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Im trying to solve a question that is similar to this one, so any help with this one will be greatly appreciated. I know that the final answer is 84 terms.

Consider the Taylor expansion:

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

Estimate how many terms are required to obtain $ln(\frac{31}{16})$ to four decimal places accuracy.

I've started off so far by saying $$ln(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$$ and $$(1+x)= \frac{31}{16} \text{ so }x=\frac{15}{16}$$

Thanks in advance.

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MathsRookie
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  • Hint: the Lagrangian remainder for the Taylor expansion of a function $f$ is $$\frac{f^{(n+1)}(\zeta)}{(n+1)!} |x-c|^{n+1} $$, provided you expand around $c$, and $\zeta \in (x, c)$. – Sean Roberson Feb 28 '18 at 01:34

2 Answers2

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Hint In an alternating series, the error is smaller then the absolute value of the first missing term.

This means that, for $x>0$, you have $$|ln(1+x)-\sum_{n=1}^N (-1)^{n+1}\frac{x^n}{n}| \leq \frac{x^{N+1}}{N+1}$$

Now, se $x=\frac{15}{16}$ and estimate for whicn $N$ do you have $$\frac{15^{N+1}}{16^{N+1}(N+1)}<0.0001$$

N. S.
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You'll need the Lagrange form of the remainder (see my hint). We're evaluating this expansion at $x = \frac{15}{16}$, and the expansion is centered at $c = 0$.

We now need to find $n$ such that

$$ \left| \frac{f^{(n+1)}(\zeta)}{(n+1)!} \left( \frac{15}{16} \right)^{n+1} \right| \leq 0.00005 $$

We can take $f^{(n+1)}(\zeta) = 1 $ as the derivatives are just terms of the expansion, and it's probably the best we can do. Now we just test values of $n$ and see what happens.

Sean Roberson
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