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This is Section 1.6 of the book "An introduction to Numerical Methods and Analysis" by James F. Epperson.

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The questions that we will be working on:

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I have done question 1. I'm having trouble with question 6, and after question 6 is done, I have to do question 7 and question 7 needs the x values from question 1.

The first confusion I have is: Do I need the Taylor series about $x_o = 0$? I'm thinking no, it should be $x_o = \frac{3}{4}$, right? Well, it doesn't have to be $\frac{3}{4}$ but since we're considering the interval $[\frac{1}{2},1]$, $x_o = \frac{3}{4}$ seems to be the best choice.

(We're focused on the $\ln(f)$ part right? Since we are assuming that we already have $\ln(2)$ and $\ln(x_o)$ up to arbitrary precision and $\ln(z) = \ln(f) + \beta \ln(2)$, so $\ln(f)$ seems to be what we're interested in, where $f \in [\frac{1}{2},1]$ and so that's why I'm thinking we take $x_o = \frac{3}{4}$.)

Let's say I don't provide the value of $x_o$ for now:

The remainder for the Taylor series of $\ln(1+x)$ about the point $x_o$ is $(-1)^{n} \int_{x_o}^{x} (x-t)^{n} \left ( \frac{1}{1+t} \right )^{n+1} \text{dt}$ and the remainder for the Taylor series of $\ln(1-x)$ about the point $x_o$ is $ - \int_{x_o}^{x} (x-t)^{n} \left ( \frac{1}{1-t} \right )^{n+1} \text{dt}$, where $t$ is between $x$ and $x_o$. We can just subtract the remainders to get the remainder of the Taylor series of $\ln \left ( \frac{1-x}{1+x} \right )$.

In part c, I'm getting $x \in [0, \frac{1}{3}]$. Now I have no idea how to continue. I did subtract the remainders, did some working but I have to go upto n=70 or something to get the error less than $10^{-16}$. The working is quite long and no doubt wrong, I was wondering if someone can help me with all these parts.

The book says that question 6 should provide faster convergence compared to the working they have done.

Sine of the Time
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Sonny Jordan
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  • $\ln(1-x)-\ln(1+x)=-2(x+\frac13x^3+\frac15x^5+...)$ has a remainder with upper bound $\frac2{2n+1}|x|^{2n+1}\frac1{1-x^2}$ for its absolute value. You get below the demanded bound if $3^{2n+1}>10^{15}$, which is the case for $n=16$, so $16$ and perhaps also $15$ terms are sufficient for the bound, not 70 or 35. (No, for $n=15$ the bound gives $1.175\cdot 10^{-16}$, which is a little too large.) – Lutz Lehmann Feb 10 '23 at 19:18
  • Thank you for replying. Why did you construct the Taylor series around $x_o = 0$? Aren't we interested in the $\ln(f)$ part where $x_o$ should be in $[\frac{1}{2},1]$? Plus, the book says to use the integral form of the remainder, I think you are using the Lagrange form? – Sonny Jordan Feb 10 '23 at 20:42
  • As you said in the question, for $z\in[1/2,1]$ you get $x\in[0,1/3]$. You could argue for $x_o=1/6$ as the center of this interval, but that would introduce more complications than just using $x_o=0$. – Lutz Lehmann Feb 10 '23 at 21:14
  • Okay, thank you. What about using the integral form of the remainder? – Sonny Jordan Feb 11 '23 at 04:51

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