# Can someone explain the Taylor's Theorem error bound?

1. Jan 24, 2016

### A.MHF

1. The problem statement, all variables and given/known data

I understand that to find the error of approximation all we have to do is:
|E(x)| = |f(x)-Tn(x)|
M*(xn+1/(n+1)!)
What's the point of this? and why does it have to be greater than or equal to |E(x)|? How would evaluating the next derivative of the desired Taylor polynomial help us?

2. Relevant equations

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3. The attempt at a solution

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2. Jan 24, 2016

### vela

Staff Emeritus
I moved your question to the discussion forum since it's more general than a homework problem.

f(x) is the exact value of the function evaluated at x. Tn(x) is an approximation of f(x). The error E(x) is, by definition, the difference between the actual value f(x) and the approximate value we get from Tn(x). So when you write $\lvert E(x) \rvert = \lvert f(x)-T_n(x) \rvert$, you're doing little more than stating the definition of E(x).

Keep in mind that we generally don't know what $f(x)$ is equal to. We need some way of calculating the answer. That's where $T_n(x)$ comes in. If you want the exact answer, you'll have to sum an infinite number of terms. But most of the time, we don't need an exact answer; we just need one that's close enough. In other words, we only need to sum a finite number of terms so that |E(x)| is small enough for our purposes. But remember, we don't know what f(x) is, so somehow we have to be able to come up with an upper bound for E(x) without knowing what f(x) is. That's where the expression for the error bound comes in.

3. Jan 24, 2016

### WWGD

This is an _estimate_ of the error; if you knew the actual error, you add it to the function and get the actual value.

4. Jan 24, 2016

### Ray Vickson

The actual error $E_n(x) = f(x) - T_n(x)$ when we expand about $x = a$ has the form
$$E_n(x) = \frac{(x-a)^{n+1}}{(n+1)!} f^{(n+1)}(\xi),$$
where $\xi$ is some unknown point in the interval between $a$ and $x$, and $f^{(n+1)}$ is the $(n+1)$st derivative of $f$. Since we generally do not know the value of $\xi$ explicitly, and also may not even know exact numerical values of the function $f^{(n+1)}$, we cannot in general compute the error exactly. However, we can bound the error a lot more easily: if we somehow find a positive number $M$ such that $|f^{(n+1)}(\xi)| \leq M$ for sure, then we know that
$$|E_n(x)| \leq \frac{M (x-a)^{n+1}}{(n+1)!},$$
so we at least know the worst-case error (which will almost always be worse than the actual error). The best value of $M$ in this regard is the maximum of $|f^{(n)}|$ on the interval between $x$ and $a$, but even that will often be unavailable or unobtainable. In that case, just choose any $M$ larger than that. Usually it is not too difficult to cook up such a value in a practical problem.

5. Jan 24, 2016

### A.MHF

Thanks guys, that was super helpful!