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A friend of mine posed a problem on a mathematics discord server.

The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.

I immediately thought of employing the binomial theorem, as what was required, actually. But, I decided to do another method, which led me to the question of how we can solve such problems without using the binomial theorem.

How can we solve this question without the binomial theorem or newton's method?

My Attempt:

Lemma: $$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ac)x+abc.$$

Note that $$(2+px)^6=(2+px)^2(2+px)^2(2+px)^2$$ $$=((px)^2+4px+4)((px)^2+4px+4)((px)^2+4px+4)$$ Therefore, in the lemma, substitute $x\mapsto (px)^2$ and $a,b,c\mapsto 4px+4$. It follows $$(2+px)^6=(px)^6+12(px+1)(px)^4+48(px+1)^2(px)^2+64(px+1)^3.$$ We can ignore the first two terms since they contain no strict $x^2$ coefficient. Thus, upon expanding $$48(px+1)^2(px)^2+64(px+1)^3$$ it follows the coefficient of $x^2$ is $240p^2$. $$240p^2=60\tag{$p>0$}$$ $\therefore p=\frac 12$.

Mr Pie
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    You write The coefficient of the expansion of $(2+px)^6$ is $60$. Do you mean The coefficient of the $x^2$ term in the expansion...? – Andrew Chin Jun 04 '20 at 04:01
  • @AndrewChin yes, that's what I meant. My apologies - that lack of specificity presumably stumped many users, lmao. – Mr Pie Jun 04 '20 at 09:23
  • You are basically explicitly expanding $(a+b)^6$, though you are doing it in two steps - first the cube $(a+b)^3$ and then substituting $(a^2, 2ab+b^2)$ for $(a,b)$ to square it. Of course you are also leaving out any terms you don't need along the way. – Jaap Scherphuis Jun 04 '20 at 09:46
  • @JaapScherphuis Eh, you are right, so imma just delete this question altogether. I mean, I don't suppose it contributes to anything, and perhaps I merely overreacted at the somewhat unorthodox method with which I approached this problem. – Mr Pie Jun 04 '20 at 10:16

2 Answers2

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This implicitly uses the Binomial Theorem (edit: no it does not), and calculus, but anyway...

Let $$p(x)=(2+px)^6=a_0+a_1x+60x^2+\mathcal{O}(x^3).$$

Differentiate twice:

$$\begin{align} p'(x)&=6(2+px)^5\cdot p=a_1+60\cdot 2x+\mathcal{O}(x^2) \\ \Rightarrow p''(x)&=6p\cdot 5\cdot(2+px)^4\cdot p=120+\mathcal{O}(x) \end{align}$$

Evaluate at $x=0$:

$$30p^2\cdot 2^4=120\Rightarrow p^2=\frac{1}{4}\Rightarrow p\underset{p>0}{=}+\sqrt{\frac14}=\frac12.$$

JP McCarthy
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  • That is very neat! (You may have made a typo at the last part, but I understand what you mean :P) It doesn't directly use the binomial theorem but I never thought about differentiating to reduce the degree of $\mathcal O$. Thank you! You answered just in time too $-$ I was ready to delete my question, lol. – Mr Pie Jun 04 '20 at 10:36
  • @MrPie it depends on how you prove that the derivative of $x^n$ is $nx^{n-1}$... I suppose it could be done with induction and so this is independent of the binomial theorem. – JP McCarthy Jun 04 '20 at 12:49
  • I'd just go with limits by first principles to prove that, upon differentiation, $x^n$ does indeed map to $nx^{n-1}$. – Mr Pie Jun 04 '20 at 23:11
  • @MrPie how do you do that not using induction nor the binomial theorem? – JP McCarthy Jun 05 '20 at 08:05
  • Well for me, it's intuitive since nearly all the terms just become $0$ given $h\to 0$. You're right, it does use the binomial theorem, but it doesn't need to be used explicitly - though that's more or less a personal thing. – Mr Pie Jun 06 '20 at 05:34
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    @mrpie I suppose you could argue that $(x+h)^n=x^n+nhx^{n-1}+\mathcal{O}(h^2)$ without the Binomial Theorem but I would just use induction. – JP McCarthy Jun 06 '20 at 07:48
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Here are three different methods. The first one is recommended, the others are just for fun and curiosity.

Combinatorial approach:

We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. Here we are looking for a positive solution $p$ of \begin{align*} 60=[x^2](2+px)^6\tag{1} \end{align*} Multiplying $(2+px)^6$ out we get a contribution to $x^2$ if and only if we select from two factors $2+px$ the term $px$ giving us $p^2x^2$ and taking from the other $4$ factors $2$ giving us $2^4=16$. Since we can select two factors from $(2+px)^6$ in $\binom{6}{2}$ ways, we obtain \begin{align*} \color{blue}{60}=[x^2](2+px)^6=\binom{6}{2}p^22^4=15p^2\cdot 16\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Algebraic approach:

A somewhat cumbersome but simple approach is to iteratively work through the linear factors $2+px$.

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=[x^2](2+px)(2+px)^5\\ &=\left(2[x^2]+p[x^1]\right)(2+px)(2+px)^4\tag{1}\\ &=\left(4[x^2]+4p[x^1]+p^2[x^0]\right)(2+px)(2+px)^3\tag{2}\\ &=\left(8[x^2]+12p[x^1]+6p^2[x^0]\right)(2+px)(2+px)^2\\ &=\left(16[x^2]+32p[x^1]+24p^2[x^0]\right)(2+px)(2+px)^1\\ &=\left(32[x^2]+80p[x^1]+80p^2[x^0]\right)(2+px)\\ &=160p^2+80p^2\tag{3}\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Comment:

  • In (1) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

  • In (2) and the following lines we determine the coefficients according to the rules from (1), i.e. \begin{align*} &\left(a[x^2]+b[x^1]+c[x^0]\right)(2+px)=2a[x^2]+(2b+ap)[x^1]+(2c+bp)[x^0] \end{align*}

  • In (3) we select the coefficients accordingly.

Complex analytic approach:

This variant shouldn't be taken too serious. It's just for fun and in fact based on the first method. We recall the residue theorem which tells us that integrating along a circle with radius one around the origin we have \begin{align*} [x^2](2+px)^6=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx \end{align*}

We obtain \begin{align*} \color{blue}{60}&=[x^2](2+px)^6\\ &=\frac{1}{2\pi i}\oint_{|x|=1}\frac{(2+px)^6}{x^3}\,dx\\ &=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{\left(2+pe^{it}\right)^6}{e^{3it}}\,ie^{it}\,dt\tag{2}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\left(2+pe^{it}\right)^6e^{-2it}\,dt\tag{3}\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}\binom{6}{2}2^4p^2\,dt\tag{4}\\ &=\frac{1}{2\pi}\binom{6}{2}2^4p^2\int_{0}^{2\pi}\,dt\\ &\,\,\color{blue}{=240p^2} \end{align*} from which $p=\frac{1}{2}$ follows.

Comment:

  • In (2) we use the substitution $x=e^{it}, dx=ie^{it}dt$.

  • In (3) we note $\int_{0}^{2\pi}e^{-kit}\,dt =0, k\in \mathbb{Z}$ which is due to Euler's identity $e^{2\pi i}=1$. So, everything vanishes in (3) besides the constant term.

  • In (4) we select the constant term in fact with the same considerations we used in the first approach.

Note: Another approach might be using shifted series multisection in order to filter the wanted coefficient of $x^2$.

Markus Scheuer
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