Positively Mister Gallagher AbsolutelyMister Shean

‘Positively Mister Gallagher. Absolutely Mister Shean.’ The Title The comedy team of Gallagher and Shean was very popular in U.S. vaudeville during the early 1900s. Their theme song was a huge hit, and its refrain began with the line quoted above, “Positively Mister Gallagher. Absolutely Mister Shean!” This seems appropriate for an exposition of infinite series that shows how absolutely convergent infinite series are almost as nice as convergent series with positive terms. I’ll use this “advice arrow” for comments about classroom implementation of some of the ideas discussed. I’ll try to record frankly whether it is likely to succeed or not, and make some other suggestions. Other Kinds of Series Power series are not the only series that are widely used. In December 2007, Google had about 1,120,000 responses to the phrase “power series” and 573,000 responses to “Fourier series.” Fourier series are infinite sums of sine and cosine functions. They were developed to understand vibrational problems in the late 1700s, and then were used by Fourier to analyze heat transfer about 50 years later. The nicest Fourier series to consider are those that are absolutely convergent. Any convergent series that can be rearranged and regrouped without changing the sum must be an absolutely convergent series. Such series are easier to discuss theoretically, and computations with partial sums are also more likely to have results that are more reliable numerically than computations with conditionally convergent series. Special Focus: Calculus Sounds and Fourier Series I’ll consider Fourier series in the context of sounds and hearing. Sound is vibration in a transmitting medium. This could be solid (you can hear through metal or wood) or liquid (through water), but usually is gaseous. The standard transmission medium is the atmosphere. A vibration is produced by, say, a string moving back and forth. The string produces a dominant tone that usually has the most energy. A picture of a string producing this dominant tone is shown to the right. What doesn’t appear, of course, is the kinetic aspect of the vibration: how fast the string oscillates. The profile of the string when it is most distorted from the neutral, rest position is k sin(x) on the interval [ 0, π ] (if the units are chosen correctly). The constant k measures the amount of energy (loudness). Natural sound production is complex, and pure tones are rarely produced by natural processes. Vibrating objects also produce overtones or harmonics, which have frequencies with integer multiples of the original frequency. The fingering of string or wind instruments can strengthen the production of some of the harmonics. Drawn to the right is an attempt to show the vibrating string producing the first overtone, which would be modeled by sin(2x) on [ 0, π ]. A real vibrating string (a guitar) or membrane (a drum) or column of air (a flute) produces a complicated collection of tones and overtones. A Web page of the Baylor College of Medicine states: The sensory organs of the eye, ear, tongue and skin are each sensitive to specific forms of energy. The nose and tongue detect chemical energy, the eye detects light energy, the skin detects heat and mechanical energy. Sound is a form of mechanical energy. Mechanical forces can be steady, like … weight … or they can vibrate, like your car when it goes over speed bumps. Sound is generated by mechanical vibrations (such as a vibrating piano string). This sets up small oscillations of air molecules that in turn cause adjacent molecules to oscillate as the sound propagates away from its source. Sound is called a pressure wave because when the molecules of air come closer together the pressure increases (compression) as they move further apart the pressure decreases (rarefaction). … The velocity of sound in air is around 1,100 ft/sec [about 1/3 km/sec] … Sound waves travel fastest in solids, slower in liquids and slowest in air. Sound vibrations extend from a few cycles per second to millions of cycles per second. Human hearing is limited to a range of between 20 to 20,000 cycles per second. ‘Positively Mister Gallagher. Absolutely Mister Shean.’ One model of sound for a vibrating string with the ends held fixed would use a finite Fourier sine series, N ∑a j =1 j sin( jx) . The dominant tone would occur with j = 1, and higher j’s would be used for the overtones. The constants aj would indicate the relative strengths or amplitudes of the sound. A model of a more general situation where the ends of the string would move as well would include cosine terms, but we’ll just consider sine series in this discussion. Please note that mathematical modeling of what happens to sound and how it is translated to nerve impulses in the human ear is not completely satisfactory, and is an object of current research. To the right is a picture of a specific finite Fourier sine series, f(x) = 8sin(x) – 4sin(2x) – 6sin(4x). The signal, f, doesn’t seem to display in any obvious way the specific amplitudes 8, –4, and –6, along with the corresponding frequencies, 1, 2, and 4. But in fact, these can be “read off” the graph using some clever natural phenomena. Inside the ear, in a structure called the cochlea, there is a collection of cells (called colloquially hair cells, but very different from and much more sophisticated than the external hair covering mammals) that vibrate sympathetically to varying frequencies of the sounds transmitted. What they do is respond to the amplitude of sound waves of appropriate frequency. The response principally depends on the length of the hair cell, but there are variations depending on their structure and location. Mathematically, the amount of response is modeled by ∫ π 0 f ( xsin(mx) ) sin(mx dx m is a positive integer. f(x) dx )where Computing Some Integrals π 2 Let’s compute ∫ f ( x ) sin ( mx ) dx for f ( x) = 8sin(x) –4sin(2 x) – 6sin(4 x). π 0 π π π π 2 2 2 2 f ( x ) sin ( mx ) dx = 8 ⋅ ∫ sin ( x ) sin ( mx ) dx − 4 ⋅ ∫ sin ( 2 x ) sin ( mx ) dx − 6 ⋅ ∫ sin 4 ( x ) sin ( mx ) dx x ∫ π 0 π 0 π 0 π 0 A standard exercise using integration by parts and half-angle formulas will show: π 0 if m ≠ n 2 sin ( nx ) sin ( mx ) dx =  when m and n are integers. The factor ∫ π 0 1 if m = n of € 2 is chosen to give an integral of 1 when m = n. π How a Physicist Would Compute These Integrals Since the integrals just mentioned are so important, another way of computing them should be mentioned. Of course (!) eix = cos( x) + isin( x)(Euler’s formula: compare the Taylor series on both sides of the equation). Then e–ix = cos(–x) + isin(–x) = Special Focus: Calculus cos(x) – isin(x) and we can subtract the equations to get eix – e–ix = 2isin(x) so that sin( x) = 21i  eix − e − ix  . Now let’s compute the antiderivative. I’ll do the n ≠ m case: ∫ sin(nx)sin(mx)dx = ∫  2i1 (e   1 2 =   2i  inx ∫ (e  1  − e−inx ) (e imx − e−imx )dx  2i  i(n +m )x ) + e−i(n +m)x − e i(n−m)x − e−i(n +m )x dx  1  2  e i(n +m ) − e−i(n +m ) e i(n−m) − e−i(n−m)  = −   − +C  2   i(n + m) i(n − m)   e i(n +m) − e−i(n +m)   e i(n−m) − e−i(n−m)  1 1 =−  +  +C 2(n + m)  2i 2i  2(n − m)   =− sin((n + m)x ) sin((n − m)x ) + +C 2(n + m) 2(n − m) Back to Fourier Series € The “hair” in the cochlea that is attuned to the mth frequency responds in proportion π 2 to ∫ f ( x ) sin ( mx ) dx , which is (for f ( x) = 8sin(x) –4sin(2x) – 6sin(4x)), π 0 0 if m ≠ 1; 1 if m = 1 π π 0 if m ≠ 2; 1 if m = 2 π 0 if m ≠ 4; 1 if m = 4 π 2 2 2 2 f ( x ) sin ( mx ) dx = 8 ⋅ ∫ sin ( x ) sin ( mx ) dx − 4 ⋅ ∫ sin ( 2 x ) sin ( mx ) dx − 6 ⋅ ∫ sin 4 ( x ) sin ( mx ) dx ∫ π 0 π 0 π 0 π 0 if m = 1 8 −4 if m = 2 π 2  Therefore ∫ f ( x ) sin ( mx ) dx =  . − 6 if m = 4 π 0   0 otherwise Here’s the product 2 ff(x (x)sin(x )sin(x)on ) the interval π [0, π]. The integral of this product over the whole interval is 8. € 2 (x)sin(3x )sin(3x )) on the interval ff(x π [0, π]. The integral of this product over the whole Here’s the product interval is 0. € ‘Positively Mister Gallagher. Absolutely Mister Shean.’ 2 (x)sin(4x )sin(3x )) on the interval ff(x π [0, π] The integral of this product over the whole 2 (x)sin(5x )sin(3x )) on the interval ff(x π [0, π]. The integral of this product over Here’s the product Here’s the product interval is -6. € the whole interval is 0. € The pictures help to confirm that the computations are both mysterious and wonderful. You can actually see the integrals turning out to be the values already found. Fourier Coefficients Let me try to describe what happens in general. First, let’s look at those functions that can be written as finite sums of constant multiplies of sine functions. If π N 2 f ( x) = ∑ am sin(mx) , then the results above are still valid: am = ∫ f ( x ) sin ( mx ) dx . π 0 m =1 These numbers are called Fourier coefficients. They are named after Jean Baptiste Joseph Fourier (1768–1830), who had an extremely eventful life that included a great deal of political turmoil. Acquainted with both Napoleon and Robespierre, certainly he was not the stereotypical remote scholar and teacher. Fourier asserted that any function could be written as a (possibly infinite) sum of sine and cosine functions. Such an infinite series is now called a Fourier series. Let’s consider f (x ) = x ( π – x ) on the interval [0, π]. If we believe ∞ that f ( x) = ∑ am sin(mx) , then previous manipulations m =1 (integrations) suggest that the value of each π am is given by the am = (π2 )∫ f ( x) sin(mx) dx . For this f ( x) we 0 can use integration by parts to get am = 83 if m is odd and mπ 0 if m is even. (I think I could do these integrations by hand, formula integrating by parts twice, but I used Maple instead because I could.) Special Focus: Calculus The tenth partial sum of the Fourier sine series for this f ( x ) is 8 π sin( x) + 278π sin(3 x) + 1258 π sin(5 x) + 3438 π sin(7 x) + 7298 π sin(9 x) . I would show the graph of this function, only most human beings couldn’t really see any distinctions between the parabola above and this graph. Here is a graph of the difference between the original function and the partial sum. Please note the vertical scale. The difference between the partial sum and the function is between –.005 and .003, not visible to me in the original scale. Almost 50 years passed before Fourier’s claims about convergence of what’s now called the Fourier series were proved for suitable functions. It is now known that if a function is twice differentiable then it will be equal to the sum of its Fourier series, but that fact is well beyond the scope of this paper. Instructional Advice! I presented the material in the preceding few pages as general background. I would be very cautious about showing it to a general audience of students (this means I wouldn’t). What follows is material I have used in class to successfully stimulate questions and help students learn. I’ll begin with an investigation that combines several aspects of calculus and can be used both in class and to construct thoughtful homework questions. I’ll return to Fourier series later. Mean Square Error Here’s a homework question I have assigned, accompanied by hoped-for versions of answers. Part (a) For x near 0, sin(x) is well approximated by its tangent line at x = 0. What is this tangent line? Answer This part should be easy for the well-educated calculus student. Since sin´ = cos, the slope of the tangent line to y = sin(x) at x = 0 is cos(0) = 1, and an equation for this tangent line is y = x. Although a picture is not requested, students may think of an image resembling what’s shown. Part (b) Approximation over an interval is preferred over approximation near a point for many purposes. One criterion for assessing the accuracy of such an approximation is mean square error. The mean square error between a straight line y = Ax going ‘Positively Mister Gallagher. Absolutely Mister Shean.’ through the origin and the function sin(x) over the interval [0, 1] is given by the 1 2 definite integral ∫ (sin(x) Ax )2 dx. dx Find the value of A that minimizes this integral. (sin( x) –− Ax) 0 Hint: Expand the integrand, compute the integral, and find the A minimizing the result. Answer: This is decidedly not a routine question for most of my students. Let’s call 1 2 f (A) the value of ∫ ((sin(x) sin( x) −– Ax Ax) dx.. We can expand (sin(x) – Ax)2 to get (sin(x)) 2 ) 2dx 0 – 2Ax sin(x) +A 2 x 2. Now integrate (I am omitting the use of integration by parts; I usually assign this problem soon after integration by parts is introduced). 1 x A2 x 3   1 ( sin ( x )) − 2 Ax sin ( x ) + A x dx =  − cos ( x ) sin ( x ) + − 2 A ( − x cos ( x ) + sin ( x )) +  2 2 3  ∫( 0 2 2 2 ) 1 2 1 1 A = − cos (1) sin (1) + − 2 A sin (1) + 2 A cos (1) + 2 2 3 2 ≈ 0.27268 − 0.60234 A + 0.33333 A So we know that f ( A ) = 0.27268 − 0.60234 A + 0.33333 A 2 , a quadratic polynomial with a positive second-degree coefficient, has a unique minimum when f ′ ( A ) = −0.60234 + 0.666666 A = 0 , and that happens when A ≈ 0.90351 . Comment: I hope when I assign this problem, and when it is subsequently discussed in class, that students may have made some connection with how they treat numerical data obtained in a science lab. Very frequently a “best” linear approximation idea is mentioned there, and sometimes the least squares method of fitting data may even be suggested (the formulas needed are frequently available on calculators). I used 1 as the upper bound of the integral instead of something π more “sine-ish” like 2 because I want numbers here. The squaring of the integrand means that differences can’t cancel, and there will be exactly one easily computable 1 minimum for f ( A). If instead we defined f ( A) as ∫ |sin(x) | sin( x)–− Ax|dx, Ax | dx there would still be 0 a minimum, but it would not be readily computable. The minimum for this definition is A ≈ 0.919, and that took some work to get. I also hope that students would think about the following pictures: A is too large: The upper chunk of area is too big. A is just right. A is too small: The lower chunk of area is too big. 0 Special Focus: Calculus Part (c) Sketch sin(x) and the straight lines found in (a) and (b) on the unit interval [0, 1]. Answer: The desired picture is displayed at right. Comment: This is not a simple problem. The “tangent line” idea includes an approximation theme: The very best closest line to a curve near a point is the tangent line. Making this precise leads to the limit of the slope, etc. But if we change the game and ask: whether there is a simple line that will give, on average, numbers that are closer to a given function over an entire interval, we may well get a different answer. Here we consider sine, the interval [0, 1], and lines through the origin. The answer is 0.90351x, certainly different from x. Back to Fourier Coefficients The Taylor polynomials of a function at a point are the partial sums of the Taylor series. These Taylor polynomials are the best—the closest polynomials of their degree to the function near that point. In this way they generalize tangent lines. What about Fourier series? If f ( x) is defined on [0, π], we previously defined the mth Fourier coefficient,am, by 2π a = ∫ f ( x )sin( mx )dx . This definition is related to mean square error. the formula m π 0 π The coefficient am is the unique A that minimizes ∫ ( f ( x) − A sin(mx)) 2 dx . To verify 0 this we start with π ∫ ( f ( x ) − A sin ( mx )) 2 dx = 0 π ∫ ( f ( x )) 0 = 2 π dx − 2 A ∫ f ( x ) sin ( mx ) dx + A 0 π π 0 0 π 2 ∫ ( sin ( mx )) 2 dx 0 2 2 ∫ ( f ( x )) dx − 2 A ∫ f ( x ) sin ( mx ) dx + A ⋅ π 2 This is minimized when π π π   d π 2 2 π ( f ( x )) dx − 2 A ∫ f ( x ) sin ( mx ) dx + A ⋅  = −2 ∫ f ( x ) sin ( mx ) dx + 2 A ⋅ = 0 , ∫  dA  0 2 2 0 0 π 2 f ( x ) sin ( mx ) dx . This clearly provides a minimum for the mean π ∫0 square error by using the second derivative test. The Fourier series of a function satisfies a different minimization criterion from Taylor series. The partial sums of the Fourier series are best mean square approximations to the function. leading to A = ‘Positively Mister Gallagher. Absolutely Mister Shean.’ What does this mean in practice? If someone wants to numerically sample a function or use a function to model the result of an experiment over a range of inputs, the Fourier series or its partial sums are likely to be more useful and more accurate on average than Taylor series. So for small perturbations near a known input, Taylor series are good. Over a range of inputs, Fourier series should be considered. Back to Absolutely Convergent Series Suppose we have an infinite series ∞ ∑a j =1 j that converges absolutely. Then consider the function defined by this Fourier sine series: f ( x ) = ∞ ∑ a sin ( jx ) . j =1 Since f ( x ) = ∞ ∞ ∞ j =1 j =1 j =1 j aj sin (ofjxsine areajalways at most 1 in absolute ) ≤∑ ∑ aj sin ( jx ) ≤and ∑ values value, we have f ( x ) = ∞ ∞ ∑ a sin ( jx ) ≤ ∑ a j =1 j j =1 ∞ j sin ( jx ) ≤ ∑ aj . So any absolutely j =1 convergent series creates a nice function on [0, π]. This function is always continuous, but verifying this assertion is not something I would do in any calculus class. Another Homework Problem Here is an ambitious homework problem that I have assigned, together with some answers. Students can solve this problem, and the problem can provide them with further understanding of convergence of an infinite series. But (as I am sure you realize) very few students will have the background given on the last few pages of this article. This is a problem about a Fourier cosine series. All the corresponding results for Fourier cosine series are true and follow from the same logic as the results for sine series. Some of the formulas in this example are easier to manipulate when cosines are used instead of sines. Please read what’s below for verification of this claim. Instructional Admission I have used the following homework problem in several classes, with varying amounts of success. That the partial sum is always so close to the true function is amazing, and I want students to understand this: A machine can’t add up infinitely many numbers, but it certainly can rapidly compute the one-hundredth partial sum. When this partial sum is graphed, it can’t be distinguished from the whole infinite series, either on screen or printed out. Special Focus: Calculus ∞ Problem stem: Define f ( x) by the sum f ( x) = ∑ 2 n n =0 100 Below is a graph of the partial sum s100 ( x) = ∑ 2 n =0 n cos( nx ) n! cos( nx ) n! . This is not a power series. of the series for 0 ≤ x ≤ 20. Part (a) Verify that the series defining f ( x) converges for all values of x. Answer: We will prove that the series converges absolutely for all values of x and because absolute convergence implies convergence, the series converges for all values of x. Since |cos(anything)|≤1, ∞ ∑ n= 0 € 2n cos( nx ) n! ≤ 2n n! ∞ . We know that ∑ n= 0 2n converges and that n! 2n = e 2 . Observing the inequalities indicates that |f ( x)|≤e 2 for all values of x. n! € Part (b) Is the apparent periodicity of the function f ( x) actually correct? If yes, explain why. Answer: Yes, f ( x) is periodic with period 2π. For integer n, cos(n(x + 2π)) = cos(nx +2nπ) = cos(nx) since cosine is 2π periodic. So all the terms in the infinite series for f ( x + 2π) are identical to the terms in the infinite series for f ( x). Part (c) Verify that the actual graph of the function is always within 0.01 of the graph shown. That is, if x is any real number, then |f ( x) –s100( x)| < 0.01. Answer: Again the observation |cos(anything)|≤1 will simplify our work. We know | f ( x) − s100 ( x) |= ∞ ∑ n =0 2n cos( nx ) n! 100 −∑ 2 n =0 n cos( nx ) n! = ∞ ∑ n =101 2n cos( nx ) n! ≤ ∞ ∑ n =101 2n n! . This infinite tail can be overestimated by a geometric series because the ratio between successive terms of the tail series is n+2 1 (the tops of the tail series terms are powers of 2, and the bottom are factorials). Here n ≥ 101, so the ratio is at most than the geometric series 2 102 = 511 . So the tail series is less ‘Positively Mister Gallagher. Absolutely Mister Shean.’ with a = 2101 (101)! and r = 511 . This is ∞ ∑ n =0 2101 (101)! ( ) 1 n 51 2101 ! = 1(101) , which has an approximate value −( 1 ) 51 of 0.27 ∙ 10-129. This result is much smaller than 0.01, so the graph shown is very much like the true graph. Not for Everyone What’s below is a discussion showing that the function itself, the sum of the whole series, can be written as a neat formula using well-known functions. This result is not obvious, and I thank my colleague, Professor Amy Cohen, for helping me with this computation. I have shown calculus students (interested students, outside of standard class time) these details. I would not do this to a whole class. What is This Function? Step 1 Euler’s formula states that eix = cos(x) +isin(x). Step 2 If we substitute nx for x, we see that einx = cos(nx) +isin(nx). Step 3 This problem is about the series ∞ ∑ n =0 2n cos( nx ) n! . But the preceding step makes me want to  ∞ n nx )   ∞ 2n sin( nx )  consider the following:  ∑ 2 cos( n!  + i  ∑ n!  .  n =0   n =0  Step 4 So we are looking at ∞ ∑ n =0 2n (cos( nx ) + i sin( nx ) ) n! , which is equal to But e  ix  n     = e , so this is the series ∑ n =0 Step 5 inx ∞ ∞ ∑ n =0 2n  eix   n!  n ∞ =∑ n =0 ix    2e    n! n . 2n  einx   n!  . Special Focus: Calculus Step 6 ∞ The exponential function is e A = ∑ n =0 the sum of the series must be e ix    2e    An n! . The series we are considering has A = 2eix , so . Step 7 Euler’s formula states that this is e 2(cos(x )+i sin (x )) = e 2 cos(x)+2i sin (x ) . Step 8 The exponential function€converts addition to multiplication, and therefore we know e 2cos( x ) + 2i sin( x ) = e 2cos( x ) e 2i sin( x ) . Step 9 2 i sin( x ) i (2sin( x )) =e Look at e . Use Euler’s formula again, replacing the x in the original formula with 2sin( x) . The result is ei (2sin( x )) = cos(2sin( x)) + i sin(2sin( x)) . Step 10 The series sum is e 2cos( x ) ei (2sin( x )) = e 2cos( x ) (cos(2sin( x)) + i sin(2sin( x)) ) = e 2cos( x ) cos(2sin( x)) + ie 2cos( x ) sin(2sin( x)) . Step 11 Compare the results of Step 3 and the preceding step. The same quantities are being described. The “real parts” (the quantities without i) should be the same, so therefore ∞ (but not clearly, definitely not clearly): ∑ n= 0 Indeed, a graph of e problem. 2cos( x ) 2 n cos(nx ) 2 cos(x ) =e cos(2sin(x )) n! cos(2sin( x)) looks the same as the graph that began the € Just One More Example... Let’s take a specific convergent positive series: ∞ ∑ j =1 1 2j . This is so nice that I even can tell you its sum, which is 1. We know if we put this series together with sines we’ll get a result that converges, and always has a value of between –1 and 1. The frequencies inside the sines do not affect convergence, since the values outside of the sine functions will still be between –1 and 1. Because absolute convergence implies convergence, nothing will go wrong…. ‘Positively Mister Gallagher. Absolutely Mister Shean.’ Here is what can go wrong. This wrongness disturbed many mathematicians in the nineteenth century. I’ll select the frequencies inside the sines in a strange way. ∞ Here we go: Define the function f by f ( x) = ∑ 21j sin(3 j x) . Some pictures of partial j =1 sums of this series and of the derivatives of these partial sums follow. Examine the pictures closely. Pay close attention to the vertical scales of the graphs, because they are nearly unbelievable! The Tenth Partial Sum and Its Derivative Graph of  d dx   ∑ Graph of 10 ∑ j =1 1 2j 10 j sin(3 x) j=1 sin(3 j x ) = 2j   10 ∑ j=1 3j cos(3 j x ) 2j € The Twentieth Partial Sum and Its Derivative Graph of  d dx   20 ∑ j =1 1 2j 20 ∑ Graph of sin(3 j x) € j=1 sin(3 j x ) = 2j   20 ∑ j=1 3j cos(3 j x ) j 2 Special Focus: Calculus The Thirtieth Partial Sum and Its Derivative Graph of  d dx   ∑ Graph of 30 ∑ j =1 1 2j 30 sin(3 j x) j=1 sin(3 j x ) = 2j   30 ∑ j=1 3j cos(3 j x ) 2j € Discussion of the Pictures Let’s deal with the three graphs on the left. They hardly differ. This is because their sum is a function f, and the error is easy to estimate. We defined f by the equation ∞ f ( x) = ∑ 21j sin(3 j x) . So we can write j =1 ∞ ∑ j =1 10 1 2j ∞ sin(3 j x) = ∑ 21j sin(3 j x) + ∑ j =1 j =11 1 2j sin(3 j x) . The error in the first graph of the partial sums is at most ∞ ∑ j =11 1 2j ∞ | sin(3 j x) |≤ ∑ j =11 1 2j = 1 210 < .001 . The error in the second graph is at most 1 1 <0.000001, < 0.000001 and the error in the third graph is 30 <0.000000001. < 0.000000001 Since the partial 20 2 2 sums are approximations of the same limit, their graphs are identical (at least to my eyes, on this scale). € The graphs on the right have a very€different aspect. I had the graphs created using Maple and this program is quite accurate. Please consider the vertical scales carefully, and deduce information about the ranges of the derivatives of these partial sums. The derivative of the tenth partial sum seems to have values between –150 and 150. The derivative of the twentieth partial sum seems to have values between –8,000 and 8,000. The derivative of the thirtieth partial sum seems to have values between –450,000 and 450,000. These graphs are not getting closer to one another, and they are not “stabilizing” in any sense. They actually seem to be getting more chaotic. What’s happening? ‘Positively Mister Gallagher. Absolutely Mister Shean.’ A Continuous Function That Is Differentiable at No Point When we try to differentiate ∞ ∑ j =1 1 2j sin(3 j x) using the rules we love to follow, the chain j rule leaves us with an unappealing multiplier, 3 j , of the function cos(3 j x). While 2 the values of cosine occupy [–1, 1], this multiplier gets large very rapidly. Things do not cancel out, and we have an example of a function that is continuous at every point and differentiable at no point! Verification of the last claim is straightforward but tedious. That such functions exist was acknowledged only with great reluctance by many authorities in mathematics during the nineteenth century. Although the example given here is usually attributed to Weierstrass or Riemann, Bolzano had similar ideas and examples several decades earlier. (Bolzano also seems to have been the first person to give a statement resembling the version of the intermediate value theorem we currently use [1817]). The idea of a continuous nowhere differentiable function was very distasteful, and the examples were widely viewed as artificial and absurdly theoretical. Most of the academic mathematicians of the late nineteenth century were quite reluctant to accept the validity of the results. But these functions really describe natural phenomena. In fact, some observations done many years earlier were consistent with the graphs shown above. In 1827, biologist Robert Brown microscopically observed the motion of pollen in water. The pollen seemed to jump bizarrely, in strange and jagged paths—very different from the smooth motion of, say, a cannon ball in a parabolic trajectory. Observation of dust particles gave the same sort of results, so the motion couldn’t be attributed to some sort of life force in the pollen grains. This movement of the particles was named Brownian motion. The doctoral thesis of Bachelier in 1900 connected Brownian motion with variations in stock and option markets. Such jagged graphs typically appear in many financial reports. One of Einstein’s famous results of 1905 explained Brownian motion using probability — the particles of dust move as a result of random molecular collisions, and the molecular motion is what we perceive as heat. The paths typically are not smooth curves, and are usually not differentiable. In the last 20 or 30 years, Brownian motion and related topics have been extensively studied by mathematicians and physicists, and nondifferentiable functions are standard tools in mathematical finance and other applications.