Re: .99999... still=/= 1

From: tinyurl.com/uh3t (rem642b_at_Yahoo.Com)
Date: 12/04/04 

Date: Sat, 04 Dec 2004 13:57:35 -0800

> From: smart1234@aol.com (S. Enterprize Company)
> A scientific calculator does approximate convergence with a certain
> number significant digits.

Just because a calculator emits a particular number of digits of
output, doesn't say anything about how accurate the represented number
is, i.e. how close to the correct answer it is. If you do a chain of
calculations, the accuracy usually gets worse and worse, yet the digits
output remains the same, looking like it's that accurate (all the way
out to the last digit shown) when in fact many of the last digits shown
are totally wrong. In fact with a long enough chain of calculations,
you can produce a displayed result which may have a whole bunch of
digits shown but not even one significant digit of accuracy.

For example, start with the exact integer 2, then take square root,
then take square root of that, etc., five square roots in a row. Now
work your way backward by multiplying the value by itself five times in
a row. The correct mathematical answer is eactly 2, but your calculator
won't show that. Instead it'll show some exact value that starts out
2.00 but then somewhere has a bunch of non-zero digits, or starts out
1.99 but then somewhere has a bunch of non-nine digits. Do you honestly
believe that the correct mathematical result is correctly shown to all
the digits displayed on your calculator?

The point I'm making is that you can't use the results given by such a
calculator to establish truth of any mathematical fact.

Now if you had a calculator that used interval arithmetic internally,
and which displayed *only* as many digits as were accurate per the
interval (lower&upper bounds on correct result), *then* you could
simply look at the output and know that all digits except the last are
accurate. But I'm not aware of any calculators that work that way.

> I think my calculator about 32 significant digits.
> 1/3 = .3333333333333333333333333333
> something like this. From this calculation on the calculator we can
> see that,
> 1/3 = .333...

No, from the output of your calculator you can't see anything like that
except in your imagination, or if you happen to already know the
correct result and are just ignoring what the calculator says.

In that case, you got lucky, the output from the calculator happened to
be accurate to the last digit shown. For a few classes of simple
calculations, such as dividing small integer by small integer, that's
generally true. But in general, even with all those 3's correct, that
still doesn't give you any proof or assurance that the rest of the
digits beyond that point will fulfill the "obvious" pattern you see
within those limited digits. Try this on your calculator:
10000000000000000001/30000000000000000004
Now you know the correct mathematical result is not 1/3, so if you see
3333333333333333333333333333 and just assume the pattern of digits of
3 continues forever then you should know you are wrong.

> Partial Sum Convergence does something like this. It uses a certain
> number of finite significant digits, then it finds a converges value.

And like the calculator, you can't really trust the value it gives you.
It might be correct out to the number of digits it shows, or it might
not be, and even if it's correct to that point you can't conclude any
pattern of digits you see will be condinued forever in the
mathematically correct result.

> This convergence value is a number that the series APPROACHES NOT
> EQUALS.

Your language there is very sloppy. FIrst, do you mean the true
mathematical limit or the approximate value given by the calculator?
Second, by "the series" do you mean any finite segment of the series,
i.e. any partial sum, or do you mean the whole series, which is defined
as the limit of partial sum as number of terms increases without bound?
If you mean exact mathematical value of limit, and value of whole
infinite series, then you are wrong because those two ways of
expression mean exactly the same thing hence are of course equal.
If you mean exact value of limit but only an individual partial sum not
the limit, then you are correct, no individual partial sum (in the
series .3 + .03 + .003 + .0003 + ... we're discussing here) exactly
equals the limit.
If you mean the value produced by your calculator, then you are wrong
because depending on which way roundoff error occurs, the result it
gives you can be greater than the mathematical limit, or smaller, or
exactly correct, and you have no way of knowing for any given problem
which is the case, so any statement you make about it definitely one
way or the other is a wrong statement.

> .999... can NEVER = 1

When any competant mathematician uses the notation .999... he/she means
the limit of the series, where the pattern of digits of 9 are continued
forever without exception. That limit of the series is defined as the
limit of partial sums. That limit is in fact exactly 1.

You seem to mean something other than the limit of partial sums when
you write that notation, and I've asked you several times to say what
exactly you mean by your use of that notation, and still you have
failed to tell me. So I ask you again, what do you mean by .999... as
you use it just above, and as you used .99999... in the Subject of this
thread. Do the two notations .999... and .99999... mean something
different to you, or are they just two different amounts of effort put
into denoting the same thing? In any case, what did *you* mean when you
wrote .99999... in the Subject field, and what did *you* mean when you
wrote .999... in the message I'm responding to now?

> Theorem 1 ( Divergence )
> If a series z_1 + z_2 + .... converges, then
> lim z_m = 0
> m-->oo

That is correct. Note that it's a one-way test. The converse is not true.
(If you don't know what "converse" means, please look it up online.)

> Hence if the series doesn't satisfy this condition, it diverges.

Correct. That's the contrapositive, which is necessarily true if the
original statement is true. (If you don't know what contrapositive
means, look it up online.)

> .999... = .9 + .09 + .... + ( 1 - 1/10^m)

Your notation is unclear. What is that extra term at the very end?
It appears to be either an error term or the individual mth term of
the sequence that is being summed to make the series.

If it's supposed to be the mth individual term, then you need to write:
> .999... = .9 + .09 + .... + ( 1 - 1/10^m) + ...
to indicate that the series doesn't stop at some point but continues on
forever. The expression you wrote with it stopping after the mth term
is *not* equal to the infinite series indicated by the left side, so
what you wrote is just plain wrong.

If you intended the partial sum, you should have written:
  .9999999...99999999 = .9 + .09 + .... + ( 1 - 1/10^m)
   |<- (m digits) ->|

If it's supposed to be the error term, that is the amount you have to
fudge a partial sum to make up all the rest of terms you haven't
included, then first of all that's not the correct error term, and the
part before that isn't a partial sum, it's the notation for the entire
series. You should have written:
  .999... = (.9 + .09 + .... + 0.000000...000000009) + 1/10^m
                                 |<- (m digits) ->|
where the first parenthesized expression is the partial sum and the
final expression is the error term. Note that if the
right side (the mth partial sum plus the mth error term) is a constant,
then the notation makes sense. In that case, the series converges if
and only if the error term approaches zero, and if it converges then
the constant right-side value is the limit. In this case, indeed the
right side is a constant (independent of m), that constant value is 1,
and the error term does indeed approach zero, so the limit of the
series is exactly 1. The only difficult part of that proof is showing
the right side has the same exact value for all values of m. That is
you must show that .9 + 1/10 equals 1 (easy), and that
(.9 + .09) + 1/100 also equals 1 (pretty easy), and that
(.9 + .09 + .009) + 1/1000 also equals 1 (not too hard), etc. etc. etc.

However this way of proving a series converges has virtually nothing to
do with the divergence test you listed above, so I you must be confused
to introduce it here.

> Z_m = ( 1 - 1/10^m)

If Z_m is supposed to be the mth term of the sequence you are summing,
then Z_m = 9/10^m, not what you wrote there. The right side of what you
wrote is the partial sum Z_1 + Z_2 + ... + Z_m, not the individual term
Z_m.

> lim 1 - 1/10^m = 1
> m -->oo

That's correct! The limit of partial sums (not individual terms) is
exactly 1, proving the series converges to 1.

> So, .999.... doesn't converge to 1.

You must be awfully confused, given that you just finished proving that
it *does* converge to 1. I suspect you are confusing the mth term with
the mth partial sum. You didn't prove the mth term approaches 1, you
proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m
which indeed approaches zero.

Relevant Pages

  • Re: Thatll Teach Em - Long Division
    ... if only so you are not totally stumped if your calculator fails. ... I remember a teacher when I was at school in the 1970s whose standard ... punishment for any misdemeanour was "cubes". ... boys to call out digits (the number of digits depended on the severity of ...
    (uk.media.tv.misc)
  • HP30S numerical anomaly
    ... I posted about a peculiar HP30S behavior over on MoHP, ... Recently while looking at the calculator forensics web site, ... functions can give substantially less than 24 digits; ...
    (comp.sys.hp48)
  • Re: Math errors in python
    ... They don't even share three digits beyond the decimal point. ... > yet not all fractions can be represented by decimals. ... Hardware designers have settled on a binary representation of floating ... And BTW, your calculator is not, in general, more accurate than the ...
    (comp.lang.python)
  • Re: FAQ 4.2 Why is int() broken?
    ... is much of an issue - 8 or 12 digits are enough. ... For example on my HP-48 calculator. ... can display 0.999999999999 exactly, ...
    (comp.lang.perl.misc)
Quantcast