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What is the Taylor series for $$g(x) = \frac{\sinh((-x)^{1/2})}{(-x)^{1/2}}$$, for $x < 0$?

Using the standard Taylor Series: $$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}$$ I substituted in $x = x^{1/2}$, since $x < 0$, it would simply be $x^{1/2}$ getting, $$\sinh(x^{1/2}) = x^{1/2} + \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} + \frac{x^{7/2}}{7!}$$ Then to get the Taylor series for $\sinh((-x)^{1/2})/((-x)^{1/2})$, would I just divide each term by $x^{1/2}$?

This gives me, $1+\frac{x}{3!}+\frac{x^2}{5!}+\frac{x^3}{7!}$

Is this correct?

Thanks for any help!

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    Comments: (i) The series does not terminate at the degree 7 term; it keeps going. (ii) "$-x^{1/2}$" means $-\sqrt{x}$ (exponents take precedence over the minus sign), which would make it impossible for negative values. You probably mean $\sqrt{-x}$, or $(-x)^{1/2}$ instead. (iii) You need to substitute $(-x)^{1/2}$, not $\sqrt{x}$, which makes no sense for negative $x$, and there is *no* simplification based on the fact that $x$ is negative.2012-06-02
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    Yes, you are right. It should be (-x)^(1/2), which I have now amended. Thanks!2012-06-02
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    @mathstudent: Is it $(-x)^{\frac{1}{2}}$ or $x^{\frac{1}{2}}$? You talk about the former but write the latter which is confusing!2012-06-03

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