# Approximating accuracy of Taylor polynomials

1. Nov 7, 2012

### Mangoes

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

Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.

$$e^x ≈ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$$

For x < 0

2. Relevant equations

Taylor's Theorem to approximate a remainder:

$$|R(x)| = |\frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}|$$

Where z is some number between c and x, n is the degree of the approximating function, and c is where the function is centered at.

3. The attempt at a solution

From the Taylor polynomial given, c = 0 and n = 3. Since f(x) is e^x, the fourth derivative is simply e^x. If I want an error of less than 0.001,

$$|R(x)| = |\frac{e^z}{4!}x^4| < 0.001$$

Not too sure about this next part, but I think that since the function above increases as x increases, which means the error increases with an increasing x, I replaced z by x since the maximum error is given by the largest z value, and the largest z value is equal to x, and I'm interested in the error bound anyways.

$$|R(x)| = |\frac{(x^4)(e^x)}{4!}| < 0.001$$

...but I'm still left with the problem that I have to find a value that's stuck in an exponent and outside of one, so I'm assuming I'm doing this wrong and I can't figure out any other way to do it.

Last edited: Nov 7, 2012
2. Nov 7, 2012

### LCKurtz

When I saw it I wasn't sure if it was a typo or relevant to your question, but as you have stated the problem it is for $x<0$, which would help with the $e^x$.

3. Nov 7, 2012

### lurflurf

You are on the right track. That is a good estimate (about 7% low). To solve your equation exactly you need the Lambert function. You could also solve it approximately in several ways. One of which is to write it
x^4e^x=24/1000
x=fourthroot(24/1000)e^-(x/4)
and iterate a few times

4. Nov 7, 2012

### Dick

If are really only worried about x<0 then you have an alternating series. There's an even easier estimate you could use that doesn't even involve e^x. It turns out to be the same error bound that LCKurtz is suggesting you could arrive at by putting in an upper bound for the exponential.

Last edited: Nov 7, 2012
5. Nov 8, 2012

### Mangoes

Wow, that's very clever.

I hadn't noticed that the polynomial is the beginning of an alternating series and you can just find the remainder of the alternating series to solve for x.

I hadn't registered the fact that x < 0 in my mind so I didn't see that you could just put 1 in place of the e^x (and you'd get the same expression as if you had used the alternating series remainder to solve for x)

That's pretty cool, thanks for the help guys.