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Given numbers rewrite to the form of series with rational terms:

(a) $\frac{\sin \sqrt{3}}{\sqrt{3}}$

(b) $\ln(\frac{8}{3})$

(c) $\frac{1}{\sqrt[3]{2}}$

I'm afraid I don't understand the order.. I have to rewrite these numbers using Taylor's formula (it's what we already had)? But how should I know at what point I should expand Taylor's formula?

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    so, $\frac{\sin\sqrt{3}}{\sqrt{3}}=\sum_{n=0}^{+\infty}(-1)^n\frac{3^n}{(2n+1)!}$ ? should I bother that expansion $\sin(x)=\sum_{n=0}^{+\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$ is in $x_0=0$ or not, and why not?2012-03-25

1 Answers 1

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There are infinitely many possible answers. The simplest approach to $\ln(8/3)$ is to note that $\ln(8/3)=-\ln(3/8)=\ln(1-5/8)$. Now use the ordinary power series for $\ln(1+x)$.

Or else we can use the following classical old trick. We find $x$ such that $\frac{8}{3}=\frac{1+x}{1-x}.$ Easily, we get $x=5/11$. Now we use the fact that $\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln(1-x).$ From the power series expansion of $\ln(1+t)$ we obtain $\ln(1+x)-\ln(1-x)=2x+\frac{2x^3}{3}+\frac{2x^5}{5}+\frac{2x^7}{7}+\cdots.$


For something like $\dfrac{1}{\sqrt[3]{2}}$, there are infinitely many possible series. If we really want to be efficient in our calculations, we might note that $\frac{125}{128}=\frac{125}{64}\frac{1}{2}.$ Moving things around and taking cube roots we obtain $\frac{1}{2^{1/3}}=\frac{4}{5}\left(\frac{125}{128}\right)^{1/3}.$ Now the power series expansion of $(1+t)^{1/3}$, with $t=-3/128$, will get us fast convergence.