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My question:

How many terms of the Maclaurin series for $\ln(1+x)$ do you need to use to estimate $\ln(1.4)$ to within $0.001$?

I feel like I'm lost and my textbook that I'm using doesn't really go into how to solve problems like these in the section Taylor series and Maclaurin. Even worse, I'm a bit confused about all of this so I'm not sure where to even proceed.

So my process is this:

  • find the general formula for this series
  • If it's alternating, find the term so that the error is less than 0.001
  • I can also use the Taylor Inequality to estimate the error.

Where in the Stewart book does it even go into problems like these?

I have found the following derivatives:

$$f'(x) = \frac{1}{1+x},\ f''(x) = \frac{-1}{(1+x)^2}, \ f'''(x) = \frac{2(1+x)}{(1+x)^4} = \frac{2}{(1+x)^3},\ f^{(4)}(x) = \frac{-6}{(1+x)^4}$$

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You don't have to compute the derivatives as you do.

Use directly

  • classical Taylor expansion :

$$\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\tfrac{x^4}{4}-...\tag{1}$$

with $x=0.4$ and

  • the theorem about alternating series saying that the error done by truncating this series is less that the module of the first discarded term (i.e., the first one not present in the truncated series).

The condition is that this first discarded $n$th term is such

$$\tfrac{0.4^n}{n}\leq 0.001$$

Due to the fact that $\tfrac{0.4^6}{6} \ < \ 0.001 \ < \ \tfrac{0.4^5}{5}$, we have $n=6$.

Thus we need $n-1=5$ terms in expansion (1).

Verification : $\ln(1.4)= 0.3364722366...$ whereas $0.4-\tfrac{0.4^2}{2}+\tfrac{0.4^3}{3}-\tfrac{0.4^4}{4}+\tfrac{0.4^5}{5}=0.3369813333...$. The error $\approx 0.0003$ is indeed less than $0.001$.

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  • $\begingroup$ Why is it a classical taylor expansion? $\endgroup$ – Jwan622 Apr 19 at 13:26
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First of all, you do need the MacLaurin series $$ \tag1\ln(1+x)=\sum_{k=0}^\infty a_kx^k.$$ Given that the derivative of $\ln(1+x)$ is $\frac1{1+x}$, and from the geometric series $$ \frac1{1+x}=\sum_{k=0}^\infty (-1)^kx^k,$$ you ought to see that $a_k=\frac{(-1)^{k-1}}k$ for $k>0$ (and $a_0=0$). In particular, for $0<x<1$, $(1)$ is an alternating series with secreasong (in absolute value) terms. Hence we can just stop when the further summands would be smaller than the desired error bound, without checking any specific error formula. In other words, we find $n$ with $\frac{0.4^n}{n}<0.001$ and then take $\sum_{k=0}^{n-1}\frac{-(-0.4)^k}k$ as our approximation of $\ln(1.4)$.

Explicitly, $$ \ln(1.4)\approx 0.4-0.08+0.021\overline3-0.0064+0.002048$$ with the next summand $-0.000682\overline6$ already small enough.

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  • $\begingroup$ Is the general form missing an x? $\endgroup$ – Jwan622 Apr 19 at 13:25

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