

Subject:
Precalculus
Category: Science > Math Asked by: mattkap123ga 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 TI83 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: funkywizardga 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 ti83 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 Ti89 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 number. From what I could glean from page 21 of the Ti 83 plus developer guide [ http://www.ti.com/calc/pdf/sdk83pguide.pdf ], the variables on the TI83 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: peterungarga 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 9digit floating point numbers, and that it uses some extra digits in each arithmetic operation with the result of each operation rounded to the nearest 9digit 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 exponent.) 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 digits. Then b = (1 + 1/n)^n becomes 1.00000001^150,000,000 = a^1.5 = 4.4816... 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: gaussianeulerga 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 n>OO i wish math wasnt so darn applicable! :) 
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