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I know that $$(1+x)^n=1+nx+\frac {n(n-1)}{2!}x^2+\frac {n(n-1)(n-2)}{3!}x^3+\frac {n(n-1)(n-2)(n-3)}{4!}x^2+\\ \dots+\frac {n(n-1)\dots(n-r+1)}{r!}x^r\text.$$ I know that this formula is valid for any real number n, provided that $|x|<1$. The formula extends infinitely for a negative value of $n$, and terminates for a positive value of $n$. Is there a formula for a case where $|x|>1$? Why is this formula applicable only for $|x|<1$?

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  • $\begingroup$ I don't have the background to disagree here, but I'm not sure you can talk about factorials of a non-positive integer without a $\Gamma$ function (and, specifically, I'm including positive non-integers here, i.e. positive rational non-integers as needing a definition of the $\Gamma$ function)... $\endgroup$
    – Jared
    Sep 9, 2021 at 8:09
  • $\begingroup$ I suggest studying (way beyond my understanding) the "real continuation" of a function: mathworld.wolfram.com/GammaFunction.html $\endgroup$
    – Jared
    Sep 9, 2021 at 8:29
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    $\begingroup$ @Kavi Rama Murthy. Why did you delete your answer ? Cheers :-) $\endgroup$ Sep 9, 2021 at 8:29
  • $\begingroup$ @Jared If the right side is replaced by an infinite sum then OP's formula is valid for all real $n$ when $|x|<1$. $\endgroup$ Sep 9, 2021 at 8:29
  • $\begingroup$ "terminates for a positive value of $n$": no, $n$ needs to be natural. In that case, any $x$ works. $\endgroup$
    – user958916
    Sep 9, 2021 at 8:33

2 Answers 2

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When $x<-1$ note that $(1+x)^{n}$ is not defined for all real $n$. For $x>1$ the series on the right is not convergent. However, you can get an expansion for $x>1$ using the fact that $(1+x)^{n}=x^{n} (1+\frac 1 x)^{n}$ and using the expansion of $(1+\frac 1 x)^{n}$ (which is valid sinec $|\frac 1 x| <1$ in this case).

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  • $\begingroup$ I feel like you're using integer $n$ here (a Lorentz series), but OP asked about real $n$. $\endgroup$
    – Jared
    Sep 9, 2021 at 8:17
  • $\begingroup$ @Jared No, I am not assuming that $n$ is an integer. I has to avoid $x <-1$ and I have corrected the answer. $\endgroup$ Sep 9, 2021 at 8:25
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    $\begingroup$ @Jared: Laurent, not Lorentz ! $\endgroup$
    – user958916
    Sep 9, 2021 at 8:36
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    $\begingroup$ @Jared Note that anyway, n! can be defined for noninteger $n$ as $n!=\Gamma(n+1)$. It's really not a problem. And not it's not "way more complicated", it's also a Taylor series. $\endgroup$ Sep 9, 2021 at 9:50
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    $\begingroup$ @Jared You don't need the $\Gamma$ function to write the Taylor expansion of $(1+x)^n$ for noninteger $n$. The (more usual?) formula involves products $n(n-1)\cdots(n-r+1)$ as in your question, which you can write using the binomial coefficient $n\choose r$ with noninteger $n$ (does not require the $\Gamma$ function, it's only an extension of the definition and a shortcut for the product). Just write down the derivatives, it's really just Taylor's formula. And there is no $n!$ anywhere, only $r!$. $\endgroup$ Sep 11, 2021 at 8:44
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The series you have written is valid for any real value of $x$ if $n$ is zero or a natural number. In this case your expansion is an identity and you get a polynomial of degree $n+1$ in $x$.

But if $n$ is negative integer or non integral, your expansion is valid only for |x|<1 and the series is infinite and convergent.

For example $$(1+x)^4=1+4x+6x^2+4x^3+x^4.$$ $$(1+x)^{4.3}=1+4.3x+(4.3*3.3)x^2/2+(4.3*3.3*3.2)x^3/6+........$$ $$(1+x)^{-4.3}=1-4.3x+(4.3*5.3)x^2/2-(4.3*5.3*6.3)x^3/6+........$$ $$(1-x)^{-1}=1+x+x^2+x^3+......., \text{if}~ |x|<1$$ $$(1+1/3)^{1/2}=1+\frac{1}{2}(1/3)-\frac{1.1}{2.2.2!}(1/3)^2+\frac{1.1.3}{2.2.2.3!}(1/3)^3+.....$$

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    $\begingroup$ What about positive reals ? $\endgroup$
    – user958916
    Sep 9, 2021 at 8:43
  • $\begingroup$ See my edir. For negative integers or non integral (positive /negatiove) values of $n$, we get an infinite series. which is valid if ${x}<1$. $\endgroup$
    – Z Ahmed
    Sep 9, 2021 at 9:36

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