Finding the first 3 terms of a power series... I am confused as to whether I can just plug numbers in or if there is more.

Question

The power series $J_0$= $$\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n}}{(n!)^22^{2n}}$$

Ive plugged in n=0,1,2 and have gotten 1-$\frac{x^2}{4}$+$\frac{x^4}{64}$

Is this all that is required?

The question is stated as "Find the first three terms of the power series for $J_0$ "

Answer 1

Yes, that's all there is to it! As you said, plugging in $n=0,1,2$, the series is;

$$\sum_{n=0}^\infty\frac{(-1)^{n}x^{2n}}{(n!)^22^{2n}}=1-\frac{x^2}{4}+\frac{x^{4}}{64}+\mathcal{O}\left(x^{6}\right)$$