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I want to find a representation of the function mentioned above, so I took into account that: $$f(x)=\sqrt{(4+x)^3}=8\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$$ and developing the binomial series for $\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$ we have to $$\left(1+\frac{x}{4}\right)^{\frac{3}{2}}=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdot\cdot\cdot$$

Now, to obtain the terms of the original series simply multiply by 8; now, how can I express the function as a series, since I have only managed to express that as a sum of terms

Any help is appreciated

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    $\begingroup$ The (infinite) sum of terms is the series. What are you looking for? $\endgroup$ Aug 1 at 2:10
  • $\begingroup$ a way of expressing the latter I have as a summation(sigma) $\endgroup$
    – Wrloord
    Aug 1 at 2:34

2 Answers 2

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Starting from $$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\,x^n \tag{1}$$ which is important and should be treated as a fundamental "building block", in my opinion
one gets by termwise integration $$ \sqrt{1-x} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(1-2n)}\,x^n \tag{2}$$ and by integrating again $$ (1-x)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}}{4^n(2n-1)(2n-3)}\,x^n. \tag{3}$$ By replacing $x$ with $-x/4$ we get $$ \left(1+\frac{x}{4}\right)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n \tag{4}$$ so $$\boxed{ \sqrt{(4+x)^3} = \sum_{n\geq 0}\frac{24\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n} \tag{5}$$

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According to the binomial series expansion we obtain \begin{align*} \left(1+\frac{x}{4}\right)^{\frac{3}{2}}&=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdots\\ &=1+\binom{3/2}{1}\frac{x}{4}+\binom{3/2}{2}\left(\frac{x}{4}\right)^2+\binom{3/2}{3}\left(\frac{x}{4}\right)^3+\cdots\\ &\,\,\color{blue}{=\sum_{j=0}^\infty\binom{3/2}{j}\left(\frac{x}{4}\right)^j} \end{align*}

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