## anonymous 4 years ago Would this be correct? $f(x)=cos(\frac{(\pi x)}{2}$ Use MacLaurin table $cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$=$1-\frac{x^2}{2!}+\frac{x^4}{4!}-frac{x^6}{6}$ $cos(\frac{(\pi x)}{2})=\sum_{n=0}^{\infty}(-1)^n\frac{\frac{(\pi x)}{2}^{2n}}{(2n)!}$=$1-\frac{\frac{(\pi x)}{2}^2}{2!}+\frac{\frac{(\pi x)}{2}^4}{4!}-frac{\frac{(\pi x)}{2}^6}{6}$

1. anonymous

$\Large \left(\frac{\pi x}{2}\right)^{2n}$

2. anonymous

The site is a bit dodgy for me today, it keeps reloading and kicking me out so I am not sure if I see the original equation correct.

3. anonymous

sorry about that. The first line should read: $f(x)=cos(\frac{\pi x}{2})$

4. anonymous

The second line was meant to be the macLaurin series for cos(x) that I found in my book. $cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$

5. anonymous

the final few lines are what I believe the answer should be. $cos\left(\frac{\pi x}{2}\right)=\sum_{n=0}^{\infty}(-1)^n\frac{\frac{(\pi x)}{2}^{2n}}{(2n)!}$

6. anonymous

Good, so you want to put wherever you see an x, you want to put the following: $\Large \left( \frac{\pi x}{2} \right)$ right? because that is your new 'x' which you can substitute into the given equation.

7. anonymous

yeah I think so. That would make. $1-\frac{\left(\frac{\pi x}{2}\right)^2}{2!}+\frac{\left(\frac{\pi x}{2}\right)^4}{4!}-\frac{\left(\frac{\pi x}{2}\right)^6}{6!}$

8. anonymous

yes, exactly! I was confused first, because the way you did $$\LaTeX$$ it in your original post it looked like you would take the Exponent $$2n$$ only in consideration for the numerator, but you also have to take it in consideration for your denominator - just like you did!

9. anonymous

Thank you! It's kinda difficult typing latex blindly in the Question box to the left.

10. anonymous

No problem, you did it all right then, I was just making sure that you noticed that (-: And I agree, there should at least be a preview function of the post!