A Tale of Two Problems (and a little history as well)
Algebra Prelude: "Geometric Series"
(1-x) (1 + x + x2 .... +xn-1 ) =
1 - xn
So 1/ (1-x) = xn / ( 1 - x)
+ 1 + x + x2 .... + xn-1
AND (1+x) (1 - x + x2 .... +(-1)n-1xn-1 ) =
1 +(-1)n-1xn
So 1/ (1+x) =(-1)n-1xn / ( 1 - x)
+ 1 - x + x2 .... +(-1)n-1 xn-1
Substitution:
`int_1^{1+h} 1 / x dx = int_0^h 1 / {1+u} du `;
Use x = 1 + u
so u = x - 1.
----------------------------------------------------------
`int_{1-h} ^1 1 / x dx = int_0^h 1 / {1-u} du` ;
Use x = 1- u
so u = 1-x and du = - dx .
Problem I : [Newton Estimates for the Natural Logarithm of 2]
What is the area of the region in the plane bounded by
the X-axis, X=1, X=2 and Y = 1/X ?
Problem II: Taylor / Maclaurin Polynomial Estimate for An Important Integral from Probability
What is the area of the region in the plane bounded by the X-axis, X=0, X=0 and `Y = e^{-X^2}`? or
what is `int_0^1 e^{-x^2} dx` ?
I. Newton:
In 1676 Isaac Newton wrote in a letter to Henry Oldenburg
on some of his applications of polynomials and series to estimating areas.
In particular he related how he estimated areas for the natural logarithm
which was also called the hyperbolic logarithm in those days.
This work was later clarified
in the book Of the Method of Fluxions and Infinite Series which was published
posthumously in 1737, ten years after Newton's death.
So...
A little background on the hyperbolic logarithm:
In 1637, Descartes published
La Geometrie as an appendix to his Discours
de la Methode. This started the investigation of geometry using arithmetic
and algebra through common measurments, coordinates, and equations that described
curves.
By about 1640, the solution to the
"area problem" for curves with equation Y n = aX m
was known by Fermat for all integer cases except when n = 1,
m = -1.
I.e., the only unsolved area problem was for Y = 1/X - the standard equation for the graph of a hyperbola.
In 1647, Gregoire de St. Vincent showed the following special property for hyperbolas:
Proposition 1: If a/b = c/d
then
the area under the hyperbola above the interval [a,b] is equal to the area under the hyperbola above the interval [c,d].
The proof of this may be discussed if time permits. You can try it as an exercise in integration substitution.
In 1649 Alfonso Antonio de Sarasa recognized this feature
in Gregoire's work and connected it to the properties of logarithms.
In particular he recognized the following
additive property of these measurements which had previously be a key feature
in the study of common logarithms (base 10):
Proposition 2: If the areas are all measured using a = 1, then
the area determined by a product
of two numbers ,
rs, is equal to the sum of the areas determined
by r and s separately.
Here's why:
We can see easily that `1/r =s/{rs}`.
So by Proposition 1 thearea under the hyperbola above the interval [1,r] is equal to the area under the hyperbola above the interval [s,rs].
But the area under the hyperbola above the interval [1,rs] can be cut into two pieces above the intervals [1,s] and [s ,rs]. so the area determined by a product
of two numbers ,
rs, is equal to the sum of the areas determined
by r and s separately.
From the calculus you've studied you should recognize this area as a definite integral of 1/x from 1 to r and from the Fundamental Theorem of Calculus, this area is precisely the logarithm base e evaluated at r.
i.e. `int_1^r 1/x dx = log_e (r)`
So
we can recognize from what you've learned in calculus that this Natural Logarithm
is the same as the logarithm with the base e. But how does one actually find these
logarithmic values. For example, how can we estimate the natural logarithm
of 2?
Using rectangles or trapezoids or even parabolas to estimate the area is not particularly accurate for estimating ln(2).
Try a quick estimate yourself with n = 10. [Demonstrate with winplot.]
RH: 0.66877 LH: 0.71877 MP: 0.69284
Newton gives this estimate using only 15 simple calculations:
0.6931471805599453
So... How did Newton make estimates for this area that were so very accurate yet easy to compute?
Newton considers the graph of 1/x and symmetrically located points on the main
axis, 1+x and 1-x with x > 0 and their related reciprocals.
Newton is familiar with the "geometric series" so he knew he could estimate 1/(1-x) and 1/(1+x) with polynomials.
He then uses two integrals related to the geometric
series to determine the related areas, (i) between the hyperbola and above the segment [1,1+x]
(red)
P=Area AFDB =
and
(ii) between the hyperbola and above the segment [1-x,1] (blue): Q=Area AFdb =
`int_0^h 1/{1-x}dx = int_0^h 1 + x + x^2 + ...+ x^{k-1} +...= h + {h^2}/2 + {h^3}/3 + ...+ {h^k}/ k + ...`
These allow the estimation of the sum and difference
of the two areas:
T = Total Area of bdDB = Q+P = `2h + 2{h^3}/3 + 2{h^5}/5 + 2{h^7}/ 7 + ...`
S = Difference of Areas Ad-AD = Q-P = `h^2 + {h^4}/2 + {h^6}/3 + {h^8}/ 4 + ...`
So we can estimate P and Q by :
Q = 1/2 [T+S] and P = 1/2[T-S].
Now Newton uses the first eight terms with h = .1
(and .2) to estimate the hyperbolic log of .9 and 1.1 (as well as .8 and 1.2).
[Tables created with Microsoft Excel]
h
0.1
.2
.01
2h
0.2
0.4
0.02
2h3/3
0.000666666666666
0.00533333333333
0.00000066666666667
2h5/5
0.000004
0.000128
0.00000000004
2h7/7
0.0000000285714286
0.00000365714285714
0.00000000000000286
2h9/9
0.0000000002222222
0.00000011377777778
2.22222222222e-19
2h11/11
0.0000000000018182
0.00000000372363636
2h13/13
0.0000000000000154
0.00000000012603077
2h15/15
0.0000000000000001
0.00000000000436907
T=Sum of Areas
0.200670695462151
0.405465108108002
0.020000666706670
h
0.1
.2
.01
h2
0.01
0.04
0.0001
h4/2
0.00005
0.0008
0.000000005
h6/3
0.0000003333333333
0.00002133333333
0.000000000000333
h8/4
0.0000000025
0.00000064
2.50000000000e-17
h10/5
0.00000000002
0.00000002048
h12/6
0.0000000000001667
0.00000000068267
h14/7
0.0000000000000014
0.00000000002341
Diff'ce of Areas
0.010050335853501
0.040821994519406
0.000100005000333
Now to find the Area of the two separate regions,P and Q, (and therefore the related
logarithms) we take 1/2 of the difference of these results and 1/2 of the
sum of these results.