## 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)

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

"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.
```