Applications of Series — Question 10

Question 10 – Solution

Step 1: Substitute into the exponential series.

$$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots.$$ Keeping terms through degree $2$ gives $$\boxed{Q_2(x)=1-x^2}.$$

Step 2: Analyze the model’s sign.

$$Q_2(x)>0\iff1-x^2>0\iff |x|<1.$$ It vanishes at $x=\pm1$ and becomes negative for $|x|>1$, whereas the true Gaussian is positive for every real $x$. This shows the model is local.

Step 3: Bound the local error.

For $|x|\le1$, the series in $x^2$ alternates with decreasing magnitudes. Therefore $$\boxed{|e^{-x^2}-(1-x^2)|\le\frac{x^4}{2}}.$$ The error is order $x^4$ near the origin.