Question

Let f(x) =√(x+1.) compute the Taylor series around a=3 for the first three terms. Use this to estimate √(4.1.) What is the error?

Answer 1

\[2+\frac{x-3}{4}-\frac{1}{64} (x-3)^2 \]
The series evaluated at 3.1 is 2.02484
\[\sqrt{4.1} = 2.02485\]
\[\text{Error}: 1*10^{-5} \]
Here is the series with an extra term if you need it.
\[2+\frac{x-3}{4}-\frac{1}{64} (x-3)^2+\frac{1}{512} (x-3)^3 \]