For which $x$ does the 1st Taylor polynomial for ln(1+x) give 2 decimal places accuracy? My work:
To find the polynomial approximation:
ln(1+0) = ln(1) = 0 so the constant term is 0.
$\frac {d}{dx} \ln(1+x) = \frac {1}{1+x}$, and at $x=0$, this is equal to 1.
So the polynomial we're looking for is $x$.
$R_1(x) = \frac {f^2(x)}{2!} x^2$
By substituting the second derivative, we obtain $R_1(x) = \frac {-1}{2(1+c)^2} x^2$
Now how do I use this remainder to determine the accuracy?
 A: Since the Taylor series for $\ln (1+x)$ is alternating, given by $x - x^2/2 + x^3/3 - \ldots$, you get that if $0 < x < 1$ then the error for the approximation of $\ln (1+x)$ by $x$ is bounded by the second term which is $x^2/2$ (because you get an alternating series of decreasing terms, so the total error is bounded by second term and in fact any expansion that ends with an even power and starts with the second power). You could also use e.g. $x^2/2 - x^3/3 + x^4/4$ as an upper bound for the error. But assuming that you wanted $x^2/2$ to be your bound for the error, then you would want $x^2/2 < 0.01$ so $x < \sqrt{0.02}$. This is not a bad bound, compared to what you would get if you bounded the error by $x^2/2 - x^3/3 + x^4/4$ or even further into the series. If you also want to bound the error by $0.01$ for negative $x$, then the series you get for the remainder is not alternating, it's all negative and given by $-|x|^2/2 - |x|^3/3- \ldots$ which in abosolute value is bounded by $|x|^2/2(1-|x|)$ by comparison to the geometric series you get when you change all denominators to $2$, so you want $|x|^2/2(1-|x|) < 0.01$ for negative $x$. You can solve this as a quadratic in $|x|$.
