Looking at the Whole

In one sentence

When you study a problem, do not stare only at one term or one condition. Step back and look at the problem as a whole: its overall structure, its overall sum, its overall symmetry. Things you cannot compute locally often cancel each other out in the whole.

The most common kinds of “whole”: pairing (add the first and the last), telescoping (neighbors cancel), substituting the whole block (treat an expression as a single letter), and constructing a function (both sides are the same function).

Example: f(x) = 4ˣ/(4ˣ + 2); find f(1/10) + f(2/10) + … + f(9/10)

Step 1 / 4

Term by term goes nowhere
$f (\frac{1}{10}) = \frac{4 ^{0.1}}{4 ^{0.1} + 2}$ … every term is irrational.

$f (\frac{1}{10}) \approx 0.3648, \dots$
See it whole: pair first and last
Count the pairs
Conclusion

Predict first

With the same $f (x) = \frac{4 ^{x}}{4 ^{x} + 2}$ , find $k = 1 \sum 2024 f (\frac{k}{2025})$ .

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

Stop 1 · Constructing a function (Vol. 2)

The equation looks like a cubic, and solving it head-on is hard. But the two sides are the same function evaluated at two points.

Example: solve (x + 1)³ + x³ + 2x + 1 = 0

Step 1 / 4

Regroup
Split $2 x + 1$ into $(x + 1) + x$ and put each piece with a cube.

$(x + 1)^{3} + (x + 1) + x^{3} + x = 0$
See one function
Use a global property of the function
Conclusion

Stop 2 · Roots of unity (Vol. 4)

You cannot compute $cos \frac{2 π}{7}$ on its own, but the seven 7th roots of unity added together give $0$ .

Example: find cos(2π/7) + cos(4π/7) + cos(6π/7)

Step 1 / 4

The whole: a sum of roots of unity
$ω = cos \frac{2 π}{7} + i sin \frac{2 π}{7}$ satisfies $ω^{7} = 1$ and $ω \neq = 1$ , so $1 + ω + \dots + ω^{6} = 0$ .

$k = 0 \sum 6 ω^{k} = \frac{ω ^{7} - 1}{ω - 1} = 0$
Take real parts
Equal in pairs
Conclusion

Stop 3 · Summing sequences (Vol. 5)

Telescoping: each term is a difference of two neighbors, so when you add everything up, everything in the middle cancels.

Each term

\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}

is a piece

[\frac{1}{k + 1}, \frac{1}{k}]

of the number line. Press “See it whole”: the pieces join end to end and cover

[\frac{1}{n + 1}, 1]

exactly.

So $k = 1 \sum n \frac{1}{k ( k + 1 )} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ . No single term shows you this result, but seen as a whole it is obvious.

Stop 4 · AM–GM and Cauchy (Vol. 6)

One problem, several solutions

Let $a, b, c > 0$ . Prove that $\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}$ (Nesbitt’s inequality).

Add 1 to each term, treat it as a whole

Add $1$ to each term: $\frac{a}{b + c} + 1 = \frac{a + b + c}{b + c}$ . With $s = a + b + c$ , the left side plus $3$ equals $s (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b})$ .

The three denominators add up to $2 s$ . By the Cauchy inequality (or the harmonic–arithmetic mean inequality), $[(b + c) + (c + a) + (a + b)] (\sum \frac{1}{b + c}) \geq 9$ , so the left side plus $3$ is $\geq \frac{9}{2}$ , that is, the left side is $\geq \frac{3}{2}$ .

CostOne small move makes all three numerators the same.

Attack each term separately

On its own, $\frac{a}{b + c}$ can be as close to $0$ as you like (when $a$ is very small), so there is no positive lower bound that works for every term.

Only when the three terms are taken together does it work: when one term is small, the other two must be large, and the whole has the lower bound $\frac{3}{2}$ .

CostYou try to bound each term on its own and find that no single term has a good lower bound: the local route is blocked.

Practice

Exercise 1Basic

Find $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{49 \times 50}$ .

Exercise 2Basic

Let $f (x) = \frac{9 ^{x}}{9 ^{x} + 3}$ . Find $f (\frac{1}{100}) + f (\frac{2}{100}) + \dots + f (\frac{99}{100})$ .

Exercise 3Advanced

Solve the equation $(2 x + 1)^{3} + x^{3} + 3 x + 1 = 0$ .

Exercise 4Challenge

Find the value of $cos \frac{π}{7} - cos \frac{2 π}{7} + cos \frac{3 π}{7}$ .

In one sentence

Example: f(x) = 4ˣ/(4ˣ + 2); find f(1/10) + f(2/10) + … + f(9/10)

Stop 1 · Constructing a function (Vol. 2)

Example: solve (x + 1)³ + x³ + 2x + 1 = 0

Stop 2 · Roots of unity (Vol. 4)

Example: find cos(2π/7) + cos(4π/7) + cos(6π/7)

Stop 3 · Summing sequences (Vol. 5)

Stop 4 · AM–GM and Cauchy (Vol. 6)

Add 1 to each term, treat it as a whole

Attack each term separately

Practice

Mistake Museum · Exhibits from this stop

Counting the middle term twice when pairing

Keeping the wrong ends after telescoping

Stops on this line