# Thread: Taylor Series and shifting indices

1. ## Taylor Series and shifting indices

For the truncated form of the Taylor series, the Nth Taylor polynomial is given in our notes as

$T_{N}(x) = \sum_{n = 0}^N \frac{f^{(n)}(0)}{n!} \cdot x^{n}$

But then later, he defines $T_{3}(x)$ for $f(x)=\sin x$

as $x-\frac{x^{3}}{6}$

Should it not be expanded to three terms, or is the notation signifying that it should be expanded until n, rather than N, reaches 3?

Also, I have a question about shifting indices. Our notes give this equality

$\sum_{n = 1}^\infty \frac{(-1)^{n}4^{n}}{n!} = \sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - 1 = e^{-4} - 1 ,$

but is it not:

$\sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - \frac{4}{n+1}?$

Thanks

2. For the first, notice that all even-numbered terms cancel since $\sin(0)=0$. For the second, you're just adding the term corresponding to $n=0$, I don't really know what you did to get to the other expression.

3. Oh ok I see now, his notes just said to take 1 away from n in the index and increase it in the summand, not very clear to be honest! Thank you for that