I'm dealing with a problem and would like a tiny bit of assistance.
Approximate the function $f(x)= 8(x+1)^{\frac 3 2}$ using a second degree Taylor polynomial at the point $a = 0$. What is the upper bound of the error at the point $x = 0.3$?
I get the following polynomial:
$$f_2(x) = 8 + 12x + 3x^2 + \frac {f'''(\xi)} {3!} x^3$$
where $a < \xi < x \implies 0 < \xi < 0.3$. The resulting error polynomial looks like this:
$$E(x) = -\frac {x^3}{2\sqrt{(\xi+1)^3}}$$
evaluating to $\frac {-0.0135} {\sqrt{(\xi+1)^3}}$ for $x = 3$. This is where it gets confusing. Since the error is negative, I'm not sure if I should maximize the denominator to get a higher value, or look at the error as sort of an absolute and minimize the denominator. Plugging in $\xi = 0$ yields $-0.0135$, and $\xi = 0.3$ yields $0.0091$.
When calculating $f(x)$ – [the approximation], I get $-0.012$, which is of course greater than $-0.0135$, but less than both $0.0135$ and $0.0091$. So I suppose my question here is really, what's my result here? The minus really throws me off.
Best Answer
There are two distinct things here, that you appear to be conflating:
Taylor's theorem says that $f(x)$ is equal to a polynomial in $x-a$ plus a remainder term, which can have various forms, $$ f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x). $$
The error in an approximation of $a$ by an expression $b$ is the absolute value of the difference, $\lvert a -b \rvert$.
So in this case, the error should be the absolute value of the difference between the function and the Taylor approximation, $$ \lvert f(x) - P_n(x) \rvert = \lvert R_n(x) \rvert. $$ Thus what you actually want is $$ \left| -\frac{x^3}{2(\xi+1)^{3/2}} \right|, $$ and the error is bounded by the maximum of this for $x=0.3$ and $ \xi \in [0,0.3]$.