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I was once taught that, for using the Taylor series to expand $ln(1+x)$ the limit/range for having an accurate approximation (when the series "converges") is $|x|<1$. So the range of values for $ln(1+2x)$ would be $-\frac{1}{2}<x<\frac{1}{2}$.

What about problems like

Expand $ln(\frac{1-2x}{1+9x^2})$ up to $x^4$ and state the range of values where it is a good approximation

I would first split it into $ln(1-2x) - ln(1+9x^2)$, expand to $x^4$ and then combine the resulting polynomials, but the ranges for each $ln$ is $|x|<\frac{1}{2}$ and $|x|<\frac{1}{3}$ respectively. Which range should I choose and why?

(Explain like I'm a high school dropout studying math again)

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  • $\begingroup$ You appear to be confusing two different things. First, there is a range of $x$ values for which a Taylor series converges. Second, there is the issue of how many terms of a series you should sum to guarantee that the result is within a given allowable error bound--the specifics of what one means by a "good" approximation. $\endgroup$ – John Wayland Bales Nov 17 '17 at 18:52
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Your first expression converges for the range $|x|<\frac 12$, the second converges for $|x|<\frac13$, so find a range of values of $x$ for which both converge. $$\left[-\frac12,\frac12\right]\cap\left[-\frac13,\frac13\right]=\left[-\frac13,\frac13\right]$$ The range where they both converge is $|x|<\frac13$.

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According to Wolframalpha.com the fourth degree Taylor polynomial expanded about $x=0$ is

$$ f(x)=\ln\left(\frac{1-2x}{1+9x^2}\right)\approx -2x-11x^2-\frac{8}{3}x^3+\frac{73}{2}x^4=P_4(x) $$

Wolfram Alpha Link

You may compare the two graphs at Desmos where there appears to be "good" approximation on the interval $[-0.1,0.1]$ if by "good" you mean with an error on the order of $1\times10^{-3}$.

Direct computation shows that

$$ \vert f(-0.1)-P_4(-0.1)\vert\approx 1\times10^{-3} $$

and also

$$ \vert f(0.1)-P_4(0.1)\vert\approx 1\times10^{-3} $$

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