I want to show that $\left | f(3)-P(3) \right |\leq \frac1{1000}$ without a calculator.

Question

Let $f(x)=\ln x$. Then the Taylor polynomial of degree 2 at $x=e$ is $P(x)=1+\frac1e(x-e)-\frac1{2e^2}(x-e)^2$

I want to show that $\left | f(3)-P(3) \right |\leq \frac1{1000}$ without a calculator. How can I show this?

Answer 1

Find the next term of the Taylor series to be $\frac {(3-e)^3}{3e^3},$ which you are ignoring. The alternating series theorem says your error is less than this. Now estimate $3-e \lt 0.3$ and convince yourself that $e^3 \gt 10$ and you will be there.

Answer 2

In general, the error of a taylor polynomial, is at most the (absolute) value of the next term, so when you have an expansion in $x=a$, you get $$ f(x)\sim f_2(x) =f(a)+f'(a)(x-a)+\frac 1{2!} f''(a)(x-a)^2 $$ and the error of it is at most $$ |f(x)-f_2(x)|\leq \left|\frac 1{3!}f^{(3)}(c)(x-a)^3\right| $$ where $c$ is some value between $a$ and $x$.

Answer 3

The Lagrange form of the remainder $R_k(x) = f(x) - P_k(x)$ of the k-th order Taylor polynomial is given by $R_k(x) = \frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1}, $ where $\xi_L$ lies between $a$ and $x.$