So, I have found the Taylor series representation of this function:
$$\sum_{n=2}^∞ \frac{(-1)^n(-1+x)^n}{(-1+n)n} +x-1$$
And I have found the normal Taylor series polynomial to the third degree:
$T_3(x)=(x-1)+ \frac12 (x-1)^2 – \frac16 (x-1)^3$
1)How do I find the error estimate between $f$ and $T_3(x)$? What does it mean to "shift"?
2)The italicized portion of the question asks what is wrong with that solution and a simple fix.. it's way off from my series representation, such that I don't know how to fix it
Best Answer
Part a: You can use the Lagrange error: $f(x)-T_3(x)=f^{(4)}(\xi) (x-a)^4/4!$ where $\xi$ is between $x$ and $a$.
Part b: What they wrote is not in the form of a Taylor series. They just multiplied the series expansion of $\ln(x)$ (written as $\ln(1+(x-1))$) with $x$ but that by itself doesn't give you a Taylor series. But you could get a Taylor series by writing $x=(x-1)+1$ and then distributing, which is presumably what they meant by shifting.