Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?
Answer
The number in question is simply
$$2^{1/2+1/4+1/8+\cdots} = 2^1 = 2$$
Yes, this works for other numbers. More interesting is if the numbers are not equal inside the radicals. For example, say you have a positive sequence element $a_n$ inside the $n$th radical. If we assume the expression converges to some value $P$, then
$$\log{P} = \sum_{k=1}^{\infty} \frac{\log{a_n}}{2^n}$$
How to compute $\cos \left( \frac{\pi}{4}(k-1) \right)$ ?
If $f$ is continuous on $I$, then: whenever $a, b \in I$, $a < b$ and $y$ lies between $f(a)$ and $f(b)$, then there exists at least one $x \in (a,b)$ such that $f(x) = y$.
Proof: Suppose $f(b) < y < f(a)$. Let $S := \{x \in [a,b] : y < f(x)\}$. Since obviously $a \in S$, then $S$ is not empty, therefore $\inf S = x_0 \in [a,b]$. Now, for every positive integer $n$, $x_0 + \frac{1}{n}$ is not a lower bound for $S$. hence, there exists $(s_n) \subseteq S$ such that $$x_0 \leq s_n < x_0 + \frac{1}{n}$$
By the squeeze rule, obviously $\lim s_n = x_0$. By continuity of $f$, $\lim f(s_n) = f(x_0)$. Hence since $f(s_n) > y$, then
\begin{align}
f(x_0) \geq y. \tag{I}
\end{align}
Now, let $t_n = \max\{b, x_0 - \frac{1}{n}\}$. Then, we have that $x_0 - \frac{1}{n} \leq t_n \leq x_0$. So, by squeeze rule, $\lim t_n = x_0$. But, we also have $t_n \notin S$, therefore, $f(t_n) \leq y$ for all $n$. The continuity of $f$ implies that $\lim f(t_n)$ =
\begin{align}
f(x_0) \leq y. \tag{II}
\end{align}
(I) and (II) gives desired result.
Is this proof correct? Is there a better way to solve this problem?
Thanks.
Answer
It’s not quite right: as you’ve defined $S$, $\inf S=a$. You want to let $x_0=\sup S$ and choose a sequence $\langle s_n:n\in\Bbb Z^+\rangle$ in $S$ such that $$x_0-\frac1n In other words, you somehow got the picture turned around at the beginning, but the logic is otherwise correct. You could also fix it by assuming that $f(a) Added: I should have said that apart from that one glitch, it’s very nicely written.
I was reading about proof by infinite descent, and proof by minimal counterexample. My understanding of it is that we assume the existance of some smallest counterexample $A$ that disproves some proposition $P$, then go onto show that there is some smaller counterexample to this which to me seems like a mix of infinite descent and 'reverse proof by contradiction'.
My question is, how do we know that there might be some counterexample? Furthermore, are there any examples of this?
Answer
Consider, for instance, the statment
Every $n\in\mathbb{N}\setminus\{1\}$ can be written as a product of prime numbers (including the case in which there's a single prime number appearing only once).
Suppose otherwise. Then there would be a smallest $n\in\mathbb{N}\setminus\{1\}$ that would not be possible to express as a product of prime numbers. In particular, this implies that $n$ cannot be a prime number. Since $n$ is also different from $1$, it can be written as $a\times b$, where $a,b\in\{2,3,\ldots,n-1\}$. Since $n$ is the smallest counterexample, neither $a$ nor $b$ are counterexamples and therefore both of them can be written as a product of prime numbers. But then $n(=a\times b)$ can be written in such a way too.
Prove ${{n+1} \choose {m+1}} = \sum_{k=m}^{n}{k \choose m}$ using a
purely combinatorial argument.
I don't think I understand how to do a combinatorial proof. I know the left side expands to:
$${{n+1} \choose {m+1}} = \frac{(n+1)!}{(m+1)!(n+1-m-1)!} = \frac{(n+1)!}{(m+1)!(n-m)!}$$
For the right side:
$$\sum_{k=m}^{n}{k \choose m} = {{k+1} \choose {m+1}} = \frac{(k+1)!}{(m+1)!(k+1-m-1)!} = \frac{(k+1)!}{(m+1)!(k-m)!}$$
Which is similar to what I got on the top.
Where do I go from here?
Answer
What you have started there would be an algebraic, rather than a combinatorial argument.
In a typical "combinatorial" argument we look at some set of things, and count the number of element in it in two different ways, yielding two different expressions for the result. Since it is the same things being counted, we now know that the two different expressions have the same value.
In this particular case, it is natural to look at "how many $(m+1)$-element subsets of $\{1,2,3,\ldots,n,n+1\}$ are there?", since that is one definition of the left-hand side $\binom{n+1}{m+1}$. We'd then try to come up with a different way to count those $(m+1)$-element subsets that naturally leads us to conclude that there must be $\sum_{k=m}^{n}\binom km$ of them.
Hint. Consider grouping the subsets according to what the largest element of each subset is.
I have two matrices
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
and $$ \begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
They are not diagonalizable. Share the same characteristic polynomial, the same trace, same determinant, eigenvalues, rank. What could I use more to say if they are similar or not?
Answer
Although using Jordan Canonical forms are probably the fastest answer, it is possible to solve this question using brute force. We want to find out if there is an invertible matrix $P$ such that
$$P \cdot
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix} =
\begin{pmatrix}
2 & 2 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix} \cdot P.$$
Denoting
$$P = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$$
this is equivalent with stating that
$$\begin{pmatrix}
2a & a + 2b & 3c\\
2d & d + 2e & 3f\\
2g & g + 2h & 3i
\end{pmatrix} =
\begin{pmatrix}
2a + 2d & 2b + 2e & 2c + 2f\\
2d & 2e & 2f\\
3g & 3h & 3i
\end{pmatrix}.$$
Comparing corresponding entries, we find that:
$$\begin{cases}
2a = 2a + 2d\\
a + 2b = 2b = 2e\\
3c = 2c + 2f\\
2d = 2d \\
d + 2e = 2e\\
3f = 2f\\
2g = 3g\\
g + 2h = 3h\\
3i = 3i
\end{cases}.$$
This is equivalent with
$$\begin{cases}
d = 0\\
a = 2e\\
c = 2f\\
0 = 0\\
e = e\\
f = 0\\
g = 0\\
h = 0\\
3i = 3i
\end{cases}.$$
And from equation 3 and 6, we also have that $c = 0$. Therefore, we find that
$$a = 2e, b = b, c = 0, d = 0, e = e, f = 0, g = 0, h = 0, i = i,$$
so we can take $b = 0$ and we find that
$$P = \begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}$$
since we had a free choice for $b, i$ but we want $P$ to be invertible, so we can't take $i = 0$ (but any other value will do) and we take $b = 0$ (so $P$ is clearly invertible).
So this is a way to find out if both matrices are similar or not (if you would have found a contradiction in the system, then they are not (or if the resulting matrix $P$ is not invertible they are also not similar). However, as I already mentioned: if you know about Jordan canonical form, always use this approach, since it is clearly much shorter!
Show that the function
$$f(x) = \begin{cases} \frac{x^2y^4}{x^4+y^8} ,& \text{if } (x,y)≠ (0,0) \\ 0, &\text{if } (x,y)= (0,0)\end{cases}$$
is Gateaux differentiable at $(0,0)$ but not continuous at $(0,0)$.
So I know how to show it is Gateaux differentiable at $(0,0)$, but I don't know how to go about showing it is not continuous...
