Thoughts on Infinity (Part 3d)

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.

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

The antiderivative of 1/(x^4+1): Part 10

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

As we’ve seen in this series, the answer is

$\displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1) + C$

Also, as long as $x \ne 1$ and $x \ne -1$ , there is an alternative answer:

$\displaystyle \int \frac{1}{x^4 + 1} dx = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + C$ .

In this concluding post of this series, I’d like to talk about the practical implications of the assumptions that $x \ne 1$ and $x \ne -1$ .

For the sake of simplicity for the rest of this post, let

$F(x) = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} ( x\sqrt{2} - 1 ) + \frac{\sqrt{2}}{4} \tan^{-1}( x \sqrt{2} + 1)$

and

$G(x) = \displaystyle \int \frac{1}{x^4 + 1} dx = \displaystyle \frac{\sqrt{2}}{8} \ln \left(\frac{x^2 + x\sqrt{2} + 1}{x^2 - x\sqrt{2} + 1} \right) + \frac{\sqrt{2}}{4} \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ .

If I evaluate a definite integral of $\displaystyle \frac{1}{x^4+1}$ over an interval that contains neither $x = 1$ or $x = -1$ , then either $F$ or $G$ can be used. Courtesy of Mathematica:

However, if the region of integration contains either $x = -1$ or $x =1$ (or both), then only using $F$ returns the correct answer.

So this should be a cautionary tale about solving for angles, as the innocent-looking $+n\pi$ that appeared several posts ago ultimately makes a big difference in the final answers that are obtained.

The antiderivative of 1/(x^4+1): Part 9

In the course of evaluating the antiderivative

$\displaystyle \int \frac{1}{x^4 + 1} dx$ ,

I have stumbled across a very curious trigonometric identity:

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < x_2$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $x_2 < x < x_1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> x_1$ ,

where $x_1$ and $x_2$ are the unique values so that

$\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ ,

$\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}$ .

I will now show that $x_1 = 1$ and $x_2 = -1$ . Indeed, it’s apparent that these have to be the two transition points because these are the points where $\displaystyle \frac{x \sqrt{2}}{1 - x^2}$ is undefined. However, it would be more convincing to show this directly.

To show that $x_1 = 1$ , I need to show that

$\tan^{-1} (\sqrt{2} - 1 ) + \tan^{-1}( \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ .

I could do this with a calculator…

…but that would be cheating.

Instead, let $\alpha = \tan^{-1} (\sqrt{2} - 1 )$ and $\beta = \tan^{-1} (\sqrt{2} + 1 )$ , so that

$\tan \alpha = \sqrt{2} - 1$ ,

$\tan \beta = \sqrt{2} + 1$ .

Indeed, by SOHCAHTOA, the angles $\alpha$ and $\beta$ can be represented in the figure below:

The two small right triangles make one large triangle, and I will show that the large triangle is also a right triangle. To do this, let’s find the lengths of the three sides of the large triangle. The length of the longest side is clearly $\sqrt{2} - 1 + \sqrt{2} + 1 = 2\sqrt{2}$ . I will use the Pythagorean theorem to find the lengths of the other two sides. For the small right triangle containing $\alpha$ , the missing side is

$\sqrt{ \left(\sqrt{2} - 1 \right)^2 + 1^2} = \sqrt{2 - 2\sqrt{2} + 1 + 1} = \sqrt{4-2\sqrt{2}}$

Next, for the small right triangle containing $\beta$ , the missing side is

$\sqrt{ \left(\sqrt{2} + 1 \right)^2 + 1^2} = \sqrt{2 + 2\sqrt{2} + 1 + 1} = \sqrt{4+2\sqrt{2}}$

So let me redraw the figure, eliminating the altitude from the previous figure:

Notice that the condition of the Pythagorean theorem is satisfied, since

$\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = 4 - 2\sqrt{2} + 4 + 2 \sqrt{2} = 8$ ,

$\left( \sqrt{4-2\sqrt{2}} \right)^2 + \left( \sqrt{4+2\sqrt{2}} \right)^2 = \left( 2\sqrt{2} \right)^2$ .

Therefore, by the converse of the Pythagorean theorem, the above figure must be a right triangle (albeit a right triangle with sides of unusual length), and so $\alpha + \beta = \pi/2$ . In other words, $x_1 = 1$ , as required.

To show that $x_2 = -1$ , I will show that the function $f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ is an odd function using the fact that $\tan^{-1} x$ is also an odd function:

$f(-x) = \tan^{-1} ( -x\sqrt{2} - 1 ) + \tan^{-1}( -x \sqrt{2} + 1)$

$= \tan^{-1} ( -[x\sqrt{2} + 1] ) + \tan^{-1}( -[x \sqrt{2} - 1])$

$= -\tan^{-1} ( x\sqrt{2} + 1 ) - \tan^{-1}( x \sqrt{2} - 1)$

$= - \left[ \tan^{-1} ( x\sqrt{2} + 1 ) + \tan^{-1}( x \sqrt{2} - 1) \right]$

$= -f(x)$ .

Therefore, $f(-1) = -f(1) = -\displaystyle \frac{\pi}{2}$ , and so $x_2 = -1$ .

The antiderivative of 1/(x^4+1): Part 8

In the course of evaluating the antiderivative

$\displaystyle \int \frac{1}{x^4 + 1} dx$ ,

I’ve accidentally stumbled on a very curious looking trigonometric identity:

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < -1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $-1 < x < 1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> 1$ .

The extra $-\pi$ and $\pi$ are important. Without them, the graphs of the left-hand side and right-hand sides are clearly different if $x < -1$ or $x > 1$ :

However, they match when those constants are included:

Let’s see if I can explain why this trigonometric identity occurs without resorting to the graphs.

Since $\tan^{-1} x$ assumes values between $-\pi/2$ and $\pi/2$ , I know that

$-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) < \frac{\pi}{2}$ ,

$-\displaystyle \frac{\pi}{2} < \tan^{-1} ( x\sqrt{2} + 1 ) < \frac{\pi}{2}$ ,

and so

$-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi$ .

However,

$-\displaystyle \frac{\pi}{2} < \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2}$ ,

and so $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )$ and $\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ must differ if $\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1)$ is in the interval $[-\pi,-\pi/2]$ or in the interval $[\pi/2,\pi]$ .

I also notice that

$-\pi< \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 ) < \pi$ ,

$-\displaystyle \frac{\pi}{2} < -\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{\pi}{2}$ ,

and so

$-\displaystyle \frac{3\pi}{2} < \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1} ( x\sqrt{2} + 1 )-\tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) < \frac{3\pi}{2}$ .

However, this difference can only be equal to a multiple of $\pi$ , and there are only three multiples of $\pi$ in the interval $\displaystyle \left( -\frac{3\pi}{2}, \frac{3\pi}{2} \right)$ , namely $-\pi$ , $0$ , and $\pi$ .

To determine the values of $x$ where this happens, I also note that $f_1(x) = x \sqrt{2} - 1$ , $f_2(x) = x \sqrt{2} + 1$ , and $f_3(x) = \tan^{-1} x$ are increasing functions, and so $f(x) = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ must also be an increasing function. Therefore, to determine where $f(x)$ lies in the interval $[\pi/2,\pi]$ ,it suffices to determine the unique value $x_1$ so that $f(x_1) = \pi/2$ . Likewise, to determine where $f(x)$ lies in the interval $[-\pi,-\pi/2]$ ,it suffices to determine the unique value $x_2$ so that $f(x_2) = -\pi/2$ .

In summary, I have shown so far that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < x_2$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $x_2 < x < x_1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> x_1$ ,

where $x_1$ and $x_2$ are the unique values so that

$\tan^{-1} ( x_1\sqrt{2} - 1 ) + \tan^{-1}( x_1 \sqrt{2} + 1) = \displaystyle \frac{\pi}{2}$ ,

$\tan^{-1} ( x_2\sqrt{2} - 1 ) + \tan^{-1}( x_2 \sqrt{2} + 1) = -\displaystyle \frac{\pi}{2}$ .

So, to complete the proof of the trigonometric identity, I need to show that $x_1 = 1$ and $x_2 = -1$ . I will do this in tomorrow’s post.

The antiderivative of 1/(x^4+1): Part 7

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

As we’ve seen in this series, the answer is

It turns out that this can be simplified somewhat as long as $x \ne 1$ and $x \ne -1$ . I’ll use the trig identity

$\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

When I apply this trig identity for $\alpha = \tan^{-1} ( x\sqrt{2} - 1 )$ and $\beta = \tan^{-1} ( x\sqrt{2} + 1 )$ , I obtain

$\tan \left[ \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) \right] = \displaystyle \frac{x \sqrt{2} - 1 + x \sqrt{2} + 1}{1 - (x\sqrt{2} - 1)(x\sqrt{2} + 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{1 - (2x^2 - 1)}$

$= \displaystyle \frac{2x \sqrt{2}}{2 - 2x^2}$

$= \displaystyle \frac{x \sqrt{2}}{1 - x^2}$ .

So we can conclude that

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + n\pi$

for some integer $n$ that depends on $x$ . The $+n\pi$ is important, as a cursory look reveals that $y = \tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1)$ and $y = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ have different graphs. (The vertical lines in the orange graph indicate where the right-hand side is undefined when $x = 1$ or $x = -1$ .

The two graphs coincide when $-1 < x < 1$ but differ otherwise. However, it appears that the two graphs differ by a constant. Indeed, if I subtract $\pi$ from the orange graph if $x < -1$ and add $\pi$ to the orange graph if $x > 1$ , then they match:

So, evidently

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) -\pi$ if $x < -1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right)$ if $-1 < x < 1$ ,

$\tan^{-1} ( x\sqrt{2} - 1 ) + \tan^{-1}( x \sqrt{2} + 1) = \tan^{-1} \left( \displaystyle \frac{x \sqrt{2}}{1 - x^2} \right) + \pi$ if $x> 1$ .

So as long as $x \ne 1$ and $x \ne -1$ , this constant $-\pi$ , $0$ , or $\pi$ can be absorbed into the constant $C$ :

However, a picture may be persuasive but is not a proof, and there are some subtle issues with this simplification. I’ll discuss these further details in tomorrow’s post.