(R. P. of San Luis Obispo, CA. 2001-01-23)
(M. M. of Gresham, OR. 2001-02-11)
Why is zero factorial equal to one?

The quantity n! (pronounced "n factorial" or "factorial n") is
defined as the product of the n integers from 1 to n.
The basic reason why 0! equals 1 is that it's merely a product of 0 factors;
Such an empty product must be equal to 1,
just like a sum of zero terms (an empty sum) must be equal to 0. Let me explain:

The product of (n+1) factors is clearly equal to the product of the first n factors
multiplied by the last one. This is "clear" to everybody when n is 2 or more.
To make this work for n=1 we have to state that a "product" consisting of a single factor
is equal to that factor.
It follows (for n=0) that a product of zero factors multiplied by any number x must
be equal to x. Therefore, the product of zero factors must be equal to 1.
(The same reasoning for sums leads to the conclusion that a sum of zero terms is equal to 0,
which is less shocking to most people than the corresponding result for empty products.)

Defining n! as a product of n factors (1,2, ... n) when n is nonzero thus implies that the
only consistent definition of 0! is 0!=1.

Another "advanced" argument is to define factorials in term of the analytic
Gamma function ( G ),
whose properties also imply a value 0!=1.

massxv2 (2002-05-13)
Raising to the Power of Zero
Diane302
(Fort Worth, TX.
2002-05-14) =
Diane Miller:
Why is any number [including 0] raised to the power of 0 equal to one?

If you multiply x^{n} by x, you obtain x^{n+1}.
So, the product of x^{0} and x is x [= x^{1}].
If x is nonzero, x^{0} must therefore be equal to 1.
Furthermore :

0^{0} = 1

This seems to bother some people offhand (including a few
textbook authors,
who should know better),
but x = 0 is not an exception to the above rule:
It's indeed true that 0^{0} = 1 .
The most fundamental explanation is that an empty product
[the product of no factor(s), which is what a zeroth power is]
cannot possibly depend on the value of any factor since
you are not using any such factor to form the "product".
Thus, the value obtained for the zeroth power of any nonzero x
must also be the correct value when x is zero.
In spite of a superficial similarity, this is not a "continuity argument"
(such analytical arguments will not work past the elementary level
where exponents are only integers, because it so happens that
the two-variable function x^{ y}
cannot be made continuous at the point x = y = 0, as discussed below).
Instead, the above argument rests purely on basic logic
[set theory and elementary algebra].

I could leave it at that and rest my case, but I know that the above
logical argument
is often unable to overcome the psychological reluctance to accept
the fundamental fact we're discussing here...
Although the mathematical case is closed,
some people may find it helpful to take a lexicographer's approach and discover that
unity is the value of zero to the power of zero which is implied in a number of
familiar mathematical contexts.
View the following examples only as a supplement to the above fundamental logic,
which illustrates
that the relevant mathematical discourse is always consistent with it
(because it's based on it)...

1

1

1

1

2

1

1

3

3

1

1

4

6

4

1

According to the binomial theorem,
(1-1)^{n}
is the alternating sum of the coefficients
in a line of Pascal's triangle.
The result is zero, except for the top line where it's equal to unity
(namely, the only nonzero coefficient in that top line)._{ }

An expression like
å a_{n}x^{n}
is understood to have value a_{o} when x = 0.

John Baez
about the set of the functions from B to A:_{ }
|A^{B}| = |A|^{ |B|}

groktr.v.grok·ked, grok·king, groks. Slang. To understand profoundly through intuition or empathy.
To assimilate everything about something to the deepest possible extent,
becoming as one with the subject of focus.
[ Stranger
in a Strange Land (1961) by Robert Anson
Heinlein ] The
American Heritage Dictionary (online)
Do you grok that?

The numerical expression x^{y} is defined only under one [or both]
of the following conditions.
(In particular, 0^{y} is not defined unless y is a
nonnegative integer.)

x is positive (in which case y could be any real or
complex number).

y is an integer (if y is a negative integer, x must be nonzero).

Lest the reader object that 0^{y}should be defined as
the limit of x^{y} as x®0^{+},
we'll point out that, when
y is the imaginary number i, we obtain exp(i Log(x)),
which keeps going around the unit circle,
without approaching any limit...

When the exponent (y) in the expression
x^{y } happens to be an integer, the base
(x) can be any number whatsoever
(positive, negative or even complex)
except that it can't be zero when y<0.
In that elementary context, it's clear that algebraic consistency imposes
the zeroth power of any number (positive, negative or complex) to be unity.
There's no reason to make an exception for zero that would introduce an arbitrary
and needless discontinuity at the origin for the function
f(x) = x^{0}, which is everywhere else equal to 1.
As we generalize the notion of exponentiation, this elementary perspective must be
retained, unless it is found to be incompatible with the more general framework.

If such a contradiction occurred,
we would have either to renounce the generalization or
introduce an unwelcome exception to the elementary concept.
Fortunately, were are not faced with this difficult choice in the case
of exponentiation. Read on.

Historically, fractional values of the exponent were introduced in the late
Middle Ages (c. 1360) by the
Frenchman Nicole Oresme (1323-1382).
However, the base x must be positive whenever fractional exponents are not ruled out!

If p and q are integers, then we may raise
a base x to the power of p/q by raising the q-th root of
x to the power of p. The q-th root is consistently defined for
positive bases as the unique positive number whose q-th power equals x.
Generalizations to negative bases could be conceived for odd values of
q but this turns out to be a useless dead end which only confuses the issue.

x^{y} is unambiguously defined only in two cases for complex numbers:
when x is real and positive (it's then equal to exp(y ln(x))
and when y is an integer (provided x is nonzero if y is negative).
If exponentiation makes any sense at all,
it must make sense for the former case of integral exponents
[unless a division by zero is involved]
and the latter case is needed too because we need the exp function
and can't possibly discriminate against any positive
bases as soon as we accept the no-so-special base e...
Curiously, you can't extend that "domain" of definition
(which is not a "domain" in the strict sense of the term)
unless you bring
Riemann surfaces [or "multi-valued" functions] into the picture,
because logarithms cannot be continuously defined any other way
for a nonpositive argument...

The nature of the essential
discontinuity of the two-variable function
f (x,y) = x^{y}
about the origin (x = y =0)
is probably best grasped by considering the curves where this
quantity is constant, for positive values of x and y near the origin [more precisely,
when x is positive and less than 1].
The cartesian equation of such a curve is y = a / ln(1/x) for some
nonnegative value of the parameter a;
all these curves include the origin in their closure!
Along any of them, the value of the function is constant: It's equal
to exp(-a), which can be essentially anything you want between
0 (excluded) and 1 (included).
This does imply that the two-variable function f does not have
a limit at the point (0,0), since you have points in any neighborhood of the
origin for which the value of f is as close as you wish from any
choice of a number in the interval [0,1].

The fact that the function
f (x,y) = x^{y}
is not continuous at the origin does not make is undefined there...
The bottom line is that zero to the power of zero must be defined to
be unity unless we're willing to rephrase
many of the theorems we all take for granted
(including the binomial theorem mentioned above).
No paradox can arise from continuity arguments because such
arguments are simply disallowed

Without any exceptions, when n is a nonnegative integer,
x^{n} denotes a product of n factors equal to x.
When n is zero, the value of x is thus disregarded and must be irrelevant...
The zeroth power of x is defined to be unity
in any monoid,
even if the base (x) is not invertible.

At this point,
some people argue that the fact that
0^{n} is zero for any n could be an equally valid argument
which contradicts the above...
This is just not so!
On one hand, fundamental logic does impose that an empty product
must be unity
(regardless of the values of the factors, since they're not used).
On the other hand, a product vanishes when at least one
of its factors is zero. This simply does not apply to a product
of zero factors.
Furthermore, if we want to keep alive the theorem
(valid in any integral domain)
that a product vanishes if and only if at least one of
its factors vanishes, we see that a product of no factors cannot
vanish: Thus, a zeroth power cannot possibly be zero.

The designers of some pocket calculators have decided to make it illegal to raise
zero to the power of zero on their machines.
This is either a misconception on their part, or an overly cautious [misguided]
approach to "fix" the fact that the function "x to the y" is not continuous
about x = y = 0,
which makes it behave erratically when previous rounding errors
have been made.
(Consider that the function equals 0 when x is 0 and y is 0.0000000001,
but it's 1 when x is 0.0000000001 and y is 0).
This is a misguided concern because the problem is with whatever prior rounding errors
occurred, not with the discontinuous function itself.
The fundamental flaw is with the so-called "floating-point" approach of such calculators
(and/or computer languages) which involves a drastic loss of accuracy when nearly
equal quantities are subtracted; It's up to the user to avoid such cases
(numerical analysis is not always easy, even with the help of a computer).
We've also been told about electronic devices which wrongly return an
"infinite" value for the zeroth power of zero
probably because of some careless internal error
propagation when attempting to take the "logarithm of zero",
as a misguided intermediate step...

chormpy (N. N. of New Zealand. 2000-10-21)
Complex Numbers
Explain what complex numbers are, in terms an idiot could understand.

Picture this. We are facing each other on open grounds and you're challenging
the very existence of negative numbers,
let alone complex numbers:

I ask politely: "Can you take two steps towards me?"

You nod and you do. So nice of you. I thank you.

Then, I ask you to move "minus two" (-2) steps towards me.

You smile, having understood what negative numbers are,
and take two steps back to your original spot._{ }

I smile back: "Can you take an imaginary
step towards me?"

You stare and say "Huh?".

However, after some thinking you take a step sideways.
Nice job !

It's ultimately a matter of convention to choose whether
a "forward" imaginary step is to the left or to the right.
However, the universal convention is that a positive imaginary step is a
step sideways to your left
(a negative imaginary step is to your right).

In other words, complex numbers are to the plane what real numbers are to the line.
They just describe position and motion in the plane the same
way real numbers do on a line.
Thus, imaginary simply means sideways...
That viewpoint was devised in 1806 by
Jean-Robert Argand (1768-1822).

Adding two complex numbers is easy:
The total number of steps taken in the "real" direction
is obviously the sum of all steps taken in the real direction. The same applies to the
imaginary direction.
Each component (real or imaginary) of the sum is
the sum of the corresponding components of the complex addends.
(In learned terms, that's a "direct sum".)

Things become only slightly more delicate if you worry about "multiplying" such "numbers"
together. However, just think about it this way:

What's the product z of two numbers x and y_{ }?
Well, it's the number z which is to
x what y is to 1. ^{ } (Isn't it?)

Picture what this means in the complex plane with, say, x=2+i
(I move two steps forward and one step to the left).
Multiplying any number y by x is like using x as a new
"unit" step.
In other words, you're now using a new "grid" where each step is of length
Ö5
(that's the length of x, because of the Pythagorean Theorem),
while the whole grid has been rotated about 26.565°,
to align x with the "forward" direction.
If you go 3 steps forward and one step right
(corresponding to the complex number y=3-i) in that new grid where do you end up?
Well, you end up on the point of the plane which, by definition, is the product of x and y.

You could have taken 7 old steps forward and one step to the left and would have
ended up at the same location.
Draw this on paper (just once in your life)
and admire the "coincidence" of the two
results, obtained with or without an intermediate grid.

Why is it that i^{ 2} = -1 ?
Well, the left of your left is your back, isn't it?

On 2000-10-22, Chormpy wrote:

Thank you for your answer, Gerard.
Although I'm still far from actually understanding,
your answer did clear things up a little.
It helped me to understand some of the other examples
and explanations of complex numbers I've found.

The existence of numbers whose squares are negative
was first put forth by
Gerolamo Cardano (1501-1576)
in his
Ars Magna
(1545). Cardano did not understand what they meant but he found them useful
to present the general solution of the cubic equation that was
revealed
to him in 1539 by
Niccolò Fontana Tartaglia
(1499-1557) under an oath of secrecy.

Tartaglia had discovered for himself the general method to solve cubic equations
in the early hours of Saturday, February 13, 1535.
At that time, even negative numbers were not commonly accepted.

The terms real number and imaginary number
(nombres réels et nombres imaginaires) were coined by
René Descartes (1596-1650) in
La Géometrie,
an appendix of Discours sur la méthode (1637).

The real linear combinations a + ib of the real unit (+1)
and the imaginary unit (i) form the
field of complex numbers
which is the two-dimensional field obtained from the real line
(the field of reals) by the general
Cayley-Dickson construction.
The reciprocal of a nonzero complex number z
is the number which gives unity when multiplied into z.
It's given by the expression:

z^{-1} =
( a + ib )^{-1} =
( a - ib )
( a^{2} + b^{2} )^{-1}

That equation expresses the reciprocal of a complex number in terms
of the reciprocal of a nonzero real number.

Fundamental Theorem of Algebra :

Arguably, a full understanding of the complex numbers
was reached only when it became clear that they form the
algebraic closure
of the real numbers.
That's what the Fundamental Theorem of Algebra means:

Every nonconstant complex
polynomial
has at least one complex root.

Elsewhere on this site, we give a nice
modern proof
of that statement. The first valid proof
(taking the Jordan
Curve Theorem for granted)
appeared in the doctoral dissertation of
Carl Friedrich Gauss (1799).

Thus, by induction on n > 0, we see that
any complex polynomial P
of positive degree n has exactly
n complex roots (not necessarily distinct) and
can be written as a product of n linear factors:

P(z) = a

n

Õ

k = 1

(z-z_{k })
[where a and z_{k} are complex numbers.]

Thinking outside the box :

Let's indulge in some metaphysics about the above introduction
of a complex realm
where planar angles and
two-dimensional curvature live:

As an extra imaginary unit i
transforms the real line into the complex plane,
so does it transform 3-dimensional space into
4-dimensional spacetime.
Time is imaginary length. Length is imaginary time...

Nobody has yet figured out (convincingly)
what it would mean to move sideways
in time. Human time remains confined to a single dimension.

Video: Complex numbers
(Chapter 5 of "Dimensions: A walk through mathematics" by Adrien Douady)

Jon
Ball (2002-10-22)
Using the
Golden Ratio (f)
to solve z^{5} = 1.
How do you express the 5 fifth roots of unity in terms of
f = ½(1+Ö5) ?

Using the fact that cos(2p/5) =
½(f-1) and the relation
f^{2} = f + 1,
it's not difficult to show that the 5 fifth roots of unity are:

1,
½ [ f-1 ± i
Ö(2+f) ],
and
½ [ -f ± i
Ö(3-f) ].

The 10 tenth roots of unity include the above and their 5 opposites...

ciderspider (Mark Barnes, UK.
2000-11-04)
Does the equation x=2^{p}
have an infinite number of complex solutions?

No, it does not.
The function 2^{z}
can only be defined as exp(ln(2) z).
Just like the exp function itself, it's
single-valued over the entire complex plane.
There's nothing to "solve", the value of x is simply
some real number: 8.824977827...

The only possible source of confusion is the use of the numerical
constant ln(2) in the above definition...
Since the extension of the ln function to complex arguments
is indeed multivalued, why not take any of the "other" values of ln(2)
and go on from there?
I won't skirt the issue, but I must first stress that,
when z is not an integer,
the expression a^{z} is never ever used by professionals
(except for fun) unless a is a positive real number.
It is then defined to be exp(ln(a) z),
where ln(a) is the (real) natural logarithm of a.

If you take "another" value of ln(2)
(say: ln(2) + 2pi )
to define your own base-2 exponential,
you simply get another single-valued function which is different from everybody else's.
You could define infinitely many such functions, but so what? e point
(p or any other point)
have no more reason to be declared "equal" than
Ö4 and -Ö4 do.

In fact, the square root function
is an introductory example of a function which, like the logarithm function, does not
present a problem for (positive) real numbers, but which cannot be generalized
to a continuous function over the whole complex plane.
As explained in the next article,
a continuous generalization of the square root function involves an
entirely new domain of definition (called a Riemann surface).
For the square root function, the Riemann surface consists of the origin and
nonzero points identified as (r,q)
where r is the [positive] distance to the origin and the "angle"
q is understood "modulo 4p",
so that (r,q) and (r,q+2p)
identify two distinct points with different square roots
(which are opposite of each other).
Loosely speaking this surface is composed of two sheets and you end up
back to the same point if you go around the origin an even number of times.

In the case of the logarithm function,
the Riemann surface has infinitely many sheets;
you may visualize it as a flattened helicoid whose nonzero points are
identified as above by a couple (r,q)
except that different values of the real number q
will always identify different points of the surface.
What this means, in concrete terms, is that whenever you use a logarithm
you must absolutely refrain from adding an arbitrary multiple of
2p to the "angle" of the argument.
This is allowed in the complex plane, but prohibited on the relevant Riemann surface
over which the continuous logarithm function is defined.

Do think about Riemann surfaces and you are safe under the umbrella of mathematical rigor.
Forget about this fundamental point and you are bound to produce a number
of false proofs, not always for a recreational purpose...

By contrast, a (recreational) equation like
x^{12} = 2^{ x} can be freely discussed in the complex realm because both sides are perfectly
well-defined for any complex value of x.
It makes sense to compare them.

The Obsolete Formula of Roger Cotes :

Before all this became clear,
Roger Cotes
(1682-1716) came up with the following formula, which is only true up to some
number of angular "turns":

ln ( cos q + i sin q )
= i q
[ modulo 2ip ]

The above uses the modern "natural" measurement of angles (in radians)
which is due to Cotes himself !
Cotes died at the age of 34 and
Isaac Newton (1643-1727) said of him:

" If he had lived, we might have known something. "

That formula is only of historical interest.
It's been superseded by the following celebrated formula by Euler (1707-1783) which
could be construed as removing all ambiguities by
taking the exponential of both sides in the above.
However, Euler's formula is best derived directly and it's much simpler in theory and in practice,
as it involves only unambiguous (i.e., "single-valued") functions:

cos q + i sin q
= exp ( i q )

silenteuphony (2003-07-20)
Generalizing the Square Root Function...
May the square root function ( Ö )
be generalized to negative numbers?

The short answer is no.
There are popular implementations (on some
handheld calculators and elsewhere) which provide
pointwise solutions to quadratic equations,
but they don't qualify as proper mathematical generalizations
of the square root function.

Such generalizations would invalidate familiar properties established in the realm
of nonnegative real numbers, where the square root of a number x is unambiguously
defined as the positive number whose square is equal to x.
Among the casualties would be one of the most trusted relations:
Öu Öv =
Ö(uv):
Indeed, if a definition of Ö(-1) could be given which was
consistent with this relation, we would have:
Ö(-1) Ö(-1) =
Ö1,
so that the square of Ö(-1) would be 1
instead of (-1)...

There is a number whose square is -1,
namely the imaginary number i [note that its opposite
-i would do just as well].
However, it is abusive to denote it Ö(-1) for a
number of reasons, including the one given above.
Unfortunately, this has not stopped a number of
otherwise distinguished authors from doing so,
in order to bypass a more proper introduction to what imaginary
and complex numbers really are.
(See above for our own attempt at such an introduction.)

What about the "square root" of a complex number?

If we insist on defining a square root
(sqrt or Ö )
as a single-valued function over the complex plane,
the best we can do is accept discontinuity
(jumping from y to -y) on some kind of curve going from the origin to infinity
(e.g., one half of a straight line).

We like to call this kind of line a cliff
(since a jump discontinuity occurs when the argument crosses such a line).
The square-root function can't be defined over the entire complex
plane without creating a cliff.

This annoying issue was cleverly resolved by Bernhard Riemann (1826-1866)
who stated essentially that the "correct" domain of sqrt was not
the complex plane itself, but (roughly) two copies of it,
properly interconnected topologically.
Each such "copy" (loosely speaking) is called a Riemann sheet and the whole thing
is the Riemann surface for the sqrt function.

This surface may be rigorously described as consisting of the origin, together with
the set of ordered pairs (r,q)
where r [the distance to the origin] is positive
and q is a real "angle"
modulo 4p
(whereas a similar definition of the ordinary single sheet complex plane
would specify that q
is "modulo 2p").
The beauty of this approach
is that sqrt is defined and continous everywhere on its two-sheet domain
(its range is the ordinary single sheet complex plane).

The "two-sheet" Riemann surface for the square root function is totally different from
the set of complex numbers.
Loosely speaking, you end up on the same point only if you travel an even
number of times around the origin.
If you wish, you may identify a point on the surface using a notation like
(r,q) where q
is between 0 and 4p,
although it's probably better to make no such restriction and state that the second number
is understood "modulo 4p" (as stated above)
so that the point (r,q)
is identical to (r,q+4kp)
for any integer k...

Points on the two-sheet Riemann surface have square roots that are ordinary complex numbers;
the square root of the point (r,q)
is defined as the complex number
(Ör) exp(iq/2).
Therefore, (r,q) and (r,q+2p)
have two different square roots that are opposite of each other.

Multiplication is well-defined:
The product of u = (a,a)
and v = (b,b) is
uv = (ab,a+b),
where a+b is understood modulo 4p.
This is how we maintain the validity of properties like
ÖuÖv =
Ö(uv).

Unfortunately, no simple "addition" is defined on this Riemann surface.

A nonzero complex number is associated with
two distinct points of the Riemann surface, which have different square roots
(opposite of each other), so
the "nice" definition of square roots over the Riemann surface does not resolve the
sign ambiguity for ordinary complex numbers.
One deep explanation for the impossibility of defining a continuous
generalization of the square root function over complex numbers is that the relevant
Riemann surface and the complex plane aren't homeomorphic
(i.e., there's no bicontinuous one-to-one correspondence between the two things).

If you choose to define Ö
on the domain of complex numbers rather than on the proper Riemann surface,
your "square root" function cannot be continuous and the "square root" of a product
is not necessarily equal to the product of the "square roots" of its factors.
There's no way around this...

A cardboard model of the Riemann surface for the sqrt function is easy to make
(but not easy to describe; the surface "goes through" itself along one line).
It's pretty too; I am keeping the one I made years ago on a shelf next to my desk,
as a constant reminder of the above fact in the realm of
complex variables.

One 3D embodiement of that Riemann surface is an unbounded version of the
self-intersecting disk. Such a surface has parametric equations:

x = r cos q
y = r sin q
z = r cos q/2

(M. M. of Salem, MA. 2000-10-11)
Two numbers have a product of 19551 and a sum of 280.
Without determining the numbers, find their difference.

If P, S and D are the product, sum and difference of the two numbers, then:

S^{2} - D^{2} = 4P

Therefore, in this case, D^{2} is
280^{2}-4´19551 or 196.
The difference D between the two numbers is thus 14.
(Don't be silly and object that it could also be -14)

You may want to prove the relation S^{2} - D^{2} = 4P
by noticing that:

FlyingHellfish (Atlanta, GA. 2002-10-08)
Find the value of D = x^{4}+y^{4}+z^{4},
given the relations at right, in particular when A=1, B=2, and C=3.

Introducing the elementary symmetric functions,
U = x+y+z, V = xy+yz+zx, and W = xyz. , we have:
A = U,
B = U^{2}-2V,
and C = U^{3}-3UV+3W.
Conversely,
U = A,
V = (A^{2}-B) / 2,
and W = (A^{3}-3AB+2C) / 6.

Since
D = U^{4}-4U^{2}V+4UW+2V^{2},
we have
D = (A^{4}-6A^{2}B+8AC+3B^{2}) / 6.
For the particular case A=1, B=2, C=3, this means D = 25/6.

The quantities x, y and z are the 3 zeroes of
X^{3}-UX^{2}+VX-W
(in the numerical case above, two of these are complex numbers).
Any symmetrical polynomial of such roots is also a polynomial in U,V,W,
and its value may thus be obtained without solving the cubic equation.
This remark may be generalized to any number of variables...

The Elementary Symmetric Functions:

For m variables, the n^{th}elementary symmetric function (s_{n })
is defined via:

s_{0} = 1
s_{1} =

å^{ }

i

X_{i}
s_{2} =

å^{ }

i<j

X_{i }X_{j }
s_{3} =

å^{ }

i<j<k

X_{i }X_{j }X_{k} etc.

If n>m, s_{n} = 0
(as there are no products of n distinct variables to sum up).
The m variables X_{1}, X_{2}, ... X_{m}
are clearly the roots of the polynomial [in x]:

m

m

Õ

(X_{i} - x)^{ }
=

å

s_{m-n} (-x)^{n}

i = 1

n = 0

A polynomial in m variables which remains unchanged under any permutation of the variables
is called symmetrical.
Any such polynomial can be expressed as a polynomial of
the above elementary symmetric functions.
Such is the case, in particular,
for the sum of the p-th powers of all the variables (S_{p }):

This last relation gave us D = S_{4} in the above case of 3 variables
(s_{4} = 0).
To extend the list in a systematic way, we observe that the following
results (known as Newton Identities or Newton-Girard Formulas)
hold for any m:

m

0 = m s_{m}^{ } +

å

s_{m-n} (-1)^{n} S_{n}

n = 1

This is true for m variables,
because each is a root of the above polynomial
(the right-hand side is thus obtained by summing m zero values of that polynomial).
This holds for less than m variables
(the result for m variables holds if some of them are set to zero)
and also for more than m variables,
because of symmetry and degree considerations which we won't go into...
For example:

Such expressions of power-sums in terms of the elementary symmetric polynomials
are known as Girard-Waring expansions
(published in 1629 by
Albert
Girard and between 1762 and 1782 by
Edward
Waring).

(S. M. of Bagdad, KY. 2000-10-18)
Find 6 numbers in continued proportion.
Their sum is 14 and the sum of their squares is 133.

Three or more numbers are said to be in "continued proportion"
when the ratio of one term to the previous one is a constant R.
This is now more commonly called a "geometric progression" of ratio R.

If A is the first of 6 such terms, their sum is A(R^{6}-1)/(R-1)
and the sum of their squares
is A^{2}´(R^{12}-1)/(R^{2}-1).
(That's assuming that R differs from 1, but it's easy to check that R=1 does not yield any
solution to the problem at hand.)
As we're told that the former is 14 and the latter is 133, we may solve this using a pair
of relations giving:

The first quantity, namely: 14 (R-1) = A (R^{6}-1)

The ratio of those two: 133 (R+1) = 14 A (R^{6}+1)

Substituting in (2) the value of AR^{6} obtained from (1)
(namely 14(R-1)+A) --or the other way
around-- we obtain the relation 9R+4A=47.
(Incidentally, this same relation would hold regardless of the length of the continued
proportion.) Either of the above equations then becomes:

9R^{7}-47R^{6}+47R-9 = 0

The obvious root R=1 is to be ruled out, as remarked at the outset
(so we could freely divide by R-1). Dividing by (R-1), there remains to solve a polynomial
equation of degree 6, namely:

9R^{6}-38R^{5}-38R^{4}-38R^{3}-38R^{2}-38R+9 = 0

Clearly, if R=K is a solution, R=1/K is another one.
Both roots correspond to the same solution
of the original problem but with the 6 numbers listed "forwards" or "backwards".
This calls for changing the unknown variable to X=R+1/R
(which gives X^{2}=R^{2}+1/R^{2}+2 and
X^{3}=R^{3}+1/R^{3}+3X);
if we have a solution for X, it's only a matter of solving a second degree equation to
recover a pair of solutions for R. Dividing the above equation by R^{3} thus gives
9(X^{3}-3X)-38(X^{2}-2)-38X-38=0, or

9X^{3}-38X^{2}-65X+38 = 0

That's still a mouthful but it's only of the third degree so we could solve it with
algebraic methods!
I hate doing this, so I'll just give the three roots approximately
(they happen to be all real):

5.41246229893, 0.469896647164, and -1.66013672387.

Now, a solution in X corresponds to a pair of real solutions in R when the equation
R^{2}-XR+1=0 has real solutions. This happens only when X^{2}-4 is positive.
Therefore only the solution X=5.412... is to be retained if we are only interested in real
solutions. This corresponds to the solution R=5.2209925253737229 (or the inverse of this to
list the numbers backwards) and A=(47-9R)/4.
The unique pair of solutions is thus composed of the following 6 numbers listed either
as below or in reverse order (last digits not guaranteed):

Now, you may check (I did!) that the sum of the above is indeed 14 and the sum of their
squares is indeed 133...

tenorboy (Todd A. Moore.
2002-05-19)
In an alley way, a 12 ft ladder leans against a building on one side;
its bottom is on the ground against another vertical building across the alley.
Similarly, a 10 ft ladder leans in the other direction across the alley...
The ladders intersect 4 ft off the ground. What's the width of the alley?

Let x be the width of the alley and ux the horizontal distance from the bottom of
the 12-ft ladder to the plumb line at the intersection;
(1-u)x is the corresponding quantity for the second ladder.
Remark that the 4-foot plumb line is equal to u times the top height of the first ladder
and (1-u) times the top height of the second one
(because of the two pairs of similar triangles involved).
In other words:

4 = u Ö(12^{2}-x^{2})
and
4 = (1-u) Ö(10^{2}-x^{2})

Eliminating u, we obtain:

1 / Ö(12^{2}-x^{2})
+
1 / Ö(10^{2}-x^{2})
= 1/4

We may use this equation directly to find the solution numerically,
with ludicrous precision:
x = 7.2575891083169677047316337322... ft.

Alternately, we may obtain a polynomial equation,
by eliminating the above two radicals:
First put one radical by itself on one side of the equation;
squaring both sides will then eliminate that first radical.
Isolating the remaining radical on one side and squaring again gives
a rational expression without radicals.
This double squaring gives a quartic [= degree 4] equation
in the variable y = x^{2}.
Because the equation is only a quartic,
it can be solved algebraically, although everybody (including myself) hates to do so.
For the record, here's the quartic:

Note that this quartic equation may include roots which do not correspond to solutions of
the original problem...
Indeed the double squaring does introduce just such a spurious solution here
(corresponding to x around 9.6668,
which is clearly not a solution of our original equation).
All told, in this age of computers and nifty scientific calculators,
it's probably best to stick with a simple equation (like the one we first gave)
rather than insist on some not-so-simple polynomial relation with a few irrelevant roots...

In the 1980's, I saw [a problem just like the above] in the French monthly magazine
Science & Vie,
staging a smart painter trying to figure out, with pencil and paper,
the width of a corridor where two opposing ladders of known lengths
cross at a height of one meter.

The question was:
Even if the scene takes place at the top floor of a high-rise building under construction (no lift)
wouldn't it be wiser for the painter to fetch the tape measure he left downstairs?

François Robert Milan, Italy

Thanks for the comment, François.
I used to be a fan of Science & Vie too.