TITLE: Let $f$ be a continuous function on $[0,2]$ such that $f(0)=f(2)$, prove the following statement.
QUESTION [0 upvotes]: I've recently come across this exercise for which I currently have no clue on how to solve.
Basically, it says that given a function $f$ that is continuous on $[0,2]$ such that $f(0)=f(2)$, you have to prove that there exist two points inside the interval, lets say $x_1,x_2\in[0,2]$ that are a unit apart, that is $x_2-x_1=1$, and that they both have the same image, that is, $f(x_1)=f(x_2)$.
The problem is that I can't even see it intuitively. Usually I tend to see these kind of problems intuitively, realize to myself they seem to be true after some thinking, and then the problem arises at the time of formalizing my thoughts. But this time, I don't even know why that is implied.
Any help will be appreciated.

REPLY [2 votes]: Consider the function $$g(x) = f(x) - f(x + 1)$$
This is a continuous function on $[0, 1]$, and $$g(0) = -g(1)$$
So either $g$ is zero at the endpoints, or it changes sign somewhere between them. Can you finish the proof?


TITLE: Definition of nowhere differentiable functions
QUESTION [1 upvotes]: Is there a formal definition for the concept of nowhere differentiable function in $\mathbb{R}$?
I know that a function can fail to be differentiable in several ways in an open interval but I am looking for references.
I am also familiar with the most common examples.

REPLY [0 votes]: Such a function exists but it is nontrivial to show that it is nowhere differentiable. The Weierstraß function
$$W:\mathbf R \to \mathbf R, \quad W(x)=\sum_{k\in\mathbf N}b^k \cos (a^k x)$$
for $0<b<1$ and $ab>1+\frac{3\pi}{2}$, where $a\in\mathbf N$, $a\neq 0$ and $b\in\mathbf R$ is continuous but nowhere differentiable. The idea to show this is by contradiction, using fourier methods and some growing conditions. If you like I can post the idea later in more detail.
The picture shows the partial sum of $W$ with $\alpha= 0.5$


TITLE: show that that for every $x \in \mathbb{R}^n$ if $\|Ax\|_2=\|x\|_2$ then $A$ is an orthogonal matrix
QUESTION [0 upvotes]: Show that for every $x \in \mathbb{R}^n$
if $\langle Ax,x\rangle=x^TAx=0$ then $A=0$
By the help of $(1)$ show that that for every $x \in \mathbb{R}^n$ if $\|Ax\|_2=\|x\|_2$ then $A$ is an orthogonal matrix

What I have tried so far:
For $(1)$ I have shown that
$$x^TAx=x^TA_+x=0$$
$$A_+=\frac{1}{2}(A+A^T)$$
since it holds for every $x \in \mathbb{R}^n$ we can write
$$0=(x+y)^TA(x+y)=x^TAx+x^TAy+y^TAx+y^TAy=x^TAy+y^TAx=x^T(A+A^T)y
$$
Then I use spectral theorem which states that for every real symmetric matrix $A$ there exists orthogonal matrix $P$ such that
$$D=PAP^T \implies A=P^TDP $$
Now using this fact I can write, considering $y=Px$
$$x^TA_+x=x^TP^TDPx=(Px)^TD(Px)=y^TDy=0$$
I have come so far. Now my question is this enough for proving $A=0$? Or there are more steps to make it clear?

For $(2)$
I honestly don't exactly know how to use $(1)$ to prove this one however here is what I have tried
$$<Ax,Ax>=\|x\|_2^2 \implies <x,A^TAx>=\|x\|_2^2$$
We know for norm-$2$ if $P$ is orthogonal $\|x\|_2=\|Px\|_2$ so
$$<Ax,Ax>=\|Px\|_2^2 \implies <x,A^TAx>=<x,P^TPx>$$
Is this enough for proving $P=A$, I guess not and that's exactly my question.

REPLY [3 votes]: Everything done so far seems too complicated.
Assume $A$ is symmetric. Start with $e_i^\intercal A e_i = 0$ implies $a_{ii} = 0.$ Next, $(e_i + e_j)^\intercal A (e_i + e_j) = a_{ii} + a_{ij} + a_{ji} + a_{jj} = a_{ij} + a_{ji}.$ This implies $A$ is also antisymmetric, so $A^\intercal = A = -A$ and $A = 0.$
What to do if $A$ is not symmetric? I think the result of part 1 is false (part 2 still is true). We will still reach that $a_{ij} + a_{ji} = 0,$ so $A$ must be antisymmetric. Consider then $A = \begin{bmatrix} 0 &-1 \\ 1 &0 \end{bmatrix}.$ Then, $(x, y) A (x, y)^\intercal = (x, y) (-y, x)^\intercal = 0.$ And $A$ is not the matrix zero.
Lemma. Any linear function that preserves norm in $\mathbf{R}^d$ is an orthogonal transformation.
Proof Let $A$ be such linear function, that is $\| A x\| = \|x\|.$ We may use polarization identity
$$
4(x, y) = \|x + y\|^2 - \|x - y\|^2.
$$
In the right hand side we can add $A,$ and by linearity we reach $(Ax, Ay) = (x, y)$ for all vectors $x, y.$ Then $(x, y) = (Ax, Ay) = (x, A^\intercal A y)$ (this is true actually whether or not $A$ is symmetric). So that $(x, (A^\intercal A - I)y) = 0$ for all $x, y$ and now we may apply the first part to the symmetric matrix $A^\intercal A - I$ to reach the conclusion. QED


TITLE: Efficient identification of an equation given inputs and outputs.
QUESTION [0 upvotes]: Looking for a more formal approach to identify factors of a given calculation and a way to approximate that equation itself.
I know of $x$ many inputs to a given calculation, and I have the output, but I don't have any good way to identify which inputs are actually used in the output. It may only use one of the inputs, it may use all of them, or anything in between. The simplest way is to modify only one input at a time but I don't have that luxury, they come as a set of values and cannot be adjusted individually (though there are many different sets of data from which to examine).
After identifying the factors, is there a good way to identify the equation itself? In this particular case (data below) I know there are two inputs but I don't have a formalised way to identify an equation for it. I have been trying various polynomial combinations and running it through excel solver to attempt to identify an equation that fits, but it has not been an efficient approach. I'd say a valid equation should be accurate up to 5 significant digits but its been difficult to do this with just trial and error.

input 1
input 2
output

61
16
0.233432051588249

61
18
0.2344106759016

61
19
0.234899689515166

61
20
0.235389300214951

65.6
16
0.239836497890295

65.6
18
0.240832674050633

65.6
19
0.241331091772152

65.6
20
0.24182885021097

74
16
0.251737360278333

74
18
0.252764823451033

74
19
0.253278555037382

74
20
0.253792286623732

77
16
0.256051361442492

77
18
0.257090428922685

77
19
0.257610417994718

77
20
0.258129496402878

REPLY [0 votes]: A fast solution could be a stepwise regression for a generalized linear model
$$z(x,y)=a_0+\sum_{i=1}^n a_i t_i$$  defining factors such as
$$\left\{x,y,\frac{1}{x},\frac{1}{y},x^2,x y,y^2,\frac{1}{x^2},\frac{1}{x
   y},\frac{1}{y^2},\frac{x}{y},\frac{y}{x},\cdots\right\}$$