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Consider all the coefficients in the expansion of $$(x^a+x^b)^n$$ where $a,b,n$ are non negative integers.

Claim: The $\color{Red}{\textit{exponent}}$ with the $\color{blue}{\textit {maximum coefficient value}}$ in the expansion is $\color{red}{\lfloor\frac{n(a+b)}2\rfloor}$
Example :
$(x^1+x^3)^4 = \sum\limits_{k=0}^4\binom{4}{k}x^{n-k}x^{3k} = x^{4} +4x^{6}+\color{blue}{6}x^{\color{red}{8}}+4x^{10}+x^{12}$

$\color{red}{\dfrac{4(1+3)}{2} = 8}$.

I feel this is a special case of central limit theorem in probability, but that theorem seems way more complicated to understand than my special case. So I'm trying to make sense of this result w/o using CLT. Any help ?

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    $\begingroup$ Well you need to add something here, because this is false with $a=1$, $b=2$, and $n=1$. $\endgroup$ – BlarglFlarg Dec 2 '18 at 17:46
  • $\begingroup$ Oh sorry when n(a+b)/2 is not an integer, both the ceil and floor will have the same coefficients.. I'll add this in the question. Ty @BlarglFlarg $\endgroup$ – rsadhvika Dec 2 '18 at 17:49
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    $\begingroup$ It is just related to the binomial coefficients characteristics, nothing to do with a and b. Look at how they are calculated with Pascal' triangle $\endgroup$ – Damien Dec 2 '18 at 17:49
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    $\begingroup$ I know that the beauty of mathematics partially relies on the fact that it can connect seemingly unrelated topics but... how do you see a connection to CLT in this problem? $\endgroup$ – Taladris Dec 3 '18 at 0:31
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    $\begingroup$ @Taladris From what I understand it is (a bit) connected because $\left(\frac{x^a+x^b}{2}\right)^n$ is the generating function of some sort of binomial distribution. (?) $\endgroup$ – Idéophage Dec 3 '18 at 4:47
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If you know the binomial expansion (which you seem to), then this follows almost immediately. The key point to remember is that the maximum value of the coefficient is obtained at $n\choose k$, where $k$ is the largest integer less than or equal to $\frac{n}{2}$. To prove this, consider the ratios $\frac{n \choose k+1}{n\choose k}$ and see for what range this is greater than $1$. Plug in your value of $k$ and you get your result.

PS: Note that your exponents must actually appear in the expansion for this to happen.

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  • $\begingroup$ Ahh I kindof see now how this can be related to the max coefficient in binomial expansion. Ty :) Pretty sure you meant this : $$$$ To prove this, consider the ratios $\frac{n \choose k+1}{n\choose k}$ and see for what range this is $\color{red}{greater ~than~ 1}$. $\endgroup$ – rsadhvika Dec 2 '18 at 18:01
  • $\begingroup$ Ah yes you’re right, I meant that the binomial coefficient itself is increasing, not the ratio. I’ll correct that. $\endgroup$ – Boshu Dec 2 '18 at 18:05
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$(x^a+x^b)^n=x^{na}(1+x^c)^n=\sum_{k=0}^n{n \choose k}x^{na+k(b-a)}$. Now, ${n \choose k}$ is maximum for $k=[n/2].$ So, answer is $na+[n/2](b-a)$ which matches with your answer if $n$ is even.

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  • $\begingroup$ $c = b-a$; I like your approach more. It is neat and to the point, but I started working on the other answer first so had to mark it best as it helped me first. Thank you so much John_Wick you're awesome :) $\endgroup$ – rsadhvika Dec 2 '18 at 21:46
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The exponents of the development run from $na$ to $nb$, in increments $b-a$. By symmetry, the requested exponent can only be

$$\frac{n(a+b)}{2}$$ when the numerator is even, and

$$\frac{n(a+b)\pm(b-a)}{2}$$

otherwise (there will be two consecutive terms with equal coefficient).

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