## Binomial expansion [2]

### Largest coefficient

B2, 2011

Show that the power of $$x$$ with the largest coefficient in the polynomial $$\displaystyle (1 + 2x/3)^{20}$$ is 8. In other words, if we write the given polynomial as $$\sum_i a_ix^i$$ then the largest coefficient $$a_i$$ is $$a_8$$.

Solution

The coefficient $$a_i = \binom{20}{i} \left(\frac{2}{3}\right)^i$$. Consider the ratio of two consecutive terms: $$a_{i+1}/a_i$$. \begin{align} \text{ratio } &= \frac{2}{3} \times \left( \frac{20!}{20-i-1!i+1!}/\frac{20!}{20-i!i!} \right) \\ &\\ &= \frac{2}{3} \cdot \frac{20-i!i!}{20-i-1!i+1!} = \frac{2(20-i)}{3(i+1)} \end{align} The ratio $$a_{i+1}/a_i > 1$$ up to $$i\leq 7$$ and strictly less than 1 for $$i>7$$. Hence, the sequence of coefficients is bitonic with the peak occurring at $$a_8$$.

### Coefficient ratio

A7, 2015

(i) By the binomial theorem $$(\sqrt{2}+1)^{10}=\sum_{i=0}^{10} C_{i}(\sqrt{2})^{i},$$ where $$C_{i}$$ are appropriate constants. Write the value of $$i$$ for which $$C_{i}(\sqrt{2})^{i}$$ is the largest among the 11 terms in this sum.

(ii) For every natural number $$n,$$ let $$(\sqrt{2}+1)^{n}=p_{n}+\sqrt{2} q_{n},$$ where $$p_{n}$$ and $$q_{n}$$ are integers. Calculate $$\lim _{n \rightarrow \infty}\left(\frac{p_{n}}{q_{n}}\right)^{10}$$.

Solution

(i) $$i=6$$. Consider the ratio:
$\frac{C_{i+1}(\sqrt{2})^{t+1}}{C_{i}(\sqrt{2})^{4}}$
This ratio is $$>1$$ till $$i=5$$ and $$<1$$ from $$i=6$$ onwards. Similar to problem B2, 2011.

(ii) 32. Using binomial expansion see that $$(\sqrt{2}-1)^{n}=\pm\left(p_{n}-\sqrt{2} q_{n}\right),$$ where the sign depends on the parity of $$n .$$ As $$n \rightarrow \infty,(\sqrt{2}-1)^{n} \rightarrow 0$$ since $$(\sqrt{2}-1)<1 .$$ Thus $$\left(p_{n}-\sqrt{2} q_{n}\right) \rightarrow 0$$ and so $$\frac{p_{n}}{q_{n}} \rightarrow \sqrt{2}$$

## Inequalities [4]

### AM-GM inequality

A4, 2011

Given positive real numbers $$a_1, a_2, \ldots , a_{2011}$$ whose product $$a_1 a_2 \cdots a_{2011}=1$$, what can you say about their sum $$S = a_1 + a_2 + · · · + a_{2011}$$?

• $$S\;$$ can be any positive number.
• $$1 \leq S \leq 2011$$.
• $$2011 \leq S \text{ and } S\;$$ is unbounded above.
• $$2011 \leq S \text{ and } S\;$$ is bounded above.
Solution

$$2011 \leq S \text{ and } S\;$$ is unbounded above.

The first inequality follows from AM-GM inequality. To see why $$S$$ is unbounded, set $$a_1=n$$, $$a_2=1/n$$ and the rest of $$a_is$$ to 1. The sum $$S>n$$ for any $$n$$.

### AM-GM inequality II

B5, 2012

Suppose $$x+x^{-1}=\frac{\sqrt{5}+1}{2}$$.

(a) For any real $$r,$$ show that $$\left|r+r^{-1}\right| \geq 2 .$$ What kind of a number is $$x$$?

Solution

Without loss of generality, we may assume that $$r > 0$$. Now use the AM-GM inequality:
$\frac{ r+\frac{1}{r} }{2} \geq \sqrt{ r \cdot \frac{1}{r} }$ since $$x+x^{-1}=\frac{\sqrt{5}+1}{2}<2$$, the number $$x$$ must be a non-real complex number.

### Combinatorial and calculus inequalities

B2b, 2021

Prove or disprove:
(i) $$2^{40} > 20!$$
(ii) $$1-1/x \leq \ln x \leq x-1$$ for all $$x>0$$

Solution (i) False.
\begin{align*} 20! &> 2\times3\times4\times5\times6\times 7\times 8\times8\times 8\ldots (\text{13 times}) \\ 20! &> 2\times3\times4\times5\times6\times 7\times (2^{3})^{13} \\ 20! &> 2^{40} \end{align*}
Comment: A similar problem was asked in mock test #6 (Problem B2).

(ii) True.
Let us prove the left hand side inequality. \begin{align*} f(x) &:= \ln x - 1 + 1/x\\ f^\prime(x) &= 1/x(1 - 1/x)\\ \end{align*} $$f^\prime < 0$$ for $$x\in(0,1)$$ and $$>0$$ for $$x\in(1,\infty)$$. This means $$f$$ is decreasing in the interval $$(0,1)$$ and increasing in the interval $$(1,\infty)$$. It reaches the minimum at $$x=1$$ which is zero. Hence $$f(x)\geq 0$$ for all $$x > 0$$.
The right hand side inequality can be proved similarly. \begin{align*} g(x) &:= x - 1 - \ln x\\ g^\prime(x) &= 1 - 1/x \end{align*} Again, $$g(x)$$ attains its minima of zero at $$x=1$$.

### Points on a sphere

B2, 2015

Let $$p, q$$ and $$r$$ be real numbers with $$p^{2}+q^{2}+r^{2}=1$$

(a) Prove the inequality $$3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3} \leq 2$$

(b) Also find the smallest possible value of $$3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3}$$.

Specify exactly when the smallest and the largest possible value is achieved.

Solution

We have \begin{align} 3p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3}&=(q+r)\left(3 p^{2}+2 q^{2}+2 r^{2}-2 q r\right)\\ &=(q+r)\left(3\left(p^{2}+q^{2}+r^{2}\right)-\left(q^{2}+r^{2}+2 q r\right)\right)\\ &=(q+r)\left(3-(q+r)^{2}\right)\\ &=x\left(3-x^{2}\right)=3 x-x^{3} \end{align} where $$x=q+r.$$

Let us examine possible values of $$x$$ in view of the constraint $$p^{2}+q^{2}+r^{2}=1$$.

We have $$2 q r \leq q^{2}+r^{2}$$ e.g. because $$(q-r)^{2} \geq 0 .$$

Adding $$q^{2}+r^{2},$$ we get $$q^{2}+r^{2}+2 q r \leq$$ $$2 q^{2}+2 r^{2} \leq 2,$$ because $$q^{2}+r^{2} \leq p^{2}+q^{2}+r^{2}=1$$. Thus $$(q+r)^{2} \leq 2$$. So $$-\sqrt{2} \leq q+r \leq \sqrt{2}$$

Note that equalities are achieved precisely when $$p=0$$ and $$q=r=\pm 1 / \sqrt{2}$$.

Thus altogether we have to find extrema of the odd function $$f(x)=3 x-x^{3}$$ over the interval $$[-\sqrt{2}, \sqrt{2}] .$$ The critical points are when $$f^{\prime}(x)=3-3 x^{2}=0,$$ i.e. $$x=\pm 1 .$$ Thus we need to see only $$f(\pm \sqrt{2})=\pm \sqrt{2}$$ and $$f(\pm 1)=\pm 2 .$$ Therefore $$-2 \leq 3 p^{2} q+3 p^{2} r+$$ $$2 q^{3}+2 r^{3} \leq 2 .$$ Moreover, $$3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3}=\pm 2$$ precisely when $$x=q+r=\pm 1$$

In each case, this gives a line segment in the $$q r$$-plane joining $$(\pm 1,0)$$ and $$(0,\pm 1) .$$ Note that both these segments lie within the circle $$q^{2}+r^{2}=1,$$ so each point on them leads to two valid points $$(p, q, r)$$ on the unit sphere.