Question

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

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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!

Answer 9

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

Answer 10

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!