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### Infinite Forum

 10/28/09 (by Ted)As a first step, we will consider the infinite sum 1/2 + 1/4 + 1/8 + 1/16 + ...  Our discussion begins having seen the following argument:  Let this infinite sum equal x.  Then 2x = 1 + 1/2 + 1/4 + ... and therefore 2x - 1 = x, implying that x = 1.  However, we remain sceptical of this decidedly ingenious deduction.  Here is an attempt at a convincing argument.Observe that, for any z different from 1 and any positive integer n, z + z2 + z3 + ... + zn = (zn+1 - z)/(z - 1).  This is true because we can multiply both sides of the equality by z - 1, and note that the product on the left hand sides telescopes as follows:                            (z - 1)(z + z2 + z3 + ... + zn) = z2 - z + z3 - z2 + ... + zn+1 - zn = zn+1 - z                        (1)Let An = 1/2 + 1/4 + ... + 1/2n and use (1) with z = 1/2 to deduce that An = (1/2n - 1/2)/(1/2 - 1) = 1 - 2-n.  It is immediately obvious therefore that An remains bounded to at most 1 as n increases.  So the infinite sum 1/2 + 1/4 + 1/8 + ..., which is obtained as the limit of An when n is allowed to grow arbitrarily, remains bounded, despite its infinitely many terms.  In particular, for any B>0 there is a finite C = -ln(A)/ln(2) such that for all k>C, |Ak - 1|k, |1 - z|0 and f(x) = 0 otherwise (try to show this; it involves some fairly involved limits).We are now in a position to apply (2) to some functions of interest.  To begin with, let f(x) =sinx.  Then f(0) = 0, f(4k)(x) = sinx, f(4k+2)(x) = -sinx, f(4k+1)(x) = cosx and f(4k+3)(x) = -cosx, which implies that a2k = f(2k)(0)/(2k)! = 0, a2k+1 = f(2k+1)(0)/(2k+1)! = 1/(2k+1)! and a2k+3 = f(2k+3)(0)/(2k+3)! = 1/(2k+3)!.  Thus                                        sinx = x - x3/6 + x5/120 - x7/5040 + ...                                                         (3)Similarly (check on your own),                                        cosx = 1 - x2/2 + x4/24 - x6/720 + ...                                                            (4)Now, let's try f(x) = ex.  Clearly, f(k)(x) = ex, and therefore                                        ex = 1 + x + x2/2 + x3/6 + x4/24 + x5/120 + x6/720 + x7/5040 + ...                (5)The similarity of (3), (4) and (5) isn't accidental.  To take advantage of it, we need a way of flipping every other sign in (5).  The imaginary unit i offers us precisely this property.  In particular                                        eix = 1 + ix - x2/2 - ix3/6 + x4/24 + ix5/120 - x6/720 - ix7/5040 + ... =                                             = (1 - x2/2 + x4/24 - x6/720 + ... ) + (ix - ix3/6 + ix5/120 - ix7/5040 + ... ) =                                              = cosx + isinxwhich is Euler's formula for the representation of the unit circle on the complex plane.  It remains for us to evaluate this expression at x equal to pi and re-arrange terms.  Now that we established Euler's identity, e to the pi i plus 1 = 0, try to answer this question: is pi to the i e plus 1 more or less than one?12/24/09 On the Riemann zeta function (by Ted)The development of power series representations and analytic functions leads naturally to the extension of function from the reals to the reals, to mappings of the complex plane onto itself.  For example, consider f(x) = lnx.  What meaning could we assign to f if x is allowed to be complex?  Using the polar coordinate representation of the complex plane, x = reiy, we can set lnx = lnr + iy.  But the Euler formula developed above shows us that lnr + i(y+2pi) is also a valid logarithm of x, because ei(y+2pi) = eiy.  In fact, lnr + i(y+2kpi) is a valid logarithm of x for all integer k, even though these values are all different.  Thus, the extension of f(x) to the complex plane is a multi-valued mapping.  This is a typical occurence in complex analysis and it gives rise to very powerful equivalence relations  that strongly constrain complex analytic functions.Moving beyond power series, let's consider the harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...  Since the 14th century this has been the archetypical example of a divergent series.  In fact, it serves to illustrate what distinguishes a convergent from a divergent series.  After all, the terms of this series, the inverse positive integers, go down to 0.  But they don't decrease quickly enough!  Here is a clear elementary argument that establishes the harmonic series' divergence:                  1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ... > 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...                                                                                             > 1 + 1/2 + 1/2 + 1/2 + ...                                                    (6)where we lower bounded 1/3 by 1/4 and 1/5, 1/6 and 1/7 by 1/8 etc.  Since the right hand side of (6) is manifestly infinite (an infinite sum of 1/2 plus 1), so must be the left hand side which dominates it.  Incidentally, where does this argument break down with 1/2 + 1/4 + 1/8 + 1/16 + ... which converges to 1 as we saw in an earlier entry?The next series that will occupy us is another major accomplishment of Euler's, and it serves to sharpen significantly our sense of balance between convergent and divergent series.  It is the sum of the reciprocals of the squares of all positive integers, 1 + 1/4 + 1/9 + 1/16 + ... In order to explore this series, we'll need a little trigonometry.  The figure above shows the unit circle.  Let w =