Date: Oct 24, 2012 9:00 AM Author: Alan Lipp Subject: RE: [ap-calculus] e NOTE:

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Erin and All,

To add to Jeff's numerical description of the limited world of finite calculator arithmetic, I like to give my students visuals of this as well. Try graphing y = (1 + 1/x)^x. In any normal window this looks like an exponential function increasing and leveling off to a horizontal asymptote at y = e. You can see this behavior nicely for x in [0, 20] and y in [0, 3].

Extend the x-range to [0, 10^6] and all you see is a horizontal line y = 2.718279.

Now graph with x in [0, 10^13] and ask yourself and your kids what is going on?

And, finally, to see the behavior you described graph in [0, 10^17]

Alan

Alan Lipp

Williston Northampton School

-----Original Message-----

From: Jeff Stuart [mailto:jeffrey.stuart@plu.edu]

Sent: Tuesday, October 23, 2012 8:08 PM

To: AP Calculus

Cc: AP Calculus

Subject: Re: [ap-calculus] e

NOTE:

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Erin:

Welcome to the world of floating point arithmetic. The problem is that calculators and most numerical software do not treat all real numbers the same way, and that except for a few special numbers, at every step of the computation process, numbers are replaced by "convenient"

approximations. For simplicity, imagine that you calculator only keeps six digits. Thus, your calculator would replace pi with 3.14159. If

you ask your calculator for (pi - 3.14159)^(-1), you will get an

error, because the first computation your calculator does is pi -

3.14159 = 0, and it then faces 0^(-1), which it knows is undefined. As another example, 1/3 stores as 3.33333 x 10^(-1), so 3 x (1/3) is computed as 3.00000 x 3.33333 x 10^(-1) =9.99999 x 10^(-1). Notice that

1 - 3(1/3) computes as 1.00000 -0.999999, and the first thing that happens in our simplified calculator is that the last 9 is dropped so the digits line up. Thus the difference computed is 0.00001, rather than 0.000001 . Further, the result is stored as 1.00000 x 10^(-4), so that neither of the first two digits is correct! This is called catastrophic cancellation. Now suppose you ask your calculator to compute (1 + 1/x )^x, where x has been chosen to have exactly six digits (or at least six digits in scientific notation). The first computation is 1/x, which is stored as a six digit approximation "a", and then your calculator computes 1 + a, which it stores as a six digit approximation

"b", and then, finally, it computes b^x. If x > 10^5, then 1/x <

10(-5), so a =< 10^(-5), meaning that as a decimal, a =0.00000##########...#000000... where at most six of the # digits are nonzero.. When the calculator computes 1 + a as 1.00000 + a, it only keeps the first FIVE decimal places of a, so 1 + a = 1.00000 rather

than 1.00000####....#0000... we

Since b = 1.00000, it is not surprising that b^x = 1. OH, by the way,

all of this happens base 2 or base 8 or base 16 rather than base base 10.

You might look up "floating point arithmetic", "IEEE floating point", and/or "truncation versus rounding".

Jeff S.

--

Professor Jeff Stuart, Chair

Department of Mathematics

Pacific Lutheran University

Tacoma, WA 98447 USA

(253) 535 - 7403

jeffrey.stuart@plu.edu

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