Convergence of Series (How NOT to Prove PI Irrational)

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)


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 

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


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 -...
              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 

                       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
                     c[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

                -----  >  2^(2m+2)

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

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 
                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 

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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