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I have the function $\dfrac 1{1+x}$ which I want to express as an infinite polynomial. I believe the correct term is Taylor Series. How do I solve this problem?

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    Binomial Theorem. Definitely the Binomial Theorem. Which is basically like an everywhere-convergent Taylor Series. – 2017-02-09
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    @BrevanEllefsen Nope. Not everywhere-convergent. – 2017-02-09
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    @SimplyBeautifulArt is right. Not everywhere convergent - my blunder. Note that the geometric series is basically a special case of the Binomial – 2017-02-09

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Use the geometric series $$\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{i=0}^n (-x)^n\quad\text{for}\quad|x|<1$$

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Brute force Taylor anyone?

$$f^{(n)}(x)=\frac{d^n}{dx^n}\frac1{1+x}=\frac{n!(-1)^n}{(1+x)^{n+1}}$$

If you want a proof, use induction.

At $x=0$, we get

$$=n!(-1)^n$$

Putting this into Taylor's theorem:

$$\frac1{1+x}=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{n!(-1)^n}{n!}x^n$$

$$\frac1{1+x}=\sum_{n=0}^\infty(-1)^nx^n$$

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    Can every infinite sum that converges to the expression in question be reduced to the geometric series? – 2017-02-09
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    @O.VonSeckendorff Yes. Would you like me to take the expansion at other points? It's obviously not too hard since I derived the general $n$-th derivative. – 2017-02-09
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    I didn't mean Taylor's series only. I meant every infinite sum that goes to the term. (Maybe it doesn't make much sense.) – 2017-02-09
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    @O.VonSeckendorff Well by infinite sum, I assumed you meant the power series, which has that polynomial like structure. And it should then be clear that it should be the Taylor series... – 2017-02-09
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    So there's is a one-to-one relationship between differentiable functions and their infinite sum representations? – 2017-02-09
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    @O.VonSeckendorff Nope. There exist $C^\infty$ functions that do not equal their Taylor series. Once you learn Lagrange remainders, you'll be able to show when a function is equal to it's Taylor series. – 2017-02-09
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    @O.VonSeckendorff your claim is true for functions in the complex plane that are differentiable though! One of the many cool thing about holomorphic function is that if they are once differentiable they are infinitely differentiable and are thus ***at least locally*** equal to their Taylor Series – 2017-02-09
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Because $1 - x + x^2 - x^3 + x^4 - \cdots = \dfrac {1}{1+x}$

It is sum of geometric series aka infinite polynomial.

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    :-) Looks good. – 2017-02-09
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Let's assume such a polynomial exist so that $\frac 1{1+x} = \sum a_i x^i$

Then $1 = (1+x)\frac1{1+x} = \sum a_i x^i + \sum a_i x^{i+1} = a_0 + \sum (a_{i+ 1} + a_i)x^i$.

So $a_0 = 1$ and $a_{i} + a_{i-1} = 0; a_{i} = -a_{i-1} \forall i > 0$.

Inductively that means $a_i = 1$ if $i$ is even and $a_i = -1$ if $i $ is odd. Or $a_i = (-1)^i$.

So the polynomial is $\sum_{i=0}^{\infty}(-1)^ix^i$. This is the geometric series one learns about in calculus. It does require a caveat that unless $|x| < 1$ the infinite sum will not converge.