AoPS Academy
, 
,
, 
, why 
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In it is perfectly OK to say that,![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
be an even positive integer. Let 
be a monic, real polynomial of degree 
; that is to say, 
for some real coefficients 
. Suppose that 
for all integers 
such that 
. Find all other real numbers 
for which 
.
be a real-valued function with 
derivatives at each point of 
. Show that for each pair of real numbers 
, 
, 
, such that
there is a number 
in the open interval 
for which
be an integer, show that there are infinite perfect squares of the form 
![$f(x) \in \mathbb{F}_p[x] - \{0\}$](http://latex.artofproblemsolving.com/f/6/4/f64f3f1429a6b9392a90365d58cd3675c94309c4.png)
let 
denote its multiplicity at 
so 
for some ![$g(x) \in \mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/0/e/b/0eb87f8c2b36ee7477413a22a212c0f97abd1957.png)
with 
. If 
mod , then 
by the product rule. We are asked to compute 
.![$f_n := \sum_{k=0}^{p-1} k^n x^k \in \mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/2/0/d/20da9a0fe29a78cc742daa7aa47e4a02e8862e15.png)
. Then 
using the fact that 
in ![$\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/e/5/a/e5a21492095c9116b100e2f7626a70cc26834e82.png)
. Since 
it follows that 
for 
. In particular 
.
, which is way neater. Here's a solution that repeats the above, i.e. 
and as usual we find the smallest with 
. Finding derivative term wise, get
(some explanation needed as 
could go to 0. We can now write the above in terms of the following polynomial:
and
Now for 
each term 
(primitive root argument), which means the sum will be for 
. For 
, only 
is left and
but 
is the coefficient of the monic polynomial 
, hence equal 1, so 
for .
, we have that![\[
(*) \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell}a_k = 0.
\]](http://latex.artofproblemsolving.com/e/e/c/eecc79ec6fa7030f90ce952e575b269b890fce1d.png)
Indeed, the above sum can be written as![\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} - \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell}.
\]](http://latex.artofproblemsolving.com/f/6/9/f692e16f90d4a3037127cc0ed9072766cd1f4f31.png)
But we have that![\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} + \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell} = \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0.
\]](http://latex.artofproblemsolving.com/9/3/4/934fa7207cc14043cc9c53588c32a89d06788c4c.png)
To see this, take some 
with 
, which is possible as 
, to see that![\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = \sum_{k\in \mathbb{F}_p^{\times}}(\alpha k)^{\ell} \implies (1 - \alpha^{\ell})\sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0 \implies \sum_{k\in \mathbb{F}_p^{\times}}k^{\ell} = 0.
\]](http://latex.artofproblemsolving.com/2/7/7/277dce0ce833132f26b5a65ecfb491991d1db38e.png)
Thus, we have that![\[
\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} - \sum_{k\in \mathbb{F}_p^{\times} - \mathbb{F}_p^{\times 2}}k^{\ell} = 2\sum_{k\in \mathbb{F}_p^{\times 2}}k^{\ell} = \sum_{m\in \mathbb{F}_p^{\times}}m^{2\ell} = 0,
\]](http://latex.artofproblemsolving.com/2/a/f/2afcb8000fe3594fec4dd7bda01741cff084576b.png)
as 
, so we can use the trick of taking some with 
as above, thus establishing 
.![\[
\sum_{k = 1}^{p - 1}a_kx^k.
\]](http://latex.artofproblemsolving.com/3/9/6/396bf7db0ab965bedfef966ff32716814bfc9d74.png)
Note that this polynomial always vanishes at 
by taking with 
. Suppose we take 
derivatives of this expression and evaluate the result at . The result will be![\[
\sum_{k = 1}^{p - 1}k(k - 1)\cdots k(k - \ell + 1)a_k.
\]](http://latex.artofproblemsolving.com/4/3/9/4399c0ed0849c089ec06d1f8f8d789245664b13c.png)
But we can expand out 
as some polynomial in with integer coefficients and degree 
. Thus, as long as 
, this expression will vanish by repeatedly applying (*). Thus, 
divides our polynomial for . To see that 
does not divide our polynomial, it suffices to show that![\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{\frac{p - 1}{2}}a_k \neq 0,
\]](http://latex.artofproblemsolving.com/4/3/9/43907045efe3ac712ade8349f4cb9073336a865e.png)
by expanding out 
as a polynomial in and applying to all . But this sum is just![\[
\sum_{k\in \mathbb{F}_p^{\times}}k^{p - 1} = p - 1\neq 0. \text{ }\square
\]](http://latex.artofproblemsolving.com/b/2/a/b2ac288a6687e1cb8e4c1183f6af29d735ef02d1.png)
and denote 
. Then 
. For 
, the multiplicity of 
in 
is 
.
th moment of 
for 
is . There are a few different reasons this is true: (number theory) plug in a primitive root and use geometric series, (group theory) the sum is a multiple of the sum of a subgroup with zero sum, (algebra) use Newton sums on the polynomial 
. as a root has degree 
, the sum of the polynomial over is .
th derivative; we get![\[
q^{(j)}(1)=\sum_{a\in\mathbb{F}_p}a^{\frac{p-1}{2}}a^{\underline{j}}
\]](http://latex.artofproblemsolving.com/1/1/c/11c73e7c442ea433e0f761131379f94f4bfc3c57.png)
where 
is the falling factorial. When 
, this is by the corollary. When 
, the corollary implies that![\[
q^{(j)}(1)=\sum_{a\in\mathbb{F}_p}a^{p-1}=-1.
\]](http://latex.artofproblemsolving.com/4/a/2/4a29600ea727d2835ba75bea8def1af7bb25186f.png)
So the first 
derivatives of 
at (a root of ) are , so has multiplicity 
in .
. Let 
denote the th derivative of 
. We have the following well known lemma.
. Then 
if and only if 
for all 
. in the form ![\[q(x)=b_0+b_1(x-1)+b_2(x-1)^2+\cdots+b_{p-1}(x-1)^{p-1}.\]](http://latex.artofproblemsolving.com/c/9/d/c9d3361ee17038dda6bdf14e2be652028b32b349.png)
In this setting, it's clear that if and only if 
for all . Differentiating times, we see that 
, and since 
, we thus have 
, so the result follows. 
![\[q^{(m)}(1)=\sum_{k=m}^{p-1}k^{\frac{p-1}{2}}k(k-1)\cdots(k-m+1)=\sum_{k=0}^{p-1}k^{\frac{p-1}{2}}k(k-1)\cdots(k-m+1).\]](http://latex.artofproblemsolving.com/c/2/c/c2c87e590c58c3bdd4ed014a8420d02fda84c0d8.png)
To evaluate this, we have the following useful lemma.
, we have ![\[\sum_{k=0}^{p-1}k^r=0.\]](http://latex.artofproblemsolving.com/a/b/2/ab2013a6cb27c81c12c48265ce99f41753f9e927.png)
Recall that we are working over .
be a primitive root mod . Then, the sum is ![\[\sum_{\ell=0}^{p-2}(g^\ell)^r=\sum_{\ell=0}^{p-2}=\frac{(g^r)^{p-1}-1}{g^r-1}=0,\]](http://latex.artofproblemsolving.com/f/b/b/fbb17c033ef9b33d5bb0f5087a3899223de523aa.png)
which works since 
, due to 
. for 
. It also implies that ![\[q^{\left(\frac{p-1}{2}\right)}(1)=\sum_{k=0}^{p-1} k^{p-1}=p-1\ne 0,\]](http://latex.artofproblemsolving.com/b/d/d/bdd7391f24dddcdad3265776484390ac697c10c4.png)
so the maximal such that is 
by lemma 1.
. The rest is calculus: the formula for a geometric sum, the fact that 
applied to 
gives 
, and the fact that if a multiplicity of a root is not zero in our field then it decreases by 1 when you differentiate. Nothing else is needed for the proof.
can be written as 
. Starting with and its multiplicity 
, keep applying until you reach and its multiplicity.
as claimed.
, and 
. Suppose is such that 
, \dots, 
has as a root, but 
does not have as a root. Then is the multiplicity of in .
will have multiplicity of one less than in 
. ![\[ q_n(1) = \sum_{k=1}^{p-1} k^n \left( \frac kp \right) = \sum_{k \text{ qr}} k^n - \underbrace{2 \sum_{k=1}^{p-1} k^n}_{=0 \text{ for } n < p-1} \]](http://latex.artofproblemsolving.com/4/5/7/457aecd2ddad136f776ddd63b41734bcef79e89d.png)
Let be a primitive root modulo . The first sum then equals ![\[ g^0 + g^{2n} + g^{4n} + \dots + g^{(p-3) n} \]](http://latex.artofproblemsolving.com/2/6/d/26d3188991610e724115d4027f57a63aa4f2dfda.png)
which equals if 
but 
otherwise. as claimed.
and let 
Additionally, define 
inductively by 
and 
To show that the sums on top evaluate to 
induct from left to right; the base case follows from the fact that the number of nonzero QRs and QNRs are equal.th sum from the left. By the inductive hypothesis, we may shift to obtain 
Therefore, since 
is a polynomial of degree 
it suffices to show that 
precisely when 
so we are done.
. Notice that including all terms with negative powers that evaluate to zero, the th derivative of is equal to 
In particular, for 
, we can always split this into sums of the form 
for some 
and 
. All these sums evaluate to zero modulo , hence 
.
, the leading term has exponent , and the sum 
evaluates to 
, and hence the process terminates. It follows by the definition of derivative that the multiplicity of in is .
.
such that
, we can explictly write 
as
we have
over 
vanishes mod unless 
.
, everything is degree less than , and the constant terms are the same, so everything vanishes. However, when , the first sum has a non-vanishing component 
, so it will not be zero as everything else vanishes. Thus, fails and we are done.
. so that 
. Then notice we can write 
as a sum of 
. In particular to preserve degrees the last term is added once. Thus the first time 
is the same as the first time 
. However, the idea is that we can rewrite this as 
which equals 
so if this is 
then we must have 
so 
. In fact for this we get 
as desired.
times. We are left with ![\[P(x) = \sum_{k = 1}^{p - 1}\frac1{k}x^k\]](http://latex.artofproblemsolving.com/7/3/8/73820e0e41295b91e9b80fbbd2440789e239aea8.png)
. Plugging in 
we get that the sum is diviisble by 
Differentiate one more time and we get ![\[\sum_{k = 1}^{p - 1}x^k = \frac{x^k - 1}{x - 1}\]](http://latex.artofproblemsolving.com/5/0/f/50ff7e0712b4141ca548defe74287312bccc26eb.png)
which obviously has 0 powers of This implies that there are exactly 
factors of 
in the original expression.
However, I quickly saw that 
and 
intersect at the point 
,
.
?
?
families, there are 
children respectively. If a random child from any of the families is chosen, what is the probability that the child has 
siblings? Express your answer as a common fraction.[/quote]
,
people, exactly 
.
,
lies on diagonal 
such that 
.
.
has 
,
,
.
.
![$q(x) \in \mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/2/4/e/24e574b86f85cc81ad02b34d6d4ec576a53a95ae.png)
be given by 
where 
mod 
divides 
in the problem is irrelevant. If you replace it by another positive integer 
then the answer is simply 
.
.
is a non-zero constant.
.
denotes congruence modulo 
if 
and 
for the set of residues and nonresidues modulo 
.
derivative of ![\[
q^{(n)}(1) = \sum_{r\in\text{QR}}\frac{r!}{(r-n)!} -\sum_{b\in\text{QNR}}\frac{s!}{(s-n)!}.
\]](http://latex.artofproblemsolving.com/0/2/1/021fa2df6d57a642373e43754119cfffc6421352.png)
We show that this is zero for 
by a showing the lemma:
for polynomials of degree less than ![\[
\sum_{r\in\text{QR}} r^n = \sum_{s\in\text{QNR}} s^n = 0
\]](http://latex.artofproblemsolving.com/b/3/4/b34a9f7b610397a13dd1af91b1c794e5741b66e5.png)
.
.
for a primitive root 
,
,
.
.
.
derivatives is simply the sum of an 
.
.
.
.
such that
.
when the denominator 
.
,
and 
for 
.
when 
or when 
.
.
for 
,
.
,