Algebra Prelude: "Geometric Series"

(1-*x*) (1 + *x* + *x*^{2} .... +*x*^{n-1} ) =
1 - *x*^{n }

So 1/ (1-*x*) = *x*^{n } / ( 1 - *x*)
+ 1 + *x* + *x*^{2} .... +* x*^{n-1}

AND (1+*x*) (1 - *x* + *x*^{2} .... +(*-*1)^{n-1}* **x*^{n-1} ) =
1 *+*(*-*1)^{n-1} *x*^{n }

So 1/ (1+*x*) =(*-*1)^{n-1} *x*^{n } / ( 1 - *x*)
+ 1 - *x* + *x*^{2} .... *+*(*-*1)^{n-1}* x*^{n-1}

(1-

So 1/ (1-

AND (1+

So 1/ (1+

Substitution:

Problem I : [Newton Estimates for the Natural Logarithm of 2]

What is the area of the region in the plane bounded by

`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 .

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` ?

So...

A little background on the

In 1637, Descartes publishedLa Geometrieas an appendix to hisDiscours 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/dthen thearea under the hyperbola above the interval[a,b] is equal to thearea 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 byrandsseparately.

Here's why:`1/r =s/{rs}`.

We can see easily that

So by Proposition 1thearea under the hyperbola above the interval[1,r] is equal to thearea 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 byrandsseparately.

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:

So... How did Newton make estimates for this area that were so very accurate yet easy to compute?

Newton considers the graph of 1/

Newton is familiar with the "geometric series" so he knew he could estimate 1/(1-

P=Area AFDB =

(ii) between the hyperbola and above the segment [1-x,1] (blue):

S = Difference of Areas Ad-AD = Q-P = `h^2 + {h^4}/2 + {h^6}/3 + {h^8}/ 4 + ...`

**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).**

h |
0.1 | .2 | .01 |

2h |
0.2 | 0.4 | 0.02 |

2h^{3}/3 |
0.000666666666666 | 0.00533333333333 | 0.00000066666666667 |

2h^{5}/5 |
0.000004 | 0.000128 | 0.00000000004 |

2h^{7}/7 |
0.0000000285714286 | 0.00000365714285714 | 0.00000000000000286 |

2h^{9}/9 |
0.0000000002222222 | 0.00000011377777778 | 2.22222222222e-19 |

2h^{11}/11 |
0.0000000000018182 | 0.00000000372363636 | |

2h^{13}/13 |
0.0000000000000154 | 0.00000000012603077 | |

2h^{15}/15 |
0.0000000000000001 | 0.00000000000436907 | |

T=Sum of Areas | 0.200670695462151 | 0.405465108108002 | 0.020000666706670 |

h |
0.1 | .2 | .01 |

h^{2} |
0.01 | 0.04 | 0.0001 |

h^{4}/2 |
0.00005 | 0.0008 | 0.000000005 |

h^{6}/3 |
0.0000003333333333 | 0.00002133333333 | 0.000000000000333 |

h^{8}/4 |
0.0000000025 | 0.00000064 | 2.50000000000e-17 |

h^{10}/5 |
0.00000000002 | 0.00000002048 | |

h^{12}/6 |
0.0000000000001667 | 0.00000000068267 | |

h^{14}/7 |
0.0000000000000014 | 0.00000000002341 | |

Diff'ce of Areas | 0.010050335853501 | 0.040821994519406 | 0.000100005000333 |

**Thus ln(1.1)= 1/2 ( 0.2006706954621511- 0.0100503358535014)
**

= 0.0953101798043248

**while ln(.9) = -(1/2)( 0.2006706954621511 + 0.0100503358535014)
**

= -0.105360516578263 .

**And ln(1.2)= 1/2 (0.405465108108002 -0.040821994519406
)
**

= 0.18232155576939546 (from Newton)

**while ln(.8) = -(1/2)(0.405465108108002 +0.040821994519406
)
**

= -0.2231435513142097 (from Newton) .

**Newton can also find ln(1.01) and ln(.99) by using h=.01
from the same tables.
**

**Then Newton finds other logarithms with great accuracy
using these.
In particular Newton calculates the estimate for **

**ln(2) = ln ( 1.44 / .72) = ln (1.44)-ln(.72) = 2ln(1.2) -[ln(.9 +ln(.8)] **

= (approx) 2(0.18232155576939546) +0.105360516578263+ 0.2231435513142097

WOW! :)

Now on to Problem II: