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I want to confirm some points about binomial series.

  1. The expression must be of the form$(1+x)^n$.

  2. $x<1$

    The part where I am confused is that while finding derivatives through first principle , i've come through cases where binomial series is applied when expression is of the form $(a+x)^n$ and it isn't mentioned if $x$ < 1.

Community
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Diaga AoS
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  • Why do you need that $x<1$? I do not quite understand what the question is. – Cornman Aug 07 '17 at 16:31
  • Well i dont know that because textbook states that proof of binomial series is beyond my scope right now but it is written that $x < 1$. And the problem i am facing is that the conditions for binomial series are not fulfiled but another book applies binomial series to solve questions. Sry for my english. – Diaga AoS Aug 07 '17 at 16:38
  • Which proof of the binomial series do you mean? Can you quote the theorem you are struggeling with? – Cornman Aug 07 '17 at 16:41
  • I am quoting as it is written : Binomial Theorem when the index $n$ is a negative integer or a fraction , then $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 +... .$ provided absolute value of $x<1$ – Diaga AoS Aug 07 '17 at 16:45
  • The Binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) holds for any real numbers x,y with $(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^ky^{n-k}$. You do not need x<1 in particular. – Cornman Aug 07 '17 at 16:50
  • I mean this : https://en.m.wikipedia.org/wiki/Binomial_series – Diaga AoS Aug 07 '17 at 17:20

1 Answers1

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HINT: Note that $$(a+x)^n = a^n \left(1+\frac{x}{a}\right)^n.$$ If $|x|<|a|$ then the expansion of $(1+y)^n$ for $|y|<1$ can be applied.

Math Lover
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  • So I cannot expand without taking $a$ common? Like $(a+x)^n = a^n + na^{n-1} x + \frac{n(n-1)}{2!} a^{n-2} x^2+ ... .$ – Diaga AoS Aug 07 '17 at 17:24
  • @DiagaAoS You can certainly do so, but you should keep in mind the condition of convergence, i.e., $|x|<|a|$. Btw, the second term in the expansion is $n a^{n-1} x$. – Math Lover Aug 07 '17 at 17:28
  • Hm the only difference is convergence . Thanks that clears up the confusion. – Diaga AoS Aug 07 '17 at 17:32