# Error of a Taylor polynomial

1. Feb 5, 2013

### wahaj

the error of a taylor series of order(I think that's the right word) n is given by
$$\frac{f^{n+1} (s)}{n!} (x-a)^n$$

I think this is right. The error in a linear approximation would simply be

$$\frac{f''(s)}{2} (x-a)^2$$

My question is what is s and how do I find it. Use linear approximation of √(47) if you need to use an example because I just did that question so it might be easier to explain.

2. Feb 6, 2013

### jbunniii

Actually, it should be
$$\frac{f^{n+1} (s)}{(n+1)!} (x-a)^{n+1}$$

I think this is right. The error in a linear approximation would simply be

$$\frac{f''(s)}{2} (x-a)^2$$

All that you know for sure is that $s$ is a number between $a$ and $x$, and it depends on $x$.

OK, you didn't specify a value for $a$, so I'll pick $a = 49$. Then, if I still remember how to take derivatives, we should have
$$f(x) = x^{1/2}$$
$$f'(x) = \frac{1}{2}x^{-1/2}$$
$$f''(x) = \frac{-1}{4}x^{-3/2}$$
so
\begin{align}f(a) + f'(a)(x-a) + \frac{1}{2}f''(s)(x-a)^2 &= (49)^{1/2} + \frac{1}{2}(49)^{-1/2}(x-49) + \frac{-1}{8}s^{-3/2}(x-49)^2\\ &= 7 + \frac{1}{14}(x-49) - \frac{1}{8}s^{-3/2}(x-49)^2\\ \end{align}
where $s$ is some number between $a$ and $x$. Setting $x=47$, we get
$$\sqrt{47} = 7 - \frac{1}{7} - \frac{1}{2}s^{-3/2}$$
where $s$ is some number satisfying $47 \leq s \leq 49$. Putting it another way, if we use only the linear approximation $7 - 1/7$, then the approximation error is
$-\frac{1}{2}s^{-3/2}$ for some $47 \leq s \leq 49$. We can bound this error by taking the worst case $s$ (the value that maximizes the absolute approximation error), which in this case is obtained by choosing $s=47$. Then the approximation error is guaranteed to satisfy
$$|\textrm{approximation error}| \leq \frac{1}{2}(47)^{-3/2} = 0.00155176...$$
Actually, we can also obtain a lower bound for the approximation error by choosing the best-case value of $s$, namely $s = 49$. The error will be at least $\frac{1}{2}(49)^{-3/2} = 0.00145772...$. Thus we in fact have
$$0.00145772... \leq |\textrm{approximation error}| \leq 0.00155176...$$
We can check this: the actual value of $\sqrt{47}$ is
$$\sqrt{47} = 6.8556546...$$
and our approximation is
$$7 - 1/7 = 6.85714285...$$
and the error is therefore $6.85714285... - 6.8556546... = 0.00148825...$ which is indeed within the computed range.

Last edited: Feb 6, 2013
3. Feb 6, 2013

### wahaj

Thanks that makes more sense than my book

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