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Q: Precalculus ( Answered,   2 Comments )
Subject: Precalculus
Category: Science > Math
Asked by: mattkap123-ga
List Price: $2.50
Posted: 01 Nov 2002 17:27 PST
Expires: 01 Dec 2002 17:27 PST
Question ID: 96015
What explains the fact that when the number 1,314,000 is entered into
the formula for compound interest {(1 + 1/n)^n} in the TI-83 graphing
calcualtor the answer is more precise than when the number 78,840,000
is entered into the same equation when it should not be?
Subject: Re: Precalculus
Answered By: funkywizard-ga on 01 Nov 2002 20:49 PST
Your calculator has a limit on how many digits of significant figures
it can store for any one variable. The larger number of the two that
you describe, when put through your formula, cannot maintain all of
its significant data inside the size of the variable that the ti-83 is
using. This would be true of any calculator or computer for that
matter, with the largest size you can use being different for each
computing platform. For instance, a Ti-89 calculator would probably
not have a problem with this calculation until the number were
increased another few orders of magnitude, but you could always find a
number that is large enough such that any calculator could not
calculate to the same level of precision as it could with a smaller

From what I could glean from page 21 of the Ti 83 plus developer guide
[ ], the variables on the
TI-83 seem to occupy 9 bytes of space, the first byte for identifying
the variable type, with the second byte indicating the power of a
floating point number (powers can range from -127 to +128), with the
remaining 7 bytes dedicated to the variables significant figures (the
maximum number of digits it can represent without losing accuracy).
They go on to say that each byte has been defined to hold two 10 base
digits (the numbers we are all familiar with) of significance, so that
gives you a total maximum value you can store (without losing
accuracy) of 14 digits.

I suspect therefore that your function's manipulation of your data at
some point uses a number that has more than 14 digits and must cut off
some of the significant figures when you are dealing with very large
numbers (in the tens of millions), leading to a loss of precision.
Even though the number you started with is not 14 digits long, some of
the intermediary numbers used when calculating the answer to your
function likely exceed 14 digits, causing the loss of accuracy.

I hope this adaquately answers your question. If it does not, please
request a clarification of the answer before rating it and I will be
happy to assist you further.
Subject: Re: Precalculus
From: peterungar-ga on 01 Nov 2002 20:21 PST
I don't know exactly how the graphing calculator does arithmetic, but
the reason why such a result can be expected is that a calculator uses
a fixed number of binary or decimal digits to represent numbers.

   Suppose for example that the calculator represents numbers as
9-digit floating point numbers, and that it uses some extra digits in
each arithmetic operation with the result of each operation rounded to
the nearest 9-digit floating number.  (If the arithmetic is done in
binary, that adds an extra complication, but the nature of the
phenomenon is the same).

If we take n = 100,000,000 the exact value of 1 + 1/n  is 1.00000001 
and the corresponding approximation to  e  is

a = 1.00000001^100,000,000  (The ^ means: what follows is an

If the power is evaluated without much further error, this should
approximate  e = 2.718281828.. to  9 digits.  It would seem that doing
99,999,999 multiplications would introduce lots of rounding errors but
the calculator uses logarithms for evaluating powers and should be
able to do it with negligible rounding errors.

Now take n = 150,000,000 which, without rounding, should give an even
better approximation to  e  than the number  a  above.  However,
rounding to  9 digits we get

1 + 1/150,000,000 = 1.0000000066666... = 1.00000001 rounded to 9

Then  b = (1 + 1/n)^n becomes 1.00000001^150,000,000 = a^1.5 =

which is way off the mark.  If  n > 200,000,000 then the rounded value
of  1 + 1/n is exactly 1,
and (1 + 1/n)^n becomes just 1.
Subject: Re: Precalculus
From: gaussianeuler-ga on 25 Nov 2002 18:36 PST
It truly saddens me that ((1 + 1/n)^n) is the "formula for compound
interest" to you, instead of the beautiful statement...
lim   (1 + 1/n)^n   =  e

i wish math wasnt so darn applicable!

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