Assigning Values

In one sentence

Assigning values means giving well-chosen numbers to objects that had nothing to do with numbers (or to the variables of an identity), so that the problem becomes a piece of arithmetic:

cell \mapsto \pm 1, x \mapsto 0, \pm 1, ω, n \mapsto the exponent of the prime p in n

Once the values are assigned, the sign of a product, the parity of a sum, a sum of coefficients and the exponent of a prime factor can all be worked out by the rules of arithmetic. A good assignment turns the quantity you care about into one you can compute.

Predict first

Fill each cell of a 5×5 table with $+ 1$ or $- 1$ . Compute the product of each row and of each column: $10$ numbers in all. Can these $10$ numbers add up to $0$ ?

Yes, for example with a few well-placed

- 1

sNoIt depends on how many

- 1

s there are

Choose an answer first; then you can see the result.

Example: for odd n, the sum of the 2n products is not 0

Step 1 / 4

If the sum were 0
Ten numbers $\pm 1$ add up to $0$ exactly when there are $5$ of $- 1$ and $5$ of $+ 1$ .

$r_{1} + \dots + r_{5} + c_{1} + \dots + c_{5} = 0 ⟺ exactly 5 of them are - 1$
Use another operation: multiply
The number of −1s is even
Contradiction

Stop 1 · Plugging values into functional equations (Vol. 2)

When an equation holds for all $x, y$ , you may plug in whatever values are most convenient: first $x = y = 0$ , then $y = 0$ .

Example: find all f: R → R with f(x)f(y) − f(xy) = x + y for all real x, y

Step 1 / 4

Let x = y = 0
$f (0)^{2} - f (0) = 0$ , so $f (0) = 0$ or $f (0) = 1$ .

$f (0) (f (0) - 1) = 0$
Let y = 0
A candidate
Substitute back

Stop 2 · Roots of unity: “sifting out” coefficients (Vol. 4)

$x = 1$ adds up all the coefficients, and $x = - 1$ separates the odd and even terms. To separate by the index mod $3$ , let $x = ω$ ( $ω^{3} = 1, ω \neq = 1$ ) and use the fact that $1 + ω^{k} + ω^{2 k}$ is $3$ when $3 ∣ k$ and $0$ otherwise.

One problem, several solutions

Find $(0 9) + (3 9) + (6 9) + (9 9)$ .

Compute directly

$1 + 84 + 84 + 1 = 170$ .

CostFastest when there are few terms, but hopeless when there are many.

Plug in roots of unity

In $(1 + x)^{9} = \sum (k 9) x^{k}$ , let $x = 1, ω, ω^{2}$ in turn and add. All the terms whose index is not divisible by $3$ cancel:

$3 [(0 9) + (3 9) + (6 9) + (9 9)] = 2^{9} + (1 + ω)^{9} + (1 + ω^{2})^{9} .$

$1 + ω = - ω^{2}$ , so $(1 + ω)^{9} = - ω^{18} = - 1$ ; likewise $(1 + ω^{2})^{9} = - 1$ . So the sum is $\frac{512 - 2}{3} = 170$ .

CostA little more complex arithmetic, in exchange for a formula that works for any n.

Stop 3 · Special values in an identity (Vol. 5)

Example: (2x − 1)⁶ = a₀ + a₁x + … + a₆x⁶. Find a₀ + a₂ + a₄ + a₆

Step 1 / 4

Let x = 1
The sum of all the coefficients.

$a_{0} + a_{1} + a_{2} + \dots + a_{6} = (2 - 1)^{6} = 1$
Let x = −1
Add
Bonus: let x = 0

Stop 4 · Partitions of a set and exponents of primes (Vol. 1)

Give each positive integer the value $v_{p} (n)$ , “the exponent of the prime $p$ in $n$ ”. The exponent of a product is the sum of the exponents: $v_{p} (ab) = v_{p} (a) + v_{p} (b)$ .

Example: can {1, 2, …, 10} be split into two groups with equal products?

Step 1 / 3

Assign values
Give each number its $v_{7}$ (the exponent of the factor $7$ ). Among $1$ to $10$ , only $7$ has a factor $7$ .

$v_{7} (7) = 1, v_{7} (k) = 0 (k \neq = 7)$
Equal products ⇒ equal exponents
A parity contradiction

Practice

Exercise 1Basic

In a 3×3 table, one cell holds $- 1$ and all the others hold $+ 1$ . What is the sum of the three row products and the three column products?

Exercise 2Basic

$(1 + x + x^{2})^{5} = a_{0} + a_{1} x + \dots + a_{10} x^{10}$ . Find $a_{0} + a_{2} + a_{4} + \dots + a_{10}$ .

Exercise 3Advanced

Plug in $x = 1, i, - 1, - i$ to find $(0 8) + (4 8) + (8 8)$ .

Exercise 4Challenge

Find $(1 10) + (4 10) + (7 10) + (10 10)$ (the sum of the binomial coefficients whose lower index is $1$ mod $3$ ).

In one sentence

Example: for odd n, the sum of the 2n products is not 0

Stop 1 · Plugging values into functional equations (Vol. 2)

Example: find all f: R → R with f(x)f(y) − f(xy) = x + y for all real x, y

Stop 2 · Roots of unity: “sifting out” coefficients (Vol. 4)

Compute directly

Plug in roots of unity

Stop 3 · Special values in an identity (Vol. 5)

Example: (2x − 1)⁶ = a₀ + a₁x + … + a₆x⁶. Find a₀ + a₂ + a₄ + a₆

Stop 4 · Partitions of a set and exponents of primes (Vol. 1)

Example: can {1, 2, …, 10} be split into two groups with equal products?

Practice

Mistake Museum · Exhibits from this stop

Substituting only x = 1

Plugging in values gives only a necessary condition

Stops on this line