# Taylor series Lagrange error

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.

• 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 $\xi \in [0,0.3]$.
• Whoops, yeah the range for $\xi$ is a mistake on my part. So when dealing with the error, I should always consider the absolute value? It happened that all of the problems I've seen now had $f(x) - P_n(x) > 0$, which is why I find this case a bit confusing. – anonra May 1 '17 at 18:20
if you look at the binomial series $$(1+.3)^{3/2} = 1 + \frac 32 (.3)+\frac 3 8 (0.3)^2 + e$$ where $$e = -\frac{3}{2^3 3!}(0.3)^3+\frac{9}{2^4 4!}(0.3)^4+\cdots$$ the point is that $e$ is an alternating series and there therefor jas the same sign and bounded by the first term $\frac 1{16}$. that is $$-\frac{1}{16} < e < 0$$ you multiply this by $8$ to get the bound for your problem.