Heuristics

Problem 1.5.5

Write an equation to represent the following statements:

a. At Mindy’s restaurant, for every four people who ordered cheesecake, there were five who ordered strudel.

b. There are six times as many students as professors at this college.

Solution

a. Let $c$ be the number of people ordering cheesecake. Let $s$ be the number of people ordering strudel. We have $\frac{c}{s} = \frac{4}{5}$.

b. Let $s$ be the number of students. Let $p$ be the number of professors. We have $s=6p$.

Problem 1.5.6

Guy wires are strung from the top of each of two poles to the base of the other.What is the height from the ground where the two wires cross?

Solution

Let $a$ and $b$ be the heights of the two poles.

Let $x$ be the horizontal distance from the base of the first pole till the projection of the point (where the two wires cross) on the ground.Let $y$ be this distance for the other pole.

Let $h$ be the height from the ground where the two wires cross. Using similar triangles, we have

$$ \begin{align*} \frac{h}{x} &= \frac{b}{x+y} \
\frac{h}{y} &= \frac{a}{x+y} \
\end{align*} $$

From the above we see that $h = \frac{ab}{a+b}$

Problem 1.5.7

A piece of paper $8$ inches wide is folded so that one corner is placed on the opposite side.Express the length of crease, $L$, in terms of the angle $\theta$ alone.

Solution

We have

$$ \begin{align*} Lcos\theta sin2\theta &= 8 \
\implies L &= \frac{4}{sin\theta cos^2\theta} \end{align*} $$

Problem 1.6.11

The product of four consecutive terms of an arithmetic progression of integers plus the fourth power of the common difference is always a perfect square.Verify this identity by incorporating symmetry into the notation.

Solution

Let $a-3d,a-d,a+d,a+3d$ be four consecutive terms of an arithmetic progression with common difference $2d$.

We have,

$$ \begin{align*} (a-3d)(a-d)(a+d)(a+3d) + 16d^4 \
= (a^2-9d^2)(a^2-d^2) + 16d^4 \
= a^4 - 10a^2d^2 + 25d^4 \
= (a^2 - 5d^2)^2 \end{align*} $$

Problem 1.10.6

a. Remove the lower left corner square and the upper right corner square from an ordinary $8-by-8$ chessboard.Can the resulting board be covered by $31$ dominos.

Solution

a. The two corner squares are of the same colour. After removing the two corner square we either have $30$ white squares or $30$ black squares. Each domino covers one white square and one black squares, so $31$ dominos cover $31$ white squares and $31$ black squares.Therefore, $31$ dominoes cannot cover the chessboard from which the corner squares have been removed.

Problem 1.10.8

Let $a_1,a_2, \dots, a_n$ represent an arbitrary arrangement of the numbers $1,2 \dots, n$. Prove that $n$ is odd, the product

$$ \begin{align*} (a_1-1)(a_2-1)\cdots(a_n-1) \end{align*} $$

is an even number.

Solution

As $n$ is odd, you have one additional $a_i$ which is odd compared to number of even $a_i$s. Therefore the above product has at least one even number.

Problem 1.10.10

Show that $x^2-y^2= a^3$ always has integral solutions for $x$ and $y$ whenever $a$ is a positive integer.

Solution

We have two cases:

1. a = 2k, where $k$ is an integer.

   From $x^2-y^2=8k^3$, We get the following equations

   $$ \begin{align*} x + y &= 4k^2 \
   x - y &= 2k \end{align*} $$

   we can see that

   $$ \begin{align*} x = k + 2k^2 \
   y = 2k^2 - k \end{align*} $$

   which are both integers.

2. a = 2k + 1, where $k$ is an integer

   From $x^2-y^2=(2k+1)^3$, we get the following equations

   $$ \begin{align*} x + y &= 4k^2 + 4k + 1 \
   x - y &= 2k + 1 \end{align*} $$

   we can see that

   $$ \begin{align*} x = 2k^2 + 3k + 1\
   y = 2k^2 + k \end{align*} $$

   which are both integers.

Problem 1.12.4

By setting $x$ equal to the appropriate values in the binomial expansion

$$ \begin{align*} (1+x)^n = \sum_{k=0}^n{n \choose k}x^k \end{align*} $$

(or one of its derivatives, etc.) evaluate each of the following:

a. $\sum_{k=1}^n k^2{n \choose k}$

b. $\sum_{k=1}^n 3^k{n \choose k}$

c. $\sum_{k=1}^n \frac{1}{k+1}{n \choose k}$

d. $\sum_{k=1}^n (2k+1){n \choose k}$

Solution

We start with the binomial expansion

a. Differentiating the binomial expansion once, we have

$$ \begin{align*} n(1+x)^{n-1} = \sum_{k=1}^n k{n \choose k}x^{k-1} \end{align*} $$

Multiplying the above by $x$ and differentiating again, we have

$$ \begin{align*} n(n-1)x(1+x)^{n-2} + n(1+x)^{n-1} = \sum_{k=1}^n k^2{n \choose k}x^{k-1} \end{align*} $$

Setting $x=1$ in the above equation we have

$$ \begin{align*} \sum_{k=1}^n k^2{n \choose k} = n(n-1)2^{n-2} + n2^{n-1} = n(n+1)2^{n-2} \end{align*} $$

b. Setting $x=3$ in the binomial expansion we have

$$ \begin{align*} \sum_{k=1}^n 3^k{n \choose k} = 4^n \end{align*} $$

c. Integrating the binomial expansion we have

$$ \begin{align*} \int_0^1 (1+x)^n dx = \frac{2^{n+1} - 1}{n+1} &= 1 + \sum_{k=1}^n \frac{1}{k+1}{n \choose k} \
\implies \sum_{k=1}^n \frac{1}{k+1}{n \choose k} &= \frac{2^{n+1} - n -2}{n+1} \end{align*} $$

d. We have

$$ \begin{align*} \sum_{k=1}^n (2k+1){n \choose k} = 2n2^{n-1} + 2^n - 1 = (n+1)2^n - 1 \end{align*} $$

Problem 1.12.5

Evaluate

$$ \begin{align*} det \begin{bmatrix} 1 & a & a^2 & a^4\
1 & b & b^2 & b^4\
1 & c & c^2 & c^4\
1 & d & d^2 & d^4 \end{bmatrix} \end{align*} $$

Solution

Replacing $d$ by $x$ in the last row, we get a polynomial $P(x)$.

$$ \begin{align*} P(x) = det \begin{bmatrix} 1 & a & a^2 & a^4\
1 & b & b^2 & b^4\
1 & c & c^2 & c^4\
1 & x & x^2 & x^4 \end{bmatrix} \end{align*} $$

$P(x)$ is a polynomial of degree $4$. Moreover, $P(a)=0$, $P(b)=0$ and $P(c)=0$, since the corresponding matrix, with $d$ replaced by $a$ or $b$ or $c$ respectively, then has two identical rows. Therefore

$$ \begin{align*} P(x) = A(x-a)(x-b)(x-c)(x-l) \end{align*} $$

where $l$ is the fourth root of $P(x)=0$.

The coefficient of $x^3$ in $P(x)=0$ is zero,so the sum of the roots $a+b+c+l=0$. Therefore $l=-(a+b+c)$. The coefficient of $x^4$ is

$$ \begin{align*} A = det \begin{bmatrix} 1 & a & a^2 \
1 & b & b^2 \
1 & c & c^2 \
\end{bmatrix} = (b-a)(c-a)(c-b) \end{align*} $$

Therefore, the value of the original determinant is given by

$$ \begin{align*} P(d) = (b-a)(c-a)(c-b)(d-a)(d-b)(d-c)(a+b+c+d) \end{align*} $$

Problem 1.12.6

a. Evaluate $\int_0^{\infty} (e^{-x}sin x)/x dx$.

b. Evaluate $\int_0^1 (x-1)/ln x dx$.

c. Evaluate

$$ \begin{align*} \int_0^{\infty} \frac{arctan(\pi x) - arctan(x)}{x} dx \end{align*} $$

Solution

We make use of differentiation under integral sign or parameter differentiation.

a. Using parameter differentiation,

$$ \begin{align*} G(k) &= \int_0^{\infty} (e^{-x}sin (kx))/x dx \
\implies \frac{dG(k)}{dk} &= \int_0^{\infty} e^{-x}cos kx dx \end{align*} $$

We have

$$ \begin{align*} \int_0^{\infty} e^{-x}cos (kx) dx &= -e^{-x}cos(kx) \bigg|_{0}^{\infty} - k\int_0^{\infty} e^{-x} sin (kx) dx \
&= 1 - k\left( -e^{-x}sin(kx) \bigg|_{0}^{\infty} \
+ k\int_0^{\infty} e^{-x} cos (kx) dx \right) \
&= 1- k^2\int_0^{\infty} e^{-x}cos (kx) dx \
\implies \int_0^{\infty} e^{-x}cos (kx) dx &= \frac{1}{k^2+1} \end{align*} $$

Hence,

$$ \begin{align*} \frac{dG(k)}{dk} = \int_0^{\infty} e^{-x}cos (kx) dx = \frac{1}{k^2+1} \end{align*} $$

The solution of the differential equation with $G(0)=0$ is $G(k) = arctan(k)$.

Therefore,

$$ \begin{align*} \int_0^{\infty} (e^{-x}sin x)/x dx = G(1) = arctan(1) = \frac{\pi}{4} \end{align*} $$

b. Using parameter differentiation,

$$ \begin{align*} H(m) &= \int_0^1 (x^m-1)/ln x dx \
\implies \frac{dH(m)}{dm} &= \int_0^1 x^m dx = \frac{1}{m+1} \end{align*} $$

The solution of the differential equation with $H(0)=0$ is $H(m) = ln(m+1)$.

Therefore,

$$ \begin{align*} \int_0^1 (x-1)/ln x dx = H(1) = ln(2) \end{align*} $$

c. Using parameter differentiation,

$$ \begin{align*} F(a) &=\int_0^{\infty} \frac{arctan(a x) - arctan(x)}{x} dx \
\implies \frac{dF(a)}{da} &= \int_0^{\infty} \frac{1}{a^2x^2+1} dx = \frac{\pi}{2a} \end{align*} $$

The solution of the differential equation with $F(1)=0$ is $F(a) = \frac{\pi}{2}ln(a)$. Therefore,

$$ \begin{align*} \int_0^{\infty} \frac{arctan(a x) - arctan(x)}{x} dx = F(\pi) = \frac{\pi}{2}ln(\pi) \end{align*} $$

Problem 1.12.7

Which is larger $\sqrt[3]{60}$ or $2+\sqrt[3]{7}$?

Solution

Let $x=a^3$ and $y=b^3$.

We have

$$ \begin{eqnarray} (\sqrt[3]{4(x + y)})^3 = 4(x+y) = 4(a^3+b^3) \label{ineq-1}\\
(\sqrt[3]{x} + \sqrt[3]{y})^3 = x + y + 3\sqrt[3]{xy}(\sqrt[3]{x} + \sqrt[3]{y}) \nonumber\\
= a^3 + b^3 + 3ab(a + b) \label{ineq-2} \end{eqnarray} $$

Subtracting \ref{ineq-2} from \ref{ineq-1}, we have

$$ \begin{align*} 4(a^3+b^3) - (a^3 + b^3 + 3a^2b + 3ab^2) \
= 3(a^3 + b^3 - a^2b - ab^2) \geq 0 \end{align*} $$

because of Muirhead’s inequality as $(3,0)$ majorizes $(2,1)$. Equality holds only when $a=b$.

Therefore,

$$ \begin{align*} \sqrt[3]{60} = \sqrt[3]{4(8 + 7)})^3 > \sqrt[3]{8} + \sqrt[3]{7} \end{align*} $$