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

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Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
\[ \Large \left(\frac{\pi x}{2}\right)^{2n} \]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
sorry about that. The first line should read: \[f(x)=cos(\frac{\pi x}{2})\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
The second line was meant to be the macLaurin series for cos(x) that I found in my book. \[cosx=\sum_{n=0}^{\infty}(1)^n\frac{x^{2n}}{(2n)!}=1\frac{x^2}{2!}+\frac{x^4}{4!}\frac{x^6}{6!}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
the final few lines are what I believe the answer should be. \[cos\left(\frac{\pi x}{2}\right)=\sum_{n=0}^{\infty}(1)^n\frac{\frac{(\pi x)}{2}^{2n}}{(2n)!}\]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yeah I think so. That would make. \[1\frac{\left(\frac{\pi x}{2}\right)^2}{2!}+\frac{\left(\frac{\pi x}{2}\right)^4}{4!}\frac{\left(\frac{\pi x}{2}\right)^6}{6!}\]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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!
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Thank you! It's kinda difficult typing latex blindly in the Question box to the left.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.1
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!
 2 years ago
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