2
$\begingroup$

I am trying to find the Taylor series of $\sin^2(4x)$ but I kept getting it wrong. The following is my work:

Apply trig identity

$$\sin^2(4x) = \frac{1-\cos(8x)}{2} $$

Use basic Taylor series which is

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

Plug the Taylor series to the trig function which

$$\sin^2(4x) = \frac12 -\frac12 \sum_{k=0}^\infty (-1)^k \frac{(8x)^{2k}}{(2k)!}$$

Next, I have to find the first three non-zero term. For the first one, all I did was plug $0$ for $k$ in the function and got $0$ but got it wrong. Can anyone help me on this problem? I would appreciate it a lot!

$\endgroup$
2
$\begingroup$

$$\sin^2(4x)$$

We know that $\sin^2(4x)=\dfrac{1-\cos8x}{2}$ and $\cos x=\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}$

$$\cos8x=\sum_{n=0}^{\infty}(-1)^n\dfrac{(8n)^{2n}}{(8n)!}$$ and we get $$\sin^2(4x)=\dfrac{1-\cos8x}{2}=\dfrac12-\dfrac12\sum_{n=0}^{\infty}(-1)^n\dfrac{(8x)^{2n}}{(2n)!}$$

The first non-zero terms are $16x^2-\dfrac{256x^4}{3}+\dfrac{8192x^6}{45}$

$\endgroup$
  • $\begingroup$ Thanks for your help but how did (2n) become (8n)? And why did you write (8n) instead of (8x) (I'm referring to the numerator)? $\endgroup$ – Niko H Mar 15 at 0:49
  • $\begingroup$ @NikoH That was a typo. Now fixed it. $\endgroup$ – Key Flex Mar 15 at 0:52
  • $\begingroup$ But when you plug in n = 0 in the equation, wouldn't you get (1/2) - (1/2) (1 * 1/1), which is 0? $\endgroup$ – Niko H Mar 15 at 0:58
  • $\begingroup$ @NikoH Yes, we do get but question asks for the "first three non-zero terms" $\endgroup$ – Key Flex Mar 15 at 1:01
  • $\begingroup$ oh! thank you much! $\endgroup$ – Niko H Mar 15 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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