My Favorite One-Liners: Part 25

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Consider the integral

\displaystyle \int_0^2 2x(1-x^2)^3 \, dx

The standard technique — other than multiplying it out — is using the substitution u = 1-x^2. With this substitution du = -2x \, dx. Also, x = 0 corresponds to u = 1, while x = 2 corresponds to u = -3. Therefore,

\displaystyle\int_0^2 2x(1-x^2)^3 \, dx = - \displaystyle\int_0^2 (-2x)(1-x^2)^3 \, dx = -\displaystyle\int_1^{-3} u^3 \, du.

My one-liner at this point is telling my students, “At this point, about 10,000 volts of electricity should be going down your spine.” I’ll use this line when a very unexpected result happens — like a “left” endpoint that’s greater than the “right” endpoint. Naturally, for this problem, the next step — though not logically necessary, it’s psychologically reassuring — is to absorb the negative sign by flipping the endpoints:

\displaystyle\int_0^2 2x(1-x^2)^3 \, dx =  -\displaystyle\int_1^{-3} u^3 \, du = \displaystyle\int_{-3}^1 u^3 \, du,

and then the calculation can continue.

My Favorite One-Liners: Part 24

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that could appear in my class in probability or statistics:

Let f(x) = 3x^2 be a probability density function for 0 \le x \le 1. Find F(x) = P(X \le x), the cumulative distribution function of X.

A student’s first reaction might be to set up the integral as

\displaystyle \int_0^x 3x^2 \, dx

The problem with this set-up, of course, is that the letter x has already been reserved as the right endpoint for this definite integral. Therefore, inside the integral, we should choose any other letter — just not x — as the dummy variable.

Which sets up my one-liner: “In the words of the great philosopher Jean-Luc Picard: Plenty of letters left in the alphabet.”

We then write the integral as something like

\displaystyle \int_0^x 3t^2 \, dt

and then get on with the business of finding F(x).

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 5

Check out this lovely integral, dubbed the Sophomore’s Dream, found by Johann Bernoulli in 1697 (Gamma, page 44):

\displaystyle \int_0^1 \frac{dx}{x^x} = \displaystyle \frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \dots.

I’ll refer to either Wikipedia or Mathworld for the derivation.

green line

When I researching for my series of posts on conditional convergence, especially examples related to the constant \gamma, the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

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

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

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

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

 

 

How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction
Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by \sec^2 x, and the substitution u = \tan x.
Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution u = \tan \theta/2.
Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle
Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter a, the magic substitution u = \tan \theta/2, and partial fractions.
Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter a, the magic substitution u = \tan \theta/2, and contour integration using the real line and an expanding semicircle.
Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

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

log2series6

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. log2series7

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 nth 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),

or

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),

or

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 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

log2series1

 

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.

log2series2

Here’s the result after 2000 terms:

log2series3

20,000 terms:

log2series4

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

log2series5We 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)

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.

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 nth 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),

or

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,

or

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,

or

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.