$$f(x) = \lim_{n\to \infty} \sum_{r=1}^n 3^{r-1}\sin^3(x/(3^r)) $$ I tried using the formula relating $\sin(3x)$ to $\sin^3(x)$ but got later stuck with a similar series who's sum I didn't know how to calculate
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1$\begingroup$ I don't understand what is being summed here. Is the equation as stated correct? $\endgroup$– TomGrubbNov 4, 2016 at 20:52
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$\begingroup$ I'm trying to but I can't get to place the infinity sign i n the limit $\endgroup$– TejusNov 4, 2016 at 20:55
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1$\begingroup$ I've edited it to try and make sense of your formula but you might want to check if this is actually what you intended to ask. $\endgroup$– FimpellizzeriNov 4, 2016 at 20:56
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1$\begingroup$ I feel like there is a good chance that this always diverges because $3^{n-1}$ grows exponentially, while the sum is bounded and does not, to my knowledge, converge to zero. $\endgroup$– Will FisherNov 4, 2016 at 21:03
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2$\begingroup$ Note that $$\sin^3\theta=\dfrac{1}{4}(3\sin\theta-\sin 3\theta)$$ and $$\sum_{r=0}^n\sin k\theta=\dfrac{\sin\left(\dfrac{n\theta}{2}\right)\sin \left(\dfrac{(n-1)\theta}{2}\right)}{\sin \left(\dfrac{\theta}{2}\right)}$$ $\endgroup$– BumblebeeNov 4, 2016 at 21:47
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1 Answer
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mrm{f}\pars{x} = \lim_{n\to \infty}\sum_{r = 1}^{n}3^{r - 1}\sin^{3}\pars{x \over 3^{r}}:\ ?}$.
\begin{align} \mrm{f}\pars{x} & = \lim_{n\to \infty}\sum_{r = 1}^{n}3^{r - 1}\sin^{3}\pars{x \over 3^{r}} = {1 \over 4}\lim_{n\to \infty}\sum_{r = 1}^{n}\bracks{% 3^{r}\sin\pars{x \over 3^{r}} - 3^{r - 1}\sin\pars{x \over 3^{r - 1}}} \\[1cm] & =\require{cancel} {1 \over 4}\lim_{n\to \infty}\left\lbrace% \bracks{\cancel{3\sin\pars{x \over 3}} - \color{#f00}{\sin\pars{x}}} + \bracks{\cancel{3^{2}\sin\pars{x \over 3^{2}}} - \cancel{3\sin\pars{x \over 3}}}\right. + \\[5mm] &\phantom{= {1 \over 4}\lim_{n \to \infty}\braces{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}} \left.\bracks{\cancel{3^{3}\sin\pars{x \over 3^{3}}} - \cancel{3^{2}\sin\pars{x \over 3^{2}}}} + \cdots + \bracks{\color{#f00}{3^{n}\sin\pars{x \over 3^{n}}} - \cancel{3^{n - 1}\sin\pars{x \over 3^{n - 1}}}}\!\!\right\rbrace \\[1cm] & = {1 \over 4}\,\lim_{n \to \infty}\bracks{-\sin\pars{x} + 3^{n}\sin\pars{x \over 3^{n}}} = \bbx{{1 \over 4}\bracks{\vphantom{\Large a}x - \sin\pars{x}}} \end{align}
answered Nov 5, 2016 at 7:03