ExplorationsHow e connects repeated growth, infinite sums and a turn around a circle

The Number Behind Continuous Growth

Provedexponential functionlimitscomplex numberscurious

Begin with 1 unit and assign a nominal growth rate of 100% over one unit of time. Adding that growth once gives 2; splitting it into two 50% increases gives 1 × 1.5 × 1.5 = 2.25. Splitting it into n equal increases gives (1 + 1/n)^n, which approaches e, about 2.71828. Follow that , then see how powers of e describe growth, decay and rotation.

Current view: Compound growth

Compare finer splits of the same 100% nominal growth rate.

Compound growthStep 0 of 10

1 growth interval

(1 + 1/1)^1 ≈ 2

Begin with 1 unit and add 100% of the current amount once. The final amount is 2 units.

01.362.72reference ≈ 2.71828183010Number of doublings of the interval count
Complete calculation transcript
All calculations are available while playback is paused. Decimals are rounded floating-point illustrations.
StepCalculation
(1 + 1/1)^1 ≈ 2
(1 + 1/2)^2 ≈ 2.25
(1 + 1/4)^4 ≈ 2.44140625
(1 + 1/8)^8 ≈ 2.56578451
(1 + 1/16)^16 ≈ 2.6379285
(1 + 1/32)^32 ≈ 2.67699013
(1 + 1/64)^64 ≈ 2.69734495
(1 + 1/128)^128 ≈ 2.70773902
(1 + 1/256)^256 ≈ 2.71299162
(1 + 1/512)^512 ≈ 2.715632
(1 + 1/1024)^1024 ≈ 2.71695573
dot
a final amount after n growth intervals
dashed reference
a numerical reference, not an error bound

Every dot is the final amount from a fresh experiment starting with 1 unit. Each horizontal step doubles the number of growth intervals; it does not add another period. The dashed line marks e as a numerical reference.

(1 + 1/1)^1 ≈ 2. The numerical reference is 2.71828183.

Step
0 / 10
Current value
2

Step by step

Use Step for one change you can inspect, or Run to watch the changes accumulate.

Speed

Your experiment

The starting amount is 1 unit and the nominal total growth rate is 100%. Each step doubles the number of intervals, from 1 to 1024, without changing that rate.

Values and plots use floating-point arithmetic. The exact identities and their conditions are explained below.

Examples to explore

Try this

Why do two 50% increases produce 2.25 rather than 2?

1/3

What is going on

Each explanation begins with a worked example and follows the same operation through intuition, formal statements and proofs. Later sections distinguish what is established from questions that remain open, so you can follow the level of detail useful to you.

Worked by handStart with 1 unit and apply two 50% increases. The second percentage is taken from the amount now present, not from the original amount.
  1. 1add 50% of 1: 1 + 0.51.5This is the first of two equal growth intervals.
  2. 1.5add 50% of 1.5: 1.5 + 0.752.25The first increase also receives growth during the second interval.
One 100% increase gives 2, but two 50% increases give 2.25. Four 25% increases give about 2.441406, so the way the growth is split matters.
Making the intervals smaller does not make the final amount grow without bound. The amounts approach the constant e, approximately 2.71828.
01

What you are seeing

Each finer split begins again with 1 unit

In the compound-growth view, n tells us how many times growth is added. Each time, the current amount is multiplied by 1 + 1/n. For n = 2, that multiplier is 1.5, applied twice. For n = 4, it is 1.25, applied four times. We hold the nominal total rate at 100%, so increasing n does not mean increasing the rate assigned to each interval.

Run and Step compare n = 1, 2, 4, 8 and so on. A dot represents the final amount from one complete experiment, not the balance halfway through it. The transcript records every calculation, including steps that have not yet been drawn.

02

Another way to build the same number

A makes successive additions smaller

The gives meaning to e raised to different powers. In the series view, start with 1, add x, then x²/2!, then x³/3!, and continue. A such as 3! means 3 × 2 × 1 = 6. At x = 1, the first sums are 1, 2, 2.5 and approximately 2.666667. They approach the same e that appeared in .

At x = 2 the sum instead approaches e², about 7.389. At x = −1 it approaches e⁻¹ = 1/e, about 0.367879. Negative powers here describe decay rather than a negative amount. The early can overshoot or even be negative, especially when x is negative; only the infinite sum is the positive exponential value.

03

How e meets π on a circle

An imaginary represents a rotation

The i is defined by i² = −1. A a + bi can be drawn at horizontal a and vertical coordinate b. On Euler’s circle, start at 1, the point (1, 0), and move counterclockwise. A measures by divided by , so a half-turn is radians and a full turn is 2π radians.

states that e raised to iθ has cos θ and sin θ:

At a half-turn, those are (−1, 0), so e^(iπ) = −1. Adding 1 gives the familiar , e^(iπ) + 1 = 0. This relates e, , i, 1 and 0 without claiming that an ordinary positive real power has become negative; the is imaginary.

04

What about 1/(1 − e)?

A familiar formula can have an unfamiliar condition

The expression 1/(1 − e) is a perfectly well-defined negative number, about −0.581977. It is not the ordinary sum of 1 + e + e² + ⋯. That series keeps adding positive terms which grow larger, so its cannot approach that negative number.

The relevant formula is 1 + r + r² + ⋯ = 1/(1 − r), and it requires |r| less than 1. Use r = 1/2 to approach 2, or r = 1/e to approach 1/(1 − 1/e), about 1.581977. The comparison view lets you try both and r = e, where the condition fails. Neither series is the -weighted series that defines eˣ.

05

Why the connections hold

The finite identities explain the limiting operations

For the geometric Sₘ = 1 + r + ⋯ + rᵐ, multiplying by r and subtracting cancels all the middle terms. Thus (1 − r)Sₘ = 1 − rᵐ⁺¹. If |r| is less than 1, the final power tends to zero, giving the claimed . If |r| is at least 1, the terms themselves do not tend to zero, which already rules out ordinary .

For , the binomial gives

Every product lies between zero and one, and for each fixed k it tends to one. The tail is bounded by the corresponding tail of ∑ 1/k!, a series: after the first few terms, each new term is at most half its predecessor. Given any desired error, first choose a small tail, then make n large enough for the remaining finitely many products to be close to one. This proves that the is ∑ 1/k! = e.

For a complex iθ, the exponential series is absolutely , so its even and odd powers can be collected separately. Using i² = −1 produces the series in the real part and the series in the imaginary part, yielding .

06

What the pictures establish

A finite rendering illustrates a rather than proving it

The laboratory uses ordinary floating-point arithmetic. Its dashed references come from the browser’s , and the circle come from and . These are numerical illustrations, not certified intervals or independently verified digits. The 16-term series can still visibly differ from eˣ, especially near the ends of the allowed range.

The statements above concern and exact identities, supported by their arguments and references. Increasing the number of drawn points cannot replace those arguments. A useful next investigation is to compare how quickly the and constructions approach e, then ask what error bound would certify a chosen number of places.

What is actually established

Every statement on this page, with its status, its exact scope, and the date that status was last checked.

Proved

As n tends to infinity, (1 + 1/n)^n tends to e.

Scope
Positive integer n, with a fresh starting amount of 1 and a nominal total rate of 100%.
Why
The binomial expansion is a sum of products bounded by 1/k!. Each fixed product approaches 1/k!, and the tails are controlled by the convergent factorial series. The detailed argument appears below.
Status checked
Proved

The infinite series ∑ xᵏ/k! equals eˣ for every real x.

Scope
An infinite series; the laboratory displays only 16 terms using floating-point arithmetic.
Status checked
Proved

For real θ, e^(iθ) = cos θ + i sin θ, so e^(iπ) + 1 = 0.

Scope
Angles in radians and the imaginary unit satisfying i² = −1.
Status checked
Proved

The geometric series 1 + r + r² + ⋯ equals 1/(1 − r) exactly when |r| < 1.

Scope
Real r and ordinary convergence of partial sums. In particular, r = e is outside this condition.
Status checked

Sources

Review notes show which bibliographic details Mathomaly has checked and which remain unresolved. Checking a publication record does not independently verify its proof.

  1. NIST Digital Library of Mathematical Functions, DLMF §4.2: Definitions, National Institute of Standards and Technology. Link

    Equations 4.2.11, 4.2.19 and 4.2.24 give the constant e, the exponential power series and its real and imaginary parts.

    Bibliographic record checked. This is not an independent verification of the proof.

    Bibliographic review:

    AI-assisted bibliographic audit. Evidence type: source text.

    Read the institutional reference, its exponential-series definition and its real/imaginary decomposition. This records source identity, not an independent proof audit.

  2. Gilbert Strang, Edwin Herman, Calculus Volume 2, §5.2: Infinite Series, OpenStax, Rice University. Link

    The geometric-series calculation states its convergence condition and derives the finite partial-sum identity.

    Bibliographic record checked. This is not an independent verification of the proof.

    Bibliographic review:

    AI-assisted bibliographic audit. Evidence type: source text.

    Read the publisher-hosted geometric-series section and its condition |r| < 1. This is bibliographic verification, not an independent proof audit.

Connected by how they work, not by sharing a topic label.