Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Question : Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Hint: consider decimal expansions of $a$ and $b$

Attempt: Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has an infinitely repeating sequence of digits.

Set $I$ is irrational numbers

$I =[x \in R: x \notin Q]$

Set $Q$ is rational numbers

$Q = [ \frac{a}{b}:a,b, \in Z$ and $b \neq 0]$

If the division process terminates, then we are done. Otherwise, since each digit of the quotient

determines in turn its successor, and since there are at most $b-1$ possible remainders when dividing

by $b$ (by the division algorithm), some digit of the remainder must show up again, forcing a sequence

of digits to repeat forever. The length of the repeating cycle is at most $b-1$

Suppose the decimal expansion of r will terminate. Therefore, for $r=0$, we have $a_1,a_2,...a_k$ where

each $a_i \in [0,1,2,3,4,5,6,7,8,9]$ and $a_k \neq 0$. Then

$ r = \frac{a_110^{k-1}+a_210^{k-2}+...+a_k}{10^k}$

satisfies the definition of a rational number. Now, suppose the decimal expansion of r has an

infinitely repeating sequence of digits that begins immediately after the decimal point:
$r=0.b_1b_2...b_k$

The sequence of digits are repeated forever. Therefore,

$10^kr=b_1b_2...b_k.b_1b_2...b_kb_1b_2...b_k$

so,

$$10^kr-r=b_1b_2...b_k$

giving

$r = \frac{b_1b_2...b_k}{10^k-1} \in Q$

Suppose $r$ has an initial sequence of digits before the repeating sequence: $r = 0.a_1a_2...a_lb_1b_2...b_k$. Then we let $r' = 0.b_1b_2...b_k$. By the previous case $r' \in Q$. It is

easy to verify that

$r=\frac{r'+a_1a_2...a_l}{10^l}$

As a result, $ r \in Q$

So my question is do I apply the decimal expansion to $c$ and $d$. Assuming I can, then the decimal

expansion of $c$ will terminate since it's rational by theorem 2.7.5.

Then,  for $r=0.c_1c_2...c_k$ where each $ c_i \in [0,1,2,3,4,5,6,7,8,9]$ and $c_k \neq 0$. Then

$ r = \frac{c_110^{k-1}+c_210^{k-2}+...+c_k}{10^k}$

Suppose $d$ is an irriational number, then the decimal expansion won't terminate.

$10^kr=d_1d_2...d_k.d_1d_2...d_kd_1d_2...d_k$

so,

$10^kr-r=d_1d_2...d_k$

giving

$r = \frac{d_1d_2...d_k}{10^k-1} \in Q$.

If we let $r = 0.c_1,c_2...c_ld_1d_2...d_k$ and $r' =0.d_1d_2...d_k$, then $r' \in Q$ and $r=\frac{r'+c_1c_2...c_l}{10^l}$

I could've sworn that this is going to work for $c$ only, but not for $d$ because as I mentioned earlier $c$'s decimal expansion will stop and $d$ will go on forever.

Using the decimal expansion for this purpose is at least cumbersome, and while we can do it by observing that $a$ and $b$ must differ at some decimal for the first time, one has to be careful with lots of $9$s and $0$s causing trouble.
Anyway you have a lot of freedon with "late" decimals, so you can make things periodic or aperiodic at will.

It is much simpler to let $\epsilon=b-a>0$ and then note that there is at least one onteger $n$ with $n>\frac1\epsilon$ and then at least one integer $m$ with $na<m<nb$ (because $nb-na>1$) and one integer $m'$ with $n(a-\sqrt 2)<m'<n(b-\sqrt 2)$. Then $c=\frac mn$ and $d=\frac {m'}n+\sqrt 2$ do the trick.

Consider the numbers $c=\dfrac{a+b}{2}$ and $d=(\sqrt{2}-1)(b-a)+a$.


Use the triangle inequality to show that $|a|+|b| \leq |a+b|+|a-b|$ How would I go about proving that

$$|a|+|b| \leq |a+b|+|a-b|$$

Using the triangle inequality?

I tried squaring both sides, yielding:

$$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$

Is it correct to move the terms to the other side?
I tried
$$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$
Then it is obvious that
$$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$
$$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$
$$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$

But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems?


Solving $x^{\log_{10}(x^2)} = 100$ $x^{\log(x^2)}=100$
Edit: I am writing $\log$ for $\log_{10}$.
I tried to solve this and can get the solutions $10$ and $10^{-1}$. But how can I get the solutions $-10$ and $-10^{-1}$? And other solutions (including complex), if exist? I'm not sure how to use the complex logarithm in this problem.
My attempt:
Note that $x \neq 0$, because we have in the equation $\log(x^2)$.
Assume $x > 0$. Then:
$x^{\log(x^2)} > 0$
$\log (x^{\log(x^2)}) = 2$
$\log(x^2)\log(x)=2$
$2\log(x)\log(x)=2$
$\log^2(x) = 1$
$\log(x) = \pm 1 \implies x=10$ or $x = 10^{-1}$.
But how about $x<0$?

\begin{align}
x^{\log x^2}&=100\\
2(\log x)^2&=2\\
(\log x)^2&=1\\
\log x&=\pm1\\
x&=10^{\pm1}
\end{align}
As this is a quadratic equation in $\log x$, given the monotonic nature of the logarithmic function, we can conclude that $10^{\pm1}$ are the only solutions.


Induction on grid Hamiltonian graph This exercise is thaken from [1]:

  1.7 A rook’s tour. Let $G$ be an $m \times n$ grid—that is, a graph with $m n$ vertices arranged in an $m \times n$ rectangle, with each vertex connected to its nearest neighbors. Assume that $m , n > 1$. Prove that $G$ is Hamiltonian if either $m$ or $n$ is even, but not if both $m$ and $n$ are odd.

I claim that for $m=1$ I can't find any Hamiltonian cycle. For $m=2$ I can find one Hamiltonian cycle.

There is a way to generalize this in order to obtain a proof by induction (on $m$, on both $m, n$)?

[1] Moore, Cristopher; Mertens, Stephan, The nature of computation, Oxford: Oxford University Press (ISBN 978-0-19-923321-2/hbk). xvii, 985 p. (2011). ZBL1237.68004.

As noted in the comments; coloring the vertices black and white in a chessboard pattern, in any path the colors of the vertices will alternate. So if $m$ and $n$ are both odd, then the color of the $mn$-th vertex is the same as the color of the first vertex, so no Hamiltonian cycle exists.

Otherwise, without loss of generality $m$ is even. Start at the top left, move along the top row (of length $n$) to the top right corner. Then move down to the bottom right corner, and zigzag up and down all the way back until you reach the top left corner. For example, if $m=6$ and $n=7$:

As already mentioned in the comments colour the tiles Black and White similar to a chessboard.

Notice that each white cell is horizontally or vertically connected to a black cell.

WLOG lets start from a white cell. Now you must go to a black cell, then a white cell and so on.

This means that

number of white cells$=$number of black cells, Hence $mn$ must be even


By considering bounds, work out V to a suitable degree of accuracy I keep getting this question in my GCSE papers, but I have no idea how to solve it, and everywhere I look there doesn't seem to be a simple answer. The general question goes like this:

$$v=\sqrt{\frac{a}{b}}$$

$a = 6.43$ correct to 2 decimal places.

$b = 5.514$ correct to 3 decimal places.

By considering bounds, work out the value to $v$ to a suitable degree of accuracy.

(Sorry about the tagging, not sure what this fitted into)

Since $a=6.43$ to two decimal places, $a$ lies between $6.425$ and $6.435$. Similarly, $b$ lies between $5.5135$ and $5.5145$. The smallest possible value of $a/b$ occurs when $a$ is as small as possible and $b$ as large as possible; the largest possible value of $a/b$ occurs when $a$ is as large as possible and $b$ as small as possible. Thus, $$\frac{6.425}{5.5145}\le \frac{a}b\le \frac{6.435}{5.5135},$$ and $$\sqrt{\frac{6.425}{5.5145}}\le v\le \sqrt{\frac{6.435}{5.5135}}\;.$$ These bounds on $v$ are approximately $1.07940$ and $1.08034$, so we know that $v$ is between $1.075$ and $1.085$ and hence that $v=1.08$ is correct to two decimal places. Can we go one place further? The best approximation of $v$ to three decimal places is clearly $1.080$, but it isn’t correct to three places, because $v$ isn’t guaranteed to be between $1.0795$ and $1.0805$: $v$ could be just a hair under $1.0795$.

Let us denote $a' = 6.43,b' = 5.514$ and $a = a'+\delta_1$ and $b = b'+\delta_2$ with $\delta_1\leq 0.01$ and $\delta_2\leq 0.001$. We know that
$$
v = \sqrt{\frac ab}.
$$

Let us denote $v' = \sqrt{\frac{a'}{b'}}$ and put $\delta = v-v'$. Note that $v,v'\geq 1$ then
$$
|\delta| = |v-v'| = \frac{|v^2-v'^2|}{v+v'} \leq\frac12|v^2-v'^2| = \frac 12\left|\frac{a'+\delta_1}{b'+\delta_2}-\frac{a'}{b'}\right|
$$
$$
=\frac 12\left|\frac{(a'+\delta_1)b' - a'(b'+\delta_2)}{b'(b'+\delta_2)}\right|\leq \frac12\cdot\frac{\delta_1}{5}\leq 0.001
$$
so you have $v\approx \sqrt{\frac{a'}{b'}}= 1.07987...$ with an error smaller than $0.001$ or up to 3 decimal places.


A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond ? A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond?

To me this seems like we can set up a fraction-like equation:

$$\frac{60 \ \text{seconds}}{1} = \frac{x \ \text{seconds}}{1/4}$$ then $x = 15$ seconds. But the answer is $58$ seconds which really makes no sense to me. Any suggestions are greatly appreciated.

I think it's easiest to work backwards: if the area doubles every second and the pond is totally covered at time $t=60$, then it must be half covered at $t=59$, and therefore one quarter covered at $t=58$.

Alternately, let $f(t)$ be the fraction of the pond's area covered at time $t\leq 60$. Then $f(t)=f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $f(0)=2^{-60}$. Therefore $f(t)=2^{-60}2^t=2^{t-60}$. Then setting $ 2^{t-60}=\frac{1}{4}$ and solving for $t$ yields $t=58$.

The area of a lily pod can be described by a function:

$$f:[0,60]\to[0,1]\\
f(x)=2^{x-60}$$

Now we have to compute the argument $x$ for which the value of the function $f(x)$ is equal to $\frac{1}{4}$:

$$\frac{1}{4}=2^{x-60}\\
\log_2 \frac{1}{4}=\log_2 2^{x-60}\\
-2 = x-60\\
x=58$$


Find $\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2<n}\Big(1-\frac{2x^2+y^2}{n^2}\Big)^{n^2}x^{2m}dxdy$ Find

$$
\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2<n}\Big(1-\frac{2x^2+y^2}{n^2}\Big)^{n^2}x^{2m}dxdy
$$

My main (so far) problem is that I do not know what I should do with $n$ which defines set over I integrate. To move limit inside I should remove it. Is it correct to move it inside anyway and just assume that integral is defined over $\mathbb{R^2}$?

Define $M_n:=\{(x,y) : 2x^2+y^2<n\}$
and $$f_n(x,y):=\chi_{M_n}(x,y)\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}.$$
Note that $f_n\geq0$ converges monotonically increasing to $f(x,y):=e^{-2x^2-y^2}x^{2m}.$ With Levi you may interchange the limit with integral and then use Fubini:
\begin{align}
&\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{2x^2+y^2<n}\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}dxdy \\
&=\sum_{m=0}^\infty \lim_{n\rightarrow\infty}\int_{\Bbb R^2}\chi_{M_n}(x,y)\left(1-\frac{2x^2+y^2}{n^2}\right)^{n^2}x^{2m}dxdy \\
&=\sum_{m=0}^\infty\int_{\Bbb R^2}e^{-2x^2-y^2}x^{2m}dxdy \\
&=\sum_{m=0}^\infty\int_{\Bbb R^2}e^{-2x^2}x^{2m}e^{-y^2}dxdy \\
&=\sum_{m=0}^\infty\int_{\Bbb R}e^{-2x^2}x^{2m}dx\int_{\Bbb R}e^{-y^2}dy \\
&=\sqrt{\pi}\sum_{m=0}^\infty\int_{\Bbb R}e^{-2x^2}x^{2m}dx.
\end{align}
Now some kind of Gamma function is left. Try to proceed from here.

Update: Well, when I transform $\int_{\Bbb R}e^{-2x^2}x^{2m}dx$ further, I'm getting something like $\approx\left(\frac{1}{2}\right)^{m+\frac{1}{2}}\Gamma(m+\frac{1}{2})$, which doesnt converge to $0$. Thus, I'm having the notion that the series doesnt exist.

Update2: Okay, an easier way of seeing this: For the summands we obtain:
\begin{align}
\int_{\Bbb R}e^{-2x^2}x^{2m}dx
\geq \int_{[1,\infty)}e^{-2x^2}x^{2m}dx
\geq \int_{[1,\infty)}e^{-2x^2}dx=:\delta>0.
\end{align}
This positive $\delta$ is independent of $m$ and thus the series over them goes to infinity.

Let $D_n$ be the set $\{(x,y)\in\Bbb R^2:2\,x^2+y^2<n\}$ and $\chi_n\colon\Bbb R^2\to\Bbb R$ its characteristic function. Then for any function $f(x,y)$ we have
$$
\int_{D_n}f(x,y)\,dxdy=\int_{\Bbb R^2}\chi_n(x,y)\,f(x,y)\,dxdy.
$$


Is the boundedness necessary to extend harmonically? "If $u$ is harmonic and bounded in the punctured disk $0<|z|< \rho$, then $u$ can be extended harmonically to the disk $|z|<\rho$ harmonically." This fact has been shown here.

My Question is: Is  the  hypothesis bounded  necessary in the above fact? Or can find counter example?

[In the answer given by Georges Elencwajg, I am unable see where he is using bounded assumption]

If a harmonic extension to the disk exists, then that extension is in particular continuous, and hence bounded on all disks $\{ z : \lvert z\rvert \leqslant r\}$ for $0 < r < \rho$, so it is a necessary condition that $u$ be bounded on some punctured disk $\{ z : 0 < \lvert z\rvert \leqslant r\}$, but it need not be bounded on the whole punctured disk $\{ z : 0 < \lvert z\rvert < \rho\}$, $u$ can be unbounded as $z$ approaches the boundary circle $\lvert z\rvert = \rho$.

In the linked answer, the boundedness is used in the "little approximation argument that you will find on page 33" that shows that $U$ coincides with $u$, I suppose. Without the boundedness, you cannot show that. I haven't looked at the linked book, but the usual approximation argument is as follows:

$v := U - u$ is a bounded harmonic function on $\{ z : 0 < \lvert z\rvert < r\}$, and on the circle $\lvert z\rvert = r$, it has boundary values $0$. Now let $\varepsilon > 0$ and consider

$$v_\varepsilon(z) = v(z) + \varepsilon \cdot \log \frac{\lvert z\rvert}{r}.$$

Then $v_\varepsilon$ is a harmonic function on $0 < \lvert z\rvert < r$, has boundary values $0$ on the circle $\lvert z\rvert = r$, and, since $v$ is bounded,

$$\lim_{z\to 0} v_\varepsilon(z) = -\infty.$$

For all sufficiently small $s > 0$, $v_\varepsilon$ is therefore strictly negative on the circle $\lvert z\rvert = s$, and by the maximum principle for harmonic functions, it follows that $v_\varepsilon(z) < 0$ for all $z$ in the annulus $s < \lvert z\rvert < r$. Letting $s \to 0$, we see that $v_\varepsilon < 0$ on the punctured disk $0 < \lvert z\rvert < r$.

This holds for all $\varepsilon > 0$, and now letting $\varepsilon \to 0$ it follows that $v(z) \leqslant 0$ for $0 < \lvert z\rvert < r$.

The same argument for $w_\varepsilon(z) = v(z) - \varepsilon\cdot \log \frac{\lvert z\rvert}{r}$ shows $v \geqslant 0$ on $0 < \lvert z\rvert < r$, so together we have $v \equiv 0$, or $U \equiv u$.

If you assume $u(z)/\ln|z|\to 0$ as $z\to 0,$ then $0$ is a removable singularity for $u.$


A more natural solution to finding the general terms of a recurrence relation in $2$ variables A high school contest math problem in a problem book:

Find the general terms of
$$a_{1}=a,\quad b_{1}=b,\quad a_{ n + 1 }=\frac { 2 a _ { n } b _ { n } } { a _ { n } + b _ { n } },\quad b_{ n + 1 }=\sqrt{ a_{n+1} b _ {n}}$$

The easiest method is to use the following trigonometric identities:

$$2 \tan \frac { \theta } { 2 } = \frac { 2 \tan \theta \sin \theta } { \tan \theta + \sin \theta } , \quad 2 \sin \frac { \theta } { 2 } = \sqrt { 2 \tan \frac { \theta } { 2 } \sin \theta }$$

We make the substitution

$$a _ { 1 } = a = c \cdot \tan \theta , \quad b _ { 1 } = b = c \cdot \sin \theta$$

Observe that

$$a_{2}=\frac{2a_1b_1}{a_1+b_1}=\frac{2c\tan\theta\sin\theta}{\tan\theta+\sin\theta}=2c\tan\frac{\theta}{2}$$
$$a_{3}=\frac{2a_2b_2}{a_2+b_2}=\frac{2(2c)\tan\frac{\theta}{2}\sin\frac{\theta}{2}}{\tan\frac{\theta}{2}+\sin\frac{\theta}{2}}=2^2c\tan\frac{\theta}{2^2}$$
$$...$$
$$a_n= 2^{n-1} c \tan \frac { \theta } { 2 ^ { n - 1 } }$$

It's possible to express the above formula in terms of $a$ and $b$, just notice

$$\theta=\cos^{-1}\left(\frac{b}{a}\right), c=\frac{a}{\tan\theta}$$

Similarly,

$$b_n=2 ^ { n - 1 } c\sin \frac { \theta } { 2 ^ { n - 1 } }$$

Although I consider the above solution as elegant, it's unnatural at first glance, as the $2$ trigonometric identities are rarely used in practice, at least for those who are mediocre in trigonometry. So my question is:

Is there an alternative solution (non-trigonometric) to the above recurrence relation that is more natural in a certain sense?

Edit: The above method has a defect: complex numbers may pop up for some real values of $a$ and $b$ in the substitution step.

There are several possible methods. A general method that I have
used for similar compound means is the following. First, we want
to find a recurrence for each sequence by itself. Thus, given
$\,a_0 = a, \,  b_0 = b, \, b_1 = \sqrt{a_1b_0}\,$
we solve for $\,a_1\,$ getting
$\,a_1 = b_1^2/b_0.\,$ Similarly $\,a_2 = b_2^2/b_1.\,$ By using
$\,a_2 = 2a_1b_1/(a_1+b_1)\,$
 we get a quadratic in $b_2$ given $b_0$ and $b_1$
 $$ b_2^2(b_0+b_1)-2 b_1^2 = 0. \tag{1} $$
Now we need an Ansatz. A little bit of numerical work suggests that the convergence of $b_n$ is linear. Thus, assume for some $\,0<r<1,\,$
without loss of generality, that $\,b_n = f(x r^n)\,$ for $\,n\ge 0\,$
where $$\,f(x) := 1 + x + c_2 x^2 + c_3 x^3 + c_4 x^4 + O(x^5). \tag{2}$$
Substitute this in equation $(1)$. The left side is
$\,(r-1)(4r-1)x + O(x^2).\,$ Solving for $\,r\,$ by the hypothesis,
$\,r=1/4.\,$ Solving for the other coefficients of $\,f(x)\,$ using equation
$(1)$ we get
$$ f(x) = 1 + x + 3/10 x^2 + 3/70 x^3 + 1/280 x^4 + 3/15400 x^5 + O(x)^6.
\tag{3}$$
Looking for the denominators leads to OEIS sequence A007019

  Denominators of coefficients of the Taylor series of sinh(sqrt(2*x))/(sqrt(2*x))

This gives us the solution
 $$ f(x) = \sinh(\sqrt{6x})/\sqrt{6x}). \tag{4}$$
Combining our results gives us
$$ b_n = c\sinh(t/2^n)/(t/2^n) \tag{5}$$

Now we do something similar for $\,a_n.\,$ Thus, given
$\,a_1 = 2 a_0 b_0/(a_0+b_0)\,$ we solve for $\,b_0\,$ getting
$\,b_0 = a_0 a_1/(2a_0 - a_1).\,$ Similarly,
$\,b_1 = a_1 a_2/(2a_1 - a_2)\,$ but since $\,b_1^2 = a_1 b_0\,$
we finally get
 $$ a_0(a_2^2 -4a_1a_2-4a_1^2)-a_1a_2^2 = 0. \tag{6}$$
This is a quadratic in $a_2$ given $a_0$ and $a_1$.

A little bit of numerical work suggests that the convergence of
 $a_n$ is linear. Thus, assume for some $\,0<r<1,\,$
without loss of generality, that $\,a_n = f(x r^n)\,$ for $\,n\ge 0\,$
where $$\,f(x) := 1 + x + c_2 x^2 + c_3 x^3 + c_4 x^4 + O(x^5). \tag{7}$$
 Substitute this in equation $(6)$. The left side is
$\, -(1-r)(1-4r)x + O(x^2).\,$ Solving for $\,r\,$ by the hypothesis,
$\,r=1/4.\,$ Solving for the other coefficients of $\,f(x)\,$ using equation
$(1)$ we get
$$ f(x) = 1 + x + 6/5 x^2 + 51/35 x^3 + 62/35 x^4 + O(x^5).\tag{8}$$
The numerators are all divisible by $3$ so this suggests looking at
$$ f(x/3) = 1 + 1/3 x + 2/15 x^2 + 17/315 x^3 + O(x^4). \tag{9}$$
A lookup of the numerators leads to OEIS squence A002430

  Numerators in Taylor series for tan(x). Also from Taylor series for tanh(x).

which gives $$ f(x/3) = \tanh(\sqrt{x})/\sqrt{x}. \tag{10}$$
Combining our results gives us
$$ a_n = c\tanh(t/2^n)/(t/2^n). \tag{11}$$

To summarize, both $\,c := \lim a_n = \lim b_n\,$ as $\,n\to\infty\,$
and $\,t\,$ depend on the initial values $\,a,b.\,$
Let $\,d := \sqrt{b^2-a^2},\,$ and $\, t := \log((b+d)/a). \,$ Then
$\, c = t\ a\ b/d.\,$

Note that if $\,0<a<b\,$ then $\,t>0\,$ is real while if $\,0<b<a\,$ then $\,t\,$ is pure imaginary. Here are numerical examples for both cases:

c=1.520691, d=1.732050, t=1.316957
 n    a_n       b_n       a_n/c    b_n/c    (t/2^n)^2
 0  1.000000  2.000000  0.657595  1.315191  1.734378
 1  1.333333  1.632993  0.876793  1.073849  0.433594
 2  1.468027  1.548315  0.965368  1.018165  0.108398
 3  1.507103  1.527570  0.991063  1.004523  0.027099
..  ........  ........  ........  ........  ........
10  1.520691  1.520692  0.999999  1.000000  0.000001

c=1.209199, d=-1.732050, t=1.047197 i
 n    a_n       b_n       a_n/c    b_n/c    (t/2^n)^2
 0  2.000000  1.000000  1.653987  0.826993  -1.096623
 1  1.333333  1.154701  1.102658  0.954929  -0.274155
 2  1.237604  1.195434  1.023459  0.988615  -0.068538
 3  1.216154  1.205759  1.005751  0.997146  -0.017134
..  ........  ........  ........  ........  .........
10  1.209200  1.209199  1.000000  0.999999  -0.000001

An algorithmic solution would be a “natural solution” in the sense that the algorithm performs the computations specified by the recurrence relations. For an example of a non-recursive model for the computation please check out the following:

"Computing the N-th Term of a Recursive Relation Using a Non-Recursive Function – A Reply to a Question at Mathematics Stack Exchange"


Checking irreducibility I have the polynomial $f(X)=X^{2n}-2X^{n}+1-p$ where $p$ is a prime number and $n\in\mathbb{N}$. I want to check whether it is irreducible or not over $\mathbb{Q}[X]$.

If $2^{2}\nmid1-p$ then $f(X)$ is irreducible by Eisenstein's Criterion. However, I can't make any progress when I consider the polynomial $f(X)=X^{2n}-2X^{n}+4r, r\in\mathbb{Z}$.

Any hints?

You are correct, it is always irreducible.

Your polynomial factors as
$AB$ where $A=X^n-(1+\sqrt{p})$ and $B=X^n-(1-\sqrt{p})$. It will suffice
to show that $A$ (and therefore $B$ also) is irreducible over $K={\mathbb Q}[\sqrt{p}]$.

Thanks to Karpilovsky’s theorem (many thanks to Bill Dubeque for
quoting it here), it will suffice to show the following points :

(1) $c=1+\sqrt{p}$ is not a $m$-th power in $K$, for any $m\geq 2$.

(2) $c$ is not of the form $-4z^4$ with $z\in K$.

Proof of (1) : suppose $1+\sqrt{p}=(x+y\sqrt{p})^m$ with $x,y\in{\mathbb Q}$.
 Then $1-p=d^m$ where $d$ is the rational number $d=x^2-py^2$. So $d$ is a rational root of the monic polynomial $X^m-(1-p)$, so $d$ is an integer. As $1-p<0$, $d$ must be a negative integer and
 $m$ is odd. Then $p=1-d^m$ is divisible by $1-d>0$, so
 $1-d$ can only be $1$ (clearly impossible) or $p$. So $1-d=p,d=1-p$ and hence
 $(1-p)^{m-1}=1$, which occurs only when $p=2$.

We then have $1+\sqrt{2}=(x+y\sqrt{2})^m$, $m$ odd and $x^2-2y^2=-1$. Each real
 number has a unique $m$-th real root, so $x+y\sqrt{2}=(1+\sqrt{2})^{\frac{1}{m}}$
 and hence $x-y\sqrt{2}=\frac{x^2-2y^2}{x+y\sqrt{2}}=(1-\sqrt{2})^{\frac{1}{m}}$. Adding
 those two last equalities, one obtains

$$
 x=\frac{(1+\sqrt{2})^{\frac{1}{m}}+(1-\sqrt{2})^{\frac{1}{m}}}{2}
 $$

Then, $r=2x$ is both rational and a sum of two algebraic integers, so it
 must be an integer. Now,

$$
 r=(\sqrt{2}-1)^{\frac{1}{m}}
 \Bigg(\bigg(\frac{\sqrt{2}+1}{\sqrt{2}-1}\bigg)^m-1\Bigg) >0
 $$

On the other hand, $\big(\frac{3}{2}\big)^3 > \sqrt{2}+1$ yields
 $(\sqrt{2}+1)^{\frac{1}{m}} \leq \frac{3}{2}$, and
 $\sqrt{2}-1 > \big(\frac{1}{2}\big)^3$ yields $1+(\sqrt{2}-1)^{\frac{1}{m}} \geq
 \frac{3}{2}$. Combining the two, we obtain $r<1$. Finally $r$ is an integer
 strictly between $0$ and $1$, which is impossible.

Proof of (2) :  $1+\sqrt{p}=-4(x+y\sqrt{p})^4$ with $x,y\in{\mathbb Q}$ is clearly impossible as a fourth power cannot be negative.

Well, the polynomial $f(X)$ can be written as $(X^n-1)^2 - p$. If $f(X)$ is reducible then there must exist a X such that :
$$ f(X) = (X^n-1)^2 - p = 0 $$ which means that
$$(X^n-1)^2 = p$$
It necessary follows that $(X^n-1)^2 ∈ N$ ergo $X^{n}-1$ is a natural number, which implies that $X ∈ N$ ... unfortunately there exists no prime number that is also a square number, otherwise it would have at least 1 more divisor different from itself and 1...so the polynomial $f(X)$ is irreducible for any $X ∈ Q.$


Quadratic form reduction on the $n$-dimensional complex space $\mathbb{C}^n$ How to reduce the quadratic form
$$\sum_{1 \leq k < l \leq n}(k + il)x_kx_l$$
to the canonical form on the $n$-dimensional complex space $\mathbb{C}^n$?
I started with its symmetric matrix
\begin{align*}
S = \frac{1}{2}\begin{pmatrix}
0 & 1 + 2i & 1 + 3i & \cdots & 1 + ni \\
1 + 2i & 0 & 2 + 3i & \cdots & 2 + ni \\
1 + 3i & 2 + 3i & 0 & \cdots & 3 + n i \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 + ni & 2 + ni & 3 + ni & \cdots & 0
\end{pmatrix},
\end{align*}
and then tried to reduce it to the diagonal matrix following the algorithmic recipe, i.e., tried reducing it by eliminating the off diagonal elements row/column after row/column: the first step is to clear the first two rows/columns by calculating $P^TSP$, where
\begin{align*}
P = \begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
1 &  1 & 0 & \cdots & 0 \\
0 &  0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 &  0 & 0 & \cdots & 1
\end{pmatrix}
\begin{pmatrix}
1 &  0 & -\frac{1 + 3i}{1 + 2i} & \cdots & -\frac{1 + ni}{1 + 2i} \\
0 &  1 & \frac{1}{2(1 + 2i)} & \cdots & \frac{1}{2(1 + 2i)} \\
0 &  0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 &  0 & 0 & \cdots & 1
\end{pmatrix}.
\end{align*}
The resultant matrix $P^TSP := (s_{ij})$ has already quite complicated entries:
\begin{align*}
& s_{kk} = \frac{k^2 - 2 - 3ki}{1 + 2i}, \; k = 3, 4, \ldots, n; \\
& s_{kl} = \frac{k - 2l + 2kl - 4 - li}{2(1 + 2i)}, \; l > k,
\end{align*}
making the subsequent calculation notoriously nasty, so I am still inconclusive.
Is my attempt on the right track to solve this problem? I think there probably should be a better and cleaner way.

This might be helpful
If we define
$$
\Phi_1:=\sum_{1\leq k<l\leq n}(k+il)x_kx_l
$$
and
$$
\epsilon_{k,l}=\left\{\begin{array}{cc}
1\textrm{ , if }k<l\\
i\textrm{ , if }k>l\\
0\textrm{ , else }
\end{array}\right\}\textrm{, }i=\sqrt{-1}
$$
then
$$
\Phi_1=\sum^{n}_{k,l=1}\epsilon_{kl}kx_kx_l=\frac{1+i}{1-i}\sum_{k<l}(k+l)x_kx_l-\frac{2i}{1-i}\sum^{n}_{k<l}kx_kx_l
$$


the use of Cartesian product I am currently studying in secondery level. I read cartesian product the other day and I found it absolutely bizarre. In my text book, there is this "order pair" which I understood fairly well and then there is cartesian product in which we multiply two sets. I don't understand the concept behind it. What is its application? How will it come to use in higher level study?

The basic point of set theory in mathematics is that it allows us to wrap up properties and treat them as if they were things.
Without sets we can say things like

$2.5$ is a rational number.
$\pi$ is not a rational number.
$36$ is a perfect square.
$(0,42)$ is an ordered pair of two whole numbers.

The wordings "is a rational number", "is a perfect square", "is an ordered pair of two whole numbers" denote properties. They are nice for talking about particular things that have or don't have the property. But our language doesn't allow us to speak about, for example, an expression whose value is a property, or a function that takes a property as input or things like that.
This is where sets come in. A set is a way to "box up" a property such that it becomes a "thing" that we can speak about together with other similar things. To a large extent, a property and a set are different ways of thinking about the same concept. If we have a property, say "is a thingamajig" we can always (with some technical exceptions that you don't need to worry about at this level) create a set from it: $$A=\{x\mid x\text{ is a thingamajig}\}$$
On the other hand, whenever we have a name or an expression for a set, this gives us a property: "is a member of such-and-such set".
Because sets are "things", they can have (or not have) properties themselves -- such as the property "is a set of real numbers (but possibly not all the real numbers)". Boxing up these properties, we get sets of sets and so forth.
Now the cartesian product $A\times B$ is neither more nor less than the set corresponding to the property "is an ordered pair of something from $A$ and something from $B$".
Thus, whenever we want to speak about ordered pairs at all, it may suddenly become useful to speak about the property of being an ordered pair. And then it immediately becomes useful to speak about the boxed-up property that corresponds to it. In fact so useful that the common way to say "is an ordered pair of something from $A$ and something from $B$" is "is in $A\times B$". This is shorter and more concise.
So cartesian products are used, or potentially used, everywhere we use ordered pairs in the first place.

also i just got to learn that we use set cartesian products in combinatorial problems .An example of this is a three course meal where there is an appetiser main course and desert.Lets say for example we have porridge,soup,lemonade and main course chicken,omlades,cooked potatoes,and desert icecream fruitpoodingand chocolate one can have soup,chicken,chocolate or soup,omlates,fruitpooding etc thus such is an example of application of set cartesian product


reordering the indices of a matrix Let $A$ be an $n \times n$ matrix of rank r. Then by reordering the indices if necessary we can bring the matrix in the form $(\frac{A_1}{A_2})$ where $A_1$ is an $r \times n$ matrix, $A_2$ is an $n-r \times n$ matrix and $A_1$ has rank r.

similarly by reordering the indices we can write the matrix $A$ as a direct sum of in decomposable matrices.

I am just thinking how to prove these two facts.

{{{{{ Note: The term irreducible is usually used instead of indecomposable.

Wikipedia: "...a matrix is irreducible if it is not similar via a permutation to a block upper triangular matrix (that has more than one block of positive size)."
(Replacing non-zero entries in the matrix by one, and viewing the matrix as the adjacency matrix of a directed graph, the matrix is irreducible if and only if the digraph is irreducible.)

PlanetMath: reducible matrix
"An n×n matrix A is said to be a reducible matrix if and only if for some permutation matrix P, the matrix PTAP is block upper triangular matrix."
If a square matrix is not reducible, it is said to be an irreducible matrix }}}}}

When we reordering the indices actually what we are doing geometrically or algebraically about the matrix?

Thanks for your valuable time and interest.

You just need to choose $r$ linearly independent rows and move them to the top. That would be left-multiplying by an $n \times n$ permutation matrix.


Coin toss game - Probability of winning Question:
Two players A and B, alternatively toss a fair coin (A tosses the coin first, then B, than A again, etc.). The sequence of heads and tails is recorded and if there is head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins. What is the probability that A wins the game?

Solution:
I was told the following solution. If $P(A),P(B)$ are the probabilities that A, B win, then we can write:
$$P(A) = P(A\mid H)P(H)+P(A\mid T)P(T)$$
where $P(H)=P(T)=1/2$ are the probabilites that A gets H, T in the first toss respectively. Then $$P(A\mid T)=P(B)=1-P(A)\tag{1}$$ and $$P(A\mid H)=1/2\cdot 0+1/2\cdot (1-P(A\mid H))\tag{2}$$
which gives $P(A)=4/9$.

I am trying to understand equation (1),(2). I understand that if A gets T then the game essentially starts over with B being the first player. But since we want to know the probability of A winning, then shouldn't we have $P(A\mid T)=1-P(B)$? As for (2) I have not been able to find a starting point. Would somebody be able to explain their derivation?

If the first toss is $T$, then $B$ cannot win on the second toss;
their chance to win after this are the same as $A$'s chance to win
was at the start of the game.
That is, after this $T$ toss the game essentially starts over,
except that now $B$ has the first toss instead of $A$;
the roles are reversed, so $A$'s chance to win ($P(A \mid T)$)
is the same as $B$'s chance to win the original game ($P(B)$).

If the first toss is $H$, there is a $\frac12$ chance of tails, in which
case $A$ loses immediately (hence the $\frac12 \cdot 0$ term
in the formula for $P(A \mid H)$),
and a $\frac12$ chance of heads, in which case
$A$ is now in the position that $B$ was in after the first head,
so $A$'s chance to win after two heads is $1$ minus their chance to win
after one head, hence the $\frac12 \cdot (1 - P(A \mid H)$ term,

if A throws the coin and get the tail then its prob to win in first throw is 1/2

if A does not get a tail so B flips and say it gets a tail then we have 1/2*1/2

if now B does not get a tail then A flips back and say it gets a tail then we have 1/2*1/2*1/2

so proceeding in this way the probability for A to win is 1/2+(1/2*1/2*1/2)+(1/2*1/2*1/2*1/2*1/2)+......... =2/3

so probability for B to win is 1-(2/3)=1/3


How to calculate $2^{\sqrt{2}}$ by hand efficiently? I've been trying to calculate $2^{\sqrt{2}}$ by hand efficiently, but whatever I've tried to do so far fails at some point because I need to use many decimals of $\sqrt{2}$ or $\log(2)$ to get a roughly good approximation.

Is it even possible to do so without facing irrational expressions like $\sqrt{2}$ or $\log(2)$ in our calculations?

EDIT

It seems like no one is paying attention to the requirements in my question at all : (  You are not allowed to use use $\log(2)$ or $\sqrt{2}$ in your answers. Use of continued fractions is allowed. Let me phrase my question in this way: Find an infinite series $\displaystyle \sum_{n=0}^{\infty}a_n$ such that $a_n \in \mathbb{Q}$. There exists at least one such series, namely, the series that is obtained by writing the decimal expansion of $2^{\sqrt{2}}$, but that series is good for nothing because if we already knew the decimal expansion of $2^{\sqrt{2}}$ then we didn't need to be after approximating $2^{\sqrt{2}}$ by using infinite series.

Look at the following series:

$\displaystyle e = \sum_{n=0}^{\infty}\frac{1}{n!} = 2 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}+\frac{1}{120}+\frac{1}{720} + \cdots$

$\displaystyle \pi = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \frac{4}{11} + \cdots$

$\displaystyle \pi = 3 + \frac{4}{2\times 3 \times 4} - \frac{4}{ 4 \times 5 \times 6} + \frac{4}{6 \times 7 \times 8} - \frac{4}{8 \times 9 \times 10} + \cdots $

Both $e$ and $\pi$ are irrational transcendental numbers. But we have found non-trivial infinite series with rational terms for them. Can someone possibly find a similar series for $2^{\sqrt{2}}$? This is something I proposed as a challenge to myself and I failed, now I wonder if someone on here could tackle it.

${\large\mbox{Hint}}$:

$$
\sqrt{\vphantom{\large A}2\,}
=
\sqrt{98 \over 49}
=
{1 \over 7}\,\sqrt{\vphantom{\large A}100 - 2}
=
{10 \over 7}\,\sqrt{\vphantom{\large A}1 - {1 \over 50}}
\approx
{10 \over 7}\,\left(1 - {1 \over 2}\,{1 \over 50}\right)
=
{10 \over 7} - {1 \over 70}
$$

What you can do is expand the function $2^x$ into its Taylor's series around x = 0 and use the first several terms of the series. Now you are evaluating a polynomial, which is much easier.

The series does not converge uniformly, so the farther x is from 0 the more terms you are going to need. But $ \sqrt 2$ is not all that far out.  You can get estimates of the remainder  (lots of textbooks discuss these remainders) and see how many terms you need to get the convergence you want.

Noted as per comment below that the coefficients of this series involve powers of log2.  However,  log2 is the sum of the alternating harmonic series, so some number of terms of that can be used to approximate it.


$2D$ Gaussian Density Reducing to $1D$ Case as Correlation Approaches $1$ I was asked the following question and thought it was interesting.
Is there an obvious way in which we can show the density of a 2D Gaussian "collapses" to a 1D density as the correlation $\rho \rightarrow 1.$ The intuition is that by having two variables $X$ and $Y$ that are perfectly correlated, we should see a function of just one of the random variables (since two of the same variable is redundant). For simplicity, we can take $\mathbb{E}(X) = \mathbb{E}(Y) = 0.$
More precisely, of course we cannot take $\rho = 1$ directly, since this makes the covariance matrix $\Sigma$ singular. But if we write the correlation as say $\rho = 1 - \epsilon$ and take $\epsilon \rightarrow 0$, can we see such a "collapse" in the density to the case of a single Gaussian random variable? I messed around with this for a while, but I couldn't get it to work.

Note that there is no reason why the covariance matrix cannot be singular. The only difference is that the distribution will not have a density with respect to the 2d Lebesgue measure. However, the general definition of a multivariate Gaussian distribution is a random vector $Z$ such that $\langle Z,v \rangle$ has a (1D) Gaussian distribution, for any vector $v$, which does assume anything about the covariance matrix. In particular, if $\rho=1$, then the $x$ and $y$ coordinates are perfectly linearly related, which implies that the distribution is of the form $Z=(X,aX+b)$ for $X$ a 1D Gaussian and some constants $a$ and $b$. Note that this distribution is concentrated on a line, and is multivariate gaussian because $\langle Z,v \rangle=(v_1+av_2)X+bv_2$ is Gaussian, for any $v\in\mathbb{R}^2$.

While it's true that as $\rho\to1$ the random variables $X$ and $Y$ get closer to  each other (in the sense $P(|X-Y|>\epsilon)\to 0$), it is not the case that the 2D density converges to a 1D density, because the 2D density remains a function of    two variables $(x,y)$ and is constrained to integrate to $1$.
As $\rho\to1$ the bulk of the 2D density gets narrower, concentrating nearer and   nearer to the line $y=x$, but at the same time the 2D density gets higher and      higher, since its peak value is $\dfrac1{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}$. So as a function of two variables, the shape of the joint density is getting more and more extreme as $\rho\to1$.     (But the marginal densities of $X$ and $Y$ remain unchanged.)


Associating the Hamiltonian vector field $X_f$ to a function $f$ is a Lie algebra homomorphism Let $(M, \omega)$ be a complex analytic symplectic variety, i.e. $\omega \in \Gamma(\Lambda^2 \Omega_M)$ is a closed holomorphic 2-form, which is everywhere non-degenerate. We can view $\omega$ as an isomorphism $\omega: \Omega_M \to T_M$, and let $\theta \in \Gamma(\Lambda^2 T_M)$ be its inverse. Then the Poisson bracket for two functions $f, g \in \mathcal{O}_M$ is defined by
$$\{f, g\} = \theta(d f, dg) = \langle \theta, df \wedge dg \rangle,$$
where $\langle -, - \rangle$ denotes the usual evaluation of a vector and a form. This is a Poisson bracket, i.e. a Lie bracket which satisfies the Leibniz rule for multiplication.
Using this, on can define a vector field $X_f$ associated to $f$ by declaring
$$X_f(g) = \{f, g\}.$$
In Systèmes Hamiltoniens Complètement Intégrables Associés aux Surfaces K3, Beauville then claims that the map $f \mapsto X_f$ is a homomorphism of Lie algebras. To check this, I tried to evaluate
$$X_{\{f, g\}}(h) = \theta(d\{f, g\}, d h) = \theta\big(d\theta(d f, d g), d h\big), \tag{$*$} $$
but I don't know how to go on from here.

I get that $\{f, g\} = X_f(g) = \mathcal{L}_{X_f}(g)$, and $[X_f, X_g] = \mathcal{L}_{X_f}(X_g)$, where $\mathcal{L}_{X_f}$ is the Lie derivative with respect to $X_f$. But to conclude $X_{\{f, g\}} = [X_f, X_g]$ we would have to know that $\mathcal{L}_{X_f}$ commutes with the mapping $g \mapsto X_g$.

Any help would be appreciated :)

Vector fields get the lie algebra structure with lie derivative.{g, f} =dg(Xf) =w(Xg, Xf).  The isomorphism of Xf and f is obsious then. Since lie bracket has the same form ( structure) as lie derivative.


Closed form of $\sum_{k=0}^{n}k\cdot2^{n-k}$ I am interested in solving the recurrence relation using iterative method. I know how to solve it using generating function and another method using solution of associated linear homogeneous recurrence relation and particular solution described in Rosen's Discrete Math book. I came up with some summation (below) but I don't know how to find its closed form. I checked WolframAlpha and it provided me with this $$\sum_{k=0}^{n}k\cdot2^{n-k}=-n+2^{n+1}-2$$ How to show this?

HINT:

$$\sum_{k=0}^nk2^{n-k}=2^n\sum_{k=0}^nk\left(\frac12\right)^k\;,$$

so you’re done if you can evaluate

$$f(x)=\sum_{k=0}^nkx^k$$

at $x=\frac12$. Start with $g(x)=\sum_{k=0}^nx^k$, differentiate, and multiply by $x$. (I’m assuming that you know a closed form for $g(x)$.)


What is the difference between gradient of divergence and Laplacian? $\nabla^2\mathbf{V}$ versus $\nabla (\nabla \cdot \mathbf{V})$. They seem pretty much the same to me. Both give back a vector.

Let me break this out in components. I let $\partial_i~=~\frac{\partial}{\partial x_i}$. Clearly the divergence of a vector ${\bf V}~=~{\bf i}V_x~+~{\bf j}V_y~+~{\bf k}V_z$ and the gradient operator $\nabla~=~{\bf i}\partial_x~+~{\bf j}\partial_y~+~{\bf k}\partial_z$ is
$$
\nabla\cdot{\bf V}~=~({\bf i}\partial_x~+~{\bf j}\partial_y~+~{\bf k}\partial_z)\cdot({\bf i}V_x~+~{\bf j}V_y~+~{\bf k}V_z)
$$
$$
=~\partial_xV_x~+~\partial_yV_y~+~\partial_zV_z~=~\sum_i\partial_iV_i.
$$
So far so good. Now let us take the divergence of this
$$
\nabla\nabla\cdot{\bf V}~=~\sum_j{\bf e}_j\partial_j\sum_i\partial_iV_i
$$
$$
=~\sum_{i,j}{\bf e}_j\partial_j\partial_iV_i~=~\sum_i{\bf e}_i\partial_i\partial_iV_i~+~\sum_{i\ne j}{\bf e}_j\partial_j\partial_iV_i.
$$
The first term on the right on the equal sign is $\nabla^2\bf V$, but the second term has mixed partials.

If instead you take the gradient of a scalar $\nabla\phi$ this is
$$
\nabla\phi~=~{\bf i}\partial_x\phi~+~{\bf j}\partial_y\phi~+~{\bf k}\partial_z\phi
$$
The divergence of this will produce $\nabla^2\phi$

The Laplacian is a scalar function and returns a scalar value.  The gradient of a function returns a vector value.


$\sin 2x - \sin^2x = 20\sin x + 10 \cos x - 1$ with $90\leq x \leq 180$ solve $\sin2x - \sin^2x = 20\sin x + 10 \cos x - 1$ with $90\leq x \leq 180$

I arrived at $2\sin x \cos x + \cos^2 x = 20\sin x + 10 \cos x $ and resulting in $(2\sin x + \cos x)(\cos x - 10) = 0$ but i am stuck at finding the x satisfying $(2\sin x + \cos x) = 0$. Maybe i was wrong somewhere. What should i do?

It's enough to solve the following equation:
$$2\sin{x}+\cos{x}=0.$$
Now, if $\cos{x}=0,$ so $\sin{x}=0,$ which gives $\sin^2x+\cos^2x=0,$ which is a contradiction.

Thus, it's $$\tan{x}=-\frac{1}{2}$$ or
$$x=-\arctan\frac{1}{2}+180^{\circ}k,$$ where $k\in\mathbb Z,$ which gives
$$x=180^{\circ}-\arctan\frac{1}{2}.$$


Did I simplify the boolean expression correctly? I wanted someone to check and make sure that I did this right, I'm hoping that I did but if I didn't, please point me in the right direction of where I went wrong

$\begin{align}&\rm y(x'z+xz')+x(yz+yz')
\\ =\quad &\rm y(x'z+xz')+xy(z+z')&& \text{Distribution}
\\ =\quad &\rm y(x'z+xz')+xy(1)&& \text{Complementation (/Inversion)}
\\ =\quad &\rm x'yz+xyz'+xy &&\text{Distribution and Identity.}
\end{align}$

You were okay up to here, then you went all agley.   Instead, the next step should be to distribute out the $y$ factor.

$\begin{align}=\quad &\rm y(x'z+xz'+x) &&\text{Distribution}
\end{align}$

From here your path to the destination should be obvious.

After that : y(x'z+xz'+x)

Replace x with xz+xz'

= y(x'z + xz'+ xz + xz')

= y(x'z + xz'+ xz)

= y(x'z + x(z'+ z))

= y(x'z + x)

= y(x'z + xz + x)

= y(z(x' + x) + x)

= y(z + x)

= yz + yx

You can check out my simplifier (There is no ads, it's firebase app)

https://cemsina.com/queen/

It works with Quine McCluskey method but you can check your answers.


How is it possible to deduce $A$ from $A\wedge B$? If I know that $A\wedge B = T$  then I can say that surely $A$ and $B$ separately are true, but if $A\wedge B = F$ then how can I infer the truth value of $A$ or the one of $B$. It seems impossible to infer the result : I can say just that  $¬A∨¬B = T$. What am I missing? Am i misuderstanding the process of making deductions?

EDIT:

Is it linked to the fact that $A\wedge B\implies A$ is a tautology? Because this is the reason given in my book. Should I consider just the case $A\wedge B = T$ ?

Looking at the truth-table is helpful for $A \wedge B$
$$
\begin{matrix}
A & B & A \wedge B \\ \hline
F & F & F \\
F & T & F \\
T & F & F \\
T & T & T
\end{matrix}
$$
If you know that $A \wedge B=F$, it could be any of the first three cases.  So you cannot infer $A$'s state or $B$'s state, but you can infer that one of them is false.

If (A $\land$ B) is false, then inferring either A or B is not valid.  In general, inferring anything from a false premise is not valid, though there do exist exceptions such as if (A <-> B) is false, and B is false, then A is true.
Often enough though, in formal systems, it's required that a formula holds true for inferences to get made from it (and the rule of inferences used always preserve truth also).  That's one reason why the axiomatic method has something to recommend to it, and sometimes gets regarded as more rigorous.


Find three numbers given their sum, product and sum of their squares Given three unknown positive integers. Is it possible to find the three numbers if we are given their
Sum->(a+b+c) = X
Product-> (abc) = Y
Sum of Squares-> (a^2 + b^2 + c^2) = Z

Here is a method you can try. Let

$s_1=a+b+c$,

$s_2=a^2+b^2+c^2$,

$p_2=ab+bc+ac$,

$p_3=abc$

$a, b, c$ are the roots of the cubic $$0=(x-a)(x-b)(x-c)=x^3-s_1x^2+p_2x-p_3$$

We don't know $p_2$ but can calculate it using $s_1^2=s_2+2p_2$, and then solve the cubic to find $a,b,c$.


Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$ Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$

I found easily the pairs $(2,2)$ and $(2,3)$ but i can't prove that these are the only pairs possibles.

Any hints?

This is not quite a full answer but it has the main ideas you'll need.

First, let us get the ability to assert a lower bound on $m$ after checking only finitely many pairs. This can be done by noting that $m^n>n!$, so by Stirling's approximation, $n<em$. Thus we can divide the problem between $m<m^*$ and $m \geq m^*$ by directly checking the cases $m=1,2,\dots,m^*-1$ and $n=1,2,\dots,\lfloor em \rfloor$.

Second, let us see what happens if $m$ is large. Similar to the above, if we assume $m>e$, then Stirling's approximation applied to $m^n>m!$ implies

$$m \leq \frac{n}{1-1/\ln(m)}.$$

So it suffices to provide an upper bound on $n$ in order to finish this case.

Third, let $r=\min \{ m,n \}$. Then $r!$ divides $m^n$ and so $r\#$ divides $m$, where $\#$ is the primorial function. Thus in particular $r\# \leq m$.

So there are two cases: $r=m$ and $r=n$. Consider $r=m$. By Bertand's postulate, if $m>2$ then there is a prime $p$ between $m/2$ and $m$, and so $m\# \geq 2p>m$. So $r=m$ can only occur for $m \leq 2$.

Now consider $r=n$. In this case $n\# \leq m$. If we assume $m \geq 8$, then $1-1/\ln(m) \leq 2$, so for $m \geq 8$ the inequality above gives

$$n\# \leq 2n.$$

Applying Bertrand's postulate twice in the same manner as the previous case implies that $n\# \geq n^2/8$. Thus $n \leq 16$.

Putting the pieces together, we divide the problem between $m<8$ and $m \geq 8$. In the case $m<8$ we have $n \leq \lfloor em \rfloor$, which is on the order of 100 cases to check. In the case $m \geq 8$ we have $n \leq 16$ and $m \leq 2n$, which is again on the order of 100 cases to check.

You can using mathematical induction to prove it, but  its not verified for n=m=1
Or make table , row contains n! and column m! Then see any numbers satisfying this equation.


Find $f$ such that $\frac{\mathrm d^2}{\mathrm dx^2}f(x)=f\left(\sqrt{x}\right)$. Which non-constant functions $f$ (if any) satisfy $\dfrac{\mathrm d^2}{\mathrm dx^2}f(x)=f\left(\sqrt{x}\right)$ for $x>0$?
I suspect there is no $f$ which satisfies the differential equation, but I cannot prove this.

I messed up in my first response, but realised this may be a lot easier after all.  Look at analytic solutions here and expand in series.

$$f(x) = \sum_{j=0}^{\infty} f_j x^j$$

$$f''(x) = \sum_{j=0}^{\infty} (j+1) (j+2) f_{j+2} x^j$$

Plug in $x^2$ in the latter and equate like coefficients.  This gives:

$$(j+1) (j+2) f_{j+2} = f_{2j}$$

and

$$f_{2j+1} = 0$$

Now, you can just read off the coefficients by starting with $j=0$ and going through the even values.

$$f_0 = 1 \cdot 2 \cdot f_2$$

$$f_4 = 3 \cdot 4 \cdot f_4$$
or
$$f_4 = 0$$

$$f_8 = 5 \cdot 6 \cdot f_6$$

$$f_{12} = 7 \cdot 8 \cdot f_8$$

Can you see how this can be consistently extended?

shouldn't you integrate twice to answer? making the answer (4/15)x^5/2 + c

http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+4%2F15x%5E%285%2F2%29


Understanding a proof that $\delta(x-x') = \mathrm{d}/\mathrm{d}x\, \theta(x-x')$ A solution to an exercise in Principles of Quantum Mechanics:

My question concerns the leap

This is apparently an application of Integration by Parts; however, it seems to me it is conflating
$$
\int g(x)\,\mathrm{d}\theta(x-x')
$$
with
$$
\int g(x) \theta'(x-x')\,\mathrm{d}x
$$
which is the form required for us to able to apply Integration by Parts.
Is this not the case?

Riemann-Stieltjes integral
Definition. The Riemann–Stieltjes integral of a real-valued function $f$ of a real variable on the interval $[a,b]$ with respect to another real-to-real function $\alpha$ is denoted by
$$\int_a^b f(x) \, \mathrm{d}\alpha(x).$$
The definition of the Riemann–Stieltjes integral can be found here.
Lemma 1. Given an $\alpha(x)$ which is continuously differentiable over $\mathbb{R}$. Then
$$\int_a^b f(x) \, \mathrm{d}\alpha(x) = \int_a^b f(x)\alpha'(x) \, \mathrm{d}x.$$
Lemma 2. Let $f$ be a Riemann-Stieltjes integrable function with respect to $\alpha$ on the interval $[a,b]$. Then $\alpha$ be a Riemann-Stieltjes integrable function with respect to $f$ on the interval $[a,b]$ and
$$\int_a^b f(x) \, \mathrm{d}\alpha(x)=\left[\alpha(x)f(x)\right]_{a}^{b}-\int_a^b \alpha(x) \, \mathrm{d}f(x).$$
A proof of this theorem can be found here.
Analyzing the leap
Choose $f(x) =g(x)$ and $\alpha(x)=\Theta(x-x')$. By Lemma 2,
$$\begin{align}\int_{-\infty}^{\infty} g(x) \, \mathrm{d}\Theta(x-x') &= \left[\Theta(x-x')g(x)\right]_{-\infty}^{+\infty}-\int_{-\infty}^\infty \Theta(x-x') \, \mathrm{d}g(x).
\end{align}$$
If we let $g(x)$ be continuously differentiable over $\mathbb{R}$, by Lemma 1 we get
$$\begin{align}\int_{-\infty}^\infty g(x) \, \mathrm{d}\Theta(x-x') &= \left[\Theta(x-x')g(x)\right]_{-\infty}^{+\infty}-\int_{-\infty}^{\infty} \Theta(x-x')g'(x) \, \mathrm{d}x.
\end{align}$$

Okay well in terms of understanding why this is the case, visualising it graphically, $\theta(x-x')$ is going to be a step from $0$ to $1$ at the point $x=x'$. Now clearly this is a discontinuity but we can visualise this as a vertical line i.e. infinite gradient at the point $x=x'$. Now if we are to plot the gradient of $\theta$ we would see that it is clearly $0$ everywhere other than at this point as it is a constant function over the two domains. As for the integral you mentioned, it is completely fine to use it in that form. the dirac delta is commonly defined as:
$$\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$
so bearing this in mind:
$$\int_{-\infty}^\infty g(x)d\theta(x-x')=\int_{-\infty}^\infty g(x)\frac{d\theta(x-x')}{dx}dx$$
$$=\int_{-\infty}^\infty g(x)\delta(x-x')dx=g(x')$$
which is effectively what this is proving


Square root of $8^3$ I'm only in intermediate algebra. I know that $\sqrt{8^3}$ is equal to $16\sqrt{2}$ but could you simply explain the process on how to get to that?

We can reduce $8^3$ to its prime factors:
$$8^3=512\implies 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$$
Because this is a square root, we are looking for groups of $2$. Similarly, if this was a cube root we would look for groups of 3:
 $$\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}\boxed{2\cdot2}2$$
We have 4 groups of 2 which we will take out of the radical:
$$2\cdot2\cdot2\cdot2\sqrt{2}$$
We can now simplify this as:
$$\boxed{16\sqrt{2}}$$


How to write "the parameter maximizing the maximum of the maximum value of two functions continuous in the domain of maximization" Say you have $f(x),g(x)$ continuous where they need to be and you want to express the following:

Give me the biggest value of $f$ for $x \leq X_f$ , give me the biggest value of $g$ for $x \leq X_g$, get the larger of the two, and return the $x$ that maximized it.

Like say $f=x$ and $g=x^2$ , and $X_f = 2$, $X_g = 1.5$, then the biggest $f$ on that interval is $f(w_f)=2$ and the biggest $g$ on that interval is $g(w_g)=2.25$, so since $2.25>2$ I want to return $w_g=1.5$.

What is the correct way to express this?

Edit: If you are wondering about the motivation, suppose $f$ and $g$ are two utility functions that are continuous for $X \leq \max(X_f,X_g)$. You know that you can only input $x \leq X_f$ into $f$ and $x \leq X_g$ into $g$ and you want to find out which one of $f$ and $g$ is the biggest over those domains and get the $x$ where the max of the max of those two functions occurs so you can plug it into your opponent's utility function.

I think it would be very difficult to express that in a single expression with no preliminary explanation. Here is a way of doing it (that is in my opinion, terrible, compared to just explaining what you mean / introducing auxiliary notation):
$$\operatorname{arg\,max}\mathopen{\bigg[}\max\mathopen{\bigg\{}\!\min(f,\infty\cdot\chi_{(-\infty,X_f]}+(-\infty)\cdot\chi_{(X_f,\infty)}),\min(g,\infty\cdot\chi_{(-\infty,X_g]}+(-\infty)\cdot\chi_{(X_g,\infty)})\mathclose{\bigg\}}\mathclose{\bigg]}$$
Given a set $X$ and a subset $S\subset X$, the function $\chi_S$ (known as the indicator function, or characteristic function, of $S$) is defined as
$$\chi_S(x)=\begin{cases}
1& \text{if }x\in S,\\
0 & \text{if }x\notin S.
\end{cases}$$
Thus,
$$\bigg[\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}\bigg](x)=\begin{cases}
\infty& \text{if }x\in S,\\
-\infty & \text{if }x\notin S,
\end{cases}$$
and thus for any function $f:X\to\mathbb{R}$,
$$\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S})(x)=\begin{cases}
\min\{f(x),\infty\}=f(x)& \text{if }x\in S,\\
\min\{f(x),-\infty\}=-\infty & \text{if }x\notin S.
\end{cases}$$
Given another subset $T\subset X$ and a function $g:X\to\mathbb{R}$, we therefore have that
$$\max\bigg\{\!\min(f,\infty\cdot\chi_S+(-\infty)\cdot\chi_{X\setminus S}),\,\min(g,\infty\cdot\chi_T+(-\infty)\cdot\chi_{X\setminus T})\!\bigg\}(x)=\begin{cases}
\max\{f(x),g(x)\} & \text{if }x\in S\text{ and }x\in T,\\
\max\{f(x),-\infty\}=f(x) & \text{if }x\in S\text{ and }x\notin T,\\
\max\{-\infty,g(x)\}=g(x)& \text{if }x\notin S\text{ and }x\in T,\\
\max\{-\infty,-\infty\}=-\infty& \text{if }x\notin S\text{ and }x\notin T.\end{cases}$$
If $x\in X$ maximizes the above function (i.e. it is an element of the $\operatorname{arg\,max}$ of it), then regardless of which of the four cases holds for that $x$, it is necessarily either an $x$ that maximizes $f(x)$ on $S$, or an $x$ that maximizes $g(x)$ on $T$.

As I interpret your question, one of the bounds $X_f$ and $X_g$ is redundant because while you vary $x$, you have to observe both inequalities $x \leq X_f$ and $x \leq X_g$ at the same time. Hence you can discard the higher bound.

Find $x_0 \leq \min(X_f, X_g)$ such that
$$
\max(f(x_0), g(x_0)) = \max_{x \leq \min(X_f, X_g)}\max(f(x), g(x)).
$$


Number of ways 6 things can be selected from a group of identical things In how many ways can a party of 6 men be selected out of 10 Hindus, 8 Muslims, and 6 Christians if the party consists of at least one person of each religion( Consider only the religion of the person?

Now, I know this problem can be solved by finding the non-negative integral solution to the given equation,

$$x+y+z=6$$
Which the answer turns out to be $\binom{6-1}{3-1}$

But I want to know is there any other method of solving this problem?

Any help would be appreciated.

No.of possibilities according to the given condition is as follows,
$$(1,1,4), (1,2,3), (2,2,2)$$

Now we find the no.of ways we can select the persons of different religion corresponding to the possibilities.

1) For $(1,1,4)$,

No.of ways of selecting the persons = $3!$.

But since there are 2 1's, each selection will have two copies.

And since we only need to consider their religion (that means persons in a particular religion are identical), we need to divide the result by 2. That gives us,

$\frac{3!}{2} = 3$ ways

For $(1,2,3)$

No.of ways of selecting the persons = $3!$.

(Notice there aren't any repeating numbers in this case!)

2) For $(2,2,2)$

No.of ways of selecting the persons = $3!$.

Now, there are 3 repeating numbers.

That means there are $3!$ copies for each selection. Applying the same logic, we divide the result by $3!$.

$\frac{3!}{3!} = 1$

Adding all the result gives us,

$3+3!+1 = 10$

The number of ways arranging this would be 10 .

And the possibilities are :

( 1 , 1 , 4)
( 1 , 2 , 3)
( 1 , 3 , 2)
( 2 , 2 , 2)
( 2 , 3 , 1)
( 2 , 1 , 3)
( 3 , 2 , 1)
( 3 , 1 , 2)
( 4 , 1 , 1)
( 1, 4 , 1)

following the pattern ( Hindu, Muslim , Christian )


Calculate the infinite sum $\sum_{1}^\infty \frac{\log{n}}{2n-1}$ I would like to calculate an asymptotic expansion for the following infinite sum:

$$\displaystyle \sum_{1}^N \frac{\log{n}}{2n-1}$$

when $N$ tends to $\infty$. I found that the asymptotic expansion for this partial sum is

$$ \displaystyle \frac{\log^2{N}}{4}+0.2282077...$$

and I would be interested in writing this constant term in an explicit way. By similarity with other sums of the same type, I believe that an explicit expression should probably include $\displaystyle \gamma$ and the first Stieltjes constant $\displaystyle \gamma_1$, but I was not able to find it.

An incomplete answer but I hope it may clarify some things.

Let us introduce
\begin{align*}
S_N&=\sum_{n=1}^N\frac{\ln n}{2n},\qquad
\bar{S}_N=\sum_{n=1}^N\frac{\ln n}{2n-1}.
\end{align*}
Now make two observations:

the sum $\displaystyle C_N:=\bar S_N-S_N=\sum_{n=1}^N\frac{\ln n}{2n\left(2n-1\right)}$ converges as $N\to \infty$.
the asymptotics of $S_N$ is known:
$$S_N=\frac14\ln^2N+\frac12\gamma_1+o\left(1\right).$$

Thus the constant we are looking for is nothing but
$$\frac12\gamma_1+C_{\infty}=\frac12\gamma_1+\sum_{n=1}^{\infty}\frac{\ln n}{2n\left(2n-1\right)}.$$
However the evaluation of the remaining infinite sum looks complicated (yet much simpler than the sum involving zeta values from another answer).

Here is a general technique to do things from scratch. You can use the integral

  $$ \int_{1}^{n} \frac{\ln(x)}{2x-1} = -\frac{1}{24}\,{\pi }^{2}-\frac{1}{2}\,\ln  \left( 2 \right) \ln  \left( 2\,n-1
 \right) - \frac{1}{2}\,{\it Li_2} \left( 1-2\,n \right) . $$

where $Li_s(z)$ is the polylogarith function. Note that you can use the asymptotic expansion for the function $Li_2(z)$ as

  $$ Li_2(z) = -\frac{1}{2}\, \left( \ln  \left( z \right) +i\pi  \right) ^{2}- \frac{1}{6}\,{\pi }^{2
}-\frac{1}{z}-O(\frac{1}{z^2}).$$

Added: Here is your constant

  $$ \frac{\pi^2}{24}- \frac{1}{4}\,  \ln^2\left( 2 \right)
\sim 0.291. $$


If G is simple and non-abelian, then $\hat{G}$ has only one element? Some extra information:
G is a group and the set $\hat{G}:=\{f:G\to S^1|f$ is a group homomorphism $\}$.
$(f_1*f_2)(g):=f_1(g)f_2(g),$ for $f_1,f_2\in\hat{G}$ and $g\in G$

We know that a simple group has no normal non-trivial subgroups. So this was my try:
If G is simple that means its only subgroups are e and G itself. But for it to have the identity element, there must be elements $a,b\in G$, s.t. $ab=e$ or $ba=e$, which cannot be and $\hat{G}$ has only one element??

If G is simple that means its only subgroups are e and G itself.

No, it means that its only normal subgroups are $\{e\}$ and $G$ itself.

Fortunately the kernel of $f$ has to be a normal subgroup, so you still know that the kernel is either $\{e\}$ or $G$ itself.

  But for it to have the identity element, there must be elements $a,b∈G$, s.t. $ab=e$ or $ba=e$, which cannot be

I cannot follow this argument.

Instead, hint: Suppose $f$ is a homomorphism with kernel $\{e\}$. Since $G$ is non-abelian, there exist $a,b$ such that $aba^{-1}b^{-1}\ne e$. But what is $f(aba^{-1}b^{-1})$?

Bonus question: Which groups other than $S^1$ will this argument also work for?

If $G$ be simple group and $f \in \hat{G}$ then $ker f$ is normal in $G$.As $G$ is simple $ker f=G$ or $ \{e_G\}$.How ever if $ker f=\{e_G\}$ then $ \hat{G}$ has a subgroup isomorphic to $G$ which is not possible as $G$ is non-abelian.So $ker f=G$ i.e. $ \hat{G}$ is trivial.


Combination with repetitions. The formula for computing a k-combination with repetitions from n elements is:
$$\binom{n + k - 1}{k} = \binom{n + k - 1}{n - 1}$$

I would like if someone can give me a simple basic proof that a beginner can understand easily.

This problem comes by many names - stars and stripes, balls and urns - it's basically a question of how to distribute $n$ objects (call them "balls") into $k$ categories (call them "urns"). We can think of it as follows.

Take $n$ balls and $k-1$ dividers. If a ball falls between two dividers, it goes into the corresponding urn. If there's nothing between two dividers, then there's nothing in the corresponding urn. Let's look at this with a concrete example.

I want to distribute $5$ balls into $3$ urns. As before, take $5$ balls and $2$ dividers. Visually:

|ooo|oo

In this order, we'd have nothing in the first urn, three in the second urn and two balls in the third urn. The question then is how many ways can we arrange these 5 balls and two dividers? Clearly: $\dfrac{(5+3-1)!}{5!(3-1)!} = \displaystyle {7 \choose 2} = {7 \choose 5}$.

We have that there are $\dfrac{(n+(k-1))!}{(k-1)! n!}$ the $n$ balls and $k-1$ dividers (since the balls aren't distinct from each other and the dividers aren't distinct from each other). Notice that this is equal to $\displaystyle {n+k-1 \choose k-1} = {n + k - 1 \choose n}$.

In distinguishing between combinations allowing repetition and those not, I think it's a question of supply of the objects being selected that's important to consider.  Without repetition is appropriate when supply is limited; with repetition when supply is unlimited.

For example, a grocery store sells 5 kinds of fruit, and you're going to purchase 3 individual fruits without restriction.  The equation would then be:
(n+r-1)! / r!(n-1)! = (5+3-1)! / 3!(5-1)! = 7! / 3!*4! = 35

Contrast this with there are five different fruits in a bowl, and you'll pick 3:
n! / r!(n-r)! = 5! / 3!(5-3)! = 5! / 3!*2! = 10

In both equations above, I'm using n = number of choices, r = number of events.


Find formula for generating function of sequence My task is to find formula for generating function of sequence $a_0, a_1...$ defined with following recurence
 $a_0=1$ and $a_n=\sum_{i=0}^{n-1} (n-i)a_i$.

I rewrote the expression
$a_n=\sum_{i=0}^{n-1} (n-i)a_i=n+(n-1)a_1+(n-2)a_2+...+a_{n-1}$
and I counted few members of sequence $a_1=1,a_2=3,a_3=8,a_4=21$ etc.

There are two answears below-wchich one i correct? Is there any way how to do some backward examination to show that it gives correct solution?

Consider the relation at $n+1$:
$$
a_{n+1}=\sum_{i=0}^{n}(n-i+1)a_i\tag1
$$
Letting $A(x)=\sum_{n\ge 0}a_nx^n$, the LHS is the coeficient of $x^{n}$ in $(A(x)-a_0)/x$, while the the RHS is a convolution, so it is the coefficient of $x^n$ in $$A(x)\cdot \sum_{k\ge 0}(k+1)x^k=A(x)\cdot (1-x)^{-2}.$$
Therefore, multiplying both sides by $x^n$ in $(1)$ and summing over $n\ge 0$, you get
$$
(A(x)-1)/x=A(x)\cdot{(1-x)^{-2}}\implies \boxed{A(x)=\frac{(x-1)^2}{1-3x+x^2}.}
$$
Here is a Wolfram Alpha sanity check.


finding the coefficient k of a joint pdf $f(x,y)$ under the multivariate normal distribution given the joint pdf $f(x,y) = $

$$k e^{-\frac12(x^2-2xy+5y^2)},$$

find $k.$

I understand that there is some trick to computing $k$ using properties of the multivariate normal distribution, but have no clue where to start.

$$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2+4y^2)} \, dx \, dy. $$

Integrate with respect to $x$ first  and note that

$$
\int_{-\infty}^\infty e^{-\frac 1 2 ((x-y)^2)} \, dx =\int_{-\infty}^\infty e^{-\frac 1 2 x^2} \,dx =\sqrt {2\pi}.
$$

Now $$\int_{-\infty}^\infty e^{-2y^2}=2\sqrt {2\pi}$$ as seen by the substitution $y=2z$. Hence $$\int_{-\infty}^\infty e^{-\frac 1 2 (x^2-2xy+5y^2)} \, dx \, dy =4\pi$$ and the value of the constant $k$ is $\dfrac 1 {4\pi}$.


If $\vec{u}+\vec{v}+\vec{w}=\vec{0}$, then triangle $ABC$ exists with $\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$ Let $ABC$ be a triangle. Then $\vec{AB}+\vec{BC}+\vec{CA}=\vec{AC}+\vec{CA}=\vec{AA}=\vec{0}.$ Is the opposite statement that if $\vec{u}+\vec{v}+\vec{w}=\vec{0}$, then triangle $ABC$ exists with $\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$ also true? Why?
I am really new to vectors and I would be very grateful if someone could explain this to me in detail. Thank you in advance!

The opposite statement is not true. It suffices to consider $A,B,C$ collinear.
Set $$\vec{AB}=\vec{u},\vec{BC}=\vec{v},\vec{CA}=\vec{w}$$
It holds $$\vec{u}+\vec{v}+\vec{w}=\vec{0},$$ but $ABC$ is not a triangle.

Take any point $A$. Define $B$ as $A+\overrightarrow{u}$ and $C$ as $B + \overrightarrow{v}$. Then
$$ C+\overrightarrow{w} = B + \overrightarrow{v}+\overrightarrow{w} = A+\overrightarrow{u}+ \overrightarrow{v}+\overrightarrow{w} = A $$
so triangle $ABC$ indeed satisfies our thesis.


Direct contour integration of the Riemann Zeta fuction at even natural numbers Is there a direct complex contour integral to evaluate
$$\int_0^\infty \frac{x^3}{e^x-1}dx$$
WITHOUT expanding $\frac1{e^x-1}$ into a series then evaluating the infinite series $\displaystyle 3!\sum_{k=1}^\infty\frac1{k^4}$.

Note: lease do NOT use the following series expression.

I know already by the Taylor series of $\frac1{1-u}$ for $|u|<1$, we have
$$\int_0^\infty \frac{x^3}{e^x-1}dx = \int_0^\infty x^3\sum_{k=1}^\infty e^{-kx}dx = \sum_{k=1}^\infty\int_0^\infty x^3 e^{-kx}dx=3!\sum_{k=1}^\infty k^{-4},$$
and then we can use contour integral to evaluate the last infinite series to obtain $\frac{\pi^4}{15}$. I do NOT want to use this series expression in the derivation.

In fact this is the Riemann Zeta function $\zeta(4)\Gamma(4).$


Express an abelian group given as finite generators and their relations as a direct sum of cyclic groups and find corresponding generators. According to page 158 of Dummit and Foote's Abstract Algebra (3rd edition):

Theorem. (Fundamental Theorem of Finitely Generated Abelian Groups) Let $G$ be a finitely generated abelian group. Then
(1) $G \cong \mathbb{Z}^r \times Z_{n_1} \times Z_{n_2} \times ... \times Z_{n_s}$, for some integers $r$, $n_1$, $n_2$, ... , $n_s$ satisfying the following conditions: (a) $r \ge 0$ and $n_j \ge 2$ for all $j$, and (b) $n_{i+1} \mid n_i$ for $1 \le i \le s-1$
(2) the expression in (1) is unique: if $G\cong \mathbb{Z}^t \times Z_{m_1} \times Z_{m_2} \times ... \times Z_{m_u}$, where $t$ and $m_1$, $m_2$, ... , $m_u$ satisfy (a) and (b) (i.e., $t \ge 0$, $m_j \ge 2$ for all $j$ and $m_{i+1} \mid m_i$ for $1 \le i \le u-1$), then $t = r$, $u = s$ and $m_i = n_i$ for all $i$.

Question: If an abelian group is expressed as finite generators and their defining relations (for example $G = \langle s,t,u,v \mid s^{4}t^{2}u^{10}v^{6} = s^{8}t^{4}u^{8}v^{10} = s^{6}t^{2}u^{9}v^{8} = e_G\rangle$), how to compute and express the group as a direct sum of cyclic groups and find the corresponding generators satisfying the relations? Could someone explain the computing procedure and show some worked examples? Or could someone point out which textbook taught such computing procedure with worked examples and at which pages/sections?
P.S.1 Although I can't find which textbook taught such computing procedure, Derek Holt gave a link to a PDF document teaching such computing procedure in his comment.
P.S.2. Could someone reply my 2nd comment of lhf's answer and explain how to find that 4 generators in $C_2 \times C_2 \times C_{\infty}$ satisfying the relations?

The tool for this is the Smith Normal Form, a kind of Gaussian elimination for PID.

For the example $G = \langle s,t,u,v \mid s^{4}t^{2}u^{10}v^{6} = s^{8}t^{4}u^{8}v^{10} = s^{6}t^{2}u^{9}v^{8} = e_G \rangle$ written additively, the matrix is
$$
\left(
\begin{array}{cccc}
 4 & 2 & 10 & 6 \\
 8 & 4 & 8 & 10 \\
 6 & 2 & 9 & 8 \\
\end{array}
\right)
$$
whose Smith Normal Form is
$$
\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 2 & 0 & 0 \\
 0 & 0 & 2 & 0 \\
\end{array}
\right)
$$
as computed by Mathematica using the code mentioned here.

This means that $G$ is generated by four elements $g_1, g_2, g_3, g_4$ such that $1g_1=0$, $2g_2=0$, $2g_3=0$, and no restrictions on $g_4$, and so $G \cong C_1 \times C_2 \times C_2 \times C_{\infty} \cong C_2 \times C_2 \times C_{\infty}$.

$g_1, g_2, g_3, g_4$ are obtained from $s,t,u,v$ by applying the matrices $P$ and $Q$ such that $PAQ$ is the diagonal matrix above.

All this is explained in several books. One is
Finitely Generated Abelian Groups and Similarity of Matrices over a Field by Christopher Norman. A shorter account appears in Jacobson's Basic Algebra I.

If G has generators {$a_1.... a_r$} where $|a_i|= \infty$ and {$b_1...b_t$} where $|b_i| = m_i$ then G = $\prod_{i=1}^r<a_i> \times \prod_{j=1}^t<b_j>$.  But each $\langle a_i \rangle \cong \mathbb Z$ so $\prod_{i=1}^r \cong \mathbb Z^r$ and each $\langle b_j \rangle \cong \mathbb Z_{m_j}$ so $G \cong \mathbb{Z}^r \times Z_{m_1} \times Z_{m_2} \times ... \times Z_{m_t}$


Alternating sum of binomial coefficients times logarithm Would like to compute the following sum: $\sum_{j=1}^m (-1)^j \begin{pmatrix} m\\ j\end{pmatrix} \log(1 - j/n)$, where $1 \leq m < n$ are fixed integers. A similar sum has already been computed on Stackexchange at Limit of alternating sum with binomial coefficient, where the logarithmic term is $\log(a+bk)$; the answer is in terms of $r = a/b$ and involves the gamma function, $\Gamma(r)$. When I apply this to my situation, I get (as part of the answer) $a=1, b=-1/n$ and therefore, I get the term $\Gamma(r) = \Gamma(-n)$ which is not defined, since the gamma function has a pole there. Please help with the correct expression for the sum in my case.

We can extend the sum to $j=0$ since the summand is $0$ for $j=0$. We have

$$
\begin{align}
\sum_{j=0}^m(-1)^j\binom mjj^l
&=\left.\sum_{j=0}^m\left(q\frac{\partial}{\partial q}\right)^lq^j\binom mj\right|_{q=-1}
\\
&=\left.\left(q\frac{\partial}{\partial q}\right)^l(1+q)^m\right|_{q=-1}
\\
&=(-1)^mm!\left\{{l\atop m}\right\}\;,
\end{align}
$$

where $\left\{{l\atop m}\right\}$ is a Stirling number of the second kind.

This vanishes for $0\le l\lt m$, so expanding the logarithm yields a power series in $1/n$ with leading term $1/n^m$:

$$
\begin{align}
\sum_{j=1}^m (-1)^j\binom mj\log(1 - j/n)
&=
-\sum_{j=0}^m (-1)^j\binom mj\sum_{l=1}^\infty\frac1l\left(\frac jn\right)^l
\\
&=
(-1)^{m+1}m!\sum_{l=m}^\infty\left\{{l\atop m}\right\}\frac{n^{-l}}l\;.
\end{align}
$$

Formally, we can also write

$$
\sum_{j=0}^m(-1)^j\binom mj\log(1 - j/n)=\log\left(1-\frac qn\frac{\partial}{\partial q}\right)(1+q)^m\;,
$$

but I'm not sure whether that's of any use.

Yesterday, I posted my answer by mistake on the other post while I was supposed to post it here. Anyways, your case is a special case of the case, we are going to deal with.

Recalling the identity of Stirling numbers of the second kind in terms of the binomial,

$$\left\{\begin{matrix} n \\ m \end{matrix}\right\} = \frac{1}{m!}\sum_{j=0}^m (-1)^{m-j}{m \choose j } j^n\,. $$

Writing the sum as
$$ \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a})$$

$$ -\ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a}) \,.$$

Expanding the $\ln(1+\frac{b k}{a})$ in terms of its Taylor series and interchanging the order of summation gives
$$ -n!\,\sum_{m=1}^{\infty}{\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \left(\frac{1}{n!}\sum\limits_{k=1}^{n} (-1)^{m-k} {n \choose k} k^m \right) =  -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} $$

$$ = -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,,$$

$$ \Rightarrow \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a+bk) = -\ln(a) -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,. $$


Why do we even need axiom schema? It has already been shown that a parameter-free form of ZFC is as strong as ZFC with parameters (hence it is not necessary to have a separate axiom for every list of parameters). Even so, comprehension remains an axiom schema because $\forall\varphi.\forall X.\exists Y.\forall y.y\in Y\iff(y\in X\land \varphi(y))$ cannot be a first-order formula if $\varphi$ is a predicate.

But then, I'm told that ZFC is single-sorted and that the only terms are sets. Furthermore, predicates can be represented as sets by way of $\varphi(x)\iff x\in S_\varphi$, where, naturally $S_\varphi=\{x:\varphi(x)\}$; and this is done specifically to avoid introducing new sorts (or types) to set theory.

That being the case, why bother with the axiom schema at all?

Why not have the single axiom $\forall P.\forall X.\exists Y.\forall y.y\in Y\iff (y\in X\land y\in P)$?

The same thing can be done with replacement.

Edit:

To clarify, my thinking goes something like this:

Technically, $\{x:\varphi(x)\}$ is, in general, a class but not a set. That being said, if $X$ is a set and $S_\varphi$ is a class, and $z\in Z\iff z\in X\land z\in S_\varphi$, then $Z$ is also a set. ZFC is single-sorted, so the distinction between set and class is not explicit. In fact, the only way to formally prove the statement "$X$ is/is not a set" is to introduce additional sorts to ZFC.

If you extend ZFC by classes (obtaining a two-sorted theory), then the axiom above can be "translated" to:

$$\forall P^\mathbf{class}.\forall X^\mathbf{set}.\exists Y^\mathbf{set}.\forall y^\mathbf{set}.y\in Y\iff(y\in X\land y\in P)$$

(which is basically the same as the second order axiom of comprehension)

The reason I didn't think this was necessary is that there oughtn't be any set $X$ such that $X$ is the intersection of a set and class (the latter of which cannot be proven to exists in ZFC), but is not the intersection of a set with another set.

Let $\varphi$ be a predicate such that $\{x:\varphi(x)\}$ is a class but not a set. Unless the class $\{x:\varphi(x)\}\cap\mathbf{set}$ is empty, then there must be some set $Y$ such that for any set $X$, $X\cap Y$ is equivalent to $X\cap\{x:\varphi(x)\}$, yes?

Why should it matter if $\{x:\varphi(x)\}$ provably exists?

$S_\varphi$ need not be a set as you defined it : take $\varphi(x) := (x=x)$ for instance. So you need to have some form of thing that tells you that when you restrict it to $X$, then it is in fact a set.

That's what the comprehension axiom scheme tells you. The same type of thing happens with replacement.

In fact it's a theorem that (assuming it's consistent) ZFC cannot be finitely axiomatized : you really do need some axiom schemes, otherwise you're strictly less powerful.

So we can't get rid of axiom schemes if we want to get ZFC, no matter how smart we are in formulating things (as long as we're doing things in the first order, and assuming ZFC is consistent)

Summary/takeaway of discussion

The axiom schema of comprehension guarantees that for any set $X$ and predicate $\varphi$, there exists a set $Y$ containing those members of $X$ for which $\varphi(x)$ holds.

In a particular extension of ZFC:

Every set is a class (but not every class is a set)
For every predicate $\varphi$, there is a class $C_\varphi$ such that $x\in C_\varphi\iff \varphi(x)$

For a given $\varphi$, the axiom of comprehension can be stated as:

$$\forall X.\exists Y.\forall y.y\in Y\iff(y\in X\land \varphi(y))$$

This is equivalent to:

$$\forall X.\exists Y.\forall y.y\in Y\iff (y\in X\land y\in C_\varphi)$$

Since the class of sets exists, if the intersection $C_\varphi$ with the class of sets is nonempty, then there is a least set $S_\varphi$ such that $\varphi(s)$ holds for all $s\in S_\varphi$ (this is not something proven within the scope of the theory, it's just a generic statement).

Since each predicate has a corresponding set, we can rewrite the above as:

$$\forall S.\forall X.\exists Y.\forall y. y\in Y\iff (y\in X\land y\in S)$$

This is equivalent to "the intersection of any two sets exists."

Moving back to ZFC, we can take away classes entirely by assuming $\{x:\varphi(x)\}$ to be the least member of the intersection of $C_\varphi$ with the class of sets (this guarantees at least one set for each predicate). If $S_\varphi$ is the least such set, then the above guarantees that for any set $X$, the set $\{x\in X:\varphi(x)\}=X\cap S_\varphi$ exists.

However, this does not prove that $S_\varphi$ exists.

So while we might be able to prove that "every subset of the integers exists" (provided that we can prove the existence of the integers), we cannot prove that "the set of odd integers exists".

This is where the "axiom of intersection," as it were, fails. While it suffices to prove that a subset of a particular set exists if there is a second set to take the intersection, it does not prove that the second set exists in the first place.

Proving the existence of the second set ultimately requires proving that for any $\varphi$, $S_\varphi$ exists - ultimately ends up being equivalent to the original axiom schema of comprehension.

For anyone who's interested, it might be possible to produce interesting "fragments" of ZFC by replacing the axiom schemata as described, and adding a "set construction axiom" (basically, an algorithm for constructing new sets from existing ones, for example $\forall x.\exists y. y=\{x\}$).

I don't know what applications this would have, if any, but it's something to think about.


If $f(x)=\int_{x-1}^x f(s)ds$, is $f$ constant? Periodic? I was thinking of periodic functions, and in particular the following type of condition:

  If a function $f:\mathbb{R}\to\mathbb{R}$ always "tends to its average", then it should be periodic.

To make things more formal, by "tending to the average" we could say something like $f(x)=\int_{x-1}^x f(s)ds$. This is only the average depending on a previous time interval of length $1$, but it seems an interesting enough property. However, the only type of functions which I could find that satisfies this property are the constant ones!

  Question: If $f:\mathbb{R}\to\mathbb{R}$ is continuous (or more generaly measurable) and $f(x)=\int_{x-1}^x f(s)ds$ for (almost) every $x\in\mathbb{R}$, then is $f$ constant (a.e.)? Periodic (a.e.)?

Here is a first try for $C^1$ functions (see edit below!): If $f$ is $C^1$ and $x$ is fixed, we can use Taylor expansion $f(s)=f(x)+O(s-x)$ (and similarly for $x-1$) to obtain
\begin{align*}
f(x+t)-f(x)&=\int_x^{x+t}f(s)ds-\int_{x-1}^{x-1+t}f(s)ds\\
&=\int_x^{x+t}f(x)+O(s-x)ds-\int_{x-1}^{x-1+t}f(x-1)+O(s-x+1)ds\\
&=t(f(x)-f(x-1))+O(t^2),
\end{align*}
so $f'(x)=f(x)-f(x-1)$. This is obviously true if $f$ is constant, but the converse is not clear to me at the moment.

Edit: From a comment and answer below, the equation $f'(x)=f(x)-f(x-1)$ has non-periodic solutions on $\mathbb{R}\setminus\mathbb{Z}$, so this should not be the way to go for $C^1$ functions. However, even in this case it is not clear that any solution of this equation will satisfy $f(x)=\int_{x-1}^x f(s)ds$, which is the question: All I can obtain, in principle, is $f(x)-f(x-1)=\int_x^{x-1}f(s)ds-\int_{x-2}^{x-1}f(s)ds$.

Here is a typical argument using characteristic equation:

Let $\alpha \in \mathbb{C}\setminus\{0\}$ solve the equation $\alpha = 1-e^{-\alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then

$$ \int_{x-1}^{x} e^{\alpha t} \, \mathrm{d}t = \frac{1 - e^{-\alpha}}{\alpha} e^{\alpha x} = e^{\alpha x}, $$

hence $f(x) = e^{\alpha x}$ is one (complex-valued) solution of the equation

$$ f(x) = \int_{x-1}^{x} f(t) \, \mathrm{d}t \tag{*}$$

If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{\alpha x}$. In particular, this tells that there exists an analytic solution of $\text{(*)}$ which is neither constant nor having real-period.

Addendum. We prove the following claim:

  Claim. There exists a non-zero solution of $\alpha = 1 - e^{-\alpha}$ in $\mathbb{C}$.

Proof. We first note that $\varphi(x) = x(1-\log x)$ satisfies $\varphi(0^+) = 0$ and $\varphi(1) = 1$. Next, let $k$ be a positive integer. Then

There exists $y \in (2k\pi, (2k+\frac{1}{2})\pi)$ such that $ \varphi(\sin(y)/y) = \cos (y) $, by the intermediate-value theorem.
Set $x = \log(\sin(y)/y)$.

We claim that $ \alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}\sin y = y $ holds. Moreover,

$$ (1-x)e^x = \varphi(\sin(y)/y) = \cos(y), $$

and so, $1 - x = e^{-x}\cos(y)$. Combining altogether,

$$ 1 - \alpha = 1 - x - iy = e^{-x}\cos(y) - ie^{-x}\sin(y) = e^{-x-iy} = e^{-\alpha}. $$

Therefore the claim follows. ////

(A careful inespection shows that this construction produces all the solutions of $\alpha = 1 - e^{-\alpha}$ in the upper half-plane.)

Let
$$f(x) = \dfrac{a_0}2+\sum\limits_{n=1}^\infty\left(a_n\cos2\pi nx + a_n\sin2\pi nx\right),$$
then
$$\int\limits_{x-1}^x f(x)\,\mathrm dx = \dfrac {a_0}2,$$
so $f(x)$ is a constant.


Find g(y) and b so that $\int_0^2 (\int_{x^3}^{4x}\sin y^2\space dy)\space dx=\int_0^b\sin y^2g(y)\space dy$ This came from a Calculus II exam:

Find $b$ and $g(y)$ so that:
$$\int_0^2 \left(\int_{x^3}^{4x}\sin y^2\space dy\right)\space dx=\int_0^b\sin y^2g(y)\space dy$$

I have no idea how to tackle this problem.

If $D=\{(x,y): 0\leq x\leq 2, x^3\leq y\leq 4x\}$ we have
$$ \iint_{D} \sin(y^2)\,dx\,dy = \int_{0}^{8}\sin(y^2)\cdot\left(\sqrt[3]{y}-\frac{y}{4}\right) \,dy $$
by Fubini's theorem, hence a solution is given by $b=8$ and $g(y)=\sqrt[3]{y}-\frac{y}{4}$.

$\hspace{2cm}$

You just have to describe the region between two curves in two different ways, as $f_1(x)\leq y\leq f_2(x)$ or as $g_1(y)\leq x \leq g_2(y)$. The region between $y=4x$ and $y=x^3$ on the interval $x\in[0,2]$ is a normal domain with respect to both the $x$ and the $y$ variable.

Formally, this is an underconstrained problem, You was wanting to seek for a value and for a function at the same time for giving an specific value, and there are infinitely many solutions.

A not so trivial guessing will be $b=7.19=\text{fresnels}^{-1}(0.46)$ and $g(y)=1$.


Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$

First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.

One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.

Well,

$$\sqrt{x^6 - 9 x^3} = x^{3} \left(1 - \frac{9}{x^{3}}\right)^{1/2}$$

and so the question is really about understanding how $\sqrt{1 - t}$ looks when $t$ is pretty small. By noticing that the derivative $\frac{d}{dt} \sqrt{t} = \frac 1 2$ when $t = 1$, we can say that

$$\left(1 - \frac{9}{x^{3/2}}\right)^{1/2} \approx \frac 1 2 \left(\frac{-9} {x^{3}}\right)$$

which leads quickly to the claimed limit of $-9/2$.

Alternatively, there is a completely standard technique of multiplying and dividing by the conjugate, which is $\sqrt{x^6 - 9x^3} + x^3$, but I find that this isn't very enlightening.

T. Bongers' answer is right, but a bit sloppy.  A better way would be to use the binomial expansion.

$$\sqrt{1-9x^{-3}} = \sum_{n=0}^\infty \frac{(1/2)^{\color{red}{n}}(-9)^nx^{-3n}}{n!} = 1 - \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots$$
Where the red thing represents a falling power.
$$\lim_{x\to+\infty}\sqrt{x^6-9x^3}-x^3=\lim_{x\to+\infty}x^3(\sqrt{1-9x^{-3}} - 1) = \lim_{x\to+\infty}x^3\left(- \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots\right) = -\frac{9}{2}$$


How many interest periods if you always pay double the interest A loan of $12,500$ is made at an effective interest rate of $8.5\%$. Payments are made at the end of each interest period. Each payment equals twice the interest due until the borrower pays off the outstanding debt with a final payment of at most $1800$. Find the number of payments $n$ and the amount of the final payment.

WHAT I'VE DONE SO FAR:

So I assumed the $n$th payment is $1800$. Then, in period $n-1$, the interest is $900$ and the principle is $900$. I dont know what to do after that. I tried to keep doubling the amount but then I got $n=4$ although the correct answer is $n=24$.

Okay, this is the amortization of a debt.

Payments are made at the end of every period, so at the time of your first payment you have a debt 12500(0.085)= 13562.5

Denote $Bn$ outstanding balance, $Pn$ principal payment and $Kn$ payment at period n and $i$ the interest.
$Bo$= 13567.5, interest for the initial period is $i Bo$ = (0.085) (13567.5)
implies $k1=2iBo$ since the paymets are twice the interest due amount.
and $B1= Bo- iBo$. One $Bo$ is to pay the debt and the other is to pay the interest.
Follow this we have: $B2= B1- iB1= Bo- 2iBo + i^3$ = $Bo * (1-2i + i^3)$
Now you see it? Binomial Theorem.

We want find n such that $Bn=Bo* (1-i)^n = 1800$ , ie , our Outstanding Balance must be 1800 to pay with a final payment
takin logarithm, n= 22.7 but 1800 AT MOST so take n=23 and the final payment is less than 1800. The answer is 23 payments and 1 final payment less that 1800.


Derivative of an exact form $\omega_t$ is still exact? Let $\omega_t$ be an exact form depending smoothly on time $t$.

Then is $\dfrac{d\omega_t}{dt}$ also exact?

$\dfrac{\omega_t - \omega_{t_0}}{t-t_0}$ is clearly exact.

However I am stuck in showing that its limit is exact.

Thanks in advance.

Using Ted's hint

Let $c$ be an any smooth k-cycle. Since $\omega_t$ is exact, $\displaystyle\int_c \omega_t =0$

It suffices to prove that $\displaystyle\int_c \lim_{t \to t_0} \dfrac{\omega_t - \omega_{t_0}}{t-t_0} = \lim_{t \to t_0}\displaystyle\int_c \dfrac{\omega_t - \omega_{t_0}}{t-t_0} $.

However this is on the world of manifolds and diffenretial forms, I am not sure it works well.

HINT: A closed $k$-form is exact if and only if its integral over every $k$-cycle is zero. (Comment: The Leibniz rule for differentiating under the integral sign applies to manifolds as well, giving
$$\frac d{dt} \int_M \omega = \int_M \frac{\partial\omega}{\partial t}.$$
The "nice" proof using dominated convergence works just as for one-dimensional integrals.)

An exact form is a differential form $\omega$ such that $\omega = df$ for some differential form $f$. Then $\frac{d\omega_{t}}{dt}$ is exact as it is the derivative of $\omega$.


Proof of $\frac{1}{(1-r)^2}$ I was given that $S=1+2r+3r^2+\cdots = \frac{1}{(1-r)^2}$ was an identity from my students, and I tried to prove directly using geometric series.

I got stuck and looked around online only to be told that taking the derivative of the series that converges to $\frac{1}{1-r}$.

I liked that proof because it was simple and easy, but unfortunately my student is in Alg II so I cannot explain her with derivatives.

My idea was to expand $S$ as $$(1+r+r^2+r^3+\cdots) +(r+r^2+r^3+\cdots)+(r^2+r^3+\cdots)+\cdots$$

but I get confused whether I should rewrite this as $$(1+r+r^2+r^3+\cdots) +r(1+r+r^2+\cdots)+r^2(1+r+\cdots)+\cdots$$

or

$$(1+r+r^2+r^3+\cdots) +((-1)+1+r+r^2+\cdots)+((-1-r)+1+r+\cdots)+\cdots$$

I was able to get the ID using the first way, but I am not sure if the step I took was valid due to some stuff I vaguely remember about manipulating an infinite series like that was not always allowed.

And on the second way I simply did not know where to go from there.

Can I have some help ?

It would also be helpful to know the criteria of when I can manipulate the order of how to add the series.

Your first approach $$1(1+r+r^2+\cdots)+r(1+r+r^2+\cdots)+r^2(1+r+r^2+\cdots)+\cdots$$ is the best one for intuition. Pulling out the common factor then gives us $$(1+r+r^2+\cdots)(1+r+r^2+\cdots)=(1+r+r^2+\cdots)^2,$$ at which point we can use the standard geometric series result for $|r|<1$.

This is perfectly fine to do, since we're dealing with absolutely convergent series, which is equivalent to unconditionally convergent (meaning we can do such rearrangements without fear of changing the result).


Induced homomorphism $p_*: H_1(D_0 - z_0)\to H_1(\mathbb{C}-0)$ by $a(z-z_0)^m$ Let $D_0 \subseteq \mathbb{R}^2$ be a closed disc with center at $z_0$ and consider the function $p: D_0 \to \mathbb{C}$ defined by $p(z):=a(z-z_0)^m$ for $z \in D_0$ with $a \neq 0$. How can I show that the induced homomorphism $p_*: H_1(D_0 - z_0) \to H_1(\mathbb{C}-0)$ is multiplication by $m$?. I know that $H_1(D_0 - z_0)$ and $H_1(\mathbb{C}-0)$ are infinite cyclic groups and that the homorphism induced by $z^m$ in $H_1(S^1) \to H_1(S^1) $ is multiplication by $m$. I think that I should work with the isomorphism between $H_1(\mathbb{C}-0)$ and $H_1(S^1)$ and the isomorphism of $H_1(D_0 - z_0)$ with $H_1(\partial D_0)$.

Thanks for any help.

Certainly we have $H_1(D_0 \setminus z_0) \approx \mathbb{Z}$ and $H_1(\mathbb{C} \setminus 0) \approx \mathbb{Z}$. However, to see what $p_*$ looks like we must specify generators $a$ of $H_1(D_0 \setminus z_0)$ and $b$ of $H_1(\mathbb{C} \setminus 0)$. There are two possible choices for both homology groups. This is equivalent to specifying isomorphisms $f : H_1(S^1) \to  H_1(D_0 \setminus z_0)$ and $g : H_1(S^1) \to H_1(\mathbb{C} \setminus 0)$. This is done in the following obvious way:

(1) There exists a strong deformation retraction $r : \mathbb{C} \setminus 0 \to S^1, r(z) = z/\lVert z \rVert$. Hence the inclusion $i : S^1 \to \mathbb{C} \setminus 0$ is a homotopy equivalence and induces the isomorphism $g = i_*$. Its inverse is $g^{-1} = r_*$.

(2) $\partial D_0$ is a strong deformation retract of $D_0 \setminus z_0$. But there is a canonical homeomorphism $h : S^1 \to \partial D_0, h(z) = z_0 + dz$, where $d$ is the radius of $D_0$. If $j : \partial D_0 \to D_0$ denotes inclusion, we take $f = (j \circ h)_*$.

You know that the map $\mu : S^1 \to S^1, \mu(z) = z^m,$ induces $\mu_*$ = multiplication by $m$.

Now we can check that $g^{-1} \circ p_* \circ f$ is multiplication by $m$. We have
$$g^{-1} \circ p_* \circ f = r_* \circ p_* \circ (j \circ h)_* = (r \circ p \circ j \circ h)_*$$
and
$$(r \circ p \circ j \circ h)(z) = r(p(z_0+dz)) = r(adz^m) = \frac{a}{\lvert a \rvert} z^m .$$
Let $\varphi = r \circ p \circ j \circ h$ and write $\frac{a}{\lvert a \rvert} = e^{i\tau}$ with $\tau \in [0,2\pi)$. A homotopy $H : S^1 \times [0,1] \to S^1$ is defined by
$$H(z,t) = e^{it\tau}z^m .$$
Then $H(z,0) = \mu(z), H(z,1) = \varphi(z)$. Hence $\mu$ and $\varphi$ are homotopic and we conclude $\varphi_* = \mu_*$.


big theta question,not sure is book mistake or my misunderstand.... I have $\Theta$-notation question from my new book example solution... the example question is:

Find $\Theta$ bound for $$f(n)={n^2\over 2} - {n\over 2}$$

and the book solution is :

$\displaystyle {n^2\over5} \le {n^2\over 2} - {n\over 2} \le n^2$, for all $n\ge 1$
so $\displaystyle{n^2\over2} - {n\over2} = \theta(n^2)$ with $c_1=\frac15$, $c_2=1$ and $n_0 = 1$.

what I don't understand is why lower bound is $\frac15$ and $n_0$ is $1$, if I put $1$ to formula it will be $\frac15 \leq 0 \leq 1$, it's not correct...

I thought the answer would be

$\displaystyle{n^2\over 4} \leq {n^2\over 2} - {n\over2} \leq n^2$, for all $n\geq2$
so $c_1=\frac14$, $c_2=1$, $n_0=2$.
could somebody explain why?

It is a typo: please put n0=2.

It is valid with n0=2


Solving an equation including $e^{-x}$ with the Lambert W function Given two functions of $x$, namely $f(x)$ and $g(x)$, where $$f(x)=x^2-4x+8$$$$g(x)=3xe^{-x}$$ the shortest distance between the graphs of the functions is sought.

I begin by defining a function $h(x)$ which returns the vertical distance between the functions, $h(x)=f(x)-g(x)$.

Then I locate the global minimum by solving $h'(x)=0$, which I solve numerically and get $x\approx 1.802$, which is what I wanted.

However, I wish to solve this algebraically. The equation $h'(x)=0$ is $$3xe^{-x}-3e^{-x}+2x-4=0$$

My first instinct is to isolate the equation in the form $a=be^b$, with $a$ as a constant term, and then use the Lambert W function to solve for $x$.

Is it possible to isolate the terms in such a way, or do I have a transcendental equation? What does this mean in terms of the number of real solutions and how should I go about approximating the solution from this point?

$h(x) = f(x)-g(x) = x^2-4x+8-3xe^{-x}$

$y=min\{h(x) , x \in \mathbb{R} \}$

Now note that $h'(x)=2x-4-3e^{-x}+3xe^{-x}=2(x-2)+3e^{-x}(x-1)$

$h'(x) = 2(x-2)+3(x-1)*e^{-x}=0$ gives $2*e^{x}(x-2)+3(x-1)=0$, cause $ e^{x}\not=0$. Solving this equation is not easy, but if you want to do this I would also suggest using the Lambert equation.


Find the two missing angles in a quadrilateral This problem originates from a student who came asking for help. After spending some time, we couldn't solve this problem using (Euclidean) geometry alone. We had to resort to trigonometry to solve this. I have since spent many hours on it and can't seem to get anywhere. Here is the problem.

Given:

Quadrilateral $ABCD$

Diagonals $\overline{AC}$ and $\overline{BD}$

$m\angle ABD=19^{\circ}$

$m\angle DBC=57^{\circ}$

$m\angle ACD=30^{\circ}$

$\overline{AB}\cong \overline{CB}$

Find:

$m\angle ADB$.

Here is a helpful digram.

It is almost immediate that $m\angle BAC \cong m\angle BCA$ and you can easily fill in the following angles, shown in green.

But $x$ and (now added) $y$ are still illusive. I have tried everything from extending lines to drawing parallel lines to looking at the circumcircles and nothing seems to work. The green angles are the ones which we can easily deduce and the missing/blank angles, I don't know.

  How can we deduce the value of the missing angle $x$ using only high school geometry?

There is obviously enough "information" here because if nothing else, you can just draw the quadrilateral very carefully and just measure the angle. Physically, the missing angle can only be one value. It is constrained. One should be able to deduce this value with a geometric proof using basic theorems without resorting to advanced theorems or even trigonometry.

The missing values are, just for giggles,

Clarification

High school geometry, at least in the USA, is "distinct" from trigonometry and it doesn't include law of sines or cosines. High school geometry tries to mimic Euclid and his "Elements" where students memorize some of the definitions and axioms and are forced to mindlessly derive theorems in an excruciatingly mind-numbing manner with cumbersome notation. High school geometry also does not include any "advanced" geometric theorems. The most advanced thing an average student might do is something like constructing a regular hexagon. My questions is, again, can this problem be solved using only the material taught in a typical high school geometry class? If yes, then how? If you believe no, then can you give a convincing argument why trigonometry is necessary?

I am hoping that this is solvable with some basic theorems and doesn't require any advanced theorems. If we allow trigonometry, then the problem is easy.

This problem is a particular case of a family of problems with broadly the same solution, so I will post this more general solution and then discuss particular instances of it.

Problem. $\angle BAC=3\angle CAD$; $\angle CBD=30^\circ$; $AB=AD$. What is $\angle DCA$?

Solution. Let $\alpha=\angle CAD$. $\triangle BDA$ is isosceles on base $BD$. Therefore $\angle DBA=\angle ADB=90^\circ-2\alpha$ and $\angle CBA=120^\circ-2\alpha$.

Let $E$ be on $BC$ such that $AE=AB$. Then $\triangle BEA$ is isosceles on base $BE$. Therefore $\angle AEB=\angle EBA=120^\circ-2\alpha$, so $\angle BAE=4\alpha-60^\circ$, so $\angle EAD=60^\circ$.

Therefore $\triangle AED$ is equilateral, so $\angle EAC=60^\circ-\alpha=\angle ACE$, so $\triangle CAE$ is isosceles on base $CA$, i.e. $CE=AE=DE$, so $\triangle CDE$ is isosceles on base $CD$. $\angle CED=2\alpha$, so $\angle DCE=90^\circ-\alpha$, so $\angle DCA=30^\circ$, which solves the problem. Note that $\angle DCA$ is independent of $\alpha$.

To adapt this to the current problem, relabel from $ABCD$ to $BCDA$ and specify $\alpha=19^\circ$.

If $\alpha$ is specified as $20^\circ$, and $\angle DBA$ as $50^\circ$, then the problem is [Langley]. $AB=AD$ is easily seen, and the proof proceeds as above. The above proof, but with angles as specified in Langley's problem, is due to J. W. Mercer.

If $\alpha$ is specified as $16^\circ$, then the problem is that at gogeometry. The point-lettering is the same, but the diagram is flipped.

[Langley] Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922, according to David Darling

Let $BC = 1$ unit.

Law of sines:

$\frac {BD}{\sin \angle BCD} = \frac {BC}{\sin \angle BDC}$.

So $BD = \frac  {\sin \angle BCD}{\sin \angle BDC}$ units.

Law of cosines

$AD^2 = AB^2 + BD^2 - 2|AB||BD|\cos \angle ABD = 1 + BD^2 -2BD\cos\angle ABD$

So $AD =\sqrt {1 + BD^2 -2BD\cos\angle ABD}$

Law of sines

$\frac {\sin \angle BAD}{BD} = \frac {\sin \angle ADB}{BA=1} = \frac {\sin \angle ABD}{AD}$.

So $\angle BAD = \arcsin ( \frac {\sin \angle ABD}{AD}BD)$

And $\angle ADB = \arcsin (\frac {\sin \angle ABD}{AD})$.