Thoughts on Infinity (Part 3f)

In recent posts, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product: while

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... = \ln 2$ ,

a rearranged series can be something completely different:

$\displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} ... = \displaystyle \frac{3}{2} \ln 2$ .

This very counterintuitive result can be confirmed using commonly used technology — in particular, Microsoft Excel. In the spreadsheet below, I typed:

=IF(MOD(ROW(A1),3)=0,ROW(A1)*2/3,IF(MOD(ROW(A1),3)=1,4*(ROW(A1)-1)/3+1,4*(ROW(A1)-2)/3+3)) in cell A1
=POWER(-1,A1-1)/A1 in cell B1
=B1 in cell C1
I copied cell A1 into cell A2
=POWER(-1,A2-1)/A2 in cell B2
=C1+B2 in cell C2

The unusual command for cell A1 was necessary to get the correct rearrangement of the series.

Then I used the FILL DOWN command to fill in the remaining rows. Using these commands cell C9 shows the sum of all the entries in cells B1 through B9, so that

$\displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} \approx 0.961544012$

Filling down to additional rows demonstrates that the sum converges to $\displaystyle \frac{3}{2}\ln 2$ and not to $\ln 2$ . Here’s the sum up to 10,000 terms… the entry in column E is the first few digits in the decimal expansion of $\displaystyle \frac{3}{2} \ln 2$ .

Clearly the partial sums are not approaching $\ln 2 \approx 0.693$ , and there’s good visual evidence to think that the answer is $\displaystyle \frac{3}{2} \ln 2$ instead. (Incidentally, the 10,000th partial sum is very close to the limiting value because $10,000$ is one more than a multiple of 3.)

Thoughts on Infinity (Part 3e)

In recent posts, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product. Here’s another classic example of this fact that’s attributed to Cauchy.

We’ve already seen in this series (pardon the pun) that

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... = \ln 2$ .

Let’s now see what happens if I rearrange the terms of this conditionally convergent series. Let

$T = \displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{11} - \frac{1}{6} \dots$ ,

where two positive numbers alternate with a single negative term. By all rights, this shouldn’t affect anything… right?

Let $s_n$ be the $n$ th partial sum of this series, so that $s_{3n}$ contains $2n$ positive terms with odd denominators and $n$ negative terms with even denominators:

$s_{3n} = \displaystyle \sum_{k=1}^{2n} \frac{1}{2n-1} - \sum_{k=1}^n \frac{1}{2n}$ .

Let me now add and subtract the “missing” even terms in the first sum:

$s_{3n} = \displaystyle \sum_{k=1}^{2n} \frac{1}{2n-1} + \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^n \frac{1}{2n}$

$s_{3n} = \displaystyle \sum_{k=1}^{4n} \frac{1}{n} - \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^n \frac{1}{2n}$

$s_{3n} = \displaystyle \sum_{k=1}^{4n} \frac{1}{n} - \frac{1}{2} \sum_{k=1}^{2n} \frac{1}{n} - \frac{1}{2} \sum_{k=1}^n \frac{1}{n}$ .

For reasons that will become apparent, I’ll now rewrite this as

$s_{3n} = \displaystyle \int_1^{4n} \frac{dx}{x} + \left( \sum_{k=1}^{4n} \frac{1}{k} - \displaystyle \int_1^{4n} \frac{dx}{x} \right)$

$- \displaystyle \frac{1}{2} \int_1^{2n} \frac{dx}{x} - \frac{1}{2} \left( \sum_{k=1}^{2n} \frac{1}{k} - \int_1^{2n} \frac{dx}{x} \right)$

$- \displaystyle \frac{1}{2} \int_1^{n} \frac{dx}{x} - \frac{1}{2} \left( \sum_{k=1}^{n} \frac{1}{k} - \int_1^{n} \frac{dx}{x} \right)$ ,

$s_{3n} = \ln(4n) - \ln 1 + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - [\ln(4n) - \ln 1]\right)$

$- \displaystyle \frac{1}{2}[\ln (2n) - \ln 1] - \displaystyle \frac{1}{2} \left( \sum_{k=1}^{2n} \frac{1}{k} - [\ln (2n) - \ln 1]\right)$

$- \displaystyle \frac{1}{2}[\ln n - \ln 1] - \displaystyle \frac{1}{2} \left( \sum_{k=1}^{n} \frac{1}{k} - [\ln n - \ln 1]\right)$

Since $\ln 1 = 0$ , $\ln(2n) = \ln 2 + \ln n$ , and $\ln(4n) = \ln 4 + \ln n = 2\ln 2 + \ln n$ , we have

$s_{3n} = 2\ln 2 + \ln n + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right)$

$\displaystyle - \frac{\ln 2 + \ln n}{2} - \displaystyle \frac{1}{2} \left( \sum_{k=1}^{2n} \frac{1}{k} - \ln 2n \right)$

$\displaystyle - \frac{\ln n}{2} - \frac{1}{2} \displaystyle \left( \sum_{k=1}^{n} \frac{1}{k} - \ln n \right)$ ,

$s_{3n}= \displaystyle \frac{3}{2}\ln 2 + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right) - \frac{1}{2}\left( \sum_{k=1}^{2n} \frac{1}{k} - \ln 2n\right) - \frac{1}{2}\left( \sum_{k=1}^{n} \frac{1}{k} - \ln n\right)$ .

I now take the limit as $m \to \infty$ :

$\displaystyle \lim_{n \to \infty} s_{3n} = \displaystyle \frac{3}{2}\ln 2 + \lim_{n \to \infty} \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right) - \frac{1}{2} \lim_{n \to \infty} \left( \sum_{k=1}^{2n} \frac{1}{k} - \ln 2n\right) - \frac{1}{2} \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{1}{k} - \ln n\right)$ .

This step reveals why I added and subtracted the integrals above: those gymnastics were necessary in order to reach a limit that converges.

As shown earlier in this series, if

$\displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right) = \gamma$ ,

the Euler-Mascheroni constant. Therefore, since the limit of any subsequence must converge to the same limit, we have

Applying these above, we conclude

$\displaystyle \lim_{m \to \infty} s_{3n} = \displaystyle \frac{3}{2}\ln 2 + \gamma - \frac{1}{2}\gamma - \frac{1}{2} \gamma = \displaystyle \frac{3}{2} \ln 2$ ,

which is different than $\ln 2$ .

Technically, I’ve only shown so far that the limit of partial sums 3, 6, 9, … is $\displaystyle\frac{3}{2} \ln 2$ . For the other partial sums, I note that

$\displaystyle \lim_{n \to \infty} t_{3n+1} = \displaystyle \lim_{n \to \infty} \left(s_{3n} + \displaystyle \frac{1}{4n+1} \right) = \displaystyle \frac{3}{2} \ln 2 + 0 = \displaystyle \frac{3}{2} \ln 2$

and

$\displaystyle \lim_{n \to \infty} t_{3n-1} = \displaystyle \lim_{n \to \infty} \left(s_{3n} - \displaystyle \frac{1}{2n} \right) = \displaystyle \frac{3}{2} \ln 2 - 0 = \displaystyle \frac{3}{2} \ln 2$ .

Therefore, I can safely conclude that

$T = \displaystyle \lim_{n \to \infty} t_n = \displaystyle \frac{3}{2} \ln 2$ ,

which is different than the original sum $S$ .

Thoughts on Infinity (Part 3d)

In yesterday’s post, I showed that

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... = \ln 2$ .

This can be (sort of) confirmed using commonly used technology — in particular, Microsoft Excel. In the spreadsheet below, I typed:

1 in cell A1
=POWER(-1,A1-1)/A1 in cell B1
=B1 in cell C1
=A1+1 in cell A2
=POWER(-1,A2-1)/A2 in cell B2
=C1+B2 in cell C2
Then I used the FILL DOWN command to fill in the remaining rows. Using these commands cell C10 shows the sum of all the entries in cells B1 through B10, so that

$1 - \displaystyle \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} - \frac{1}{10} \approx 0.645634921$

Filling down to additional rows demonstrates that the sum converges to $\ln 2$ , albeit very slowly (as is typical for conditionally convergent series). Here’s the sum up to 200 terms… the entry in column E is the first few digits in the decimal expansion of $\ln 2$ .

Here’s the result after 2000 terms:

20,000 terms:

And finally, 200,000 terms. (It takes a few minutes for Microsoft Excel to scroll this far.)

We see that, as expected, the partial sums are converging to $\ln 2$ , as expected. Unfortunately, the convergence is extremely slow — we have to compute 10 times as many terms in order to get one extra digit in the final answer.

Thoughts on Infinity (Part 3c)

Define

$S = \displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ...$

By the alternating series test, this series converges. However,

$\displaystyle \sum_{n=1}^\infty |a_n| = \displaystyle \sum_{n=1} \frac{1}{n}$ ,

which is the divergent harmonic series which was discussed earlier in this series. Therefore, the series $S$ converges conditionally and not absolutely.

To calculate the value of $S$ , let $s_n = \displaystyle \sum_{k=1}^n \frac{(-1)^{k-1}}{k}$ , the $n$ th partial sum of $S$ . Since the series converges, we know that $\displaystyle \lim_{n \to \infty} s_n$ converges. Furthermore, the limit of any subsequence, like $\displaystyle \lim_{n \to \infty} s_{2n}$ , must also converge to $S$ .

If $n$ is even, so that $n = 2m$ and $m$ is an integer, then

$s_{2m} = \displaystyle \sum_{k=1}^n \frac{(-1)^{k-1}}{k}$

$= \displaystyle \sum_{k=1}^{m} \frac{1}{2k-1} - \sum_{k=1}^m \frac{1}{2k}$

$= \displaystyle \left( \sum_{k=1}^{2m} \frac{1}{k} - \sum_{k=1}^m \frac{1}{2k} \right) - \sum_{k=1}^m \frac{1}{2k}$

$= \displaystyle \sum_{k=1}^{2m} \frac{1}{k} - 2 \sum_{k=1}^m \frac{1}{2k}$

$= \displaystyle \sum_{k=1}^{2m} \frac{1}{k} - \sum_{k=1}^m \frac{1}{k}$ .

For reasons that will become apparent, I’ll now rewrite this as

$s_{2m} = \displaystyle \int_1^{2m} \frac{dx}{x} + \left( \sum_{k=1}^{2m} \frac{1}{k} - \displaystyle \int_1^{2m} \frac{dx}{x} \right) - \displaystyle \int_1^m \frac{dx}{x} - \left( \sum_{k=1}^m \frac{1}{k} - \int_1^m \frac{dx}{x} \right)$ ,

$s_{2m} = \ln(2m) - \ln 1 + \displaystyle \left( \sum_{k=1}^{2m} \frac{1}{k} - [\ln(2m) - \ln 1]\right)$

$- [\ln m - \ln 1] - \displaystyle \left( \sum_{k=1}^m \frac{1}{k} - [\ln m - \ln 1]\right)$ .

Since $ln 1 = 0$ and $\ln(2m) = \ln 2 + \ln m$ , we have

$s_{2m} = \ln 2 + \ln m + \displaystyle \left( \sum_{k=1}^{2m} \frac{1}{k} - \ln 2m \right) - \ln m - \displaystyle \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right)$

$= \ln 2 + \displaystyle \left( \sum_{k=1}^{2m} \frac{1}{k} - \ln 2m \right) - \left( \sum_{k=1}^m \frac{1}{k} - \ln m\right)$ .

I now take the limit as $m \to \infty$ :

$\displaystyle \lim_{m \to \infty} s_{2m} = \ln 2 + \displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^{2m} \frac{1}{k} - \ln 2m\right) - \lim_{m \to \infty} \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right)$ .

This step reveals why I added and subtracted the integrals above: those gymnastics were necessary in order to reach a limit that converges.

In yesterday’s post, I showed that if

$t_m = \displaystyle \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right)$

$\displaystyle \lim_{m \to \infty} t_m = \displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right) = \gamma$ ,

the Euler-Mascheroni constant. Therefore, the limit of any subsequence must converge to the same limit; in particular,

$\displaystyle \lim_{m \to \infty} t_{2m} =\displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^{2m} \frac{1}{k} - \ln 2m \right)= \gamma$ .

Applying these above, we conclude

$\displaystyle \lim_{m \to \infty} s_{2m} = \ln 2 + \gamma - \gamma$ ,

$S = \ln 2$ .

Thoughts on Infinity (Part 3b)

The five most important numbers in mathematics are $0$ , $1$ , $e$ , $\pi$ , and $i$ . In sixth place (a distant sixth place) is probably $\gamma$ , the Euler-Mascheroni constant. See Mathworld or Wikipedia for more details. (For example, it’s astounding that we still don’t know if $\gamma$ is irrational or not.)

In yesterday’s post, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product. In tomorrow’s post, I’ll present another classic example of this phenomenon due to Cauchy. However, to be ready for this fact, I’ll need to see how $\gamma$ arises from a certain conditionally convergent series.

Separately define the even and odd terms of the sequence $\{a_n\}$ by

$a_{2n} = \displaystyle \int_n^{n+1} \frac{dx}{x}$

and

$a_{2n-1} = \displaystyle \frac{1}{n}$ .

It’s pretty straightforward to show that this sequence is decreasing. The function $f(x) = \displaystyle \frac{1}{x}$ is clearly decreasing for $x > 0$ , and so the maximum value of $f(x)$ on the interval $[n,n+1]$ must occur at the left endpoint, while the minimum value must occur at the right endpoint. Since the length of this interval is $1$ , we have

$\displaystyle \frac{1}{n+1} \cdot 1 < \displaystyle \int_n^{n+1} \frac{dx}{x} < \displaystyle \frac{1}{n} \cdot 1$ ,

$a_{2n+1} < a_{2n} < a_{2n-1}$ .

Since the subsequence $\{a_{2n-1}\}$ clearly decreases to $0$ , this shows the full sequence $\{a_n\}$ is a decreasing sequence with limit $0$ .

By the alternating series test, this implies that the series

$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} a_n$

converges. This limit is called the

Since this series converges, that means that the limit of the partial sums converges to $\gamma$ :

$\displaystyle \lim_{M \to \infty} \sum_{n=1}^M (-1)^{n-1} a_n = \gamma$ .

Let’s take the upper limit to be an odd number $M$ , where $M = 2N-1$ and $N$ is an integer. Then by separating the even and odd terms, we obtain

$\displaystyle \sum_{n=1}^{2N-1} (-1)^{n-1} a_n = \displaystyle \sum_{n=1}^{N} (-1)^{2n-1-1} a_{2n-1} + \sum_{n=1}^{N-1} (-1)^{2n-1} a_{2n}$

$= \displaystyle \sum_{n=1}^N a_{2n-1} - \sum_{n=1}^{N-1} a_{2n}$

$= \displaystyle \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^{N-1} \int_n^{n+1} \frac{dx}{x}$

$= \displaystyle \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x}$ .

Therefore,

$\displaystyle \lim_{N \to \infty} \left( \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right) = \gamma$ .

With this interpretation, the sum can be viewed as the sum of the $N$ rectangles in the above picture, while the integral is the area under the hyperbola. Therefore, the limit $\gamma$ can be viewed as the limit of the blue part of the above picture.

In other words, it’s an amazing fact that while both

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

and

$\displaystyle \int_1^\infty \frac{dx}{x}$

diverge, somehow the difference

$\displaystyle \lim_{N \to \infty} \left(\sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right)$

converges… and this limit is defined to be the number $\gamma$ .

Thoughts on Infinity (Part 3a)

Last summer, Math With Bad Drawings had a nice series on the notion of infinity that I recommend highly. This topic is a perennial struggle for math majors to grasp, and I like the approach that the author uses to sell this difficult notion.

Part 3 on infinite series and products that are conditionally convergent discusses a head-scratching fact: according to the Riemann series theorem, the commutative and associative laws do not apply to conditionally convergent series.

An infinite series $\displaystyle \sum_{n=1}^\infty a_n$ converges conditionally if it converges to a finite number but $\displaystyle \sum_{n=1}^\infty |a_n|$ diverges. Indeed, by suitably rearranging the terms, the sum can be changed so that the (rearranged) series converges to any finite value. Even worse, the terms can be rearranged so that the sum converges to either $\infty$ or $-\infty$ . (Of course, this can’t happen for finite sums, and rearrangements of an absolutely convergent series do not change the value of the sum.)

I really like Math With Bad Drawing’s treatment of the subject, as it starts with an infinite product for $\pi/2$ :

The top line is correct. However, the bottom line has to be incorrect since $\pi/2 > 1$ but each factor on the right-hand side is less than 1. The error, of course, stems from conditional convergence (the terms in the top product cannot be rearranged).

Conditional convergence is typically taught but glossed over in Calculus II since these rearrangements are such a head-scratching topic. I really like the above example because the flaw in the logic is made evidence after only three steps.

In tomorrow’s post, I’ll continue with another example of rearranging the terms in a conditionally convergent series.

Thoughts on Infinity (Part 2b)

Here’s Part 2 on the harmonic series, which is extremely well-written and which I recommend highly. Here’s a brief summary: the infinite harmonic series

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

diverges. However, if you eliminate from the harmonic series all of the fractions whose denominator contains a 9, then the new series converges! This series has been called the Kempner series, named after the mathematician who first published this result about 100 years ago.

Source: http://smbc-comics.com/index.php?id=3777

To prove this, we’ll examine the series whose denominators are between 1 and 8, between 10 and 88, between 100 and 888, etc. First, each of the terms in the partial sum

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$

is less than or equal to $1$ , and so the sum of the above eight terms must be less than $8$ .

Next, each of the terms in the sum

$\displaystyle \frac{1}{10} + \frac{1}{11} + \dots + \frac{1}{88}$

is less than $\displaystyle \frac{1}{10}$ . Notice that there are $72$ terms in this sum since there are 8 possibilities for the first digit of the denominator (1 through 8) and 9 possibilities for the second digit (0 through 8). So the sum of these 72 terms must be less than $\displaystyle 8 \times \frac{9}{10}$ .

Next, each of the terms in the sum

$\displaystyle \frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{888}$

is less than $\displaystyle \frac{1}{100}$ . Notice that there are $8 \times 9 \times 9$ terms in this sum since there are 8 possibilities for the first digit of the denominator (1 through 8) and 9 possibilities for the second and third digits (0 through 8). So the sum of these $8 \times 9 \times 9$ terms must be less than $8 \times \displaystyle \frac{9^2}{100}$ .

Continuing, we see that the Kempner series is bounded above by

$\displaystyle 8 + 8 \times \frac{9}{10} + 8 \times \frac{9^2}{10^2} + \dots$

Using the formula for an infinite geometric series, we see that the Kempner series converges, and the sum of the Kempner series must be less than $8 \times \displaystyle \frac{1}{1-9/10} = 80$ .

Using the same type of reasoning, much sharper bounds for the sum of the Kempner series can also be found. This 100-year-old article from the American Mathematical Monthly demonstrates that the sum of the Kempner series is between $22.4$ and $23.3$ . For more information about approximating the sum of the Kempner series, see Mathworld and Wikipedia.

It should be noted that there’s nothing particularly special about the number $9$ in the above discussion. If all denominators containing $314159265$ , or any finite pattern, are eliminated from the harmonic series, then the resulting series will always converge.

Thoughts on Infinity (Part 2a)

Here’s Part 2 on the harmonic series, which is extremely well-written and which I recommend highly. Here’s a brief summary: the infinite harmonic series

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

diverges. This is a perennial head-scratcher for students, as the terms become smaller and smaller yet the infinite series diverges.

To show this, notice that

$\displaystyle \frac{1}{3} + \frac{1}{4} > \displaystyle \frac{1}{4} + \frac{1}{4} = \displaystyle \frac{1}{2}$ ,

$\displaystyle \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \displaystyle \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \displaystyle \frac{1}{2}$ ,

and so on. Therefore,

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > \displaystyle 1 + \frac{1}{2} + \frac{1}{2} = 2$ ,

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} +\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \displaystyle 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \displaystyle \frac{5}{2}$ ,

and, in general,

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^n} > \displaystyle 1+\frac{n}{2}$ .

Since $\displaystyle \lim_{n \to \infty} \left(1 + \frac{n}{2} \right) = \infty$ , we can conclude that the harmonic series diverges.

However, here’s an amazing fact which I hadn’t known before the Math With Bad Drawings post: if you eliminate from the harmonic series all of the fractions whose denominator contains a 9, then the new series converges!

I’ll discuss the proof of this fact in tomorrow’s post. Until then, here’s a copy of the comic used in the Math With Bad Drawings post.

Source: http://smbc-comics.com/index.php?id=3777

Inverse Functions: Logarithms and Complex Numbers (Part 30)

Ordinarily, there are no great difficulties with logarithms as we’ve seen with the inverse trigonometric functions. That’s because the graph of $y = a^x$ satisfies the horizontal line test for any $0 < a < 1$ or $a > 1$ . For example,

$e^x = 5 \Longrightarrow x = \ln 5$ ,

and we don’t have to worry about “other” solutions.

However, this goes out the window if we consider logarithms with complex numbers. Recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$ , with $\theta$ in the appropriate quadrant. This is analogous to converting from rectangular coordinates to polar coordinates.

Over the past few posts, we developed the following theorem for computing $e^z$ in the case that $z$ is a complex number.

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$ . Then we define

$\log z = \ln r + i \theta$ .

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to $r e^{i \theta}$ . However, this complex logarithm doesn’t always work the way you’d think it work. For example,

$\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i$ .

This is analogous to another situation when an inverse function is defined using a restricted domain, like

$\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3$

$\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi$ .

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

$\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0$ ,

but

$\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i$ .

green line

This material appeared in my previous series concerning calculators and complex numbers: https://meangreenmath.com/2014/07/09/calculators-and-complex-numbers-part-21/

Calculators and complex numbers (Part 24)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$ , with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

Theorem. If $z = x + i y$ , where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$ . Then we define

$\log z = \ln r + i \theta$ .

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$ . Then we define

$z^w = e^{w \log z}$

Technical point: for the latter two definitions, these are the principal values of the functions. In complex analysis, these are usually considered multiply-defined functions. But I’m not going to worry about this technicality here and will only consider the principal values.

This is the last post in this series, where I state some generalizations of the Laws of Exponents for complex numbers.

In yesterday’s post, we saw that $z^{w_1} z^{w_2} = z^{w_1 + w_2}$ as long as $z \ne 0$ . This prevents something like $0^4 \cdot 0^{-3} = 0^1$ , since $0^{-3}$ is undefined.

Theorem. Let $z \in \mathbb{C} \setminus \{ 0 \}$ , $w \in \mathbb{C}$ , and $n \in \mathbb{Z}$ . Then $(z^w)^n = z^{wn}$ .

As we saw in a previous post, the conclusion could be incorrect outside of the above hypothesis, as $\displaystyle \left[ (-1)^3 \right]^{1/2} \ne (-1)^{3/2}$ .

Theorem. Let $u \in \mathbb{R}$ and $z \in \mathbb{C}$ . Then $(e^u)^z = e^{uz}$ .

Theorem. Let $x, y > 0$ be real numbers and $z \in \mathbb{C}$ . Then $x^z y^z = (xy)^z$ .

Again, the conclusion of the above theorem could be incorrect outside of these hypothesis, as $(-2)^{1/2} (-3)^{1/2} \ne \left[ (-2) \cdot (-3) \right]^{1/2}$ .

green line

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.