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How do I find Maclaurin series for the function:

$$\sqrt[3]{\sin(x^3)}$$

The answer should be:

$$ x - \frac {x^7}{18} - \frac {{x}^{13}}{3240} + o(x^{13})$$

I tried:

$$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ...$$

So, I changed $x$ to $x^3$ to get:

$$\sin(x^3) = x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} - \frac {x^{21}}{7!} + ...$$

But, I'm stuck when it comes to power of 1/3:

$$\sqrt[3]{x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} - \frac {x^{21}}{7!} + ...} = a_0+a_1x+a_2x^2+a_3x^3+...$$

vonbrand
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Bek Abdik
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    looks like you can cube both sides and compare the coefficients. $a_0$ is zero and $a_1$ is 1 ... and so on – S L Feb 23 '13 at 08:56
  • Yeah, I found $a_0$ in that way, but, other coefficients are taking a lot of time. – Bek Abdik Feb 23 '13 at 08:59
  • Woops mathematica calculated it to $a3^3 + 6 a2 a3 a4 + 3 a1 a4^2 + 3 a2^2 a5 + 6 a1 a3 a5 + 6 a0 a4 a5 + 6 a1 a2 a6 + 6 a0 a3 a6 + 3 a1^2 a7 + 6 a0 a2 a7 + 6 a0 a1 a8 + 3 a0^2 a9$ – S L Feb 23 '13 at 09:06
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    and all even coefficients are zero. – S L Feb 23 '13 at 09:11
  • sorry, how can we prove it? – Bek Abdik Feb 23 '13 at 09:15
  • comparing the coefficients of $x^5$ you can show that $a_3$ is zero, and comparing coefficients of $x^7$ you can show that $a_5$ is zero. And find the coefficients of $x^9$ – S L Feb 23 '13 at 09:22

1 Answers1

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If you factor out $x^3$ from cubic root you'll get $$ x\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} $$ Now, use power series expansion for cubic root $$ \sqrt[3]{1-x} = 1-\frac x3-\frac {x^2}9+o(x^2) \\ \sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} = 1-\frac 13\left(\frac{x^6}{3!}-\frac{x^{12}}{5!} \right)-\frac 19\left(\frac {x^{12}}{3!3!}+o(x^{12}) \right ) = \\ \\= 1-\frac {x^6}{3\cdot 3!}+\left ( \frac 1{3\cdot5!}-\frac 1{9\cdot 3! \cdot 3!}\right)x^{12}+o(x^{12}) = 1-\frac {x^6}{18}-\frac {x^{12}}{3240}+o(x^{12}) $$ After multiplying to $x$ you'll get your answer.

Kaster
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