Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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

1 Answer 1

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:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.