y = x^(1/x)
James
the xth root of x:
y = x^(1/x)
peaks when x = e.
Just out of interest, why is this? Can anyone offer proof? I know that
this is probably baby stuff for most of you... sorry if I waste your time :)
Thanks,
James
Gyan Doshi
James wrote:
> y = x^(1/x)
> peaks when x = e.
> Just out of interest, why is this? Can anyone offer proof?
With regards to your 2nd question.
Do you know differential calculus ?
If so, first take natural log (ln) of the function
ln y = (1/x) ln x
d(ln y) = (1/x^2) (1 - ln x) .... [1]
For x < e
ln x < 1 --> hence [1] is positive throughout (-inf, e)
In other words, it is increasing from (-inf, e).
For x > e
ln x > 1 --> hence [1] is negative throughtout [e, inf)
In other words, it is decreasing from [e, inf)
For x = e
ln x = 0 --> hence [1] is zero.
In other words, a maxima or minima is attained.
Since [1] is positive before and negative after 'e' for reals --> 'e'
is a maxima.
Since [1] implies monotonic behaviour before and after, it is a global
maximum.
Technically, there's still a crucial step required. (Linking the
derivative of the
log of the function to the function's derivative. I don't know if the
similar
behaviour is implied automatically.)
Gyan
Marko Marin
James wrote:
Write your function in exponential-logarithmic form:
y = exp(1/x*ln(x)), there ln(x) is natural logarithm and is defined for x > 0.
then differentiate using chain rule:
y´ = exp(1/x*ln(x)) * (1/^x^2 - 1/x^2*ln(x))
the extremal points (maxima, minima or terrace points) occur where is the
derivate zero,
y' = 0, now because exponential function is always > 0 ==>
1/x^2 * (1 - ln(x)) = 0, for x = e
now is this the maxima?
for 0 < x < e the derivate is positive because 1 - ln(x) > 0 in that interval,
that means that your function is monotonic increasing in that interval.
for e < x < +oo the derivate is negative because 1 - ln(x) < 0 in that interval,
that means that your functions is monotonic decreasing in that interval.
now observe that lim x -> +oo y' = 0, so the points x = e is global maximum
so function x^(1/x) attains it's higest value at point x = e.
This has interesting consequences, namely we can prove that the most "economic"
number base is e.
But that another story...
Cheers, Marko
Norm Dresner
enough to do that.
Look at x^(1/x) intuitively. The mantissa of the expression is definitely
an increasing function of x ;-) and if the mantissa were a constant, i.e.
2^(1/x) it's clearly a decreasing function. The only question then is which
dominates. [This isn't anywhere near rigorous but it's intuitively
satisfying and that's all I promised.] I can believe -- and most
mathematically inclined people will too -- that the only two "reasonable"
behaviors are increasing if the increasing mantissa dominates or (since it's
clearly increasing near x = 1) peaking and eventually decreasing (albeit
slowly since it can't go below 1.0000+) for large enough x. I can offer
several reasons why it peaks, but they all involve calculus, not intuition,
and that's another course.
Norm
John Prussing
>y = x^(1/x)
>Thanks,
> James
Others have responded as to why. My comment is that this is a
way of showing that e^PI > PI^e, without actual computation.
Once you know that e^(1/e) > PI^(1/PI), simply raise
the inequality to the power (e*PI).
--
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
John E. Prussing
Dept. of Aeronautical & Astronautical Engineering
University of Illinois at Urbana-Champaign
http://www.uiuc.edu/~prussing
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Dave L. Renfro
[sci.math Sun, 27 May 2001 07:30:22 -0500]
<http://forum.swarthmore.edu/epigone/sci.math/whenspedy>
wrote (in part)
> James wrote:
>
>> y = x^(1/x)
>> peaks when x = e.
>>
>> Just out of interest, why is this? Can anyone offer proof?
ln y = (1/x) ln x
d(ln y) = (1/x^2) (1 - ln x) .... [1]
[snip details]
> Since [1] is positive before and negative after 'e' for
> reals --> 'e' is a maxima. Since [1] implies monotonic
> behaviour before and after, it is a global maximum.
>
> Technically, there's still a crucial step required. (Linking the
> derivative of the log of the function to the function's
> derivative. I don't know if the similar behaviour is implied
> automatically.)
Taking d/dx of both sides of (ln y) = (1/x)*(ln x) gives you
(dy/dx)*(1/y) = (1/x^2) (1 - ln x), so
dy/dx = y*(1/x^2)*(1 - ln x) [2]
Your analysis isn't affected since y = x^(1/x) is positive
everywhere it's defined (i.e. for x > 0).
Here's a list of similar examples I came up with one afternoon
a few years ago. If anyone has similar ones (whose max/min's
can be obtained explicitly), I'd be interested. [Besides those
that can be obtained from one of the functions below by replacing
x with ax, or by replacing one occurrence of x with ax and another
occurrence of x with bx (a and b are positive constants).]
y = x^x local min @ x = e^(-1)
y = x^(1/x) local max @ x = e
y = x^[x*ln(x)] local max @ x = e^(-2), local min @ x = 1
y = (ln x)^(ln x) local min @ x = e^(1/e)
y = (ln x)^[-(ln x)] local max @ x = e^(1/e)
y = (1 + ln x)^(ln x) local min @ x = 1
y = [ln(ln x)]^[ln(ln x)] local min @ x = exp[e^(1/e)]
y = (ln x)^[ln(ln x)] local min @ x = e
y = [(ln x) - ln(ln x)]^x local min @ x = e
Dave L. Renfro
David W. Cantrell
[snip]
> Your analysis isn't affected since y = x^(1/x) is positive
> everywhere it's defined (i.e. for x > 0).
Hmm. I suppose that one might be tempted to say that there is an
absolute maximum at x = e. [Furthermore, if one extended by limit
as x -> 0+, thus taking f(0) = 0, one might also be tempted to say
that there is an absolute minimum there.] But such could be
considered as incorrect unless we explicitly restrict the domain to
nonnegative values. The default domain could be taken to be larger,
saying that f(x) = x^(1/x) is defined for some x < 0.
For example, f(-1) = -1 and f(-1/2) = 4. It all depends, I suppose,
on precisely what sort of exponentiation ^ is taken to represent.
> Here's a list of similar examples I came up with one afternoon
> a few years ago. If anyone has similar ones (whose max/min's
> can be obtained explicitly), I'd be interested. [Besides those
> that can be obtained from one of the functions below by replacing
> x with ax, or by replacing one occurrence of x with ax and another
> occurrence of x with bx (a and b are positive constants).]
When dealing with functions of this sort, it is often nice to modify
them, extending their domains, by using absolute values as
appropriate (and sometimes extending by limit). A few examples
of such are given below.
> y = x^x local min @ x = e^(-1)
y = |x|^x ditto, but also local max @ x = - e^(-1)
> y = x^(1/x) local max @ x = e
y = |x|^(1/x) ditto, but also local min @ x = - e
> y = (ln x)^(ln x) local min @ x = e^(1/e)
y = | ln|x| |^ln|x| for nonzero x; 0 for x = 0
local min @ x = +/- e^(1/e), absolute min @ x = 0
local max @ x = +/- e^(-1/e)
I hope these examples convey the idea.
Regards,
David Cantrell
--
-------------------- http://NewsReader.Com/ --------------------
Usenet for the Web
Paul V. S. Townsend, M.Sc.
This equation has no solution in real numbers for y < e^(1/e). This constant
works out at 0.6922...... but having first found the 0.69 bit I jumped to
the conclusion that it was ln(2) and suspected a bug in the algorithm when I
found I could go a little below ln(2).
In fact x^x goes through my 0.6922... as a minimum for x=1/e.
Is there any practical application for the x^x and similar functions?
James <a@b.c> wrote in message news:9eqosf$j48$1...@plutonium.btinternet.com...
David W. Cantrell
> I discovered a similar singularity years ago, when playing round with
> y=x^x. This equation has no solution in real numbers for y < e^(1/e).
> This constant works out at 0.6922......
Two errors. The first is a typo: You meant to say something like
e^(-1/e). The second error is more serious: The equation y=x^x does
have solutions in real numbers for some y < e^(-1/e). As a simple
example, 1/4 = x^x has the solution x = -2.
Oscar Lanzi III
Let x_1 = (1+1/n)^n and x_2 = (1_1/n)^(n+1). Then f(x) = x^(1/x) works
out to be the same for both values of x. Now as n --> infinity, x_1
approaches e from below and x_2 approaches e from above, so the function
value is equal for two values of x that lie arbitrarily close to e and
on opposite sides of e. If f(e) is defined (clearly the case) it must
be an extremal value.
--OL.
Dave L. Renfro
[sci.math 28 May 2001 17:44:12 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/whenspedy>
wrote (in part)
> The default domain could be taken to be larger,
> saying that f(x) = x^(1/x) is defined for some x < 0.
> For example, f(-1) = -1 and f(-1/2) = 4. It all depends, I suppose,
> on precisely what sort of exponentiation ^ is taken to represent.
> When dealing with functions of this sort, it is often nice to
> modify them, extending their domains, by using absolute values as
> appropriate (and sometimes extending by limit). A few examples
> of such are given below.
Your last comment -- Yes, you're correct. But I don't think this
will give rise to any non-explicitly obtainable local extrema if
you didn't have any before you do this. [This is certainly the case
when replacing x with |x|, as this just reflects the graph across
the y-axis. I'm not quite as sure when x's are replaced with |x|,
ln(x)'s are replaced with |ln|x||, etc.]
Your first comment -- The typical situation in elementary math
texts (i.e. precalculus and calculus) is that when considering
U^V for a VARIABLE expression V, you impose the restriction that
U is nonnegative.
Here are couple of reasons for this restriction:
(1) How is (-2)^[sqrt(2)] to be understood? Here's how these books
explain what 2^[sqrt(2)] means, or at least a method for
computing 2^[sqrt(2)] in terms of things already studied:
You pick a sequence of positive rational numbers r_n --> sqrt(2)
and then let 2^[sqrt(2)] be the limit of 2^(r_n). [Of course,
these books have to mention that you get the same limit
if different rational sequences approaching sqrt(2) are used,
the proof of which is left for a later math course.]
However, with (-2)^[sqrt(2)], you're faced with the problem
that (-2)^(r_n) -- depending on how you choose the sequence
r_n that approaches -2 -- is defined for all for all values
of n, for some values of n, or for no values of n.
(2) To me, a more serious problem is that x^(1/x) is NOT A FUNCTION
for negative x's, even when x is restricted to values such as
-1 or -3/2, UNLESS some "operational instructions" BEYOND what
are needed for x > 0 are imposed. For example, suppose x = -3/2.
Then x also equals -6/4. However, (-3/2)^(-2/3) is presumably
defined whereas (-6/4)^(-4/6) is presumably not defined--UNLESS
we specifically stipulate in the definition of f(x) = x^(1/x)
that when x is a negative rational number we evaluate x^(1/x)
by first reducing x to lowest terms.
I suppose you could argue that it's understood when faced with
(-6/4)^(-4/6) that you reduce the exponent to lowest terms
before determining whether or not this is defined. But I bet
you can't find this mentioned explicitly in any precalculus
or calculus book. These books don't deal with the evaluation
of x^y when x is a negative real number and when y is a negative
rational number. Rather, they deal with the evaluation of x^y
when x is a negative real number and when y is a specific
"fractional presentation" of a negative rational number.
Dave L. Renfro
Martin Cohen
I saw this proof a number of years ago - can't remember for sure where
(maybe in "What is Mathenatics").
The only assumption is that e^x >= 1+x for all real x
with equality only when x = 0.
Substitute (x-e)/e for x. Then
e^((x-e)/e) >= 1 + ((x-e)/e) = x/e
or e^(x/e - 1) >= x/e
or e^(x/e) >= x
or e^(1/e) >= x^(1/x)
with equality only when x = e.
Martin Cohen
David W. Cantrell
> [sci.math 28 May 2001 17:44:12 GMT]
> <http://forum.swarthmore.edu/epigone/sci.math/whenspedy>
>
> wrote (in part)
>
> > The default domain could be taken to be larger,
> > saying that f(x) = x^(1/x) is defined for some x < 0.
> > For example, f(-1) = -1 and f(-1/2) = 4. It all depends, I suppose,
> > on precisely what sort of exponentiation ^ is taken to represent.
>
> > When dealing with functions of this sort, it is often nice to
> > modify them, extending their domains, by using absolute values as
> > appropriate (and sometimes extending by limit). A few examples
> > of such are given below.
>
> Your last comment -- Yes, you're correct. But I don't think this
> will give rise to any non-explicitly obtainable local extrema if
> you didn't have any before you do this.
I'm not certain about that. Note that I was _not_ advocating replacing
_all_ occurrences of x by |x| in all cases. In my first example, I had
suggested considering |x|^x, not |x|^|x|, instead of x^x.
In fact, this is related to my first comment. Defining ^ as I would for
reals, it's not straightforward to obtain a graph of y = x^x using a
normal computer algebra system (CAS) so that points like (-1,-1) and
(-2,1/4) are included. [Note: Gary Tupper's excellent program GrafEq
does include such points. See
<http://www.peda.com/grafeq/gallery/rogue/xx_exponential.html>.]
What you can do with a normal CAS, however, is to graph both y = |x|^x
for all x and y = - |x|^x for negative x. Then, the desired graph of
y = x^x should be contained in the union of those. [I said "should be",
rather than "is", simply because most CASs will still not give you the
point (0,1).]
> [This is certainly the case when replacing x with |x|, as this just
> reflects the graph across the y-axis. I'm not quite as sure when
> x's are replaced with |x|, ln(x)'s are replaced with |ln|x||, etc.]
>
> Your first comment -- The typical situation in elementary math
> texts (i.e. precalculus and calculus) is that when considering
> U^V for a VARIABLE expression V, you impose the restriction that
> U is nonnegative.
Maybe, I'm not sure. [I remember having taught out of a calculus text
where the author did not impose such a restriction and, since this led
to some (probably unintended) complications in certain exercises, I
essentially imposed such a restriction myself when making the homework
assignment.]
> Here are couple of reasons for this restriction:
>
> (1) How is (-2)^[sqrt(2)] to be understood?
[snip of stuff with which I agree]
> (2) To me, a more serious problem is that x^(1/x) is NOT A FUNCTION
> for negative x's, even when x is restricted to values such as
> -1 or -3/2, UNLESS some "operational instructions" BEYOND what
> are needed for x > 0 are imposed. For example, suppose x = -3/2.
> Then x also equals -6/4. However, (-3/2)^(-2/3) is presumably
> defined whereas (-6/4)^(-4/6) is presumably not defined--UNLESS
> we specifically stipulate in the definition of f(x) = x^(1/x)
Preferably, such stipulation would already have been made in the
definition of ^ itself.
> that when x is a negative rational number we evaluate x^(1/x)
> by first reducing x to lowest terms.
>
> I suppose you could argue that it's understood when faced with
> (-6/4)^(-4/6) that you reduce the exponent to lowest terms
> before determining whether or not this is defined. But I bet
> you can't find this mentioned explicitly in any precalculus
> or calculus book.
I only have a few precalculus books around. As it happens, the first
one I looked in (_Algebra and Trigonometry with Analytic Geometry_
by Karl Smith) says "Definition of Rational Exponents -- Let x = m/n
in the notation b^x. Let m and n be any natural numbers with no
common factors and b be any real number for which b^(1/n) is defined.
Then b^(m/n) = (b^(1/n))^m and b^(-m/n) = 1/b^(m/n)."
> These books don't deal with the evaluation
> of x^y when x is a negative real number and when y is a negative
> rational number. Rather, they deal with the evaluation of x^y
> when x is a negative real number and when y is a specific
> "fractional presentation" of a negative rational number.
In which case I would argue that they are not really dealing with a
binary operation ^ at all, but rather some sort of ternary operation
f(x,m,n) which, by abuse of notation IMO, they might write as x^(m/n).
I don't seriously disagree with any of your comments above. You
have more-or-less elaborated on what I was getting at when I said
"It all depends, I suppose, on precisely what sort of exponentiation ^
is taken to represent."
Regards,
Dave L. Renfro
[sci.math 30 May 2001 04:09:58 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/whenspedy>
wrote (in part)
> I'm not certain about that. Note that I was _not_ advocating
> replacing _all_ occurrences of x by |x| in all cases. In my
> first example, I had suggested considering |x|^x, not |x|^|x|,
> instead of x^x.
Yes, you're correct--my mistake. I guess I was wondering how
the graph of y = (ln x)^(ln x) compares with y = |ln|x||^(ln|x|),
and related substitutions of absolute values (to ensure no negative
bases and no negative inputs to ln's), and lost my focus.
> In fact, this is related to my first comment. Defining ^ as
> I would for reals, it's not straightforward to obtain a graph
> of y = x^x using a normal computer algebra system (CAS) so that
> points like (-1,-1) and (-2,1/4) are included. [Note: Gary
> Tupper's excellent program GrafEq does include such points. See
> <http://www.peda.com/grafeq/gallery/rogue/xx_exponential.html>.]
> What you can do with a normal CAS, however, is to graph both
> y = |x|^x for all x and y = - |x|^x for negative x. Then, the
> desired graph of y = x^x should be contained in the union of
> those. [I said "should be", rather than "is", simply because
> most CASs will still not give you the point (0,1).]
I would argue that these are misguided, unless things are done
from a complex analysis point of view (e.g. defining x^y as
exp[y*(ln x)] ). One is either not dealing with a function
(i.e. the output varies according to whether you use -3/5 or
-6/10) or, when the base is negative and rational, it has to be
understood that we express the base as a quotient of integers
in lowest terms first.
As far as I can tell, precalculus and calculus books only NEED
negative bases when an explicit expression of the form W^(p/q)
is being dealt with, where q > 0 and p are fixed integers. That is,
p and q are part of the definition of the expression, and we're not
thinking of p/q as a rational number. [To use what used to be
called "new math" terminology, the former involves the NUMERALS
p, q, and p/q, whereas the latter involves the NUMBER p/q.]
One of the things done is to interchange W^(p/q) with [sqrt[q](W)]^p.
Typically, in intermediate algebra and college algebra texts, or at
least it used to be the case, this was done BEFORE the notion of
a function had been introduced. (1) Positive integer roots and
integer powers are introduced (with even roots of negative numbers
being undefined), (2) the expression W^(p/q) for appropriate
integers p and q [Note: NOT the rational number p/q, although
authors, students, and/or teachers often overlook this point.] is
defined as a certain power-radical expression, (3) their properties
are investigated, and then (4) this is used to solve equations
(e.g. x^8 - x^4 - 6 = 0, which is quadratic in x^4).
> I only have a few precalculus books around. As it happens, the
> first one I looked in (_Algebra and Trigonometry with Analytic
> Geometry_by Karl Smith) says "Definition of Rational
> Exponents -- Let x = m/n in the notation b^x. Let m and n be
> any natural numbers with no common factors and b be any real
> number for which b^(1/n) is defined. Then b^(m/n) = (b^(1/n))^m
> and b^(-m/n) = 1/b^(m/n)."
I'd have to know the context of this definition to know whether
he's defining b^(m/n) for the pair of natural numbers (m,n) or
whether he's defining b^(m/n) for the rational number m/n.
But it does appear that he's defining b^(m/n) for the rational
number m/n.
Personally, I don't see the necessity for defining b^(m/n) when
b < 0 and m/n is a rational number (as opposed to m/n being a
rational numeral). As I've already mentioned, in algebra you work
with integer powers, then you work with roots, and then you work
with a pair of these applied successively (i.e. the cube root of
2^6). Then you apply these notions to solving polynomial and
radical equations. Finally, when it comes time to make the jump
from algebra (symbol manipulation and equation solving) to
functions, and you want to define b^x for numbers x (rather than
numerals x), you restrict yourself to b > 0. You don't HAVE to do
it this way (it appears that Smith doesn't), but this seems the most
reasonable compromise to me, as I don't know of any necessity to
consider b^x for b < 0 and x a rational number at this level.
[I say "compromise" because, from a mathematical standpoint, all
of this is easily taken care of via the standard definition of
b^x as exp[x*ln(b)] in complex analysis, an option not available
at the precalculus level.]
In practice, of course, these issues aren't relevant. Students do
the symbol pushing well and succeed in the class, or they don't
symbol push well and have trouble. I think the only reason this topic
(e.g. what is the graph of x^x when x < 0) is coming up so much in
recent years is because of graphing calculators. It's easy to enter
x^x, look at the graph, and then (inappropriately) interpret what
the calculator gives you as mathematics.
> In which case I would argue that they are not really dealing with
> a binary operation ^ at all, but rather some sort of ternary
> operation f(x,m,n) which, by abuse of notation IMO, they might
> write as x^(m/n).
This is a good summary of what they're actually doing, although some
people (authors, students, and/or teachers) overlook this matter
entirely while other people (authors--Smith above, students, and/or
teachers) manage to define x^(m/n) for x < 0 and m/n a rational
number in a consistent way (replace m/n by p/q where p,q have no
common factors, ...).
Dave L. Renfro