Download Lagrange`s Attempts to Formalize The Calculus

Joel Hermansen
2/26/16
Math 161
Lagrange’s Attempts to Formalize The Calculus
Joseph-Louis Lagrange sought to simplify and formalize the Calculus through
the use of algebraic manipulation. Lagrange’s main focus was on algebra, but he
tried to assist in the quest for making Calculus more rigorous. Lagrange defined
differentials and fluxions in terms of coefficients in expansion series, depending
particularly on Taylor’s series. “Lagrange proceeded therefore to make Taylor’s
series fundamental in his work, assuming implicitly that all functions allow of such
an expansion” (Boyer, 1959, p. 252). Lagrange’s series-based approach seemed at
first to be solid in practice. This is how he tackled the derivative:
Supposing f(x+i) is an infinite series in i of the form
f(x+i) = f(x) + ip(x) + (i^2)q(x) + (i^3)r(x) +…
where p,q,r… represent “’new functions of x, derived from the primitive function x
and independent of the indeterminate I’” (Dunham, 2005, p. 73). As this series goes
on into infinity, the primes are closer to zero. Here is a practical application of
Lagrange’s formula above:
Suppose f(x) = 3x^2 and we want to find f ‘ (x).
Then 3(x + i)^2 = 3x^2 + ip(x) + (i^2)q(x) + (i^3)r(x) +…
 ip(x) + (i^2)q(x) + (i^3)r(x) +… = 3(x + i)^2 – 3x^2
(subtracting 3x^2 from both sides of the equation)
 i[p(x) + iq(x) + (i^2)r(x) +…] = 3x^2 + 6ix +3i^2 – 3x^2
(Remember that i in this case does not mean an imaginary number. The notation
sometimes uses h in place of i.)
 i[p(x) + iq(x) + (i^2)r(x) +…] = 6ix + 3i^2
(since 3x^2 – 3x^2 = 0)
 p(x) + iq(x) + (i^2)r(x) +… = 6x + 3i
(dividing both sides of the equation by i)
At this point, after the equation is simplified, Lagrange would say the i equals zero.
In this case, our f ‘ (x) equals 6x. Simply by looking at the function, anyone who has
been educated in modern day Calculus can easily verify that 6x is indeed the
derivative of 3x^2.
Lagrange’s approach toward making the Calculus more rigorous, however,
had not been thoroughly tested. Other theorists and mathematicians were able to
poke holes in Lagrange’s series-based method. How could we know with certainty
that every function would follow the same manner of simplification and canceling of
i’s as in the example above? Cauchy proved that Lagrange’s method did not work in
every case. He showed that “the function f(x) = {e^(-1/x^2) if x ≠ 0, 0 if x = 0
and all of its derivatives are zero at x=0. Consequently, as a power series about the
origin, f(x) = 0 + 0*x + 0*x^2 + 0*x^3 +…= 0, which in turn means that, if we begin
with f and write it as a series, we end up with a different function than we started
with! As a series, we would find it impossible to distinguish between f above and the
constant function g(x) = 0” (Dunham, 2005, p. 74). Hence, piecewise functions such
as this could not be assessed through Lagrange’s method. Although there were
problems with Lagrange’s power series approximations, Cauchy and others still
used and adapted some of his ideas, and Lagrange developed a number of proofs
concerning derivatives. “For instance, to prove that a function with positive
derivative on an interval is increasing there, to prove the mean-value for
derivatives, and to obtain the Lagrange remainder for the Taylor series” (Grabiner,
1983, p. 191). Grabiner also notes that Cauchy worked off of Lagrange’s efforts
coupled with his own to develop a more precisely defined approach with
derivatives. “Cauchy brought Lagrange's inequality properties and proofs together
with a definition of devised to make those techniques rigorously founded”
(Grabiner, 1983, p. 191-192).
Lagrange’s work in making approximations with power series expansion was
not ultimately accepted, but his work with inequalities was helpful in the process of
making the Calculus more rigorous (not completely rigorous). “Lagrange invented a
new approximation method using continued fractions, and by extremely intricate
inequality-calculations, gave necessary and sufficient conditions for a given iteration
of the approximation to be closer to the result than the previous iteration”
(Grabiner, 1983, p. 188). By expansion of fractions, Lagrange was able to
approximate the irrational numbers of quadratic equations. Lagrange’s theorem on
continued fractions states that “the continued fraction expansion of any quadratic
irrational is periodic from some point on, and, conversely, every periodic continued
fraction represents a quadratic irrational” (Olds, 1963, p. 89). Periodic relates to
how a number ends up repeating itself. An example is when you look at some
fractions in decimal form on a calculator. If you divide 1 by 3, the calculator will
show .33333333333. If you saw this periodic type of repeating numbers when a
fraction was already expanded, you would know that you were dealing with an
irrational number. Conversely, by expanding an irrational number, you would
expect at some point to see a pattern of repeating numbers. For instance, expanding
the square root of 2 results in a continued fraction [1, 2, 2, 2, …], where 1 is the
biggest integer that comes from the square root of 2, and the following 2’s represent
smaller and smaller expanded fractions and are periodic. With each layer of the
continued fraction, we get a closer and closer approximation to the actual number.
The periodic patterns are important because, when we reach that point, we don’t
need to expand any further. Expansions on square roots are unique in that they are
infinite. To expand a square root, you start by taking the largest perfect square
within the number under the radicand, taking the square of that number, and adding
it to a variable (an unknown). For example, take the square root of 5:
sqrt(5) = sqrt(4) + x = 2 + x
This makes sense because the square root of 5 will be 2 plus some unknown
amount. Next, you can square both sides of the equation:
5 = (2 + x)^2 = 4 + 4x + x^2
The end goal here is to put the equation in terms of x, so we keep playing with the
numbers until we can present what x equals:
1 = 4x + x^2 (subtracted 4 from both sides)
1 = x(4 + x)
x = 1/(4 + x)
(rearranging, factoring out x)
(divided both sides by 4 + x)
From this point, you have successfully transformed the square root into the form of
a fraction that can be expanded. Recall from the beginning of this process that the
square root of 5 equals the square root of 4 plus x. Now that we know what x equals,
we can say that the square root of 5 equals 2 plus 1/(4 + x). Expanding out from
here, we get sqrt(5) = 2 + 1/(4 + 1/(4 + 1/(4 + x))). We can keep expanding out from
here forever. The expansion of the square root never stops. Thus, the expansion of
the square root of 5 can be represented with [2, 4, 4, 4…]. A general formula for
expanding a square root is this:
1) Let sqrt(n) = a + x, assuming n is not a perfect square and a^2 is less than n.
2) Square both sides.  n = a^2 + 2ax + x^2
3) Subtract any values that do not have multiples of x to isolate x.
4) Factor out x.  n – a^2 = x(2a + x)
5) Divide both sides by (2a + x)  x = (n – a^2)/(2a + x)
6) Substitute x into original equation  sqrt(n) = a + (n – a^2)/(2a + x) =
a + (n – a^2)/(2a + (n – a^2)/(2a + x)) =…
(http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Fibonacci/cfINTRO.html)
Such an infinite expansion of a square root raises many questions. Firstly,
how can the square root of a number expand infinitely? What does that mean about
square roots? Are square roots infinitely large, infinitely small, or neither? Why do
some numbers expand infinitely while other numbers expand only to a finite
position? The expanding towards infinity seems to be related to precision. For
instance, say that I want to split $100 evenly between two buddies and myself. That
would be 1/3 of $100 for each of us. If we each get 30% of $100, then we will each
get $30. That leaves $10. Now, say we each take 33.3% of the $100. That would give
each of us $33.30, leaving only 10 cents left. Adding more and more 3’s after the
decimal gives us a more and more precise number.
Lagrange attempted to simplify the Calculus through Algebra, and in some
ways he did contribute toward a more formalized Calculus. He was responsible for
some of today’s Calculus notations. For lack of a better term, in Lagrange’s work
with expansions, he was “approaching” some key concepts that are very much a part
of today’s rigorous Calculus. Although Lagrange was not successful in formulating a
more rigorous Calculus, his work was not futile. Through his attempts to simplify
and formalize the Calculus, he was able to develop many useful theorems and
formulas for Algebra.
References
Boyer, C.B. (1959). The History of the Calculus and its Conceptual Development.
New York, NY: Dover.
Dunham, W. (2005). The Calculus Gallery: Masterpieces from Newton to Lebesgue.
Princeton, NJ: Princeton University Press
Grabiner, J.V. (1983). Who Gave You the Epsilon? Cauchy and the Origins of Rigorous
Calculus. The American Mathematical Monthly, 90(3), 185–194.
http://doi.org/10.2307/2975545.
Olds, C.D. (1963). Continued Fractions. New York, NY: Random House, Inc.
Surrey University. (1996-2016). Retrieved from a website:
http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Fibonacci/cfINTRO.html.