Question about Peano form of the remainder Let $f(x)$ be a real-valued function defined on a closed interval [a, b], differentiable on the open interval (a, b) $n-1$ times.

$x_0$ belongs to [a, b]. Suppose that we have:$$f(x_0)=0,f'(x_0)=0,f''(x_0)=0,...,f^{(n-1)}(x_0)=0$$
Also suppose that the function admits the nth derivative at $x_0$. In this case my book says that is:$$f(x)=\frac{(x-x_0)^n}{n!}[f^{(n)}(x_0)+\sigma(x)]$$
where:$$\begin{matrix} \lim_{x \to x_0}\sigma(x)=0 \end{matrix}$$
It seems that in my book's proof must be (because it isn't completely clear ):
$$\begin{matrix} \lim_{x \to x_0}f^{(n)}(x)\end{matrix}=f^{(n)}(x_0)$$
It means that $f^{(n)}(x)$ is continuous at $x=x_0$...but in the theorem it isn't mentioned that $f^{(n)}(x)$ is continuous at $x=x_0$; I know that the first $(n-1)$th derivatives are continuous in $x_0$ and $f^{(n)}(x_0)$ exists. So is the continuity of $f^{(n)}(x)$ at $x=x_0$ necessary to prove the theorem?

Proof's book:

If $x\ne x_0 $ and $n>1$, according to Cauchy's mean value theorem, we have:$$\frac{f(x)}{\phi(x)}=\frac{f^{(n-1)}(\eta)}{\phi^{(n-1)}(\eta)}$$

where $\eta$ is a suitable interior point of the interval $[x_0,x]$ and $\phi(x)=(x-x_0)^n$.
Since:$$\phi^{(n-1)}(\eta)=n!(\eta-x_0)$$$$f^{(n-1)}(x_0)=0$$the last formula I wrote becomes:$$\frac{f(x)}{(x-x_0)^n}=\frac{1}{n!}\cdot\frac{f^{(n-1)}(\eta)-f^{(n-1)}(x_0)}{\eta-x_0}$$
Now:
$$\begin{matrix} \lim_{x \to x_0}(n!\frac{f(x)}{(x-x_0)^n}) \end{matrix}=\begin{matrix} \lim_{x \to x_0}\frac{f^{(n-1)}(\eta)-f^{(n-1)}(x_0)}{\eta-x_0} \end{matrix}=f^{(n)}(x_0)$$I'm stuck here: in the second limit; I think that writing means that $f^{(n)}(x)$ is continuous at $x=x_0$. Am I right?
Let's keep on with the proof; let:$$\sigma(x)=n!\frac{f(x)}{(x-x_0)^n}-f^{(n)}(x_0)$$ for $x\ne x_0 $ and $\sigma(x_0)=0$ we have:
$$\begin{matrix} \lim_{x \to x_0}\sigma(x) \end{matrix}=0$$
Finally:$$f(x)=\frac{(x-x_0)^n}{n!}[f^{(n)}(x_0)+\sigma(x)]$$QED

My doubt is: is the continuity of $f^{(n)}(x)$ at $x=x_0$ necessary to prove the theorem? If it isn't can someone explain me why the second limit is valid?

Any help would be greatly appreciated.


The sum of $\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$ I don't have an idea how to calculate this sum. I knew that $\sum_{k=0}^{\infty}\zeta(2k+2)-1 = \frac{3}{4}$. I also knew $\sum_{k=1}^{\infty}\frac{\zeta(2k)k-k}{{2k^2+k}}=\frac{1}{2}(3-ln(4\pi))$. Thank you very much for help. The sum to calculate is $$\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$$

Write $\zeta(2k+2) - 1 = \sum_{n=2}^\infty n^{-2k-2}$ and interchange order of the two sums.  I get
$$ \sum_{n=2}^\infty \frac{\text{arctanh}(1/n)}{n} = \sum_{n=2}^\infty \frac{\ln((n+1)/(n-1))}{2n} $$
However, I still don't have a closed form.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1}:\ {\Large ?}}$.

  Note that $\pars{~\mbox{see}\ \color{#000}{\textbf{6.3.14}}\ \mbox{in A & S Table}~}$
  $\ds{\left.\vphantom{\Large A}
\Psi\pars{1 + z}\right\vert_{\ \verts{z} < 1} = -\gamma +
\sum_{n = 2}^{\infty}\pars{-1}^{n}\,\zeta\pars{n}z^{n - 1}}$ where $\ds{\Psi}$ and $\ds{\gamma}$ are the Digamma Function and de Euler Constant, respectively.

Then, with $\ds{\verts{z} < 1}$,
\begin{align}
\Psi\pars{1 + z} & = -\gamma +
\sum_{k = 2}^{\infty}\pars{-1}^{k}\,\zeta\pars{k}z^{k - 1}
\\
\Psi\pars{1 - z} & = -\gamma -
\sum_{k = 2}^{\infty}\,\zeta\pars{k}z^{k - 1}
\end{align}
which leads to
\begin{align}
\Psi\pars{1 + z} - \Psi\pars{1 - z} & =
\sum_{k = 2}^{\infty}\bracks{\pars{-1}^{k} + 1}\,\zeta\pars{k}z^{k - 1} =
\sum_{k = 0}^{\infty}2\,\zeta\pars{2k + 2}z^{2k + 1}
\\[2mm]
\mbox{and}\
{\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}} & =
\sum_{k = 0}^{\infty}\bracks{\zeta\pars{2k + 2} - 1}z^{2k}
\end{align}

  Integrating over $\ds{\left[0,1\right)}$:

\begin{align}
\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} & =
\int_{0}^{1}\bracks{%
{\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\braces{%
{\bracks{\Psi\pars{z} + 1/z} - \Psi\pars{1 - z} \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\braces{%
-\,{\bracks{\Psi\pars{1 - z} - \Psi\pars{z}} - 1/z \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\bracks{%
-\,{\pi\cot\pars{\pi z} - 1/z \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {2 \over 1 - z^{2}}}\dd z
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z -
{1 \over 2}\int_{0}^{1}{\dd z \over 1 + z}
\end{align}

$$
\bbx{\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} =
{1 \over 2}\ \underbrace{\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z}
_{\ds{\mbox{Numerically}\ \approx 2.0463}}\ -\
{1 \over 2}\,\ln\pars{2} \approx 0.6766}}
$$


Equations involving arithmetic functions, totatives and even perfect numbers I've deduced simple relationships that satisfy each even perfecf number (even numbers $n$ for which $\sum_{d\mid n}d=2n$) and now I wondered about related conjectures.
For each integer $m\geq 1$ we denote the sum of divisors function $\sum_{d\mid m}d$ as $\sigma(m)$, and the Euler's totient function as $\varphi(m)$.
Claim 1. It's easy to prove that each even perfect number $n$ satisfies $$\sigma(n)=\frac{1}{2}\left(1+8\varphi(n)+\sqrt{1+8n}\right).\tag{1}$$
Claim 2. Thus each even perfect number satisfies also $$4n=1+8\varphi(n)+\sqrt{1+8n}.\tag{2}$$

Question.  I would like to know what work can be done about the following conjectures (prove it or provide us what calculations/reasonings can be done, or refute these finding a counterexample):
C1) If an integer $m\geq 1$ satifies $(1)$, then $m$ is an even perfect number.
C2) Similarly to previous conjecture, each integer $l\geq 1$ satisfying $(2)$ is an even perfect number.
Many thanks.

I've tested that seems there are no counterexamples in my experiments, say us $\leq 10^5$.
I don't know if these equations are in the literature, thus answer as a reference request this question adding the articles where there are information about these equations and I try to search and read those from the literature.

This is not a complete answer, just some comments that are too long to fit in the appropriate section.

From the paper On the Ratio of the Sum of Divisors and Euler's Totient Function I, we have the following inequality:
$$\sum_{p \mid n}{\log\bigg(1+\frac{1}{p}\bigg)} \leq \log\bigg(\frac{\sigma(n)}{\phi(n)}\bigg) = \sum_{p \mid n}{\log\bigg(\frac{p^2 - p^{1 - \nu_p(n)}}{(p-1)^2}\bigg)} \leq 2\sum_{p \mid n}{\log\bigg(1 + \frac{1}{p-1}\bigg)}.$$

From the paper On the Ratio of the Sum of Divisors and Euler's Totient Function II, we have that if $\sigma(N) = {a}\cdot{\phi(N)}$ and $a = 4$, then $\omega(N) = 2$ and $N$ is of the form ${2^{q-2}}{M_q}$ where $M_q = 2^q - 1$ is a prime.

I am guessing that you can use these results for showing that if $n$ satisfies
$$\sigma(n) = \frac{1}{2}\cdot\bigg(1+8\phi(n)+\sqrt{1+8n}\bigg),$$
then $n$ must be an even perfect number.


Locally Bounded Functional Equation $f(x+y) = f(x) + f(y)$ and Continuity Let $f$ be a real-valued function on $\mathbb{R}$ s.t. $f(x+y) = f(x) + f(y)$ for all $x,y$ reals. Suppose there are reals $c$ and $M$ s.t. $|f(x)| \leq M $ for all $x$ in $[-c,c]$. Show that $f$ is continuous. I am able to show that $f$ must take the form $f(x) = xf(1)$ for $x$ rational, but am having trouble showing this holds for the irrationals as well. Hints appreciated!

You can show that $f(0)=0$ using the additivity property of $f$.

Suppose $f$ is not continuous at $0$.
Then there is an $\epsilon>0$ so that for any $\delta>0$, there is an $x_\delta$ satisfying both $|x_\delta|<\delta$ and $|f(x_\delta)|>\epsilon$.

Let $N$ be a positive integer so that $N\epsilon>M$ and take  $\delta=c/N$.  Now choose $x_\delta$ satisfying both $|x_\delta|<\delta$ and $|f(x_\delta)|>\epsilon$.

Then $Nx_\delta$ is in $(-c,c)$ and
$$|f( N {x_\delta })| =N |f(x_\delta )|>N\epsilon>M,$$
a contradiction.

Thus,
$f$ is  continuous at $0$; and, by the additivity of $f$, on $\Bbb R$ as well.

Indeed, this is easily proven using sequences: Let $y\in\Bbb R$ and suppose $y_n\rightarrow y$. Then $y_n-y\rightarrow 0$.  Since $f$ is continuous at $0$, it follows that $f(y_n-y)$ converges to 0. By the graces of the additivity of $f$, it follows that
$f(y_n)$ converges to $f(y)$; whence, $f$ is continuous at $y$.

(Note, now with continuity in hand, you can obtain that $f(\alpha x)=\alpha f(x)$ for all $\alpha$ by considering rationals $q_\alpha$ converging to $\alpha$).

Use the definition of continuity:

$\frac{f(x+\epsilon)-f(x)}{ \epsilon}=\frac{f(\epsilon)}{\epsilon}\leq \frac{M}{\epsilon}  \forall  \epsilon \in [-c,c]$

Update:  $\forall  \epsilon \in [-c,c]-\{0\}$

Update 2:
here is the definition:
$f$ is continuous at $x$ iff $\lim_{\epsilon\to x} \frac{f(x+\epsilon)-f(x)}{ \epsilon} $ is defined.

Update 3: all above is about the differentiability and not continuity; and its a working proof too


If $a,b,c$ are in AP, $b,c,d$ are in GP and $c,d,e$ are in HP, then $a,c,e$ are in ($a\neq b\neq c\neq d\neq e>0\in\mathbb R$) If $a,b,c$ are in AP, $b,c,d$ are in GP and $c,d,e$ are in HP, then $a,c,e$ are in? (HP/GP/AP/None)($a\neq b\neq c\neq d\neq e>0\in\mathbb R$)

So $a+c=2b\tag{1}$
$ce=d^2\tag2$
$\frac1c+\frac1e=\frac2d\tag3$
But I cannot see any pattern between the three numbers. Is there any?

Just write everything in terms of $a$ and $b$:
$$c = 2b -a$$
$$d = \frac{(2b-a)^2}{b}$$
$$\frac{1}{e} = \frac{2}{d} - \frac{1}{c}$$
Therefore, substituting $c$ and $d$ from above:
$$e = \frac{(2b-a)^2}{a}$$

Now it is easy to see that the sequence $a, c, e$ is a G.P:
$$(a,c,e) = \Big(a, 2b-a, \frac{(2b-a)^2}{a}\Big)$$
therefore
$$\sqrt{ae} = c.$$

Solution:
a,b,c are in AP.So,
b=(a+c)/2
or,2b=a+c
or,c=2b-a............(1)
also
b,c,d are in GP.so
c^2=bd
(2b-a)^2÷b=d.............(2)
and
c,d,e are in HP
d=2ce/(c+e)
or,c+e/ce=2/d
or,1/e+1/c=2/d
or,1/e=2/d-1/c..........(3)
from equation 1 2 and  3, we have
1/e=2b/(2b-a)^2 - 1/2b-a
or,e=(2b-a)^2/a......4
now if a,c,e are in GP
c^2=ae
substituting the value of c and e
(2b-a)^2=a.{(2b-a)^2}/a
therefore. (2b-a)^2=(2b-a)^2
so, a,c,e are in GP


logic symbol for 'unlike, differing from' I'd like to express 'unlike, differing from', in the most 'academic/professional' fashion using logic symbol(s) or mathematical operators. The descriptions of the corresponding Unicode block seem to show that there's not one symbol for it.

Maybe it's more of a formal semantic/linguistic issue, if so please let me know.

Example: Unlike set B, set A contains X (or X belongs to A).

There is nothing unprofessional about using words like "unlike" in your mathematical narrative. If you were working on a modal logic with a modality for "likeness" and "unlikeness" you might be justified in adopting a new symbol for it, but otherwise it is best to stick with well-known notation.

Try the following,

$$\forall x
\forall y
( x \neq y \leftrightarrow
\left.
\left(
\neg x \neq x \wedge
\left(
\begin{array}{c}
( y \neq x \rightarrow x \neq y ) \\
\wedge \\
( x \neq y \rightarrow y \neq x ) \\
\end{array}
\right)
\right)
\right)$$

$$\forall x \forall y ( x = y \leftrightarrow \exists z( x = z \wedge z = y ) ) $$

$$\forall x \forall y ( x \neq y \rightarrow ( x = x \wedge y = y ) )$$

$$\forall x ( x = x \rightarrow \neg x \neq x )$$

$$\forall x \forall y ( \neg x = y \rightarrow x \neq y ) $$

In the classical logic of identity, there is an objection to what is called the priniciple of identity of indiscernibles.  A singular term is said to convey "self-identity".  So, "unlikeness" is neither a primitive or a simple negation (if it were, identity would follow from the inability to show that two singular  terms differ).  The second sentence above had been developed by Tarski.  Taken as a definition for a non-eliminable identity, the definiens provides a semantic warrant for uses of the sign of equality.  Because a diversity relation cannot convey "self-identity", a true diversity relation must imply the self-identity of its terms.  If an individual term is self-identical, it ought not be diverse from itself.  And, if an identity statement is false, a corrresponding diversity relation ought to be true.  Note, however, that a conditional is used rather than a biconditional in order to respect the objection to the identity of indiscernibles.

Now, consider the eliminable defined equivalence relation obtained from the set-theoretic primitive,

$$\forall x
\forall y
( x \approx y \leftrightarrow (
\forall z ( x \in z \leftrightarrow y \in z )
\wedge
\forall z ( z \in x \leftrightarrow z \in y ) )$$

If the sentences,

$$\forall x \forall y (
\forall z( x \in z \leftrightarrow y \in z ) \rightarrow \neg x \neq y )$$

$$\forall x \forall y(
\exists z \neg( z \in x \leftrightarrow z \in y )
\rightarrow \neg x = y )$$

$$\forall x \forall y
( \forall z( z \in x \leftrightarrow z \in y )
\rightarrow
\forall z (  x \in z \leftrightarrow y \in z ))$$

are taken as axioms, one can prove

$$\forall x \forall y ( x \approx y \rightarrow x = y )$$

$$\forall x \forall y( \neg x \approx y \rightarrow x \neq y )$$

on the basis of the usual notion of extensionality in set theory as expressed through the third axiom.

However, to finish what has been done here, the sentence

$$\forall x \forall y ( \exists z ( x = z \wedge z = y ) \rightarrow x = y )$$

would have to be introduced into the logical schema with adjustments to the quantifier rules.

Philosophers would object to this on the basis that such logic rules would suggest what they call "contingent identity".  This is fundamentally a metaphysical objection.  So it is important to remember that the situation with set theory is such that the standard axiom of extension is not an eliminiable, defined equivalence.  That axiom is structured specifically to defer questions about identity to the first-order predicate logic.  That set theory is formulated in this way is typically attributed to the influence of Skolem.


Exponential function - with one point in fraction form Question: determine the exponential function when given 2 points: (-1,2) and (3/2,1/2). The examples I've found are with integers, not fractions, and I'm not sure how/if that effects the answer...

One way to do this problem is using the formula $y=ab^x$. Now if $y=2$ then $x=-1$ so that gives $2=ab^{-1}$. And also if $y=0.5$ then $x=1.5$ which gives $0.5=ab^{1.5}$. When you divide first equation by second, the $a$ cancels and you are left with $4=b^{-2.5}$ from which you can find $b$. This can be written as $\frac{1}{b^{2.5}}=4$ or $b^{2.5}=\frac{1}{4}$. Raising both sides to the power $2.5$ gives $b=\frac{1}{32}$. You can now backsub to find $a$


Point inside closed region between intersected circles I have circles (in my example: 3 circles) intersected to each other. There is a closed region between those circles. I have a point K in this closed region. What is the property that fits only for point inside this region. I have one property: Distance between K and each circle centre is greater than the circle radius. However, this property also fits for point outside this closed region, such as K' in my figure. Thanks for any help.

Yes. Distance between point and center should be greater than max of radii, and should be inside the triangle formed by circle centers.


Continuity of conjugate of $z$: $f(z)=\bar z$ How to prove that the function $\;f(z)=\bar z\;$
is continuous on the whole plane?

Hint: $$ |z - z_0| = |\bar{z} - \bar{z_0}|. $$

Then use Weierstrass $\epsilon$-$\delta$ definition of continuity at $z_0$.

   We want to show that for every $\epsilon>0$ there is $\delta>0$ such that if $|z-z_0|<\delta$, then $|f(z)-f(z_0)| = |\bar{z}-\bar{z_0}| = |z-z_0| < \epsilon$. We can take $\delta$ equal to $\epsilon$ itself.

$$f(z+\delta)=\overline{z+\delta}=\overline z+\overline\delta=f(z)+\epsilon.$$
It suffices to set
$$\overline\delta=\epsilon.$$


Algebra with finite summations Let $X$ and $Y$ be two random variables that can take on $n$ values.

How do I show: $$\frac {n\sum X_{i}Y_{i}-\sum X_{i} \sum Y_{i}}{n\sum X_{i}^2-(\sum X_{i})^2}=\frac{\sum (X_{i}-\bar X)(Y_{i}-\bar Y)}{\sum (X_{i}-\bar X)^2}$$

With $\bar X$ and $\bar Y$ the means of the random variables $X$ and $Y$ respectively.

Using the formulas that Daniel McLaury indicated:

$$\frac{E(XY)-E(X)E(Y)}{E(X^2)-E(X)E(X)}=\frac{\sum X_{i}Y_{i}P(X_{i},Y_{i})-\sum X_{i}f(X_{i})\sum Y_{i}g(Y_{i})}{\sum X_{i}^2 f(X_{i})-\sum X_{i}f(X_{i})\sum X_{i}f(X_{i})}$$

With $f(X)$ the PDF of $X$ and $g(X)$ the PDF of $Y$.
Since the equation only holds for uniformly distributed random variables:

$$=\frac {\frac{1}{n}\sum X_{i}Y_{i}-\frac{1}{n^2}\sum X_{i}\sum Y_{i}}{\frac{1}{n}\sum X_{i}^2-\frac{1}{n^2}\sum X_{i}\sum X_{i}}$$

We multiply numerator and denominator by $n^2$ to get the given equation:

$$=\frac {n\sum X_{i}Y_{i}-\sum X_{i} \sum Y_{i}}{n\sum X_{i}^2-(\sum X_{i})^2}$$

$$\therefore \frac{E(XY)-E(X)E(Y)}{E(X^2)-E(X)E(X)}=\frac {n\sum X_{i}Y_{i}-\sum X_{i} \sum Y_{i}}{n\sum X_{i}^2-(\sum X_{i})^2}=\frac {Cov(X,Y)}{Var(X)}=\frac {\sum (X_{i}-\bar X)(Y_{i}-\bar Y) P(X_{i},Y_{i})}{\sum (X_{i}-\bar X)^2f(X_{i})}=\frac {\frac{1}{n}\sum (X_{i}-\bar X)(Y_{i}-\bar Y)}{\frac{1}{n}\sum (X_{i}-\bar X)^2}=\frac{\sum (X_{i}-\bar X)(Y_{i}-\bar Y)}{\sum (X_{i}-\bar X)^2}$$

Edit: as far as I know, $E(XY)=\sum \sum X_{i}Y_{j}P(X_{i},Y_{j})$, so either I'm missing something or the study book put a mistaken $i$ subscript for Y in the left-hand side of the numerator.

Edit 2: For every $X_{i}$ there is a unique $Y_{i}$ associated. The variables are dependent, we cannot include cases $X_{i}Y_{j}$ where $i\ne j$ (we could, but the associated intersection probability would be 0 for all cases).

Up to some power of $n$, the numerators on each side are the covariance of $X$ and $Y$ and the denominators on both sides are the variance of $X$.  The equivalence is a result of the famous formula
$$Var[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$$
and the analogous result for variances.

This is something you want to be able to spot in general; it's maybe the most useful formula in all of statistics and probability.


When do the multiples of two primes span all large enough natural numbers? It is well-known that given two primes $p$ and $q$, $pZ + qZ = Z$  where $Z$ stands for all integers.
It seems to me that the set of natural number multiples, i.e. $pN + qN$ also span all natural numbers that are large enough. That is, there exists some $K>0$, such that
$$pN + qN = [K,K+1,...).$$

My question is, given $p$ and $q$, can we get a upper bound on $K$?

$K = pq + 1$ if $\mathbb{N} = \{ 1, 2, 3, ... \}$, and $K = pq - p - q + 1$ if $\mathbb{N} = \{ 0, 1, 2, 3, ... \}$.  This is known as the coin problem, or Frobenius problem (and you only need $p, q$ relatively prime).  It frequently appears on middle- and high-school math competitions.

Edit:  I completely misremembered how hard the proof is.  Here it is.  If $n$ is at least $pq+1$, then the positive integers

$$n-p, n-2p, ... n-qp$$

have distinct residue classes $\bmod q$, so one of them must be divisible by $q$.  On the other hand, it's not hard to see that $pq$ itself cannot be written in the desired way.

This follows easily by Bézout's lemma (as Qiaochu notes, x and y only need to be coprime to generate the unit ideal (indeed, this is the proper generalization of the term "coprime")).

We can ever so-slightly strengthen this to show that $\mathbf{Z}$ is a PID (this is the true content of Bézout's lemma).


The n-th prime is less than $n^2$? Let $p_n$ be the n-th prime number, e.g. $p_1=2,p_2=3,p_3=5$. How do I show that for all $n>1$, $p_n<n^2$?

In Zagier's the first 50 million prime numbers a very elementary proof is given that for $n > 200$ we have

$$\pi(n) \ge \frac23 \frac{n}{\log n}$$

where $\pi(x)$ is the number of primes below $x$ (as well as a bound in the other direction). In fact, it already holds for $n \ge 3$, as can be directly checked.

Suppose that $p_n > n^2$, then for this $n$ we have $\pi(n^2) < n$, but that violates the bound already for $n = 5$.

You may use Bertrand's Postulate, which states that, for any natural number $n>1$, there exists a prime number $p$ such that $n<p<2n$; and induction.

First note that the result does not hold for $p_1=2$. However it does hold that $p_2=3<2^2=4$.

Now following an induction argument let's assume that for some fixed $n>1$ it holds that $p_n < 2^n$. Then by Bertrand's Postulate clearly $p_n < p_{n+1} < 2p_n$. So $p_{n+1}<2p_n < 2^{n+1}$.


What is the largest interval containing $0$ so that $f(x)=(x+1)^2$ has an inverse? What is the largest possible interval containing $0$ such that $f$ has an inverse on that interval, where $f(x)=(x+1)^2$?

I have tried to find the inverse and the derivative.

$f(x)$ is $0$ when $x=-1$.

$f$ derivative is $0$ when $x=-2$; is this the local minimum?

$f$ inverse is $0$  when $x=1$; is this the local maximum?

For a function to admit a restricted inverse, one has to check injectivity (if a function is not injective in $(a,b)$ it for sure does not admit an inverse function in that interval).

Now, check that the function you gave is injective in the interval $(-\infty,-1]$ (for example check that the sign of the derivative there is always negative, therefore the function is decreasing, hence injective). Analogously, it is injective in $[-1,\infty)$ (change a bit the previous argument to conclude so).

Then, if I had to choose the “largest” interval containing zero for which f admits an inverse I’d go with $[-1,\infty)$, which geometrically corresponds to the right “half” of the parabola.

This is a parabola centered at $(x,y)=(-1,0)$. Notice that both branches are separately invertible. Consider the interval $[-1,\infty)$. This contains $0$ and $f$ is invertible. For any other interval, say, $[a,\infty)$ with $a<-1,$ the function is not one to-one.


If $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable. In all the books that I've checked the use of roots of unity (as hypothesis) is very crucial to prove that if $f(x)\in F[x]$ is solvable by radicals then its Galois group $Gal(E/F)$ is solvable, but the professor said that it's not needed that the field must have enough  roots of unity. I think this is curious, and I want to prove it. She also gave three steps to prove this and I'm struggling with them.

  (1) Let $F=B_0\subseteq B_1\subseteq\dots\subseteq B_t$ a radical extension with $E\subseteq B_t$ and $B_t$ the splitting field of $f(x)\in F[x]$ with $deg(f(x))=m$. We defined $\mathcal E$ as the splitting field of $h(x)=x^m-1$ over $E$ and $\mathcal F=F(\Psi)$, where $\Psi$ are all the $m$th roots of unity. Prove that $\mathcal E$ is the splitting field of $f(x)$ over $\mathcal F$, and there for $Gal(\mathcal E/\mathcal F)$ is solvable.

I don't see why do we need that function $h(x)$ if what we are trying to do is to take the roots of unity out. Also, for $\mathcal E,\mathcal F$ we are basically adding the same elements, $\Psi$, to $E,F$, but by hypothesis we already have that $E=F(\alpha_1,\dots,\alpha_m)$ with $\alpha_i$ the roots of $f(x)$, and $\mathcal E= E(\Psi)=F(\alpha_1,\dots,\alpha_m,\xi_1,\dots,\xi_m)$ so if anything we are adding things and $f(x)$ can still get decomposed into linear factors, so what are we suppose to do?

  (2) Prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)\cong Gal(\mathcal F/F)$ and $Gal(\mathcal E/F)$ is solvable.

To prove that $Gal(\mathcal E/\mathcal F)\unlhd Gal(\mathcal E/ F)$, we show that $\mathcal F/F$ is a Galois extension: lets take $f(x)\in F[x]$, then by definition all the roots of $f(x)$ are in $\mathcal F$, then it is a splitting field, hence $\mathcal F/F$ is a Galois extension.
Now we can talk about the quotient $Gal(\mathcal E/ F)/Gal(\mathcal E/\mathcal F)$ and $Gal(\mathcal F/F)$, but I can't find the right isomorphism to fit this, and I'm starting to think that maybe this isn't right.
 And to prove $Gal(\mathcal E/F)$ is solvable, I wanted to prove that there exists a radical extension, but to use that I would need that $F$ is of characteristic $0$, and I don't have as a hypothesis.

  (3) Prove that $Gal(\mathcal E/E) \unlhd Gal(\mathcal E/F)$ and $Gal(\mathcal E/ F)/Gal(\mathcal E/ E)\cong Gal(E/F)$, and conclude that $Gal(E/F)$ is solvable.

The prove is pretty much the same as before, isn't it?

(p.s. sorry for the long post)

EDIT: I think I came up with te proof for (1). We know that $\alpha_i\in\mathcal E$ and $\mathcal F\subseteq\mathcal E$ then by the definition of $\mathcal E$ it is the splitting field of $f(x)$ over $\mathcal F$. Now lets take the tower of fields: $\mathcal F=B_0(\Psi)\subseteq B_1(\Psi)\subseteq\dots\subseteq B_t(\Psi)$ with $\mathcal E\subseteq B_t(\Psi)$ then it is a radical extension and $\mathcal E/\mathcal F$ is Galois, where $\mathcal F$ contains all the roots of unity, then $Gal(\mathcal E/\mathcal F)$ is solvable

For (2),(3) I still can't figure oit how to prove the isomorphism.

"Exercise 5.27 on page 468 shows how to eliminate this hypothesis" on Rotman's "A First Course in Abstract Algebra"
http://www.math.uiuc.edu/~kapovich/417-05/book.pdf
And with Diana Avella, you can assume that char(F) is not [Bi+1 : Bi] for all i.


Prove or disprove: There exists $T \colon \mathrm M_2(\mathbb R) \to\mathbb{R}^3$ which is 1-1. Prove or disprove the claim:
  There exists $T \colon \mathrm M_2(\mathbb R) \to\mathbb{R}^3$ which is 1-1.

I have no clue how to start this problem. We only learned in class about determining whether a defined linear transformation is 1-1 or onto.

Since you tagged your question with linear-algebra, I am assuming you want $T$ to be linear.

You get from the linearity of $T$ that $\mathrm {Im}(T)$ is a vector space of $\mathbb R^3$, and since $T$ is 1-1, you have

$$4=\dim \mathrm M_2(\mathbb R)=\dim \mathrm{Im} (T).$$

Which is a contradiction since $\dim \mathbb R^3=3$.

Thus, such a $T$ can not exist.

The dimension of $\mathcal M_2(\Bbb R)$ is 4 so by the rank nullity theorem

$$\dim \ker T=\dim \mathcal M_2(\Bbb R)- rank T\ge 4-3=1$$
so $T$ can't be 1-1.


Differential equation with separable, probably wrong answer in book I have a differential equation:

$$ \frac{dy}{dx} = y \log(y)\cot(x)$$

I'm trying solve that equation by separating variables and dividing by $y\log(y)$:

$$ dy = y \log(y) \cot(x) dx$$

$$ \frac{dy}{y \log(y)} = \cot(x) dx$$

$$   \cot(x) -  \frac{dy}{y \log(y)} = 0   $$

Where of course $ y > 0 $ regarding to division

Beacuse:

$$ \int \frac{dy}{y \log(y)} = \ln | \ln(y) | +C $$

and:

$$ \int \cot(x) dx  = \ln| \sin(x) | + C$$

So:

$$ \ln| \sin(x) | - \ln | \ln(y) | = C $$

$$ \ln \lvert\frac{\sin(x)} {\ln(y)} \rvert  = C $$

$$ \frac{\sin(x)}{\ln(y)} = \pm e^{C} $$

$d = \pm e^{C} $

$$ \sin(x) = d  \ln(y) $$

$$ \frac{\sin(x)}{d} = \ln(y)$$

$$ e^{\frac{\sin(x)}{d}} = y$$

This is my final answer. I have problem because in book from equation comes the answer to exercise is:

$$ y = e^{c \sin(x)}$$

Which one is correct?

I will be grateful for explaining Best regards

Note that
$$ \int \frac{du}{u} = \ln |u| + c_1 = \ln |u| + \ln |c| = \ln |cu|.$$

Starting from your result:

$$ \cot x\, dx =\frac{dy}{y \ln  y} $$
$$\implies \frac{d(\sin x)}{\sin x} = \frac{d(\ln y)}{y \ln  y}$$

we integrate both sides:

$$ \int \frac{d(\sin x)}{\sin x} \, dx = \int \frac{d(\ln y)}{\ln  y}\, dx$$
$$\implies \ln |a\sin x| = \ln |b\ln  y| $$
$$\implies a\sin x = b\ln  y, \quad(1)$$
$$\implies \ln  y = c\sin x$$
$$\implies y = e^{c\sin x}$$
where $a, b, c$ are arbitrary constants. Note that $(1)$ includes all possible solutions including the case $a=0$.

NOTE:
$$\ln|\frac{\sin x}{\ln y}|=C$$
Can also be written as
$$\ln|\frac{\sin x}{\ln y}|=\ln C$$
(Because both terms are constant.)

This shall give you $$\sin x =C \ln y$$
Which will give you:
$$y=e^{c \sin x}$$ where $c=\frac{1}{C}$. I will prefer this solution because in your answer d=0 won't be giving you any solution but Note that c=0 will do. {Credits to that comment}


What's wrong in my example about Q-matrix? I find the next property about Q-matrices:
$$ \text{Let Q a Q-Matrix, then}
\frac{d^k P(0)}{dt^k}=Q^k
 ; \text{ for }  k=0,1,2,\cdots.$$ I was trying to verify the property with the next example:
$Q=\begin{pmatrix}
-1 & 1 \\
1 & -1
\end{pmatrix}$ then
$$P(s)=\begin{pmatrix}
e^{-s} & e^s \\
e^s & e^{-s}
\end{pmatrix}
$$
$$\frac{d^k P(0)}{ds^k}=\begin{pmatrix}
(-1)^ke^{-0} & e^0 \\
e^0 & (-1)^ke^{-0}
\end{pmatrix}=\begin{pmatrix}
(-1)^k & 1 \\
1 & (-1)^k
\end{pmatrix}\neq Q^k$$
But it contradicts the proposition!! I want to know what is wrong in my example or interpretation.

If $P(s)$ and $Q$ are as above, then we have

$\frac{d^k P(0)}{ds^k}=Q^k$ only for $k=1,3,5,...$.


Why are all non-prime numbers divisible by a prime number? In Euclid's infinite prime numbers proof, the logic is as follows:

Assume a set $S$ of all prime numbers in existence is finite (there are a finite amount of primes)

Then there must be a greatest prime $p$

$$n = (2 \cdot 3 \cdot 5\cdots p) + 1$$

$n > p$, and under the proof's assumption, $n$ cannot be prime.*

This is where the logic confuses me. Why is it that given that the if a number is not prime, then it is automatically divisible by a prime. I can't think of an example to contradict, but that's not proof that there exists no number that is not prime and non divisible by primes.

One can prove this for all integers greater than $1$ by induction: we know that $2$ is a prime. Now for ou inductive step assume that for all $i<n$, $i$ is either prime or divisible by a prime. Case 1: $n$ is prime; we're done. Case 2: $n$ is composite, so $ab = n$ for $a, b < n$. So each of those is divisible by a prime. We're done.

Essentially the same argument shows that all integers greater than $1$ can be written as a product of primes.

Let $n$ be a composite (i.e., non-prime) integer.

According to the Fundamental Theorem of Arithmetic, $n$ has a unique prime factorization:

$$ n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$$

where the $p_{i}$ are distinct prime numbers and the $\alpha_{i}$ are positive integers.

Choose any of the $p_{i}$, say $p_{j}$.

Now, $p_{j} | p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, and so,

$$p_{j}|n.$$

Therefore, $n$ is divisible by a prime number. (Actually $k$ of them in this case.)


Can we prove that axioms do not contradict? We construct many structures by chosing a set of axioms and deriving everything else from them. As far as I remember we never proved in our lectures that those axioms do not contradict. So:

Is it always possible to prove that a set of axioms does not contain a contradiction? (So if not could we perhaps find a contradiction in a structure that we definde by axioms someday?) Or is it proven that doing so is impossible?

Proved from what?

Even logic is formalized in a mathematical fashion, so it has a mathematical backbone. We need to first assume that first-order logic does not include a contradiction. Then we need to decide on which system we are going to use to develop our mathematics. Arithmetical systems will suffice for syntactical manipulations of strings, and set theoretical systems will suffice for semantical proofs using models.

Whatever meta-theory, in which we formalized the logic and the proofs and whatnot, you are working in, if it is sufficiently strong, it will not prove its own consistency. This means that along with the assumption that first-order logic does not contain a contradiction, we implicitly assume that our meta-theory is consistent, whatever it might be. And the more we need, the more consistency we will have to prove.

Goedel proved that a theory which is both consistent, recursively enumerable, and interprets arithmetics, cannot prove its own consistency (and is therefore not complete). This means that we really have to assume the consistency of our meta-theory (which is usually a theory satisfying these conditions).

Of course, we can prove the consistency of arithmetic, using set theory, and the consistency of set theory using an even stronger set theory, and so on. But that's not what you're looking for in an answer. The answer, simply, is that at some point, you have to take some things for granted. The question is how much you are willing to take for granted, how much logic and foundational theories you are familiar with, to transform an argument form one system to another.

You can prove contradiction by constructing results which stem from them that contradict.  Right? If A leads to Z and B leads to NOT Z then something is obviously fishy.  Or equally showing that C leads to both Y and NOT Y simultaneously.

The inability to show contradiction does not prove anything.  Indeed even the ability to demonstrate consistency in any given set of logic does not prove that the logic is valid.

Logics are constructed by logicians that are self-consistent but contradict other forms of logics, each based on its own set of axioms.   Thus it is unfair and even irrational to judge one train of thought based in one set of logical principles against that of another. A logic can only be disproved through self-inconsistency, or inconsistency with the real world if you want to go there, but not through inconsistency between "worlds" of logic.

Even if you do show that a premise results in a contradiction. Does that necessarily prove that the premise is false? Ultimately though, and I think you have failed to realize this (many people do), but the Law of Non-Contradiction that prohibits contradiction in logic is itself an axiom.

Id love to ask a question. What should we do when mathematics or logic contradicts reality?

I want to point out the fact that reality has already contradicted mathematics and logic.  Take Euclidean geometry, for example. It makes the assumption that space is flat (Euclidean).  Most of modern mathematics is based atop of these fundamental assumptions about geometry, including the all-important calculus. All of modern physics, too.  Except one day Einstein comes along and shows that real space is curved. What can we conclude but all of the mathematics we've been using all along is built on flat-space assumptions and is faulty at the fundamental level - at least to the real world we apply it in.  How do you rectify that?  There is the good-old "accurate enough" argument, and of course we have invented other forms of geometry, but... convolution is the key to success.

The law of non-contradiction is pretty steadfast in logic and mathematics. But in quantum physics - the reality from which the entire universe is built - allows a certain degree of contradictory truths. And gad-zooks! The law of non-contradiction might actually be fallible or non-applicable in certain contexts of reasoning.

If you can appreciate any of this from a pure mathematical or pure logical perspective, or the significance it has on physics... just imagine the implications in philosophy, religion, morality.


Does the Definition of a Formal Proof in Mendelson's Book Allow for the Use of Theorem/Derived Axiom Schema in Formal Proofs? The relevance of this question concerns deciding whether or not a sequence of well-formed formulas in a text will qualify as a formal proof.

Mendelson's Introduction to Mathematical Logic on p. 25 of the 5th edition reads when defining a formal system:

"There is a finite set R$_1$, ..., R$_n$ of relations among wfs, called rules of inference.  For each R$_i$, there is a unique positive integer j such that, for every set of j wfs and each wf B, one can effectively decide whether the given j wfs are in the relation R$_i$ to B, and, if so, B is said to follow from or to be a direct consequence of the given wfs by virtue of R$_i$.

A proof in L is a sequence B$_1$, ..., B$_k$ of wfs such that, for each i, either B$_i$ is an axiom of L or B$_i$ is a direct consequence of some of the preceding wfs in the sequence by virtue of one of the rules of inference of L.

A theorem of L is a wf B of L such that B is the last wf of some proof in L."

He then sets up a formal system, and describes a few others in the exercises, though those details don't matter to this question.  One of his references is S. C. Kleene's Introduction to MetaMatheamtics where on p. 82 Kleene writes:

"The relation of 'immediate consequence' is defined thus.  A formula is an immediate consequence of one or two other formulas, if it has the form shown below the line, while the other(s) have the form(s) shown above the line in 2, 9, 12."

2, in context, states a rule of modus ponens.  9 states a one premise universal quantifier rule, and 12 a one premise existential quantifier rule.  Kleene then explains his definition of a formal proof more in detail partially by referring to the notion of an immediate consequence.

The phrases 'immediate consequence' and 'direct consequence' both suggest to me that a substitution instance of a formal theorem would qualify as a formal theorem also, but re-reading those definitions suggests that I may have read into things.  That said though, other authors like Prior, Bergmann, Lukasiewicz, and papers I've seen, suggest that substitutions in theorems in axiomatic systems coheres with the definition of a formal proof (and soundness works out).  But, there's also something called The Pushback Lemma which sounds like it says that for any substitution and detachment proof where we have substitutions in the formal theorems, there can get found from that formal proof a formal proof which only has substitutions in the primitive axioms or first axiom schemata.

So, does Mendelson's definition forbid substitutions in theorems in formal proofs, or does it allow substitutions in a formal theorem?

By the very definition you gave, clearly a proof can only use the given rules of inference and hence cannot invoke theorems in the meta-system. A common meta-theorem is the deduction theorem, namely that $A \vdash B$ if and only if $\vdash A \to B$, which a proof in $L$ is not allowed to use. Similarly a proof in $L$ cannot invoke theorem schemas by definition. For instance you can prove that, given any binary predicate symbol $P$, if $L \vdash \exists x\ \forall y\ ( P(x,y) )$ then $L \vdash \forall y\ \exists x\ ( P(x,y) )$. But you cannot use it in a proof.
$
\def\prov{\square}
$

In general, unless the formal system $L$ includes some kind of meta-logic that allows you to use meta-theorems, a proof is literally a sequence of deductions using only the inference rules. It is dangerous to have too strong internal meta-logic, because if $L$ has decidable proof validity and interprets arithmetic then:

  If $L \vdash A$ then $L \vdash \prov_L A$, for every sentence $A$ over $L$.

But if $L$ is $Σ_1$-sound then it cannot have:

  (INVALID) $L \vdash A \to \square_L A$, for every sentence $A$ over $L$.

This is because any such $L$ has 'internal completeness' in the sense that, given any sentence $A$ over $L$, we have that $L$ proves "$\prov A \lor \prov \neg A$", which is a $Σ_1$-sentence and hence must be true for $\mathbb{N}$, and the witness can be decoded into a proof of either "$A$" or "$\neg A$" over $L$, which contradicts Godel's first incompleteness theorem.

It forbids substitutions in theorems in formal proofs (NO formal proofs exist in the book... not a single one).  But, that is not generally how the axiomatic method has gotten used historically speaking.


Can anyone give me a hint how to start this problem? Let $X$ be a random variable such that $P(X \leq 0)=0$ and let $\mu =E(X)$ exist. Show that $P(X \geq 2\mu) ≤ \frac{1}{2}$. I don't really know how to start this one... so any hint will be appreciated.

Hint: For all $t \geq 0$ the inequality $X \geq t\,1_{X \geq t}$ holds almost surely.

$P(X \geq 2\mu) = P\left(z \geq \dfrac{2\mu-\mu}{\sigma}\right)= P\left(z\geq \dfrac{\mu}{\sigma}\right)$, but $P(X \leq 0) = 0 \Rightarrow P\left(z \leq \dfrac{0-\mu}{\sigma}\right) = 0 \Rightarrow P\left(z \geq \dfrac{\mu}{\sigma}\right) = P\left(z \leq -\dfrac{\mu}{\sigma}\right) = 0 \leq \dfrac{1}{2}$.


Prove if f is positive and increasing on $[a, b]$ then $L_n ≤ A ≤ R_n.$ (riemann sum) Prove if f is positive and increasing on $[a, b]$ then for all $n\ge 0$ we have $L_n \le A \le R_n$. (Riemann sum)

Let $A$ denote the actual area.

Let $L_n$ denote the left Riemann sum.

Let $R_n$ denote the right Riemann sum.

So far what I did:

$$\int_{a}^{b} f(a) dx \leq \int_{a}^{b} f(x) dx \leq \int_{a}^{b} f(b)$$

But after this I am stuck on the proof?

Just write the definitions

  $$L_n = \sum_{i=0}^{n-1}f(x_i){b-a\over n}\le \sup_m L_m = A= \inf_m R_m \le \sum_{i=1}^n f(x_i){b-a\over n}$$

here the $\sup=\limsup$ by monotonicity of $f$.

and if you're not familiar with the $\sup$ notation, then try this

$$L_n =\sum_{i=0}^{n-1}f(x_i){b-a\over n}=\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}f(x_i)\,dx \le \sum_{i=1}^{n-1}\int_{x_i}^{x_{i+1}}f(x)\,dx $$
$$= \int_a^b f(x)\,dx$$
$$=\sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x)\le \sum_{i=1}^n\int_{x_{i-1}}^{x_i}f(x_i)\,dx =\sum_{i=1}^nf(x_i){b-a\over n}=R_n$$

For partition $x_0<x_1<\cdots<x_n$

Left Riemann sum: $ = \sum_\limits{i=1}^{n} f(x_{i-1})(x_{i}-x_{i-1})$

Right Riemann sum: $ = \sum_\limits{i=1}^{n} f(x_i)(x_i-x_{i-1})$


What is $\frac{1}{\infty}$? Ok. When I say "infinity," I mean an infinitely large number (such as $9999999\ldots$)

So we all know the common proof for $0.999\ldots = 1$.

If not, here it is:

$$
\begin{align*}
x &= 0.999\ldots \\
10x &= 9.999\ldots \\
10x-x &= 9x \\
9.999\ldots - 0.999\ldots &= 9
\end{align*}
$$

Therefore, $9x = 9, x = 1$.

Using this logic, it can be proved that $0.000\ldots1 = 0$ because

$$
\begin{align*}
1 - 1 &= 0 \\
1 - 0.999\ldots &= 0.000\ldots1
\end{align*}
$$

and since $1 = 0.999\ldots$, that means that $0 = 0.000\ldots1$.

Now, if we were to take any number and divide it by an infinitely large number, then the answer would eventually consist of $0.000000\ldots1$ (or at least have an infinite series of zeros before a number other than one). Since $0.000\ldots1 = 0$, this must mean that $\frac{1}{\infty} = 0$.

Is this correct?

(Even though the mathematics is incorrect, +1 for a well-asked question!)

Your core misunderstanding seems to be the definition of $0.999\dots$. By definition, that number equals $\frac9{10} + \frac9{10^2} + \frac9{10^3} + \cdots$, an infinite series (which equals $1$, by the proof you gave). But my point is that it has a specific mathematical definition; it's not just "notation for an idea".

On the other hand, you seem to be thinking of $0.000\dots01$ as an infinite string of $0$s followed by a $1$. Such an object does not have any mathematical meaning, and so reasoning with it will not be mathematically valid. Alternatively, you might be thinking of that expression as "what happens when I take $0.000\dots01$, with a finite number (call it $n$) of $0$s, and let $n$ get larger and larger". While I wouldn't use that notation for this idea, the idea itself is perfectly valid and leads to the true mathematical fact
$$
\lim_{n\to\infty} \frac1{10^n} = 0.
$$
In either case, your equation $1 - 0.999\dots = 0.000\dots1$ is not correct, because the number $0.999\dots$ does not have any last $9$ that would yield a $1$ after the subtraction.

We don't know, and it's depending on quite a lot of factors, too. Like, what infinity are you using? Bear in mind that $\lim\limits_{n\to \infty}n$ results into $\infty$ but $\lim\limits_{n \to \infty}n^2$ does not equal  the vanilla $n$-variant, and both get dirrefent answers. Also, you cannot get rational results using infinities. We can get close, though, using limits:
$$\lim\limits_{x \to \infty} \frac{1}{x} = 0$$
That basicly means that if we get $x$ reeaally big, $\frac{1}{x}$ will approach 0. We can say it won't actually reach 0 though, not even when $x = \infty$. Otherwise we'd get $\frac{1}{\infty} = 0$ resulting in both $1 = \infty * 0$ and $\frac{1}{0} = \infty$, which both are incorrect statements (the first one because $a = a * 1 = a * 0 * \infty = 0 * \infty = 1$, which is incorrect, and the second one because we don't divide by $0$). Furthermore, the "mistake" made by the proof that $0.999... = 1$ is that they don't use limits (for further explanation on that, look at the question I asked earlier on my profile, lots of answers there), and, you should keep in mind that you don't know if $\infty = 10^x$, so $\frac{1}{\infty}$ does not necessarily have to be $0.000...01$. That should answer most of your questions.


$A$ can walk certain distance in $20$ days when he rests $6$ hours a day Problem : $A$ can walk certain distance in $20$ days when he rests $6$ hours a day. How long will he take to walk the distance twice as fats and resting twice as long each day ?

Solution :  let distance be $x$

time taken without rest  $=20 -5 =15$ days

speed =$ \frac {x}{15}$

new speed = $ \frac {2x}{15}$

distance =  $x$

time  taken without rest=  $ \frac {15x}{2x}$

=7.5

time taken with rest $=12 *7 =84$ hours

$3.5$days

total time taken $=7.5 +3.5=11$ days

But this answer is not correct.

Help me

When he rests $6$ hours a day, that would mean he would walk $24-6=18$ hours a day, or $\frac{18}{24}=\frac{3}{4}$ days everyday. He does this for $20$ days, so the total time he walks would be $20 \times \frac{3}{4} = 15$ days.

If the distance he walks is $x$, his speed would be $\frac{x}{15}$.

Then his new speed is $\frac{x}{15}\times{2}=\frac{2x}{15}$

He walks for the same distance, so the product between the new speed and the new time would still be $x$: (remember: $\frac{\text{distance}}{\text{time}}=\text{speed}$)

$$
\frac{x}{t}=\frac{2x}{15} \\x \times 15 = 2x \times t \\ t = \frac{15x}{2x}=\frac{15}{2}
$$
This means it would take him $\frac{15}{2}$ days of nonstop walking to walk the same distance.

But he rests twice more than he did before, so he would walk $24-6\times2 = 12$ hours a day, or $\frac{12}{24}=\frac 1 2$ day everyday. That is, he would walk $\frac{1}{2}$ of some number of days, and at the end, he would have walked for $\frac{15}{2}$ days. Then:

$$
\text{Total days} \times \frac{1}{2} = \frac{15}{2} \\
\text{Total days} = 15 \text{ days}
$$

Hope this helps.

20 x 18 x 1(let speed is 1km/hr)  x 2(distance twice) = D(Number of days) x 12 x 1(speed)

D = 60 days


How to prove any $N \times N$ matrix $M$, satisfying $M^2 = \lambda^2 I_{N\times N}$, $\lambda\neq 0$ is diagonalizable. I want to prove following things.

Any $N \times N$ matrix $M$, satisfying $M^2 = \lambda^2 I_{N\times N}$, with $\lambda\neq 0$ is diagonalizable.

Can you give me some hint or proof of this?

My trail starts from determinant
\begin{align}
0=det(M^2 - \lambda^2 I) = det(M - \lambda I) det(M+ \lambda I)
\end{align}
thus i conclude that either $det(M - \lambda I)=0$ or $ det(M+ \lambda I)=0$.
But i don't know how to go further..

As $M^2$ is in diagonal form, then we can write it's diagonalization $M^2=I(\lambda^2 I)I=Q\Lambda Q^{-1}$. Then we can write $M^2 = I(\lambda I)(\lambda I)I=Q\sqrt{\Lambda}\sqrt{\Lambda}Q^{-1}$. Then we can also write $M^2=Q\sqrt{\Lambda}Q^{-1}Q\sqrt{\Lambda}Q^{-1} = I(\lambda I)II(\lambda I)I=MM$. So from this derivation we see that $M=\lambda I$ with possible eigenvalues $\pm \lambda$.


$1$ man can eat $1$ apple in $1$ day.How many apples can six men eat in six days? $1$ man can eat $1$ apple in $1$ day.How many apples can six men eat in six days?

My approach:

$1$ men can eat $1$ apple in $1$ day.

Now, 6 men can eat in 6 days 6 apples only assuming each men eat $1$ apple a day.

Am I right in my approach?

no. 1 man can eat 1 apple in 1 day therefor 1 man can eat 6 apples in 6 days. therefor 6 men can eat 36 apples in 6 days....

The following piece by Lewis Carroll first appeared in ``The Monthly
Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_,
edited by John Fisher, Bramhall House, 1973.

                             Cats and Rats

If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100
   rats in 50 minutes?

This is a good example of a phenomenon that often occurs in working
   problems in double proportion; the answer looks all right at first, but,
   when we come to test it, we find that, owing to peculiar circumstances in
   the case, the solution is either impossible or else indefinite, and needing
   further data.  The 'peculiar circumstance' here is that fractional cats or
   rats are excluded from consideration, and in consequence of this the
   solution is, as we shall see, indefinite.

The solution, by the ordinary rules of Double Proportion, is as follows:

   6 rats   :  100 rats  \
                          >   :: 6 cats : ans.
  50 min.   :    6 min.  /

Then:  ans = (100)(6)(6)/(50)(6) = 12

But when we come to trace the history of this sanguinary scene through all
   its horrid details, we find that at the end of 48 minutes 96 rats are dead,
   and that there remain 4 live rats and 2 minutes to kill them in: the
   question is, can this be done?

Now there are at least four different ways in which the original feat,
   of 6 cats killing 6 rats in 6 minutes, may be achieved.  For the sake of
   clearness let us tabulate them:
      A.  All 6 cats are needed to kill a rat; and this they do in one minute,
          the other rats standing meekly by, waiting for their turn.
      B.  3 cats are needed to kill a rat, and they do it in 2 minutes.
      C.  2 cats are needed, and do it in 3 minutes.
      D.  Each cat kills a rat all by itself, and take 6 minutes to do it.

In cases A and B it is clear that the 12 cats (who are assumed to come
   quite fresh from their 48 minutes of slaughter) can finish the affair in
   the required time; but, in case C, it can only be done by supposing that 2
   cats could kill two-thirds of a rat in 2 minutes; and in case D, by
   supposing that a cat could kill one-third of a rat in two minutes.  Neither
   supposition is warranted by the data; nor could the fractional rats (even
   if endowed with equal vitality) be fairly assigned to the different cats.
   For my part, if I were a cat in case D, and did not find my claws in good
   working order, I should certainly prefer to have my one-third-rat cut off
   from the tail end.

In cases C and D, then, it is clear that we must provide extra cat-power.
   In case C less than 2 extra cats would be of no use.  If 2 were supplied,
   and if they began killing their 4 rats at the beginning of the time, they
   would finish them in 12 minutes, and have 36 minutes to spare, during which
   they might weep, like Alexander, because there were not 12 more rats to
   kill.  In case D, one extra cat would suffice; it would kill its 4 rats in
   24 minutes, and have 24 minutes to spare, during which it could have killed
   another 4.  But in neither case could any use be made of the last 2
   minutes, except to half-kill rats---a barbarity we need not take into
   consideration.

To sum up our results.  If the 6 cats kill the 6 rats by method A or B,
   the answer is 12; if by method C, 14; if by method D, 13.

This, then, is an instance of a solution made indefinite by the
   circumstances of the case. If an instance of the impossible be desired,
   take the following: If a cat can kill a rat in a minute, how many would be
   needed to kill it in the thousandth part of a second?  The mathematical
   answer, of course, is 60,000, and no doubt less than this would not
   suffice; but would 60,000 suffice?  I doubt it very much.  I fancy that at
   least 50,000 of the cats would never even see the rat, or have any idea of
   what was going on.

Or take this: `If a cat can kill a rat in a minute, how long would it be
   killing 60,000 rats?'  Ah, how long, indeed!  My private opinion is that
   the rats would kill the cat.


How do I write x in terms of y in a quartic equation I have a few quartic equations that are in the general form, $$y = ax^4 + bx^3 + cx^2 + dx + e$$Is there any way I can write $x$ in terms of $y$?

For example, one of the equations is:
$$y = -0.000439x^4 + 0.052x^3 - 1.56x^2 + 17.57x +14.46$$

And I need to separate the $x$ variable.
Is it possible?
Thanks for the help.

Since y(x) is not generally invertible, there's no closed form for x in terms of y. To show it is not invertible, use the horizontal line test, which fails for some choices of the constants.


Partial fractions of $\frac{1}{x^5-x^3}$ Is the following correct?

$$\frac{1}{(x^5-x^3)}=\frac{Ax+B}{(x^2-x)}+\frac{Cx+D}{(x^2-x)^2}+\frac{E}{x}$$


Is the determinant of $\left( \begin{array}{rr} A & B\\ -B & A \end{array}\right) $ non-negative? I'd like to know if $$\left(
\begin{array}{rr} A & B\\
-B & A \end{array}\right)
$$
has non-negative determinant, where $A,B$ are $n \times n$ real matrices, not necessarily commuting.
I'm trying to show almost complex structure $J$ admits an orientation given by $\{e_1,\cdots ,e_n,Je_1,\cdots Je_n \}$. To show this orientation is well defined, I need to pick another such basis and show the transition matrix has positive determinant. And above is the transition matrix.
Thanks in advance for anyone who could help.
Note that in here the answer is only provided when $A$ and $B$ commute.

$\newcommand{\overbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}$
As noted in the comment, taking determinants on both sides of the identity
\begin{align*}
\begin{pmatrix} I_{(n)} & iI_{(n)} \\ 0 & I_{(n)} \end{pmatrix}
\begin{pmatrix} A & B \\ -B & A \end{pmatrix}
\begin{pmatrix} I_{(n)} & -iI_{(n)} \\ 0 & I_{(n)} \end{pmatrix} =
\begin{pmatrix} A - iB & 0 \\ -B & A + iB \end{pmatrix}
\end{align*}
yields
\begin{align*}
\det\begin{pmatrix} A & B \\ -B & A \end{pmatrix} &= \det(A + iB)\det(A - iB) = \det(A + iB)\det(\overbar{A + iB}) \\
&= \det(A + iB)\overbar{\det(A + iB)} = |\det(A + iB)|^2 \geq 0,
\end{align*}
where we used the condition that $A$ and $B$ are real matrices, and for any $C \in \mathbb{C}^{n \times n}$, $\det(\overbar{C}) = \overbar{\det(C)}$, which can be easily verified by the definition of determinant.

We know:
det (X+Y) ≥ det(X) + det (Y)
this means: det(A^2 + B^2) ≥ det(A^2) + det(B^2)
which is ==> |A^2 + B^2| ≥ |A^2| + |B^2|; EQN 1
Using another property of matrices: |A|^n = |A^n|
Thus, RHS of EQN 1 = |A|^2 + |B|^2
Also, determinant of any nxn real matrix always exists and is a real number. So, square of a real number will always be non-negative!
Since RHS is a non-negative quantity and LHS is greater than equal to RHS, LHS WILL always be a non-negative quantity too!
Hope this addresses your concerns..


Improper integral from 1 to infinity $\Rightarrow$ integrated function converges towards zero? Let $f: [1, \infty) \to \mathbb{R}$ be a continuous function such that the improper integral $$\int_1^\infty f(x) \ dx$$ exists. Show or disprove that $\lim \limits _{x \to \infty} f(x) =0$.

Our professor said that there are counterexamples but after frying my brain out I still could not find any. It does sound logical.

It does sound plausible, especially by analogy with infinite series. But yes, there are counterexamples. Here's one:

Consider a continuous function that is zero, except for triangles centered on the integers, of height 1 and area going like $1/2^n$. Then the integral will exist but the function itself doesn't converge to zero.

Added: A rough diagram (graph of function with is zero  for all $x \geq 1$ except for two sides of isosceles triangles of height 1, base length $2^{-(n-2)}$ centered over each integer $n \geq 2$; the area of the triangle over the integer $n$ is $2^{-(n-1)}$.)

Then for such a function,
$$\lim_{x\to\infty} \int_1^x f(x) \ dx = \lim_{N\to\infty} \sum_{n=2}^N 2^{-(n-1)} = 1$$

however $\lim_{x\to\infty} f(x)$ does not exist.

The counterexample can be made more exotic by allowing it be unbounded: let the isosceles triangle over integer $n$ have height $n/2$ and base length $(n2^{n-3})^{-1}$; each triangle's area remains $2^{-(n-1)}$.

You may even have positive continuous unbounded functions with an existing improper integral.

You can have a look here with some pictures.


Real analysis using field axioms to prove Prove that $(-x)(-y)=xy$ using field axiom theorems

My attempt

I know that $-(xy)$ is the additional inverse for $xy$, and I know $xy + -(xy) = 0$ (equation $1$)

I assumed that $(-x)(-y)=xy$ and substitute into eqn $1$ and I get

$(-x)(-y)+ -(xy) =0 $

so now I just have to show that this equation is still true to justify my claim that $xy=(-x)(-y)$

But how do I move on from here??

list of axioms : https://kevinbinz.files.wordpress.com/2014/11/group-theory-group-marriage-1.png

by associativity and commutativity of multiplication
we get

$(-x)(-y) = (-1.x)(-1.y) = (-1.-1)(x.y)$

now remains to prove $-1.-1 = 1$

consider $-1 + -1.-1  = -1.1 + -1.-1 = -1(-1 + 1) = 0 $


Computing the derivative of square root of a matrix maybe this is an idiot question, however I could not figure out how to solve it. Let $X =M_n(\mathbb{R})$ be the space of $n \times n$ matrix over the reals, then there exists two open neighborhoods of the identity, $U$ and $V$, such that the function $\phi: V \longrightarrow U$, $\phi(A) = \sqrt{A}$ is well defined and is differentiable at the identity $I$.Furthermore,  what's $d\phi(I)(T)$ ? I was thinking in inverting the matrix $A = I - B$ by the usual $\sum_i B^{i}$ and then, somehow, find the unique square root.

Thanks in advance.

This looks like a straightforward application of inverse function theorem. Consider
$$\psi \colon M_n(\mathbb{R}) \to M_n(\mathbb{R}); A \mapsto A^2$$
Then $\psi$ is continuously differentiable everywhere with derivative $d\psi(A)(T) = AT+TA$. At $A = I$, this is $T \mapsto 2T$, which is invertible.
Now conclude by the inverse function theorem (statement taken from wikipedia):

If the total derivative of a continuously differentiable function $F$ defined from an open set $U$ of $\mathbb{R}^n$ into $\mathbb{R}^n$ is invertible at a point $p$ (i.e., the Jacobian determinant of $F$ at $p$ is non-zero), then $F$ is an invertible function near $p$. That is, an inverse function to $F$ exists in some neighborhood of $F(p)$. Moreover, the inverse function $F^{-1}$ is also continuously differentiable.
Further,
$$J_{F^{-1}}(F(p)) = [J_F(p)]^{-1}$$
where $[\cdot]^{-1}$ denotes matrix inverse and $J_G(q)$ is the Jacobian matrix of the function $G$ at the point $q$.

Those answers don't look quite right. Let $A$ and $B=A^{1/2}$ be $n$ by $n$ matrix functions on $\mathbb{R}$ with derivatives $A_1, B_1.$ Since $A=B^2, A_1/2=BB_1+B_1B$. One has to solve this for $B_1$.


Proof that the derivative of a constant is zero I know that the derivative of a constant is zero, but the only proof that I can find is:
given that $f(x) = {x}^{0}$,
$$ f'(x) = \lim_{h\to 0} \frac {f(x+h) - f(x)}{h} $$
$$ f'(x) = \lim_{h\to 0} \frac {{(x+h)}^{0} - {x}^{0}}{h} $$
and then because $ {(x+h)}^{0} - {x}^{0} = 1 - 1 = 0 $, then
$$ f'(x) = \lim_{h\to 0} \frac {0}{h} = 0 $$
This seems sort of fishy to me, however, as if you plug in 0 for h in the limit, you get an indeterminate. It might just be me, though, but it just doesn't seem entirely right to me. Is this proof actually fine, and there another proof that the derivative of a constant is zero, or is this the only one?

The "$x^0$" really plays no role; besides, it is $1$ and any other constant works with the same proof.

In any case, as you write, if $f(x)=c$ for some number $c$, then
$$
\frac{f(x+h)-f(x)}h=\frac{c-c}h=0.
$$
So, the Newton quotient is already zero before taking the limit as $h\to0$.

I had the same question, in an effort to understand the essence of the rule.  I can't say this is rigidly correct, or even correct, but this is the most simple explanation I can come up with.  It's nearly totally algebraic, and builds a calculus expression from an algebra expression using the "method of increments".

Note: I'm using the triangle here. 'Δy/Δx' refers simply to the change in y in relation to the change in x. I use this because dy/dx seems less intuitive and a little like black magic.

(1) y = C; which is equivalent to y = C + 0(x)
(2) Using the method of increments, a change in x corresponds to a certain change in y, so: y + Δy = C + 0(x + Δx)
(3) Since y equivalent to C + 0(x) ...
(4) C + 0(x) + Δy = C + 0(x) + 0(Δx)
(5) Algebraic manipulation: remove 'C + 0(x)' from both sides of the equation.
(6) Then divide both sides of the equation by Δx.
(7) You end up with Δy/Δx = 0(Δx/Δx).
(8) So, Δy/Δx = 0
(9) which is to say dy/dx = 0, meaning there is 0 change in y when there is a change in x. Normally in calculus, we're concerned with the dy/dx ratio in the case of  dx (i.e., Δx) approaching 0, but that's irrelevant in this case.


The speed of the top of a sliding ladder A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.

Is my solution wrong? How come I can't get the correct answer?

x=horizontal axis , y=vertical axis
Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s
Y=2A/X
dy/dx=-2A/x^2.....A=6,x=3
     =-12/9

dy/dt=dy/dx * dx/dt
     =-12/9 * 0.4
     =-0.533333

You can't use an area relation because we're not told what $\dfrac{dA}{dt}$ is.  If you draw out the situation, we see that we can actually use a different relation to solve this problem.

The equation that you want to use comes directly from the Pythagorean theorem; in particular, if $x$ is the distance the base of the ladder is from the wall and $y$ is the the height of the ladder along the wall, then for the 5 m long ladder, we have that $x^2+y^2 = 25$.

Differentiating this will then give you the appropriate equations involving $\dfrac{dx}{dt}$ (i.e. the rate the ladder is sliding away from the wall) and $\dfrac{dy}{dt}$ (i.e. the rate the ladder is sliding down the wall).  You can then plug in all the known values at the appropriate instant in time (when $x=3$, $y=\ldots$ [which can be found using the Pythagorean theorem], and $dx/dt = 0.4\text{ m/s}$) and then solve for $\dfrac{dy}{dt}$.

Hopefully this is enough information to help you get the correct solution. :-)

This problem can be solved 'numerically' without resorting to other methods. First we have:

$$
w^2 + g^2 = L
$$
$$
(w + f)^2 + (g + p)^2 = L
$$

where $w$ is the wall, $g$ is the ground, $L$ is the ladder squared, $f$ is the fall rate and $p$ is the pull rate. Note that the units of f and p are metres - time is implicit. Also note that we do not presume that the fall rate is negative - it should come out that way. Equating, canceling common terms and solving the quadratic in $f$ yields:

$$
f = \sqrt{w^2 - 2gp - p^2} - w
$$

Now we can calculate a series of increasingly accurate approximations of the fall rate if the pull rate is $0.4 m/s$. Here is a Python function that does this:

def fallrate(wall, ground, pullrate):
  for i in [2**n for n in range(15)]:
    print(i * ((wall**2 - 2 * ground * (pullrate/i) - (pullrate/i)**2)**.5 - wall))
  input("\nEnter to exit. ")

fallrate(4, 3, 0.4)

Note that if we halve the distance we use to calculate the pull rate we will get a rate for half the distance. We must therefore double the result for the purpose of comparison - hence the i * etc at the start of the approximation formula. The fall rate seems to converge on $-0.3 m/s$. This approach can be seen as a type of calculus - if we divide the rates on both sides of the approximation formula by $i$ we are in effect turning them into infinitesimals; and it agrees with the result given by Eff worked out using conventional methods.


You get $20$ cents for each correct answer, but pay $10$ ten cents for each incorrect one. After $30$ questions, you have a profit of $\$1.80$ ... An algebra word problem question in my textbook:

A kind teacher gives you $20$ cents for every question you get right, but you have to pay the teacher $10$ cents for every question you get wrong. After $30$ questions you have made a profit of $\$1.80$.
A) Form an equation with $x$ representing the number of questions you got right.
B) Solve your equation to find how many questions you got right.

I realise this is a very rudimentary question for a Year 10/11 Maths class, but I’m homeschooled with no teaching guidance whatsoever and this question type has no explanation in my textbook. So I super appreciate the help.

What are we looking for ?
the number R of right answers and the number W of wrong answers.
As We have TWO  unknowns, we need TWO equations
What do we know ?
The total number of question, which is  $ 30$
so $$R + W= 30$$
For each good answer, we win $ 20 $ cents. For $  R$ answers. we get $ 20 R$ and for $ W$ answers, we loose $
10 W$ cents which means that we win $ -10W$.
In the end, we have
$$R+W=30$$
$$20R-10W=180$$
Using system of equations, we can derive the values of R and W. We can use substitution, by creating an equation for R from the first equation:
$$R = 30 - W$$
Using this, we can substitute this for R in the second equation:
$$20(30 - W) -10W = 180$$
Now, we have an equation with only W,and we can solve for it:
$$600 - 20W - 10W = 180$$
$$600 - 30W = 180$$
$$-30W = -420$$
$$W = 14$$
Now that we know $$W = 14$$, we can substitute into the equation $$R + W = 30$$ to find that $$R = 16$$. Thus, these are your equations and answers. Since the question explicitly asks to use the variable $x$, use $x$ instead of $R$.

If you answer $x$ many questions right, with that you've earned $20x$ cents.
But this means you also got $30-x$ wrong, for which you paid back $10$ cents each, i.e. a total of $10(30-x)=300-10x$ cents.
So your "profit" in general (in cents) is $$20x - (300-10x)  =20x - 300 + 10x = 30x - 300 = 30(x-10)$$
That answers A.
If this equals $\$1.80 = 180$ cents we have to solve $$180 = 30(x-10) \iff 6 = (x-10) \iff x=16$$
And with $16$ right and $14$ wrong we indeed earn $\$3.20$ and pay $\$1.40$ so the profit is correct.


Proof of the identity $\sum_{k=0}^{n}\binom {2k}{k}\binom {2(n-k)}{n-k}=4^n$ Prove the following identity :
  $$\sum_{k=0}^{n}\binom {2k}{k}\binom {2(n-k)}{n-k}=4^n$$

I am sorry, this topic is very new to me and so is this website.  I have looked at all the "duplicates" of this question on this website, but none have had an answer that I can comprehend.  I thought I was able to just plug each one into the binomial theorem and arrive at the answer, but my professor stated that I am using the theorem incorrectly.  I do not have an intuitive understanding of proofs yet.  Any direction will help!  Thank you.

I am helping you on this one. I am trying to make reference to this equality:
$$\sum_{k=0}^n {{n}\choose{k}} = \sum_{k=0}^n {{n}\choose{k}} 1^k \cdot1^{n-k} = (1+1)^{k+n-k}=2^n$$
So, by the similar fashion (I will explain why it is the multiplicative)
$$ \sum_{k=0}^n {{2k}\choose{k}} \cdot{{2(n-k)}\choose{n-k}} = \sum_{k=0}^n {{2k}\choose{k}}1^k1^k \cdot {{2(n-k)}\choose{n-k}} 1^{n-k}1^{n-k} = (1+1)^{k+k} \cdot (1+1)^{(n-k)+(n-k)} = 2^{2k+2n-2k}=2^{2n}=4^n$$
This actually leads to the good end. But I will establish why or improve the solution.

It seems like the following is problematic.....
$$\sum_{k=0}^n {{2k}\choose{k}} \cdot{{2(n-k)}\choose{n-k}} = \sum_{k=0}^n  \frac{(2k)!}{k!k!} \cdot \frac{[2(n-k)]!}{(n-k)!(n-k)!} = \sum_{k=0}^n \frac{(2n)!}{{{2n}\choose{2k}}} \cdot \Big( \frac{{{n}\choose{k}}}{n!} \Big)^2$$


$n$th derivative of $\frac{1}{\sqrt{1-2x}}$ I was asked to find the nth derivative of the given  function:
$$\frac{1}{\sqrt{1-2x}}$$

I tried to use binomial theorem, finally with some effort I could find a way to express the function in the following form:

$$\frac{1}{\sqrt{1-2x}}=\sum_{k=0}^{∞}{{2k}\choose{k}}\left(\frac{1}{2}\right)^{k}x^{k}$$

then I found a formula for the nth derivative of the given function:

$$\frac{d^{n}}{dx^{n}}\left(\frac{1}{\sqrt{1-2x}}\right)=\sum_{k=0}^{∞}{{2k}\choose{k}}\left(\frac{1}{2}\right)^{k}\left(\prod_{m=0}^{n-1}\left(k-m\right)\right)x^{\left(k-n\right)}$$
 which is non-zero for $k\le n$

Also I wanted to know what is the radius of convergence of this expression then I used ratio test such that:

$$\lim\limits_{k \to ∞}\left|\frac{a_{k+1}}{a_{k}}\right|=\lim\limits_{k \to ∞}\left|\frac{\ x^{\left(k+1\right)}\left(2k+2\right)!\left(\frac{1}{2}\right)^{\left(k+1\right)}\ }{\left(\left(k+1\right)!\right)^{2}}\cdot\frac{\left(\left(k\right)!\right)^{2}}{x^{k}\left(2k\right)!\left(\frac{1}{2}\right)^{\left(k\right)}}\right|=$$$$\lim\limits_{k \to ∞}\left|\frac{x\left(2k+2\right)\left(2k+1\right)\left(\frac{1}{2}\right)}{\left(k+1\right)^{2}}\right|=$$$$\lim\limits_{k \to ∞}\frac{1}{2}\left|\frac{4k^{2}+6k+2}{k^{2}+2k+1}\cdot x\right|=$$$$\lim\limits_{k \to ∞}\frac{1}{2}\left|4x\right|$$

hence the expression is valid for $\left|x\right|<\frac{1}{2}$, the same radius of convergence is for the derivative of the given function.

my question is that: is there any way which calculate the nth derivative of the function for all the domain of the function?


how to prove $\prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}}$? i found this relation whilst trying to evaluate the norm (over $\mathbb{Q}$) of $1-\zeta$ for $\zeta$ a primitive $p$-th root of unity ($p$ supposed prime)
$$
\prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}}
$$
as yet i have no means of proving it. any suggestions?

Kindly see this article page number $5$.


A six-sided die is rolled 5 times. The roller wins if the last roll is the same as one of the previous rolls. What is the probability of winning this game?

I tried some but the closest I get was $\frac 16 \cdot \frac 16$, but it's not right.

Write down the results of the five rolls and then read them backwards.  You win if the "first" result is ever repeated, and you lose if it's not.  The probability of losing is $(5/6)^4$ -- i.e., a given roll does not repeat the "first" result with probability $5/6$, and there are $4$ "subsequent" rolls -- so the probability of winning is

$$1-\left(5\over6\right)^4={1296-625\over1296}={671\over1296}$$

I am not gonna go with massive calculations rather simple logic.

I think the probability will change with each dice roll.

On the first roll it doesn't matter what comes up as one cannot win on first roll(according to the rules you stated).
Now on second roll the probability of winning will be 1/6 as the probability of repeating what came up on first roll is 1/6.
On the third roll however the probability of winning shall increase to 2/6 and so on.


Better to memorize logarithm rules? And how? Do people good at math totally memorize these logarithm rules below? If so, are there good mnemonics for this? I'm bad at math and I only memorize these rules really vaguely by rote, thus when needed, I'm not sure if I remembered them correctly. So I always end up checking if they are true by testing them assigning numbers like 2 to m and n on the fly. Is this something smarter people would think is a stupidly inefficient process to go through? I wonder how other people deal with this?

$\log_b (mn) = \log_b(m) + \log_b(n)$
$\log_b(m/n) = \log_b(m) – \log_b(n)$
$\log_b(m^n) = n \cdot \log_b(m)$

You're asking questions and seeing if you can do things more efficiently -- that's what people who are good at math do.

I always forgot these when I first learned them. It might help to have a mnemonic. Anything that helps you remember is a mnemonic -- I find that getting an intuitive understanding or a feeling of why something is true helps me remember it.

Here's something that might help: think about $\log(n)$ as being the number of zeroes after $n$. (This is true if $n$ is a multiple of $10$ and the base of the log is $10$).

$\log_b (mn) = \log_b(m) + \log_b(n)$

If I multiply a number with three zeroes ($1,000$) by a number with two zeroes ($100$), how many zeroes will the answer have? Five: $100,000$.

$\log_b(m/n) = \log_b(m) – \log_b(n)$

Same deal as the previous, except with division. If I take $1,000$ and divide it by $100$, I get $10$. Three zeroes - two zeroes = one zero.

$\log_b(m^n) = n \cdot \log_b(m)$

If my number gets raised to the $n$-th power, how many more zeroes will it have?
$$100^3 = 100 \cdot 100 \cdot 100 = 1,000,000$$

so $2$ zeroes to the $3$rd power is six zeroes. That's $n$ times as many.

Of course these laws also work for numbers that are not multiples of $10$. In fact, $10$ wasn't special to the above. But starting from the simplest case makes it easier to remember.

EDIT: Anything that helps you remember a good mnemonic, just want to stress that point. Something like "$\log$ takes the power down and puts it in the front" works perfectly well. I like the above because it helps me remember why the laws are true.

YES OF COURSE, people memorize those rules.  As far as "how" goes I don't know how to answer - you just do it.  One thing that ought to help is that it's all the same rule, really, because division is the inverse of multiplication and exponentiation is repeated multiplication.  Maybe it will help you to think of it that way.


Forming 4-digit odd numbers under certain rules How many four-digit odd numbers can be formed such that every $"3"$ in the number is followed by a $"6"$?

A) 108
B) 2592
C) 2696
D) 2700

A "3" need not be immediately followed by a "6", and each "3" must be uniquely associated with a "6", i.e. multiple occurrences of a "3" cannot share the same "6".

Hint: firstly take a look at the possibilities without $"3"$: you have $8$ choices for the 1st digit, $9$ choices for the 2nd digit, $9$ choices for the 3rd digit, and finally $4$ choices for the last digit (in order to be odd, the number must end with $1,3,5,7 \text{ or } 9$) $\Rightarrow$ at least $8\cdot 9\cdot 9\cdot 4 = 2592$ numbers, so answers A and B can't be right...then try by examinate the number of numbers with only 1 $"3"$, 2 $"3"$, and so on...


Upper bound of a quadratic, given the upper bound of variable Problem: Let $p(x)$ be a quadratic polynomial $p(x)$ = $ax^2+bx+c$ such that $|p(x)|$ $≤$ $1$ for |$x$| $≤$ $1$.
Prove that
$|cx^2+bx+a|$ $≤$ $2$ for |$x$| $≤$ $1$. (Source: Challenge and Thrill of Pre-College Mathematics, Page 201, Problem 16)
My attempt: I tried $|cx^2+bx+a|$ $≤$ |$a+b+c$| = $|p(1)|$ for $|x|$ $≤$ 1, but after that it gets complicated. More specifically, the upper bound for the quadratic asked actually comes out to be $|p(1)|$ $≤$ 1, and so the given statement is proved. However I'm thrown off by why the problem states $≤$ $2$ and not $1$.
Any hints would be appreciated, but please don't post a definite solution. In the case I solve it using the hints given, I'll post my own answer below. If I cannot, I'll edit this question asking for a solution.

Employing a rather common strategy for such problems, from $p(x)=ax^2+bx+c$, we get $c=p(0), b = \frac12\left[p(1)-p(-1)\right]$ and $a = \frac12\left[p(1)+p(-1)\right]-p(0)$.
Hence with $q(x) = cx^2+bx+a$, we get
$$q(x) = p(0)x^2+\tfrac12\left[p(1) - p(-1)\right]x +\tfrac12\left[p(1)+p(-1)\right]-p(0)$$
$$\qquad = p(0)(x^2-1)+\tfrac12p(1)(x+1)-\tfrac12p(-1)(x-1)$$
$$\implies |q(x)| \leqslant |p(0)||x^2-1|+\tfrac12|p(1)||x+1|+\tfrac12|p(-1)||x-1|$$
$$\qquad \leqslant (1-x^2)+\tfrac12(1+x)+\tfrac12(1-x)=2-x^2\leqslant 2$$
--
P.S.  Take $p(x)=2x^2-1$ and you will notice this bound is tight.

Define the class of quadratic polynomials P2 whose members having sum of all abs coefficients being less than or equal to 1:
P2={p(x)=ax^2 +bx+c |  |a|+|b|+|c|<=1}
Show that

If p(x) ε P2 then |p(x)|<=1 for all |x|<=1

Proof
|p(x)|<= |a||x|^2+|b| |x|+ |c|<= |a|+|b|+|c|<=1 for all |x|<=1

If p ε P2 then q ε P2 where q(x)= cx^2 + bx +a then q ε P2

Proof
|q(x)|<= |c||x|^2 + |b| |x|+ |a|<= |c|+|b|+|a| <= 1 for all |x|<=1
It follows that |q(x)| <=2 for all |x| <=1
QED


Can we sum a series over noninteger bounds? Are there any ways to generalize series to noninteger bounds? For example, is there any way to make sense of
$$
f(\nu)=\sum_{k=\nu}^\infty a_k
$$
for $\nu\in\Bbb C$? I would expect that if such a generalization exists, it should obey have the same properties as the case where $\nu$ is an integer.  In particular, I would expect it to obey the recurrence relation
$$
f(\nu)-f(\nu+1)=a_\nu.
$$

We find in chapter 2 Sums of  Concrete Mathematics
by R.L. Graham, D.E.  Knuth  and O.  Patashnik the following statement (see formula 2.4):

Formally, we write
\begin{align*}
\sum_{P(k)} a_k\tag{2.4}
\end{align*}
as an abbreviation for the sum of all terms $a_k$ such that $k$  is an integer satisfying a given property $P(k)$. (A property $P(k)$ is any statement about $k$ that can be either true or false.)

We see the index $k$ is an integer. Sometimes we have the situation that the upper limit of a sum is not an integer. In this situation we take the greatest integer less or equal to the upper limit (floor function).
\begin{align*}
\sum_{k=0}^{\frac{n}{2}} a_k=\sum_{k=0}^{\color{blue}{\left\lfloor\frac{n}{2}\right\rfloor}} a_k=a_0+a_1+\cdots+a_{\left\lfloor\frac{n}{2}\right\rfloor}
\end{align*}
Similarly the lower limit is sometimes not an integer. In this situation we take the smallest integer greater or equal to the lower limit
(ceiling function).
\begin{align*}
\sum_{k=\frac{m}{2}}^{\frac{n}{2}} a_k=\sum_{k=\color{blue}{\left\lceil\frac{m}{2}\right\rceil}}^{\left\lfloor\frac{n}{2}\right\rfloor} a_k=a_{\left\lceil\frac{m}{2}\right\rceil}+a_{\left\lceil\frac{m}{2}\right\rceil+1}+\cdots+a_{\left\lfloor\frac{n}{2}\right\rfloor}
\end{align*}

If the lower limit of the index is equal to $e$, we have consequently
\begin{align*}
\sum_{k=e}^\infty a_k=\sum_{k=\left\lceil e\right\rceil}^\infty a_k=\sum_{\color{blue}{k=3}}^\infty a_k
\end{align*}

I guess you could write something like
$$
\sum_{n = 0}^\infty f(e + n)
$$


$x = 1+2^p$ $y= 1+2^{-p}$ Finding the $y$ in terms of $x$. $$x = 1+2^p$$ $$y= 1+2^{-p}$$ Find the $y$ in terms of $x$.

What if I combine these 2 equations?

$$x+y = 1 + 2^p + 2^{-p}$$

It seems like that there will be no solution from there.

Put $2^p$ from one equation to other

$y=1+2^{-p} = 1 + \cfrac{1}{x - 1}$

write $$x-1=2^p$$ and $$y-1=2^{-p}$$ by multiplication we obtain $$(x-1)(y-1)=1$$ since $$2^p\cdot 2^{-p}=1$$


Is distributional Derivative of $\delta^{(2)}(-x)=-\delta^{(2)}(x)$ Is distributional Derivative of

$\delta^{(2)}(-x)=-\delta^{(2)}(x)$ ??

or

$\delta^{(2)}(-x)=\delta^{(2)}(x)$ ??

I know that $\delta(-x)= \delta(x)$ and $\delta^{(1)}(-x)=-\delta^{(1)}(x)$. How to show similar property for the second derivative?

Thanks.

With the definition of the derivative $\langle\delta^{(n)},phi\rangle=(-1)^n\langle\delta,\phi^{(n)}\rangle$ and $\langle\delta(-x),\phi\rangle=\langle\delta,\phi(-x)\rangle$ where $\phi$ is a test function we should get $$\delta^{(1)}(-x)=-\delta^{(1)}x)$$ and $$\delta^{(2)}(-x)=\delta^{(2)}(x)$$ and this is consistent with what happens to functions


Does a System to Automatically (Mathematically) Generate Arbitrary Computer Programs Really Exist? Has anyone actually developed a Program Synthesis system for creating computer programs automatically from a non-procedural specification that is taken from a fairly robust system of specifications?  For example, I might say "Check if X is a  factor of Y." or "List all prime numbers between X and Y." and out pops a program or two written in PHP that does just that.

I would actually pay money for an example that shows that the answer is yes.

Let's say you have relations for less than and multiplication, and you give a wff of logic.  So it can be (exists A)MUL(A,X,Y) where MUL(A,B,C) iff A x B = C, for the first specification above.  For the second specification with output of a number (the prime numbers) instead of TRUE or FALSE, we could say variable N means output all values of N such that the wff is TRUE.

I have heard the claim that the answer is yes, but articles give no real examples of its being used.  Either there is no example, or what they call an example merely displays a program and claims that it was or can be generated by some system that is imagined or even described in detail, but the example lacks any details at all.

Can someone give a complete example of a system, a specification, the resulting program(s) and step-by-step exactly how that program is automatically constructed?

And if an example is given, could it be a simple program with only a few steps to create, rather than dozens of pages of formulas that would take months to go through and is suspect of being obfuscation hiding the fact that it is not genuine?

Depending on what exactly "non-procedural specification" means, prolog might match your requirements.

Prolog is a so-called declarative programming language where you don't specify that steps that an algorithm perform, but rather describe it's result in (a version of) first-order logic. There are compilers for prolog, which translate this declarative description into a procedural one.

Prolog, being just a programming language, does not avoid the problem of having to write a computer program.  You are just writing it in one language and translating that into another programming language.

What is needed is to input a formula of logic that is true iff the output satisfies the desired conditions.  For example, (exists A)A*I=J means I is a factor of J and is a valid specification for a program to follow.  It means to input I and J, and output TRUE or FALSE depending on whether or not I is a factor of J or not.

Have anybody developed a system to take a formula of logic and create computer programs that meet its requirement?  Professors say yes but none has even been published.


Finding the first three terms of a geometric sequence, without the first term or common ratio. Given a geometric sequence where the $5$th term $= 162$ and the $8$th term $= -4374$, determine the first three terms of the sequence.

I am unclear how to do this without being given the first term or the common ratio. please help!!

We have that $162 = a_{1}r^{4}$ and $-4374 = a_{1}r^{7}$ by the formula $a_{n} = a_{1}r^{n-1}$.

Then solving for $a_{1}$ in both equations and setting them equal to one another,
$$\frac{162}{r^{4}} = \frac{-4374}{r^{7}}$$

You can then solve for $r$ (your common ratio), and subsequently $a_{1}$ (your first term). You then have all of the information you need.

I had a similar question and tried to solve it alone i used this way

since the initial value is f(1) so u use this equation f(n) = ar^n-1

since u have f(5) as initial value or actually (known value) then we have to change it to this

f(n) = a5 * r (n-5)  , so we know that the 8 th term f(8) = -4734
so :  f(8) = a5 * r(8-5)   and f(5)= 162

-4734 = 162 *r^3

-47324/162 = r^3

r^3 = -27 then calculate ......


What's the difference between "relation", "mapping", and "function"? I think that a mapping and function are the same; there's only a difference between a mapping and relation. But I'm confused. What's the difference between a relation and a mapping and a function?

Mathematically speaking, a mapping and a function are the same. We called the relation
$$
f=\{(x,y)\in X\times Y : \text{For all $x$ there exists a unique $y$ such that $(x,y)\in f$} \}
$$
a function from $X$ to $Y$, denoted by $f:X\to Y$. A mapping is just another word for a function, i.e. a relation that pairs exactly one element of $Y$ to each element of $X$.

In practice, sometime one word is preferred over another, depending on the context.

The word mapping is usually used when we want to view $f:X\to Y$ as a transformation of one object to another. For instance, a linear mapping $T:V \to W$ signifies that we want to view $T$ as a transformation of $v\in V$ to the vector $Tv\in W$. Another example is a conformal map, which transforms a domain in $\Bbb C$ to another domain.

The word function is used more often and in various contexts. For example, when we want to view $f:X\to Y$ as a graph in $X\times Y$.

Good question. I can give you a simple example.

  You can generally think of relations as functions in a higher dimensional space, intersected with a zero plane.

So for example, lets take a parabola and a circle to demonstrate this. Imagine the parabola

$$ f(x) = x^2 + 3x $$

This is clearly a function of x, because if you give me an x value, I can give
you the corresponding value of f(x) - a mapping is really just another name
for a function. If we want to graph it, we can let the y value
equal the output of $f$, so we would get this graph:

On the other hand, if we graph a circle, like:

$$x^2+y^2=4$$

Its graph is given by:

Now this is fundamentally different to the function. If you wanted the y value at x = 0,
I can't point to a unique value, I'd have to say y = 2 or y = -2. A function can only
have one output. So we have to call this a relation.

Higher Dimensions

However, we can use a clever trick for this circle. We can rewrite it as:

$$ x^2 + y^2 - 4 = 0 $$

Which is obviously the same thing, but on the left hand side, notice that we now
have a function of (x,y), so we can think of this like:

$$ g(x, y) = 0 $$

In a higher dimension, this would be the intersection between the shapes:

$$ z = g(x, y) $$

and

$$ z = 0 $$

Which I've shown below:

Notice that same circle hiding in plain sight.

Key takeaway (tl;dr)

  Relations are functions in a higher dimension, intersected with a zero plane
  in the higher dimension.

(Edit) More Precisely

I agree that I could have been more precise with the things I've written above. So hopefully I can be more clear by what I meant here. The usual definition is:

  A relation is a pairing between elements of two sets, which are not necessarily unique. A function forces each member of the domain to have only one "partner" in the output set (or codomain).

So for example, here are two example sets. On the left we have a relation that is a function, and on the right we have a relation that is not a function.

However, under a particular circumstance (that I will discuss below), we can actually turn that relation into a function. First, we can take the cartesian product between the two sets X and Y on the right portion of the image above as shown, this will produce every possible pair of tuples (x, y):

Now under the assumption that there exists another function to map these tuples to another set z, (which is the role of z = g(x, y) and z = 0 = 0(x, y) (using zero as a function name), the relation can be viewed as a function that maps all of these tuples into the same point as I've shown below:

Now of course, the reason I bothered to subtract everything to have zero on the other side, was so that the element on the output set that I was mapping to was 0 in the set of real numbers.

However, I am assuming that there exists a function $g(x, y)$, to map from $X \times Y$ to $Z$, which at least for a function of real numbers to real numbers; should always be true.

I hope this clears things up. I appreciate your comments :)


How to show that $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}=9/2$? $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}$

I need to prove that this limit equals to $\frac{9}{2}$.
Can someone give me a step by step solution?

EDIT: I am sorry. The $x$ goes to $0$, not $1$.

If you allow Taylor expansions, recall that

$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$

Thus,

$$\sin(3x)=3x-\color{red}{\frac92}x^3+\mathcal O(x^5)$$

Thus,

$$\begin{align}\frac{3x-\sin(3x)}{x^3}&=\frac{\frac92x^3+\mathcal O(x^5)}{x^3}\\&=\frac92+\mathcal O(x^2)\\&\to\frac92\end{align}$$


Can I use sups and infs on an increasing function on a closed, bounded interval and state that it is continuous? In Royden's book, we are given the following proposition:

Let $C$ be a countable subset of the open interval $(a,b)$. Then there is an increasing function $f$ on $(a,b)$ such that it is continuous only at points in $(a,b) \sim C$.

Then, he gives a problem which is similarly stated as

Let $C$ be a countable subset of the nondegenerate closed, bounded invterval $[a,b]$. Show that there is an increasing function on $[a,b]$ that is continuous only at points in $[a,b] \sim C$.

This problem seems very trivial because I could define a function $g(x)$ where $g = f$ on $(a,b)$, and $g(a) = \inf \{ f(x) : x \in (a,b) \}$, if $C$ does not contain the endpoint $a$. The same construction would be used on $b$, but with a supremum. Then what would remain to be shown is that using this definition of $f$ for the endpoints, is continuous on at either endpoint. Is this sort of construction the right idea?

Your question does not mention Royden's construction in the open interval case; that makes it a bit confusing, since whether the construction extends easily to the closed interval case depends on what the open interval construction is.

I don't have a copy of Royden's book, so I don't know what his construction is.  But there is a standard construction here: the case in which $C$ is finite is a very easy one, so suppose $C$ is countably infinite and let $C = \{c_n\}_{n=1}^{\infty}$ be an enumeration.  Then we define

$f(x) = \sum_{n \ \mid \ c_n \leq x} 2^{-n}$.

This function is (weakly) increasing and has a jump discontinuity at each point of $C$.  (If you want a function which is strictly increasing, just add the function $g(x) = x$ to it.)  Moreover, this function has $\lim_{x \rightarrow a^+} f(x) = 0$ and $\lim_{x \rightarrow b^-} f(x) = 1$.  So if you extend it $[a,b]$ by taking the value $0$ at $a$ and $1$ at $b$, then yes, this construction works to answer your question.

Note also that the extension to $[a,b]$ follows the recipe that you give in your question.  However, for this to work we needed $f$ to be bounded on $(a,b)$.  There are certainly going to be functions which satisfy the property on $(a,b)$ but are not bounded: e.g. add to $f$ a function like the arctangent which has limit $-\infty$ at $a$ and $\infty$ at $b$, and then the extension procedure would not work (unless you allow your functions to take values in $[-\infty,\infty]$, which seems unlikely).

Let us start with the open interval (a,b).

(1)  suppose the number of points in C is finite.  Obviously we can make f a function which takes a jump at any point of C, but is continuous otherwise.  Then all the points in S = (a,b) ~ C have a neighborhood in which they are continuous by definition, so it is done.

(2) Next, consider C to be a countable set which is not dense in (a,b). We can use the same f as above, one which jumps at each element of C but is continuous elsewhere.  Then again, all the points in (a,b) ~ C are in open neighborhoods on which f is continuous by definition, so it is continuous at each such point.

(3) The issue is the C which is dense in (a,b). Since C is countable, it can be described as the limit a sequence of finite sets: $R_1 \subset R_2 \subset ...$.  On each $R_n$ we can define $f_n$ as we did in part 1, so that it is continuous on $S_n = (a,b) ~ R_n$ and discontinuous on $S_n$.  Let us only specify in addition that the jump chosen, which we will call q, should be the same for each $R_n$.  We claim the {$f_n$} converge and will call that limit function f.

Why the convergence.  Because each $f_{n+1}$ differs from $f_n$ only at a finite number of points, and contains all the points of discontinuity of $f_n$.

We can hope that this f is our desired function, one which is discontinuous at all elements of C and continuous on S = C ~ (a,b).  Certainly it is continuous at all elements of S, because every $f_n$ is continuous there.  The danger is that it might have become continuous on C as well. However, because the jump at r $\in R_n$ is always q, there is always a sequence of $r_n \in C$ such that $r_n \rightarrow r$ but $f(r_n)$ does not converge to f(r).

Now let us look at the closed interval [a,b].  Since we are constructing f, we can set f(a) to be anything we want, whether a $\in$ C or not.  If we specify f(a), the difficulty is at the point b.  That is, by using a constant jump of q to make f discontinuous on C, we have defined an f which $\rightarrow \infty$ at b.

To deal with this, we are going to have to drop the constant q, and replace it with a sequence {$q_n$} which $\rightarrow 0$ as $r_n$ goes to b. The resulting function f will still be discontinuous at every point of S except possibly b itself.  If b is not in S, then f is okay.

But b could be in S.  I do not have an answer to this.  Perhaps someone else can answer.


Why is it considered incorrect to use the variable of integration as a boundary of integration? Recently, a question was asked about limits of integration and the OP made the following "typo":
$$I =  \int_0^x f(x) \, dx$$
as opposed to
$$I =  \int_0^x f(t) \, dt $$
I commented that this typically frowned upon and can make for some odd results. It occured to me however, that I wasn't sure why this was formally incorrect, but that I had just been taught that it shouldn't be done. I couldn't think of a reason why it's considered best practices to differentiate between variable of integration and variables in the limits of integration..

Any thoughts / reasoning behind this?

Let's consider a simple case:
$\int_0^x x\; dx$.

Applying the usual method,
$\int_0^x x\; dx
= \dfrac{x^2}{2}\big|_0^x
=\text{?}
$

The lower term is
$\dfrac{x^2}{2}$
evaluated at $x=0$,
and this is obviously $0$.

The upper term is
more problematical.
It is
$\dfrac{x^2}{2}$
evaluated at $x=x$,
and it is unclear what this means.

Suppose the integral was
$\int_0^{2x} x\; dx$.
The upper term would then be
$\dfrac{x^2}{2}$
evaluated at $x=2x$,
and it is even more unclear what this means.

Good notation makes ideas clear
and helps avoid mistakes;
as shown here,
using the same variable for
the variable of integration
and a limit of integration
can make things unclear.

In addition to what Marty said i think the latter integral (too lazy to type!) is considered wrong mainly because the limits of integration are a function of x as well.
A big contradiction to the equality of former integral and latter integral is when you differentiate the integral.
Just try that thing out in both the cases we get totally different answers.
latter integral is a f(x,t) which has a different and more complicated differentiating form.

There might be many more reasons to this but yeah you should get the idea.

If however the limits are independent of x then you can always do that x --> t shift .


How to calculate the number of automorphisms of a given graph? How do determine the number of isomorphisms that a graph has to itself?

For instance, suppose we have the following graph:

How do I determine how many isomorphisms there are from G itself?

Here are all 48 automorphisms generated by Sage:

$$\begin{gathered}
  {\text{() }} \hfill \\
  {\text{(d f)(e g)(h k) }} \hfill \\
  {\text{(d e) }} \hfill \\
  {\text{(a b) }} \hfill \\
  {\text{(b m) }} \hfill \\
  {\text{(f g) }} \hfill \\
  {\text{(a m b) }} \hfill \\
  {\text{(a b)(f g) }} \hfill \\
  {\text{(a b)(d f)(e g)(h k) }} \hfill \\
  {\text{(a b m) }} \hfill \\
  {\text{(b m)(d e) }} \hfill \\
  {\text{(d e)(f g) }} \hfill \\
  {\text{(b m)(d f)(e g)(h k) }} \hfill \\
  {\text{(d g e f)(h k) }} \hfill \\
  {\text{(d f e g)(h k) }} \hfill \\
  {\text{(b m)(f g) }} \hfill \\
  {\text{(a b)(d e) }} \hfill \\
  {\text{(d g)(e f)(h k) }} \hfill \\
  {\text{(a m) }} \hfill \\
  {\text{(a m b)(f g) }} \hfill \\
  {\text{(b m)(d e)(f g) }} \hfill \\
  {\text{(a m b)(d f)(e g)(h k) }} \hfill \\
  {\text{(a b)(d f e g)(h k) }} \hfill \\
  {\text{(a m b)(d e) }} \hfill \\
  {\text{(a b m)(d f)(e g)(h k) }} \hfill \\
  {\text{(b m)(d f e g)(h k) }} \hfill \\
  {\text{(a b m)(d e) }} \hfill \\
  {\text{(a b)(d e)(f g) }} \hfill \\
  {\text{(b m)(d g e f)(h k) }} \hfill \\
  {\text{(a b m)(f g) }} \hfill \\
  {\text{(a b)(d g e f)(h k) }} \hfill \\
  {\text{(a b m)(d g e f)(h k) }} \hfill \\
  {\text{(a b)(d g)(e f)(h k) }} \hfill \\
  {\text{(a m b)(d e)(f g) }} \hfill \\
  {\text{(a m b)(d f e g)(h k) }} \hfill \\
  {\text{(a b m)(d f e g)(h k) }} \hfill \\
  {\text{(a m)(d f)(e g)(h k) }} \hfill \\
  {\text{(a m)(d e) }} \hfill \\
  {\text{(b m)(d g)(e f)(h k) }} \hfill \\
  {\text{(a m b)(d g e f)(h k) }} \hfill \\
  {\text{(a m)(f g) }} \hfill \\
  {\text{(a b m)(d e)(f g) }} \hfill \\
  {\text{(a m b)(d g)(e f)(h k) }} \hfill \\
  {\text{(a m)(d e)(f g) }} \hfill \\
  {\text{(a m)(d f e g)(h k) }} \hfill \\
  {\text{(a b m)(d g)(e f)(h k) }} \hfill \\
  {\text{(a m)(d g e f)(h k) }} \hfill \\
  {\text{(a m)(d g)(e f)(h k)}} \hfill \\
\end{gathered} $$

Considering the fact, that isomorphic graph $H$ differs from the an initial graph $G$ by a permutation of columns of adjacency matrix, we can possibly have $n!$ of such permutations, where $n$ is a number of vertices of both graphs.