Mathematics Asked on December 8, 2020
We know the binomial expansion of $(1+x)^n$ when $n$ is a fractional index or negative index. Why can we not expand $(a+b)^n$ directly when $n$ is a fractional or negative index? Instead of expanding directly we first take ‘$a$‘ common and write it as $a^n(1+b/a)^n$ and then we expand $(1+b/a)^n$ and multiply with $a^n$. It is written in my textbook that when $n$ is a fractional or negative index ‘$a$‘ must be equal to $1$ in $(a+b)^n$. But the result is same if we expand $(a+b)^n$ directly when $n$ is a fractional or negative index or first expand $(1+b/a)^n$ and then multiply $a^n$. I expanded $(a+x)^n$ by Taylor theorem also, the result is same. Then why can we not use binomial theorem to expand $(a+b)^n$ directly when $n$ is a fractional or negative index?
You can, in a way. The generalized binomial theorem affords a definition of $binom{n}{k}$, for $ninBbb C$ and integer $kge0$, such that$$(1+b/a)^n=sum_{kge0}binom{n}{k}(b/a)^k,$$or equivalently$$(a+b)^n=sum_kbinom{n}{k}a^{n-k}b^k,$$provided $|a|>|b|$. Note this modulus requirement prevents us exchanging $a,,b$ on the RHS, even though the LHS is symmetric. (Another issue with exchanging the exponents is that $binom{n}{k},,binom{n}{n-k}$ are in general no longer both defined, let alone equal, unless e make sure to write the definition of binomial coefficients in terms of Gamma functions rather than factorials & Pochhammer symbols.) Note also that our summing over all non-negative integers $k$ also holds when $n$ is a non-negative integer, because in that case any $k>n$ yields $binom{n}{k}=0$. This case also lets us drop the constraint $|a|>|b|$ altogether, so its presence when $n$ isn't a non-negative integer is very important.
Answered by J.G. on December 8, 2020
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP