The Number e

Where does the special number e (2.718281828…) come from? The special number π (pi, 3.141582653…) is easy to understand, it’s the ratio between a circle’s circumference and diameter. But the number e is a bit more mysterious.

To see where it comes from, imagine a bank account that pays 100% interest annually. Basically, deposit a dollar and after one year get two dollars back. But that result actually depends on how often the interest is calculated and added in (compounded).

If we deposit one dollar and the interest is compounded once annually, then after one year, the calculation is:

$\displaystyle{1.00}+({1.00}\times{1.0})={2.00}$

The original one dollar plus 100% interest (the original dollar times 1.0). The result, as mentioned above, is two dollars.

But suppose we compound the interest twice a year, once at six months and again at twelve months. Since we calculate twice, we only calculate half the interest rate, 50% each time. The first impression might be that 50% + 50% = 100%, so splitting the interest in two doesn’t matter. But the first interest calculation is added to the original amount, so the second one calculates 50% of a larger amount.

The first calculation at six months:

$\displaystyle{1.00}+({1.00}\times{0.5})={1.50}$

And then the second at twelve months:

$\displaystyle{1.50}+({1.50}\times{0.5})={2.25}$

So, compounding twice annually gives us back 2.25 dollars for every 1.00 dollar deposited.

Let’s see what happens if we compound three times annually (every four months). We calculate three times, so each time we use one-third the interest rate. Which means the second and third calculations involve progressively larger amounts.

The first calculation at four months:

$\displaystyle{1.00}+({1.00}\times{0.333\ldots})={1.33}$

Then the second at eight months:

$\displaystyle{1.33}+({1.33}\times{0.333\ldots})={1.77}$

Lastly, the third at twelve months:

$\displaystyle{1.77}+({1.77}\times{0.333\ldots})={2.37}$

Compounding three times annually gives us back 2.37 dollars on our 1.00 deposit. Clearly, compounding more often increases the amount we earn. (As an exercise for the reader: Compound the interest quarterly.)

Here’s how our original 1.00 deposit grows when we compound monthly (twelve times):

Starting Amount: 1.00
${1.00}+({1.00}\times{0.08\overline{333}})={1.08}$
${1.08}+({1.08}\times{0.08\overline{333}})={1.1736\overline{111}}$
${1.17}+({1.17}\times{0.08\overline{333}})={1.2714}\dots$
${1.27}+({1.27}\times{0.08\overline{333}})={1.3773}\dots$
${1.37}+({1.37}\times{0.08\overline{333}})={1.4921}\dots$
${1.49}+({1.49}\times{0.08\overline{333}})={1.6164}\dots$
${1.62}+({1.62}\times{0.08\overline{333}})={1.7511}\dots$
${1.75}+({1.75}\times{0.08\overline{333}})={1.8971}\dots$
${1.90}+({1.90}\times{0.08\overline{333}})={2.0552}\dots$
${2.06}+({2.06}\times{0.08\overline{333}})={2.2264}\dots$
${2.23}+({2.23}\times{0.08\overline{333}})={2.4120}\dots$
${2.41}+({2.41}\times{0.08\overline{333}})={2.6130}\dots$
Final Amount: 2.61

Now the return on the 1.00 deposit after one year is 2.61 dollars. Compounding more often definitely results in a higher return. Compounding daily (365 times) results in an annual return of 2.71 dollars.

[Note an important assumption in these calculations: That the bank does not round off to the nearest cent when compounding. The amount represented in the bank account at one given time may be rounded off in how it’s displayed or in how much can be withdrawn. The resolution for interacting with the account is pennies, but the interest calculation is assumed to have greater precision (the highest possible; ideally infinite).]

Given the introduction above and the numeric progression from 2.00 to 2.25, to 2.37, to 2.61, and finally to 2.71, it may be apparent that things are converging on the mysterious number e.

Which, if interest is compounded continually, is correct.

We have here the quintessential notion of growth, of exponential expansion. As something grows larger and larger, it grows larger faster and faster.

Consider the general formula used above:

$\displaystyle\textsf{amount}+(\textsf{amount}\times\textsf{interest})=\textsf{result}$

Where interest is the interest rate (such as 1.0 for 100% interest or 0.25 for 25%). To make things simpler later, we re-write it like this:

$\displaystyle\textsf{amount}+\left(\frac{\textsf{amount}}{\textsf{interest}}\right)=\textsf{result}$

And now interest is expressed as the number of compound intervals (rather than a decimal fraction as it is above). For example, when calculating the monthly rate:

$\displaystyle\textsf{amount}+\left(\frac{\textsf{amount}}{{12}}\right)=\textsf{result}$

Because there are twelve compound intervals.

To see why this is helpful, first consider what happens if we deposited more than one dollar. To start simple, suppose we deposited ten dollars. If the interest was calculated annually (as in the first example above), then we expect twenty dollars — the original plus 100% interest:

$\displaystyle{10.00}+\left(\frac{10.00}{1}\right)={20.00}$

There is just one interest calculation (at the end of the year), so the denominator in the interest fraction is just 1, and we have our final result.

But notice that we can factor out the ten-dollar deposit amount:

$\displaystyle{10.00}\times\left({1}+\left(\frac{1}{1}\right)\!\!\right)={20.00}$

Which restores the idea of having a single dollar but allows calculating the interest on any actual amount. More importantly, we can take note of a pattern:

$\displaystyle\textsf{amount}\times\left({1}+\left(\frac{1}{\textsf{intervals}}\right)\!\!\right)=\textsf{result}$

Which is important two ways. Firstly, we have a generalized template, the (1+1/n) part, with a single parameter, n, the number of compound intervals. Secondly, that we multiply this by whatever amount currently exists in the account.

So, consider the second example above where we compound twice annually:

$\displaystyle{1.00}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{{2}}\right)\!\!\right)\!\!\times\!\!\left(\!{1}\!+\!\!\left(\frac{1}{{2}}\right)\!\!\right)=\textsf{result}$

And the case where we compound three times annually:

$\displaystyle{1.00}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{{3}}\right)\!\!\right)\!\!\times\!\!\left(\!{1}\!+\!\!\left(\frac{1}{{3}}\right)\!\!\right)\!\!\times\!\!\left(\!{1}\!+\!\!\left(\frac{1}{{3}}\right)\!\!\right)=\textsf{result}$

From left-to-right, we evaluate each term, starting with the original amount. The growing sum becomes the input to the next term to be evaluated (as if we were doing an initial calculation with a starting amount).

Next note that, since the template terms repeat, we can collapse them into a single instance of the template with a power telling us how many times the term appears. For instance, in the twice annual case:

$\displaystyle{1.00}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{{2}}\right)\!\!\right)^{2}=\textsf{result}$

And in the thrice annual case:

$\displaystyle{1.00}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{{3}}\right)\!\!\right)^{3}=\textsf{result}$

Which means the quarterly case is:

$\displaystyle{1.00}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{{4}}\right)\!\!\right)^{4}=\textsf{result}$

And so on. This gives us a general formula:

$\displaystyle\textsf{amount}\!\times\!\left(\!{1}\!+\!\!\left(\frac{1}{n}\right)\!\!\right)^{n}=\textsf{result}$

If the amount is set to 1.0, then this converges on the number e, (2718281828…) as n approaches infinity. Which gives us one way of defining the number e:

$\displaystyle{e}=\lim_{{n}\rightarrow\infty}\left[\left({1}+\left(\frac{1}{n}\right)\!\!\right)^{n}\right]$

But there are a number of other ways to discover this value lurking in mathematics! One of them is the exp(x) function, which calculates powers of e.

The Exponential Function

The exponential function is identical to raising e to some power:

$\displaystyle\exp(x)={e}^{x}$

It’s formally defined:

$\displaystyle{e}^{x}=\exp(x)=\sum_{n=0}^{\infty}\frac{\;\;x^n}{n!}$

Which expands to the conceptually infinite series:

$\displaystyle\exp(x)=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\cdots$

Or, evaluating what we can:

$\displaystyle\exp(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\cdots$

What’s great about this definition is that for x we can plug in anything with a multiplication operation — complex numbers, even matrices. Each appearance of x here has a positive integer power, so each appearance is just x times itself some number of times (including zero and one). This makes the exponential function widely applicable.

It’s common in complex math where it’s one side of Euler’s Formula:

$\displaystyle\exp(i\theta)={e}^{i\theta}=\cos(\theta)+{i}\sin(\theta)$

This is the source of Euler’s Identity, which has been called the most beautiful equation in math:

$\displaystyle\exp(i\pi)={e}^{i\pi}+1=0$

See the Beautiful Math post for an overview. Also see Circular Math and Sideband #70: The exp Function for more on the exponential function.

More to the point here, if we set x=1, (in other words, calculating e¹) then:

$\displaystyle\exp(1)={e}^{1}={e}={1}+{1}+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}\cdots\frac{1}{n!}$

Which evaluates to the number e. Note that both these methods involve infinite sequences, so they evaluate closer and closer to the correct value depending on the number of iterations.

Note also that it takes billions of iterations for the first technique to converge to a reasonably accurate value. This one converges much faster.

Exponential Derivatives

It’s worth digging into the mechanics of taking the derivative of the exponential function. It begins with the basic fact that the derivative of the exponential function is the exponential function:

$\displaystyle\frac{d}{dx}\,{e}^{x}={e}^{x}$

The slope of e^x at x is always just e^x. Note this is true when the exponent is just x. (We’ll sort of see why below.) If the exponent is an expression containing x, then the expression is a function of x, and the chain rule applies.

The chain rule is:

$\displaystyle\frac{d}{dx}\,f(g(x))={f'}(g(x))\cdot{g'}(x)$

If a function, f, depends on a sub-function, g, multiply the derivative of the outer function times the derivative of the inner function.

To see how this works, we can see what happens when we apply the chain rule to the basic exponential function (which we know derives to itself). We’ll treat the plain x in the exponent as a function, g(x), that just returns x:

Applying the chain rule and setting u=g(x) to simplify the exponential:

$\displaystyle\frac{d}{dx}\,{e}^{g(x)}=\frac{d}{du}\,{e}^{u}\,\cdot\,\frac{d}{dx}\,g(x)={e}^{u}\,\cdot\,\frac{d}{dx}\,{x}$

Since the derivative of x is just 1, we end up with, as expected, the original exponential. Rearranged a bit, this is a basic formula for deriving the exponential function:

$\displaystyle\frac{d}{dx}\,{e}^{u}=u'\;{e}^{u}$

Where u is some expression containing (at least one occurrence of) x. Just take the derivative of that expression and multiply it to the exponential.

With that understanding, let’s try this:

$\displaystyle\frac{d}{dx}\,{e}^{2x}=\frac{d}{dx}\,\left[{2x}\right]{e}^{2x}={2}\cdot{e}^{2x}$

Because the derivative of 2x is just 2.

Here’s one with a square of x, such as appears in the Gaussian exponential function:

$\displaystyle\frac{d}{dx}\,{e}^{x^2}=\frac{d}{dx}\,\left[{x}^{2}\right]{e}^{x^2}={2x}\cdot{e}^{x^2}$

Because the derivative of x² is 2x.

If there was a constant in front of the exponential, it just gets multiplied by the derived exponent:

$\displaystyle\frac{d}{dx}\,{ae}^{x^2}=\frac{d}{dx}\,\left[{x}^{2}\right]{ae}^{x^2}={2ax}\cdot{e}^{x^2}$

We can break this down as an instance of the product rule, which is:

$\displaystyle\frac{d}{dx}\,f(x)\,g(x)=f(x)g'(x)+f'(x)g(x)$

But since a is a constant, it derives to zero, so:

$\displaystyle\frac{d}{dx}{ae}^{x^2}\!=\!\left[{a}\cdot\frac{d}{dx}{e}^{x^2}\right]\!\!+\!\!\left[\frac{d}{dx}{a}\cdot{e}^{x^2}\right]\!=\!\left[2ax\cdot{e}^{x^2}\right]\!\!+\!\!\left[0\cdot{e}^{x^2}\right]$

Which gives us the result shown above.

Finally, note that, since a constant derives to zero, an exponential function with a constant exponent also derives to zero:

$\displaystyle\frac{d}{dx}\,{e}^{k}=\frac{d}{dx}\,\left[{k}\right]{e}^{k}={0e}^{k}=0$

Hopefully this provides an idea about how to derive more involved exponential functions!

Getting Complex

Noting that the template discussed above resembles the form of a complex number, (a+bi), we can extend the template as (1+i/n) to make a definition of the number e for the complex numbers:

$\displaystyle{e}=\lim_{{n}\rightarrow\infty}\left[\left({1}+\left(\frac{i}{n}\right)\!\!\right)^{n}\right]$

Returning to the exp(x) function, which can take complex numbers as input, if we set x=i, then:

$\displaystyle\exp(i)={e}^{i}={1}+{i}+\frac{-1}{2}+\frac{-i}{6}+\frac{1}{24}+\frac{i}{120}\cdots\frac{i^n}{n!}$

This evaluates to the complex number (0.540… + 0.841…i), which is the point on the complex unit circle at an angle of one radian (just over 57 degrees). Now, this value can be understood as:

$\displaystyle{e}^{i}={e}^{({i}\times{1})}=(\sim\!{0.540},\,\sim\!{0.841}{i}))$

Which highlights that the exponent i results in a rotation of one radian. A full 360° rotation is 2π radians (that is: 6.28… radians), and a half rotation of 180° is therefore just π radians (3.14… radians), so if we replace the “1” in the equation above with π:

$\displaystyle{e}^{({i}\times{\pi})}$

We should get a point halfway around the complex unit circle at an angle of π radians. Which is where -1 on the real number line intersects the circle, 180° around from +1, and indeed:

$\displaystyle{e}^{{i}\pi}={-1}$

Which is (one way of writing) Euler’s beautiful identity. [See the Beautiful Math post for details.

And of course:

$\displaystyle{e}^{{i}{0}}={+1}$

Because x⁰=1 for any x, including x=e, but also because an angle of zero radians is where the real number line intersects the unit circle at a value of +1.

Pascal’s Triangle

Getting down in the weeds, here are the first eight expansions on the (1+1/n) template:

$\displaystyle\left(\frac{1}{n^0}+\frac{1}{n^1}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{2}{n^1}+\frac{1}{n^2}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{3}{n^1}+\frac{3}{n^2}+\frac{1}{n^3}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{4}{n^1}+\frac{6}{n^2}+\frac{4}{n^3}+\frac{1}{n^4}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{5}{n^1}+\frac{10}{n^2}+\frac{10}{n^3}+\frac{5}{n^4}+\frac{1}{n^5}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{6}{n^1}+\frac{15}{n^2}+\frac{20}{n^3}+\frac{15}{n^4}+\frac{6}{n^5}+\frac{1}{n^6}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{7}{n^1}+\frac{21}{n^2}+\frac{35}{n^3}+\frac{35}{n^4}+\frac{21}{n^5}+\frac{7}{n^6}+\frac{1}{n^7}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{8}{n^1}+\frac{28}{n^2}+\frac{56}{n^3}+\frac{70}{n^4}+\frac{56}{n^5}+\frac{28}{n^6}+\frac{8}{n^7}+\frac{1}{n^8}\right)$

Note that the denominators, as powers of n, get increasing (very) large, and therefore the terms get increasing (very) small. Note also that all denominators all have exponents to highlight the exponential pattern. Typically, the first term is shown as just 1 (because n⁰=1) and the second as just x/n (no exponent needed because n¹=n).

The pattern of the numerators might look familiar. The numerators for a given n are the same values found in Pascal’s Triangle on the nth row from the top (counting the top as row zero).

Because both the numerators and denominators follow a regular pattern, we have a simple general formula for generating the sequence for any value of n:

$\displaystyle\sum^{n}_{x=0}\frac{\textsf{PT}(n,x)}{n^x}$

Where PT(n,x) is function that returns the value of the xth member of the nth row of Pascal’s Triangle. (Basically, nth row, xth column.)

Filling in the above and setting n to the iteration count at that point (1, 2, 3…):

$\displaystyle\left(1\!+\!\frac{1}{1}\right)={2.0}$

$\displaystyle\left(1\!+\!\frac{2}{2}\!+\!\frac{1}{4}\right)={2.25}$

$\displaystyle\left(1\!+\!\frac{3}{3}\!+\!\frac{3}{9}\!+\!\frac{1}{27}\right)={2.\overline{370}}$

$\displaystyle\left(1\!+\!\frac{4}{4}\!+\!\frac{6}{16}\!+\!\frac{4}{64}\!+\!\frac{1}{256}\right)={2.4414\ldots}$

$\displaystyle\left(1\!+\!\frac{5}{5}\!+\!\frac{10}{25}\!+\!\frac{10}{125}\!+\!\frac{5}{625}\!+\!\frac{1}{3125}\right)={2.48832\dots}$

$\displaystyle\left(1\!+\!\frac{6}{6}\!+\!\frac{15}{36}\!+\!\frac{20}{216}\!+\!\frac{15}{1296}\!+\!\frac{6}{7776}\!+\!\frac{1}{46656}\right)={2.521626\ldots}$

$\displaystyle\left(\frac{1}{1}\!+\!\frac{7}{7}\!+\!\frac{21}{49}\!+\!\frac{35}{343}\!+\!\frac{35}{2401}\!+\!\frac{21}{16807}\!+\!\frac{7}{117649}\!+\!\frac{1}{823543}\right)\!=\!{2.5465\ldots}$

$\displaystyle\left(\frac{1}{1}\!+\!\frac{8}{8}\!+\!\frac{28}{64}\!+\!\frac{56}{512}\!+\!\frac{70}{4096}\!+\!\frac{56}{32768}\!+\!\frac{28}{262144}\!+\!\frac{8}{2097152}\!+\!\frac{1}{16777216}\right)\!=\!{2.56578\ldots}$

Which shows the slow convergence on 2.71828… (the number e).

Note that, since fractions always evaluate to either an exact decimal or a repeating decimal, the coefficients here, as well as their sum, eventually enter a repeating pattern of digits.

In comparison, here are the first eight expansions of the complex form of the growth formula, (1+i/n):

$\displaystyle\left(\frac{1}{n^0}+\frac{1}{n^1}{i}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{2}{n^1}{i}-\frac{1}{n^2}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{3}{n^1}{i}-\frac{3}{n^2}-\frac{1}{n^3}{i}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{4}{n^1}{i}-\frac{6}{n^2}-\frac{4}{n^3}{i}+\frac{1}{n^4}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{5}{n^1}{i}-\frac{10}{n^2}-\frac{10}{n^3}{i}+\frac{5}{n^4}+\frac{1}{n^5}{i}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{6}{n}{i}-\frac{15}{n^2}-\frac{20}{n^3}{i}+\frac{15}{n^4}+\frac{6}{n^5}{i}-\frac{1}{n^6}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{7}{n^1}{i}-\frac{21}{n^2}-\frac{35}{n^3}{i}+\frac{35}{n^4}+\frac{21}{n^5}{i}-\frac{7}{n^6}-\frac{1}{n^7}{i}\right)$

$\displaystyle\left(\frac{1}{n^0}+\frac{8}{n^1}{i}-\frac{28}{n^2}-\frac{56}{n^3}{i}+\frac{70}{n^4}+\frac{56}{n^5}{i}-\frac{28}{n^6}-\frac{8}{n^7}{i}+\frac{1}{n^8}\right)$

The pattern is similar, but using complex numbers means the terms interact a bit differently (because of complex multiplication). Again though, the denominators get large by increasing powers (and the terms therefore increasingly small).

Note also the pattern off pluses and minuses. I showed eight expansions to make that pattern clearer. It’s a simple one: two pluses alternate with two minuses. One way to calculate it is:

$\displaystyle\mathsf{sign}({n})=\mathfrak{Re}({i}^{n})+\mathfrak{Im}({i}^{n})$

Where n is the same as the exponent used in the denominators. The powers of i, starting at i⁰, alternate +1, +i, -1, –i in a repeating pattern, and the sum of their real (ℜ𝔢) and imaginary (ℑ𝔪) parts yields the +1, +1, -1, -1 repeating sequence for the terms.

A sequence generator also needs to make every other term a multiple of i. These two differences from the real sequence make the generator a bit more complicated:

$\displaystyle\sum^{n}_{x=0,2,4\ldots}\left[\frac{\textsf{sign}(x)\!\cdot\!\textsf{PT}(n,x)}{n^x}\,+\,\frac{\textsf{sign}(x\!\!+\!\!1)\!\cdot\!\textsf{PT}(n,x\!+\!1)}{n^{x\!+\!1}}{i}\right]$

Again, PT(n,x) is a function that returns a row from Pascal’s Triangle. The sign(n) function, from just above, determines whether a given term is positive or negative. In this case, the x value jumps in steps of two, and each step generates two terms, real and imaginary. Note that the second step is skipped on the last iteration if n is an odd number.

Completing the first six by setting n to the iteration count at that point:

$\displaystyle\left(1\!+\!\frac{i}{1}\right)=({1}+{i})$

$\displaystyle\left(1\!+\!\frac{2}{2}{i}-\!\frac{1}{4}\right)=\left(\frac{3}{4}+{i}\right)$

$\displaystyle\left(1\!+\!\frac{3}{3}{i}-\!\frac{3}{9}+\!\frac{1}{27}{i}\right)=\left(\frac{2}{3}+\frac{26}{27}{i}\right)$

$\displaystyle\left(1\!+\!\frac{4}{4}{i}-\!\frac{6}{16}-\!\frac{4}{64}{i}+\!\frac{1}{256}\right)=\left(\frac{161}{256}+\frac{15}{16}{i}\right)$

$\displaystyle\left(1\!+\!\frac{5}{5}{i}-\!\frac{10}{25}-\!\frac{10}{125}{i}+\!\frac{5}{625}+\!\frac{1}{3125}{i}\right)=\left(\frac{76}{125}+\frac{2876}{3125}{i}\right)$

$\displaystyle\left(1\!+\!\frac{6}{6}{i}-\!\frac{15}{36}-\!\frac{20}{216}{i}+\!\frac{15}{1296}+\!\frac{6}{7776}{i}-\!\frac{1}{46656}\right)=\left(\frac{27755}{46656}\!+\!\frac{1177}{1296}{i}\right)$

Those fractional results and the expansion make things get a bit unwieldy (which is why I limited the list to the first six). Ignoring the expansions and adding a final conversion to more familiar decimal (complex) numbers:

$\displaystyle\left({1}+\frac{i}{n}\right)^{1}=({1}+{i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{2}=\left(\frac{3}{4}+{i}\right)=({0.75}+{i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{3}=\left(\frac{2}{3}+\frac{26}{27}{i}\right)=({0.\overline{666}}+{0.\overline{962}i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{4}=\left(\frac{161}{256}+\frac{15}{16}{i}\right)=({0.62890625}+{0.9375i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{5}=\left(\frac{380}{625}+\frac{2876}{3125}{i}\right)=({0.608}+{0.92032i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{6}=\left(\frac{27755}{46656}\!+\!\frac{1177}{1296}{i}\right)=({0.594}\ldots+{0.908}\dots{i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{7}=\left(\frac{9848}{16807}\!+\!\frac{740536}{823543}{i}\right)=({0.586}\ldots+{0.899}\dots{i})$

$\displaystyle\left({1}+\frac{i}{n}\right)^{8}=\left(\frac{9722113}{16777216}\!+\!\frac{233919}{262144}{i}\right)=({0.579}\ldots+{0.892}\dots{i})$

Just as the real series converges on 2.71828… this complex series converges on (0.540… + 0.841…i). And as with the real series, this series also converges fairly slowly.

∇