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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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    Do you know series expansion for $\frac{\sin x}x$, $\log(1+x)$ and $(1+x)^\alpha$?2012-03-25
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    yes, so I should for example treat $\ln(8/3)$ like $\ln(1+x)$ and expand it with $x=5/3$?2012-03-25
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    This won't work exactly this way. The Taylor expansion of $\ln (1+x)$ at $0$ has a radius of convergence of $1$, so we can't naively input $x=5/3$ in the formula.2012-03-25
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    so how should I solve this task? and I have to expand into series with rational terms, is it relevant?2012-03-25
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    You could for example write $\log\frac 83 = \log 2 + \log \frac 4 3$ and expand these two terms separately.2012-03-25
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    This, or $\log\frac83=-\log(1-\frac58)$.2012-03-25
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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.