Sequences and series.

Hey there.

I'm wondering if anyone here will be able to assist with the following issues:

1. I'm currently working on questions asking me to:


- $\displaystyle Find \: \alpha \: so \: that \: a_n \: = \: O(n^\alpha)$

- A sequence is then given such as $\displaystyle n^2 \: + \: n \: + \:1 $

I'm curious as to what type of questions these are, they are listed under "Order of magnitude" in my notes but searching for this returns no useful resources.

2. MacLaurin Series

-Lastly, I'm struggling in creating the MacLaurin Series for the function:

$\displaystyle \frac{1}{x^2 + 1} $

-I have tried differentiating but this becomes quite difficult after the third derivative or so. I am guessing there is another method to solve this?


Any help concerning either of these issues would be greatly appreciated.

Thank you.

Quote:

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Originally Posted by track02

-Lastly, I'm struggling in creating the MacLaurin Series for the function: $\displaystyle \frac{1}{x^2 + 1} $

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Write:

$\displaystyle f(x)=\dfrac{1}{1+x^2}=\dfrac{1}{1-(-x^2)}$

and use:

$\displaystyle \displaystyle\sum_{n=0}^{+\infty}t^n=\dfrac{1}{1-t}\quad (|t|<1)$

Regards.

Fernando Revilla

Thanks for the assistance.

Correct me if I am wrong, but by treating $\displaystyle x^2 $ as $\displaystyle (-x^2) $ you are allowing the use of sigma notation for an infinite geometric series?

Which will have the same effect as using a MacLaurin Series?

Also, does anyone have any idea where I can find some information regarding my first issue, I've been searching and still nothing!

Thanks again.

For large n we have
$\displaystyle
a_n=n^2.
$

$\displaystyle
O(n^{\alpha})
$
means that

$\displaystyle
\displaystyle { \lim \frac{a_n}{n^{\alpha}}=const
}
$

that is that these functions have the same order of magnitude.

Quote:

---

Originally Posted by track02

Correct me if I am wrong, but by treating $\displaystyle x^2 $ as $\displaystyle (-x^2) $ you are allowing the use of sigma notation for an infinite geometric series?

---

Right.

Quote:

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Which will have the same effect as using a MacLaurin Series?

---

Right, as a consequence of unicity of the Taylor expansion.

Regards.

Fernando Revilla