Study Companion: Programme F.7 — Binomials

You already trust two rules half-consciously: that (a+b)^2=a^2+2ab+b^2 has a "2" in the middle, and that choosing 3 things from 8 is somehow different from arranging them. Both come from the same forced idea — when a choice is made repeatedly, the number of ways to make it multiplies, and when some of those ways look identical afterward, you have overcounted and must divide. Every rule in this programme, from n! to the binomial coefficients to the number e, is that one idea applied to a different picture.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in 5 units instead, because the programme's many topics are really 5 ideas.

How to use this companion

Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

Unit 01: Counting Arrangements Forces Factorials

The problem

A production line has 5 assembly steps — drill, bond, cure, inspect, pack — that must be run in some order, one after another, with no step repeated. How many different orders are there?

First attempt

Just list them: drill-bond-cure-inspect-pack, drill-bond-cure-pack-inspect, ... By the time the list has a few dozen entries it is clear this will not finish by hand, and it gives no way to answer the same question for 8 steps or 12.

The picture

Draw the choice as a tree instead of a list. The first step can be any of the 5 tasks: 5 branches. Whichever task went first, the second step is any of the remaining 4 tasks: from each of the 5 branches, 4 more branches. From each of those, the third step has 3 choices. Only the first three levels are drawn here, and only one branch is followed out at each level:

Notice what happens to the branches. Under "drill", the second level offers bond, cure, inspect and pack, but never drill again: a task, once used, is gone. Every branch at a level splits into the same number of sub-branches, so the first three levels already hold 5\times4\times3=60 paths. The fourth level splits each of those in 2, and the fifth step is forced. Each full path through the tree is one ordering, and the number of paths is the product of the branch counts at each level — the multiplication principle.

Only the branch count at each level matters for the total, so the tree compresses to a row of five slots, one per step, with the number of choices written over each:

Check it with numbers

For 9 steps the same tree gives 9\times8\times\cdots\times1=362\,880, and for 11 steps 11\times10\times\cdots\times1=39\,916\,800 — both easy on a calculator once the pattern is named.

The rule it gives

This repeated product is called n-factorial:

Peeling off just the first factor gives a recursive form that is more useful than it looks:

Read this at n=1: 1!=1\times0!. Since 1!=1, this forces

not as a convention chosen for convenience, but as the one value consistent with the recursion. (The book states 0!=1 without this argument — see the reference sheet — but the recursion is why it has to be true.)

The recursive form also settles ratios of factorials by cancelling the common tail: for n=4,

Caution. 2n! and (2n)! are not the same object: 2n! means 2\times(n!), doubling the factorial, while (2n)! is the factorial of the doubled number, a much larger product. At n=3:

Worked example

Simplify \frac{(2n+1)!}{(2n-1)!}. Peel the top factorial back two steps:

so the (2n-1)! cancels and

Where this shows up

A test rig runs 8 distinct calibration checks in sequence, and an engineer suspects the order matters (a warm-up effect, say). Testing every order means 8!=40\,320 runs. At 15 minutes per run of all 8 checks, that is 40\,320\times15=604\,800 minutes, which is 10\,080 hours, or 420 days of continuous rig time. With 5 checks it would be 5!\times15=1800 minutes, or 30 hours: a few shifts. The factorial is why exhaustive order testing is a sound plan at 5 steps and an impossible one at 8, and why test plans past a handful of steps fix one order, or sample a few, instead of trying them all.

Narration spine. Five empty slots stand for the five assembly steps, and the choice count drops in above each in turn — 5, 4, 3, 2, 1, one slot per level of the tree. The counts gather into the product 5\times4\times3\times2\times1=120, which is then named n!. The recursion n!=n\times(n-1)! is read at n=1 to show that 0!=1 is forced, not chosen. A ratio of factorials cancels at n=4 to 20, and the trap 2n! against (2n)! is struck out at n=3: 12 against 720.

Unit 02: Choosing Without Order Means Dividing Out the Repeats

The problem

A control panel has 8 numbered mounting holes, and a redundancy design calls for exactly 3 identical backup sensors to be installed, at most one per hole. The sensors carry no numbers, and no hole is preferred. How many different installations are there?

First attempt

Reuse Unit 01's tree: place a first sensor in any of 8 holes, a second in any of the remaining 7, a third in any of the remaining 6:

But the sensors are identical. The tree counted which hole first, which hole second, which hole third as different outcomes even when the same three holes end up filled — so 336 overcounts the real answer.

The picture

Draw the board as holes, filled (●) or empty (○). Two of the tree's paths — "hole 2 then hole 5 then hole 7" and "hole 5 then hole 2 then hole 7" — draw the identical picture ○●○○●○●○. In fact every one of the 3!=6 orders in which those same three holes could have been filled draws that same picture, because the sensors carry no label to tell the orders apart.

So the tree's count of 336 is exactly 3!=6 times too large: every genuine installation was drawn 6 times over, once per ordering of the 3 identical items. Divide it out.

Check it with numbers

Writing 8\times7\times6 as \frac{8!}{5!} gives the general form \frac{n!}{(n-r)!\,r!}, and here

A useful cross-check: choosing which 3 holes get a sensor is the same information as choosing which 5 holes are left empty, so {}^{8}C_{3} must equal {}^{8}C_{5} — and indeed \frac{8!}{3!\,5!}=\frac{8!}{5!\,3!}=56 too.

The rule it gives

The combinatorial coefficient — the book's name for the count of ways to place r identical items in n distinct locations — is

(also written \binom nr). Three properties follow, each provable two ways.

1. {}^{n}C_{n}={}^{n}C_{0}=1: there is exactly one way to fill every location and exactly one way to fill none. Algebraically, {}^{n}C_{n}=\frac{n!}{0!\,n!}=1 and {}^{n}C_{0}=\frac{n!}{n!\,0!}=1, both using 0!=1 from Unit 01.
2. {}^{n}C_{n-r}={}^{n}C_{r}: choosing the r filled locations is the same act as choosing the n-r empty ones (the cross-check above). Algebraically, put n-r in place of r in the formula: {}^{n}C_{n-r}=\frac{n!}{\big(n-(n-r)\big)!\,(n-r)!}=\frac{n!}{r!\,(n-r)!}={}^{n}C_{r}, the same two factorials in the denominator, in the other order.
3. Pascal's rule: {}^{n}C_{r}+{}^{n}C_{r+1}={}^{n+1}C_{r+1}. Counting proof: to place r+1 identical items among n+1 locations, single out one particular location. Either it is left empty — then all r+1 items go in the remaining n locations, {}^{n}C_{r+1} ways — or it is filled — then the other r items go in the remaining n locations, {}^{n}C_{r} ways. These cases don't overlap and cover every possibility, so they add. Algebraic proof: factor \frac{n!}{(n-r-1)!\,r!} out of both terms, ${}^{n}C_{r}+{}^{n}C_{r+1}=\frac{n!}{(n-r-1)!\,r!}\left(\frac{1}{n-r}+\frac{1}{r+1}\right)=\frac{n!\,(n+1)}{(n-r)!\,(r+1)!}={}^{n+1}C_{r+1}.Both proofs describe the same fact; the counting proof says *why*, the algebraic proof confirms it symbolically. Atn=5,r=2:{}^{5}C_{2}+{}^{5}C_{3}=10+10=20={}^{6}C_{3}$.

Pascal's rule is exactly what builds Pascal's triangle, the table of {}^{n}C_{r} with row n and column r both numbered from 0. Drawn centred, as below, each entry is the sum of the two entries diagonally above it: {}^{n+1}C_{r+1} sits under {}^{n}C_{r} (up and to the left) and {}^{n}C_{r+1} (up and to the right). Every edge is 1 (property 1). The book draws the same numbers left-justified, one column per value of r; in that layout the two parents are the entry directly above and the one above and to the left.

Add along each row: 1,\ 2,\ 4,\ 8,\ 16,\ 32. Every row sum is double the one before, so row n sums to 2^n. Pascal's rule already says why: building row n+1 uses each entry of row n exactly twice, once for each of the two entries below it, so the total doubles. Unit 03 finds the same total a second way.

Worked example

9 identical umbrellas are to be issued from a rack of 15 numbered hooks. The number of ways is

Where this shows up

A reliability engineer must pick 3 of 8 candidate sensor models for a redundancy group, and qualify each candidate trio on a vibration rig. Marking 3 of the 8 models "selected" is placing 3 identical ticks in 8 distinct boxes, so there are {}^{8}C_{3}=56 trios. At 2 hours of rig time per trio that is 56\times2=112 hours, or 14 eight-hour shifts. Budgeting from the ordered count instead, 336\times2=672 hours or 84 shifts, would book six times the rig time needed, to test the same 56 trios over and over.

Narration spine. Eight holes appear, and three identical sensors are placed one at a time: 8, 7, then 6 choices, 336 ordered placements. The board those placements draw is shown, with the note that all 6 orderings of the same three holes draw it, and 336 is divided by 3!=6 to give 56, which becomes the boxed {}^{n}C_{r}=\frac{n!}{(n-r)!\,r!}. The symmetry {}^{8}C_{3}={}^{8}C_{5}=56 follows. Pascal's rule is argued by singling out one hole, excluded or included, and checked as {}^{5}C_{2}+{}^{5}C_{3}=10+10=20={}^{6}C_{3}. Pascal's triangle is then built row by row, each entry arriving as the sum of its two parents.

Unit 03: The Binomial Theorem Is a Counting Argument

The problem

Multiplying out (a+b)^2=a^2+2ab+b^2 is routine. Multiplying out (a+b)^6 by hand, bracket by bracket, is not — yet its answer has a completely predictable shape. Where do the coefficients 1,6,15,20,15,6,1 come from without grinding through the algebra?

First attempt

Multiply (a+b)^3=a^3+3a^2b+3ab^2+b^3 by (a+b) one term at a time to reach (a+b)^4, and it works — (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4 — but this only produces the next row from the previous one. It does not explain why the coefficients are what they are, and repeating it six times to reach (a+b)^6 is exactly the tedium the problem is trying to escape.

The picture

Write (a+b)^n as n separate brackets multiplied together:

Expanding this product means picking, from each bracket, either its a or its b, and multiplying the n picks together. A term with a^{n-r}b^r comes from picking b out of exactly r of the n brackets (and a from the rest) — and the number of ways to choose which r brackets contribute a b is exactly {}^{n}C_{r}. It is the same counting question as Unit 02's sensors in holes: the n brackets are the numbered holes, and each b is an identical sensor.

The square and the cube show this happening. A square of side a+b, cut at distance a along each edge, falls into four pieces:

Each piece is one pick from the "width" bracket times one pick from the "height" bracket. The two ab rectangles are the {}^{2}C_{1}=2 ways to take b from exactly one of the two brackets, so (a+b)^2=a^2+2ab+b^2.

For (a+b)^3, cut a cube of side a+b at distance a along all three edges, and slice it horizontally into a bottom layer of thickness a and a top layer of thickness b. Looking down on each layer:

There are eight pieces: one main cube a^3; three slabs a^2b, each with two sides a and thickness b; three rods ab^2, each of length a and cross-section b\times b; and one corner cube b^3. Each piece is a pick of a or b along each of the three edge directions — width, depth and height are the three brackets — so the number of pieces of each kind is the number of ways to choose which directions contribute a b:

So (a+b)^3=a^3+3a^2b+3ab^2+b^3, and the coefficients 1,3,3,1 are {}^{3}C_{0},{}^{3}C_{1},{}^{3}C_{2},{}^{3}C_{3}: row 3 of Pascal's triangle.

Multiplying by one more (a+b) to reach (a+b)^4 takes every a^{n-r}b^r term and splits it into an a^{n-r+1}b^r part and an a^{n-r}b^{r+1} part — which is precisely Pascal's rule combining two neighbouring coefficients. For example, the a^2b^2 term of (a+b)^4 collects 3a^2b\times b and 3ab^2\times a, so its coefficient is {}^{3}C_{1}+{}^{3}C_{2}=3+3=6={}^{4}C_{2}. That is why the "multiply by (a+b)" method of the first attempt always reproduces Pascal's triangle: it is Pascal's rule in disguise.

Check it with numbers

Let a=2,b=1, so (a+b)^4=3^4=81 directly. Term by term, using row 4 of Pascal's triangle (1,4,6,4,1):

Setting a=b=1 in the general expansion gives every term equal to {}^{n}C_{r}, so the sum of row n is (1+1)^n=2^n — the doubling noticed in Unit 02, now seen a second way: row n counts the ways of placing any number of identical items in n locations, and each location is independently filled or empty, 2\times2\times\cdots\times2=2^n ways.

The rule it gives

Any single term without expanding the rest — the (r+1)th term, since r starts at 0 — is the general term:

A negative or fractional second term is handled by substitution, not a new rule: write (3-2x)^4 as \left(3+[-2x]\right)^4 and use b=-2x throughout, so every b^r picks up the sign (-1)^r automatically: (3-2x)^4=81-216x+216x^2-96x^3+16x^4, the odd powers of x negative. The same substitution handles descending powers of x, where b is a fraction such as -3/x.

Worked example

Find the 8th term of \left(2-\frac x3\right)^{12}. The 8th term has r=7 (since r starts at 0), with a=2, b=-x/3, n=12:

Where this shows up

A steel cube is to be machined to side L=20 mm, but it comes off the machine \delta=0.1 mm oversize on every edge. How much extra material is there? The cube dissection answers it directly, with a=L and b=\delta:

The three slabs, each 20\times20\times0.1=40 mm^3, carry 120 mm^3; the three rods, each 20\times0.1\times0.1=0.2 mm^3, carry 0.6 mm^3; the corner cube is 0.001 mm^3. Keeping only the slab term 3L^2\delta is off by 0.601 mm^3, about 0.5\% of the answer, which is why an inspector estimates extra volume as 3L^2\delta and ignores the rest: a 0.5\% oversize on each edge (0.1/20) makes about a 1.5\% oversize in volume, and the 3 is {}^{3}C_{1}, the number of slabs. F12 Unit 02 uses exactly this expansion, with the rods and the corner shrinking away, to find the power rule.

Narration spine. A square of side a+b splits into four regions, read straight off as a^2+2ab+b^2. Four brackets (a+b)(a+b)(a+b)(a+b) appear, and choosing b from 2 of them is shown to be one of {}^{4}C_{2}=6 ways; row 4 of Pascal's triangle agrees, and a=2, b=1 checks it as 16+32+24+8+1=81=3^4. The general expansion and the general term are boxed, a negative second term is handled by substitution in (3-2x)^4, and the 8th term of \left(2-\frac x3\right)^{12} is worked. Last, multiplying by one more bracket merges two neighbours, {}^{3}C_{1}+{}^{3}C_{2}={}^{4}C_{2}: Pascal's rule again, the same fact as the counting argument.

Unit 04: Sigma Notation Is Shorthand for a Pattern, Not a New Idea

The problem

A batch counter starts at 3 and rises by 3 with every unit produced: 3, 6, 9, ... What is the 20th reading, and what is the running total after 20 batches?

First attempt

Write out all 20 terms and add: 3+6+9+\cdots+60. It gives the right answer eventually, but a formula with n left as a symbol — "the sum after n batches" — cannot be reached by listing.

The picture

Stack two copies of the staircase 1+2+\cdots+n, one upside down against the other. Every column, top plus bottom, has height n+1, and there are n columns:

n columns, each adding to n+1.

Two copies of the staircase make an n\times(n+1) rectangle, so 2(1+2+\cdots+n)=n(n+1).

Check it with numbers

The same two staircases at n=20 make a 20\times21 rectangle of 420 unit squares, so 1+2+\cdots+20=\frac{420}{2}=210. Every reading of the counter is 3 times its batch number, so the 20 readings are 3 times the 20 numbers 1,2,\ldots,20:

The 20th reading is 3\times20=60; the running total after 20 batches is 630.

The rule it gives

is shorthand for a sum whose typical term is f(r), the counting number r running in integer steps between the values below and above \Sigma. Two manipulation rules follow directly from ordinary addition — nothing new is being assumed. A factor common to every term can be taken outside, as the 3 was above; and a sum of two-part terms can be regrouped into two sums:

The staircase argument gives the one closed form the programme relies on:

Together, the two rules and \sum r sum any general term of the form "multiple of r plus a constant". For \sum_{r=1}^{n}(6r+5), Rule 2 splits it into \sum6r+\sum5, and Rule 1 takes the 6 outside and turns \sum5 into 5n:

At n=2 this reads 2(14)=28, matching 11+17=28 term by term.

General terms are built the same way whatever the pattern: an even sequence is 2r, an odd one is 2r-1 or 2r+1 depending where it starts, and alternating signs come from (-1)^r (positive on even r) or (-1)^{r+1} (positive on odd r). Counting can also start at r=0, in which case the nth term sits at r=n-1.

Beyond the book: sums of squares and cubes. These are not part of this programme, and the staircase does not extend to them, but a different trick does: telescoping. Expand (r+1)^3-r^3=3r^2+3r+1 and add this identity for r=1,2,\ldots,n. On the left, almost everything cancels:

Each middle cube appears once with a plus sign and once with a minus sign. On the right, the total is 3\sum r^2+3\sum r+n. Setting the two totals equal and using \sum r=\frac{n(n+1)}2,

because (n+1)^3-1-n=(n+1)\big((n+1)^2-1\big)=n(n+1)(n+2), and n+2-\frac32=\frac{2n+1}2. Dividing by 3:

The same move one power up, (r+1)^4-r^4=4r^3+6r^2+4r+1, gives (n+1)^4-1=4\sum r^3+6\sum r^2+4\sum r+n. Substituting the two sums already found, 6\sum r^2=n(n+1)(2n+1) and 4\sum r=2n(n+1), and taking out the common factor n+1:

so

The cube sum is the square of the plain sum for every n, because the two sides are the same expression. At n=5: \sum r=15, \sum r^2=55=\frac{5\cdot6\cdot11}{6}, and \sum r^3=225=15^2 — a check that confirms the algebra, not a substitute for it.

Worked example

Find the general term and the sum of the first 10 terms of -1+2-3+4-\cdots. The size at position r is r; the sign alternates, negative on odd r, so the general term is (-1)^r r. Summed in pairs, (-1+2)+(-3+4)+\cdots+(-9+10)=5\times1=5.

Where this shows up

A solid concrete stair of 8 steps is cast against a loading dock. Each riser is 0.15 m, each tread 0.3 m, and the stair is 1.2 m wide. Seen from the side it is exactly the staircase of the picture: the column under step r is r risers tall, so its volume is (0.3\ \text{m})(1.2\ \text{m})(0.15r\ \text{m})=0.054r\ \text{m}^3, and the whole stair is

The tempting shortcut treats the side view as a triangle, half of 2.4 m (8 treads) by 1.2 m (8 risers), times the width: \frac12\times2.4\times1.2\times1.2=1.728 m^3. That is short by 0.216 m^3 — one ninth of the pour — because each step sticks out above the triangle's slope by half a block, and 8 half-blocks are 4 blocks, 4\times0.054=0.216 m^3. It is the difference between \frac{n^2}2 and \frac{n(n+1)}2, and it is the difference between ordering enough concrete and not. At 2400 kg/m^3 the stair weighs 1.944\times2400=4665.6 kg, about 4.7 t, the load the slab beneath it must carry.

Narration spine. The batch counter's readings 3, 6, 9, \ldots, 60 are written, and listing all twenty is shown to give no formula. A staircase of bars of heights 1 to 6 is built, a flipped copy is stacked on it, and a brace marks the n\times(n+1) rectangle the two make, from which \sum r=\frac{n(n+1)}2 is read off. At n=20 that is 210, so the batch total is 3\times210=630, and the 20th reading is 60. Rule 1 and Rule 2 are stated and applied together to \sum(6r+5)=n(3n+8), checked at n=2. Last, the beyond-the-book sums of squares and cubes are shown with their n=5 check.

Unit 05: Where the Number e Comes From

The problem

A quantity that compounds — interest credited more and more often within the same year, or a capacitor's charge updated in ever-smaller time slices — seems like it should grow without bound as the compounding gets more frequent. Does it?

First attempt

Compute \left(1+\frac1n\right)^n for a few values of n and watch:

It is rising, but far more slowly than "without bound" would suggest — each jump in n buys a smaller gain than the last. But a slow rise is not a ceiling: nothing in four numbers rules out a total that keeps creeping up forever. That needs an argument.

The picture

Expand \left(1+\frac1n\right)^n with the binomial theorem from Unit 03, a=1, b=1/n. The term with r=2 is {}^{n}C_{2}\frac1{n^2}=\frac{n(n-1)}{2!\,n^2}=\frac{1}{2!}\left(1-\frac1n\right), and every other term tidies up the same way:

Term r is \frac1{r!} times r-1 bracket factors \left(1-\frac1n\right)\left(1-\frac2n\right)\cdots, each between 0 and 1 and closer to 1 the larger n is. So as n grows, each term climbs toward its own ceiling \frac1{r!}:

(The whole sum includes the terms beyond r=4, which are too small to show.) Two things can be read off this picture with certainty. Every entry in a column rises as n grows, and a larger n also brings extra terms (at n=2 the expansion stops at r=2), so \left(1+\frac1n\right)^n always rises. And every entry is at most its ceiling, so

An intuition, not a proof: it is tempting to let n\to\infty in every term at once and conclude that \left(1+\frac1n\right)^n becomes the infinite sum \sum\frac1{r!}. But the number of terms grows with n as well, and infinitely many small shortfalls could, in principle, add up to something that does not vanish. The bound below is what makes the conclusion safe, and The rule it gives closes the argument.

Check it with numbers

Add the ceilings one at a time:

The running totals are

They rise, but can they rise forever? Shrinking terms alone do not settle it: 1+\frac12+\frac13+\frac14+\cdots also has shrinking terms, yet \frac13+\frac14=\frac7{12}>\frac12, then \frac15+\frac16+\frac17+\frac18=\frac{533}{840}>\frac12, and so on in blocks forever, so its total passes any number you name. The terms \frac1{r!} do better: compare each with the halving sequence 1,\frac12,\frac14,\frac18,\ldots

So the running total through \frac1{5!}, which is 2.7166667, is at most 1+1+\frac12+\frac14+\frac18+\frac1{16}=2.9375.

The rule it gives

Beyond the book: why the sum is finite. The book says only that the sum "can be shown" to be finite; the halving comparison shows it. For r\ge1, r!=1\times2\times3\times\cdots\times r has r-1 factors from 2 upward, each at least 2, so

The halving sum 1+\frac12+\frac14+\cdots never reaches 2: each term covers exactly half the gap still left to 2, so after k terms the total is 2-\frac1{2^{k-1}}. Hence every running total of \sum\frac1{r!} is less than 1+2=3. A total that keeps rising but can never pass 3 must level off at some number no larger than 3, and that number is called e:

Why the limit and the series agree. The picture gave the upper side: \left(1+\frac1n\right)^n is at most 1+1+\frac1{2!}+\cdots+\frac1{n!}, which is below e. So the compounded total rises, never passes e, and therefore levels off at some value no larger than e — which answers the unit's question: it does not grow without bound. For the lower side, keep only a fixed number of terms, say the first five. For every n\ge4, \left(1+\frac1n\right)^n is at least those five terms, and because there are only five of them, letting n\to\infty in each one separately is now legitimate: they approach 1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}=2.7083333. So the limit is at least 2.7083333, and the same argument with any fixed number of terms puts it at or above every running total of \sum\frac1{r!}. Caught between every running total and e itself, the limit is e:

The same expansion with x in place of 1 gives the general series, stated here and reconciled with the derivative (F12 Unit 05) once differentiation is available:

Because the terms shrink quickly, a numerical value can be found to a stated accuracy by adding terms only until the next term is too small to affect the required decimal places — the stopping rule.

Worked example

Evaluate e^{-0.25} to 3 dp. With x=-0.25, the terms alternate in sign and shrink fast:

The running totals are

Because the signs alternate and the sizes shrink, the running totals zig-zag, and the true value is trapped between any two consecutive ones. After 0.7786458, the terms still to come pair up as (+0.0001628-0.0000081)+\cdots, and each pair is positive because its first, larger member is positive, so the true value is above 0.7786458. After 0.7788086, the remaining pairs (-0.0000081+0.0000003)+\cdots are each negative, so the true value is below 0.7788086. Both bounds round to 0.779, so

and the next term, -0.0000081, is far too small to move the third decimal place: the stopping rule, now with a guarantee.

Where this shows up

Charge a 100\ \mu\text{F} capacitor to 9 V from a battery, then disconnect the battery and let the capacitor discharge through a 100\ \text{k}\Omega resistor, with a multimeter across it. The voltage follows V(t)=V_0e^{-t/RC}, and since ohms times farads give seconds, the time constant is RC=(100\times10^{3}\ \Omega)(100\times10^{-6}\ \text{F})=10 s. After t=2.5 s, \frac{t}{RC}=0.25, and the meter should read

about 78% of the starting voltage after a quarter of a time constant. The series alone predicts it, with no calculator "e^x" key, and a stopwatch and the meter's display let you watch the reading fall through 7.01 V on schedule.

Narration spine. A live readout of \left(1+\frac1n\right)^n climbs as a tracker sweeps n through 1, 2, 12 and 365, rising by less each time. The binomial expansion appears with its bracket factors highlighted, and a readout of 1-\frac1n slides from 0.5 toward 1 as n is sent up, so each term climbs toward its ceiling \frac1{r!}; the bridge e=\sum\frac1{r!} is written. The running totals 1,\ 2,\ 2.5,\ 2.6666667,\ 2.7083333,\ 2.7166667 build up, the series for e^x follows, and e^{-0.25} is added term by term to 0.779. The beat closes on the discharging capacitor at 0.779\,V_0 after a quarter of a time constant.