# Validity of binomial expansion for any power

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 $$|x|<1$$?

• 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)... Sep 9, 2021 at 8:09
• I suggest studying (way beyond my understanding) the "real continuation" of a function: mathworld.wolfram.com/GammaFunction.html Sep 9, 2021 at 8:29
• @Kavi Rama Murthy. Why did you delete your answer ? Cheers :-) Sep 9, 2021 at 8:29
• @Jared If the right side is replaced by an infinite sum then OP's formula is valid for all real $n$ when $|x|<1$. Sep 9, 2021 at 8:29
• "terminates for a positive value of $n$": no, $n$ needs to be natural. In that case, any $x$ works.
– user958916
Sep 9, 2021 at 8:33

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).

• I feel like you're using integer $n$ here (a Lorentz series), but OP asked about real $n$. Sep 9, 2021 at 8:17
• @Jared No, I am not assuming that $n$ is an integer. I has to avoid $x <-1$ and I have corrected the answer. Sep 9, 2021 at 8:25
• @Jared: Laurent, not Lorentz !
– user958916
Sep 9, 2021 at 8:36
• @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. Sep 9, 2021 at 9:50
• @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 $r!$. Sep 11, 2021 at 8:44

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

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