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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. – Arturo Magidin Jun 2 '12 at 2:25
Yes, you are right. It should be (-x)^(1/2), which I have now amended. Thanks! – JackReacher Jun 2 '12 at 7:36
@mathstudent: Is it $(-x)^{\frac{1}{2}}$ or $x^{\frac{1}{2}}$? You talk about the former but write the latter which is confusing! – Gigili Jun 3 '12 at 7:38

As Arturo pointed out in a comment, It has to be $(-x)^{\frac{1}{2}}$ to be defined for $x<0$, then you have:

$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}+\dots$$

Substituting $x$ with $(-x)^{\frac{1}{2}}$ we get:

$$\sinh (-x)^{\frac{1}{2}} = (-x)^{\frac{1}{2}} + \frac{({(-x)^{\frac{1}{2}}})^3}{3!} + \frac{({(-x)^{\frac{1}{2}}})^5}{5!} + \frac{({(-x)^{\frac{1}{2}}})^7}{7!}+\dots$$

Dividing by $(-x)^{\frac{1}{2}}$:

$$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 + \frac{({(-x)^{\frac{1}{2}}})^2}{3!} + \frac{({(-x)^{\frac{1}{2}}})^4}{5!} + \frac{({(-x)^{\frac{1}{2}}})^6}{7!}+\dots$$

And after simplification:

$$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!}+\dots$$

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