Building a Taylor series for the Runge function

Question

I'm being asked to consider a Runge function $$f(x) = \frac{1}{1+a^2x^2}$$ for arbitrary $a$, and $x \in [-1,1]$, and then compute a truncated Taylor series expansion of the form $$\sum_{k=0}^{N}a_kx^k$$ where the $a_k$ are "suitable coefficients".

I'm a bit confused on how I could go about constructing that sum. It doesn't even look like a Taylor series, so I'm not exactly sure what's expected. Even if I were to try to build a standard Taylor series, thhe derivatives of $f$ get messy very quickly.

Can someone shed some light on what this sum is, exactly?

Answer 1

It is known that $$\frac{1}{1+u}=1-u+u^2-u^3+\dots\qquad |u|<1.$$

With $u=(ax)^2$, we get $$\frac{1}{1+a^2x^2}=\frac{1}{1+(ax)^2}=1-(ax)^2+(ax)^4-(ax)^6+\dots\qquad |(ax)^2|<1. $$ Equivalently, we get $$\frac{1}{1+a^2x^2}=\sum_{n=0}^\infty (-1)^n(ax)^{2n},\qquad |x|<\frac{1}{|a|}$$

Answer 2

The point of creating a Taylor Series to approximate the Runge function is to show that as you increase the order of your approximation, the error can sometimes increase at certain sample points of your approximation, which is contrary to what you would intuitively think. Imagine collecting data on the contraction of metal at different temperatures and you have 10 points that are equal distance from each other on the horizontal axis (equidistant nodes in the article attachment). If the order of your polynomial is too high you can cause the approximation to be farther away from the actual curve at certain nodes rather than closer, so the high order of approximation causes divergence away from the desired curve. I posted a link to a useful article below. The article will have a more detailed explanation.

http://www.tlu.ee/~tonu/Arvmeet/Runge's%20phenomenon.pdf

Hope this helps!

Truth_Seeker24

Quote for the day: If learning were a science, then everyone would be humble

Answer 3

Hint: Observe \begin{align} \frac{1}{1-x} = \sum^\infty_{n=0} x^n \ \ \ |x|<1 \end{align} then it follows \begin{align} \frac{1}{1+a^2x^2} = \sum^\infty_{n=0}(-1)^na^{2n}x^{2n} \ \ \ \ |x|<\frac{1}{|a|}. \end{align}

Answer 4

Since a is being chosen by you then the $a_k$ terms should just be given by you as say a = 2

$\sum_{k=0}^{N}a_kx^k$ where $a_k$ given by Jacky Chong would be: $\sum^\infty_{n=0}(-1)^n(2)^{2n}x^{2n}$