College Math Teaching

February 5, 2016

More fun with selective sums of divergent series

Just a reminder: if \sum_{k=1}^{\infty} a_k is a series and c_1, c_2, ...c_n ,, is some sequence consisting of 0’s and 1’s then a selective sum of the series is \sum_{k=1}^{\infty} c_k a_k . The selective sum concept is discussed in the MAA book Real Infinite Series (MAA Textbooks) by Bonar and Khoury (2006) and I was introduced to the concept by Ferdinands’s article Selective Sums of an Infinite Series in the June 2015 edition of Mathematics Magazine (Vol. 88, 179-185).

There is much of interest there, especially if one considers convergent series or alternating series.

This post will be about divergent series of positive terms for which lim_{n \rightarrow \infty} a_n = 0 and a_{n+1} < a_n for all n .

The first fun result is this one: any selected x > 0 is a selective sum of such a series. The proof of this isn’t that bad. Since lim_{n \rightarrow \infty} a_n = 0 we can find a smallest n such that a_n \leq x . Clearly if a_n = x we are done: our selective sum has c_n = 1 and the rest of the c_k = 0 .

If not, set n_1 = n and note that because the series diverges, there is a largest m_1 so that \sum_{k=n_1}^{m_1} a_k \leq x . Now if \sum_{k=n_1}^{m_1} a_k = x we are done, else let \epsilon_1 = x - \sum_{k=n_1}^{m_1} a_k and note \epsilon_1 < a_{m_1+1} . Now because the a_k tend to zero, there is some first n_2 so that a_{n_2} \leq \epsilon_1 . If this is equality then the required sum is a_{n_2} + \sum_{k=n_1}^{m_1} a_k , else we can find the largest m_2 so that \sum_{k=n_1}^{m_1} a_k + \sum_{k=n_2}^{m_2} a_k \leq x

This procedure can be continued indefinitely. So if we label \sum_{k=n_j}^{m_{j}} a_k = s_j we see that s_1 + s_2 + ...s_{n} = t_{n} form an increasing, bounded sequence which converges to the least upper bound of its range, and it isn’t hard to see that the least upper bound is x because x-t_{n} =\epsilon_n < a_{m_n+1}

So now that we can obtain any positive real number as the selective sum of such a series, what can we say about the set of all selective sums for which almost all of the c_k = 0 (that is, all but a finite number of the c_k are zero).

Answer: the set of all such selective sums are dense in the real line, and this isn’t that hard to see, given our above construction. Let (a,b) be any open interval in the real line and let a < x < b . Then one can find some N such that for all n > N we have x - a_n > a . Now consider our construction and choose m large enough such that x - t_m > x - a_n > a . Then the t_m represents the finite selected sum that lies in the interval (a,b) .

We can be even more specific if we now look at a specific series, such as the harmonic series \sum_{k=1}^{\infty} \frac{1}{k} . We know that the set of finite selected sums forms a dense subset of the real line. But it turns out that the set of select sums is the rationals. I’ll give a slightly different proof than one finds in Bonar and Khoury.

First we prove that every rational in (0,1] is a finite select sum. Clearly 1 is a finite select sum. Otherwise: Given \frac{p}{q} we can find the minimum n so that \frac{1}{n} \leq \frac{p}{q} < \frac{1}{n-1} . If \frac{p}{q} = \frac{1}{n} we are done. Otherwise: the strict inequality shows that pn-p < q which means pn-q < p . Then note \frac{p}{q} - \frac{1}{n} = \frac{pn-q}{qn} and this fraction has a strictly smaller numerator than p . So we can repeat our process with this new rational number. And this process must eventually terminate because the numerators generated from this process form a strictly decreasing sequence of positive integers. The process can only terminate when the new faction has a numerator of 1. Hence the original fraction is some sum of fractions with numerator 1.

Now if the rational number r in question is greater than one, one finds n_1 so that \sum^{n_1}_{k=1} \frac{1}{k} \leq r but \sum^{n_1+1}_{k=1} \frac{1}{k} > r . Then write r-\sum^{n_1+1}_{k=1} \frac{1}{k} and note that its magnitude is less than \frac{1}{n_1+1} . We then use the procedure for numbers in (0,1) noting that our starting point excludes the previously used terms of the harmonic series.

There is more we can do, but I’ll stop here for now.

January 26, 2016

More Fun with Divergent Series: redefining series convergence (Ces├áro, etc.)

Filed under: analysis, calculus, sequences, series — Tags: , , — collegemathteaching @ 10:21 pm

This post is more designed to entertain myself than anything else. This builds up a previous post which talks about deleting enough terms from a divergent series to make it a convergent one.

This post is inspired by Chapter 8 of Konrad Knopp’s classic Theory and Application of Infinite Series. The title of the chapter is Divergent Series.

Notation: when I talk about a series converging, I mean “converging” in the usual sense; e. g. if s_n = \sum_{k=0}^{k=n} a_k and lim_{n \rightarrow \infty}s_n = s then \sum_{k=0}^{\infty} a_k is said to be convergent with sum s .

All of this makes sense since things like limits are carefully defined. But as Knopp points out, in the “days of old”, mathematicians say these series as formal objects rather than the result of careful construction. So some of these mathematicians (like Euler) had no problem saying things like \sum^{\infty}_{k=0} (-1)^k = 1-1+1-1+1..... = \frac{1}{2} . Now this is complete nonsense by our usual modern definition. But we might note that \frac{1}{1-x} = \sum^{\infty}_{k=0} x^k for -1 < x < 1 and note that x = -1 IS in the domain of the left hand side.

So, is there a way of redefining the meaning of “infinite sum” that gives us this result, while not changing the value of convergent series (defined in the standard way)? As Knopp points out in his book, the answer is “yes” and he describes several definitions of summation that

1. Do not change the value of an infinite sum that converges in the traditional sense and
2. Allows for more series to coverge.

We’ll discuss one of these methods, commonly referred to as Cesàro summation. There are ways to generalize this.

How this came about

Consider the Euler example: 1 -1 + 1 -1 + 1 -1...... . Clearly, s_{2k} = 1, s_{2k+1} = 0 and so this geometric series diverges. But notice that the arithmetic average of the partial sums, computed as c_n = \frac{s_0 + s_1 +...+s_n}{n+1} does tend to \frac{1}{2} as n tends to infinity: c_{2n} = \frac{\frac{2n}{2}}{2n+1} = \frac{n}{2n+1} whereas c_{2n+1} = \frac{\frac{2n}{2}}{2n+2} =\frac{n}{2n+2} and both of these quantities tend to \frac{1}{2} as n tends to infinity.

So, we need to see that this method of summing is workable; that is, do infinite sums that converge in the previous sense still converge to the same number with this method?

The answer is, of course, yes. Here is how to see this: Let x_n be a sequence that converges to zero. Then for any \epsilon > 0 we can find M such that k > M implies that |x_k| < \epsilon . So for n > k we have \frac{x_1 + x_2 + ...+ x_{k-1} + x_k + ...+ x_n}{n} = \frac{x_1+ ...+x_{k-1}}{n} + \frac{x_k + x_{k+1} + ....x_n}{n} Because k is fixed, the first fraction tends to zero as n tends to infinity. The second fraction is smaller than \epsilon in absolute value. But \epsilon is arbitrary, hence this arithmetic average of this null sequence is itself a null sequence.

Now let x_n \rightarrow L and let c_n = \frac{x_1 + x_2 + ...+ x_{k-1} + x_k + ...+ x_n}{n} Now subtract note c_n-L =  \frac{(x_1-L) + (x_2-L) + ...+ (x_{k-1}-L) +(x_k-L) + ...+ (x_n-L)}{n} and the x_n-L forms a null sequence. Then so do the c_n-L .

Now to be useful, we’d have to show that series that are summable in the Cesàro obey things like the multiplicative laws; they do but I am too lazy to show that. See the Knopp book.

I will mention a couple of interesting (to me) things though. Neither is really profound.

1. If a series diverges to infinity (that is, if for any positive M there exists n such that for all k \geq n, s_k > M , then this series is NOT Cesàro summable. It is relatively easy to see why: given such an M, k then consider \frac{s_1 + s_2 + s_3 + ...+s_{k-1} + s_k + s_{k+1} + ...s_n}{n} = \frac{s_1+ s_2 + ...+s_{k-1}}{n} + \frac{s_k + s_{k+1} .....+s_{n}}{n} which is greater than \frac{n-k}{n} M for large n . Hence the Cesàro partial sum becomes unbounded.

Upshot: there is no hope in making something like \sum^{\infty}_{n=1} \frac{1}{n} into a convergent series by this method. Now there is a way of making an alternating, divergent series into a convergent one via doing something like a “double Cesàro sum” (take arithmetic averages of the arithmetic averages) but that is a topic for another post.

2. Cesàro summation may speed up convergent of an alternating series which passes the alternating series test, OR it might slow it down. I’ll have to develop this idea more fully. But I invite the reader to try Cesàro summation for \sum^{\infty}_{k=1} (-1)^{k+1} \frac{1}{k} and on \sum^{\infty}_{k=1} (-1)^{k+1} \frac{1}{k^2} and on \sum^{\infty}_{k=0} (-1)^k \frac{1}{2^k} . In the first two cases, the series converges slowly enough so that Cesàro summation speeds up convergence. Cesàro slows down the convergence in the geometric series though. It is interesting to ponder why.

January 14, 2016

Trimming a divergent series into a convergent one

Filed under: calculus, induction, sequences, series — Tags: , , — collegemathteaching @ 10:28 pm

This post is motivated by this cartoon
harmonic-series

which I found at a Evelyn Lamb’s post on an AMS Blog, this fun Forbes math post by Kevin Kundson and by a June 2015 article in Mathematics Magazine by R. John Ferdinands called Selective Sums of an Infinite Series.

Here is the following question: start with a divergent series of positive terms which form a decreasing (non-increasing) sequence which tends to zero, say, \sum^{\infty}_{k =1} \frac{1}{k} . Now how does one select a subset of series terms to delete so as to obtain a convergent series? The Kundson article shows that one can do this with the harmonic series by, say, deleting all numbers that contain a specific digit (say, 9). I’ll talk about the proof here. But I’d like to start more basic and to bring in language used in the Ferdinands article.

So, let’s set the stage: we will let \sum a_k denote the divergent sum in question. All terms will be positive, a_{k} \geq a_{k+1} for all k and lim_{k \rightarrow \infty} a_k = 0 . Now let c_k represent a sequence where c_k \in \{0,1\} for all k ; then \sum c_ka_k is called a selective sum of \sum a_k . I’ll call the c_k the selecting sequence and, from the start, rule out selecting sequences that are either eventually 1 (which means that the selected series diverges since the original series did) or eventually zero (just a finite sum).

Now we’ll state a really easy result:

There is some non-eventually constant c_k such that \sum c_ka_k converges. Here is why: because lim_{k \rightarrow \infty} a_k = 0 , for each n \in \{1,2,3...\} one can find a maximal index n_j, n_j \notin \{n_1, n_2, ...n_{j-1} \} so that \frac{1}{2^n} > a_{n_j} . Now select c_k = 1 if k \in \{n_1, n_2, n_3,... \} and c_k =0 otherwise. Then \sum \frac{1}{2^k} > \sum c_ka_k and therefore the selected series converges by comparison with a convergent geometric series.

Of course, this result is petty lame; this technique discards a lot of terms. A cheap way to discard “fewer” terms (“fewer” meaning: in terms of “set inclusion”): Do the previous construction, but instead of using \frac{1}{2} use \frac{M}{M+1} where M is a positive integer of choice. Note that \sum^{\infty}_{k=1} (\frac{M}{M+1})^k = M

Here is an example of how this works: Consider the divergent series \sum \frac{1}{\sqrt{k}} and the convergent geometric series \sum (\frac{1000}{1001})^k Of course \frac{1000}{1001} < 1 so c_1 = 0 but then for k \in \{2,3,....4169 \} we have (\frac{1000}{1001})^k > \frac{1}{\sqrt{k}} . So c_k = 1 for k \in \{2,3,4,....4169 \} . But c_{4170} = 0 because (\frac{1000}{1001})^{4170} < \frac{1}{\sqrt{4170}}. The next non-zero selection coefficient is c_{4171} as (\frac{1000}{1001})^{4170} > \frac{1}{\sqrt{4171}} .

Now playing with this example, we see that \frac{1}{\sqrt{k}} > (\frac{1000}{1001})^{4171} for k \in \{4172, 4173,....4179 \} but not for k = 4180 . So c_k = 0 for k \in \{4172,....4179 \} and c_{4180} = 1 . So the first few n_j are \{2, 3, ....4169, 4171, 4180 \} . Of course the gap between the n_j grows as k does.

Now let’s get back to the cartoon example. From this example, we’ll attempt to state a more general result.

Claim: given \sum^{\infty}_{k=1} c_k \frac{1}{k} where c_k = 0 if k contains a 9 as one of its digits, then \sum^{\infty}_{k=1} c_k \frac{1}{k} converges. Hint on how to prove this (without reading the solution): count the number of integers between 10^k and 10^{k+1} that lack a 9 as a digit. Then do a comparison test with a convergent geometric series, noting that every term \frac{1}{10^k}, \frac{1}{10^k + 1}......, \frac{1}{8(10^k) +88} is less than or equal to \frac{1}{10^k} .

How to prove the claim: we can start by “counting” the number of integers between 0 and 10^k that contain no 9’s as a digit.

Between 0 and 9: clearly 0-8 inclusive, or 9 numbers.

Between 10 and 99: a moment’s thought shows that we have 8(9) = 72 numbers with no 9 as a digit (hint: consider 10-19, 20-29…80-89) so this means that we have 9 + 8(9) = 9(1+8) = 9^2 numbers between 0 and 99 with no 9 as a digit.

This leads to the conjecture: there are 9^k numbers between 0 and 10^k -1 with no 9 as a digit and (8)9^{k-1} between 10^{k-1} and 10^k-1 with no 9 as a digit.

This is verified by induction. This is true for k = 1

Assume true for k = n . Then to find the number of numbers without a 9 between 10^n and 10^{n+1} -1 we get 8 (9^n) which then means we have 9^n + 8(9^n) = 9^n (8+1) = 9^{n+1} numbers between 0 and 10^{n+1}-1 with no 9 as a digit. So our conjecture is proved by induction.

Now note that 0+ 1 + \frac{1}{2} + ....+ \frac{1}{8} < 8*1*1

\frac{1}{10} + ...+ \frac{1}{18} + \frac{1}{20} + ...+ \frac{1}{28} + \frac{1}{30} + ...+ \frac{1}{88} < 8*9*\frac{1}{10}

\frac{1}{100} + ...\frac{1}{88} + \frac{1}{200} + ....\frac{1}{888} < 8*(9^2)\frac{1}{100}

This establishes that \sum_{k=10^n}^{10^{n+1}-1} c_k \frac{1}{k} < 8*(9^k)\frac{1}{10^k}

So it follows that \sum^{\infty}_{k=1} c_k \frac{1}{k} < 8\sum^{\infty}{k=0} (\frac{9}{10})^k = 8 \frac{1}{1-\frac{9}{10}} = 80 and hence our selected sum is convergent.

Further questions: ok, what is going on is that we threw out enough terms of the harmonic series for the series to converge. Between terms \frac{1}{10^k} and \frac{1}{10^{k+1}-1} we allowed 8*(9^k) terms to survive.

This suggests that if we permit up to M (10-\epsilon)^k terms between 10^k and 10^{k+1}-1 to survive (M, \epsilon fixed and positive) then we will have a convergent series. I’d be interested in seeing if there is an generalization of this.

But I am tried, I have a research article to review and I need to start class preparation for the upcoming spring semester. So I’ll stop here. For now.:-)

December 22, 2015

Multi leaf polar graphs and total area…

Filed under: calculus, elementary mathematics, integrals — Tags: , — collegemathteaching @ 4:07 am

I saw polar coordinate calculus for the first time in 1977. I’ve taught calculus as a TA and as a professor since 1987. And yet, I’ve never thought of this simple little fact.

Consider r(\theta) = sin(n \theta), 0 \theta \ 2 \pi . Now it is well know that the area formula (area enclosed by a polar graph, assuming no “doubling”, self intersections, etc.) is A = \frac{1}{2} \int^b_a (r(\theta))^2 d \theta

Now the leaved roses have the following types of graphs: n leaves if n is odd, and 2n leaves if n is even (in the odd case, the graph doubles itself).

3leafedrose

4leafrose

6leafedrose

So here is the question: how much total area is covered by the graph (all the leaves put together, do NOT count “overlapping”)?

Well, for n an integer, the answer is: \frac{\pi}{4} if n is odd, and \frac{\pi}{2} if n is even! That’s it! Want to know why?

Do the integral: if n is odd, our total area is \frac{n}{2}\int^{\frac{\pi}{n}}_0 (sin(n \theta)^2 d\theta = \frac{n}{2}\int^{\frac{\pi}{n}}_0 \frac{1}{2} + cos(2n\theta) d\theta =\frac{\pi}{4} . If n is even, we have the same integral but the outside coefficient is \frac{2n}{2} = n which is the only difference. Aside from parity, the number of leaves does not matter as to the total area!

Now the fun starts when one considers a fractional multiple of \theta and I might ponder that some.

October 29, 2015

The Alternating Series Test: the need for hypothesis

Filed under: calculus, series — Tags: — collegemathteaching @ 9:49 pm

It is well known that if series \sum a_k meets the following conditions:

1. (a_k)(a_{k+1}) < 0 for all k
2. lim_{k \rightarrow \infty} a_k = 0
3. |a_k| > |a_{k+1} | for all k

the series converges. This is the famous “alternating series test”.

I know that I am frequently remiss in discussing what can go wrong if condition 3 is not met.

An example that is useful is 1 - \frac{1}{\sqrt{2}} + \frac{1}{3} - \frac{1}{\sqrt{4}} + ...+\frac{1}{2n-1} - \frac{1}{\sqrt{2n}} .....

Clearly this series meets conditions 1 and 2: the series alternates and the terms approach zero. But the series can be written (carefully) as:

\sum_{k=1}^{\infty} (\frac{1}{2k-1} - \frac{1}{\sqrt{2k}}) .

Then one can combine the terms in the parenthesis and then do a limit comparison to the series \sum_{k=1}^{\infty} \frac{1}{k} to see the series diverges.

July 13, 2015

Trolled by Newton’s Law of Cooling…

Filed under: calculus, differential equations, editorial — Tags: , — collegemathteaching @ 8:55 pm

From a humor website: there is a Facebook account called “customer service” who trolls customers making complaints. Though that isn’t a topic here, it is interesting to see Newton’s Cooling Law get mentioned:

newtonscoolinglaw

June 9, 2015

Volumes of n-balls: why?

Filed under: calculus, geometry — Tags: , , — collegemathteaching @ 11:38 am

I talked about a method of finding the hypervolume of an n-ball in a previous post. To recap: the volume of the n-ball is that “hypervolume” (I’ll be dropping “hyper”) of the region described by \{(x_1, x_2,...x_n) | x_1^2 + x_2^2 + ...x_n^2 \leq R^2 \} .

The formula is: V_n = \frac{\pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2} + 1]} R^n

Here, we’ll explore this topic further, both giving a different derivation (from Greg Huber’s American Mathematical Monthly paper) and make some informal observations.

Derivation: the argument I present can be tweaked to produce a formal induction argument that, if the volume is V_n, it is proportional to the n’th power of the radius R .

Now note that if the surface area of the n-1 sphere is given by W_{n-1} = w_{n-1}R^{n-1} we have, from the theory of differentials, \frac{dV_n}{dR} = W_{n-1}  . Think of taking a sphere and adding just a bit \Delta R to its radius; you obtain a shell of thickness \Delta R all the way around the sphere which is roughly equal to the surface area times \Delta R

sphereshell

So we can rewrite this as V_n = \int^R_0 W_{n-1} dr = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr

To see what comes next, we first write this same quantity in two different ways:

\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1} dx_i = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr  = \int \int \int ...\int^R_{0} r^{n-1} J(\theta_1, ..\theta_{n-1}) dr .

The first integral is integral in rectangular coordinates within the boundary of the n-1 sphere. The rightmost integral is the same integral in generalized spherical coordinates (see Blumenson’s excellent Monthly article) where the first iterated integrals are those with angular limits with J being the angular volume element. The middle integral is the volume integral. All are equal to the volume of the n-ball. The key here is that the iterated integrals evaluated over the entire n-1 sphere are equal to w_{n-1} .

Now integrate e^{-r^2} over the region bounded by the sphere r^2 = x_1^2 + x_2^2 + ...x_n^2 , noting that e^{-r^2} = e^{-x_1^2}e^{-x_2^2}...e^{-x_n^2} :

\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i  = w_{n-1}\int^R_0e^{-r^2}r^{n-1} dr  = \int \int \int ...\int^R_{0} r^{n-1}e^{-r^2}J(\theta_1, ..\theta_{n-1}) dr

Equality holds between the middle and right integral because in “angle/r” space, the r and angular coordinates are independent. Equality between the leftmost and rightmost integrals holds because this is a mere change of variables.
So we can now drop the rightmost integral. Now take a limit as R \rightarrow \infty :

\int \int \int ...\int_{R^{n}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i  = (\int^{\infty}_{-\infty} e^{-x^2} dx)^n = w_{n-1}\int^{\infty}_0e^{-r^2}r^{n-1} dr

The left integral is just the n-th power of the Gaussian integral and is therefore \pi^{\frac{n}{2}} and a substitution r = \sqrt{u} turns this into \frac{w_{n-1}}{2}\int^{\infty}_{0} u^{\frac{1}{2} -1}e^{-u}du = w_{n-1} \frac{1}{2}\Gamma[\frac{1}{2}] (recall \Gamma[x] = \int^{\infty}_{0} t^{x-1}e^{-t} dt ).

So w_{n-1} = \frac{2 \pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2}]} and hence, by integration, v_n = \frac{2 \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}]}= \frac{ \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}+1]}

Now v_n =V_n when R = 1 . \frac{v_n}{2^n} can be thought of as the percentage of the cube with vertexes (\pm 1, \pm 1, ... \pm 1) that is taken up by the inscribed unit sphere.

Now we set R = 1 at look at a graph of hypervolume vs. n:

The first graph is the ratio of the volume taken up by the ball verses the hypervolume of the hyper cube that the ball is inscribed in.

volpercentcube

Next we see the the hypervolume peaks at n = 5 (max is between 5 and 6 )and then starts to decline to zero. Of course there has to be an inflection point somewhere; it turns out to be between n = 10 and n = 11.

volumenballgraph

Now we plot the hyperarea of the n-1 sphere vs. the hypervolume of the ball that it bounds; we see that more and more of the hypervolume of the ball is concentrated near the boundary as the dimension goes up.

volvsareawithn

For more: see the interesting discussion on Division by Zero.

May 31, 2015

And a Fields Medalist makes me feel better

Filed under: calculus, editorial, elementary mathematics, popular mathematics, topology — Tags: — collegemathteaching @ 10:30 pm

I have subscribed to Terence Tao’s blog.

His latest post is about a clever observation about…calculus: in particular is is about calculating:

\frac{d^{k+1}}{dx^{k+1}}(1+x^2)^{\frac{k}{2}} for k \in \{1, 2, 3, ... \} . Try this yourself and surf to his post to see the “slick, 3 line proof”.

But that really isn’t the point of this post.

This is the point: I often delight in finding something “fun” and “new to me” about an established area. I thought “well, that is because I am too dumb to do the really hard stuff.” (Yes, I’ve published, but my results are not Annals of Mathematics caliber stuff.:-) )

But I see that even the smartest, most accomplished among us can delight in the fun, simple things.

That makes me feel better.

Side note: I haven’t published much on this blog lately, mostly because I’ve been busy updating this one. It is a blog giving notes for my undergraduate topology class. That class was time consuming, but I had the teaching time of my life. I hope that my students enjoyed it too.

May 11, 2015

The hypervolume of the n-ball enclosed by a standard n-1 sphere

I am always looking for interesting calculus problems to demonstrate various concepts and perhaps generate some interest in pure mathematics.
And yes, I like to “blow off some steam” by spending some time having some non-technical mathematical fun with elementary mathematics.

This post uses only:

1. Integration by parts and basic reduction formulas.
2. Trig substitution.
3. Calculation of volumes (and hyper volumes) by the method of cross sections.
4. Induction
5. Elementary arithmetic involving factorials.

The quest: find a formula that finds the (hyper)volume of the region \{(x_1, x_2, x_3,....x_k) | \sum_{i=1}^k x_i^2 \leq R^2 \} \subset R^k

We will assume that the usual tools of calculus work as advertised.

Start. If we done the (hyper)volume of the k-ball by V_k  we will start with the assumption that V_1 = 2R ; that is, the distance between the endpoints of [-R,R] is 2R.

Step 1: we show, via induction, that V_k =c_kR^k where c_k is a constant and R is the radius.

Our proof will be inefficient for instructional purposes.

We know that V_1 =2R hence the induction hypothesis holds for the first case and c_1 = 2 . We now go to show the second case because, for the beginner, the technique will be easier to follow further along if we do the k = 2 case.

Yes, I know that you know that V_2 = \pi R^2 and you’ve seen many demonstrations of this fact. Here is another: let’s calculate this using the method of “area by cross sections”. Here is x^2 + y^2 = R^2 with some y = c cross sections drawn in.

crosssections

Now do the calculation by integrals: we will use symmetry and only do the upper half and multiply our result by 2. At each y = y_c level, call the radius from the center line to the circle R(y) so the total length of the “y is constant” level is 2R(y) and we “multiply by thickness “dy” to obtain V_2 = 4 \int^{y=R}_{y=0} R(y) dy .

But remember that the curve in question is x^2 + y^2 = R^2 and so if we set x = R(y) we have R(y) = \sqrt{R^2 -y^2} and so our integral is 4 \int^{y=R}_{y=0}\sqrt{R^2 -y^2}  dy

Now this integral is no big deal. But HOW we solve it will help us down the road. So here, we use the change of variable (aka “trigonometric substitution”): y = Rsin(t), dy =Rcos(t) to change the integral to:

4 \int^{\frac{\pi}{2}}_0 R^2 cos^2(t) dt = 4R^2 \int^{\frac{\pi}{2}}_0  cos^2(t) dt therefore

V_2 = c_2 R^2 where:

c_2 = 4\int^{\frac{\pi}{2}}_0  cos^2(t)

Yes, I know that this is an easy integral to solve, but I first presented the result this way in order to make a point.

Of course, c_2 = 4\int^{\frac{\pi}{2}}_0  cos^2(t) = 4\int^{\frac{\pi}{2}}_0 \frac{1}{2} + \frac{1}{2}cos(2t) dt = \pi

Therefore, V_2 =\pi R^2 as expected.

Exercise for those seeing this for the first time: compute c_3 and V_3 by using the above methods.

Inductive step: Assume V_k = c_kR^k Now calculate using the method of cross sections above (and here we move away from x-y coordinates to more general labeling):

V_{k+1} = 2\int^R_0 V_k dy = 2 \int^R_0 c_k (R(x_{k+1})^k dx_{k+1} =c_k 2\int^R_0 (R(x_{k+1}))^k dx_{k+1}

Now we do the substitutions: first of all, we note that x_1^2 + x_2^2 + ...x_{k}^2 + x_{k+1}^2 = R^2 and so

x_1^2 + x_2^2 ....+x_k^2 = R^2 - x_{k+1}^2 . Now for the key observation: x_1^2 + x_2^2 ..+x_k^2 =R^2(x_{k+1}) and so R(x_{k+1}) = \sqrt{R^2 - x_{k+1}^2}

Now use the induction hypothesis to note:

V_{k+1} = c_k 2\int^R_0 (R^2 - x_{k+1}^2)^{\frac{k}{2}} dx_{k+1}

Now do the substitution x_{k+1} = Rsin(t), dx_{k+1} = Rcos(t)dt and the integral is now:

V_{k+1} = c_k 2\int^{\frac{\pi}{2}}_0 R^{k+1} cos^{k+1}(t) dt = c_k(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt)R^{k+1} which is what we needed to show.

In fact, we have shown a bit more. We’ve shown that c_1 = 2 =2 \int^{\frac{\pi}{2}}_0(cos(t))dt, c_2 = 2 \cdot 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt = c_1 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt and, in general,

c_{k+1} = c_{k}c_{k-1}c_{k-2} ....c_1(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt) = 2^{k+1} \int^{\frac{\pi}{2}}_0(cos^{k+1}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k-1}(t))dt .....\int^{\frac{\pi}{2}}_0(cos(t))dt

Finishing the formula

We now need to calculate these easy calculus integrals: in this case the reduction formula:

\int cos^n(x) dx = \frac{1}{n}cos^{n-1}sin(x) + \frac{n-1}{n} \int cos^{n-2}(x) dx is useful (it is merely integration by parts). Now use the limits and elementary calculation to obtain:

\int^{\frac{\pi}{2}}_0 cos^n(x) dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_0 cos^{n-2}(x)dx to obtain:

\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{3}{4})\frac{\pi}{4} if n is even and:
\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{4}{5})\frac{2}{3} if n is odd.

Now to come up with something resembling a closed formula let’s experiment and do some calculation:

Note that c_1 = 2, c_2 = \pi, c_3 = \frac{4 \pi}{3}, c_4 = \frac{(\pi)^2}{2}, c_5 = \frac{2^3 (\pi)^2)}{3 \cdot 5} = \frac{8 \pi^2}{15}, c_6 = \frac{\pi^3}{3 \cdot 2} = \frac{\pi^3}{6} .

So we can make the inductive conjecture that c_{2k} = \frac{\pi^k}{k!} and see how it holds up: c_{2k+2} = 2^2 \int^{\frac{\pi}{2}}_0(cos^{2k+2}(t))dt \int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}

= 2^2 ((\frac{2k+1}{2k+2})(\frac{2k-1}{2k})......(\frac{3}{4})\frac{\pi}{4})((\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......\frac{2}{3})\frac{\pi^k}{k!}

Now notice the telescoping effect of the fractions from the c_{2k+1} factor. All factors cancel except for the (2k+2) in the first denominator and the 2 in the first numerator, as well as the \frac{\pi}{4} factor. This leads to:

c_{2k+2} = 2^2(\frac{\pi}{4})\frac{2}{2k+2} \frac{\pi^k}{k!} = \frac{\pi^{k+1}}{(k+1)!} as required.

Now we need to calculate c_{2k+1} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt c_{2k} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}

= 2 (\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......(\frac{4}{5})\frac{2}{3}\frac{\pi^k}{k!} = 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \frac{\pi^k}{k!}

To simplify this further: split up the factors of the k! in the denominator and put one between each denominator factor:

= 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \pi^k Now multiply the denominator by 2^k and put one factor with each k-m factor in the denominator; also multiply by 2^k in the numerator to obtain:

(2) 2^k (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \pi^k Now gather each factor of 2 in the numerator product of the 2k, 2k-2…

= (2) 2^k 2^k \pi^k \frac{k!}{(2k+1)!} = 2 \frac{(4 \pi)^k k!}{(2k+1)!} which is the required formula.

So to summarize:

V_{2k} = \frac{\pi^k}{k!} R^{2k}

V_{2k+1}= \frac{2 k! (4 \pi)^k}{(2k+1)!}R^{2k+1}

Note the following: lim_{k \rightarrow \infty} c_{k} = 0 . If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume (2R)^n . Then consider the ratio \frac{2^n R^n}{c_n R^n} = 2^n \frac{1}{c_n} ; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the 2^n corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of dimensions increases.

One also notes that for fixed radius R, lim_{n \rightarrow \infty} V_n = 0 as well.

There are other interesting aspects to this limit: for what dimension n does the maximum hypervolume occur? As you might expect: this depends on the radius involved; a quick glance at the hyper volume formulas will show why. For more on this topic, including an interesting discussion on this limit itself, see Dave Richardson’s blog Division by Zero. Note: his approach to finding the hyper volume formula is also elementary but uses polar coordinate integration as opposed to the method of cross sections.

May 4, 2015

Hitting the bat with the ball….the vector calculus integral theorems….

Filed under: calculus, editorial, vector calculus — Tags: , — collegemathteaching @ 4:43 pm

When I was a small kid, my dad would play baseball with me. He’d pitch the ball and try to hit my bat with the ball so I could think I was actually hitting the ball.

Well, fast forward 50 years to my vector calculus final exam; we are covering the “big integral” theorems.

Yeah, I know; it is \int_{\partial \Omega} \sigma = \int_{\Omega} d \sigma but, let’s just say that we aren’t up to differential forms as yet.:-)

And so I am giving them classical Green’s Theorem, Stokes’ Theorem and Divergence Theorem problems….and everything in sight basically boils down to integrating a constant over a rectangle, box, sphere, ball or disk.

I am hitting their bats with the ball; I wonder how many will notice.:-)

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