}{4}, \dfrac{\pi}{4} \right] \]

and

\[ \dfrac{\pi}{4} < 1, \]

so

\[\begin{align} |R_n(x) | &\le \left(\dfrac{\pi}{4}\right)^{n+1} \frac{1}{(n+1)!} \\ &< \frac{1}{(n+1)!}. \end{align}\]

Now you need to make sure that the error is small enough, which means you need that

\[ \frac{1}{(n+1)!} < \frac{1}{100},\]

which is much easier to calculate. In fact if you take \(n=4\) you get that

\[ \frac{1}{(4+1)!} = \frac{1}{5!} = \frac{1}{120} < \frac{1}{100}.\]

That might make you think that you need a \(4^{\text{th}}\) degree Maclaurin polynomial, but you already know that the even terms of the Maclaurin polynomial are zero! So do you pick \(n=3\) or \(n=5\) to make sure the error is small enough since the Maclaurin polynomial is the same for \(n=3\) and \(n=4\)? If you want an absolute guarantee that the error is going to be small enough, use \(n=5\).

If you check the actual errors,

\[ \begin{align} \left|T_3\left(\dfrac{\pi}{16}\right) - \sin \left(\dfrac{\pi}{16}\right) \right|&= \left| \frac{1}{1!}\left(\dfrac{\pi}{16}\right) + \frac{-1}{3!}\left(\dfrac{\pi}{16}\right) ^3 - \sin \left(\dfrac{\pi}{16}\right) \right| \\ &= \left|\dfrac{\pi}{16} - \dfrac{1}{6}\left(\dfrac{\pi}{16}\right) ^3 - \sin \left(\dfrac{\pi}{16}\right) \right| \\ &\approx 0.0000024, \end{align}\]

which is quite a bit smaller than you needed!

Would it have been small enough if you had taken \(n=1\)? In that case

\[ \begin{align} \left|T_1\left(\dfrac{\pi}{16}\right) - \sin \left(\dfrac{\pi}{16}\right) \right|&= \left| \frac{1}{1!}\left(\dfrac{\pi}{16}\right) - \sin \left(\dfrac{\pi}{16}\right) \right| \\ & \approx 0.00126, \end{align}\]

so even that is smaller than the error you were given. The problem of course is doing the approximation without using a calculator!