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I do not know how to prove that $$\frac{1}{x^2+2x+8}$$ is analytic on R.

What I need to do is let "a" be any real number, and try to make this function become a power expansion at a.

e.g. $\frac{1}{x}$ = $\sum_{k=0}^i (-1)^n \frac{1}{a^(n+1)}(x-a)^n$, where i goes to infinity.

This one is to let $\frac{1}{x}$ = $\frac{1}{a+(x-a)}$ and then keep going we will get the equation above.

But $\frac{1}{x^2+2x+8}$ does not even have real root..., which makes it tricky and I have no idea how to go. I tried many ways to change the denominator but all failed.

Please help... Thanks!

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    Do you know complex numbers? If so, you can still make a partial fraction decomposition using the non-real zeros. – 2017-02-28
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    But I need to make it contain a (x-a) in the end... and the number before (x-a) has to be real. I just don't feel that the non-real zeros will gone then. – 2017-02-28
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    Write it Taylor expansion at $a$ and show that it radius of convergence is not zero and that the remainder vanishes. Alternatively you can show that $f^{-1}$ is analytic and with it radius of convergence greater than zero, then $f$ it is also analytic in it domain with radius of convergence greater than zero. – 2017-02-28
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    But the key is that I have problem writing it in Taylor expansion at a. I can only change it into something like $\frac{1}{7}*\frac{1}{1-(-1/7)(x+1)^2}$. – 2017-02-28
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    @Similarity as Daniel said use partial fraction decomposition, into it complex roots. After each fraction can be written as a geometric series. From here manage to write it Taylor expansion around any point $a\in\Bbb R$ and show that it radius of convergence is positive. – 2017-02-28
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    And after using the def of the convergence of geometric series I found that $\lvert x-a\rvert$ < $\lvert \frac{1}{-a-1+\sqrt{7}i}\rvert$. How come a inequality have i in it!? – 2017-02-28

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Polynomial functions are analytic on $\mathbf R$. Reciprocal of an analytic function is analytic on every open set contained in its domain, and $x^2+2x+8$ has no real root…

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    Just not being sure if I can use this theorem and finish this proof in 3 secs... – 2017-02-28
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    What theorems do you have on analytic functions? This one is very basic. Of course, I only gave a sketch. – 2017-02-28
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    The professor only provides definition... Just like a small topic before uniform convergence. So I guess he wants us to treat this case in a way like $\frac {1}{x}$ – 2017-02-28
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    Then you can use substitution in the expansion of $\dfrac1{1+u}$ or $\dfrac1{1-u}$, setting $u=\dots$ – 2017-02-28
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    Yeah I finally solved it in a similar form like this. Thanks! – 2017-02-28
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Following the comments, suppose that $q$ is a polynomial of second degree with two complex roots, then

$$\frac1{q}=\frac{a_1}{x-z}+\frac{a_2}{x-\bar z}$$

for some $a_1,a_2,z\in\Bbb C$. Then

$$\frac{a_1}{x-z}=-\frac{a_1}{z}\cdot\frac1{1-x/z}=-\frac{a_1}z\sum_{k=0}^\infty (x/z)^k,\quad |x|<|z|$$

Then for any $x\in\Bbb R$ exists some $c\in\Bbb R$ such that $x=r-c$ and $|r|<|c+z|$, hence

$$\frac{a_1}{x-z}=\frac{a_1}{r-(c+z)}=-\frac{a_1}{c+z}\cdot\frac1{1-\frac{r}{c+z}}=-\frac{a_1}{c+z}\sum_{k=0}^\infty \left(\frac{r}{c+z}\right)^k=\\=-\frac{a_1}{c+z}\sum_{k=0}^\infty \left(\frac{c+x}{c+z}\right)^k,\quad |c+x|<|c+z|$$

From here is trivial to see that $1/q$ is analytic in all $\Bbb R$, provided that $x-z\neq 0$ for all $x\in\Bbb R$.