0
$\begingroup$

I'm having a lot of trouble with the following function:

$$f(x)= \frac{cos(6x^2)-1}{x^3} $$

I am trying to find a Maclaurin series representation for it which I will eventually need to compare it to it's Taylor series to find a higher derivative. I began with the Maclaurin series:

$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!} $$

$$cos(6x^2)= \sum_{n=0}^\infty \frac{(-36)^nx^{4n}}{(2n)!} $$

At this point is where I get confused, because of the fact that I have to subtract one. I tried subtracting 1 from both sides then dividing by x to the third power, but it got very messy. I then thought of splitting the initial function into:

$$f(x)= \frac{cos(6x^2)}{x^3} - \frac {1}{x^3} $$

So I ended up with:

$$\frac{cos(6x^2)-1}{x^3}= -\frac{1}{x^3} \sum_{n=0}^\infty \frac{(-36)^nx^{4n-3}}{(2n)!} $$

But I don't know if this is right, and I do not know how to compare this to the power series for the same function when there is the term in front of the sum.

Any help would be appreciated!

Thanks!

  • 0
    Hint: The first term in the expansion for cosine is $1$, so that cancels. From there, divide the power series by $x^3$. – 2017-01-31
  • 1
    Subtracting one from $\cos(6x^2)$, with the power series you gave above, yields $\sum_{n=1}^{\infty}\frac{(-36)^nx^{4n}}{(2n)!}$. That seems fine to me - just divide by x^3. – 2017-01-31
  • 0
    @SimplyBeautifulArt Thank you to you both, you've helped me solve this bit! Just another question to follow. When it comes time to compare my Maclaurin series to the Taylor series of the same function, is it valid to compare them even if my Maclaurin series would start at 1? – 2017-01-31
  • 0
    Yes, of course. – 2017-02-01

1 Answers 1

0

$f(x)$ is an odd function, hence if $f(x)$ is continuous at the origin the wanted limit is simply $0$.
On the other hand $1-\cos(6x^2) = 2\sin^2(3x^2)$, hence $f$ is clearly continuous at the origin.