Approximation by a polynomial degree 3 centered at a=2?

Question

Approximate $\:\mathbf{f(x) = 1/\sqrt{x^3 + 1}}\:$ by a polynomial of degree 3 centered at a = 2.

What exactly is this question asking when it says "approximate"? What is the end result supposed to be I'm so lost

I found the derivatives(up to the 3rd) and I substituted x with 2, but now what?

I'd appreciate any help!

Answer 1

The taylor polynomial of $f$ of dgree $3$ at $2$ is given by

$$f(2)+\frac{f'(2)}{1!}(x-2)+\frac{f''(2)}{2!}(x-2)^2+\frac{f'''(2)}{3!}(x-2)^3.$$

Answer 2

"Approximate" means: make use of some mathematical tool that you have mastered, or studied, that will give you a good approximation of a certain function. The most known method is Taylor series. So what you have to do is make use of Taylor Series. You know that

$$f(x) \approx \sum_{k = 0}^{+\infty} \dfrac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

Where $(k)$ stands for the derivative of $f$ of order $k$, evaluated in the given center $x_0$.

In your case, you find the three derivatives of $f$ as Fred wrote, centered at $x_0 = 2$ (your $a$ is my $x_0$).

Answer 3

To approximate a function usually means that you want to construct another function that 1) should be easier to compute and 2) takes values reasonably close to the original function. [The criterion 1) might be different, but this is rare.]

A method to obtain this is to start with a candidate expression (such as a polynomial) with adjustable parameters, and find the values of the parameters that minimize some measure of the average error. So you are trying a family of functions, and try to find the "best" of them.

Another method (probably the one meant to be used here) is to match the function closely in the neighborhood of a given $x$. This is done by ensuring that the value and derivatives (up to some order) of the function and its approximation are equal at the given $x$. This is the so-called Taylor approximation.

On the picture, the given function in blue, and Taylor approximations of increasing degree.