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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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  • $\begingroup$ Binomial Theorem. Definitely the Binomial Theorem. Which is basically like an everywhere-convergent Taylor Series. $\endgroup$ Feb 9, 2017 at 1:30
  • $\begingroup$ @BrevanEllefsen Nope. Not everywhere-convergent. $\endgroup$ Feb 9, 2017 at 1:31
  • $\begingroup$ @SimplyBeautifulArt is right. Not everywhere convergent - my blunder. Note that the geometric series is basically a special case of the Binomial $\endgroup$ Feb 9, 2017 at 2:33

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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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  • $\begingroup$ Can every infinite sum that converges to the expression in question be reduced to the geometric series? $\endgroup$
    – OFRBG
    Feb 9, 2017 at 1:42
  • $\begingroup$ @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. $\endgroup$ Feb 9, 2017 at 1:43
  • $\begingroup$ 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.) $\endgroup$
    – OFRBG
    Feb 9, 2017 at 1:45
  • $\begingroup$ @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... $\endgroup$ Feb 9, 2017 at 1:46
  • $\begingroup$ So there's is a one-to-one relationship between differentiable functions and their infinite sum representations? $\endgroup$
    – OFRBG
    Feb 9, 2017 at 1:48
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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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    $\begingroup$ :-) Looks good. $\endgroup$ Feb 9, 2017 at 1:39
  • $\begingroup$ No convergence criteria? $\endgroup$
    – NoChance
    May 2, 2023 at 10:42
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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.

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  • $\begingroup$ Good that you included the convergence criteria! $\endgroup$
    – NoChance
    May 2, 2023 at 10:42
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$$\frac1{1+x}$$

$$=\frac{1-x}{1-x^2}$$

$$=\frac{(1-x)(1+x^2)}{1-x^4}$$

$$=\frac{(1-x)(1+x^2)(1+x^4)}{1-x^8}$$

At the $n^{th}$ step, we have

$$=\frac{(1-x)\prod\limits_{k=1}^n (1+x^{2^k})}{1-x^{2^n}}$$

As $n\to\infty$ with $|x|<1$, this becomes

$$=1-x+x^2-x^+\dots+(-x)^{k+1}+\dots+$$

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