Let $f$ be monotonic increasing on $[a,b]$ If $\int_a^bf'(x)dx = f(b)-f(a)$ Show $f$ is absolutely continuous I tried to prove it by the definition of absolutely continouse:

Let $\{(a_i,b_i)\}_{i=0}^n$ be disjoint sub-interval on $[a,b]$

By Reimenn integral,

$\int_a^bf'(x)dx = \sum_{i=0}^nf'(t_i)(b_i-a_i)$ ,for any $t_i \in[a_i,b_i]$

Since f is differentiable, there exist $c_i\in[a_i,b_i] $ such that $\frac{f(b_i)-f(a_i)}{b_i-a_i} = f'(c_i)$
......

And then I think I am on the wrong direction. What else can I know from $\int_a^bf'(x)dx = f(b)-f(a)$

Help me please

Extend $f$ to be $f(a)$ for $x\leq a$ and $f(b)$ for $x\geq b$. Since $f$ is
increasing, for every $h>0$ we have%
\begin{align*}
\int_{a}^{b}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\,dx &  =n{\biggl \{}%
\int_{b}^{b+\frac{1}{n}}f(x)\,dx-\int_{a}^{a+\frac{1}{n}}f(x)\,dx{\biggr \}}\\
&  \leq n\frac{f(b)-f(a)}{n}=f(b)-f(a).
\end{align*}
It follows from Fatou's lemma, that
$$
0\leq\int_{a}^{b}f^{\prime}(x)\,dx\leq\liminf_{n\rightarrow\infty}\int_{a}%
^{b}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\,dx\leq f(b)-f(a).
$$
With the same proof if $a<y<b$ you get
\begin{align*}
\int_{a}^{y}f^{\prime}(x)\,dx  & \leq f(y)-f(a),\\
\int_{y}^{b}f^{\prime}(x)\,dx  & \leq f(b)-f(y).
\end{align*}
Adding these two inequalities gives
$$
\int_{a}^{y}f^{\prime}(x)\,dx+\int_{y}^{b}f^{\prime}(x)\,dx\leq
f(y)-f(a)+f(b)-f(y)=f(b)-f(a).
$$
Since by hypothesis $\int_{a}^{b}f^{\prime}(x)\,dx=f(b)-f(a)$, it follows that
both inequalities must be equalities, that is,
$$
f(y)=f(a)+\int_{a}^{y}f^{\prime}(x)\,dx
$$
for every $y\in\lbrack a,b]$. Now you need to prove that if $g\geq0$ is
integrable, then the function
$$
h(y)=\int_{a}^{y}g(x)\,dx
$$
is absolutely continuous.

the function is differentiable on that interval, so it has to be continuous. and continuous functions are absolutely continuous on an interval. you could use the definition of absolute continuity to prove that a continuous function is absolutely continuous on an interval.


What does it mean to have a lone plus sign in the exponent/superscript (Modified Weiszfeld algorithm) I was skimming through Vardi & Zhang's paper on "A modified Weiszfeld algorithm for the Fermat-Weber location problem".

It is available at http://www.stat.rutgers.edu/home/cunhui/papers/43.pdf

If you open the paper and scroll down to page 3, you'll find equation 12, which I found to be odd. There is a lone plus sign in the exponent/superscript. Does anyone know what that means? Could it be that this simply a typo and that the "+" sign refers to an addition operation instead?

Many thanks for your help!
Nicolas

I have come across this notation in a different context (page 80, eq. (4.40)), where it was defined as

$$ x^+ \equiv \max(x,0) $$

And correspondingly

$$ x^- \equiv \min(x,0) $$

That is, the $+$ or $-$ indicates that the value is clamped to non-negative or non-positive numbers. Next to the other term in that equation, it seems likely that the same meaning is intended here.

A similar notation occurs in Equation (19), which suggests it's not a typo.  In both instances, there is a convention mentioned in the following text, modifying the computation under certain circumstances.  I suspect the + notation is just an indication that the expression is not always to be evaluated literally.


Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.

Taking the contrapositive, we suppose $x+y+z<\sqrt 3$ and want to prove
$$xy+yz+zx<\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)<1$$
Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z<3$ and we want to prove
$$(xy+yz+zx)^2(x^2+y^2+z^2)<27$$

And that's a simple AM-GM:
$$(xy+yz+zx)^2(x^2+y^2+z^2)\le\left(\frac{2(xy+yz+zx)+x^2+y^2+z^2}3\right)^3=\left(\frac{(x+y+z)^2}3\right)^3<27$$

Sketch indicates the truth of the proposition. Blue boxes : $x+y+z > \sqrt{3}$.
Blue with purple centers   $$xy + yz + zx > \frac{1}{\sqrt{x^2+y^2+z^2}}$$
The surfaces indicate where equality holds.


Limit of improper integrals of uniformly convergent function I've got a problem.

Let  $g(t)\ge0$ and it has improper integral on interval $[A, B)$.
Furthermore, sequence of integrable functions $f_{n}(t)$ is uniformly convergent do $f(t)$ on every subinterval $[a,b]$ of $[A,B)$ and
$0 \le f_{n}(t) \le g(t)$ for every $t\in[A,B)$

Prove that

$$\lim_{n\to\infty} \int_{A}^{B^{-}}f_{n}(t)dt = \int_{A}^{B^{-}}f(t)dt$$

As I suppose our goal is to show that the difference

$$|\int_{A}^{B^{-}}f_{n}(t)dt-\int_{A}^{B^{-}}f(t)dt| < \epsilon$$

I've tried to do it like this

$$|\int_{A}^{B^{-}}f_{n}(t)dt-\int_{A}^{B^{-}}f(t)dt| =  |\int_{A}^{b}f_{n}(t)dt-\int_{A}^{b}f(t)dt + \int_{b}^{B^{-}}f_{n}(t)dt - \int_{b}^{B^{-}}f(t)dt| \le$$

$$\le \int_{A}^{b}|f_{n}(t)-f(t)|dt + |\int_{b}^{B^{-}}(f_{n}(t)-f(t))|$$

From the uniformly convergence we can make this first integral infitesimal $(b-A)\epsilon$, but don't know how to estimate the second one. As I think it has to be connected with function $g(t)$, but is it correct to say it is lower than $2\int_{b}^{B^{-}}g(t)dt$, which we can make infinitesimal, because of convergence of improper integral $\int_{A}{B^{-}}g(t)dt$.

Edit:

Okey, that's my last try

From uniform convergence of $f_{n}(t)$ to $f(t)$
We can make the difference $|f_{n}(t)-f(t)| < \epsilon_1 = \frac{\epsilon}{3(b-A)}$ then

$$\int_{A}^{b}|f_{n}(t)-f(t)|dt < (b-A)\cdot\epsilon_1 = \frac{\epsilon}{3}$$

Next step, it is known that our improper integral of $g(t)$ exists, so

$$\lim_{b\to B}\int_{b}^{B}g(t)dt = 0$$

Hence, we can make $|\int_{b}^{B}g(t)dt| < \epsilon_2 = \frac{\epsilon}{3}$

To sum up

$$\int_{A}^{b}|f_{n}(t)-f(t)|dt + |\int_{b}^{B^{-}}(f_{n}(t)-f(t))| \le (b-A)\cdot\epsilon_1 + 2\epsilon_2 = 3\cdot\frac{\epsilon}{3} = \epsilon$$

I'm just cleaning up the original poster's answer, and formalizing Daniel Fischer's good hints in the comments, for anyone who wants all the details.

The limit we're trying to prove is equivalent to showing,
$\left| \int_A^{B-} f-f_n \right| < \epsilon$ for all $n$ sufficiently large.

We have
$\left| \int_A^{B-} f-f_n \right| \le \left| \int_A^{b} f-f_n \right| + \left| \int_b^{B-} f-f_n \right|$.

By the bounds on $f$ and $f_n$ with respect to $g$, we have $|f-f_n| \le g(x)$, so,

$\left| \int_b^{B-} f-f_n \right| \le \int_b^{B-} |f-f_n| \le \int_b^{B-} g < \lim_{c \uparrow B} \int_b^c g $. (The last inequality follows from $g\ge0$.)

Now choose $b$ sufficiently large such that $\int_b^{B} g < \epsilon/2$ (as guaranteed by the integrability of $g$).

With $b$ now fixed, we have for all $n$ sufficiently large that, by uniform convergence of $f_n \to f$,
$\left| \int_A^b f-f_n \right| < \epsilon/2$. We can conclude that for all $n$ sufficiently large,
$\left| \int_A^{B-} f-f_n \right| < \epsilon/2 + \epsilon/2$. Taking the limit in $n$ gives,
$\left| \int_A^{B-} f-\lim_{n \to \infty} \int_A^{B-} f_n \right| < \epsilon$. Since $\epsilon$ is arbitrary,
$\int_A^{B-} f = \lim_{n \to \infty} \int_A^{B-} f_n$.

~~~

(Edited slightly to include $\int_b^{B-} g < \lim_{c \uparrow B} \int_b^c g$; why the downvote? also including a comment is more helpful for the community, and for those making good faith efforts.)


Best Fake Proofs? (A M.SE April Fools Day collection) In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen.

I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.

$$x^2=\underbrace{x+x+\cdots+x}_{(x\text{ times})}$$
$$\frac{d}{dx}x^2=\frac{d}{dx}[\underbrace{x+x+\cdots+x}_{(x\text{ times})}]$$
$$2x=1+1+\cdots+1=x$$
$$2=1$$

$1=2$: A Proof using Complex Numbers
This supposed proof uses complex numbers. If you're not familiar with them, there's a brief introduction to them given below.
The Fallacious Proof:

Step 1: $-1/1 = 1/-1$
Step 2: Taking the square root of both sides: $\sqrt{-1/1} = \sqrt{1/-1}$.
Step 3: Simplifying: $\sqrt{-1} / \sqrt{1} = \sqrt{1} / \sqrt{-1}$
Step 4: In other words, $i/1 = 1/i$.
Step 5: Therefore, $i / 2 = 1 / (2i)$,
Step 6: $i/2 + 3/(2i) = 1/(2i) + 3/(2i)$,
Step 7: $i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) )$,
Step 8: $(i^2)/2 + (3i)/2i = i/(2i) + (3i)/(2i)$,
Step 9: $(-1)/2 + 3/2 = 1/2 + 3/2$,
Step 10: and this shows that $1=2$.

See if you can figure out in which step the fallacy lies.


Find all points $C$ such that area of $ABC$ is $3$ In an exercise I'm asked the following:

Consider the points $A=(0,0,0)$ and $B=(1,0,0)$. What are all the points $C$ such that the triangle $ABC$ has area 3?

I did the following:
let $C=(a,b,c)$, then the triangle has area equal to:
$$\frac{1}{2} ||B\times C||$$
so we have that $$\frac{1}{2}\sqrt{c^2 +b^2}=3$$
Giving:
$$c= \pm \sqrt{36 -b^2}$$
So all the points $C$ are:
$$C=(t,\lambda, \pm \sqrt{36 -\lambda^2}
), \ \ t,\lambda \in \mathbb R$$
My question is: Is this correct? It seems a little bit odd that there are infinite points that satisfy this condition.

Let C be in plane $z=0$
$$\dfrac12 \cdot1 \cdot h = 3 \to h=6 $$
So rotating by radius $ \sqrt{y^2+z^2} =6$ around x-axis gives all points on a torus as the locus required.

Think about this possibility: the points $A$ and $B$ create the base of your triangle along the x-axis. An obvious starting point would be $C = (\frac{1}{2}, 6, 0)$ which creates an isosceles triangle that has an area of 3 (draw and check). Rotating this coordinate around the x-axis would leave you to create infinitely many possibilities for coordinates that create triangles with an area of 3 (provided in this case we are working in $\mathbb{R}$).
SO YES, it is possible to have infinitely many points.
The above example does NOT give ALL possible points which is where we need other tools, enter Linear Algebra.


Maximum Likelihood Estimator of parameters of multinomial distribution Suppose that 50 measuring scales made by a machine are selected at random from the production of the machine and their lengths and widths are measured. It was found that 45 had both measurements within the tolerance limits, 2 had satisfactory length but unsatisfactory width, 2 had satisfactory width but unsatisfactory length, 1 had both length and width unsatisfactory. Each scale may be regarded as a drawing from a multinomial population with density
$$
\pi_{11}^{x_{11}}
\pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}
$$
Obtain the maximum likelihood estimates of the parameters.
I have tried this by the following way:
the likelihood function is
\begin{align}
L & =L(\pi_{11},\pi_{12},\pi_{21},(1-\pi_{11}-\pi_{12}-\pi_{21})) \\[8pt]
& =\prod_{i=1}^{50}[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}] \\[8pt]
& =[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}}
\pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}]^{50} \\[8pt]
& =[\pi_{11}^{45}\pi_{12}^{2}
\pi_{21}^{2}(1-\pi_{11}-\pi_{12}-\pi_{21})^{1}]^{50} \\[8pt]
& =\pi_{11}^{2250}\pi_{12}^{100}
\pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}
\end{align}
Taking logarithm of the likelihood function yields,
\begin{align}
L^*& =\log L=\log \left[\pi_{11}^{2250} \pi_{12}^{100}
\pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}\right] \\[8pt]
& =2250\log [\pi_{11}]+100\log [\pi_{12}]+100\log [\pi_{21}]+50\log (1-\pi_{11}-\pi_{12}-\pi_{21})
\end{align}
Now taking the first derivative of $L^*$ with respect to $\pi_{11}$
$\frac{\partial L^*}{\partial \pi_{11}}$
$=\frac{2250}{\pi_{11}}-\frac{50}{(1-\pi_{11}-\pi_{12}-\pi_{21})}$
setting $\frac{\partial L^*}{\partial \pi_{11}}$ equal to $0$,
$$\frac{\partial L^*}{\partial \hat\pi_{11}}=0$$
$$\Rightarrow\frac{2250}{\hat\pi_{11}}-\frac{50}{(1-\hat\pi_{11}-\hat\pi_{12}-\hat\pi_{21})}=0$$
$$\Rightarrow \hat\pi_{11}=\frac{45(1-\hat\pi_{12}-\hat\pi_{21})}{44}$$
$\bullet$Are the procedure and estimate of $\pi_{11}$ correct?
$\bullet$I have another question that if it is multinomial then where the term $\binom{n}{x_{11}x_{12}x_{21}x_{22}}=\binom{50}{45,2,2,1}$?

Consider a positive integer $n$ and a set of positive real numbers $\mathbf p=(p_x)$ such that $\sum\limits_xp_x=1$. The multinomial distribution with parameters $n$ and $\mathbf p$ is the distribution $f_\mathbf p$ on the set of nonnegative integers $\mathbf n=(n_x)$ such that $\sum\limits_xn_x=n$ defined by
$$
f_\mathbf p(\mathbf n)=n!\cdot\prod_x\frac{p_x^{n_x}}{n_x!}.
$$
For some fixed observation $\mathbf n$, the likelihood is
$L(\mathbf p)=f_\mathbf p(\mathbf n)$ with the constraint $C(\mathbf p)=1$, where $C(\mathbf p)=\sum\limits_xp_x$. To maximize $L$, one asks that the gradient of $L$ and the gradient of $C$ are colinear, that is, that there exists $\lambda$ such that, for every $x$,
$$
\frac{\partial}{\partial p_x}L(\mathbf p)=\lambda\frac{\partial}{\partial p_x}C(\mathbf p).
$$
In the present case, this reads
$$
\frac{n_x}{p_x}L(\mathbf p)=\lambda,
$$
that is, $p_x$ should be proportional to $n_x$. Since $\sum\limits_xp_x=1$, one gets finally $\hat p_x=\dfrac{n_x}n$ for every $x$.

We can then say: (i.e., Lemma)
If $n_i>0$, for every $i$, and subject to $0<p_i<1$. Then $$f(p_1,p_2,..., p_q)= \sum^q_{i=1} n_i \log_e{p_i} $$  ...  the only expression with  $p_i 's $ we are interested in the reduced  log likelihood (p).
So, the maximum of $f(p_1,p_2,..., p_q)$ is achieved at the points $(\hat{p_i}, \hat{p_i}, ... , \hat{p_q})$ where $\hat{p_i}= \frac{n_i}{n} $ for every i.


proof by induction: $7^{2n} - 2^n$ is always divisible by $47$ as per the title, i need to prove by induction that the expression "$7^{2n} - 2^n$" is always divisible by $47$.

base case is fine: $7^{2*1} - 2^1 = 47$.

Induction step gets lost quickly, any ideas?

step 1: $7^{2(n+1)} - 2^{(n+1)}$

step 2:  $7^2 * 7^{2n} - 2^1 * 2^{n}$

step 3: ???

step 4: profit.

Step 1 is not “$7^{2(n+1)} - 2^{n+1}$.”  You're thinking in terms of expressions and you want to think in terms of sentences.  This is one half of the conclusion to an if-then statement you want to prove:

  For any $n$, if $7^{2n} - 2^n$ is divisible by $47$, then $7^{2(n+1)} - 2^{n+1}$ is also divisible by $47$.

Step 0 is the base case, which you have shown.

Step 1 is “Suppose there exists a positive integer $n$ such that $7^{2n} = 2^n$.”

Step 2 is: you need to show that $7^{2(n+1)} - 2^{n+1}$ is divisible by $47$.

You've got a few answers on how to show that.  Essentially it's the old add-and-subtract-the-same-thing trick.
\begin{align*}
    7^{2n+2} - 2^{n+1} &= 7^2\cdot 7^{2n} - 2\cdot 2^{n}
\\&= 7^2 \cdot 7^{2n} - 7^2 \cdot 2^n + 7^2 \cdot 2^n - 2\cdot 2^n
\\&= 49 (7^{2n} - 2^{n}) + 47 \cdot 2^n
\end{align*}
If $7^{2n} - 2^{n} = 47k$ for some integer $k$, then
$$
    49 (7^{2n} - 2^{n}) + 47 \cdot 2^n = 49\cdot 47k + 47 \cdot 2^n = 47(49k + 2^n)
$$
in other words, also a multiple of $47$.

Step 3 is “Therefore, by induction, $7^{2n} - 2^n$ is divisible by $47$ for all positive integers $n$.”

If you follow these steps, and apply them to all proofs by induction, you will indeed profit.


the symbol for translation, transformation, or conversion What is the symbol for demonstrating syntactic conversion (transformation or translation)? For example, I want to show a calculation sequence, from $ \neg ( A \wedge B ) $ to $ \neg A \vee \neg B  $. Now I just use $ \vdash $: $ \neg ( A \wedge B ) \vdash \neg A \vee \neg B  $. Is there a suitable symbol to replace $ \vdash $?

Thank you.

Kejia

P :⇔ Q means P is defined to be logically equivalent to Q. This could be a suitable substitution, but I believe what you are using is best suited, since it is literally translated as "is derived from."

The symbol $\Rightarrow$ (or simply =>)
 would be my answer for a symbol for transformation, but I think it would be better if it were a single character symbol.


Number of $m$-dimensional subspaces of $V$ is same as number of $(n-m)$-dimensional subspaces. Let $V$ be an $n$-dimensional vector space over a finite field $F$. For $ 0\le m \le n$, the number of $m$-dimensional subspaces of $V$ is same as the number of $(n-m)$-dimensional subspaces.

I tried using the following argument:

Corresponding to every $m$ dimensional subspace, we can find a complementary subspace of dimension $(n-m)$.
So,
$$
\text{number of subspaces of dimension }m\le \text{number of subspaces of dimension }(n-m)
$$

Reversing the argument, we get they are equal.

Is this correct?

HINT: Denote by $\operatorname{Sub}(V)$ the set of subspaces of the vector space $V$. Use the map
$$\begin{matrix} \operatorname{Sub}(V) &\longrightarrow &\operatorname{Sub}(V^*)\\
U & \longmapsto & U^{\bot}
\end{matrix}$$
Where $U^{\bot} = \{ \phi \in V^* : \phi (U) = \{ 0 \} \}$. Since $V$ and $V^{**}$ are canonically isomorphic, you can consider an analogous map
$$\operatorname{Sub}(V^*) \longrightarrow \operatorname{Sub}(V^{**}) \ "=" \ \operatorname{Sub}(V)$$
And note that these two are inverse each other.

To conclude, you should know that $\dim U^{\bot} = n- \dim U$.

Problem: How do you choose the complementary subspace? Maybe you choose the same complement for different subspaces. Then your estimate does not work...

Example: $\mathbb F_2^2$, the subspace $\langle (1,1)\rangle$ is complementary to $\langle (0,1)\rangle$ and $\langle (1,0)\rangle$. Why should you use a different one?

Solution: Using a nondegenerate bilinear form $b$, you can find $b$-orthogonal complements. This way your choice of complement becomes unique.

In the example above with the product $(a,b)\cdot(c,d)=ac+bd$, the space $\langle (0,1)\rangle$ is orthogonal to $\langle (1,0)\rangle$ and $\langle (1,1)\rangle$ to itself. Thus there are as many 1-dimensional subspaces as (2-1)-dimensional subspaces. (ok, the statement is stupid but the correspondence works)


How to prove this is true? The question is:

Show that $$\log_2(n!)\in O(n \log_2(n)).$$

I'm guessing I'll have to use principle of simple induction for this one. But how would I go about writing the proof for this? Should I use proof by cases?

What will be my first assumption? Usually the other proofs I wrote, there was a domain assumption and I assume x in set of integers, etc. Not sure what to write for this one.

Thanks for your help!

$$0 \le \log n! = \log \prod\limits_{k=1}^n k = \sum\limits_{k=1}^n \log k\le \sum\limits_{k=1}^n \log n = n\log n$$

First you should prove that $n!=O((n \ln n)^2).$ It implies $\ln n!=O(n \ln n)$.


Find the sum of all values of $x$ which satisfy the following systems of equations: Systems of equations are as follows:
$$\left\lbrace\begin{aligned}
y&=x^2-5x+5\\
z&=x^2-12x+35\\
y^z&=1\\
x,&y,z \in\mathbb{R}\\
\end{aligned}\right.$$

I got the obvious $x=5,7,1$ and $4$ but apparently there are other answers as the sum should equal to $20$ and the answers I got sum to $17$; and if $z=0$, $y$ can be anything and if $y=1$, $z$ can be anything.

Another possibility: $y=-1$ and $z$ is an even integer yields the unique $(x,y,z) = (3,-1,8)$ and now your $x$ values do sum to $20$.

There are infinitely many values of $x$ satisfying this system. For if you write the second equation as $z=x^2-5x+5-7x+30,$ and substitute for $x^2-5x+5$ using the first equation, we get $$z=y-7x+30.$$ Finally, the last equation implies $z=0$ or $y=1.$

If $z=0,$ then $y=7x-30$ and if $y=1,$ then $z=31-7x.$ Thus we have the families of solutions $(\lambda,7\lambda-30,0)$ and $(\mu, 1, 31-7\mu).$ The last case is when $z=0$ and $y=1,$ which gives the singleton $(31/7, 1, 0).$ Thus, you literally can't add all the values of $x$ satisfying the relation as $\lambda$ and $\mu$ can be any real number.


Proving a vector space has an uncountable basis I am required to prove that a vector space $V=C[0,1]$ (the space of continuous functions over complex field) has an uncountable basis. The approach I took was to show that there is a subspace $W$ of $V$ with an uncountable basis and concluded that the vector space $V$ itself must have an uncountable basis.
Is this approach correct? I'm just unsure about the last step, if you have a subspace with an uncountable basis does it imply that the space itself must have an uncountable basis?

The answer to your general question is yes. If $W$ is a subspace of $V$ and $W$ has an uncountable basis $B\subseteq W$ then $B$ is still linearly independent as a subset of $V$. So $B$ can be extended to a basis for $V$ which means $V$ also has an uncountable basis.

Let W be the set of all polynomials.  Any finite subset must have a polynomial of largest degree so cannot be a basis.


Number of Permutations with given Cycle Types Let $n_1,\ldots ,n_k$ be the sizes of the 1st, 2nd,...,kth cycles in a permutation of $n$ elements. How is the number of permutations with this cycle structure and $n_1+\cdots + n_k=n$ equal to $${{n-1}\choose{n_1-1}}(n_1-1)!{{n-n_1-1}\choose{n_2-1}}(n_2-1)!\cdots{{n-n_1-\cdots-n_{k-2}-1}\choose{n_k-1}}(n_{k-1}-1)!$$

I understand that the chosen cycles can be permuted and that the elements of each cycle can be permuted, but I don't understand where all of the minus ones come from.

All of the $-1$'s come from the fact that the first element is chosen for you, in one way or another. For instance, when writing $\binom{n-1}{n_1-1}$, this is the number of ways to choose the elements in the first cycle. We have $-1$ everywhere because we already know that $1$ is part of that cycle. We are choosing the remaining $n_1-1$ elements fform the remaining $n-1$ in the set.

Likewise, $(n_1-1)!$ counts the number of different cycles you can make with those $n_1$ elements, because you put $1$ in front and then permute the remaining $n_1-1$ elements freely.

Next, with $n_2$, we choose the first element not chosen by $n_1$ (which can only be done one way) as well as $n_2-1$ elements freely from the remining $n-n_1-1$ elements. And so on.

Note that this way of counting sees $(123)(45)$ as having different cycle structure from $(12)(345)$, with the first having $n_1 = 3, n_2 = 2$ and the second having $n_1 = 2, n_2=3$.


System of equations with $3^x+4^x+5^x$ Solve over reals:

\begin{align}
3^x + 4^x + 5^x &= 2^{x} 3^{x-1} y \\
3^y + 4^y + 5^y &= 2^{y} 3^{y-1} z \\
3^z + 4^z + 5^z &= 2^{z} 3^{z-1} x
\end{align}

I searched for it, but can't find anything similar. I don't really know how to begin. Maybe divide by $2^{x}3^{x-1}$ and then what? Can somebody give me a starting idea? Thank you.

Define
$$
f(x) = \frac{3^x+4^x+5^x}{2^x3^{x-1}}
$$
The system is: $f(x) = y, f(y) = z, f(z)= x$.

The function $f(x)$ decreasing, so the function $f(f(f(x)))$ is decreasing.  Thus, there is at most one solution for $f(f(f(x))) = x$.  As Ilya noted, $3$ is that solution.

One of the answers is $x = 3 , y = 3 , z = 3$


Wolfram Derivation Error I'm trying to derive the equation

$$y = (2x-6)^4$$

I thought that it would be

$$\frac{dy}{dx} = 8(2x-6)^3$$

Wolframalpha says $dy/dx = 64(x-3)^3$

Who's correct? I thought it would be a simple calc-1 chain rule.

$$\dfrac{dy}{dx}=8(2x-6)^3=8(2(x-3))^3=8\cdot(2^3(x-3)^3)=64(x-3)^3$$

As demostrated above, from the result you obtained to that obtained by Wolfram Alpha. As you can see, they are both equal.

There is no error. There is a common fact of $2$ in $(2x-6)$. That means we can write
\begin{eqnarray*}
(2x-6)^4 &\equiv& \left[2(x-3)\right]^4 \\ \\
&\equiv& 2^4\cdot(x-3)^4 \\ \\
&\equiv& 16(x-3)^4
\end{eqnarray*}
If you differentiate this then you will get what $64(x-3)^3$ as Wolfram Alpha gives you.

Alternatively, if you differentiate first then you have
\begin{eqnarray*}
8(2x-6)^3 &\equiv& 8\left[2(x-3)\right]^3 \\ \\
&\equiv& 8\cdot 2^3\cdot(x-3)^3 \\ \\
&\equiv& 64(x-3)^3
\end{eqnarray*}


Are proofs by "maximality" equivalent to proofs by induction? I apologize for the lack of proper terminology; I have zero experience in this field.

What I mean by "proof by maximality": One way to show that a set $A$ has a certain property $p$ is to assume there is a largest proper subset $X \subsetneq A$ which verifies $p$ and then show that there is another subset of $A$ larger than $X$, $Y\supsetneq X$, which also verifies $p$, a contradiction. Then one concludes that $A$ verifies $p$.

Now, these kinds of proofs seem strikingly similar to proofs by induction, where one assumes that a proposition holds for $n$ (i.e. $n$ is the largest number that verifies it) and then proves that it must hold for $n+1$ as well and therefore, by the principle of induction, it is true for all the elements of $\mathbb{N}.$ The "base case" is similar as well, since in proofs by "maximality" one first has to show that there exists a subset verifying the property $p$.

I was wondering if there is a link between these two proof techniques and, if at all, how one can be turned into the other. I am having trouble trying to formalize this. Where can I learn more about it?

Well, not familiar with your argument. But the induction axiom goes in the direction of maximality.

Let ${\Bbb N}_0$ be the set of natural numbers. Let $N$ be a subset of ${\Bbb N}_0$ such that $0\in N$ and with each $n\in N$ we have $n+1\in N$. Then $N = {\Bbb N}_0$.


How to scale data for a log graph I'm writing a program to "printer plot" some data (printing "*" characters), and I'd like to scale the data on the vertical axis logarithmically (ie, semi-log chart).  I have data values ranging from zero to about 10000, and I'd like to scale the data to fit on 30-40 lines.
For a linear plot it's fairly simple -- divide 10000 by 30 (number of lines) and then, for each row being printed, calculate row# * (10000/30) and (row# + 1) * (10000/30), then print a "*" if the value in the data vector is between those two values.
How do I do the same conceptual thing with log scaling?  (I realize that on a log chart you never really have zero, but let's say I want to plot the data on a 2 decade chart spread over the 30 lines.)
I've got the following C code:
float max = 10000;
int xDim = 200;
int yDim = 30;
for (int yIndex = yDim - 1; yIndex >= 0; yIndex--) {

    --<clear "buffer">--

    float top = exp((float) yDim);
    float upper = exp((float) yIndex + 1);
    float lower = exp((float) yIndex);
    float topThresh = (max / top) * upper;
    float bottomThresh = (max / top) * lower;

    for (int xIndex = 0; xIndex < xDim; xIndex++) {
        if (data[xIndex] > bottomThresh && data[xIndex] <= topThresh) {
            buffer[xIndex] = '*';
        }
    }

    --<print "buffer">--
}

This does print something resembling a semi-log chart, but all the data (five decades, more or less) is compressed into the top 12 or so lines (out of 30), and obviously I've not factored in the number of decades I want to display (since I can't figure out how to do that).
So how do I scale to plot two decades?
Update
I've discovered that if I divide the arguments to the three exp calls with a value of about 6.5 (or multiply by about 0.15), I get close to two decades on the graph.  But I have no idea how to compute this value.  It kind of made sense to do log(pow(10,2)) (ie, natural log of 10 raised to the number of decades desired), but that yields a value near 4 (and it's moving in the wrong direction).  How should I compute the value?
Update #2
If I multiply the arguments to the exp calls times the value (log(10) * decades)/yDim I get the right result.

If you take the log of all your values, you can then treat it just like a linear scale.  Using base 10 logs, if your data ranges from 10 to 100,000, taking logs makes it range from 1 to 5.  Then scale that as you have before, presumably giving 10 lines to each unit to make 40 lines in your plot.  Then plot you * where the log of the value belongs, so if you have a value of 2,000, the log of that is 3.3 and you would plot it on the 24th line.


Markov chains transition identity proof Let $X = (X_{n})_{n \geq 0}$ a Markov chain proof that

$$\mathbb{P}(X_{n + 2} = j \mid X_{n} = i) = \sum_{l \in \mathcal{S}} \mathbb{P}(X_{n + 2} = j, X_{n + 1} = l \mid X_{n} = i)$$

Note: $\mathcal{S}$ denotes the state space.

Simple Proof without using the Markov Property:

Using the fact that $\Omega = \bigcup_{l \in \mathcal{S}} \{ X_{n + 1} = l\}$, this is a disjoint union, then

$$    \begin{align}
      \mathbb{P}(X_{n + 2} = j \mid X_{n} = i) &= \mathbb{P}(\{ X_{n + 2} = j \} \cap \Omega \mid X_{n} = i)  \\
      &= \mathbb{P}\left(\{ X_{n + 2} = j\} \cap \left[ \bigcup_{l \in \mathcal{S}} \{ X_{n + 1} = l\} \right] \mid X_{n} = i \right) \\
      &= \mathbb{P}\left(\bigcup_{l \in \mathcal{S}} \left[ \{ X_{n + 1} = l\} \cap \{ X_{n + 2} = j\}  \right] \mid X_{n} = i \right) \\
      &= \sum_{l \in \mathcal{S}} \mathbb{P}\left( \{ X_{n + 1} = l\} \cap \{ X_{n + 2} = j\}  \mid X_{n} = i \right) \\
      &= \sum_{l \in \mathcal{S}} \mathbb{P}\left(X_{n + 1} = l,X_{n + 2} = j \mid X_{n} = i \right)
    \end{align}
$$


: Can the canonical theorem on Riemann integrability be taken as the definition of continuity? Can the canonical theorem on Riemann integrability (namely, that a bounded real-valued function defined on a closed interval is Riemann-integrable if, and only if, its discontinuities constitute a set of measure zero) be taken as the definition of continuity? If so, it would then be a theorem that a function f is continuous at a limit point a in its domain if, and only if, the limit of f(x) as x goes to a is f(a).

edit (2.Sep.2013, CST, MERCA)

I am not interested in defining continuity this way. I'm only interested in whether it is POSSIBLE to define continuity this way. That is, just as there is a (long) list of statements equivalent to the Axiom of Choice, so also was I wondering whether this is on the list of statements equivalent to the definition of continuity, that's all. I (weakly) conjecture that it is indeed on the list, and so I'm holding out for an answer that proves this. An answer that simply says that the answerer does not see how it can be done is not sufficient for me to give up my conjecture, but a counterexample certainly would.

The (Riemann)-Lebesgue Criterion gives a criterion for a bounded function to be almost everywhere continuous.  So you could take that as the definition of almost everywhere continuity for bounded functions if you want, although I don't see any particular advantage in doing so.  (Why do you want to define continuity in terms of Riemann integrability anyway?)

If $f$ is continuous at $c$ and we modify the value of $f$ at $c$ (while keeping every other value unchanged) then the modified function cannot be continuous at $c$.  The same holds if we modify the value of $f$ at any finite set of points on its domain.

On the other hand, if $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable, then if we change the values of $f$ at any finite set we do not disturb the Riemann integrability or the value of the integral.  This is a nice elementary exercise involving the definition of the Riemann integral: the idea is to take an arbitrarily small subinterval around each modified point.

These diametrically opposite properties make it hard to see how one could possibly define continuity in terms of Riemann integrability.

You have given a definition of Riemann integrability for bounded functions, which states that the function f can be discontinuous on a set of measure 0.  So if it is discontinuous, it is discontinuous, even though it is perfectly integrable.  Your definition of continuity is correct, but there is no guarantee that f will meet that definition -- it might, or it might not.

However $\int f$ (i.e. any function F whose derivative is f where f is continuous) can be defined to be continuous although it will not be everywhere differentiable.

In general integration makes functions smoother, so that a slightly discontinuous function has a continuous integral F (in the sense of described above); a continuous function will yield an F that is differentiable and so on.


What exactly is a local expansion? From time to time, I come across authors writing about performing a local expansion of a complex function $f$ at some point. This usually means performing a singular expansion (such as Laurent or Puiseux) or something more subtler – e.g., performing a singular expansion for one argument of a multivariate analytic function.

I have wondered if this term indeed means the same as "singular expansion" or possibly arbitrary "series expansion". I have not been able to find any text, in which the notion of local expansion is actually defined: it just usually strikes as bolt out of the blue. And in fact, I just cannot see any motivation for such term, as (at least in my understanding) almost all series expansions, possibly except some with infinite radius of convergence, are inherently valid only locally.

For these reasons, I would like to ask what exactly is meant by a local expansion and in case it is a synonym of some other term, what is the usual motivation of using this term instead of the other one. Thanks a lot.

Defining things locally is an extremely useful and ubiquitous modus operandi in modern mathematics. A function having what you call a 'local expansion' is one example. This appears in the definition of an analytic function:

  an analytic function is a function that is locally given by a convergent power series
  (Wikipedia)

This means that although the function itself may not be given by a convergent power series, we can always find a small set around every point in its domain such that it is equal to a convergent power series when restricted to this set.

I think you might call this a 'local expansion'.


Combinations (Hand Shakes) I have a question as such:

You and a group of 9 Friends are playing a game of basketball. At the end of the game, if each player shakes hands with every other player, how many handshakes will there be?

At first I interpreted this question logically,
if you have 9 friends, you yourself will shake hands with 9 other people (1 less because you don't include yourself).

The other 9 friends will shake hands with 9 other people.
Thus being 9*9 = 81 handshakes.

But the answer is 45

( i.e $10\choose2$)

which logically to me, makes absolutely no sense(if every person if shaking everyone else's hand)
Any clarification would be much appreciated.

There is one handshake for each pair of players, and there are $\binom{10}2$ pairs of players. Your calculation of of $9\cdot9$ is off in two respects. First, there are $10$ players, not $9$, and each shakes hands $9$ times, so your reasoning should result in $10\cdot9=90$ handshakes, not $9^2=81$. But you’ve counted each handshake twice, once for each of the two players involved in it, so you have to divide that $90$ by $2$. And when you do that, you do indeed get $45=\binom{10}2$.

Answer is 36. Because on each hand shakes we tempt to include ourselves, but we should not be doing that. Instead we should calculate one less hand shakes. Therefore re by the equation n*(n-1)/2 we could find the solution. That is n=9 and therefore the answer is 36.


What's wrong with this logarithm calculation? We know that $\displaystyle\log_a(xy)=\log_ax+\log_ay$.

Consider the following:

  $$\displaystyle\ln(1)=\displaystyle\ln((-1)\times(-1))=\displaystyle\ln(-1)+\displaystyle\ln(-1)
$$

$\displaystyle\ln(1) $ is a completely valid statement, but I'm not sure if $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ is.

$\displaystyle\ln(-1) $ doesn't exist, but $\displaystyle\ln((-1)\times(-1)) $ does, and if I plug $\displaystyle\ln((-1)\times(-1)) $ into my calculator it gives a 0 as the answer (which is correct), but if I plug in $\displaystyle\ln(-1)+\displaystyle\ln(-1) $ it says that there is a domain error (which there is). So my question is, what is wrong with the highlighted equation?

$$\displaystyle\log_a(xy)=\log_ax+\log_ay$$

The above identity holds only for $x,y>0$ and $a\not =1$, $a>0$.

Basically, once you go to taking the log of negative numbers, and therefore getting complex results, you are into the realm of multi-valued functions. Cf. $arcsin(\theta)$ etc.
Euler's identity says $exp(i\pi) = -1$ But also, for example, $exp(-i\pi) = -1$. So by taking logs of both sides, we get $log(-1) = (2n - 1)i\pi$ for an arbitrary integer value of n, and it is effectively by convention that we take $n = 0$ to get the principal value of the log function..

Then the effect of writing $log(1) = log(-1) + log(-1)$ simply amounts to:
$$0 = (2m - 1)i\pi + (2n - 1)i\pi$$which reduces to choosing m, n such that $m + n = 1$.

This is the same order of paradox as saying $\sqrt4 = 2$ and $\sqrt4 = -2$. By convention we take the principal (positive) value throughout; but there are circumstances where we have to break the convention to get a sensible result.


How do you solve the area of a trapezoid using diagonals The height of a trapezoid is $10$ cm. The lengths of the two diagonals of the trapezoid are $30$ cm and $50$ cm. Calculate the area of the trapezoid.

On the homework I solved this using $${D_1D_2\over 2}$$ and my teacher marked me wrong. So I don't know what I did wrong. Please help. I know I can only use the formula if the diagonals are $90$ degrees. But how do I check that ?

the formula $\frac{d_1\cdot d_2}2$ is the area of the special quadrilateral called rhombus but not the trapezium..

If you want to find the area of the trapezium the like the above mentioned using the Pythogoras theorem. Find the sides and form an equation and with that solve the linear equation then you can solve the area of the trapezium..
finally you will get

area$=\frac{h(a+b)}2$


Invariance Principle Question A circle is divided into six sectors. Then the numbers $1, 0, 1, 0, 0, 0$ are written into the sectors (counter-clockwise say). You may increase two neighboring numbers by $1$. Is it possible to equalize all numbers by a sequence of such steps?

So, $a_1 = 1, a_2 = 0, a_3 = 1, a_{4, 5, 6} = 0$.

The invariant they find is:

$$S = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 = 2$$

If you increase $a_k$ by $1$ then you have:

$a_{2, 4, 5, 6} = 1, a_{1, 3} = 2$

$$S = 2$$ Still.

Then he says, the goal, $S = 0$ cannot be reached.

How does that solve the issue?

$$S=0 \implies a_1 + a_3 + a_5 = a_2 + a_4 + a_6$$

How does that do anything?

Your target is to have all six numbers the same, i.e. $$a_1 = a_3 = a_5 = a_2 = a_4 = a_6$$ which would then imply $$a_1 + a_3 + a_5 = a_2 + a_4 + a_6$$ and so $$S=0$$ which we know is impossible from the starting point as $S$ is a constant $2$ under the permitted change.

The argument in words is that if the sum of the values in the even positions and the sum of the values in the odd positions start as being different, and all you can do is add the same numbers to odd and to even positions, then the sums will remain different, and therefore the individual values cannot all be made the same.

I fail to see how $a_1−a_2+a_3−a_4+a_5−a_6$ is an invariant. Additionally I think I found a way to equalize the sectors. It is as follows:

$(1, 0, 1, 0, 0, 0) \rightarrow (1, 1, 1, 0, 0, 1)  \rightarrow (2, 1, 1, 0, 1, 1) \rightarrow (2, 2, 1, 1, 1, 1)\rightarrow (2, 2, 1, 2, 1, 2) \rightarrow (2, 2, 2, 2, 2, 2)$

Where the tuples of numbers are the values in the circular sectors written counter clockwise. It's possible I'm miss understanding the problem though.


Calculating the infinite series $1-\frac13+\frac15-\frac17+\frac19-\frac1{11}\cdots$ My main question is the one in the title, however I was also wondering, in general when it comes to infinite series, how can you find out whether the series converges to a value or not? And can you tell if there will be something strange about it? What I mean is that you would expect the sum of all natural numbers to be infinity, but it is -1/12. Is there a way of knowing if something like this will happen?

Sorry for the ton of questions! And thank you for any answers :)

Recall that
$$
\frac{1}{1+x^2}=1-x^2+x^4+\cdots
$$
If you integrate from $0$ to $1$
$$
\begin{align}\int_0^1\frac{1}{1+x^2} dx& =\int_0^11-x^2+x^4-x^6+\cdots\,dx\\
& =1-\frac13+\frac15-\frac17+\cdots
\end{align}
$$ but $$\int_0^1\frac{1}{1+x^2} dx=\arctan 1-\arctan 0 = \frac\pi4, $$
then
$$
1-\frac13+\frac15-\frac17+\frac19-\cdots=\frac\pi4
$$

$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+ \dots =\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$$

Have you got taught the Dirichlet's criterion?

Use this,and you prove that your infinite sum converges.


Large deviation for gaussian distribution Given a random variable $X$ that is $N(0,1)$ distributed and a sequence $(X_i)$ of iid distributed $N(0,1)$ random variables(copies of $X$) and I am supposed to calculate $P(X \ge 5)$ by means of large deviations.

Hence, I calculated the rate function $\gamma^*(l) = \frac{l^2}{2}$. And now I am stuck. Is the random variable $Z:=\frac{e^{-\gamma^*(5)} }{(2\pi)^{\frac{n}{2}}}e^{- \frac{X_1^2+...+X_n^2}{n}}$ now my estimator for this event, I am a little bit confused. Basically, I want to apply this theory here : wikipedia reference

If anything is unclear, please let me know.

If $X$ is standard normal, then, when $x\to+\infty$,
$$
P(X\geqslant x)\sim\frac{\mathrm e^{-x^2/2}}{x\sqrt{2\pi}},
$$
in the sense that the ratio of the LHS and the RHS converges to $1$. For $x=5$ this suggests that $P(X\geqslant5)$ might be close to
$$
\frac{\mathrm e^{-12.5}}{5\sqrt{2\pi}}\approx2.97\cdot10^{-7},
$$
while the exact value is
$$
P(X\geqslant5)\approx2.87\cdot10^{-7}.
$$
Not an ounce of large deviations here. Large deviations in this context would yield the cruder estimate
$$
P(X\geqslant x)=\exp\left(-\frac12x^2+o(x^2)\right),
$$
or, equivalently,
$$
\lim_{x\to+\infty}\frac{\log P(X\geqslant x)}{x^2}=-\frac12.
$$

I think what you're after is

$$\Pr(\bar X > 5),$$

correct?

Note that for standard normal distributions the cumulant generating function is

$$\lambda(\theta) = \log E\bigl(\exp(\theta X_1)\bigr)
 = \log \Bigl[ \int_{-\infty}^\infty \exp(\theta x) \exp(-x^2/2) dx \big/ \sqrt{2\pi}\Bigr] =
 \theta^2/2.$$

So the Legendre-Fenchel transformation value is $I(5)=12.5$.

Hence

$$\Pr( \bar X >5) \approx \exp(-12.5n).$$

If that's not what you're after then please clarify.


Compound Interest Formula I was trying to explain $e$ to someone and I ran across this idea. So, $e$ is typically defined as $$e =\lim_{n\to \infty} (1+1/n)^n $$
Which comes from the formula for compound interest
$$P(t) = P_0\left(1+\frac{r}{n}\right)^{nt}.$$
I could not explain why it is that we divide by $n$. The typical answer is that you simply "divide" the interest rate across each compounding period, but of course means that the borrower will end up paying more. This just seems like trickery on the part of the lender. Does anybody have any idea why this formula in particular? Was it simply to trick lenders?

$e$ also then becomes kinda flukey. How would you justify that
$$ \lim_{n\to \infty} \left( 1+\frac{x}{n}\right)^n$$
is an exponential function to some base let alone to the base $e$?

As
$$
e = \lim_{n \to \infty} \left( 1 + \frac1n \right)^n,
$$
and as for any real number $r > 0$, we have
$$
\lim_{n \to \infty} \frac{n}{r} = \infty = \lim_{n \to \infty} n \  \mbox{ and } \ \lim_{n \to \infty} \frac{r}{n} = 0 = \lim_{n \to \infty} \frac1n,
$$
so we obtain
$$
\lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^n = \lim_{n \to \infty} \left( \left( 1 + \frac{r}{n} \right)^{\frac{n}{r}}  \right)^r = e^r.
$$

Hence we also conclude that for any real numbers $r > 0$ and $t > 0$, we have
$$
\lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^{nt} = \lim_{n \to \infty} \left( \left( 1 + \frac{r}{n} \right)^{\frac{n}{r}}  \right)^{rt} = e^{rt}.
$$

Thus if $P_0$ is the initial amount, $r > 0$ is the interest rate compounded $n$ times a given period of time, then after $t$ such periods of $t$, our amount will be
$$
P_0 \times \left( 1 + \frac{r}{n} \right)^{nt},
$$
and so if our interest at rate $r$ is compounded continuously, then our amount after $t$ number of periods will be
$$
P_0 e^{rt}.
$$

Hope this helps.

There is a formula for case of f(x)^g(x)
When f(x) tends to 1 then
We can write e^((f(x)-1)*g(x))
It is applicable everywhere ..
It is a odd formula but very helpful


Does $\mu(x+B)= \mu(B)$ for all balls $B$ imply that $\mu$ equals the Lebesgue measure (up to scaling)? Suppose that $\mu$ is a measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ such that $\mu(K)<\infty$ for any compact set $K$ and
$$\mu(x+B) = \mu(B) \tag{1}$$
for all $x \in \mathbb{R}^d$ and open balls $B$. Does this imply that $\mu$ equals up to a constant the Lebesgue measure $\lambda^d$, i.e. does it hold that $\mu = c \lambda^d$ for some constant $c \in [0,\infty]$?

My attempts:

It is well-known that if $(1)$ holds for all ($d$-dimensional) rectangles, then the claim holds true. So if we knew that any rectangle can be covered by disjoint balls, then we would be done. However, as far as I know, this is actually not possible. (See this question)
Define $$\mathcal{D} := \{A \in \mathcal{B}(\mathbb{R}^d); \forall x: \mu(x+A) = \mu(A)\}$$ and show that $\mathcal{D}$ is a Dynkin system. If we knew this, then the fact that the balls are contained in $\mathcal{D}$ would imply $\mathcal{D} = \mathcal{B}(\mathbb{R}^d)$. My problem: Since $\mu$ is (in general) not a finite measure, I don't see how to prove the implication $A \in \mathcal{D} \implies A^c \in \mathcal{D}$. I also considered defining $$\mathcal{D}_R := \{A \in \mathcal{B}(B(0,R)); \forall x: \mu(x+A) = \mu(A)\};$$ then it is easy to show that $\mathcal{D}_R$ is a Dynkin system, but, unfortunately, we cannot conclude that the balls are contained in $\mathcal{D}_R$ (since the intersection $B \cap B(0,R)$ is in general not a ball).

Any ideas, counterexamples,...?

Edit: @NateEldredge suggested as a counterexample the measure $$\mu(A) := \sum_{q \in \mathbb{Q}} \delta_q(A)$$ in case that $\mu$ does not need to be finite on compact sets.

Yes, this is true. The following is based on the proof of uniqueness of Haar measure in Hewitt and Ross (if I remember correctly):

The general idea is to write an arbitrary $C_c$ function as an infinite linear combination of characteristic functions of balls. Rigorously, this amounts to approximation by convolution.

I denote the usual Lebeague measure by $\lambda$, and integration against this measure is written $dx$.

For $\epsilon >0$, let $f_\epsilon = \chi_{B_\epsilon}/\lambda(B_\epsilon)$, where $\chi_{\dots}$ denotes the indicator function of the set in the index and $B_\epsilon$ is the ball with radius $\epsilon$ around the origin.

Let $h\in C_c$ be arbitrary. It is a classical fact that the convolution $f_\epsilon \ast h$ converges locally uniformly to $h$, has support in a fixed compact set (for $0<\epsilon<1$) and is uniformly bounded. Using dominated convergence, we get $f_\epsilon \ast h \to h$ in $L^1(\nu)$ or every locally finite measure $\nu$.

Hence,

$$
\begin{eqnarray*}
\int h d\mu &=&\lim_\epsilon \int f_\epsilon \ast h d\mu\\
&=& \lim_\epsilon \int\int f_\epsilon (y) h(x-y)\, dy \, d\mu (x)\\
&\overset{z=x-y}{=} &\lim_\epsilon \int \int f_\epsilon (x-z) h(z)\,dz \, d\mu (x)\\
&\overset{Fubini}{=}& \int h(z)\int f_\epsilon (x-z)\,d\mu (x)\, dz\\
&=& \lim_\epsilon \int h(z)\, dz \cdot \frac{\mu (B_\epsilon )}{\lambda (B_\epsilon )}.
\end{eqnarray*}
$$

Fix some $h_0\in C_c$ with $h\geq 0$ and $h\not\equiv 0$.  Then the above calculation shows

$$
c:=\frac{\int h_0 \,d\mu}{\int h_0dx}=\lim_\epsilon \frac{\mu (B_\epsilon)}{\lambda(B_\epsilon)}
$$
(in particular, the limit exists and we take the limit $\epsilon \downarrow 0$), as well as

$$
\int h\,d\mu =c \int h \,dx
$$
for all $h \in C_c$. This easily implies the claim.

[ edit- this is wrong. ]

$\newcommand{\ball}{\mathbf{B}_N} \DeclareMathOperator{\Leb}{Leb}$Actually, my comment is not 100% correct. We need to also assume that  we assign (say) $c=\mu([0,1]^d)<∞$.  From this we easily get that
$$ \mu (A) = c\Leb(A)$$
for any rectangle $A=∏_i[a_i,b_i)$ such that $b_i,a_i ∈ \mathbb{Q}$.
 Let
\begin{align} P :=& \{\text{all such rational rectangles $A$}\} \\
L_N:=&\{A∈\mathcal{B}(ℝ^d) : \mu (A∩ \ball) = c\Leb(A∩ \ball) \}
\end{align}
where $\ball$ is the ball around $\mathbf{0}$ of radius $N<∞$. (The proof might be easier if you use a cube around 0 of radius $N$)
We can check that  $P$ is a π-system, $L_N$ is a $\lambda$-system and $P ⊂ L_N$. Thus by Dynkin's $\pi-\lambda$ Theorem,
$$ \mathcal{B}(ℝ^d) = \sigma( P ) ⊂ L_N $$

So we have $\mu (A∩ \ball) = c\Leb(A∩ \ball)$ for any $N$; sending $N→∞$ gives the result.

I think this is reasonably standard, I adapted it from my unsourced written notes for the 1D case.

(response to comments)In the 1D case: we can get $$\mu[0,k) = \mu [0,1) + … + \mu[k-1,k)= c \text{Leb}[0,k)$$ by translation invariance and $\sigma$-additivity. Then $$\mu [0,k) = q\mu[0,k/q) = c \text{Leb}[0,k)$$ for any $q\in\mathbb{N}$, again by TI and $\sigma$-additivity. Now translate by a rational coordinate; $k,q$ was arbitrary, so this proves that $P⊂ L_N$ in the 1D case.

The same proof works in $d$ dimensions by first 'fixing' the last $d-1$ coordinates. the same construction yields
$$\mu[a_1,b_1)\times[0,1)^{d-1} = c \text{Leb}[a_1,b_1)\times[0,1)^{d-1}$$
for an arbitrary rational $a_1,b_1$. Repeating once for each dimension yields the result.


Explaining the trigonometric addition formulas via composition of rotations I am generally dissatisfied with the way trigonometric addition formulas like
$$
\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)
$$
are derived in high school textbooks. There are numerous proofs, some of which are short but unintuitive, some of which introduce unnecessary calculations, many with restrictions on $\alpha$, $\beta$ and $\alpha + \beta$, like having to lie between $0$ and $\frac{\pi}{2}$ radians.

To me, the addition formulas are simply a coordinatization of the observation that composing a rotation by $\alpha$ with a rotation by $\beta$ yields a rotation by $\alpha + \beta$. Hence in my opinion the proper way to prove such a formula would be to observe that
$$
\begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix} = \begin{bmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \end{bmatrix} \cdot \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix}
$$ then work out the product on the right to obtain the formulas by equating entries. However, such a proof is beyond the scope of a high-school textbook, as high schoolers - if they know about matrices at all - are rarely taught the link between composition of linear transformations and matrix multiplication.

Is there a way to salvage the essence of this proof - that the formulas are merely a way of expressing that a rotation by $\alpha+ \beta$ can be obtained by composing rotations by $\alpha$ and $\beta$ - without using linear algebra? Of course complex numbers are also out of the question.

Can we use vectors? I assume so, since rotation is a transformation in vectors.

The $\alpha$ rotation sends $a\vec{I}+b\vec{j}$ to $A\vec{I}+B\vec{j}$ with $A:=a\cos\alpha-b\sin\alpha,\,B:=a\sin\alpha+b\cos\alpha$. If you compute where three second rotation sends the latter vector, you can complete the proof. It's the same logic as in the matrix case, but you don't have to explain the theory thereof.

You might take refuge to complex numbers instead, by means of the Euler relation $\exp(i\phi)=\cos(\phi)+i\sin(\phi)$ and the fundamental property of the $\exp$ function:

$\cos(\alpha+\beta)+i\sin(\alpha+\beta)=\exp(i(\alpha+\beta))=\exp(i\alpha)\cdot\exp(i\beta)=$
$=(\cos(\alpha)+i\sin(\alpha))\cdot(\cos(\beta)+i\sin(\beta))=$
$=(\cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta))+i(\cos(\alpha)\cdot\sin(\beta)+\sin(\alpha)\cdot\cos(\beta))$

And finally use therefrom the real resp. imaginary part separately.
This is how you'd get the trigonometric addition theorems.

--- rk


Calculate supreme of $((T+T^*)f,f)$ Let $0<\theta <\pi $, and continuous linear operator $T:L^2(0,1)\to L^2(0,1)$  $Tf(x):=\int _0^x e^{i\theta} f(t)dt$

Then, what is $\sup_{\{||f||\leq 1\}}\int_0 ^1 (T+T^*)f(t) \bar{f(t)}dt$ ?

My idea :$T+T^*$ is symmetry, so $\int (T+T^*)f(t)\bar{f(t)}dt$ is real value.
I calculated $\int_0 ^1 (T+T^*)f(t) \bar{f(t)}dt=\int \int \cos{\theta}f(t)\bar{f(x)}+i\sin{\theta} \bar{f(x)}(\chi _{[0,x]}(t)-\chi_{[0,t]}(x))dtdx$ by Fubini's theorem, but I don't know how to find sup.
I guess supremum is $\sqrt{2}\cos{\theta}$

The adjoint of $Tf = e^{i\theta}\int_{0}^{x}f(t)dt$ is $T^*f = e^{-i\theta}\int_{x}^{1}f(t)dt$. Both $T$ and $T^*$ are compact operators and, therefore $T+T^*$ is selfadjoint and compact. Therefore,
$$
       \sup_{\|f\|=1}\langle (T+T^*)f,f\rangle
$$
is the largest eigenvalue of $T+T^*$. The eigenvalue problem for $T+T^*$ is

$$            e^{i\theta}\int_{0}^{x}f(t)dt+e^{-i\theta}\int_{x}^{1}f(t)dt=\lambda f
$$
Any such $f$ must satisfy $\lambda f(0)=e^{-i\theta}\int_{0}^{1}f(t)dt$ and $\lambda f(1)=e^{i\theta}\int_{0}^{1}f(t)dt$, or

$$
           e^{i\theta}f(0)-e^{-i\theta}f(1)=0.
$$
$$
          f(1)=e^{2i\theta}f(0).
$$
And, $f$ must satisfy the ODE
$$
        e^{i\theta}f(x)-e^{-i\theta}f(x)=\lambda f'(x)
$$

$$
          2i\sin\theta f(x)=\lambda f'(x)
$$

$$
             \frac{f'(x)}{f(x)}=\frac{2i\sin\theta}{\lambda}
$$

$$
        f(x) = Ce^{2i\sin(\theta) x/\lambda}
$$
The endpoint condition is satisfied for $C\ne 0$ iff

$$
       e^{2i\sin(\theta)/\lambda}=e^{2i\theta}
$$
$$
            \sin(\theta)/\lambda=\theta+\pi n
$$
$$
            \lambda = \frac{\sin(\theta)}{\theta+\pi n}.
$$
So the supremum is the largest value of $\lambda$ given above for $n=0,\pm 1,\pm 2,\cdots$.

1) The set of which you are computing supremum of, is a subset of the real numbers.

2) Therefore, the supremum is the maximum of the closure.

3) Every point in the spectrum of $T+T^*$ is in that subset.

4) Since $T+T^*$ is normal, then its spectral radius is $\|T+T^*\|$.

5) $((T+T^*)f,f)\leq \|T+T^*\|$ for $\|f\|=1$.

It follows that the supremum that you need is $\|T+T^*\|$.


How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work because the derivative of $2x^2-4x+3$ is not a multiple of $(x-1)^{-2}$ and the derivative of $x-1$ is not a multiple of $2x^2-4x+3$.

Note that

$$2x  \color{red}{-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}} \color{green}{-2+K} = 2x \color{red}{-\frac{1}{x-1}} \color{green}{+C}.$$

Well both the answers are correct. The thing is what is asked in the question. The answers are exactly same.


Another question about $x_0$ in the Taylor series When we talk about Taylor series, we say it's around point $x_0$. It's in the Taylor series formula:

$$f(x) = f(x_0) + (x-x_0)f'(x_0) + \frac{(x-x_0)^2f''(x_0)}{2} + \frac{(x-x_0)^3f'''(x_0)}{6} +  + ...$$

What I don't understand what is this point $x_0$?

We're given Taylor formulas for $\sin(x)$, $\cos(x)$, etc, and they don't have this point $x_0$ in the formula? It's somehow chosen to be $0$ so it disappears, but what does that mean?

Doesn't that mean the Taylor series works only around $0$? But then we somehow can use the formula for any $x$, even $100$ or $100,000$, and it still gives the right result?

What happens if we choose $x_0$ to be $1$? or $25$? That would change the Taylor formula for the primitive functions but would it still work for any $x$?

I'm very confused.

In Taylor's formula, we have to choose an initial point $x_0$, in which the sum actually converges to the function. In any other point the sum may converge or diverge, depending on the function.

For example, by evaluating Taylor's formula for $\sin(x)$, $\cos(x)$ and also $e^x$, it doesn't matter which point $x_0$ you choose, and the sum will converge everywhere to the function (if I am not mistaken, there is a theorem that if the sum converges at $x$, it converges to the value of the function at $x$).

However, in other functions (like $\ln(1+x)$), the sum may not converge always; Taylor's formula for $\ln(1+x)$ converges $\Leftrightarrow$ $|x|<1$ (and also $x=1$, if I remember correctly).

To sum up, you may choose any point $x_0$ you want; but it is usually convinient to choose it as $0$ (because it is easier to calculate the derivatives).

You can develop a Taylor series of a function at any point of the domain/interval where it is nice enough (continuously n-differentiable...). The formulae well known for sine cosine exp are at $x_0=0$ but you could using the general formula get a Taylor expansion of sine at $x_0=\pi$ for example or any other real by the way.


Real analysis- open and closed set I just signed up on this website because I can't find the way to solve these questions anywhere.

Let $( ℝ , d)$ be a metric space , where $d(x,y)=|x-y|$ , $∀ x,y ∈ ℝ$.
Show that the singleton $\{x\}$ is closed set.

A metric space is often imagined as a box with a lot of points in it.

Where each pair of points has a distance assigned to it $d(x,y)$. The distance is assumed to be positive and only zero when $x=y$(aka if you have different points their distance is positive).

Along with a notion of distance comes a notion of neighboorhoods or open balls. An open ball $B(x,r)$ is simply the set of points $z$ in the space that have distance $d(x,z)$ less than $r$.

An open set is a set $V$ with which for every point $x$ in $V$ you can find an open ball of $x$ contained inside of $V$.

Closed sets $C$ however are defined to be the sets that contain all of their limit points. In the picture the black area around C is also meant to be apart of it. Limit points of $C$ are points that always have an open ball "touching" $C$. So if a point isn't a limit point it has an open ball that doesn't touch $C$.

Also note that since every limit point is in $C$, any point not in $\color{red}{C}$ () or is in $\color{blue}{R/C}$ has an $\color{purple}{\textrm{open ball}}$ not "touching" $\color{red}{C}$

With all that said. Assume we don't know that $\{x\}$ is closed but want to show that it is closed (I have drawn the set $\{x\}$ as a white circle containing $x$). We need to show that every point $y$ $\color{blue}{\textrm{outside}}$ of $\{x\}$ has an $\color{purple}{\textrm{open ball}}$ that doesn't "touch" $\{x\}$. Since for any singular point $y$ outside $\{x\}$ we know the distance between $x$ and $y$ we simple make an $\color{purple}{\textrm{open ball}}$ around $y$ with radius less than half the distance between $x$ and $y$.

Is it correct ? @Student If not what to change?


Knights and knaves again on an island In an island there are 3 inhabitants, one of which is a knight (who always tells the truth) and the other two are Jokers who randomly decide whether to tell truth or lie. The 3 men have the numbers 1, 2 and 3 on their t-shirt. You need to find at least one person who can give you some information, but the problem is that you do not know who is who. You can ask use one question to one of them, which must have a number as an answer, in order to determine the knight.
  What question will you ask?

I have never dealt with any such type of problem.
Obviously the question must be something related to the numbers they have on their t-shirt.

Any help will be appreciated :)

(this is not homework - I wish I were 17, even if I had to bear with tons of homework!! Unfortunately I am 64!)

How about this question:

  If you have decided to tell the truth for this question, please answer with the number of the knight, and if you have decided to lie, please answer with the number of one of the jokers.

Truth tellers will respond to the first clause truthfully, giving you the number of the knight, and liars will respond to the second clause falsely, giving you the number of a someone who is not a joker.

As far as I can tell, this is a "legal" question. The question certainly needs to refer to the joker's decision to be truthful; otherwise, the Joker's two possible states would give different answers, meaning you cannot determine the truthfulness of their answer.

Assuming even Jokers answer in the way we can consistently assign truth values to their answers, we can ask, for example,

Let $u$ be the number of Knight. Let $v$ be the number you will answer to my question with. Let $x$ be defined as
$$x = \begin{cases}
u, & \text{if your answer to this question will be true}\\
v, & \text{if your answer to this question will be false}
\end{cases}$$
 What is $x$?

Assume person we ask will decide to provide true answer (they are Knight or Joker who answers true). Then $x$ is number of knight. As answer is true, the answer will be $x$ - ie number of knight.

Assume person we ask will decide to provide false answer, and their answer will be $v$. Then $x = v$, and their answer will be true, not false! So they can't consistently provide false answer.

So both Knight and Joker will answer to this question with number of Knight.


Where does the following formula come from? Where does the following formula come from?

For a Laurent polynomial $f(z)=\sum a_j z^j$ and a positive integer $n$ we have $$\sum_{k\equiv \alpha\pmod n} a_k=\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega).$$

I hope someone can answer this question or give some refferences about it ! Thanks a lot!

first we shall see that:

$\frac1n\sum_{\omega:\omega^n=1} \omega^{-\alpha}f(\omega) = \frac1n\sum_{\omega:\omega^n=1}\omega^{-\alpha }\sum_{j}a_j\omega^j = \frac1n\sum_{\omega:\omega^n=1}\sum_{j}\omega^{j-\alpha }a_j =$

$ = \frac1n\sum_{j}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j$

I'm leaving for you to think why we can preform the last move (change the order).
also put notice that $\omega^{j-\alpha} = 1 \iff j \equiv \alpha \ (mod \ n)$ and also because there are n roots of unity in $\mathbb{C}$ we get:

$ \frac1n\sum_{j}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j  = \frac1n\sum_{j \equiv \alpha (mod \ n)}\sum_{\omega:\omega^n=1}a_j + \frac1n\sum_{j \neq \alpha (mod \ n)}\sum_{\omega:\omega^n=1}\omega^{j-\alpha }a_j = \sum_{j \equiv \alpha (mod \ n)} \frac{n}{n} a_j + \frac1n\sum_{j \neq \alpha (mod \ n)}a_j\sum_{\omega:\omega^n=1}\omega^{j-\alpha }$

but we know that $\forall \ 0<m< \ n: \sum_{\omega:\omega^n=1}\omega^{m } = 0 $ and therefor we get: $\frac1n\sum_{j \neq \alpha (mod \ n)}a_j\sum_{\omega:\omega^n=1}\omega^{j-\alpha } = 0 $ (as $0<j-\alpha < n$) and were done (-:

This formula comes from Thomas Simpson's series Multiseciton theory. Speak in the concrete，
a multisection of the series of an  analytic function

$$f(z) = \sum_{n=0}^\infty a_n\cdot z^n.$$

has a  closed-form expression  in terms of the function   $f(x)$:

$$\sum_{m=0}^\infty a_{qm+p}\cdot z^{qm+p} = \frac{1}{q}\cdot \sum_{k=0}^{q-1} \omega^{-kp}\cdot f(\omega^k\cdot z),$$

where $\omega = e^{\frac{2\pi i}{q}}$ is a  Primitive nth root of unity primitive ''q''-th root of unity .  This solution was first discovered by  Thomas Simpson.


Are all vectors matrices? It's pretty clear to me that all vectors are matrices (either 1 x n or n x 1).  But there is some discussion about this at work.  We're wondering whether the expression "Vectors/Matrices" in a software GUI can be shortened to "Matrices".  (The shorter the expression, the better, because of space restrictions in the GUI.)

So, is every vector a matrix or is there some subtlety in the definitions that I don't know about?

Every finite dimensional vector space of dimension $n$ is isomorphic to a vector space in which the vectors are $n$-tuples (whether written as "row vectors", $1\times n$ 'matrices', or "column vectors", $n\times 1$ matrices). That is, under a suitable interpretation, you can think of them as vectors.

However,

It is often useful to think of the vectors differently for specific vector spaces; and
For infinite dimensional vector spaces, you cannot do it unless you extend your definition of "matrix" to include matrices that have an infinite number or rows or an infinite number of columns.

For example: consider the vector space of all polynomials of degree at most $3$ with real coefficients, with the usual polynomial addition for vector addition, and usual multiplication by real numbers as the scalar multiplication. This is a vector space in which the vectors are not matrices: they are polynomials of the form $a+bx+cx^2+dx^3$ with $a,b,c,d\in\mathbb{R}$. However, you can "think of them" as tuples $(a,b,c,d)$ because the vector operations occur "componentwise" in any case: when you add two polynomials, $a+bx+cx^2+dx^3$ and $r+sx+tx^2+ux^3$, you get $(a+r) + (b+s)x + (c+t)x^2+(d+u)x^3$; when you multiply by scalars, again each coefficient gets multiplied independently. That is, there is what we call an "isomorphism" between the vector space of all polynomials of degree at most $3$ and the vector space $\mathbb{R}^4$ that makes the polynomial $a+bx+cx^2+dx^3$ correspond to the tuple $(a,b,c,d)$; under this correspondence, the image of a sum is the sum of the images, and the image of a scalar multiple is the scalar multiple of the image (it is what we call a linear transformation, and it is both one-to-one and onto).

However, even though we can think of the polynomials as $4$-tuples (by doing the "translation" described above), there are situations where it is far more useful to think of the vectors as polynomials. For example, the function that takes a polynomial $p(x) = a+bx+cx^2+dx^3$ as an input and returns $p(3)$ as the output is a very nice function: we have that $(p+q)(3) = p(3)+q(3)$ and $(\alpha p)(3) = \alpha(p(3))$ (again, a linear transformation). It is very easy to think of it in terms of polynomials, but it is more annoying to think of it in terms of tuples. Or consider the function that sends a polynomial $p(x)$ to the polynomial you get by taking $p(x) + 3p'(x) + 2p''(5)$ (derivatives). Again, easy to handle as polynomials, more annoying as tuples (but certainly possible).

Or consider the collection of all continuous, differentiable functions $f\colon \mathbb{R}\to\mathbb{R}$ that satisfy the following differential equation:
$$xf''(x) + \cos(x)f'(x) + 2f(x) = 0.$$
The collection of all such functions forms a vector space under the usual addition and scalar multiplication of functions; but even though it is isomorphic to a vector space of tuples (in fact, to $\mathbb{R}^2$), it's better to think of it in terms of functions than in terms of tuples for purposes of actually dealing with the differential equation.

Going to point 2 above: consider the vector space of all polynomials with the usual polynomial addition. Unless you want to allow infinite tuples (which you can, but they tend to be difficult to work with), you cannot represent arbitrary polynomials with $1\times n$ matrices and still keep represent all polynomials. Or to make things even worse: consider the vector space $V$ of all functions $f\colon \mathbb{R}\to\mathbb{R}$, with pointwise addition and scalar multiplication; you would need a tuple that has an many entries as there are real numbers to represent it as a "matrix". It is far better to think of the vectors as functions, not as row or column vectors.

So, as long as you stick to finite dimensional vector spaces, the answer is "up to suitable translation, yes, you can think of all vectors as tuples". However, there are in general lots of translations and no good reason to prefer one over the other, so that you would need to be careful to specify in each case how you are translating. And in many situations, it is better to think of the vectors as they are given rather than as tuples, because the kinds of things you want to do with them are easier to do if they are polynomials/functions/other kinds of objects than if they are tuples.

to my knowledge all vectors are either nx1 or 1xm  matrices.


Curves triangular numbers. Sometimes you have to deal with this equation: $X^2+aX+Y^2+bY=Z^2+cZ$

$a,b,c$ - integer coefficients.

I wrote below - to start a particular solution of Diophantine equations.

To do this, use the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$

I turned solutions such.

$X=(k-1)((a+b)k+a-c)s^2-(ck-b)ps$

$Y=k((a+b)k+c-a-2b)s^2-(ck+a-c)ps$

$Z=(2ck^2+(a-b-2c)k+c-a)s^2-((a+b)k-b)ps$

And more:

$X=(c-a-b)p^2+(k(3c-2a-2b)+2a+b-2c)ps+(k-1)((2c-a-b)k+a-c)s^2$

$Y=(c-a-b)p^2+(k(3c-2a-2b)+a-c)ps+k((2c-a-b)k+a-c)s^2$

$Z=(c-a-b)p^2+(k(4c-3a-3b)+2a+b-2c)ps+(2k-1)((2c-a-b)k+a-c)s^2$

Probably will have to write a more general formula. But I think that's not a problem.

Here is another question. When trying to solve the more general equation:

$nX^2+aX+qY^2+bY=jZ^2+cZ$

Always turns in decisions that need to share the decision itself by type: $w=n+q-j$

That is, the formula is written as a fraction. Certainly at Pell's equation infinitely many solutions. And of course no matter how large the number it was not always a chance that the fraction reduced. But still not clear how all of these decisions to allocate whole?  Although there may be another idea to solve this equation and all that there is no need to do?

The coefficients should have this appearance. Because they will appear as unknown for other equations.

Solving this equation turned out rather unexpected result.

As is well known equation of the form: $aX^2+cY^2=jZ^2$

as you know is not always the solution in integers.

But such an equation: $aX^2+bX+cY^2+qY=jZ^2+dZ$

will always have a solution integers.
At least you can always write a formula.

That's quite unexpected and it is not clear result.

Decisions will always be the coefficients $b,q,d$ - at least one non-zero.


How to understand the lectures on the fly as much as possible? My problem is, that during the lectures my mind just goes blank and I cannot follow for a few seconds. Consequently I will have trouble catching up with what was just explained. This repeats more and more as the time goes on.

So the information I am hearing just turns into white-noise as the lecture proceeds. Obviously this is something that other people also have to struggle with. But I feel (there is no other way to tell) that this issue is more hindering to me than to others.

Maybe you others have your ways to cope with that. I am curious.

All the best

One thing that helps me is taking pictures of the board periodically throughout the lecture. This allows me to spend less time frantically copying notes as to not miss anything, and spend more time actually listening and processing the information. Also, I would assume that your professor doesn't mind questions during lecture. If something catches you off guard and pulls you out of it (your brain might just be stopping to see if things are making sense), I would just raise your hand and ask the professor to re-explain it. Most of the time, there are other students in the class who are hung up on the same exact point.

Understanding on the fly is possible when you already know intimately what is being lectured. This intimate knowledge comes after you have spent time and effort in something called study: a process whereby a pre-concept (a new notion of which one can have some hypotheses about their meaning thanks to intuition coming from prior knowledge) tranforms into a concept through an activity of check of its coungrence with respect to all concepts already owned by the learner. So my suggestion is not to attend lectures unless they are given by authorities, and in this case do it after having studied.


Let $a,b \in \mathbb R$ and $f(x)=a\cos x+b\cos3x$. Prove that $|b|\le 1$. Let $a,b \in \mathbb R$ and $f(x)=a\cos x+b\cos3x$. It is known that $f(x)>1$ has no real solutions. Prove that $|b|\le 1$. We can write the given equality as
\begin{align}
&f(x)=(a-3b)\cos x+4b\cos^3x\\[2ex]
\Rightarrow\quad&f'(x)=(3b-a)\sin x-12b\cos^2x\sin x=0\\[1ex]
\Rightarrow\quad&\sin x=0,\quad \cos^2x=\frac {3b-a}{12b}
\end{align}
Therefore $f(x)$ has minima or maxima when $\sin x=0$, so $\cos x=\pm 1$ or when $\cos x=\pm\sqrt{\frac {3b-a}{12b}}$. Hence $f(x)=\pm(a+b)$ or $\pm\frac 23(3b-a)$.

Now what should I do?

By plugging in for $x$, we obtain:
$$f(0)=b+a,\qquad f\left(\frac{2\pi}{3}\right)=b-\frac{a}{2}$$
Since $f$ is continuous, it takes all values between $b+a$ and $b-a/2$ (regardless of which is larger/smaller). In particular, it takes on the value $b$. Since $f(x+\pi)=-f(x)$, it follows that $f$ also takes on the value $-b$. Since $f(x)>1$ has no real solutions, it follows that neither $b$ nor $-b$ exceed 1, i.e. $|b|\leq 1$.

Setting $a=1,b=1.01,c=1.01$, we obtain one real solution, $\cos x=0.885831$ or $x=0.482515$, from equation (1):

$$f(x)=c>1\tag{1}$$

where
$$f(x):=(a-3b)\cos x+4b\cos^3x\tag{2}$$

Thus it seems to me that the condition $f(x)>1$ has no real solution is not enough to deduce that $|b|<1$.


Show that if $x_n\rightarrow\ x_0^{-}$ and $y_n\rightarrow\ x_0^{+}$ then $ \lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$ Let $f: \Bbb[a,b]\rightarrow\Bbb{R}$ be differentiable in a point $a<x_0<b$. Use Taylor's expansion with the remainder of $f$ around $x_0$ to show that if $ x_n\rightarrow\ x_0^{-}$, and  $ y_n\rightarrow\ x_0^{+}$ then:
  $$ \lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$$

I know that:

$f(x_n)= f(x_0)+f^\prime(x_0)(x_n-x_0) + R_1 (x_n) $

and:

$f(y_n)= f(x_0)+f^\prime(x_0)(y_n-x_0) + R_1 (y_n)$

Note: $R_1$ is the remainder.

and then by substracting:

$ \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0) + \frac{R_1(x_n)-R_1(y_n)}{x_n-y_n}$

but It seemed that I couldn't continue from here to anywhere, can someone please show the right way to do this?

A nice exercise in Rudin's Principles of Mathematical Analysis (Exercise 5.19) states that, if $\alpha_n \to 0$ and $\beta_n \to 0$, then
$$
D_n = \frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}
$$
converges to $f'(0)$ under each of the following assumptions:

$\alpha_n < 0 <\beta_n$;
$0<\alpha_n<\beta_n$ and $\{\beta_n/(\beta_n-\alpha_n)\}_n$ is bounded;
$f'$ is continuous.

Here you are in case 1. You can write
$$
D_n = \frac{f(\beta_n)-f(0)}{\beta_n-0}\frac{\beta_n}{\beta_n-\alpha_n} + \frac{f(\alpha_n)-f(0)}{\alpha_n}\frac{-\alpha_n}{\beta_n-\alpha_n}.
$$
Then you remark that $0<\frac{\beta_n}{\beta_n-\alpha_n}<1$ and $-1<\frac{\alpha_n}{\beta_n-\alpha_n}<0$. You should find your way from here.

By the way, case 3 is easily covered by some MVT or Taylor's theorem.

You need to use an estimate for the remainder. For exemple

$$R_1(x) = (x-x_0)\epsilon(x)$$

with $$\lim_{x\to x_0}\epsilon(x) = 0$$

Or in other words, $R_1(x) = o(x-x_0)$


Combinatorics question, combinations involved in filling up of vacancies. To fill $12$ vacancies, there are $25$ candidates of which $5$ are from
  scheduled castes. If $3$ vacancies are reserved for scheduled caste
  candidates while the rest are open to all, find the number of ways in
  which selections can be made?

According to me it should be $^5C_3×^{22}C_9$ because we first find the combinations for scheduled castes then multiply it with the combinations for all the remaining candidates (including the scheduled castes).

However, answer given is $10 \times ^{20}C_9+ 5 \times ^{20}C_8 + ^{20}C_ 7$.

Why is my reasoning incorrect?

Edit: I have realised my error, now how do I eliminate it and get the right answer?

The reasoning is incorrect because you're counting multiple times. Let's focus only on the "scheduled caste" section at the moment. Call the "scheduled caste" members A, B, C, D, E. These two cases are the same:

A, B, C make it through reservation and D, E make it through an open slot; rest same
A, D, E make it through reservation and B, C make it through an open slot; rest same

You have considered them as distinct.

This would be clearer with a simpler problem: if you would get rid of all non-"scheduled" members make a selection of five from the "scheduled" group, with three reserved and two open. There is in effect only one way to do so, but with your method it is 5C3 * 2C2.

EDIT: To eliminate OP's error, just notice the simple fact that the number of people making it is $3$, $4$ or $5$ from the "scheduled" community. If we separated the entry process into the two classes reserved and non-reserved, we'd be counting multiple times because it doesn't matter what process a "scheduled" person is selected by as long as the selections are the same. Instead, we separate people into classes scheduled and non-scheduled. There are three cases: 3 from "scheduled", rest from open only; 4 from "scheduled", rest from open only and 5 from "scheduled", rest from open only.

So  we simply calculate $\binom{5}3 \cdot \binom{20}{9} + \binom{5}4 \cdot \binom{20}8 + \binom{5}5 \cdot \binom{20}7$ and we're done.

There are more candidates than vacancies.  The way the answer is structured is that possible ways the selection could be made is to fill the 3 vacancies from 5 SC, and the remaining 9 from Non-SC, 4 SC are placed and the remaining 8 from Non_SC and all 5 Scs are placed and the remaining 7 from Non-SC.  I would kind of agree with you that the 3 vacancies are filled from 5 SCs and the remaining nine vacancies are filled with 22 open candidates.  In the textbook solution, it assumes that all 5 SC candidates are placed which is not mentioned in the problem.


On the rate of convergence of nested radicals The sequence of nested radicals $\sqrt {a+\sqrt{a+\sqrt{a+...}}}$ is defined by
  $x_1=\sqrt a$,   $x_{n+1}=\sqrt{a+x_n}$,   where $a>0$.
Here are three questions:

(a) Show that $x=\displaystyle\lim_{n\to\infty} x_n=\frac{1+\sqrt{1+4a}}{2}$.
(b) Prove $\displaystyle x-x_n\sim \frac{C_a}{(2x)^n}$ for some $C_a>0.$
(c) For $a=2$, what is the value of $C_2$ in (b)?

My attempt:
(a) is easy. It is easy to prove $\{x_n\}$ is monotone and bounded, hence $x=\displaystyle\lim_{n\to\infty} x_n$ exists. Then we have the equation $x^2-x-a=0$.
For (b), I start by writing:
\begin{align}
x-x_n &= \frac{x^2-x_n^2}{x+x_n} \\
&= \frac{(x+a)-(x_{n-1}+a)}{x+x_n} \\
&= \frac{x-x_{n-1}}{x+x_n}
\end{align}
By induction,
$$x-x_n= \frac{x}{\prod_{k=1}^n (x+x_k)}$$
and I got stuck at this step.
For (c), once we prove $\displaystyle x_n=2\cos\left(\frac{\pi}{2^{n+1}}\right)$, it is not hard to see the constant $C_2$ is equal to $\displaystyle \frac{\pi^2}{4}$. This is a standard example of trigonometric substitution.

Any idea on how to tackle (b)?

Notation: $f(n)\sim g(n)$ if and only if $\displaystyle\lim_{n\to \infty} \frac{f(n)}{g(n)}=1$.

Let $x_n:=x-h_n$, so that $$(x-h_{n+1})^2=a+x-h_n$$ which simplifies to $$h_{n+1}^2-2xh_{n+1}+h_n=0$$ \begin{align*}h_{n+1}&=x-\sqrt{x^2-h_n}\\
&=\frac{h_n}{2x}(1+\frac{h_n}{4x^2}+\frac{h_n^2}{8x^4}+\cdots)\\
\end{align*}

So multiplying out gives $$h_n=\frac{h_1}{(2x)^n}\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$ so the required constant is $$C_a=h_1\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$

This constant need not have a closed formula but it exists by standard results on convergence of products.

With $y_n:=x-x_n$, we can rewrite the recurrence as

$$y_{n+1}=x-\sqrt{a+x-y_n}.$$

As $y_n$ tends to zero, we have

$$y_{n+1}=x-\sqrt{a+x-y_n}=\frac{x^2-a-x+y_n}{x+\sqrt{a+x-y_n}}=\frac{y_n}{x+\sqrt{a+x-y_n}}\approx\frac{y_n}{2x}$$

and by induction,

$$y_n\sim\frac{C_a}{(2x)^n}$$ for some constant

$$C_a=\lim_{n\to\infty}y_n(2x)^n.$$

Convergence is ensured if for some $n$, $x+\sqrt{a+x-y_n}>1$, guaranteeing a geometric decay of $y_n$.


As $a \in\left\{ 1,2,3 \right\}~~$ is held, can I write $f(a=1)=k~~$ rather than $\text{As}~a=1, f(a)=f(1)=k~~$? $$  x \in\left\{1,2,3\right\}  $$
$$  f(x):=\text{function which returns certain value depending on argument }x $$
Assume the below statements .
$$ \text{As }x=1 ~,~  $$
$$  f(x)=f(1)=k $$
Can I write it as the below notation?
$$  f(x=1)=k $$
I want to know whether teachers subtract of scores or not as the above notation.
The purpose to write as my way is just I want to save the time.

I think that you want to underline the fact that $f$ Is a function on $x$ and that if $x=1$ $f(1)=k$, I Would writer $f(x)|_{x=1}$


Infinitely number of primes in the form $4n+1$ proof Question: Are there infinitely many primes of the form $4n+3$ and $4n+1$?

My attempt: Suppose the contrary that there exist finitely many primes of the form $4n+3$, say $k+1$ of them: $3,p_1,p_2,....,p_k$

Consider $N = 4p_1p_2p_3...p_k+3$, $N$ cannot be a prime of this form. So suppose that $N$=$q_1...q_r$, where $q_i∈P$

Claim: At least one of the $q_i$'s is of the form $4n+3$:

Proof for my claim: $N$ is odd $\Rightarrow q_1,...,q_r$ are odd $\Rightarrow q_i \equiv 1\ (\text{mod }4)$ or $q_i ≡ 3\ (\text{mod }4)$

If all $q_1,...q_r$ are of the form $4n+1$, then $(4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1$

Therefore, $N=q_1...q_r = 4m+1$. But $N=4p_1..p_k+3$, i.e. $N≡3\ (\text{mod }4)$, $N$ is congruent to $1\ \text{mod }4$ which is a contradiction.

Therefore, at least one of $q_i \equiv 3\ (\text{mod }4)$. Suppose $q_j\equiv 3\ (\text{mod }4)$

$\Rightarrow$ $q_j=p_i$ for some $1\leq i \leq k$ or $q_j =3$

If $q_j=p_i≠3$ then $q_j$ | $N = 4p_1...p_k + 3 \Rightarrow q_j=3$ Contradiction!

If $q_j=3$ ($\neq p_i$, $1\leq i \leq k$) then $q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t$ for some $1 \leq i \leq k$ Contradiction!

In fact, there must be also infinitely many primes of the form $4n+1$ (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me?

Regards

Suppose $n>1$ is an integer.  We define $N=(n!)^2 +1$. Suppose $p$
is the smallest prime divisor of $N$. Since $N$ is odd, $p$ cannot
be equal to $2$. It is clear that $p$ is bigger than $n$
(otherwise $p \mid 1$).  If we  show that $p$ is of the form
$4k+1$ then we can repeat the procedure replacing  $n$ with  $p$
and we produce an infinite sequence of primes of the form $4k+1$.

We know that $p$ has the form $4k+1$ or $4k+3$.    Since $p\mid N$
we have $$ (n!)^2 \equiv -1 \ \ (p)  \ . $$   Therefore $$
(n!)^{p-1} \equiv (-1)^{ \frac{p-1}{2} } \ \ (p) \ . $$
Using Fermat's little Theorem  we get  $$ (-1)^{ \frac{p-1}{2} } \equiv 1 \ \ (p) \ . $$
If $p$ was of the form $4k+3$ then $\frac{p-1}{2} =2k+1$ is odd
and therefore we obtain $-1 \equiv 1 \ \ (p)$ or $p \mid 2$ which
is a contradiction since $p$ is odd.

Actually there is an even more elegant proof that shows that this holds true for all primes>2:

∀ primes,p ∈P & p>2,∃ k∈Q| k=(p-1)/4  or k= (p+1)/4

Proof:
By definition every 4th integer, k ≥ 4, must have 2^2 (2 squared) as a factor.  This means that every other even integer > 2 must have 2^2 as a factor.

For all primes p>2, the prime itself is odd and therefore p-1 and p+1 are both even.

Since p-1 and p+1 represent consecutive even integers, one and only one of them must have 2^2 as factors and thus (p±1)/4 is an integer.  ∎

This applies to all primes greater than 2.


Proof by "infinite induction" Prove that $\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2$.

We can check this is true for n=0,1,2,3,4. Since the right side is a polynomial of degree 4, and the left side is a sum of monomials whose degree is <= 4, then if both polynomials coincide for five points, they must be the same.

My question: is this proof rigorous? I'm concerned about the left side sum.

No, your proof is not rigorous. It is actually wrong. The point is that you don't know a priori that the left-hand-side is a polynomial of degree $4$ on $n$.

As an example of your "reasoning", consider the equality
$$
\sum_{i=1}^n i = 2n-1.
$$
This is of course false (the actual formula on the right should be $n(n+1)/2$. But, for $n=1$ and $n=2$, both sides agree ($1$ and $3$ respectively), and the right hand side is a polynomial of degree $1$, so two points would suffice to determine it, if we were allowed to reason as you did.

Yes this is rigorous because the a basic fact about polynomials is that the discrete integral of a degree $d$ polynomial is a degree $d+1$ polynomial.


Distance between sets implies completness of a metric space Given the metric space $(X,d)$ suppose that for every nonempty, closed $A,B\subseteq X$ with $$A\cap B=\varnothing, \mathrm{dist}(A, B)>0.$$
  Show that  $(X,d)$ is complete.

But given the definition of distance as $$\operatorname{dist}(A,B)=\inf\{d(a,b):a \in A,b \in B\}$$
and given that  $$A\cap B=\varnothing$$
and the fact that $A,B$ are closed
we have that  $$\mathrm{dist}(A, B)=0\iff A\cap B \neq \varnothing$$

Am I missing something?

As Micapps pointed out in the comments, in $\mathbb{Q}$ with the usual metric, $[0,\sqrt{2}), (\sqrt{2},5]$ are disjoint closed sets with distance zero, so your last statement is not true.

To show the problem, we show the contrapositive. Let $\{s_n\}_{n\in \mathbb{N}}$ be a non-convergent Cauchy sequence of distinct points (which thus has no accumulation point). For each $s_{2n}$, let $B_{2n}$ be an open ball about $s_{2n}$ that does not include any of the other $s_i$ ($i \neq 2n$). (In particular, none of these open balls include any odd-indexed $s_i$.)

Let $A = \{s_{2n} : n \in \mathbb{N}\}$ and $B = \{x : x\notin B_{2n}$ for any $n\}$, i.e. the complement of the union of the $B_{2i}$. Then $A,B$ are disjoint closed sets with distance $0$.

I don't think this statement is true.
For example, the interval $(0,1] \in \mathbb{R}$ seems to has the property from the hypothesis but clearly is not complete.
But I'm not sure if the property is really true on this space.


Understanding a definition I'm reading this article on Circuit Theory by John Baez
https://ncatlab.org/johnbaez/show/Circuit+theory,
and I'm having trouble in understanding how a map is defined.  In the "Cochain Complexes From Circuits" section he defines a map $r$ which should be the canonical isomorphism between the Hilbert space of chains $C_i(\Gamma)$ of a circuit and its dual $C^i(\Gamma)$.  He says that this map is given by the relation
$$
a(\beta)=\langle r(a),\beta\rangle,
$$
where $a\in C_i(\Gamma)$ and $\beta\in C^i(\Gamma)$.
What I don't understand is:  What is $a(\beta)$?  That is, how can I evaluate a chain in a cochain?  Moreover, what is the angle bracket between two cochains, as in the second member?
So, what is this function $r$ that he is defining?

I think I've got it.  With $a(\beta)$ he really meant $\beta(a)$, so the relation is
$$
\beta(a)=\langle r(a), \beta\rangle,\quad\text{for each $a\in C_i(\Gamma)$ and $\beta\in C^i(\Gamma)$.}
$$
This is equivalent to saying that $r(b)=\langle b,\cdot\rangle$ for each $b\in C_i(\Gamma)$, which is the standard way of defining an isomorphism given an inner product.
In fact, for each $b\in C_i(\Gamma)$,
$$
\eqalign{
(rb)(a)&=\langle b,a\rangle\cr
&=\langle a,b\rangle\cr
&=\langle r(a),r(b)\rangle.
}
$$
So, upon defining $\beta=rb$, we have $\beta(a)=\langle r(a),\beta\rangle$.


prove that two integers are in the same remainder class mod n provided that they have the same remainder when divided by n? prove that two integers are in the same remainder class mod n provided that they have the same remainder when divided by n.

Let $a,b \in \mathbb{Z}$, and since they have the same remainder when divided by n, this implies that $\exists s,t \in \mathbb{Z} $ such that $a=sn+r,b=tn+r \implies$ $ a \equiv r \ mod \ n,b \equiv r \ mod \ n \implies a \equiv b \equiv r \ mod \ n$.therefore, $$a,b \in \mathbb{Z} /n \mathbb{Z} $$

Is this proof valid?

Formal answer.

Definition: Two integers are equivalent$\mod n$ if they leave the same remainder when divided by $n$. We say $a \equiv b \mod  n$ if $a$ and $b$ are equivalent.

So, when you've proved that they have the same remainder, you can easily conclude that they're equivalent.


Extreme Value Theorem For any function, given an interval with closed/open brackets, does a function have to have a min and max for the Extreme Value Theorem to apply? For example, does the extreme value theorem apply for the function f(x) = |x|?
In my opinion, I believe not because this function doesn't have a discrete maximum value. Is this thinking correct?
In addition, if at some point of any function is discontinuous within a interval, does it automatically mean that the application of the EVT is invalid? For example, take a simple floor function, since it's not continous, then it doesn't apply.

The EVT applies exactly when the hypothesis I mentioned in my comment exist, and there is no redundancy, meaning all of them must apply in order for the theorem to work. Importantly, it has nothing to do with differentiability.
$f(x)=\lvert x\rvert$  is indeed continuous so, pick a bounded, closed interval (say, $[a,b]$) then indeed, the EVT applies. Namely, the extreme values are $0$ if $ab<0$ or $min(\lvert a\rvert,\lvert b\rvert)$ else, for the minimum; $max(\lvert a\rvert,\lvert b\rvert)$ for the maximum.
Again, for a discontinuous function such as the floor function the EVT does not apply, which doesn't mean that, by chance, the extreme values are not attained, which indeed they are, for any bounded interval.

A variant of this is Rolle's theorem (with is proved using the Extreme value theorem). If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists some $c$ in the open interval $(a, b)$ such that $f'(c)=0$. This means when you take two points of a function that satisfies this criteria (and produce equal $y$ values) there will be a relative min/max between those points.
However, if you take $f(x)=|x|$ (as an example $[a,b]=[-1,1]$) differentiability fails at an interior point of the interval (when $x=0$ the function isn't differentiable). Only when $(a,b)$ is differentiable and $f$ is continuous on $[a, b]$, we can deduce the curvature of the function (applies for Rolle's theorem).
So, Rolle's Theorem does not apply for $|x|$ because the function is not differentiable at $0$ (differentiation yields a step function). That is if you are trying to find a point when $f'(x)=0$. Hopefully, this is still useful for this problem or others down the line. However, the Extreme Value Theorem may yield a global minimum at $x=0$ because $f$ is continuous over all real numbers and our interval is bounded/closed.


Time taken to traverse distance, respecting acceleration and decceleration I am working on project simulating vehicle transport.
My current function that calculates time required to get from A to B is simple

$time = dist/speed$

I need to change this function, so that it will account for acceleration and decceleration. I found some solutions, that work if the car is allowed to get to full speed, but I also need to account for the cases in which the distance is too short to attain maximum speed. I know ,this question is little bit below math.stackexchange difficulty, and I'm sorry for that, but I do not have slightest idea where to start.  If you can point me in the general direction of the solution, I would be extremely grateful.

A vehicle must travel a distance $D$ and we wish to find the time it takes $T$ with the following assumptions:

a) The vehicle starts at rest and must end at rest

b) It accelerates at a constant rate $a$ and decelerates at a constant rate $d$ ($d$ is negative)

c) When (or if) it achieves its maximum speed $V_{max}$ it cruises at this speed until it must decelerate

Two formulas used in the following: $$V_{final}=V_{initial}+at$$

$$s=V_{initial} \, t+ \frac{1}{2}at^2$$

where $V$ is speed, $t$ is time and $s$ is distance.

Let's call $V_1$ the vehicle's speed after accelerating for a time $t_1$. As it starts at rest, we have $$V_1 = at_1$$

Let's call $V_2$ the vehicle's speed after decelerating for a time $t_2$. As it ends at rest, we have $V_2 = 0$, which means $$0= V_1 + dt_2$$

Combining these two equations we get $$t_2= -\frac a d t_1$$

Now let's look at the distances covered. For the acceleration phase we find $$s_1=\frac{1}{2}at_1^2$$

and for the deceleration phase we find $$s_2=V_1t_2+ \frac{1}{2}dt_2^2$$

Inserting the equation for $t_2$ found above, we can convert the last equation to $$s_2=-\frac{1}{2}\frac{a^2}{d}t_1^2$$

Let's assume for the moment that the vehicle never reaches $V_{max}$. This means there is only an acceleration phase and a deceleration phase. Which means $D = s_1 + s_2$. Which means $$D = \frac{1}{2}at_1^2(1- \frac a d)$$

Extracting $t_1$ from this equation we find $$t_1 = \sqrt{\frac{2D}{a(1-\frac a d)}}$$

We are now ready to give the final answers. Using $t_1$ above, calculate the speed achieved during the acceleration phase ($V_1=at_1$) and compare it with $V_{max}$.

1. $V_1 \lt V_{max}$

We are done. The vehicle never reaches its maximum speed and $$T = t_1+ t_2$$

2. $V_1 \ge V_{max}$

The vehicle does reach its maximum speed. The time it takes to reach this speed will be $$t_{max}=\frac{V_{max}}{a}$$

This means the distance it covers during the acceleration and deceleration phases will be $$s = \frac{1}{2}at_{max}^2(1- \frac a d)$$

The time it spends cruising at $V_{max}$ must then be $$t_{cruise}=\frac{D-s}{V_{max}}$$

Giving the total time as $$T = t_{max}(1- \frac a d)+ t_{cruise}$$

If uniform acceleration (or deceleration) , start (or end) speed are known then we have time $t$

$$ t = (v-u)/a = (\sqrt {2 a s + u^2}-u)/a $$

where $u,v$ are speeds at start and end. If $a>=<0$ or positive acceleration, then  $ v>=<u. $


A good way to retain mathematical understanding? What is a good way to remember math concepts/definitions and commit them to long term memory?

Background:

In my current situation, I'm at an undergraduate institution where I have to take a lot of core classes and easy math classes because there are no higher level courses available. Therefore, I have to study math on my own. It worked out fairly well, but I have a big problem when there's a lapse in my studying. For example, in February, I was working on transfer college applications, so I was unable to study math. Right now, I have less than a month left in school, so I need to work on all my term papers. It seems that when I take these extended breaks from studying math, like 2+ weeks, I have trouble going back, and occasionally I have to re learn a lot of the material. Previously, I was able to understand the definitions, theorems, etc and solve some of the exercises. Now, I need to look back on the definitions to understand what I wrote down! It's frustrating that I'm regressing in my mathematical education when I'm not actually studying math.

Teach.

Seriously it is absolutely the number one way to learn.

Take a look at http://www.supermemo.com/

This system (and the method it describes) is based on repetition over a changing interval of time - short time intervals first, then each interval longer than the preceding interval. The author is rigorous in his approach to analyzing how memory is reinforced.
What is repeated is putting items INTO memory, and and TAKING THEM OUT (recalling them). Both actions are components of memory, that is, being able to "memorize" and being able to "recall" from memory.
The continual use of this technique also results in a second-order effect: an improved ability to memorize things in a shorter time, and with fewer repetitions.

DISCLAIMER: I have no relationship to the author of this method, nor do I know the author personally. For me this was just a useful discovery.


Why in an inconsistent axiom system every statement is true? (For Dummies) I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?

I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.

This previous question is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.

Thanks a lot in advance.

I'll try to say this all in plain English:

Let's say we decide to accept the following two facts: (1) "I am a fish", and (2) "I am not a fish". Just keep those in mind.

Now let's pick any old statement, say: (3) "You can fly". Now let's prove that the statement is true!

Alright, we've already accepted that (1) "I am a fish". Of course, any time I have a true statement P, I can make a new true statement by making the statement "P or Q is true." Because to check if an 'or' statement is true, I only need to check that one of them is true. (If I tell you "My name is Dylan OR I can spit fire," you don't need to wait around with a fire extinguisher to tell if that statement is true. It's true because the first part of it is true).

So by this logic, the statement (4) "I am a fish or you can fly" must be true (since the first part is true.)

OK, but now let's say, in general, I have some 'or' statement "P or Q" and I know for a fact that the whole statement is true. If I also know that P is false then I can conclude that Q is true. Right? Because an 'or' statement is true if and only if at least one of the statements inside it is true, so if I rule out one of them the other one must be true. (So if I always tell the truth and I tell you that you have a billion dollars in your bank account OR I just ate a sandwich, you can check your bank account and quickly conclude that I just ate lunch... unless you're very wealthy.)

Alright, so far so good. We know the statement "I am a fish or you can fly" is definitely true. But wait, we also know that the statement "I am a fish" is false (remember, it's one of the things we assumed in the very beginning!). So that means, by what we just talked about, that the statement "You can fly" must be true.

So voilà! Using the magic of a contradictory system, we've proven you can fly!

"inconsistent" means that if you ask the same question twice, you get a different answer. To say that you get the answer "true" every time is "consistent". The definition of "negative proposition" (the opposite of a proposition) is meant for the purpose of getting a negative result in the event the proposition gets a positive. So if we want to make a system which is "consistent" and gives the same result for a proposition and its negation, we cannot do it within the accepted meaning of the definitions "consistent" and "negation"