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).
table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
, prove that 
.
is an involution, 
can be or .
yields that 
for some 
and 
for some 
. is even then has to be even, and vice versa. If is odd then 
. So our solution is 
even and 
odd. This works upon checking.
the assertion of the following F.E., i claim that 
works for all integers 
where either 
are odd or 
are both even.
Also 
gives 
so infact is an involution (i.e. 
).
, now set 
then by 
we get that 
for all even .
we get 
for all integers so 
for all odd , now if 
is odd then notice that 
for all odd . so odd, if is even, but 
odd then 
for all even so which cant happen so is also even in this case. Hence our claim is true and we are done 
.
such that for all integers and 
, ![\[ f(2x+f(y))+f(f(2x))=y. \]](http://latex.artofproblemsolving.com/b/0/d/b0d16e87a8b30ebec5997ed12254094cdc3125d5.png)
and 
and 
as per our claim.
or 
are odd that 
for even and 
for odd where either 
are both even or 
These solutions clearly work.
, we get that . Letting only 
, we have 
so 
Thus, is an involution.
(and ).
Letting 
gives 
Since involutions are both injective and surjective, we can de-nest this into 
Thus, for all even , we have 
, we have that 
Applying to both sides and using the fact that is an involution, 
Thus, by induction, if 
, then 
for all odd integers 
.
If is even, then this equation is clearly true since 
is even. Then, if is even, then we have that all even inputs lead to even outputs. Hence, if 
is odd, then odd also maps to even, contradicting surjectivity. Thus, if is even, then must also be even. Otherwise, if is odd, then we must have 
Thus we are done.
![\[ f(x) = \begin{cases} a - x & x \equiv 0 \pmod 2 \\ b - x & x \equiv 1 \pmod 2 \end{cases} \]](http://latex.artofproblemsolving.com/e/a/e/eae545d48ed07867af7cc41795addd69bed920e6.png)
where and are either both even, or 
. It can be checked that all of these work, so we prove they're the only solutions. be the given assertion.
.
.
.
implies takes the above form for some and , so we'd be done if we could show the parity condition. If is odd, then plug in to deduce ; if is odd, plug in to deduce 
. This finishes the problem.
from 
and 
from 
. Then our condition can be rewritten as![\[y=f(2x+f(y))+2x.\]](http://latex.artofproblemsolving.com/c/e/7/ce76b9222ff3afaf771c95071e5060b3c0f2e981.png)
based on parity:
and with our involution 
. We get the following solution, which can be easily tested:![\[\boxed{f(x) = \begin{cases} -x+c & \text{for even } x \\ -x+d & \text{for odd } x \end{cases} \quad \text{where }a,b \text{ both even or } a=b.}\]](http://latex.artofproblemsolving.com/d/0/9/d09d5876e59d532e5fe1b962b0786afd6ac667ae.png)
for odd and 
for even where 
are integers such that either 
or 
. With some effort, we can verify that these all work.
. From 
we have 
. Then from 
, we have 
, which implies that is bijective. This also lets us rewrite the original equation as 
.
, we have 
, which implies that there exists integers such that when is odd and when is even. From , we see that:
is even, then must also be even, as if is odd then 
, a contradiction. Any pair 
with and both even will work, as mentioned above. is odd, then 
, so , which also works.
then 
to get 
Thus, is an involution. The original equation is now
Then we can separate into cases even and odd like this :
If is even, must also be even, as if is odd then 
contradiction. is odd, then 
because 
which works. Thus, our answer is
for even and 
for odd where 
or and are both even. It is easy enough to do cases and check that these solutions work. denote the assertion. gives us , so by , we get ![\[f(f(y))+f(f(0))=f(f(y))=y,\]](http://latex.artofproblemsolving.com/f/f/9/ff9c87585b1956edb167a10a89e886f008d42b18.png)
meaning is an involution, and is therefore bijective. Now, We have ![\[f(2x+f(y))+f(f(2x))=f(2x+f(y))+2x=y,\]](http://latex.artofproblemsolving.com/4/a/a/4aa0f07711a1dbea416bba54f4339f89065d715a.png)
giving 
. Thus, for even , we have 
. We also have 
, giving 
for odd . Let 
and 
, so we can write for even and for odd . is even in our original equation. Then, we have ![\[f(2x+f(y))+f(f(2x))=f(2x-y+c)+2x=y,\]](http://latex.artofproblemsolving.com/0/0/4/0042bafb3e8a539f524ab0399da223496c2336db.png)
so if is even, then this works. Otherwise, if is odd, then we must have . is odd in our original case. Then, we have ![\[f(2x+f(y))+f(f(2x))=f(2x-y+d)+2x=y,\]](http://latex.artofproblemsolving.com/2/2/d/22ddd8cdc51bb54652839bf9e8b5589508bbdf97.png)
so if is even then this works. Otherwise, we must have ., or and are both even, as desired.
to get 
. Putting 
gives 
. This implies is bijective. be such that 
. Putting 
gives us that 
since 
giving us that 
. be such that 
. Putting 
gives us that 
. is even and is odd or 
in order to satisfy the given equation.
where 
satisfy the above conditions.
is the assertion then 
. Then 
. We also have 
which when gives us 
. Thus our solutions are determined by 
and 
, say write it as 
when is odd and 
when is even. Now tells us that 
. Therefore, can be even or . A similar analysis on 
tells us that is even or . Thus, we have 
when is odd and when is even for some 
such that they are both even or equal.
denote the assertion. Then 
Hence 
so 
is bijective. Thus the assertion becomes 
Thus 
If is odd, we have that 
This holds also when is odd. Hence, the solutions are 
where either are both even or 
if is some odd integer, or 
if is even and 
if is odd, and 
are even integers. denote our assertion. From 
we see that 
From 
we get that 
using the fact that By rearranging our given equation, and taking on both sides, we see that 
Let 
Note that by plugging in 
, 
Now we consider two seperate cases.
we plug it back into our original equation to see that 
and thus we have that 
Thus, 
and hence 
Thus, we can write 
clearly we can't say anything, and both solutions are valid.
.
with 
for some fixed positive integer 
for constants 
and 
for some constant 
.
for these two classes of functions. For 
,
For the other function, note that if 
and if 
So we have shown that these two classes of functions work.
denote the assertion. 
Thus, 
Therefore, we have that the original functional equation can be rewritten as 
so 
Since 
Now, note that from plugging in 
into this, 
and plugging in 
into this, 
Therefore, it follows that there exists a constant 
for all even 
.
but 
but 
.
.
or 
where either 
or if 
then both of them are even. It is easy to check that all such solutions indeed work, so it remains to show that the given functional equation implies that 
and 
Hence, we have 
so we can plug this into 
Now, applying 
Then, let us fix 
and let 
for some 
Then, we get
Therefore, as we vary 
(so that 
for some constant 
This implies the existence of two constants 
such that
Now, we use the fact that 
If 
we get
and it is easy to see that 
being odd implies the same thing. On the other hand, if 
so no conclusion can be found about 
and the same thing holds for