Re: doubt on indeterminate forms


"Roger Stafford" <ellieandrogerxyzzy@xxxxxxxxxxxxxxxxxxxxxx> wrote in
message
news:ellieandrogerxyzzy-2404061235350001@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
In article <e2emcv$dq5$1@xxxxxxxxxxxxxxxxxx>, "Steven Lord"
<slord@xxxxxxxxxxxxx> wrote:

"nithya" <nithya.ramakrishnan@xxxxxxxxx> wrote in message
news:ef3188f.-1@xxxxxxxxxxxxxxxxxxx
all the maths text books say that anything other than zero when
raised to zero give 1 as the answer and that 0 raised to 0 is an
indeterminate form.
but in matlab when 0^0 is typed 1 is returned as the answer instead
of an error message

does anybody have an answer for that

This is the most commonly asked question, by a wide margin, over in the
sci.math newsgroup. Here's their FAQ article that explains why 0^0 is
commonly taken to be 1:

http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

Now that I think about it, 0^0 may not be the most common question on
sci.math ... I forgot about the other common sci.math question, whether
0.99999..... is equal to 1. Even if it isn't the most common question, 0^0
is one of the big ones.

--
Steve Lord
slord@xxxxxxxxxxxxx
-----------------------
Nithya, Rune, and Steve, I have had to revise my ideas on the 0^0
question since reading the article by Alex Lopez-Ortiz in sci.math
referred to Saturday by Steve at

<http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/>.

It was an eye-opener.

That 0^0 is indeterminate remains a fact, since, for example,

lim, x->0, (exp(-log(t)/x^2)) ^ (x^2) = 1/t

for any t > 1, even though this is a 0^0 situation. (In fact, the
expression is identically 1/t.)

However, I no longer contend that 0/0 and 0^0 are qualitatively alike.

Indeed. It's been a little while since I read IEEE 754, and I could be
wrong, but I believe that exponentiation isn't addressed specifically in the
standard, while division is. This seems to be supported by page 196 of
David Goldberg's article and page 3 of Professor Kahan's notes linked from
this page:

http://grouper.ieee.org/groups/754/

It is quite difficult to come up with an example of 0^0 that fails to
converge to 1, whereas with 0/0 it happens easily. The above article
mentioned that f(x)^g(x), with f and g both approaching zero and both
analytic at x = 0 must have a limit of 1, and this is fairly easy to
demonstrate. In the above example, exp(-log(t)/x^2) possesses an
essential singularity at x = 0, meaning that it cannot be expanded in a
power series about that point, so it has to be considered as exhibiting
fairly aberrant behavior there.

In the binomial expansion of (x+y)^n for n an integer, if, say, y is set
to 0, the expansion would give x^n*0^0 which ought to be equal to x^n and
would create an inconvenience if a special case had to be made for the
zero value.

When I first read that FAQ article a while ago, this was the one that was
most convincing for me -- I can accept that 0^x is not equal to 0 when x=0
if it makes the binomial theorem work as we expect for that case.

For me, though, the most convincing argument of all is to imagine using
matlab to investigate the behavior of x^y in the immediate vicinity of the
point (0,0). Suppose we did

[x,y] = meshgrid(0:1e-8:1e-5,0:1e-8:1e-5);

for a very small 1e-5 by 1e-5 square in the upper-right neighborhood of
(0,0). Then imagine doing a 'plot3' of z = x.^y to a high resolution in
z. (I have to imagine it because my version of matlab doesn't permit this
many elements in an array.) Of course, along the y-axis for the thousand
points with y > 0 and x = 0, we will get z = 0. But for all other of the
million-odd points in the square, z lies between 1 and 0.9998 in value!
There is a virtually flat plateau of values very near 1 over the whole
square except for the single line at x = 0. This dramatically illustrates
that x would have to approach 0 very much faster than y for x^y to
approach anything but a 1 as the point (0,0) is approached.

So, Steve, I hereby absolve MathWorks of any carelessness in setting 0^0
equal to 1, and they have my apology for saying so. They have good
practical reasons for doing this in spite of its indeterminacy. However,
if this hasn't been done already, it would be good to place a footnote in
matlab manuals and in the help files for the '^' operation, explaining
briefly why this has been done. Otherwise it is bound to remain a puzzle
to students of calculus who have learned of the indeterminacy.

We could, but this is only the second or third time I can remember being
asked this in almost 5 years here at The MathWorks, and I'd worry a bit
about that note confusing more people than it helped.

--
Steve Lord
slord@xxxxxxxxxxxxx


.

Relevant Pages

  • Re: doubt on indeterminate forms
    ... Steve Lord ... with f and g both approaching zero and both ... matlab to investigate the behavior of x^y in the immediate vicinity of the ... million-odd points in the square, z lies between 1 and 0.9998 in value! ...
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  • Re: doubt on indeterminate forms
    ... raised to zero give 1 as the answer and that 0 raised to 0 is an ... but in matlab when 0^0 is typed 1 is returned as the answer instead ... Here's their FAQ article that explains why 0^0 is ... Steve Lord ...
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