Someone on the internet proposed a novel approach to proving the irrationality of pi. Figuring out why this approach doesn't work led to some interesting discussion and eventually to the construction of continuous but non-differentiable functions, and the characteristic function of the rationals. Amos Watson wrote:Consider the infinite series sin^2 x + sin^2 x^2 + sin^2 x^4 + sin^2 x^8 ... Expand each term of the series into its MacLaurin series, and combine like powers of x... The absolute value of each coefficient in the expansion of sin^2 x^2 is less than half that of the previous coefficient,... so all the coefficients are positive. [Also], ...all the powers of x are even, so the sum is greater than zero for all x>0. But if pi were rational and k was an integer multiple of pi, this series would converge to 0 for x=k. Thus we have a contradiction, and our supposition that pi is irrational is false. QED?Your assertion is that the power series expansion of the "function" defined by f(x) = sin^2(x) + sin^2(x^2) + sin^2(x^4) + sin^2(x^8) + ... has no negative coefficients, and therefore cannot equal zero for any positive x. Of course, this assumes that f(x) is well-defined, and it implies that f(x) is not only positive, but monotonically increasing. However, if you examine the partial sums f_1(x) = sin^2(x) f_2(x) = sin^2(x) + sin^2(x^2) f_3(x) = sin^2(x) + sin^2(x^2) + sin^2(x^4) etc. you find that these functions, rather than becoming monotonic, are actually oscillating more and more wildly as you add terms (and this oscillation intensifies as x increases). Also, they pass arbitrarily close to zero for many values of x. In fact, the only reason we can say that none of these partial sums ever equals zero for some x>0 is becase we know pi is irrational. If you examine the power series expansions of these partial sums you find that f_j(x) starts to have negative coefficients beginning with the 2^(j+1)th power of x, and these negative terms are by no means negligible; they are responsible for the oscilatory behavior of the functions. It's tempting to just forget about all those negative terms up there, and assume that the limit of all these wildly oscillating partial sums is somehow a nice monotonic "function" f_inf(x), but that isn't valid, because those partial sums don't converge. Notice that the expansion of general term sin^2(x^n) is x^2n - (1/3)x^(4n) + (2/45)x^(6n) - ... and this is not well-defined as n -> inf. For one thing, it implies infinite frequency. Thus, as you add more sin^2 terms the resulting partial sum jumps around wildly. As a result, there is no well-defined limit function f_inf(x). By the way, you could have made the same argument with successive powers of x, rather than exponential powers of x. In other words, using the series sin^2(x) + sin^2(x^2) + sin^2(x^3) + sin^2(x^4) + ... The partial sums of this series also give non-negative coefficients up to a point, but then negative terms start to show up. Like your sequence, this one also has no well-defined limit. Amos Watson wrote:I would like to point out, though, that for any fixed x (specifically we can choose x=k), you can carry out the partial sum...even as you did it....far enough to ensure that it is greater than 0. After that, the oscillations are tempered by the factorially increasing denominators. Indeed, the point where the oscillations begin can be chosen such that the value of the first negative oscillation will be less than the value of the previous partial sum, thus ensuring convergence > 0, or divergence to positive infinity.That isn't true. Each component of the form sin^2(x^n) oscillates between 0 and 1 for all values of x, no matter how large x becomes. In fact, it oscillates between 0 and 1 more rapidly and x increases. Of course, you have to go to higher and higher terms of the series to drive the sum to zero as x increases. Those are precisely the terms you are throwing away when you pass from a well-defined sequence of (non-convergent) partial sums to the "limit" function, which you have arbitrarily chosen to define as the limit of just the positive terms in the partial sums. Amos Watson wrote:I will grant you that the ability to rearrange terms does need a little stronger rigorous underpinning. (I said I was 99% confident that it was valid.) However, I am 99% confident that it is valid, and proving so is actually not as interesting to me as the rest of the proof, although I would like to see a proof either way if you have one.Let me see if I can think of an example to illustrate... Okay, let's define the sequence of functions s_n(x) = 2x^n - x^(n+1), so we have s_0(x) = 2 - x s_1(x) = 2x - x^2 s_2(x) = 2x^2 - x^3 s_3(x) = 2x^3 - x^4 etc. I think you will agree that 2 is a root of each of these functions. In other words, we have s_n(2) = 0 for every value of n. Now consider the infinite sum S(x) = s_0(x) + s_1(x) + s_2(x) + s_3(x) + .... Clearly we must have S(2)=0, because each of the terms vanishes at x=2. However, the sum is given by S(x) = 2 - x + 2x - x^2 + 2x^2 - x^3 + 2x^3 - x^4 + 2x^4 - x^5 + 2x^5 -... etc. ---------------------------------------- 2 + x + x^2 + x^3 + x^4 + x^5 + .... Therefore, we've shown that S(x) = 2 + x^2 + x^3 + x^4 + x^5 + ... but since S(2)=0 this implies that 0 = 2 + 4 + 8 + 16 + 32 + ... What should we conclude from this? Have we proven that the number 2 cannot exist? Clearly no amount of rearranging terms is going to make the above "equality" true. The error in the above reasoning is exactly the same as in your proposal, namely, it takes a (non- convergent) sequence of partial sums that are each dominated by their high-order negative terms, and attempts to define the limit of the sequence as just the limit of the low-order positive terms. Of course, it is undeniably true that you can push the first negative coefficient out as far as you want simply by increasing the number of sine terms. Just for fun, here are the initial terms of the sum of sin^2(x^k) for k=1,2,...,n: 2 2 209 2 x^2 + --- x^4 + --- x^6 + --- x^8 + ----- x^10 3 45 315 14175 20788 4 423648224 2 + ------ x^12 + -------- x^14 + --------- x^16 + ----------- x^18 467775 42567525 638512875 97692469875 1309458148 4 + ------------- x^20 + ---------------- x^22 + ... 9280784638125 2143861251406875 The point is that lurking behind all these positive terms are infinitely many negative terms, and those terms can't be neglected. Now let's consider the possibility that AW's proof could be repaired by replacing all the terms of the form sin^2(x^(2^n)) with terms of the form sin^2(x^(2^n))/(2^n). In other words, consider the infinite sum sin^2(x^2) sin^2(x^4) sin^2(x^8) S(x) = sin^2(x) + ---------- + ---------- + ---------- + ... 2 4 8 Since the numerators are all between 0 and 1, we know this series converges to a real value in the range 0 to 2 (inclusive) for all x. (These bounds correspond to the cases when all the sin's are 0 and all are 1, respectively.) Now consider the power series expansion of this function: 1 2 101 2 S(x) = x^2 + --- x^4 + --- x^6 + ---- x^8 + ----- x^10 6 45 1260 14175 10393 4 204729517 + ------ x^12 + -------- x^14 + ---------- x^16 + ... 467775 42567525 5108103000 As with the series expansion of AW's original function, it appears that all the coefficients of this expansion are positive. To prove this, let c[k] denote the absolute value of the coefficient of x^k in the expansion of sin(x)^2. For odd k we have c[k]=0, and for even k we have 2^(k-1) c[k] = ------- k! Now let C[k] denote the coefficient of x^k in the infinite sum sin^2(x) + sin^2(x^2)/2 + sin^2(x^4)/4 + ... It follows that C[2j+1] = 0 for all j and the values of C[2j] are as shown below: c[2] = C[2] c[2]/2 - c[4] = C[4] c[6] = C[6] c[2]/4 - c[4]/2 - c[8] = C[8] c[10] = C[10] c[6]/2 - c[12] = C[12] c[14] = C[14] c[2]/8 - c[4]/4 - c[8]/2 - c[16] = C[16] c[18] = C[18] c[10]/2 - c[20] = C[20] c[22] = C[22] c[6]/4 - c[12]/2 - c[24] = C[24] c[26] = C[26] c[14]/2 - c[28] = C[28] c[30] = C[30] c[2]/16 - c[4]/8 - c[8]/4 - c[16]/2 - c[32] = C[32] and so on. Clearly for k = 2m(2^n) where m is odd we have C[k]=c[k] if n=0, and if n>0 we have c[2m] c[4m] c[8m] C[k] = ----- - ------- - ------- - ... - c[k] 2^n 2^(n-1) 2^(n-2) Therefore, we need to show that for any positive integers m,n we have the inequality c[2m] c[4m] c[8m] ----- > ------- + ------- + ... + c[2m(2^n)] 2^n 2^(n-1) 2^(n-2) Recall that c[j] = 2^(j-1) / j!, so this inequality can be written as 2^(2m-n-1) 2^(4m-n) 2^(8m-n+1) 2^(m2^(n+1)-1) ---------- > --------- + ---------- + ... + -------------- (2m)! (4m)! (8m)! (2(2^n)m)! Multiplying through by 2^n gives 2^(2m-1) 2^(4m) 2^(8m+1) 2^(16m+2) 2^(m2^(n+1)-1+n) -------- > ------ + -------- + --------- + .. + ---------------- (2m)! (4m)! (8m)! (16m)! (2(2^n)m)! Clearly each term on the right is (much) less than half of the preceeding term, so the inequality is certainly true if the first term on the right is less than half the left hand side. In other words, the inequality is certainly true when (4m)! ----- > 2^(2m+2) (2m)! which is true for all m > 1. For m=1 the first term on the right is greater than half the left hand side. However, by inspection we can verify that 16 512 262144 1 > ---- + ----- + ------ + ... 24 40320 16! so it follows that the inequality is always satisfied, and therefore all of the C[k] coefficients are non-negative. So have we proved that pi is irrational?? Well, no, but now the problem is more subtle than in AW's original example. To illustrate the problem, consider the following simplified version. If PI is rational then there exists a positive integer k that is an integer multiple of 2PI. It follows that k^2, k^3, ... and all higher powers of k are also integer multiples of 2PI, and so cos(k^n)=1 for all n. Thus, if we define the function cos(x^2) cos(x^4) cos(x^8) C(x) = cos(x) + -------- + -------- + -------- + .... 2 4 8 we have C(k) = 1 + 1/2 + 1/4 + 1/8 +... = 2. However, recall that cos(x) = 1 - x^2/2! + x^4/4! - ..., from which it follows that the expansion of C(x) is 1 / 1 \ 1 /1 1 \ 1 / 1 \ C(x) = 2 - -( --- )x^2 - -( -- - -- )x^4 - -( --- )x^6 2 \ 1!/ 4 \1! 3!/ 6 \ 5!/ 1 /1 1 1 \ 1 / 1 \ - -( -- - -- - -- )x^8 - --( --- )x^10 8 \1! 3! 7!/ 10 \ 9!/ 1 /1 1 \ - --( -- - --- )x^12 - ... 12 \5! 11!/ Obviously the coefficients of all odd powers of x are zero. The coefficient of x^N where N = 2m(2^n), m odd, is just 1/N! if n=0, and for n>0 the coefficient is given by 1 / 1 1 1 \ - - ( ------------ - ------------ - ... - ------------ ) N \ (2m2^0 - 1)! (2m2^1 - 1)! (2m2^n - 1)! / It's easy to see that the quantites in the braces are always positive, because each subtractive term is less (usually much less) than 1/3 the previous term. Therefore, the coefficients of C(x) are all negative (after the constant 2). This seems to imply that C(x) is monotonically decreasing from 2 for all x>0, which would contradicts the existance of an integer k that is an integer multiple of 2PI. However, notice that the overall function should always be in the range -2 to +2, whereas the series seems to be unbounded in the negative direction as x increases. The same problem applies to the earlier series expansion for the "geometric" sum of squared sines. The maximum value the series could possibly have is 2, corresponding to the case when all the sine's are 1, but the "power series expansion" of this function increases without limit as x increases. This obviously indicates that these series doesn't accurately represent the respective functions. John Rickard wrote:The error in the proof is actually in the assumption that, because sin^2(x^2) sin^2(x^4) sin^2(x^8) S(x) = sin^2(x) + ---------- + ---------- + ---------- + ... 2 4 8 converges to a real value in the range 0 to 2 for all x, its power series expansion 1 2 101 2 x^2 + --- x^4 + --- x^6 + ---- x^8 + ----- x^10 + ... 6 45 1260 14175 must also converge (and to the same value) for all x. Given that all the coefficients are positive, it's clear that this power series does not converge to S(x) for x > sqrt(2), since the first term is already too big! (In fact, I think the power series diverges for x > 1.)Yes. The same problem applies to the simpler function cos(x^2) cos(x^4) C(x) = cos(x) + -------- + -------- + ... 2 4 What we've succeeded in constructing here are continuous but nowhere differentiable functions, somewhat similar to Koch's snowflake curve. Notice that the function C(x) has a component of oscillation on every scale, and every frequency. This is the kind of "pathological function" that was considered somewhat disturbing back around the turn of the century. Evidently the earliest example was found by Cellerier in 1860: sin(kx) sin(k^2 x) sin(k^3 x) f(x) = -------- + ---------- + ---------- + ... k k^2 k^3 where k was a big positive integer. Because he applied the exponentiation to the factor k (instead of x), this function is pathological for any x, whereas our C(x) function is fairly inncouous for x<1. Nevertheless, for x>1 the C(x) function is, if anything, even more pathological than Cellerier's. Unfortunately for Cellerier, his example wasn't published until 1890, by which time Weierstrass had given (in 1872) his "canonical" example of a continuous but nowhere differentiable function cos(k pi x) cos(k^2 pi x) f(x) = cos(pi x) + ----------- + ------------- + ... q q^2 where q>1. Kline comments that "The historical significance of this discovery...was great. It made mathematicians all the more fearful of trusting intuition or geometrical thinking." It's interesting to explore the mechanism by which the sum of a convergent set of strictly convergent power series fails to converge. I'd look at it like this: For any given x the function cos(x^(2^n))/(2^n) yields a convergent power series for any fixed integer n. However, the number of terms required to give convergence for a given x>1 increases exponentially with n. Therefore, the number of terms required for convergence of ALL the cosines is unbounded. Another, possibly more intuitive, way of looking at this is to realize that the leading terms of each cosine series (for x>1) are a relatively large and erratic, and the terms of each successive cosine are twice as widely spaced (because the argument is x^(2^n)). As a result, at any given point in the power series the erratic leading terms are very over-represented. There's no point at which we can truncate the series and have a representative (convergent) contribution from most of the cosines. Adding more terms, attempting to force convergence, just introduces the flakey low-order terms of even higher order cosines, which will require us to go to even higher powers to resolve, and so on. It may seem puzzling at first that, although each cosine expansion converges in a finite number of terms for any given x, the sum of all these expansions does not converge for any x>1. This is essentially the same phenomenon that sometimes bothers people in regard to Cantor's diagonalization proof, when they argue that the diagonalization procedure could be applied to a list of all rational numbers to yield another rational number not on the list. Of course, we know that the number produced in this way would not be rational, but people sometimes have difficulty understanding this because the new number is generated solely based on the digits of a set of purely rational numbers. It's basically the "Door Paradox": Suppose you need to make a door big enough to allow the tallest possible person to pass through. If every person's height is finite (but there is no upper limit to a person's height) how big must the door be? The "paradox" is that you evidently need an infinitely tall door to make room for a bunch of people whose heights are all finite. This transition between the "finite-but-unbounded" and the "infinite" seems to be the point where our intuition is most easily confused. On the other hand, in fairness to intuition, it actually is possible to establish a connection between limits of trig functions and the distinction between rational and irrational numbers. For example, Dirichlet defined the function f(x) = lim [m->inf] lim [n->inf] { cos(m!*pi*x) }^(2n) and proved that f(x) equals 1 for all rational x and 0 for all irrational x.

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