Question

Let E_1 = \ln\!\left(x+1\right) - P_1(x) = \ln\!\left(x+1\right) - (x). Using a calculator or computer, graph E_1 for -0.1 \leq x \leq 0.1, and notice what shape the graph is. Then use the Error Bound for Taylor polynomials to find a formula for the maximum error, as a function of x, in this case:
|E_1(x)| \le

Answer 1

@myininaya

Answer 2

The error bound for an \(n\)th degree Taylor polynomial approximation (centered about \(x=x_0\)) is
\[|E_n(x)|=|f(x)-P_n(x)|\le\frac{M}{(n+1)!}|x-x_0|^{n+1}\]
where \(|f^{(n+1)}(x)|\le M\).

You're given a polynomial \(P_1(x)=x\) (i.e. \(n=1\)), and \(x_0=0\), so
\[|E_1(x)|=|\ln(x+1)-x|\le\frac{M}{2!}|x|^2\le \frac{M(0.1)^2}{2}\]
To find this value of \(M\), you need to determine the upper bound for the second derivative of \(\ln(x+1)\) in the given interval.