How Leibniz Might Have Anticipated Euler
In 1671 James Gregory mentioned in correspondence that he had found
the inverse tangent of x equals the area under the curve of 1/(1+t^2)
between t=0 and t=x. He didn't include his derivation, but it's not
too difficult to see how, with the ideas of slopes and quadratures
of functions that Gregory and others were developing at the time,
this result could be found. Beginning with
sin(u)
t = tan(u) = ------
cos(u)
we know that the slope of t versus u (in modern notation) is the
derivative
dt
-- = sin(u) (-1/cos(u)^2) (-sin(u)) + (1/cos(u)) cos(u)
du
sin(u)^2 + cos(u)^2
= --------------------- = tan(u)^2 + 1
cos(u)^2
Therefore, since t = tan(u), we have dt/du = t^2 + 1, and so
du 1
-- = ---------
dt 1 + t^2
To invert the function t = tan(u) we need to express t in terms of
u, which we can do by multiplying the above equation by dt and
integrating both sides
/ / 1
| du = | ------- dt
/ / 1 + t^2
which is Gregory's result, i.e., we have
x
/ 1
arctan(x) = | ------- dt
/ 1 + t^2
t=0
Indeed, Leibniz derived this same result independently just a few
years later (around 1674). Now, by simple division of polynomials
he also knew that
1
------- = 1 - t^2 + t^4 - t^6 + t^8 - ...
1 + t^2
Furthermore, although this was prior to development of calculus proper
by Newton and Leibniz, several people (including Fermat, Cavalieri,
Pascal, etc) had already noticed that the area under the curve t^n
from t=0 to x is (x^(n+1))/(n+1). Thus Gregory had found an infinite
series for the inverse tangent of x:
x^3 x^5 x^7
arctan(x) = x - --- + --- - --- + ....
3 5 7
(Some evidence suggests that this series was actually known in India
by about 1500, but the discoverer's identity is not known, nor do
we know by what method the result was derived. See Rajagopal, On
Medieval Kerala Mathematics, 1986.) A little later, thinking along
these same lines, Leibniz noted that since arctan(1)=pi/4 we have
the remarkable relation
pi 1 1 1 1 1
---- = 1 - --- + --- - --- + --- - --- + ...
4 3 5 7 9 11
It's interesting to consider what Leibniz (or Gregory) might have done
with this result. Unburdened by any concerns about re-arranging the
terms of conditionally convergent series, they might well have noticed
that the sum could be expressed (at least formally) as a product of
geometric series in inverse primes
pi / 1 1 \ / 1 1 \ / 1 1 \
-- = ( 1 - - + - - ... )( 1 + - + -- + ... )( 1 - - + -- - ... )...
4 \ 3 9 / \ 5 25 / \ 7 49 /
Infinite products were certainly not unknown at this time. For
example, Wallis had already given the closely related infinite
product
pi 2 4 4 6 6 8 8
-- = - - - - - - - ...
4 3 3 5 5 7 7 9
Anyway, notice that each geometric series has either alternating signs
or is strictly additive, accordingly as p is congruent to -1 or +1
(mod 4). Of course, each of these geometric sums converges, and is
given by 1/(1 +- 1/p), so throwing cauchy.. er, caution, to the wind,
we could express Leibniz's series as the infinite product
pi / 1 \ / 1 \ / 1 \ / 1 \ / 1 \
-- = ( ------- )( ------- )( ------- )( -------- )( -------- )..
4 \1 + 1/3/ \1 - 1/5/ \1 + 1/7/ \1 + 1/11/ \1 - 1/13/
where the product is taken over all odd primes p, and the sign in the
denominator is + or - depending on whether p is congruent to -1 or +1
modulo 4. This isn't the same as Wallis's product, as can be seen
from the first few factors
pi 3 5 7 11 13 17
-- = - - - -- -- -- ...
4 4 4 8 12 12 16
Numerically this product appears to converge, albeit very slowly. In
any case, considering that Leibniz could digest the notion that
1+2+4+8+.. = -1 based on the geometric series 1 + x + x^2 + x^3 +..
equaling 1/(1-x), it's hard for me to imagine him being squeamish
about this product.
Now here's the interesting part (to me, least). Suppose we ask what
happens if we reverse the signs in our infinite product. In other
words, what if we use plus signs for primes congruent to -1 (mod 4),
and minus signs for primes congruent to +1 (mod 4)? This gives the
infinite product
/ 1 \ / 1 \ / 1 \ / 1 \ / 1 \
( ------- )( ------- )( ------- )( -------- )( -------- )..
\1 - 1/3/ \1 + 1/5/ \1 - 1/7/ \1 - 1/11/ \1 + 1/13/
which, if we expand it into a sum, is
1 1 1 1 1 1 1 1 1 1
1 + --- - --- + --- + --- + --- - --- - --- - --- + --- + --- +...
3 5 7 9 11 13 15 17 19 21
where the sign of 1/n is plus or minus accordingly as n has an even
or odd number of prime divisors (counting multiplicities) congruent
to 1 (mod 4). It's easy to form the conjecture based on numerical
evidence that this series (and the infinite product) converges on
pi/2, i.e., exactly twice the former series (and product).
Now, I don't know of any trigonometric interpretation of this series,
analagous to Gregory's arctangent expansion for Leibniz's series, but
there might be one. Also, at the risk of introducing some actual
math into the discussion, I might just mention Dirichlet's celebrated
class number formula
h pi inf / D \ 1
------------ = SUM ( ) ---
2 sqrt(|D|) n=1 \ n / n
where the quantity in parentheses is 0 if n has a common factor with
2D, and otherwise it is the Jacobi symbol. With D=-1 this gives
Leibniz's series, showing that the class number h equals 1 for D=-1.
On the other hand, with D=-5 the class number is 2, and Derichlet's
formula gives
pi 1 1 1 1 1 1 1 1
------- = 1 + --- + --- + --- - --- - --- - --- - --- + --- + ...
sqrt(5) 3 7 9 11 13 17 19 21
where the sign of 1/n is positive if n = 1,3,7,9 (mod 20), and
negative if n=11,13,17,19 (mod 20). But I digress.
Returning to our hypothetical Leibniz and his reversal of the signs
in the infinite product that formally corresponds to his famous
series for pi/4, notice what happens if we multiply his two results
together. We immediately have
pi^2 / 1 \ / 1 \ / 1 \ / 1 \
---- = ( --------- )( --------- )( --------- )( ---------- )...
8 \1 - 1/3^2/ \1 - 1/5^2/ \1 - 1/7^2/ \1 - 1/11^2/
1 1 1 1 1 1 1
= 1 + --- + --- + --- + --- + ---- + ---- + ---- + ...
3^2 5^2 7^2 9^2 11^2 13^2 15^2
which is the sum of the inverses of all the ODD square numbers.
Furthermore, since it's obvious that the sum of the inverses of the
EVEN squares is simply 1/4 times the sum of the inverses of ALL the
squares, we have
S(all) = S(odd) + S(even) = S(odd) + S(all)/4
and so S(all) = (4/3) S(odd). (Of course, this also follows simply
from multiplying the product involving the odd primes by the factor
1/(1 - 1/2) = 4/3.) Therefore, our hypothetical (and rather reckless)
Leibniz has found that the sum of the inverses of all the squares is
pi^2 / 6, anticipating Euler and solving the problem that defeated
not only the historical Leibniz (judging from his lack of response
to Oldenberg's query) but also his disciples the Bernoullis, who after
much effort were able to prove only that the sum of the inverse squares
converges to a value less than 2, but not to determine the value. It
remained for Euler to finally solve the problem, using means that were,
if anything, even more reckless and haphazard than the reasoning above
(although, to be fair, he did later tidy things up).
For more on this topic, see Factoring Zeta.
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