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Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? Using the Cauchy-Schwarz Inequality, we have
$$
\begin{align}
1
&=\left(\int_n^{n+1}1\,\mathrm{d}x\right)^2\\
&\le\left(\int_n^{n+1}x\,\mathrm{d}x\right)\left(\int_n^{n+1}\frac1x\,\mathrm{d}x\right)\\
&=\left(n+\frac12\right)\log\left(1+\frac1n\right)
\end{align}
$$
which means that
$$
\left(1+\frac1n\right)^{n+1/2}\ge e
$$
This hints that $\left(1+\frac1n\right)^{n+1/2}$ might be decreasing.
In this answer, it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. The proofs use Bernoulli's Inequality. However, applying Bernoulli to $\left(1+\frac1n\right)^{n+1/2}$ is inconclusive.
Attempt to show decrease:
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}
&=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\
&\ge\left(1+\frac{2n}{n^2-1}\right)\frac{n-1}{n+1}\\[6pt]
&=1-\frac{2}{(n+1)^2}
\end{align}
$$
Attempt to show increase:
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{2n+1}}{\left(1+\frac1{n-1}\right)^{2n-1}}
&=\left(1-\frac1{n^2}\right)^{2n}\frac{n+1}{n-1}\\
&\ge\left(1-\frac2n\right)\frac{n+1}{n-1}\\[6pt]
&=1-\frac{2}{n(n-1)}
\end{align}
$$
Neither works.
Without resorting to derivatives, is there something stronger than Bernoulli, but similarly elementary, that might be used to show that $\left(1+\frac1n\right)^{n+1/2}$ decreases?
| Preliminaries: A couple of extensions to Bernoulli's Inequality.
Bernoulli's Inequality says that $(1+x)^n$ is at least as big as the first two terms of its binomial expansion. It turns out, at least for $n\in\mathbb{Z}$, that a sharper inequality can be obtained using any partial sum with an even number of terma.
Theorem $\bf{1}$: for $m\ge1$, $n\ge0$, and $x\gt-1$,
$$
(1+x)^n\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag1
$$
Proof (Induction on $n$): $(1)$ is trivial for $n=0$. Assume $(1)$ is true for $n-1$, then
$$
\begin{align}
(1+x)^n
&=(1+x)(1+x)^{n-1}\tag{1a}\\[9pt]
&\ge(1+x)\sum_{k=0}^{2m-1}\binom{n-1}{k}x^k\tag{1b}\\
&=\sum_{k=0}^{2m-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]x^k+\binom{n-1}{2m-1}x^{2m}\tag{1c}\\
&\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: factor
$\text{(1b)}$: assumption for $n-1$
$\text{(1c)}$: multiply sum by $1+x$
$\text{(1d)}$: Pascal's Rule
Thus, $(1)$ is true for $n$.
${\large\square}$
Theorem $\bf{2}$: for $m\ge1$, $n\ge0$, and $x\gt-1$,
$$
(1+x)^{-n}\ge\sum_{k=0}^{2m-1}\binom{-n}{k}x^k\tag2
$$
Proof (Induction on $n$): Note that another way of writing $(2)$ is
$$
(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\le1\tag{2a}
$$
$\text{(2a)}$ is trivial for $n=0$. Assume $\text{(2a)}$ is true for $n-1$, then
$$
\begin{align}
&(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k(1+x)\tag{2b}\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k{\textstyle\left[\binom{n+k-1}{k}-\binom{n+k-2}{k-1}\right]}x^k-{\textstyle\binom{n+2m-2}{2m-1}}x^{2m}(1+x)^{n-1}\tag{2c}\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-2}{k}x^k-\binom{n+2m-2}{2m-1}x^{2m}(1+x)^{n-1}\tag{2d}\\[9pt]
&\le1\tag{2e}
\end{align}
$$
Explanation:
$\text{(2b)}$: factor
$\text{(2c)}$: multiply sum by $1+x$
$\text{(2d)}$: Pascal's Rule
$\text{(2e)}$: assumption for $n-1$
Thus, $\text{(2a)}$ is true for $n$.
${\large\square}$
Note that for positive integer exponents, Bernoulli's Inequality is the case $m=1$ of Theorem $1$, and for negative integer exponents, it is the case $m=1$ of Theorem $2$.
Answer: Use the case $m=2$ of Theorem $1$:
$$
\begin{align}
&\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}\\
&=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\
&\ge\left(1+\frac{2n}{n^2-1}+\frac{2n(2n-1)}{2\left(n^2-1\right)^2}+\frac{2n(2n-1)(2n-2)}{6\left(n^2-1\right)^3}\right)\frac{n-1}{n+1}\\
&=1+\frac{n^2+n+6}{3(n-1)(n+1)^4}\\[9pt]
&\ge1
\end{align}
$$
That is, $\left(1+\frac1n\right)^{n+1/2}$ is decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259175",
"timestamp": "2023-03-29T00:00:00",
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Find $\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx$ I came across this integral while i was working on a tough series.
a friend was able to evaluate it giving:
$$\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$ using integral manipulation. other approaches are appreciated.
| Start with breaking the denominator
$$I=\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$
For the first integral, use $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ and for the second integral, use the identity $\frac{\arctan x}{1+x^2}=\frac12\sum_{n=0}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$ we have
$$I=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln^2x\ dx-\frac12\sum_{n=0}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n}\ln^2x\ dx$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n+1)^3}$$
$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^3},\quad H_{2n}=H_{2n+1}-\frac{1}{2n+1}$$
$$=\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}$$
Substitute
$$\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
and
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$
we obtain that
$$I=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving DLP by Baby Step, Giant Step I want to solve the DLP $6\equiv 2^x\pmod {101}$ using Baby Step, Giant Step.
$$$$
I have done the following:
We have that $n=\phi (101)=100$, since $101$ is prime.
$m=\lceil \sqrt{100}\rceil=10$.
For each $j\in \{0,1,\ldots , 9\}$ we calculate $(j,2^j)$:
\begin{align*}j \ \ \ & & \ \ \ 2^j \\ 0 \ \ \ & & \ \ \ 1 \\ 1 \ \ \ & & \ \ \ 2 \\ 2 \ \ \ & & \ \ \ 4 \\ 3 \ \ \ & & \ \ \ 8 \\ 4 \ \ \ & & \ \ \ 16 \\ 5 \ \ \ & & \ \ \ 32 \\ 6 \ \ \ & & \ \ \ 64 \\ 7 \ \ \ & & \ \ \ 27 \\ 8 \ \ \ & & \ \ \ 54 \\ 9 \ \ \ & & \ \ \ 7\end{align*}
It holds that $2^{-10}=1024^{-1}\equiv 65$.
For each $i\in \{0,1,\ldots , 9\}$ we calculate $6\cdot 65^i$:
\begin{align*}i \ \ \ & & \ \ \ 6\cdot 65^i \\ 0 \ \ \ & & \ \ \ 6 \\ 1 \ \ \ & & \ \ \ 87 \\ 2 \ \ \ & & \ \ \ 100 \\ 3 \ \ \ & & \ \ \ 36 \\ 4 \ \ \ & & \ \ \ 17 \\ 5 \ \ \ & & \ \ \ 95 \\ 6 \ \ \ & & \ \ \ 14 \\ 7 \ \ \ & & \ \ \ 1\end{align*}
Therefore for $i=1$ and $j=0$we get the same result, and so \begin{equation*}\log_26=7\cdot 10+0=70\end{equation*}
Is everything correct?
| Using An Introduction to Mathematical Cryptography, J. Hoffstein, J. Pipher, J. H. Silverman, let's use Shank's Babystep, Giantstep Algorithm:
*
*Let $G$ be a group and let $g \in G$ be an element of order $N \ge 2$. The following algorithm solves the discrete logarithm problem in $\mathscr{O}(\sqrt{N} . \log N)$ steps.
*$(1)$ Let $N$ be the multiplicative order of $p$ and then $n = 1 + \lfloor \sqrt{N} \rfloor$, so in particular $n \gt \sqrt{N}$.
*$(2)$ Create two lists,
List $1: g, g^2, g^3, \ldots, g^n$,
List $2: h, h g^{-n}, h g^{-2n},h g^{-3n},\ldots, hg^{-n^2}$.
*$(3)$ Find a match between the two lists, say $g^{i} = h g^{-j\cdot n}$.
*$(4)$ Then, $x = i + jn$ is a solution to $g^x = h \pmod {p}$.
For this problem, we want to perform the algorithm to solve for $x$ in
$$2^x\equiv 6\pmod {101}$$
We have
$$g = 2, h = 6, p = 101$$
For $p = 101$, we have order $N = \phi{(101)} = 100$ in $\mathbb{F}^*_{101}$, so set $$n = 1 + \lfloor \sqrt{N} \rfloor = 1 + \lfloor\sqrt{100}\rfloor = 11$$
Set
$$u = g^{-n}\pmod{101} = 2^{-11} \pmod{101} = 83$$
Create the table
$$\begin{array}{|c|c|c|} \hline
\text{k} & g^k & h u^k \\ \hline
1 & 2 & 94 \\ \hline
2 & 4 & 25 \\ \hline
3 & 8 & 55 \\ \hline
4 & 16 & 20 \\ \hline
5 & 32 & 44 \\ \hline
6 & 64 & 16 \\ \hline
7 & 27 & 15 \\ \hline
8 & 54 & 33 \\ \hline
9 & 7 & 12 \\ \hline
10 & 14 & 87 \\ \hline
11 & 28 & 50 \\ \hline
\end{array}$$
From the table, we find the collision at $16$, so we have
$$x = i + jn = 4 + (6)(11) = 70$$
Hence, $x = 70$ solves the problem $2^x \equiv 6 \pmod{101}$ in $\mathbb{F}^*_{101}$.
Aside: Here is a calculator and nice description of the algorithm to verify or you can solve this using Wolfram Alpha.
| {
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maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$
Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$
How do i solve it Help me plesse
| From the restriction $x^2+y^2>0$, we get that $x,y$ are not both zero.
If $x=0$, then $K={\large{\frac{1}{4}}}$.
Suppose $x\ne 0$.
Letting $t={\large{\frac{y}{x}}}$, and following your approach, we get
$$(4K-1)t^2+Kt+(K-1)=0$$
which has a real solution for $t$ if and only $K={\large{\frac{1}{4}}}$ or the discriminant
$$K^2-4(4K-1)(K-1)$$
is nonnegative.
Equivalently, either $K={\large{\frac{1}{4}}}$ or
$$-15K^2+20K-4\ge 0$$
With the restriction $K\ne{\large{\frac{1}{4}}}$, the quadratic inequality solves as
$$\frac{10-2\sqrt{10}}{15}\le K\le \frac{10+2\sqrt{10}}{15},\;\;K\ne{\large{\frac{1}{4}}}$$
Noting that
$$\frac{10-2\sqrt{10}}{15}<\frac{1}{4}<\frac{10+2\sqrt{10}}{15}$$
it follows that
*
*The minimum value of $K$ is ${\large{\frac{10-2\sqrt{10}}{15}}}\approx .2450296455$.$\\[8pt]$
*The maximum value of $K$ is ${\large{\frac{10+2\sqrt{10}}{15}}}\approx 1.088303688$.
| {
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If $p$, $q$, $r$ and $s$ are four sides of a quadrilateral, then find the minimum value of $\frac{p^2+ q^2 + r^2}{s^2}$ with logic If $p$, $q$, $r$ and $s$ are four sides of a quadrilateral then find the minimum value of $\frac{p^2+ q^2 + r^2}{s^2}$ with logic.
Please help me with this.
| By the triangle inequality and by C-S we obtain:
$$\frac{p^2+q^2+r^2}{s^2}>\frac{p^2+q^2+r^2}{(p+q+r)^2}=\frac{(1+1+1)(p^2+q^2+r^2)}{3(p+q+r)^2}\geq\frac{(p+q+r)^2}{3(p+q+r)^2}=\frac{1}{3}.$$
The equality does not occur, but easy to see that $\frac{1}{3}$ is an infimum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine if there exist rational number a and irrational number A such that $A^3+aA^2+aA+a=0$. Determine if there exist a rational number a and irrational number A such that $A^3+aA^2+aA+a=0$. If so, can we say something about them? Are there infinitely many of them?
| For any integer $a$ except $0$ or $1$, the polynomial $x^3 + a x^2 + a x + a$ has no rational roots. Any rational root $A$ would have to be an integer (by Gauss's lemma, or the Rational Root Theorem). Now $A^3 + a A^2 + a A + a = 0 $ means
$$a = - \frac{A^3}{A^2 + A + 1} = -A + 1 - \frac{1}{A^2 + A + 1}$$
which, if $A$ is an integer, is not an integer unless $A = 0$ (corresponding to $a=0$) or $A = -1$ (corresponding to $a = 1$): otherwise $A^2 + A + 1 = (A + 1/2)^2 + 3/4 > 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$?
If $\tan x=3$, then what is the value of
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$
So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\over5\cos^2{x}-5\sin^2{x}+8\sin{x}\cos{x}}$$
Now I thought of changing the top part to $\sin{x}$ and bottom part to $\cos{x}$ hoping to somehow get $\tan{x}$ in this way, but I ultimately got just
$${3-6\sin^2{x}-4\sin{x}\cos{x}\over-5+10\cos^2{x}+8\sin{x}\cos{x}}$$
Had really no ideas what to either do after this, seems pretty unusable to me. Was there possibly a mistake I made in the transformation or maybe another way of solving this?
| The answer is $\displaystyle \frac 94$.
Alternative method.
I like this half angle identity: $\displaystyle \tan \frac 12 y = \frac{\sin y}{1 + \cos y}$
So $\displaystyle 3 = \tan x = \frac{\sin 2x}{1 + \cos 2x}$, giving $\displaystyle \sin 2x = 3 + 3\cos 2x$.
Substituting that into the original expression transforms it into:
$\displaystyle \frac{-3\cos 2x - 6}{17\cos 2x + 12}$
and using $\displaystyle \cos 2x = 2\cos^2x - 1$, that can be re-written:
$\displaystyle \frac{-6\cos^2x - 3}{34\cos^2x -5}$
Finally, going back to $\tan x = 3$, note that $\displaystyle 1+ \tan^2x = 10$, so $\displaystyle \sec^2x = 10$, so $\displaystyle \cos^2x = \frac 1{10}$.
Putting that into the expression yields the value $\displaystyle \frac 94$, which is the required answer.
| {
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| The big error is that $4x^2+16x+16=4(x+2)$ is the same as $4x^2+12x+8=0.$ You somehow got $4x^2+12x+14=0.$ Did you treat $4(x+2)$ as the same as $4x+2?$ The equation $4x^2+12x+8=0$ has $x=-1$ and $x=-2$ as roots.
There's an earlier error where you write: $3x+7=(1-\sqrt{x+2})^2.$ The right side should be $(1+\sqrt{x+2})^2,$ but your later expansion somehow yields the correct value - so two errors led to a correct expression.
It's easier, when you have $2x+4=2\sqrt{x+2},$ if you divide by $2$ before squaring, and get: $x+2=\sqrt{x+2}.$
One quick way to simplify it from the start is to set $y=x+2.$ Then $3y+1=3x+7.$ So you have a slightly simpler equation:
$$\sqrt{3y+1}-\sqrt{y}=1\\
\sqrt{3y+1}=1+\sqrt{y}\\
3y+1 = 1+2\sqrt{y}+y\\
2y=2\sqrt{y}\\
y=\sqrt{y}\\
y^2=y\\
y=0,1$$
You have to go back and check each $y$ in the original equation, then take $x=y-2$ for each solution $y.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Writing a matrix as a product of two matrices Consider the matrix
$$ A = \begin{pmatrix}
0 & y & -x\\
y & y^2 & -xy\\
-x & -xy & x^2
\end{pmatrix}. $$
Is it possible to find matrices $X = X(x)$ and $Y=Y(y)$ such that $A = XY$ (or $A = YX$)?
A possibly unrelated observation of mine is that if we consider the vector $v = \begin{pmatrix}y\\-x\end{pmatrix}$, then we can write $A$ in block form as
$$ A = \begin{pmatrix}
0 & v^t\\
v & vv^t
\end{pmatrix}, $$
which allows us to write
$$ A = \begin{pmatrix}
1 & 1\\
0 & v
\end{pmatrix}\cdot
\begin{pmatrix}
-1 & 0\\
1 & v^t
\end{pmatrix}, $$
but this is not really what I want since now the factors depend on both $x$ and $y$.
EDIT: As suggested in the comments, setting $z = -x$ yields
$$ A = \begin{pmatrix}
0 & y & z\\
y & y^2 & yz\\
z & yz & z^2
\end{pmatrix}. $$
| Becasue of the symmetry, if it is possible with $A=YX$ then it is possible with $A=XY$ too. So without loss of generality let us assume that we can write $A(x,y)=X(x)Y(y)$ for some matrix-valued functions $X$ and $Y$.
Now setting $x=-1$ we get
$$ X(-1)Y(y)\begin{pmatrix}1-w\\0\\w\end{pmatrix} = \begin{pmatrix} 0 & y & 1 \\ y & y^2 & y \\ 1 & y & 1 \end{pmatrix} \begin{pmatrix}1-w\\0\\w\end{pmatrix} =
\begin{pmatrix} w \\ y \\ 1\end{pmatrix}$$
and these vectors clearly span $\mathbb R^3$ when we allow $y$ and $w$ to vary. So $X(-1)$ must have full rank.
A similar argument from the other side shows that $Y(1)$ must have full rank too.
But $A(-1,1)$ has rank $2$ and cannot be the product of invertible $X(-1)$ and $Y(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of Bound for Growth of Divergent Trajectory in $3x+1$ Problem In this paper, Lagarias makes the following claim in section 2.7 (Do divergent trajectories exist?).
Context
$$T(x) = \left\{ \begin{array}{rl} \dfrac{3x + 1}{2}, & 2 \nmid x \\ \dfrac{x}{2}, & 2 \mid x \end{array} \right.$$
$$\begin{align*} \tag{2.30} \lim_{k \to \infty} |T^{(k)}(n_0)| = \infty \end{align*}$$
Claim
If a divergent trajectory $\{T^{(k)}(n_0) : 0 \leq k < \infty\}$ exists, it cannot be equidistributed $\pmod{2}$. Indeed if one defines
$N^*(k) = |\{j : j \leq k \mathrm{\ and\ } T^{(j)}(n_0) \equiv 1 \pmod{2}\}|$,
then it can be proved that the condition (2.30) implies that
$$\begin{align*} \tag{2.31} \liminf_{k \to \infty} \dfrac{N^*(k)}{k} \geq (\log_2 3)^{-1} \approx .63097 \end{align*}$$
Question
How can this statement be proved?
Difficulty
It seems like the author may be ignoring the $+1$ term under the assumption that the factors will dominate. (I've seen this assumption made often for heuristic arguments for the truth of the Collatz conjecture.) I don't see how such an assumption can be justified.
Given any length $n$ sequence of $n - k$ zeros and $k$ ones, we can find an $x \in \mathbb{N}$ such that
$$T^n(x) = \dfrac{3^k x + m}{2^n}$$
where
$$3^k - 2^k \leq m \leq 2^{n-k}(3^k - 2^k)$$
Now, suppose, for example, $n = 2k$. Then, we have the bound
$$T^n(x) \leq \dfrac{3^k x + 2^{n-k}(3^k - 2^k)}{2^n} = \left(\dfrac{3}{4}\right)^k x + \left(\dfrac{3}{2}\right)^k - 1$$
and the exponential "$+1$" term dominates for large $n$. Now, of course, $m$ won't always be as large as possible, but even if we look at "random" $m$, that only introduces a constant factor in front of the exponential.
Additional Questions
Is the proof of this statement difficult? Is that why the author doesn't include it? Is there a paper containing a proof?
| I contacted the author, and he was kind enough to write up a proof for me. I have attempted to simplify his proof for presentation here. I also use some notation without explanation to reduce clutter; the meanings should be clear. The trick is to use an apparently well known result from lattice theory.
Proposition 1 (Lattice Theory rotation trick)
If $b_1, b_2, \ldots, b_\ell$ are real numbers such that
\begin{align*}
b_1 + b_2 + \cdots + b_\ell = r \ell
\end{align*}
($\ell \geq 2$) then the lattice path
\begin{align*}
(0, 0), (1, b_1), (2, b_1 + b_2), \ldots, (\ell, b_1 + b_2 + \cdots + b_\ell) = (\ell, r \ell)
\end{align*}
has a cyclic forward shift by some $k$ with $0 \leq k \leq \ell - 1$
\begin{align*}
(0, 0), (1, b_{k+1}), (2, b_{k+1} + b_{k+2}), \ldots, (\ell, b_{k+1} + b_{k+2} + \cdots + b_\ell + b_1 + b_2 + \cdots + b_k) = (\ell, r \ell)
\end{align*}
so that
\begin{align*}
b_{\overline{k+1}} + b_{\overline{k+2}} + \cdots + b_{\overline{k+j}} \leq jr
\end{align*}
for all $1 \leq j \leq \ell$, where $\overline{k+i} \equiv k + i \pmod{\ell}$ and $1 \leq \overline{k+i} \leq \ell$.
Proof
Let $k$ be the smallest index such that the point $(k, b_1 + b_2 + \cdots + b_k)$ is not above the line $y = rx$ and the distance between this point and the line is maximum. Notice that $k \leq \ell - 1$ by the extreme value theorem.
Corollary 2
If $n_1 < n_2 < \ldots < n_k$ and $r = n_k/k$, then there is some $\hat{k}$ with $1 \leq \hat{k} \leq k - 1$ such that
\begin{align*}
n_{k-\hat{k}+1} - n_j \geq (k - \hat{k} + 1 - j) r
\end{align*}
for all $1 \leq j \leq k - \hat{k}$.
Proof
Apply Propositon 1 to the sequence
\begin{align*}
b_1 = n_k - n_{k-1}, b_2 = n_{k-1} - n_{k-2}, \ldots, b_{k-1} = n_2 - n_1, b_k = n_1
\end{align*}
Lemma 3 (bound on additive term)
For odd $x \in \mathbb{N}$, suppose that
\begin{align*}
T^n(x) = \dfrac{3^k}{2^n}x + e(x, k)\;\;\;\;\; e(x, k) = \sum_{i=0}^{k-1} \dfrac{3^i}{2^{n_k - n_{k-1-i}}}
\end{align*}
where $r = n/k \geq \log_2 3$. Then there is a $1 \leq \hat{k} < k$ such that
\begin{align*}
e(x,k-\hat{k}) \leq \dfrac{1}{2^r - 3}
\end{align*}
Proof
Let $r = n/k = (\log_2 3)(1 + \delta)$, where $\delta > 0$, and apply Corollary 2 to $0 = n_0 < n_1 < \cdots < n_{k-1}$ to find an index $\hat{k}$ such that $1 \leq \hat{k} < k$ and
\begin{align*}
n_{k-\hat{k}} - n_{j-1} \geq (k - \hat{k} - j + 1)r
\end{align*}
for all $1 \leq j \leq k - \hat{k}$. Then,
\begin{align*}
2^{n_{k-\hat{k}} - n_{i-1}} \geq 2^{(k-\hat{k}-i+1)r} = 3^{(k-\hat{k}-i+1)(1+\delta)}
\end{align*}
and
\begin{align*}
e(x, k - \hat{k}) & = \sum_{i=1}^{k-\hat{k}} \dfrac{3^{k-\hat{k}-i}}{2^{n_{k-\hat{k}}-n_{i-1}}} \\
& \leq \sum_{i=1}^{k-\hat{k}} \dfrac{3^{k-\hat{k}-i}}{3^{(k-\hat{k}-i+1)(1 + \delta)}} \\
& = \dfrac{1}{3} \sum_{i=1}^{k-\hat{k}} \dfrac{1}{3^{(k-\hat{k}-i+1)\delta}} \\
& \leq \dfrac{1}{3^{1+\delta} - 3} \\
& = \dfrac{1}{2^r - 3}
\end{align*}
Remark
The key fact about Lemma 3 is that the bound is independent of both $x$ and $e(x,k)$, depending only on the value of $r = n/k$. It achieves this by making the choice $\hat{k}$ that depends on $x$ and showing that such a choice must exist for every $x$ with $r \geq \log_2 3$. This is the ``trick."
Theorem 4
If $x, T(x), T^2(x), \ldots$ is a divergent trajectory with
\begin{align*}
k(x, n) = |\{T^j(x) \equiv 1 \pmod{2} : 0 \leq j < n\}|
\end{align*}
then
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} \geq \log_3 2
\end{align*}
Proof
Since $x$ has a divergent trajectory, there must be a sequence $y_0 < y_1 < y_2 < \cdots$ such that $y_j = T^{n_j}(x)$, for some natural numbers $n_0 < n_1 < n_2 < \cdots$, and such that $T^n(y_j) > y_j$ for all $n \in \mathbb{N}$. For each (fixed) $y_j$, we have
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} = \liminf_{n \to \infty} \dfrac{k(x, n) - k(x, n_j)}{n - n_j} = \liminf_{n \to \infty} \dfrac{k(y_j, n)}{n}
\end{align*}
Now, suppose that
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x, n)}{n} < \log_3 2
\end{align*}
Then, there is some constant $c$ such that
\begin{align*}
\dfrac{k(x, n)}{n} \leq \dfrac{1}{c} < \log_3 2
\end{align*}
infinitely often, and, in particular, for each $y_j$, there is always an $n$ such that
\begin{align*}
\dfrac{k(y_j, n)}{n} \leq \dfrac{1}{c} < \log_3 2
\end{align*}
Let $d = c - \log_2 3 > 0$. Then, $n \geq k(y_j, n)(\log_2 3 + d)$ implies
\begin{align*}
2^n \geq 3^{k(y_j, n)} 2^{kd} \geq 3^{k(y_j, n)} 2^d \iff \dfrac{3^{k(y_j, n)}}{2^n} \leq 2^{-d}
\end{align*}
Let $r = n/k(y_j, n) \geq c$. Applying Lemma 3, we have an $n^*$ such that
\begin{align*}
T^{n^*}(y_j) \leq 2^{-d} y_j + \dfrac{1}{2^r - 3} \leq 2^{-d} y_j + \dfrac{1}{2^c - 3}
\end{align*}
where the values $c$ and $d$ are constant across all the $y_j$ (i.e., for each $y_j$, there is an $n^*$ such that the bound holds). Since $2^{-d} < 1$ and $y_j$ grow unbounded, there is a
\begin{align*}
y_j > \dfrac{1}{(2^c - 3)(1 - 2^{-d})}
\end{align*}
at which point
\begin{align*}
T^{n^*}(y_j) < y_j
\end{align*}
contradicting our construction of the $y_j$ and proving
\begin{align*}
\liminf_{n \to \infty} \dfrac{k(x,n)}{n} \geq \log_3 2
\end{align*}
Remark
If you find any errors in the above, it is almost certainly from my attempt to simplify the proof I was given and not from the author.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Burnside's Lemma on octagon, using one of two colors on each side I am trying to find why the number of colorings of a regular octagon there are such that each side is colored either red or blue is 10. As well as this condition, each color must be used once. Essentially, out of the 8 edges, 4 are red and 4 are blue. If an octagon can be rotated to become another octagon, then they are considered the same.
My work on this is such: there are ${8 \choose 4} = 70$ possible combinations, forgetting about rotation. If a possible octagon is rotated clockwise once, there is no way it can be considered the same. If it is rotated twice, there are two ways it can be considered the same - a RBRBRBRB or BRBRBRBR color scheme. Similarly, if it is rotated $3$, $5$, or $7$ times, there are $0$ "fixed" combinations. If it is rotated $4$ or $6$ times, there are $2$ fixed points. The answer should then be $\frac{70+0+2+0+2+0+2+0}{8} = 9.5$. However, this is obviously not right, being a non-integer. The correct answer is $10$, found through writing a quick computer program.
Can anyone help me find where I messed up?
| We may apply PET here since we require the cycle index $Z(C_8)$ of the
cyclic group $C_8$ anyway in order to apply Burnside. We have
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}$$
With $n=8$ this works out to
$$Z(C_8) = \frac{1}{8} a_1^8 + \frac{1}{8} a_2^4
+ \frac{1}{4} a_4^2 + \frac{1}{2} a_8.$$
We get
$$[R^4 B^4] Z(C_8; R+B) \\ =
\frac{1}{8} [R^4 B^4] (R+B)^8
+ \frac{1}{8} [R^4 B^4] (R^2+B^2)^4
+ \frac{1}{4} [R^4 B^4] (R^4+B^4)^2
\\ + \frac{1}{2} [R^4 B^4] (R^8+B^8)
\\ = \frac{1}{8} {8\choose 4}
+ \frac{1}{8} [R^2 B^2] (R+B)^4
+ \frac{1}{4} [R B] (R+B)^2
\\ = \frac{1}{8} {8\choose 4} + \frac{1}{8} {4\choose 2}
+ \frac{1}{4} {2\choose 1}
= \frac{35}{4} + \frac{3}{4} + \frac{1}{2} = 10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276521",
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"answer_count": 2,
"answer_id": 1
} |
Compute the following sum in closed form : $\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}$ $$\text{Find : }\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}$$
I know that $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-4x^{2}}}$ so $\displaystyle\sum_{n=1}^{\infty}n\binom{2n}{n}x^{2n}=\frac{2x^2}{\sqrt{1-4x^{2}}}$.
But I don't know he to complete this work because I find hypergeometric function.
| As remarked in the OP,
\begin{equation}
\sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-4x^{2}}}
\end{equation}
or, by changing $x\to x/2$,
\begin{equation}
\sum_{n=0}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-x^{2}}}
\end{equation}
We use the decomposition
\begin{equation}
\frac{n}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 4n+1 \right)}=\frac{1}{12}\frac{1}{2n-1}-\frac{1}{4}\frac{1}{2n+1}+\frac{1}{3}\frac{1}{4n+1}
\end{equation}
Then, for $\left|x\right|<1$, as the three series converge absolutely
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{nx^{2n}}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 4n+1 \right)}=\frac{1}{12}S_--\frac{1}{4}S_++\frac{1}{3}S_2
\end{equation}
where
\begin{align}
S_+&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{ 2n+1}\\
S_-&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n-1}\\
S_2&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{4n+1}
\end{align}
With the notation
\begin{equation}
f(x)=\frac{1}{\sqrt{1-x^{2}}}
\end{equation}
by integrating the above series, we have
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n}&=f(x)-1\\
S_+=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n+1}&=\int_0^1 \left[f(x)-1\right]\,dx\\
&=\frac{\pi}{2}-1
\end{align}
A simple transformations is necessary forevaluating $S_-$:
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n-2}&=\frac{f(x)-1}{x^2}\\
S_-=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n-1}&=\int_0^1\frac{f(x)-1}{x^2}\,dx\\
&=1
\end{align}
For the third term, changing $x\to x^2$ in the series before integrating
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{4n}&=f(x^2)-1\\
S_2=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{4n+1}&=\int_0^1\frac{dx}{\sqrt{1-x^4}}-1\\
&=\frac{\Gamma(\frac{5}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}-1\\
&=\frac{\left[\Gamma\left( 1/4\right)\right]^2 }{4\sqrt{2\pi}}-1
\end{align}
This classical integral can be found for example here.
Putting all the results together gives
\begin{equation}
\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}=-\frac{\pi}{8}+\frac{\left[\Gamma\left( 1/4\right)\right]^2 }{12\sqrt{2\pi}}
\end{equation}
which is also identical to the result given by @ChipHurst in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$
How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$
where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{n^3}$ is the trilogarithm and $G=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}$ is Catalan's constant
Trying the algebraic identity $\ 4ab=(a+b)^2-(a-b)^2\ $ where $\ a=\ln(1-x)$ and $b=\ln(1+x)\ $is not helpful here and the integral will be more complicated.
Also, applying IBP or substituting $x=\frac{1-y}{1+y}$ is not that useful either.
All approaches are appreciated.
| Different approach:
Start with subbing $x\mapsto \frac{1-x}{1+x}$
$$\small{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx=\ln2\underbrace{\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx}_{-G}-\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx}\tag1$$
where
$$\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{x\mapsto 1/x}$$
$$=\underbrace{\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx}_{2\ \text{Im}\operatorname{Li}_3(1+i)}-\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}dx}_{\pi^3/16}$$
$$\Longrightarrow \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx=\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx+\text{Im}\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}\tag2$$
Plug $(2)$ in $(1)$ we obtain
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)
$$
How do I proceed further as I am stuck with the last infinite series.
| First, it must be:
$$S=\sum_{n=2}^\infty\frac{^nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\\
\color{blue}{=\sum_{n=2}^\infty\frac{(n+1)-n}{2(n+1)(n-2)!}=\frac12\left[\sum_{n=2}^\infty\frac{1}{(n-2)!}-\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}\right]=}\\
=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big).$$
Second, evaluating the second series is more difficult:
$$\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}=\sum_{n=2}^\infty\frac{n^2(n-1)}{(n+1)!}=\\
\sum_{n=2}^\infty\frac{(n+1)(n^2-2n+2)-2}{(n+1)!}=\sum_{n=2}^\infty\frac{(n+1)n(n-1)-(n+1)n+2(n+1)-2}{(n+1)!}=\\
\sum_{n=2}^\infty\left[\frac{1}{(n-2)!}-\frac{1}{(n-1)!}\right]+2\sum_{n=2}^\infty\left[\frac{1}{n!}-\frac{1}{(n+1)!}\right]=\\
\left[\frac1{0!}-\frac1{1!}+\frac1{1!}-\frac1{2!}+\frac1{2!}-\frac1{3!}+\cdots\right]+2\left[\frac1{2!}-\frac1{3!}+\frac1{3!}-\frac1{4!}+\frac1{4!}-\frac1{5!}+\cdots\right]=2.$$
Thus, following lab bhattacharjee's method is more efficient.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
change of base for $M_{22}$ I have been stuck on this question for a while now. I can easily do the change of basis matrix if the entries in the basis are vectors as opposed to a matrix.
Let
$$B_1 = \{\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
,\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
,\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}\}
\quad and \quad B_2 = \{\begin{bmatrix}
1 & 1 \\
0 & -1
\end{bmatrix}
,\begin{bmatrix}
1 & 0 \\
1 & -1
\end{bmatrix}
,\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}\}$$
be two bases for $span(B_1)$ in $M_{22}$, where the usual left to right ordering is assumed. Find the transition matrix (change of coordinate/change of basis matrix) $P_{B1\rightarrow B2}$
| Take the first vector of $B_1$ (in this case, by vector I mean the matrix) and write it as a linear combination of the elements in $B_2$, like this:
$$\begin{pmatrix} 1&1 \\ 1&-1\end{pmatrix}=0\begin{pmatrix} 1&1 \\ 0&-1\end{pmatrix}+1\begin{pmatrix} 1&0 \\ 1&-1\end{pmatrix}+1\begin{pmatrix} 0&1 \\ 0&0\end{pmatrix}$$
the scalars are those that go in the first column of the desired matrix. Repeat the same process with the rest of the elements of $B_1$
$$\begin{pmatrix} 0&1 \\ 1&0\end{pmatrix}=-1\begin{pmatrix} 1&1 \\ 0&-1\end{pmatrix}+1\begin{pmatrix} 1&0 \\ 1&-1\end{pmatrix}+2\begin{pmatrix} 0&1 \\ 0&0\end{pmatrix}$$
$$\begin{pmatrix} 0&-1 \\ 1&0\end{pmatrix}=-1\begin{pmatrix} 1&1 \\ 0&-1\end{pmatrix}+1\begin{pmatrix} 1&0 \\ 1&-1\end{pmatrix}+0\begin{pmatrix} 0&1 \\ 0&0\end{pmatrix}$$
and put the scalars in their respective columns. Thus
$$P=\begin{pmatrix} 0&-1&-1 \\ 1&1&1 \\ 1&2&0\end{pmatrix}$$
| {
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $C$ such that $C^TAC=$diag$(I_k,-I_l,O)$ (Diagonal form) Let $A=\begin{pmatrix} 1 & 2 & 2 & 0 \\ 2 & 1 & 0 & 2 \\ 2 & 0 & 1 & 2 \\ 0 & 2 & 2 & 1 \end{pmatrix} \in M_4(\mathbb{R})$
I want to find a matrix $C$, such that $C^TAC=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$, since there are four eigenvalues, three positive and one negative.
I used the algorithm
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
and got:
$D=\begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ \frac{2}{3} & -\frac{4}{3} & 1 & 0 \\ -\frac{8}{7} & \frac{2}{7} & \frac{2}{7} & 1 \end{pmatrix}$
So: $DAD^T=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -3 & 0 & 0 \\ 0 & 0 & \frac{7}{3} & 0 \\ 0 & 0 & 0 & \frac{15}{7} \end{pmatrix}$
But I want $C^TAC=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$
So I multiplied on the far left and far right by $\begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 & 0 \\ \frac{2}{3}\sqrt{\frac{3}{7}} & -\frac{4}{3}\sqrt{\frac{3}{7}} & \sqrt{\frac{3}{7}} & 0 \\ -\frac{8}{7}\sqrt{\frac{7}{15}} & \frac{2}{7}\sqrt{\frac{7}{15}} & \frac{2}{7}\sqrt{\frac{7}{15}} & 0 \end{pmatrix}$ and it's transpose, but the result isn't
$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$.
So which matrix do I have to multiply on the far left and far right and how can I find such a matrix?
| You’re well on the way to a solution, having done the hard part: you’ve found a matrix $D$ such that $DAD^T$ is a diagonal matrix. Now you just have to massage this diagonal matrix the desired form. There are two things that you’ll need to do, in either order:
*
*Move the negative element on the diagonal down to the lower-right corner.
*Scale the diagonal elements so that they’re equal to $\pm1$.
For the first of these tasks, look for a permutation matrix $P$ that will move the second column of a $4\times4$ matrix all the way to the right when you right-multiply by $P$. There are many choices, but a simple transposition will do the trick. Left-multiplying by $P^T$ will move the second row to the bottom, so the net effect on your diagonal matrix will be to move the negative element in the second row and column to the bottom right.
For the second of these tasks, find another diagonal matrix $S$ such that both left- and right-multiplying by this matrix will scale the elements of your diagonal matrix appropriately.
The matrix $C$ that you seek is the product of these two matrices and $D^T$, taken in an appropriate order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
\dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\
\dfrac{1}{n!} & \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!}
\end{pmatrix}
=\dfrac{1}{n!}.
$$
| Hints
Prove it by induction.
At each step, expand by minors along the top row.
At the end, think about the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logarithmic inequality (looking for a better solution) $1+\sqrt{17-\log_{x}{2}} \cdot \log_{2}{x^7} \geq \log_{2}{x^{27}}$
Let $t = \log_{2}{x}$. Then we get (taking account of the fact that $x>0$ and $x \ne 1$
$$1+\sqrt{17-\frac{1}{t}}\cdot 7t \geq 27t$$ or
$$\sqrt{17-\frac{1}{t}} \cdot 7t \geq 27t-1 \tag{1}$$
It's clear that $17-\frac{1}{t} \geq 0$ or else the root is not defined so either $t<0$ or $t\geq \frac{1}{17}$.
Then I consider two cases: when $t<0$ and we can divide by $t$ and reverse the inequality or when $t\geq \frac{1}{17}$ and we can divide by $t$ safely. This way, I am able to get the correct answer.
Do you think this is the most rational approach to the problem?
At first, I mistakenly simplified the inequality by dividing the left part by $\sqrt{t}$ but this is obviously not correct for all values of $t$.
| As you wrote, we have to have
$$t< 0\qquad\text{or}\qquad t\ge\frac{1}{17}\tag1$$
Now, we have
$$\begin{align}&1+\sqrt{17-\frac{1}{t}}\cdot 7t \geq 27t
\\\\&\iff 7t\sqrt{17-\frac{1}{t}} \geq 17t-1+10t
\\\\&\iff 7t\sqrt{17-\frac{1}{t}}\geq t\left(17-\frac 1t\right)+10t
\\\\&\iff t\left(17-\frac 1t\right)-7t\sqrt{17-\frac 1t}+10t\le 0
\\\\&\iff t\left(\left(17-\frac 1t\right)-7\sqrt{17-\frac{1}{t}}+10\right)\le 0
\\\\&\iff t\left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\le 0\tag2\end{align}$$
Now, let us consider $(1)$.
When $t< 0$, we have
$$\begin{align}(2)&\iff \left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\color{red}{\ge} 0
\\\\&\iff \sqrt{17-\frac{1}{t}}\le 2\qquad\text{or}\qquad \sqrt{17-\frac{1}{t}}\ge 5
\\\\&\iff 17-\frac 1t\le 4\qquad\text{or}\qquad 17-\frac 1t\ge 25
\\\\&\iff 17t-1\ge 4t\qquad\text{or}\qquad 17t-1\le 25t
\\\\&\iff t\ge -\frac 18\end{align}$$
When $t\ge\frac{1}{17}$, we have
$$\begin{align}(2)&\iff \left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\le 0
\\\\&\iff 2\le \sqrt{17-\frac{1}{t}}\le 5
\\\\&\iff 4\le 17-\frac 1t\le 25
\\\\&\iff 4t\le 17t-1\le 25t
\\\\&\iff t\ge\frac{1}{13}\end{align}$$
So, we get
$$-\frac 18\le t <0\qquad\text{or}\qquad t\ge\frac{1}{13}$$
Hence, the answer is
$$\color{red}{2^{-1/8}\le x< 1\qquad\text{or}\qquad x\ge 2^{1/13}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $ab + cd = -1; ac + bd = -1; ad + bc = -1$ over the integers I am trying to solve this problem:
Solve the system of equations
\begin{align}
\begin{cases}
ab + cd = -1 \\
ac + bd = -1 \\
ad + bc = -1
\end{cases}
\end{align}
for the integers $a$, $b$, $c$ and $d$.
I have found that the first equation gives $d = \dfrac{-1-cd}{a}$, which gives $a\neq0$. Other than that, I don't know where to start.
Tips, help or solution is very appreciated. Thanks!
| Hint:
Squaring the equations gives
\begin{align}
\begin{cases}
a^2b^2 +2abcd+ c^2d^2 = 1 \\
a^2c^2 +2abcd +b^2d^2 = 1 \\
a^2d^2 + 2abcd+b^2c^2 = 1
\end{cases}
\end{align}
Subtracting them yields
\begin{align}
\begin{cases}
(a^2-b^2)(c^2-d^2)=0 \\
(a^2-c^2)(b^2-d^2)=0 \\
(a^2-d^2)(b^2-c^2)=0
\end{cases}
\end{align}
Now consider some cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:
Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$
We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so this is only possible when $x^3+y^3=4$ which no $x,y \in \mathbb{Z}$ satisfy.
Now I don't know if that helps at all solve the second bit. We have $z^6 = x^3+y^3-3$. I know that $x^3+y^3 = (x+y)(x^2-xy+y^2)$.
| Any solution for $x^3+y^3-z^6-3=0$ is also a solution mod $7$. Therefore there are no solutions to the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What $x$ makes $\frac{x}{(a^2 + x^2)}$ maximum? The problem (Calculus Made Easy, Exercises IX, problem 2 (page 130)) is:
What value of $x$ will make $y$ a maximum in the equation
$$y = \frac{x}{(a^2 + x^2)}$$
I successfully differentiate, equate to zero, and wind up with
$$x^2 = a^2$$
Which gives me the answer of
$$x = a$$
This is correct. But why isn't $x = -a$ also correct?
| We are given
$$y = \frac{x}{a^2 + x^2}$$
where $a$ is a constant.
Differentiating with respect to $x$ using the Quotient Rule yields
\begin{align*}
y' & = \frac{1(a^2 + x^2) - x(2x)}{(a^2 + x^2)^2}\\
& = \frac{a^2 + x^2 - 2x^2}{(a^2 + x^2)^2}\\
& = \frac{a^2 - x^2}{(a^2 + x^2)^2}
\end{align*}
Setting the derivative equal to zero yields the critical points $x = \pm a$.
We can apply the First Derivative Test.
First Derivative Test. Assume $f$ is continuous on a closed interval $[u, v]$ and $f$ is differentiable everywhere in the open inteval $(u, v)$ except possibly at $c$.
(a) If $f'(x) > 0$ for all $x < c$ and $f'(x) < 0$ for all $x > c$, then $f$ has a relative maximum at $x = c$.
(b) If $f'(x) < 0$ for all $x < c$ and $f'(x) > 0$ for all $x < c$, then $f$ has a relative minimum at $x = c$.
If $a = 0$, then $y = \dfrac{1}{x} \implies y' = -\dfrac{1}{x^2}$, so the function has no critical points and no relative extrema.
Assume $a \neq 0$. Since it has not been specified whether $a > 0$ or $a < 0$, the critical points occur at $x = -|a|$ and $x = |a|$. If we perform a line analysis on the derivative, we see that $y'$ changes from negative to positive at the critical point $x = -|a|$ and from positive to negative at the critical point $x = |a|$.
Thus, by the First Derivative Test, the function has a relative maximum at $x = |a|$ and a relative minimum at $x = -|a|$. If it is specified that $a > 0$, you can replace $|a|$ by $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?
$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$
Thank you in advance.
| You have to solve this inequality system:
$$\left\{\begin{matrix}
& \\k \neq 1 \: \land \: k\neq0
& \\k(k+1)\geqslant0
& \\1+\frac{2}{k}+\frac{3}{4k^2}\geq 0
& \\ \frac{\sqrt{k(k+1)}}{k-1} \geq 0
& \\\left (\frac{\sqrt{k(k+1)}}{k-1}\right )^2\leq \left ( 1+\frac{2}{k}+\frac{3}{4k^2}
\right )^2
\end{matrix}\right.$$
In this way you wil get the solutions of the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
| Make up the frequency and cumulative frequency table:
$$\begin{array}{c|c|c}
x&f&F\\
\hline
1&1&1=2^1-1\\
2&2&3=2^2-1\\
4&4&7=2^3-1\\
8&8&15=2^4-1\\
\vdots&\vdots&\vdots\\
256&256&511=2^{9}-1\\
512&512&1023=2^{10}-1\\
1024&1024&2047=2^{11}-1\\
\vdots&\vdots&\vdots\\
2^n&2^n&2^{n+1}-1
\end{array}$$
So, your approach was efficient to notice that $a_{2^n}=2^n$. Hence, $a_{1024}=\color{red}{a_{1025}}=\cdots =a_{2047}=\color{red}{1024}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 6
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Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$
what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2x(1+\sin^2 2x))$
$16\sin^4 (2x)+68\sin^3 2x+34\sin^2 2x+68\sin 2x+68\sin^2 4x+16=0$
How do i solve it Help me please
| You're not required to find all solutions, just to find how many there are.
Let $u=\sin(2x)$. Then the trigonometric equation in $x$ becomes a polynomial equation in $u$:
$$
0 = (1 - u)^4 + 2 u^2 - 17 (1 + u)^4 = -2 (8 u^4 + 36 u^3 + 47 u^2 + 36 u + 8)
$$
Now plot this function of $u$ and see how many solutions are in the interval $[-1,1]$ so that $u$ can be the sine of something.
The graph below tells us that there is exactly one solution in $[-1,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| Hint: $$\left( \frac{\sqrt{6} + \sqrt{2}}{2}\right)^2 = \frac{6 + 2 \sqrt{6} \sqrt{2} + 2}{4} = ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sqrt{a\pm bi\sqrt c}$ I recently encountered this problem
$$\sqrt{10-4i\sqrt{6}}$$
To witch I set the solution equal to $a+bi$ squaring both sides leaves
$${10-4i\sqrt{6}}=a^2-b^2+2abi$$
Obviously $a^2-b^2=10$ and $2abi=-4i\sqrt{6}$, using geuss and check, the solution is $a=\sqrt12, b=\sqrt2$
But I was wondering if there is a faster way to solve these types of problems or a method that doesn't involve guess and check (since it can get tedious at times) for the basic form
$$\sqrt{a\pm bi\sqrt c}=x+yi$$ Where you're solving for $x$ and $y$.
I've attempted but failed since it gets pretty messy. All formula will be very much appreciated. (not all equations of that form can reduce)
| We have that
$$a^2-b^2=10 \quad \textrm{and} \quad 2ab=-4\sqrt 6$$
Now, squaring both equalities and addem up we get
$$(a^2+b^2)^2=(a^2-b^2)^2 +(2ab)^2=10^2+(-4\sqrt 6)^2 =196$$
$$\Rightarrow \quad a^2+b^2=14$$
and using again the first equality we obtain
$$a^2=12 \quad \textrm{and} \quad b^2=2$$
or
$$a=\pm 2\sqrt 3 \quad \textrm{and} \quad b=\pm \sqrt 2$$
but $ab<0$ (by the second equality), that is, $a$ and $b$ are the opposite sign. Thus, the solutions are given by
$$a= 2\sqrt 3 \quad \textrm{and} \quad b=- \sqrt 2$$
or well
$$a=-2\sqrt 3 \quad \textrm{and} \quad b=\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$ I am trying to find a closed form for this infinite series:
$$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$
Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.
I found this integral representation of S:
$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx$$
Sketch of a proof:
Recall the integral representation of the harmonic numbers: $H_n=\displaystyle\int_{0}^{1}\frac{1-x^n}{1-x}dx$
By plugging it into the definition of S and interchanging the order of summation between $\displaystyle\sum$ and $\displaystyle\int$ (justified by the uniform convergence of the function series $\displaystyle\sum\left(x\to\frac{4^n}{n^2{2n\choose n}}\frac{1-x^n}{1-x}\right)$, because $\forall x\in[0,1],\frac{1-x^n}{1-x}<n$), we get:
$$S=\int_{0}^{1}\frac{1}{1-x}\sum\limits_{n=1}^{\infty}\frac{4^n(1-x^n)}{n^2{2n\choose n}}dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
Using the result $\displaystyle\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}=\frac{\pi^2}{2}$.
At that point, we will rely on the taylor series expansion of $\arcsin^2$:
$$\arcsin^2(x)=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}x^{2n}, |x|<1$$
Out of which we get $\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$
So,
$$S=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-2\arcsin^2\left(\sqrt{x}\right)\right)dx$$
Which, through the substitution $u=\sqrt{x}$, gives the integral representation above.
But beyond that, nothing so far. I tried to use the integral representation of $\frac{H_n}{n}$ to switch the order of summation, but it didn't lead anywhere. Any suggestion?
| From here, we have
$$\frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n=1}^\infty\frac{(2z)^{2n-1}}{n{2n \choose n}}$$
substitute $z=\sqrt{y}$, we get
$$\sum_{n=1}^\infty\frac{4^ny^n}{n{2n \choose n}}=2\sqrt{y}\frac{\arcsin\sqrt{y}}{\sqrt{1-y}}$$
Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $1$ and using the fact that $-\int_0^1 y^{n-1}\ln(1-x)\ dy=\frac{H_n}{n}$, we get
\begin{align}
\sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n \choose 2}}&=-2\int_0^1\frac{\arcsin\sqrt{y}}{\sqrt{y}\sqrt{1-y}}\ln(1-y)\ dy\overset{\arcsin\sqrt{y}=x}{=}-8\int_0^{\pi/2}x\ln(\cos x)\ dx\\
&=-8\int_0^{\pi/2}x\left\{-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}\right\}\ dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x\cos(2nx) dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right)\\
&=\pi^2\ln2+2\pi\sum_{n=1}^\infty\frac{(-1)^n\sin(n\pi)}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^n\cos(n\pi)}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\\
&=\pi^2\ln2+0+2\sum_{n=1}^\infty\frac{(-1)^n(-1)^n}{n^3}-2\operatorname{Li}_3(-1)\\
&=\pi^2\ln2+2\zeta(3)-2\left(-\frac34\zeta(3)\right)\\
&=\pi^2\ln2+\frac72\zeta(3)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| By your work the coefficient before $x$ it's $1$, which ends.
Also, $$\frac{x+6}{x^2-x-6}=\frac{x+6}{(x-3)(x+2)}=\frac{x-3+9}{(x-3)(x+2)}=$$
$$=\frac{1}{x+2}+\frac{9}{5}\left(\frac{1}{x-3}-\frac{1}{x+2}\right)=\frac{\frac{9}{5}}{x+3}+\frac{-\frac{4}{5}}{x+2},$$
which gives $A+B=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating 2 non-definite integrals Calculate the following:
A) $\int \sqrt{3x^4 +x^6 +9x^2} \, dx$
B) $\int \sqrt[3]{{\frac{1}{x^2 +1}}}\, dx$
A) I managed to write $\int x \sqrt{3x^2 +x^4 +9} \, dx$, but then I didn't know what to do because of the square root, even with integration by parts.
B) I tried substituting $u=x^2 +1$ so $du=2x\,dx$ but then I don't have any $x$ for $du$.
| In the integral
$$\int x \sqrt{x^4+3x^2+9} \ dx$$
substitute $u=x^2$, $du=2xdx$ to get
$$\frac{1}{2} \int \sqrt{u^2+3u+9} \ du = \frac{1}{2} \int \sqrt{ \left( u+\frac{3}{2} \right)^2 + \frac{27}{4} } \ du$$ and to continue, substitute
$$u+\frac{3}{2}= \frac{3\sqrt 3}{2} \tan \theta$$
you will need to remember the integral of $\sec^3 \theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial Fractions: Why does this shortcut method work? Suppose I want to resolve $1/{(n(n+1))}$ into a sum of partial fractions. I solve this by letting $1/{(n(n+1))} = {a/n} + {b/(n+1)}$ and then solving for $a$ and $b$, which in this case gives $a=1$ and $b=-1$.
But I learnt about a shortcut method. It says suppose $1/{(n(n+1))} = {a/n} + {b/(n+1)}$, then find $a$ by finding the value which makes its denominator in the RHS equal to $0$ and computing the LHS with the $0$ term (or $a$'s denominator in RHS) removed so we get $a = {1/(0+1)} = 1$ [as $n=0$], and we get $b = {1/(-1)} = -1$ [as $n+1=0$].
Another example, if I am not clear, suppose $$\frac{1}{n(n+1)(n+2)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2};$$ then
$$
\begin{eqnarray}
a &=& \frac{1}{(0+1)(0+2)}=\frac{1}{2}, \\
b &=& \frac{1}{(-1)(-1+2)}=-1, \\
c &=& \frac{1}{(-2)(-2+1)}=\frac{1}{2}.
\end{eqnarray}
$$
Why does this shortcut method work?
| Let's take your example. We have \begin{align}\frac{1}{n(n+1)(n+2)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2}&=\color{red}{\boxed{\frac an+\frac{b(n+2)+c(n+1)}{(n+1)(n+2)}\cdot\frac nn}}\quad(\text{group together}\,b,c)\\&=\color{blue}{\boxed{\frac b{n+1}+\frac{a(n+2)+cn}{n(n+2)}\cdot\frac{n+1}{n+1}}}\quad(\text{group together}\,a,c)\\&=\color{green}{\boxed{\frac c{n+2}+\frac{a(n+1)+bn}{n(n+1)}\cdot\frac{n+2}{n+2}}}\quad(\text{group together}\,a,b)\end{align} so we get $$\color{red}{\frac a{\color{black}{\boldsymbol{n}}}=\frac{1-n[b(n+2)+c(n+1)]}{n(n+1)(n+2)}\implies \color{red}a=\frac{1-\color{black}{\boldsymbol{n}}\boldsymbol{[b(n+2)+c(n+1)]}}{(n+1)(n+2)}}\\\phantom{2cm}\\\color{blue}{\frac b{\color{black}{\boldsymbol{n+1}}}=\frac{1-(n+1)[a(n+2)+cn]}{n(n+1)(n+2)}=\frac{1-\color{black}{\boldsymbol{(n+1)}}\boldsymbol{[a(n+2)+cn]}}{n(n+2)}}\\\phantom{2cm}\\\color{green}{\frac c{\color{black}{\boldsymbol{n+2}}}=\frac{1-(n+2)[a(n+1)+bn]}{n(n+1)(n+2)}=\frac{1-\color{black}{\boldsymbol{(n+2)}}\boldsymbol{[a(n+1)+bn]}}{n(n+1)}}$$ Notice that in each case, when you set $n,n+1,n+2=0$ respectively, the terms in bold disappear, so you get $$\color{red}{a=\frac{1-0}{(0+1)(0+2)}=\frac12}\\\color{blue}{b=\frac{1-0}{(-1)(-1+2)}=-1}\\\color{green}{c=\frac{1-0}{-2(-2+1)}=\frac12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Inverse of a structured matrix of sines. Suppose I have a matrix $P$ defined by
$$
P =\begin{pmatrix}
\sin(\frac{\pi}{n+1}) & \sin(\frac{2\pi}{n+1}) & \cdots & \sin(\frac{n\pi}{n+1}) \\
\sin(\frac{2\pi}{n+1}) & \sin(\frac{4\pi}{n+1}) & \cdots & \\
\vdots & & \ddots & \\
\sin(\frac{n\pi}{n+1}) & \cdots & & \sin(\frac{n^2\pi}{n+1})
\end{pmatrix}
$$
and suppose I wish to find its inverse.
I claim that $P^2 = \frac{n+1}{2} I_n$ and hence $P^{-1} = \frac{2}{n+1} P$ and indeed computing this in Matlab for a variety of $n$ it appears to be true, but I've been struggling to show this rigourously. I suspect I'm just very rusty with my trig manipulations but any help would be appreciated.
So far, I have that
$$
(P^2)_{kl} = \sum_{j=1}^n \sin(\frac{kj\pi}{n+1})\sin(\frac{lj\pi}{n+1})
$$
which looks somewhat similar to the stuff you get in fourier anyalysis where $\sin$ and $\cos$ form orthogonal polynomials, but I've been struggling to adapt it to this sum situation.
From here I've been trying to use that
$$
(P^2)_{kl} = \frac{1}{2}\sum_{j=1}^n \cos\left(\frac{(k-l)j}{n+1} \pi\right) - \cos\left(\frac{(k+l)j}{n+1} \pi\right)
$$
but I really haven't made much progress. Even the diagonal case where $k = l$ seems to not work out as immediately as I hoped, since I'm a bit lost on what to do with the second term. I suspect I am missing something obvious...
| I managed to solve it. It was a bit of a headache but here is my proof.
For $k, l \in \mathbb{N}$ and $n \in \mathbb{N}$ consider the sum
\begin{equation}
A_{kl} = \sum_{j=1}^n \sin \left(\frac{k \pi}{n+1}j\right) \sin \left(\frac{l \pi}{n+1}j\right).
\end{equation}
Claim:
\begin{equation}
A_{kl} = \frac{n+1}{2}\delta_{kl}.
\end{equation}
We start by noting that
\begin{equation}
A_{kl} = \frac{1}{2} \sum_{j=1}^n \cos \left(\frac{(k-l)\pi}{n+1} j\right) - \cos \left(\frac{(k+l)\pi}{n+1} j\right)
\end{equation}
and letting $\theta_\pm = \frac{(k\pm l)\pi}{n+1}$
we have
\begin{equation}
A_{kl} = \frac{1}{2} \sum_{j=1}^n \cos( j\theta_- ) - \cos(j\theta_+ ).
\end{equation}
We now recall the following trigonometric identity due to Lagrange:
\begin{equation}
\sum_{j=1}^n \cos(j \theta) = \frac{1}{2} \left[\frac{\sin((n+\frac{1}{2})\theta)}{\sin(\frac{1}{2}\theta)}-1 \right]
\end{equation}
which holds when $\theta \neq 0$.
Note as well that
\begin{align*}
\sin((n+\frac{1}{2})\theta) &= \sin((n+1)\theta - \frac{1}{2}\theta) \\
&= \sin((n+1)\theta)\cos(\frac{1}{2}\theta) - \cos((n+1)\theta)\sin(\frac{1}{2}\theta).
\end{align*}
and hence the Lagrange identity reduces to
\begin{equation}
\sum_{j=1}^n \cos(j \theta) =\frac{1}{2} \left[ \sin((n+1)\theta)\cot(\frac{1}{2}\theta) - \cos((n+1)\theta) - 1\right].
\end{equation}
We start by considering the diagonal case, when $k = l$. In this situation $\theta_- = 0$ and $\theta_+ = 2\pi k$. Then
\begin{align*}
A_{kk} &= \frac{1}{2} \sum_{j=1}^n (1 - \cos(j\theta_+)) \\
&= \frac{n}{2} - \frac{1}{4} \left[\frac{\sin((n+\frac{1}{2})\theta_+)}{\sin(\frac{1}{2}\theta_+)}-1 \right] \\
&= \frac{n}{2}-\frac{1}{4}\left[\sin(2k\pi)\cot(\frac{k\pi}{n+1}) - \cos(2 k \pi) - 1\right] \\
&= \frac{n}{2} - \frac{1}{4}\left[ -2 \right] \\
&= \frac{n+1}{2},
\end{align*}
since $\sin(2k\pi) = 0$ and $\cos(2k\pi) = 1$ for all $k$.
If we now consider $k \neq l$, then the same method follows but we have to expand both terms.
\begin{align*}
A_{kl} &= \frac{1}{4} \left[\left[\frac{\sin((n+\frac{1}{2})\theta_-)}{\sin(\frac{1}{2}\theta_-)}-1 \right] - \left[\frac{\sin((n+\frac{1}{2})\theta_+)}{\sin(\frac{1}{2}\theta_+)}-1 \right] \right] \\
&= \frac{1}{4}\left[\sin((k-l)\pi)\cot(\frac{1}{2}\theta_-) - \cos((k-l)\pi) - \sin((k+l)\pi)\cot(\frac{1}{2}\theta_+) + \cos((k+l)\pi)\right] \\
&= \frac{1}{4}\left[\cos((k+l)\pi) - \cos((k-l)\pi) \right]\\
&= \frac{1}{4} \left[\cos{k\pi}\cos{l\pi} - \sin{k\pi}\sin{l\pi} - \cos{k\pi}\cos{l\pi} - \sin{k\pi}\sin{l\pi}\right] \\
&= -\frac{1}{2}\sin{k\pi}\sin{l\pi} \\
&=0
\end{align*}
since $\sin(m \pi) = 0$ for all $m \in \mathbb{Z}$.
| {
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"url": "https://math.stackexchange.com/questions/3314129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$ In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$.
What's the strategy to approach such problems. Writing expansion seems tedious here.
| Referring to Jack Crawford's comment above, there are three possible ways to get $x^4$:
$$\underbrace{1\times1\times\cdots\times 1}_{9}\times x^4\\
\underbrace{1\times1\times\cdots\times 1}_{8}\times x\times x^3\\
\underbrace{1\times1\times\cdots\times 1}_{6}\times x\times x\times x \times x$$
The first is the combination: ${10\choose 1}=10$.
The second is the permutation: $P(10,2)=\frac{10!}{8!}=90$.
The third is again combination: ${10\choose 4}=210$.
Hence: $10+90+210=310$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Domain of $7^{\log_7(x^2-4x+5)}$
If $$7^{\log_7(x^2-4x+5)}=x-1$$ then $x$ may have values...
My attempt: $$x^2-4x+5=x-1$$
So, $$x^2-5x+6=0$$
So, $$x=2,3$$
To check the domain of log, $$x^2-4x+5>0$$ i.e., $$(x-2+i)(x-2-i)>0$$ That gives me, $x<2-i$ and $x>2+i$. Is this a valid way of writing domain here? If No, how to write the domain of $7^{\log_7(x^2-4x+5)}$?
Also, if I put the value of $x$ as $2$ or $3$ in the given equation, it satisfies, but if I compare it with the inequalities $x<2-i$ or $x>2+i$, then I am not able to get a satisfactory answer.
| Note that $$x^2-4x+5 = (x-2)^2 +1 >0 $$ for all x, so there is no problem with logarithm and we have
$$7^{log_7(x^2-4x+5)}=x^2-4x+5$$
Therefore $$ 7^{log_7(x^2-4x+5)}=x-1 \iff x^2-5x+6=0$$
Thus the solutions are $x=2$ and $x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$ Find a solution $(2n+1)x \equiv -7 \pmod 9$
I’m sure this is trivial but I still have doubts about it.
I know the equation has solution for certain $n \in \mathbb {Z}$. Actually I have tried a few and got a similar results (with Diophantine equations ). I wonder if there’s general solution for the equation without changing the n for an integer.
Thanks in advance.
| Since $2n + 1$ 'cycles through' the modulo $9$ residues, the problem is reduced to solving
$$\tag 1 x'x \equiv 2 \pmod 9$$
This is equivalent to $x'x = 9k +2$ and we need only look for solutions
$$ 0 \le x' \lt 9 \text{ and } 0 \le x \lt 9$$
We represent both $x'$ and $x$ in $\text{base-}3$ format,
$$\tag 2 x' = a' + b'3 \text{ and } x = a + b3 \quad \text{with } a',b',a,b \in \{0,1,2\}$$
Multiplying,
$$ x'x = a'a + (a'b+ab')3 + bb'3^2$$
Since $a'a + (a'b+ab')3 \le 28 \lt 29 = 2 + 3 \times 9$, we segment the work into 3 parts.
Part 1: $a'a + (a'b+ab')3 = 2$
$\quad$ Ans: [$x' = 1$ and $x = 2$] OR [$x = 1$ and $x' = 2$]
Part 2: $a'a + (a'b+ab')3 = 11$
$\quad$ Ans: [$x' = 4$ and $x = 5$] OR [$x = 4$ and $x' = 5$]
Part 3: $a'a + (a'b+ab')3 = 20$
$\quad$ Ans: [$x' = 7$ and $x = 8$] OR [$x = 7$ and $x' = 8$]
We only work out the details for Part 3:
Since $3 \nmid 20$, $\,3 \nmid 19$ and $3 \nmid 16$, if we have any solutions at all we must have
$\quad a'a = 2$
$\quad (a'b+ab') = 6$
If we set $a' = 2$ and $a = 1$ we get $2b + b' = 6$. So $b = 2$ and $b' =2$. So $x' = 2 + 2 \times 3 = 8$ and $x = 1 + 2 \times 3 = 7$. Up to an interchange, there can be no other solutions.
| {
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"url": "https://math.stackexchange.com/questions/3315462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$
| The expression
(2x+1)/(x-1)(x-2). = A / ( x-1). + B/(x-2) is discontinuous at x= 1 and at x= 2 but identy for restall value of x.
Once we write as 2x +1 = A (x-2) + B (x-1) since left hand side is continuous thus Right-hand side will be having same behaviour .hence we can put x= 2 and x= 1 to find A and B
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sec A-\cos A=1$, then determine the value of $\tan^2\frac A2$ This is what I tried
$\sec A=\frac{1}{\cos A}$, so the equation becomes
$1-\cos^2A=\cos A$
If we solve the above quadratic equation, we the values of $\cos A$ as $\frac{-1\pm \sqrt5}{2}$
Therefore, $\tan\frac A2$ becomes
$$\sqrt \frac{3-\sqrt 5}{1+\sqrt 5}$$
Squaring that value, the answer remains meaningless
The options are
A) $\sqrt 5+ 2$
B) $\sqrt 5-2$
C) $2-\sqrt5$
D) $0$
Since the options are not matching, where am I going wrong?
| Let $t=\tan^2\dfrac A2$. Then $\cos A=\dfrac{1-t}{1+t}$.
$\dfrac{1+t}{1-t}-\dfrac{1-t}{1+t}=1$
$4t=1-t^2$
$(t+2)^2=5$
$t=-2+\sqrt5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$
Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}}{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}} $$
$$ = \frac{ (\sqrt[3]{16} - 1) [(3 + \sqrt[3]{4}) - \sqrt[3]{2}]}{ (3 + \sqrt[3]{4})^{2} - 2^{2/3}} $$
$$ = \frac{ 3 \sqrt[3]{16} - 3\sqrt[3]{4} + \sqrt[3]{2} + 1}{ (9 + 5 \sqrt[3]{4} + \sqrt[3]{16}) } $$
From here on I don't know how to continue.
I can let $a = \sqrt[3]{2}$, but still cannot do anything.
| Let $x=\sqrt[3]2$ then we have $${x^4-1\over x^2+x+3}={x^6-x^2\over x(x^3+x^2+3x)}={4-x^2\over x(2+x^2+3x)}= {(2-x)(2+x)\over x(x+2)(x+1) }$$
$$ = {2-x\over x^2+x}= {(2-x)(x-1)\over x(x+1)(x-1)}= {(2-x)(x-1)\over x^3-x}$$
$$= {(2-x)(x-1)\over 2-x} = x-1$$
Edit: but other solution is much nicer then this one.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find value of $(\cos\frac{2\pi}{7})^ {\frac{1}{3}} + (\cos\frac{4\pi}{7})^ {\frac{1}{3}} + (\cos\frac{8\pi}{7})^ {\frac{1}{3}} $ This question was on my list. I was trying to apply the $n$-th roots of unity, but other ideas are welcome. I also tried Newton's sums, but it's not working.
I searched around here and I didn't find a similar one, but if they do, just say that I delete the topic.
| This cubic-root sum, which has the inscrutable value $\sqrt[3]{(5-3\sqrt[3]{7})/2}$, was discovered over a hundred years ago. Outlined below is an elementary evaluation of the sum.
Note that $\cos\frac{2\pi}7$, $\cos\frac{4\pi}7 $ and $\cos\frac{8\pi}7 $ are the roots of
$$x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{8}=0$$
With the short-hands,
$$c_1=\left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_2=\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_3=\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}}
$$
we have
$$c_1^3+c_2^3+c_3^3 = -\frac{1}{2},\>\>\>
c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 = -\frac{1}{2},\>\>\>
c_1^3c_2^3c_3^3 = \frac{1}{8}\tag{1}$$
Let
$$A=c_1+c_2+c_3, \space\space\space B=c_1c_2+c_2c_3+c_3c_1$$
and evaluate
$$A^3 = c_1^3+c_2^3+c_3^3 +3AB-3c_1c_2c_3,$$
$$B^3= c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3+3c_1c_2c_3 AB-3(c_1c_2c_3)^2
$$
Plugging (1) into above expressions to get
$$A^3=-2+3AB,\space\space\space B^3=-\frac{5}{4}+\frac{3}{2}AB\tag{2}$$
Next, evaluate
$$(2AB-3)^3=8A^3B^3-36A^2B^2+54AB-27$$
and use the results (2) to get the equation satisfied by $A$,
$$\left( \frac{2A^3-5}{3}\right)^3 =-7\tag{3}$$
Finally, solve (3) and we obtain the sum,
$$A = \left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}} +\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}}+\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}} =
\left[ \frac{1}{2}\left(5-3\cdot 7^{\frac{1}{3}}\right) \right]^{\frac{1}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate how many ways you can give $7$ children $7$ identical candies My try:
Calculate how many ways you can give $7$ children $7$ identical candies if each child got at most 2 candies.
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7=7 \text{ for } x_i \in \left\{ 0,1,2\right\}$$
$$[t^7](1+t+t^2)^7[t^7](\frac{1-t^3}{1-t})^7=[t^7](1-t^3)^7 \sum {n+6 \choose 6}t^n$$
$$\begin{array}{|c|c|c|c||c|c|}
\hline
\text{first parenthesis} & \text{ways in the first} & \text{ways in the second }\\ \hline
\text{1} & 1 & { 13 \choose 6} \\ \hline
{t^3} & { 1 \choose 1} & { 10 \choose 6} \\ \hline
{t^6} & { 7 \choose 2} & { 7 \choose 6}\\ \hline
\end{array}$$
Sollution:$${ 7 \choose 2}{ 7 \choose 6}+{ 7 \choose 1}{ 10 \choose 6}+{ 13 \choose 6}=3333$$
But I checked it in Mathematica and I get $393$. So can you check where the error is?
| As was commented, the only issue is the sign of the middle term must be negative. To see it, express the first binomial as a sum too:
$$[t^7](1-t^3)^7 \sum {n+6 \choose 6}t^n=[t^7]\sum_{k=0}^7{7\choose k}(-t^3)^k \sum_{n=0}^{\infty} {n+6 \choose 6}t^n=\\
\underbrace{{7\choose 0}{13\choose 6}}_{k=0,n=7}-\underbrace{{7\choose 1}{10\choose 6}}_{k=1,n=4}+\underbrace{{7\choose 2}{7\choose 6}}_{k=2,n=1}=393.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Angles between vectors of center of two incircles I have two two incircle between rectangle and two
quadrilateral circlein. It's possible to determine exact value of $\phi,$ angles between vectors of center of two circles.
| There are constraints that $A$ and $B$ must fullfill, i.e., the following system of equations coming from Pythagoras theorem applied to certain right triangles :
$$\begin{cases}(B-2)^2+2^2=(A/2)^2\\(B-1)^2+(A/2)^2=(B+1)^2\end{cases}$$
giving $A=8 \sqrt{2}$ and $B=8$.
If now we take equations as in the partial solution you gave (a good idea) :
$$\tan(\theta) = \dfrac{2}{A/2} = \dfrac{\sqrt{2}}{4}\tag{1}$$
and
$$\tan(\phi + \theta) = \dfrac{B-1}{A/2} = \dfrac{7}{4 \sqrt{2}}\tag{2}$$
Equation (2) can also be written :
$$\dfrac{\tan(\phi) + \tan(\theta)}{1-\tan(\phi)\tan(\theta)} = \dfrac{7\sqrt{2}}{8}\tag{3}$$
Let $T:=\tan(\phi)$. (3) is equivalent to :
$$\dfrac{T + \sqrt{2}/4}{1-T\sqrt{2}/4} = \dfrac{7\sqrt{2}}{8}\tag{4}$$
giving $T=\dfrac{10 \sqrt{2}}{23}$. Therefore
$\phi=\arctan(\dfrac{10 \sqrt{2}}{23})\approx 31.5863 \ \text{degrees}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with $x^{6} - 2 = 0$ compute roots in $\mathbb{C}$
I have problem with simple equation $x^{6} - 2 = $ compute roots in
$\mathbb{C}$
I will try compute roots of $x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=(x-2^{1/6})(x^{2}+2^{1/6}x+2^{1/3})(x^{3}+\sqrt{2})$, but this not looks good.
Maybe is better solution.
Any suggestions?
| As commented above, you used $a^3-b^3=(a-b)(a^2+ab+b^2)$. You can also use $a^3+b^3=(a+b)(a^2-ab+b^2)$ and easily solve the quadratic equations:
$$x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=\color{red}{(x-2^{1/6})}\color{green}{(x^{2}+2^{1/6}x+2^{1/3})}\color{blue}{(x+2^{1/6})}\color{purple}{(x^{2}-2^{1/6}x+2^{1/3})} \Rightarrow \\
\color{red}{x_1=2^{1/6}},\color{blue}{x_2=-2^{1/6}},\\
\color{green}{x^2+2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{-2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{-2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2},\\
\color{purple}{x^2-2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
I tried many times, and finally I used Muirhead, but it failed!
$\begin{split}L.H.S.-R.H.S.&=8(1-a)(1-b)(1-c)-abc\\&=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$
But as $(3,0,0)$ majorizes $(2,1,0)$ but $(2,1,0)$ majorizes $(1,1,1)$, so it fails.
Could someone help? Any help is appreciated!
| Also, we can use Muirhead here.
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Thus, $$x=a+b+c-2a=2(1-a)>0.$$
Similarly, $y>0$ and $z>0$ and we need to prove that
$$8xyz\leq(x+y)(x+z)(y+z)$$ or
$$\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$$ which is true by Muirhead.
Also, we can use AM-GM:
$$\prod_{cyc}(x+y)\geq\prod_{cyc}2\sqrt{xy}=8xyz.$$
| {
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Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.
I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither could think of any other ways to proceed.
How to proceed?
Thank you.
| You are asking for
$$ x^3 + 5 x^2 - 19 x + 21 = (y+1)^3 $$
For large enough positive $x,$ (you need to find out explicit lower bound for $x$),
$$ (x+1)^3 < x^3 + 5 x^2 - 19 x + 21 < (x+2)^3 $$
and cannot be a cube. Then check the small values of $x$ remaining.
so, $x^3 + 3 x^2 + 3x + 1 < x^3 + 5 x^2 - 19 x + 21,$ or
$0 < 2 x^2 -22x + 22$ or $x^2 - 11 x + 11 > 0.$ This one is true for $x \geq 10,$ either draw a picture or...
The other one is $x^3 + 5 x^2 - 19 x + 21 < x^3 + 6 x^2 + 12 x + 8,$ or $0 < x^2 + 31 x - 13.$ This one is true for integers $x \geq 1.$
So, check the original problem for $x = 0,1,2,3,4,5,6,7,8,9.$
| {
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Finding solution to a double integral Well, here it is
$$\int\int_{[0,1]^{2}}\sqrt{1+x^2+y^2}\mathrm dx\mathrm dy\tag*{}$$
Maybe there's a really nice way of doing it, hopefully someone knows. Good luck!
| We can do this integral in polar coordinates by recognizing a symmetry - divide the square in half by the line $y=x$. The integral on the top half will equal the integral on the bottom half, so we will do one of the integrals and multiply its value by 2.
$$\iint_{[0,1]^2}\sqrt{1+x^2+y^2}dA = \int_0^{\pi/4} \int_0^{\sec\theta}2r\sqrt{1+r^2}drd\theta = \frac{2}{3}\int_0^{\pi/4} (1+\sec^2\theta)^{3/2} - 1 d\theta$$
$$= \frac{2}{3}\int_0^{\pi/4} (1+\sec^2\theta)^{3/2} d\theta - \frac{\pi}{6}$$
Now focusing on the integral left over, do the following substitution $$1+\sec^2\theta = 2\cosh^2 t$$ $$\sec^2\theta \tan\theta d\theta = 2\cosh t \sinh t dt \implies d\theta = \frac{\sqrt{2}\cosh t}{\cosh 2t}dt$$ The integral becomes:
$$\frac{8}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{\cosh^4 t}{\cosh 2t}dt = \frac{8}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{\cosh^2 t + \cosh^2 t \sinh^2 t}{\cosh 2t}dt$$ $$ = \frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} \frac{2+ 2\cosh 2t + \sinh^2 2t}{\cosh 2t}dt
$$$$= \frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} 2\text{ sech } 2t + 2 + \tanh 2t \sinh 2tdt$$
Integrating tanh sinh by parts, we get
$$\frac{2}{3}\int_{0}^{\cosh^{-1}(\sqrt{3/2})} (\text{sech } 2t + 2)dt + \frac{1}{3}\sinh 2t = \frac{1}{3}\left[\tan^{-1}(\sinh 2t) + 4t + \sinh 2t \right]_{0}^{\cosh^{-1}(\sqrt{3/2})}$$ $$= \frac{1}{3}\left[\tan^{-1}(2u\sqrt{u^2-1}) + 4\cosh^{-1}(u) + 2u\sqrt{u^2-1}\right]_{1}^{\sqrt{3/2}} = \frac{1}{3}\left[\tan^{-1}(\sqrt{3}) + 4\cosh^{-1}(\sqrt{3/2}) + \sqrt{3}\right]$$
Simplifying and subtracting off the term from earlier, we get
$$\frac{4}{3}\cosh^{-1}\left(\sqrt{\frac{3}{2}}\right) + \frac{1}{\sqrt{3}} - \frac{\pi}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Regarding perfect squares Is there any positive integer $n > 2$ such that $(n - 1)(5n - 1)$ is a perfect square? It is observed that $(n - 1)(5n - 1)$ is of the form $4k$ or $4k+ 1$.
Affirmative answers were given by Pspl and Mindlack (by providing some examples).
Now my question is the following:
Is there any characterization of positive integer $n$ such that $(n - 1)(5n - 1)$ is a perfect square?
| If $m^2=(n-1)(5n-1)$, then $5m^2=5(n-1)(5n-1)=(5n-3)^2-4$.
Write this as $x^2-5y^2=4$, for $x=5n-3$ and $y=m$.
Write this as $N\left(\frac{x+y\sqrt5}{2}\right)=1$. Then $\frac{x+y\sqrt5}{2}=\left(\frac{1+\sqrt5}{2}\right)^{2k}$, since $\frac{1+\sqrt5}{2}$ is a fundamental unit of norm $-1$.
Thus, $\frac{x+y\sqrt5}{2}=\left(\frac{3+\sqrt5}{2}\right)^{k}$. We just need to find those $x$ such that $x \equiv -3 \equiv 2 \bmod 5$.
Write $\frac{x_k+y_k\sqrt5}{2}=\left(\frac{3+\sqrt5}{2}\right)^{k}$. Then $x_{k+1}=\frac{3x_k+5y_k}{2}$ and $y_{k+1}=\frac{x_k+3y_k}{2}$, with $x_0=2, y_0=0$.
Then, $x_{k+2}=\frac{7x_k+15y_k}{2}$ and $y_{k+2}=\frac{3x_k+7y_k}{2}$. This gives $x_{k+2} \equiv x_k \bmod 5$. Thus $x_{2k} \equiv 2 \bmod 5$, since $x_0 = 2$.
Thus, the solutions of $m^2=(n-1)(5n-1)$ are exactly $n=\frac{x_{2k}+3}{5}$ and $m=y_{2k}$.
Equivalently, $n_k=\frac{u_{k}+3}{5}$ and $m=v_{k}$, where $\frac{u_k+v_k\sqrt5}{2}=\left(\frac{1+\sqrt5}{2}\right)^{4k}=\left(\frac{7+3\sqrt5}{2}\right)^{k}$, which gives $u_{k+1}=\frac{7u_k+15v_k}{2}$ and $v_{k+1}=\frac{3u_k+7v_k}{2}$, with $u_0=2, v_0=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How To Use The Steps on Page 149 in Calculus Made Easy To Solve Chapter XIV Example 12 Please how do I take the derivative of
$ y = \left(\frac{1}{a^x}\right)^{ax} $
from Calculus Made Easy Chapter XIV Example 12 using the steps used
to solve
$y=a^x$
on page 149.
The steps are
\begin{align*}
y & = a^x\\
\log_ey & = x\log_e a\\
x & =\frac{\log_ey}{\log_ea} = \frac{1}{\log_ea}\, x \,\log_ey\\
\frac{dx}{dy} & =\frac{1}{\log_ea}\, x \, \frac{1}{a^2 \, x\, \log_ea}\\
\frac{dy}{dx} & = \frac{1}{\frac{dx}{dy}} = a^x \, x\, \log_ea
\end{align*}
These were my steps and were I got stuck
\begin{align*}
y & = \left(\frac{1}{a^x} \right)^{ax}\\
y & = a^{-ax^2}\\
\log_ey & = \log_ea^{-ax^2}\\
\log_ey & = -ax^2\log_ea\\
\frac{\log_ey}{\log_ea} & = -ax^2
\end{align*}
| A good place to start is to consider that
$$\bigg(\dfrac{1}{a^x} \bigg)^{ax} = (a^{-x}) ^{ax} = a^{-ax^2}.$$
Once you get there, apply logarithmic differentiation to get
\begin{array}[rcl]
$y & =&a^{-ax^2}\\
\ln y& =& -ax^2 \ln a\\
\dfrac{1}{y}\cdot\dfrac{dy}{dx}&=&-2a\ln a \cdot x\\
\dfrac{dy}{dx}&=& -2a\ln a \cdot a^{-ax^2}\cdot x
\end{array}
...assuming you are solving for the derivative of y with respect to x.
Edit: Following the steps in your book, we can do the following:
\begin{array}[rcl]
$y & = & a^{-ax^2} \\
\ln y & = & -ax^2 \ln a \\
-ax^2 & = & \dfrac{1}{\ln a}\cdot \ln y \\
-2ax \cdot \dfrac{dx}{dy} & = & \dfrac{1}{\ln a}\cdot \dfrac{1}{y} \\
\dfrac{dx}{dy} & = & -\dfrac{1}{2ax\cdot\ln a}\cdot \dfrac{1}{y} \\
\dfrac{dy}{dx} & = & -2ax\cdot \ln a\cdot y \\
\dfrac{dy}{dx} & = &-2a\ln a\cdot a^{-ax^2}\cdot x \\
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338038",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Determining convergence of $\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$ I've tried to solve this problem about convergence:
$\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$
it's supposed to be solved using ratio, root tests or by testing the limit of the sumand. Anyways, I've tried both 3 and I had no success:
I get to a point where I'm stucked on:
$$lím_{n\rightarrow \infty} \left[\frac{n+1}{n} \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3\right]$$
Any suggestions? what would you usually do with therms like $(n+2i)^3$? I tried assuming non imaginary n values (because of the sum) and converting to polar form: $\sqrt{n^2+4}e^{i tan^{-1}(2/n)}$.
Also I tried expanding:
$$\left|\left(\frac{n+2i}{n+1+2i}\right)^3\right| = \frac{|n+2i||n+2i||n+2i|}{|n+1+2i|^3} = \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3$$
Gelp
| Hint:
$$\left| \frac{3n i^n}{(n+2i)^3} \right| =\frac{3n}{\sqrt{(n^2+4)^3}}<\frac{3n}{\sqrt{(n^2)^3}} = \frac{3n}{n^3}=\frac{3}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341578",
"timestamp": "2023-03-29T00:00:00",
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A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8?
I have found by computation that the condition that the 8th power of a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 + 2 b c + d^2)^2 + ((a^2 + b c)^2 +
b c (a + d)^2)^2=1, \qquad b (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad c (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 +
4 a b c d + 4 b c d^2 + d^4)=0, \qquad b c (a + d)^2 (a^2 + 2 b c +
d^2)^2 + (b c (a + d)^2 + (b c + d^2)^2)^2=1
$$
and the condition for invertibility is $ad\neq bc$. If the 4th power is not the identity, then no power that is not a multiple of 8 is not the identity (because we could cancel out to either get that the first power is the identity or that the second power is the identity, both lead to contradiction). That is another cumbersome condition to write out.
I hope somebody can suggest a nicer way.
| We can use the same approach but reduce drastically the complexity of the system in the entries $a, b, c, d$ if we instead look for a square root of some matrix with order $4$.
The matrix $A = \pmatrix{0&-1\\1&0}$ satisfies $A^2 = -I$ and so has order $4$ over any field of characteristic not $2$. In particular, if we can find a matrix $B$ such that $B^2 = A$, then $B$ will have order $8$.
Writing $B = \pmatrix{a&b\\c&d}$, the condition is equivalent to the system \begin{align} a^2 + bc &= 0\\ b(a + d) &= -1\\ c(a + d) &= 1\\ d^2 + bc &= 0 \end{align} The second equation implies that $b = \pm 1$, and by negating the matrix (which preserves the property that $B^2 = I$) we may assume $b = 1$, and substituting gives $a + d = -1$ and $c = -1$. The first and fourth equations give that $a, d = \pm 1$, and then the third equation gives that $$a = d = 1 ,$$ yielding the solution $$\pmatrix{1&1\\-1&1} .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard inequality :$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$ I have a hard problem this is it :
Let $a,b,c>0$ such that $a^ab^bc^c=1$ then we have :
$$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$$
I try to use Jensen's ienquality applied to the function $f(x)=\frac{1}{x^2}$ but it doesn't works for all the values .
if we apply Am-Gm it's like Jensen's inequality so I forget this ways .
Maybe If we apply Karamata's inequality but I didn't found the right majorization .
I try also to use Muirhead inequality but without success .
So I'm a bit lost with that if you have a hint or a full answer I will be happy to read your work .
| First, I will ignore the constraint $a^ab^bc^c=1$ to sketch a possible solution for the aforementioned inequality, under an alternative constraint:
Set $x=\frac{1}{a^2+b^2}$ and $x=\frac{1}{b^2+c^2}$ and $z=\frac{1}{c^2+a^2}$.
From the aritmetic-geometric inequality
$$
x^2+y^2+z^2\geq 3(x^2y^2z^2)^{\frac{1}{3}},
$$
it remains to show that $$(x^2y^2z^2)^{\frac{1}{3}}\geq \frac{1}{4},$$
By noting that
$$ (a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2=\frac{1}{x^2y^2z^2},$$ the previous inequality to be shown becomes
$$
(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2 \leq 64.
$$
or equivalently
$$ (a^2+b^2)(b^2+c^2)(c^2+a^2)\leq 8,$$
if we take the square on both sides of the inequality.
Noteworty, if $(a,b),(b,c)$ and $(a,c)$ belong to the closure of the disc $\mathbb{T}$ of radius $\sqrt{2}$:
$$ \overline{\mathbb{T}}=\{(u,v)\in \mathbb{R}^2~:~\sqrt{u^2+v^2}\leq \sqrt{2}\},$$
the previous inequality is always satisfied.
| {
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How to convert the equation to a standard form (paraboloid)? I am given the equation: $9x^2 + 4y^2 + z = 3$
The standard equation of a paraboloid parallel to z-axis is: $$\frac{z-z_0}{c} = \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2}$$
I think the given equation should be a paraboloid due to the $z$ component, then I try to put it in a std. form.
$9x^2+ 4y^2 + z = 3$
$\frac{9x^2}{3} + \frac{4y^2}{3} + \frac{z}{3} = 1$
$3x^2 + \frac{4}{3}y^2 = 1 - \frac{z}{3}$
And then, I am confused since this is obviously not the std form. The one and also the negative value of z is not std.
I would appreciate any tips or someone pointing me the right way!
| Note $1 - \frac{z}{3} = \frac{3 - z}{3} = \frac{z - 3}{-3}$. Thus, $z_0 = 3$ and $c = -3$, along with $x_0 = y_0 = 0$, $a = \frac{1}{\sqrt{3}}$ and $b = \frac{\sqrt{3}}{2}$, allows your equation to be in standard form.
| {
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Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:
*
*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$
*
*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, contradiction since $15$ is not prime
*$y \equiv 1\mod 2\Rightarrow y+1\geq 2^2 \Leftrightarrow y+1=2*3, z+1=2^3$ or $y+1=2^2*3, z+1=2^2 \Leftrightarrow y=5, z=7$ or $y=11, z=3$
And since the equation is symmetric the solutions are the permutations of the latter ones. Similarly, through casework, we find the same solutions if $x \equiv 1\mod 2$ (if I haven't made a mistake). My question is if there exists a more simple way to solve this problem besides lots of casework (and if there exists another triplet that satisfies the given equation).
| $144 = 2^4*3^2$
If $x,y,z$ are prime then $a=x+1,b=y+1,c=z+1 \ge 3$ so we will only consider factors at least $3$. If, wolog, $x+1, y+1 \ge 3$ then $z+1 \le \frac {144}9 = 16$.
So we need to only consider triplets of factors between $3$ and $16$.
The factors of $144$ are of the form $2^4*3^2$ and are $1,2,4,8,16, 3,6,12,24,48,9,18, 72, 144$ and in order of size we are considering only the factors $3,4,6,8,9,12,16$. As we want these to be one more than primes, we can't have $16$ or $9$.
So the factors we may have are $3,4,6,8,12$. Now lets find the triplets by listing them in order. wolog $a \le b \le c=\frac {144}{ab}$.
We have $a,b,c =$
$3,3,16$ no good! $16$ not in our list
$3,4,12$
$3,6,8$
$3,8, *erk*$ we have $c = \frac {144}{ab} < b$ so that's it for $a = 3$. $a=4$ next.
$4*4*9$ no good. $9$ not on our list.
$4,6,6$ And that's it. We've "hit the middle" for $a=4$. Should we try $a=6$ next?
$6,6,*erk*$ we have $c = \frac {144}{ab} < b$ so we've hit the wall.
So ignoring permutations $\{x,y, z\} = \{2,3,11\},$ or $\{2,5,7\}, \{3,5,5\}$
| {
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Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$
Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
Now, AM-GM again
$$ a^{2}+b^{2}+c^{2} + 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} ... (1)$$
Then $a+b+c = 3 \ge 3 \sqrt[3]{abc} \implies 1 \ge \sqrt[3]{abc}$. Also
$$a^{2} + b^{2} + c^{2} \ge 3 \sqrt[3]{(abc)^{2}}$$
multiply by $1 \ge \sqrt[3]{abc}$ and will get
$$ a^{2} + b^{2} + c^{2} \ge 3 abc ... (2)$$
subtract $(1)$ with $(2)$ and get
$$ 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} - 3 abc$$
$$ 3 + 3 abc \ge \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} $$
$$ \frac{3abc}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} \ge 1 - \frac{3}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
How to continue..?
| Another way.
Your inequality is a sixth degree. We can reduce this degree by the Bacteria's method.
Indeed, by C-S, Murhead, Rearrangement and SOS(here it's also a Tangent Line method) we obtain: $$\sum_{cyc}\frac{a}{b^2+1}=\sum_{cyc}\frac{a^2(a+c)^2}{a(a+c)^2(b^2+1)}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(b^2+1)}=$$
$$=\tfrac{3(a+b+c)\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=\frac{3}{2}+\tfrac{3\left(2(a+b+c)\left(\sum\limits_{cyc}(a^2+ab)\right)^2-\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)\right)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2+4a^3c^2+8a^3bc-14a^2b^2c)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\tfrac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2-2a^3c^2+8(a^3bc-a^2b^2c)+6(a^3c^2-a^2b^2c))}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}\geq$$
$$\geq\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b+2a^4c-4a^3b^2-2a^3c^2)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a^5+3a^4b-4a^3b^2-2a^2b^3+2ab^4)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}a(a-b)(a^3+4a^2b-2b^3)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}\left(a(a-b)(a^3+4a^2b-2b^3)-\frac{3(a^5-b^5)}{5}\right)}{2\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}=$$
$$=\frac{3}{2}+\frac{3\sum\limits_{cyc}(a-b)^2(2a^3+19a^2b+16ab^2+3b^3)}{10\sum\limits_{cyc}a(a+c)^2(9b^2+(a+b+c)^2)}\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx$ The answer of this integral is
$$\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx$$
$$=\frac{2}{\sqrt{3}}arctan(\frac{2}{\sqrt{3}}(x+\frac{1}{x})+\sqrt{3})+C$$
But I can't figure out how could I solve it. I tried to use partial fraction, but the denominator
$$x^4+3x^3+5x^2+3x+1$$
can't be easily factored.
I also tried to use WolframAlpha to know how to solve it, but it can't give a useful answer for this integral:
| \begin{align}
&\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx\\
=&\int \frac{ 1 - \frac{1}{x^2}}{x^2+3x+5+\frac{3}{x}+\frac{1}{x^2}}dx
=\int \frac{ d\left( x+\frac{1}{x}\right) }{\left( x+\frac{1}{x}+\frac{3}{2}\right)^2 + \frac 34 } \\
=&\ \frac{2}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}\left(x+\frac{1}{x}\right)+\sqrt{3}\right]+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Use coordinate method to solve a pretty hard geometry problem Here is a hard geometry problem for my homework.
Let $D$ be a point inside $\Delta ABC$ such that $\angle BAD=\angle BCD$ and $\angle BDC=90^\circ$. If $AB=5,BC=6$ and $M$ is the midpoint of $AC$, find the length of $DM$.
(Taken from HK IMO Prelim 2012)
The teacher wants us to use the coordinate geometry to solve this problem. It does not appear too hard, but there is a big problem.
I set the coordinates of $A,B,C,D,M$ as in the diagram. Then, we can get the following statement:
$$\begin{cases} b^2+c^2=36 \\ x^2+\left(b+y\right)^2=25 \\ \sin\angle BAD=\sin\angle BCD \end{cases}$$
We need to find the value of $\sin\angle BAD$ and $\sin\angle BCD$
$$\sin\angle BCD=\dfrac{BD}{BC}=\dfrac{b}{6}\\ \text{The area of }\Delta ABD=\dfrac{1}{2}\left(AB\right)\left(AD\right)\sin\angle BAD \\ \dfrac{bx}{2}=\dfrac{5\sqrt{x^2+y^2}\sin\angle BAD}{2} \\ \sin\angle BAD=\dfrac{bx}{5\sqrt{x^2+y^2}}\\ \dfrac{b}{6}=\dfrac{bx}{5\sqrt{x^2+y^2}} \\ \sqrt{x^2+y^2}=\dfrac{6}{5}x \\ x^2+y^2=\dfrac{36}{25}x^2 \\ y^2=\dfrac{11}{25}x^2 \\ y=\dfrac{\sqrt{11}}{5}x$$
Then, when I proceed further, the expression gets very complicated. I felt puzzled, so I stopped. I found a geometric solution, which gives the answer of $\dfrac{\sqrt{11}}{2}$.
I hope you guys can help me solve this question using coordinate method. Thank you!
| Let $\angle BDA = \alpha$ and apply the sine rule to the triangle $\triangle ABD$,
$$\frac{\sin \alpha}{\sin\angle BAD}=\frac{5}{6\sin\angle BCD}$$
Given that $\beta = \angle BAD = \angle BCD$, we immediately get
$$\sin \alpha = \frac 56$$
Now, express the coordinates of $A(-x,-y)$ as follows,
$$x=AB\sin \angle ABD = 5\sin (\alpha + \beta)$$
$$y=AB\cos \angle ABD - BD = -5\cos (\alpha + \beta) - 6 \sin\beta$$
Then, the coordinates of the point $M(x_m,y_m)$ are
$$x_m = \frac 12 (CD - x)=\frac 12 [6\cos\beta - 5\sin (\alpha + \beta)]$$
$$y_m = -\frac 12 y = \frac 12 [5\cos (\alpha + \beta) + 6 \sin\beta]$$
As a result,
$$DM^2 = x_m^2 + y_m^2 = \frac 14 (36+25-60\sin\alpha)=\frac {11}{4}$$
Note that the unknown angle $\beta$ cancels out in the process. Finally, the length of $DM$ is
$$DM = \frac{\sqrt{11}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find all $n$ for which $3n^2+3n+1$ is a perfect square. Find all natural numbers $n$ for which $3n^2+3n+1$ is a perfect square.
I used discriminant method but failed. Then I found upper and lower bounds of this expression:
Lower:$(n+1)^2$
Upper:$(2n)^2$
But, this too does not seem to be useful. Please help me.
| Cool. I'll just fully answer the question.
Starting with $ \ m^2=3n^2+3n+1\iff (2m)^2-3(2n+1)^2=1 \quad $. Lets call $ \ p=2m, \ \ $ $q=2n+1 \ \ $. So it's now $$p^2-3q^2=1$$
By a quick inspection the smallest solution is: $(p_0,q_0)=(2,1) \quad $. It can be used to find all others. Use it to write the number $1$ in a funny/arbitrary way, and multiplying our equation with this special $1$ will lead to the rest of the solutions. The back-substitution, $(p,q,p_0,q_0) \to (2m,2n+1,2,1)$, will wait until the end to elucidate how this algorithm can be applied to other situations involving pell-type equations:
$$
\begin{align}
p_0^2-3q_0^2&=1\\
(p_0-q_0 \sqrt 3)(p_0+q_0 \sqrt 3)&=1\\
(p_0-q_0 \sqrt 3)^2(p_0+q_0 \sqrt 3)^2&=1^2=1\\
\bigg[(p_0^2+3q_0^2)-(2p_0q_0) \sqrt 3\bigg]\bigg[(p_0^2+3q_0^2)+(2p_0q_0)\sqrt 3\bigg]&=1\\
\text{now multiply $p^2-3q^2=1$ by this "$1$" in the following way (factor it first):} & \\
(p-q \sqrt 3) \cdot \bigg[(p_0^2+3q_0^2)-(2p_0q_0) \sqrt 3\bigg]& \cdot \\
(p+q \sqrt 3) \cdot \bigg[(p_0^2+3q_0^2)+(2p_0q_0) \sqrt 3\bigg]&=1\\
&\vdots \\
\underbrace{\bigg[(p_0^2+3q_0^2)p+(2\cdot 3p_0q_0)q\bigg]^2-3\bigg[(2p_0q_0)p +(p_0^2+3q_0^2)q\bigg]^2=1}_{=p^2-3q^2=1}\\
\end{align}
$$
Interpretable as
$$(p_k,q_k) \xrightarrow{k \to k+1} \bigg((p_0^2+3q_0^2)p+(2\cdot 3p_0q_0)q \ \ , \ (2p_0q_0)p +(p_0^2+3q_0^2)q\bigg)$$
Evaluating
$$
\begin{cases}
p_k&=2m_k\\
p_0=2 \implies m_0&=1\\
q_k&=2n_k+1 \\
q_0=1 \implies n_0&=0\\
\end{cases}
$$
Thus
$$(2m_k,2n_k+1) \xrightarrow{k \to k+1} \bigg( (7)(2m_k)+(12)(2n_k+1) \ \ , \ (4)(2m_k) +(7)(2n_k+1)\bigg)$$
Or, finally,
$$(m_k,n_k) \xrightarrow{k \to k+1} \bigg( 7m_k+12n_k+6 \ \ , \ 4m_k +7n_k+3\bigg)$$
Which is exactly @S.Dolan's ordered pair.
Also expressable as
$$
\begin{pmatrix}
m_k \\
n_k
\end{pmatrix}
\xrightarrow{T}
\begin{pmatrix}
7 & 12 \\
4 & 7
\end{pmatrix}
\cdot
\begin{pmatrix}
m_k \\
n_k
\end{pmatrix}
+
\begin{pmatrix}
6 \\
3
\end{pmatrix}
$$
if you're into that sort of thing...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the smallest positive integer x where 149|(x^2-69^3) $$
\text{Find the smallest positive integer x where 149|}\left( \text{x}^2-69^3 \right)
$$
I almost do not spot any clue about this question.
It is from PUMaC-CHINA, 2019.8.17
| This follows the comment by J. W. Tanner and the answer by sirous.
All congruences are mod $149$.
We have $x^2-69^3 \equiv x^2+36 =x^2+6^2 \equiv (x-6j)(x+6j)$, if $j^2\equiv -1$.
Write $149 = 7^2 + 10^2$. Then $10j \equiv 7$ and so $j\equiv 105$.
Thus $x^2-69^3 \equiv (x-34)(x+34)$ and so $34$ is the smallest positive solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$ When $a=1,2,3,...,2010,2011$ the roots of the equation $x^2 -2x-a^2-a=0$ are $(m_1,n_1 ), (m_2,n_2 ), (m_3,n_ 3),..., (m_{2010},n_{2010} ), (m_{2011},n_{2011 }) $ respectively. Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2010}} +\frac{1}{n_{2010}} +\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$
**My attempt **$$$$
$ \frac{1}{m_a} +\frac{1}{n_a}$can be manipulated to give
$$\frac{m_a+n_a}{m_an_a}$$then w.l.o.g $m_a\ge n_a$
And since$$m_a=2\frac{2+2\sqrt{a^2+a+1}}{2}$$
And $$n_a= 2\frac{2-2\sqrt{a^2+a+1} }{2}$$
We get
$$\frac{m_a+n_a}{m_an_a}=\frac{-2}{a(a+1)}$$
But I dont know what to do next,
Suggestions and solutions would be appreciated
Taken from the 2011 IWYMIC
| Hint:
$$
\frac{-2}{a(a+1)} = \frac2{a+1} - \frac 2a
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this inequality with Big O term? Let $s= s_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))$
where $\zeta_0 = \left(\frac{3\pi}{4}\right)^{2/3}$ and $s_0 = b^{-2/3}\zeta_0.$ Note that here $b$ is a parameter.
We define $\omega(s) = \exp\left(\frac{2}{3}s_{+}^2\right)$ where $s_{+} = \max(0,s).$ I want to show that for some constant $C$ we have that,
$$\omega(s) \leq C\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right)\leq C\exp\left(-\frac{1}{2\sqrt{b}}\right)\Sigma_{b}^{-1},$$
where $\Sigma_{b}^{-1}=\exp\left(\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\right).$
For the second inequality does not make sense to me as $\exp\left(-\frac{1}{2\sqrt{b}}\right)\leq 1$, but I might be wrong.
For the first inequality,
$$\frac{2}{3}s_{+}^{2}\leq \frac{2}{3}|s|^2\leq \frac{4}{3}|s_0|^2+\frac{8}{3}(\zeta_0^{-1}b^{-1/3} + C\zeta_0^{-1}b^{2/3})$$
$$=\frac{4}{3}\left(\frac{3\pi}{4b}\right)^{4/3}+\frac{8}{3}\left(\left(\frac{4}{3\pi\sqrt{b}}\right)^{2/3} + C\left(\frac{4b}{3\pi}\right)^{2/3}\right)$$
where I used the inequality $|a+b+c|^2\leq 2|a|^2+4(|b|^2+|c|^2).$ I cannot see how I reduce this expression to derive the first inequality. Any hints would be much appreciated.
| The second inequality never holds.
The first inequality is equivalent to $\frac 23s_+^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$ for some constant $D$. Since $\frac{\pi}{2b}-\frac{1}{\sqrt{b}}\ge -\frac 1{4\pi}$ for each $b\ge 0$, if $D\ge -\frac 1{4\pi}$ then the inequality holds for $s_+=0$. So it remains to consider an inequality
$\frac 23s^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$
$\frac 23\left(b^{-2/3}\zeta_0-\zeta_0^{-1/2}b^{-1/6}(1+\mathcal{O}(\sqrt{b}))\right)^2\le \frac{\pi}{2b}-\frac{1}{\sqrt{b}}+D$
For big $b$ this inequality can fail. For instance, when $\mathcal{O}(\sqrt{b})$ term equals $-1-\sqrt{b}$, then when $b$ tends to infinity, the left hand side of the inequality tends to infinity, whereas the right hand side is bounded. For small $b$ the inequality fails too. Indeed, when $b$ tends to zero, the left hand side of the inequality grows as $b^{-4/3}$, whereas the right hand grows as $b^{-1}$.
| {
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"url": "https://math.stackexchange.com/questions/3368169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts ,
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But it got me no where.
Any hints.
| *
*Divide the numerator by $x^2$ and you have $1+\frac{1}{x^2}$
*Divide the denominator by $x^2$ and you have $x^2+\frac{1}{x^2} =\bigl(x-\frac{1}{x}\bigr)^{2}+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Compute in a closed form the following sum : $\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$ Today Im going to find the closed form of :
$\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$
My attempt :
We know that : $\Gamma(z)=\int_0^{+\infty}t^{n-1}e^{-t}dt$
So : $\Gamma^{4}(n+\frac{3}{4})=\int_{[0,+\infty[}(xyzt)^{n-\frac{1}{4}}e^{-x-y-z-t}dxdydzdt$
But the problems in : $\sum_{n=1}^{+\infty}\frac{x^{n}}{(4n+3)^{2}(n!)^{4}}$ I think related with hypergeomtric function
if there some trick to compute this original sum drop here , thanks!
| As one could expect, the result must involve hypergeometric function.
A CAS gave for the infinite summation
$$\frac{\Gamma \left(\frac{3}{4}\right)^4 \left(3136 \left(\,
_5F_4\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4};1,1,1,\frac{7}{4};1\right)-1\right)-243 \,
_6F_5\left(\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4};2,2,2,\frac{11}{4},\frac{11}{4};1\right)\right)}{28224}$$ and its numerical representation is $0.0211403036686719835443455214070$.
Inverse symbolic calculators do not identify this number but it seems to be very close to the positive root of
$$ 43 x^2+141 x-3=0 \implies x=\frac{1}{86} \left(\sqrt{20397}-141\right)\approx 0.02114030330$$
Edit
In order to keep it, I shall mention that $\color{red}{\text{David H}}$ greatly simplified the expression to
$$\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}=\frac{4 \pi ^4}{9 \Gamma \left(\frac{1}{4}\right)^4}\left(\,
_6F_5\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{
4};1,1,1,\frac{7}{4},\frac{7}{4};1\right)-1 \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest possible value of $x+y$?
The book says "the greatest value of $4$ is divided by $6$, which produces a remainder of $4$. The greatest value of $y$ is when $2$ is divided by, which produces a remainder of $2$. Therefore, the greatest value of $x+y$ is $6$."
I think what's throwing me off is the phrase "multiple of $4$" bc it makes me think that any multiple of $4$ can be divisible by $6$ (i.e. $24/6 = 4$). The books answer doesn't use multiples, just the $4$ and $2$, respectively. I don't understand how this works. Can someone please clarify?
| In order to get to the book's conclusion, you can test out a few numbers:
$4$ leaves remainder $4$ when divided by $6$.
$8$ leaves remainder $2$ when divided by $6$.
$12$ leaves remainder $0$ when divided by $6$.
$16$ leaves remainder $4$ when divided by $6$.
Then you can observe the possible remainders are $0, 2$ and $4$. This is because $16$ is $12$ more than $4$, a multiple of $6$, so the remainders will follow the same pattern after $16$.
There is another way to think about this pattern. When you add $4$ to a number, the remainder will be the same as if you subtract $2$ from the number, which explains the 'decreasing by $2$ pattern'.
Now try this with small multiples of $2$ and find the remainder when they are divided by $3$. This gives the book's answer of $4 + 2 = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the solution to the following differential equation: $ \frac{dy}{dx} = \frac{x - y}{xy} $ The instructor in our Differential Equations class gave us the following to solve:
$$ \frac{dy}{dx} = \frac{x - y}{xy} $$
It was an item under separable differential equations. I have gotten as far as $ \frac{dy}{dx} = \frac{1}{y} - \frac{1}{x} $ which to me doesn't really seem much. I don't even know if it really is a separable equation.
I tried treating it as a homogeneous equation, multiplying both sides with $y$ to get (Do note that I just did the following for what it's worth)...
$$ y\frac{dy}{dx} = 1 - \frac{y}{x} $$
$$ vx (v + x \frac{dv}{dx}) = 1 - v $$
$$ v^2x + vx^2 \frac{dv}{dx} = 1 - v $$
$$ vx^2 \frac{dv}{dx} = 1 - v - v^2x$$
I am unsure how to proceed at this point.
What should I first do to solve the given differential equation?
|
We write the differential equation as
\begin{align*}
xyy^\prime=x-y\tag{1}
\end{align*}
and follow the receipt I.237 in the german book Differentialgleichungen, Lösungsmethoden und Lösungen I by E. Kamke.
We consider $y=y(x)$ as the independent variable and use the substitution
\begin{align*}
v=v(y)=\frac{1}{y-x(y)}=\left(y-x(y)\right)^{-1}\tag{2}
\end{align*}
We obtain from (2)
\begin{align*}
v&=\frac{1}{y-x}\qquad\to\qquad x=y-\frac{1}{v}\\
v^{\prime}&=(-1)(y-x)^{-2}\left(1-x^{\prime}\right)=\left(\frac{1}{y^{\prime}}-1\right)v^2
\end{align*}
From (1) we get by taking $v$:
\begin{align*}
\frac{1}{y^{\prime}}=\frac{xy}{x-y}=\left(y-\frac{1}{v}\right)y(-v)=y-y^2v\tag{3}
\end{align*}
Putting (2) and (3) together we get
\begin{align*}
v^{\prime}=\left(y-y^2v-1\right)v^2
\end{align*}
respectively
\begin{align*}
\color{blue}{v^{\prime}+y^2v^3-(y-1)v^2=0}\tag{4}
\end{align*}
and observe (4) is an instance of an Abel equation of the first kind.
| {
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} |
Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cdot 10^n}-1+3^{8\cdot 10^n}-1+\cdots +3^{10^n}-1+10\right),\end{align*}$$
but I don't see how this proves the equality in the question. I got only
$$3^{10^{n}}\equiv 1\pmod{3^{10^{n-1}}-1}.$$
Perhaps someone here can explain.
| Use induction
Basis
$$3^{100}\equiv (3^{10})^{10}$$
$$\equiv 59049^{10}$$
$$\equiv 49^{10}$$
$$\equiv 2401^5$$
$$\equiv 1\pmod {100}$$
Induction hypothesis
$$\frac{3^{10^{n}}-1}{10^n}\in\mathbb Z$$
Inductive step
$$\frac{3^{10^{n+1}}-1}{10^{n+1}}$$
$$=\frac{3^{10^{n}}-1}{10^n}\times \frac{1+\sum_{k=1}^9 3^{10^nk}}{10}$$
But of the terms multiplied are integers, as the first one directly follows from induction hypothesis and second one by $$\forall 1\leq k\leq 9, 10|10^n|(3^{10^n}-1)|(3^{10^nk}-1)$$
$$\Longrightarrow 3^{10^nk}\equiv 1\pmod {10}\forall 1\leq k\leq 9$$
$$\Longrightarrow 1+\sum_{k=1}^9 3^{10^nk}\equiv 0\pmod {10}$$
Hence proved
| {
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"url": "https://math.stackexchange.com/questions/3373359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
how to solve $\operatorname{rem}(6^{15},17)$ without using a calculator. I am trying to solve $\operatorname{rem}(6^{15}, 17)$.
I know that we have to use congruences but don't know how to go on.
$6 ≅ 6 \mod 17$??
Can anyone please point me in the right direction?
Do I have to use CRT in here?
| A low tech solution:
$6^2 = 36 \equiv 2 \pmod{17}$ ($34$ is a multiple of $17$).
So $6^4 = (6^2)^2 \equiv 2^2 = 4 \pmod{17}$
Hence: $6^8 = (6^4)^2 \equiv 4^2 = 16 \equiv -1 \pmod{17}$
Also: $6^{15}=6^1 \cdot 6^2 \cdot 6^4 \cdot 6^8$ ($15$ is $1111$ in binary), so modulo $17$ this becomes
$6 \times 2 \times 4 \times -1 = -48 \equiv -48 + 3\times 17 = 3$
So $3$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan x - \sin x} = 1$$
But the correct answer is $2$. Where am I wrong$?$
| @Surb identified your error with the choice$$f_1=\tan(\tan x),\,g_1=\tan x,\,f_2=\sin(\sin x),\,g_2=\sin x.$$One method that would work is to use$$\tan x=x+\frac13 x^3+o(x^3),\,\sin x=x-\frac16 x^3+o(x^3)$$together with$$x+cx^3+c(x+cx^3)^3=x+2cx^3+o(x^3),$$viz.$$\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\tan(x+\frac13 x^3+o(x^3))-\sin(x-\frac16 x^3+o(x^3))}{\frac12x^3+o(x^3)}\\=\frac{(x+\frac23 x^3)-(x-\frac13 x^3)+o(x^3)}{\frac12x^3+o(x^3)}=2+o(1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Question aboutPartial Fractions for example:
$$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$$
first method is:
$${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$$
but it is hard and take much time to find A,B,C
second method(Substituting the roots, or "zeros") is:
\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}
it is better and easier method for Partial Fractions. My question is why this method works? can you prove this method?
| It works from there
$$f(x)={x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$$
indeed
$$f(0)=4=A(2)(-2)\implies A=-1$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The slope of the hyperbola $b^2 x^2 - a^2y^2 = a^2 b^2$ at the upper end of its right-hand latus rectum is $4/3$. What is the eccentricity? How to approach this type of problem?
The slope of the curve $b^2 x^2 - a^2y^2 = a^2 b^2$ at the upper end of its latus rectum to the right of the origin is $4/3$. What is the eccentricity of the curve?
I get the equation of derivative of $y'= \dfrac{xb^2}{ya^2}$. I still don't have a point of $x$ and $y$ for latus rectum to insert in equation $e=\sqrt {a^2 + b^2}/a$
| As mentioned by @Blue, the latus rectum is a vertical line through the focus $(c,0) \equiv (ae,0)$.
The abcissa of the latus rectum is $ae = \sqrt{a^2+b^2}.$ So, to find the $y$ coordinates of its terminii, we have $$b^2a^2e^2 - a^2y^2 = a^2b^2$$ $$\implies y^2 = b^2(e^2-1)$$
$$\implies y' = \frac 43 = {aeb^2\over b\sqrt{e^2-1} a^2} =\frac ba\cdot{e\over\sqrt{e^2-1}}$$
Note that $$e^2 = 1+{b^2\over a^2}$$
$$\implies \sqrt{e^2-1} = \frac ba$$
$$\implies \boxed{e = \frac 43}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$ Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$
I tried using Induction:
For the Base Step $n=2$ we have:
$$x=\sqrt{1+\frac{1}{\sqrt{2}}}+\sqrt{1-\frac{1}{\sqrt{2}}}$$
Then we get:
$$x^2=2+\sqrt{2}\lt 4$$
So $x \lt 2$
Now Let $P(n)$ is True, We shall need to prove $P(n+1)$ is also True
We have $P(n+1)$ as:
$$\left ( 1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}+\left ( 1-\frac{(n+1)^{\frac{1}{n+1}}}{n+1} \right )^\frac{1}{n+1}$$
Now i tried to use the fact that:
$$f(x)=x^{\frac{1}{x}}$$ is a Monotone Decreasing $\forall x \ge e$
Hence $\forall n \ge 3$ we have:
$$(n+1)^{\frac{1}{n+1}} \lt n^{\frac{1}{n}} \tag{2}$$ and also
$$\frac{1}{n+1} \lt \frac{1}{n} \tag{3}$$
Multiplying $(2),(3)$ We get:
$$1+\frac{(n+1)^{\frac{1}{n+1}}}{n+1}\lt 1+\frac{n^{\frac{1}{n}}}{n}$$
Can we proceed from here?
| Note that by Bernoulli inequality in the form
$$(1+x)^a<1+ax, \quad 0<a<1$$
which can be easily proved by induction, we obtain
$$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<1+\frac{n^{\frac{1}{n}}}{n^2}+1-\frac{n^{\frac{1}{n}}}{n^2} =2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$ Determine $x+y$ where $x$ and $y$ are real numbers such that $(2x+1)^2+y^2+(y-2x)^2=\frac{1}{3}$
I used the quadratic equation to get $$x=\frac{y-1\pm\sqrt{-2y-3y^2-\frac{5}{3}}}{4}$$
But I don’t see how that helps, hints and solutions would be appreciated
Taken from the 2006 IWYMIC
| Hint: Show that $\frac{1}{3}$ is the unique minimum of the function $f(x,y)=(2x+1)^2+y^2+(y-2x)^2$. Then find the $(x,y)$ where this minimum occurs.
Your expression under the radical is also incorrect. It should be $-3y^2-2y-\frac{1}{3}$, or $-\frac{1}{3}(9y^2+6y+1) = -\frac{1}{3}(3y+1)^2$. Thus there is exactly one value of $y$ for which the expression under the radical is nonnegative.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time I've tried it so I'm stuck at a point...
So $$x_1+x_2+x_3 = 10$$
Subject to the condition that :
$$1\leq x_1 \leq8$$
$$1\leq x_2 \leq8$$
$$1\leq x_3 \leq8$$
As each beggar can get at maximum 8 blankets and at minimum, 1.
So the number of ways must correspond to the coefficient of $x^{10}$ in:
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)^3$
= coeff of $x^{10}$ in $x^3(1-x^7)^3(1-x)^{-3}$
= coeff of $x^{10}$ in $x^3(1-x^{21}-3x^7(1-x^7))(1-x)^{-3}$
= coeff of $x^{10}$ in $(x^3-3x^{10})(1+\binom{3}{1}x + \binom{4}{2}x^2+...+ \binom{12}{10}x^{10})$
= $\binom{9}{7} - 3 = 33$
Is this right? From here I get the answer as $\binom{9}{7} - 3 = 33$ but the answer is stated as $36$. I don't understand where I'm making a mistake
| Try stars and bars. You have $10$ stars for the $10$ blankets:
$**********$
Now you can use $2$ bars to split this into $3$ sections. For example
$**|*******|*$
would mean beggar $1$ gets $2$ blankets, beggar $2$ gets $7$ blankets, and beggar $3$ gets $1$ blanket
Since each beggar should get at least $1$ blanket, we can't put the bars on the outside of the stars, and you also can't have the two bars between the same two stars. In other words, you need to choose $2$ out of the $9$ in-between locations, giving $9 \choose 2$ possible ways to do this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \left(9^x+4\cdot 3^x+1\right)-(4\cdot 3^x+1)>0$$
$$a\cdot \left(9^x+4\cdot 3^x+1\right)>4\cdot 3^x+1$$
$$a>\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$$
Now if a is greater than the maximum value of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then inequality will be true for all x.
So if we can find the range of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then we can say a should be greater than the maximum value in the range.
Let's assume y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$ and substitute $3^x$ with $t$.
$$y=\frac{4t+1}{t^2+4t+1}$$
$$yt^2+4ty+y=4t+1$$
$$yt^2+4t(y-1)+y-1=0$$
We want to have real values of t satisfying the equation, so $D>=0$
$$16(y-1)^2-4*y*\left(y-1\right)>=0$$
$$4(y-1)(4y-4-y)>=0$$
$$4(y-1)(3y-4)>=0$$
$$y\in \left(-\infty,1 \right] \cup \left[\frac{4}{3},\infty\right)$$
So I am getting maximum value tending to $\infty$ for y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$
I am not able to understand where am I making mistake.
Official answer is $a\in \left[1,\infty\right) $
| There is an easier way :-):
Let $t = 3^x$, noting that we need $t>0$. The equation becomes
$at^2+4(a-1)t +(a-1) > 0$ for all $t>0$.
First note that if the above is true for all $t >0$ then we must have (taking
the limit as $t \downarrow 0$) that $(a-1) \ge 0$ (note $\ge$ not $>$).
In
particular, $a \ge 1$ must hold.
The $\min$ of the left hand side (over all $t$) can be found using:
$2at + 4(a-1) = 0$, or $t^* = -2 {a-1\over a}$.
Since $t^* \le 0$, we see (because it is a quadratic) that the
left hand side is an increasing function of $t$ for $t \ge 0$ and the
value at $t=0$ is $a-1$.
In particular, if $t >0$ we have $at^2+4(a-1)t +(a-1) > a-1\ge 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$
for substitution $y = (x-1)^{2}$ we have:
$y^{2} - 4y + 7 = 0 $
$y_{0} = 2+i\sqrt{3}$
$y_1 = 2 - i\sqrt{3}$
and we have
$y_{0}^{1/2} + 1 = x_{0} $, $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$, $-y_{1}^{1/2} +1 = x_{3} $
I am not sure what is good results.
Please check my solution.
EDIT:
The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $
EDIT2:
I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2.
But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2.
| Solving algebraically is perhaps not too messy if one is happy to have roots in terms of the solution to a cubic:-
Using the substitution $t=x-1$ we have $$ t^{4}- 4t^{2}-4t+3=0$$ and therefore $$ (t^{2}- 2)^2=4t+1.$$
For any $z$, $$ (t^{2}- 2+2z^2)^2=4z^2t^2+4t+4z^4-8z^2+1.$$
Let $z$ be a solution (found by Cardan's method) of the cubic in $z^2$
$$4z^6-8z^4+z^2-1=0$$
Then $$ (t^{2}- 2+2z^2)^2=(2zt+\frac{1}{z})^2.$$
This is a quadratic in $t$ giving, for example, $t=z+\sqrt {2-z^2+\frac{1}{z}}$ and $t=z-\sqrt {2-z^2+\frac{1}{z}}$.
(One has to be careful choosing the correct signs for whatever root of the cubic is chosen- the signs I have used give real roots for the real root of the cubic.)
| {
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Solve with eigenfunction expansion $\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} + e^{-t} + e^{-2t} \cos \frac{3\pi x}{L}$
Q: Solve the following non-homogeneous problem:
\begin{align*}
\frac{\partial u}{\partial t} &= k \frac{\partial^2 u}{\partial x^2} + e^{-t} + e^{-2t} \cos \frac{3\pi x}{L} \\
\end{align*}
With the following boundary and initial value conditions:
\begin{align*}
\frac{\partial u}{\partial x}(0,t) &= 0 \\
\frac{\partial u}{\partial x}(L,t) &= 0 \\
u(x,0) &= f(x) \\
\end{align*}
Assume that $2 \neq k(3\pi/L)^2$. Use the method of eigenfunction expansion. Look for the solution as a Fourier cosine series. Assume appropriate continuity.
TEXTBOOK GIVEN ANSWER:
\begin{align*}
u &= \sum\limits_{n=0}^\infty A_n(t) \cos \frac{n \pi x}{L} \\
A_{n,n \neq 0, n\neq 3}(t) &= A_n(0) e^{-k \left( \frac{n \pi}{L} \right)^2 t} \\
A_0(t) &= A_0(0) + 1 - e^{-t} \\
A_3(t) &= A_3(0) e^{-k \left( \frac{3 \pi}{L} \right)^2 t} + \frac{e^{2t} - e^{-k \left( \frac{3 \pi}{L} \right)^2 t}}{k \left( \frac{3 \pi}{L} \right)^2 - 2} \\
A_0(0) &= \frac{1}{L} \int_0^L f(x) \, dx \\
A_{n \ge 1}(0) &= \frac{2}{L} \int_0^L f(x) \cos \frac{n \pi x}{L} \, dx \\
\end{align*}
My work: I fully understand how to get most of the answer, but I am stuck on the rest. Specifically, I don't understand the $n=0, n=3$ cases.
First, the part I fully understand:
We look for the solution as a Fourier cosine series. The right hand side is an even extended and periodized version of $u(x,t)$ along the $x$ variable.
\begin{align*}
u(x,t) &\sim A_0 + \sum\limits_{n=1}^\infty A_n(t) \cos \frac{n \pi x}{L} \\
\end{align*}
The problem says we can assume appropriate continuity which means that we can assume the original, non-periodized $u(x,t)$ is continuous. We can apply term by term partial differentiation to both $x$ and $t$. For $t$, we are not periodized by $t$, the given function is continuous, so that's all we need. For $x$, we are dealing with the even extended, periodized version of a continuous function, which must be continuous, so we can use term by term partial differentiation there as well.
\begin{align*}
\frac{\partial u}{\partial x} &\sim - \sum\limits_{n=1}^\infty \frac{n \pi}{L} A_n(t) \sin \frac{n \pi x}{L} \\
\frac{\partial u}{\partial t} &\sim \sum\limits_{n=1}^\infty A'_n(t) \cos \frac{n \pi x}{L} \\
\end{align*}
We also assume that $\frac{\partial u}{\partial x}$ is continuous. Since we are given that $\frac{\partial u}{\partial x}(0,t) = 0 = \frac{\partial u}{\partial x}(L,t)$, then we can see that the periodized version along the $x$ variable must also be fully continuous so that we can again use term by term differentiation to get:
\begin{align*}
\frac{\partial^2 u}{\partial x^2} &\sim - \sum\limits_{n=1}^\infty \left(\frac{n \pi}{L}\right)^2 A_n(t) \cos \frac{n \pi x}{L} \\
\end{align*}
Now plugging that back in to the PDE:
\begin{align*}
\frac{\partial u}{\partial t} &= k \frac{\partial^2 u}{\partial x^2} + e^{-t} + e^{-2t} \cos \frac{3\pi x}{L} \\
\sum\limits_{n=1}^\infty A'_n(t) \cos \frac{n \pi x}{L} &= k \left( - \sum\limits_{n=1}^\infty \left(\frac{n \pi}{L}\right)^2 A_n(t) \cos \frac{n \pi x}{L} \right) + e^{-t} + e^{-2t} \cos \frac{3\pi x}{L} \\
\end{align*}
That will be satisfied if the summation terms on the left/right side are all equal and the term on the right hand side outside the summation is zero:
\begin{align*}
A'_n(t) &= -k \left(\frac{n \pi}{L}\right)^2 A_n(t) \\
A_n(t) &= A_n(0) e^{-k \left( \frac{n \pi}{L} \right)^2 t} \\
\end{align*}
Solving for the coefficients with the initial condition:
\begin{align*}
u(x,t) &= \sum\limits_{n=0}^\infty A_n(t) \cos \frac{n \pi x}{L} \\
u(x,0) = f(x) &= \sum\limits_{n=0}^\infty A_n(0) \cos \frac{n \pi x}{L} \\
A_0(0) &= \frac{1}{L} \int_0^L f(x) \, dx \\
A_{n \ge 1}(0) &= \frac{2}{L} \int_0^L f(x) \cos \frac{n \pi x}{L} \, dx \\
\end{align*}
ok, so far so good. All of that lines up perfectly with the given answer. But I am unsure about how to get the rest.
From earlier, I said if the terms on the right hand side out side the summation came to zero, it would satisfy the PDE. That would mean that:
\begin{align*}
e^{-t} + e^{-2t} \cos \frac{3\pi x}{L} &= 0 \\
e^{t} &= -\cos \frac{3\pi x}{L} \\
\end{align*}
I am not sure how to use this. I also don't see what to do with the problem given that $2 \neq k(3\pi/L)^2$
| Letting
$$u \sim \sum_{n \ge 0} A_{n}(t) \cos \left( \frac{n \pi x}{L} \right)$$
then substituting into the PDE (and noting that $1 \equiv \cos(0 \pi x/L)$) yields
$$\sum_{n \ge 0} A_{n}'(t) \cos \left( \frac{n \pi x}{L} \right) = -k \sum_{n \ge 0} \left( \frac{n \pi}{L} \right)^{2} A_{n}(t) \cos \left( \frac{n \pi x}{L} \right) + e^{-t} \color{red}{\cos \left( \frac{0 \pi x}{L} \right)} + e^{-2t}\cos \left( \frac{3 \pi x}{L} \right)$$
Equating terms in $\cos$ for $n = 0, 1, \dots$ and letting $C_{n}$ be the constants of integration, we have
\begin{align}
A_{0}'(t) = -k(0)+e^{-t} &\implies A_{0}(t) = -e^{-t} + C_{0} \\
&\implies A_{0}(0) = -1+C_{0} \\
&\implies A_{0}(t) = A_{0}(0) + 1 - e^{-t} \\\\
A_{1}'(t) = -k \left( \frac{\pi}{L} \right)^{2} A_{1}(t) &\implies A_{1}(t) = A_{1}(0) e^{-k \left( \frac{\pi}{L} \right)^{2} t} \\\\
A_{2}'(t) = -k \left( \frac{2 \pi}{L} \right)^{2} A_{2}(t) &\implies A_{2}(t) = A_{2}(0) e^{-k \left( \frac{2 \pi}{L} \right)^{2} t} \\\\
A_{3}'(t) = -k \left( \frac{3 \pi}{L} \right)^{2} A_{3}(t) + e^{-2t} &\implies A_{3}(t) = \frac{e^{-2t}}{-2 + k \left( \frac{3 \pi}{L} \right)^{2}} + C_{3} e^{-k \left( \frac{3 \pi}{L} \right)^{2} t} \\
&\implies A_{3}(0) = \frac{1}{-2 + k \left( \frac{3 \pi}{L} \right)^{2}} + C_{3} \\
&\implies A_{3}(t) = \frac{e^{-2t} - e^{-k \left( \frac{3 \pi}{L} \right)^{2} t}}{-2 + k \left( \frac{3 \pi}{L} \right)^{2}} + A_{3}(0) e^{-k \left( \frac{3 \pi}{L} \right)^{2} t}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that this sequence is monotonic increasing?
Let $k>1$ and define a sequence $\left\{a_{n}\right\}$ by $a_{1}=1$ and $$a_{n+1}=\frac{k\left(1+a_{n}\right) }{\left(k+a_{n}\right)}$$
(a) Show that $\left\{a_{n}\right\}$ is monotonic increasing.
Assume $a_n \geq a_{n-1}$. Then,
$$a_{n+1} = \frac{k(1+a_n)}{k+a_n} \geq \frac{k(1+a_{n-1})}{k+a_n}....$$
But I get hung up on the $a_n$ in the denominator. I cannot replace it with $a_{n-1}$ since $a_n \geq a_{n-1}$. Is there a trick to get around this?
| Turn the question of whether $(a_n)$ is monotone increasing into an inequality purely in terms of a single term $a_n$. In particular,
$$a_n \le a_{n+1} = \frac{k(1 + a_n)}{k + a_n}.$$
Simplifying, making the temporary assumption that $k + a_n > 0$,
$$a_n(k + a_n) \le k(1 + a_n) \iff a_n^2 - k \le 0 \iff a_n \in [-\sqrt{k}, \sqrt{k}].$$
Addressing this assumption, it is extremely easy to show $a_n > 0$ for all $n$ by induction, hence $k + a_n > k > 1 > 0$. Thus, you really just need to show $a_n \le \sqrt{k}$.
You can now show this by induction. In fact, you can bundle the positivity proof in as well. That is, you can show that $0 \le a_n \le \sqrt{k}$ for all $n$ by induction.
Let's begin. Since $k \ge 1$, note that $0 \le 1 \le \sqrt{k}$, hence $0 \le a_1 \le \sqrt{k}$.
Now, assume $0 \le a_n \le \sqrt{k}$. Then,
\begin{align*}
a_{n+1} &= \frac{k(1 + a_n)}{k + a_n} \\
&= \frac{k + k^2 + ka_n - k^2}{k + a_n} \\
&= \frac{k(k + a_n)}{k + a_n} + \frac{k - k^2}{k + a_n} \\
&= k - \frac{k^2 - k}{k + a_n}.
\end{align*}
Note that $k^2 - k > 0$, hence $x \mapsto k - \frac{k^2 - k}{k + x} = \frac{k(1 + x)}{k + x}$ is increasing for $x > -k$. So, since $-k < 0 \le a_n \le \sqrt{k}$, we have
$$\frac{k(1 + 0)}{k + 0} \le \frac{k(1 + a_n)}{k + a_n} \le \frac{k(1 + \sqrt{k})}{k + \sqrt{k}} \implies 1 \le a_{n+1} \le \sqrt{k},$$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If a real matrix has a complex eigenvalue on the unit circle, when is it a root of unity? Let $A$ be a real matrix with integer entries, and suppose $z$ is a complex eigenvalue of $A$ with $|z|=1$. As shown in this answer, $z$ need not be a root of unity (i.e. there need not exist an $m$ with $z^m = 1$). Under which conditions on $A$ can this be guaranteed? What if all of $A$'s entries are either $1$ or $0$?
| all 1 and 0 is not good enough. This first characteristic polynomial is the one given by Jose Carlos Santos in your earlier question If an eigenvalue of an integer matrix lies on the unit circle, must it be a root of unity?
$$
\left(
\begin{array}{cccc}
1&1&1&0 \\
1&1&0&1 \\
0&1&0&0 \\
0&0&1&0 \\
\end{array}
\right)
$$
$$ x^4 - 2 x^3 - 2x + 1 = \left(x^2 - (1 + \sqrt 3)x+1 \right) \left(x^2 - (1 - \sqrt 3)x+1 \right)$$
Characteristic polynomial can be solved by dividing by $x^2,$ then writing as a quadratic function of $x + \frac{1}{x}$
Here is another one, the characteristic polynomial is once again palindromic. The roots can be found by the same trick.
$$
\left(
\begin{array}{cccc}
1&1&1&1 \\
1&1&1&0 \\
0&1&0&1 \\
1&1&0&0 \\
\end{array}
\right)
$$
$$ x^4 - 2 x^3 - 2 x^2 - 2x + 1 = \left(x^2 - (1 + \sqrt 5)x+1 \right) \left(x^2 - (1 - \sqrt 5)x+1 \right) $$
parisize = 4000000, primelimit = 500000
? m = [1,1,1,0;1,1,0,1;0,1,0,0;0,0,1,0]
%1 =
[1 1 1 0]
[1 1 0 1]
[0 1 0 0]
[0 0 1 0]
? charpoly(m)
%2 = x^4 - 2*x^3 - 2*x + 1
?
?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$? Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$?
I know that $\frac{1}{\sqrt{3+\sqrt{3}}} = \frac{\sqrt{3-\sqrt{3}}}{\sqrt6}$. So I just need to know whether $\sqrt6 \in L$. Since $\sqrt 3 = (\sqrt{3+\sqrt{3}})^2 - 3$, I only need to know if $\sqrt 2 \in L$. I would guess it is not, but how to show it? If I suppose that $\sqrt 2 \in L$ and aim for a contradiction:
It is clear that $\mathbb{Q}(\sqrt 3) \subset L$, and if $\sqrt 2 \in L$, then $\mathbb Q(\sqrt 2) \subset L$ as well. Then $K = \mathbb Q (\sqrt 2, \sqrt 3) \subset L$. Since both $K$ and $L$ are degree $4$ over $\mathbb Q$, this would imply they are equal. But then I'm back at square one.
| Use the tracial method, for example. The trace map on a number field $Q(\alpha)$ assigns to each $\beta \in \mathbb Q(\alpha)$ the quantity $\sum_{\sigma} \sigma(\beta)$ where $\sigma(\beta)$ are all the conjugates of $\beta$.
Suppose that $\sqrt 2 \in L$. Then $\sqrt 3 \in L$ implies that $\mathbb Q(\sqrt{3+\sqrt 3}) = \mathbb Q(\sqrt 2 ,\sqrt 3)$.
Write $\sqrt{3 + \sqrt 3} = a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$. Taking the trace of both sides gives $2a = 0$(note that $\sqrt{3+\sqrt 3}$ has minimal polynomial $x^4 - 6x^2+6$ hence has trace zero) hence $a = 0$.
Next multiply by $\sqrt 2$ to get $2b + c \sqrt 6 + 2d \sqrt 3 = \sqrt{6+2\sqrt 3}$. Once again the trace of the RHS is seen to be zero(find the minimal polynomial, it is easy!), so $b = 0$ is obtained on taking trace.
We are left with $\sqrt{3+\sqrt 3} = c \sqrt 3 + d \sqrt 6$. Multiplying by $\sqrt 3$ gives $\sqrt{9+3\sqrt 3} = 3c + 2d \sqrt 2$. Again the trace of the LHS is zero(easy again!) so we get $c = 0$.
Finally we are left with $\sqrt{3 + \sqrt{3}} = d \sqrt 6$, here squaring gives $\sqrt 3 = 6d^2 - 3$, an obvious problem. Hence, we are done.
Note that $\sqrt{3 - \sqrt 3}$ is a conjugate of $\sqrt{3+\sqrt 3}$, therefore the above shows that this extension is not normal, hence not Galois.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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More infinite nested square roots Let us consider the following nested square root:
$$A=\sqrt{b_1+\sqrt{b_2+ \sqrt{b_3+\cdots}} }.$$
Consider the following three cases:
*
*$b_k = k^2 + k -1 \Rightarrow A = 2.$
*$b_k = (k+2)^4 - 4(k+2)^3+5(k+2)^2-4(k+2)+1 \Rightarrow A = 4.$
*$b_k = 4^{k-1} \Rightarrow A = 2.$
The last case (the methodology) is discussed in my previous MSE question, here. These are very particular cases of a methodology that can create many similar formulas for nested square roots, nested cubic roots, or continued fractions. The first formula could be related to the Ramanujan's infinite radicals (see here) though I am not sure - this Wikipedia article is quite confusing. Also the third case is not new, indeed it validates my approach.
This does not seem to be a very advanced topic, though convergence issues are not simple in some cases. For instance, the case $b_1 = 1, b_2 = 7, b_{k+1} = 3b_k(b_k -1)$ if $k>2$, using nested cubic roots rather than nested square roots, does converge but not to the simple value predicted by my scheme.
Before going any deeper into this, I'd like to have people look into the first two formulas and try to prove the result. The first two cases involve an algorithm with $b_k = a_{k+2}, c_{k+1}= (c_k - a_k)^2$ if $k>2$, $c_2 = 4, a_k = \lfloor c_k - k^\alpha \rfloor$ if $k>1$, and $a_1 =1$. The first case corresponds to $\alpha =1$, the second case to $\alpha = 2$.
| For the first case, we can start with $\:2 = \sqrt{1+\sqrt{9}} = \sqrt{(1^2+1-1)+\sqrt{(2+1)^2}}\:$. Then we repeatedly apply the identity
$$(k+1)^2 = (k^2+k-1) + (k+2) = (k^2+k-1) + \sqrt{(k+2)^2}$$
So we get
$$\begin{align}
2 = \sqrt{1+\sqrt{9}} &= \sqrt{(1^2+1-1)+\sqrt{(2+1)^2}}\\[1.5ex]
&= \sqrt{(1^2+1-1)+\sqrt{2^2+2-1 + \sqrt{(3+1)^2}}}\\[1.5ex]
&= \sqrt{(1^2+1-1)+\sqrt{2^2+2-1 + \sqrt{3^2+3-1 +\sqrt{(4+1)^2}}}}\\[1.5ex]
&= \; ...
\end{align}$$
$$\color{white}{text}$$
In the second case, the identity $\:(x-1)^4 = (x^4-4x^3+5x^2-4x+1) + x^2\:$ leads to
$$\begin{align}
\color{white}{text}\\
(k+1)^4 &= \left[(k+2)^4-4(k+2)^3+5(k+2)^2-4(k+2)+1\right] + (k+2)^2\\[1.5ex]
&= b_k + \sqrt{\left[(k+1)+1\right]^4}\\
\color{white}{text}\\
\end{align}$$
Then, starting with $\:4 = \sqrt{7+\sqrt{81}} = \sqrt{b_1+\sqrt{(2+1)^4}}\:$, we repeatedly apply the above identity and get
$$\begin{align}
4 &= \sqrt{7+\sqrt{81}}\\[1.5ex]
&= \sqrt{b_1+\sqrt{(2+1)^4}}\\[1.5ex]
&= \sqrt{b_1+\sqrt{b_2 + \sqrt{(3+1)^4}}}\\[1.5ex]
&= \sqrt{b_1+\sqrt{b_2+\sqrt{b_3 + \sqrt{(4+1)^4}}}}\\[1.5ex]
&= \; ...
\end{align}$$
$$\color{white}{text}$$
[P.S. Would appreciate any suggestions to make those last lines in each proof look better; or is three the highest number of nested radicals that MathJax can handle well?]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor $x^{35}+x^{19}+x^{17}-x^2+1$ I tried to factor $x^{35}+x^{19}+x^{17}-x^2+1$ and I can see that $\omega$ and $\omega^2$ are two conjugate roots of $x^{35}+x^{19}+x^{17}-x^2+1$. So I divide it by $x^2+x+1$ and the factorization comes to the following
$$(x^2+x+1)(x^{33}-x^{32}+x^{30}-x^{29}+x^{27}-x^{26}+x^{24}-x^{23}+x^{21}-x^{20}+x^{18}-x^{16}+2x^{15}-x^{14}-x^{13}+2x^{12}-x^{11}-x^{10}+2x^{9}-x^{8}-x^{7}+2x^6-x^5-x^4+2x^3-x^2-x+1)$$
I couldn't go further. My question is is it end here or there is a simple way to do further?
| Can one find the factors by hand? Well, perhaps with a bit of guessing.
In the big factor
$$\begin{align}p_{33}(x)&=x^{33}-x^{32}\\&+x^{30}-x^{29}\\&+x^{27}-x^{26}\\&+x^{24}-x^{23}\\&+x^{21}-x^{20}\\&+x^{18}\\&-x^{16}+2x^{15}-x^{14}\\&-x^{13}+2x^{12}-x^{11}\\&-x^{10}+2x^{9}-x^{8}\\&-x^{7}+2x^6-x^5\\&-x^4+2x^3-x^2\\&-x+1,\end{align}$$
you may notice a pattern that is almost perfect, namely we almost always can group the monomials as $x^{n+1}-x^n$ (or, in particular in lower degrees, as $x^{n+2}-2x^{n+1}+x^n=(x^{n+2}-x^{n+1})-(x^{n+1}-x^n)$.
Only $x^{18}$ does not fit into this. And if we want to insist on the $x^{n+2}-2x^{n+1}+x^n$ pattern for all low degrees, then also $-x+1$ dose not fit any more.
Thus
$$p_{33}(x)=(x^{18}-x+1)+ (x-1)\cdot x^2\cdot p_{30}(x)$$
for some degree $30$ polynomial $p_{30}$, which happens to be nice in that it suspiciously only has coefficients $\in\{-1,0,1\}$. Of course, such an additive composition is usually of no help to finding a factorization.
Out of despair, we check if $p_{30}$ is a multiple of the first dummand $x^{18}-x+1$ and by chance succeed:
$$\begin{align}p_{30}(x)&=x^{30} + x^{27} + x^{24} + x^{21} + x^{18} - x^{13} + x^{12} - x^{10} + x^9 - x^7 + x^6 - x^4 + x^3 - x + 1\\
&=(x^{18}-x+1)(x^{12} + x^9 + x^6 + x^3 + 1)
\end{align}$$
and that last factor is easily recognized as $\frac{x^{15}-1}{x^3-1}$, which might be considered motivational.
However, $p_{30}$ is not what we wanted to factor. Nevertheless, our pipe-dream made us arrive at
$$p_{33}(x)=(x^{18}-x+1)\cdot p_{15}(x)$$
with $$\begin{align}p_{15}(x)&=1+(x-1)x^2\frac{x^{15}-1}{x^3-1}\\&=1+x^2\frac{x^{15}-1}{x^2+x+1}
\\&=x^{15} - x^{14} + x^{12} - x^{11} + x^9 - x^8 + x^6 - x^5 + x^3 - x^2 + 1.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\lim_{x\to 4} \left(\frac{\sqrt {2x-1}}{\sqrt {x-3}}\right) = \sqrt 7$ using $ \varepsilon - \delta$
Prove that $$\lim_{x\to 4} \left(\frac{\sqrt {2x-1}}{\sqrt {x-3}}\right) = \sqrt 7$$ using $\varepsilon - \delta$.
We find $\delta$ such that
$0<|x-4| <\delta$
$$\left|\frac{\sqrt {2x-1}}{\sqrt {x-3}}-\sqrt 7\right|= \left|\frac{\sqrt {2x-1}-\sqrt{7x-21}}{\sqrt {x-3}}\right|$$
I know that we should get to $|x-4|$ but i dont know how
| We want to show that $\lim\limits_{x\to 4} \dfrac{\sqrt{2x-1}}{\sqrt{x-3}}=\sqrt{7}$. By the $\epsilon-\delta$ limit definition, this means $\forall \epsilon >0,\exists \delta >0 \space(0<|x-4|<\delta \Rightarrow \left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |<\epsilon)$. Simplifying, we obtain $\left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |=
\left |\dfrac{\sqrt{2x-1}-\sqrt{7x-21}}{\sqrt{x-3}} \right |$. Let $\delta = \dfrac{3}{4}$, then $|x-4|<\delta\Rightarrow 3.25<x<4.75\Rightarrow\sqrt{x-3}>\sqrt{3.25-3}=\sqrt{0.25}=0.5\Rightarrow \left |\dfrac{\sqrt{2x-1}-\sqrt{7x-21}}{\sqrt{x-3}} \right |<\left |\dfrac{(\sqrt{2x-1}-\sqrt{7x-21})\cdot(\sqrt{2x-1}+\sqrt{7x-21})}{0.5(\sqrt{2x-1}+\sqrt{7x-21})} \right|=\left |\dfrac{-5(x-4)}{0.5(\sqrt{2x-1}+\sqrt{7x-21})} \right |< \left |\dfrac{-5(x-4)}{0.5(\sqrt{5.5}+\sqrt{1.75})} \right |<\left| \dfrac{-5(x-4)}{0.5(2+1)}\right|=\dfrac{10}{3}|x-4|$.
Take $\delta = \min\{\dfrac{10}{3}\epsilon,\dfrac{3}{4}\}$ and we have that $\left |\dfrac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right |<\dfrac{10}{3}|x-4|<\dfrac{3}{10}\cdot\dfrac{10}{3}\epsilon=\epsilon$, as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Without calculus, show that the sign of $(1+a)^{1+x}-a^{1+x}-1$ matches the sign of $x$, for any positive $a$ Let $a$ be a positive real constant. Consider the function
$$
f(x) = (1+a)^{1+x}-a^{1+x}-1
$$
We have for any choice of $a$, that $f(x) >0$ for $x>0$ and $f(x) <0$ for $x<0$, which can be proved using calculus.
My question:
How would a proof go without using calculus?
| For positive $x$ we have
$$
\begin{align}
\frac{1+a^{1+x}}{(1+a)^{1+x}} &= \left( \frac{1}{1+a}\right)^{1+x} + \left( \frac{a}{1+a}\right)^{1+x}\\
&=
\frac{1}{1+a} \left( \frac{1}{1+a}\right)^{x} + \frac{a}{1+a}\left( \frac{a}{1+a}\right)^{x} \\
&< \frac{1}{1+a} + \frac{a}{1+a} = 1 \, .
\end{align}
$$
For negative $x$ the same holds with $<$ replaced by $>$.
| {
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$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{4}$. The possible triples are:
$$ 2, 2^{3}, 5^{4} $$
$$ 2^{2}, 2^{2}, 5^{4} $$
$$ 2^{3}, 2, 5^{4} $$
$$ 2^{4}, 2^{4}, 5^{4} $$
$$ 5, 5^{3}, 2^{4}$$
$$ 5^{2}, 5^{2}, 2^{4}$$
$$ 5^{3}, 5, 2^{4}$$
The smallest sum is $5^{2} + 5^{2} + 2^{4}$. Are there better approaches?
| Well. "no zero" means none of the $a,b,c$ have both factors of $2$ and $5$ so you either have two numbers that are powers of $2$ and one that is a power of $5$ or you have $2$ that are powers of $5$ and one that is a power of $5$.
CASE 1: $\{a,b,c\} = \{2^k, 2^{4-k}, 5^4\}$.
So $a + b + c = 2^k + 2^{4-k} + 625$.
We have to minimize $2^k + 2^{4-k}$.
By AM-GM $2^k + 2^{4-k} \ge 2\sqrt{2^k2^{4-k}} = 2\sqrt{2^4} = 2*4=8$ with equality holding if and only if $2^k = 2^{4-k}$.
So $\{a,b,c\} = \{4,4,625\}$ and $a + b +c =633$. Is the minimum sum for this case.
CASE 2: $\{a,b,c\} = \{2^4, 5^{4-k},5^{k}\}$
Same argument of minimizing $a+b+c$ via $AM.GM$ $a=5^2, b=5^2, c=16$ and $a + b + c = 66$.
So the min is $\{a,b,c\} = \{25,25, 16\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
| As you got (for $x\neq0$):
$$ x+\sqrt{x} + \sqrt{x^2-x} = m\sqrt{x} $$
$$ x+\sqrt{x} + \sqrt{x}\sqrt{x-1} = m\sqrt{x} $$
Setting $ u = \sqrt{x} $ ($ u\neq0 $) we get:
$$ u^2+u+u\sqrt{u^2-1} = mu $$
$$ u+1+\sqrt{u^2-1} = m $$
$$ \sqrt{u^2-1} = (m-1) - u$$
Squaring both sides:
$$ u^2-1 = (m-1)^2-2u(m-1)+u^2 $$
We get:
$$ u = \frac{(m-1)^2+1}{2(m-1)} $$
where $m\neq1$ and $x$ becomes:
$$ x = \Bigg(\frac{(m-1)^2+1}{2(m-1)}\Bigg)^2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving indefinite integral $\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$ How can we solve this integration?
$$\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}}dx$$
I tried to make the following substitution
$$1+x^2=w^2$$
but this substitution complicated the integral.
| Like the answer of @AmerYR, taking $1+x^2=u^2$
$$I=\int \frac{1+\sqrt{1+x^2}}{x^2+\sqrt{1+x^2}} dx= \int \sec u ~du + \int \frac{sec u}{\tan^2 u+ sec u} du$$ $$ \implies I =\log (\tan u+ sec~ u)+J(u)$$
Next, using $\sin u=\frac{2 \tan(u/2)}{1+\tan^(u/2)},~~ \cos u= \frac{1-\tan^(u/2)}{1+\tan^2(u/2)}$ and $t=\tan(u/2), dt= sec^2 (u/2) du/2$, we get
$$J=\int \frac{\cos u ~du} {\sin^2 u + \cos u}= \frac{(1-t^2)/(1+t^2)}{4t^2/(1+t^2)^2+(1-t^2)/(1+t^2)} \frac{2dt}{(1+t^2)}$$
$$\implies J= 2 \int \frac{t^2-1}{t^4-4t^2-1} dt= 2 \int \frac{t^2-1}{(t^2+a^2)(t^2-b^2)} dt = 2 \int \left(\frac{A}{t^2-a^2} + \frac{B}{t^2+b^2} \right) dt$$
$$\implies 2 \left(\frac{A}{2a} \log \frac{t-a}{t+a} + \frac{B}{b} \tan^{-1} \frac{t}{b} \right) +C$$
Here $a=\sqrt{2+\sqrt{5}}, b=\sqrt{\sqrt{5}-2}$, $A=\frac{\sqrt{5}+1}{2\sqrt{5}}$, $B=\frac{\sqrt{5}-1}{2\sqrt{5}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a much stronger version of AM-GM for three variables Here, @MichaelRozenberg stated the following inequality without proof:
Theorem. For all non-negative $a,b,c\in\mathbb R$,
$$a^6+b^6+c^6-3a^2b^2c^2\geq16(a-b)^2(a-c)^2(b-c)^2.$$
In my answer below I give a complete brute-force proof. However, more elegant proofs are welcome.
| I will use the Buffalo method: By symmetry of the inequality, we can assume without loss of generality that $a\le b\le c$. So we can substitute $a=x,b=x+y,c=x+y+z$ for some non-negative reals $x,y,z$.
Putting this in our inequality and expanding results in
$$\color{green}{12 x^4 y^2 + 12 x^4 y z + 12 x^4 z^2 + 28 x^3 y^3 + 42 x^3 y^2 z + 54 x^3 y z^2 + 20 x^3 z^3 + 27 x^2 y^4 + 54 x^2 y^3 z + 87 x^2 y^2 z^2 + 60 x^2 y z^3 + 15 x^2 z^4 + 12 x y^5 + 30 x y^4 z + 60 x y^3 z^2 + 60 x y^2 z^3 + 30 x y z^4 + 6 x z^5} + \color{blue}{2 y^6 + 6 y^5 z - y^4 z^2 - 12 y^3 z^3 - y^2 z^4 + 6 y z^5 + z^6}\geq 0.$$
The green part is automatically greater or equal than $0$ because the $x,y,z$ are non-negative. Also,
$$\color{blue}{2 y^6 + 6 y^5 z - y^4 z^2 - 12 y^3 z^3 - y^2 z^4 + 6 y z^5 + z^6}=y^6+(y-z)^2(y+z)^2(y^2+6yz+z^2)\geq0.$$
This achieves a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.
| Here is my solution:
$$\int_0^\frac{\pi}{2}\frac{\sin^2 x}{(b^2\cos^2 x+a^2\sin^2 x)^2}dx=\int_0^\frac{\pi}{2}\frac{\tan^2 x\sec^2 x}{(b^2 +a^2\tan^2 x)^2}dx\overset{\tan x=t}=\int_0^\infty \frac{t^2}{(b^2+a^2 t^2)^2}dt$$
$$\overset{at=bp}=\frac{1}{a^3b}\int_0^\infty\frac{p^2}{(p^2+1)^2} dp=\frac{1}{a^3b} \left(\int_0^\infty \frac{1}{p^2+1}dp- \underbrace{\int_0^\infty \frac{1}{(p^2+1)^2}dp}_{p=\tan \theta }\right)$$
$$=\frac{1}{a^3b}\left(\arctan p\bigg|_0^\infty-\int_0^\frac{\pi}{2}\frac{1+\cos(2\theta)}{2}d\theta\right)=\frac{1}{a^3b}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{4a^3b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\sum_{k = 1}^{\infty} k^{-2} = \pi^2 / 6$ using Fourier analysis I'm doing a homework for my analysis class, and a problem says the following:
Let $f : \mathbb{T}^1 \to \mathbb{C}$ be the function
$$f(\theta) = \begin{cases} 1 & 0 < \theta < \pi \\ 0 & - \pi < \theta < 0 \end{cases}$$
Compute the Fourier transform of $f$. Finally, it says to use the fact that the Fourier transform is an isometry from $L^2 \left( \mathbb{T}^1 \right) \to \ell^2$ to show that $\sum_{k = 1}^{\infty} k^{-2} = \pi^2 / 6$.
I've computed the Fourier transforms of $f$ as
$$\hat{f}(k) = \begin{cases}
\frac{1}{2} & k = 0 \\
0 & \textrm{$k \neq 0$ even} \\
- \frac{i}{\pi k} & \textrm{$k$ odd}
\end{cases}$$
If I understand correctly, what it wants me to do next is observe that since $f \mapsto \hat{f}$ is a unitary operator from $L^2 \left( \mathbb{T}^1 \right)$ to $\ell^2$, we have
$$\pi = \| f \|_{L^2}^2 = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2 = \frac{1}{4} + \frac{1}{\pi^2} \sum_{\textrm{$k$ odd}} \frac{1}{k^2}$$
Now there's an example in the text that does something with a different function to get that $\sum_{\textrm{$k \geq 1$} odd} k^{-2} = \frac{\pi^2}{8}$, then does some arithmetic trickery to get that $\sum_{k = 1}^{\infty} k^{-2} = \frac{\pi^2}{6}$. But whenever I do this calculation on the $f$ shown above, I get different numbers. I've already sunk a lot of time into this. Do I have an arithmetic error somewhere that I need to iron out, or am I on the wrong track?
Thanks.
EDIT: In this text, the Fourier transform is defined as $$\hat{f}(k) = (2 \pi)^{-1} \int_{- \pi}^{\pi} f(\theta) e^{-k i \theta} \mathrm{d} \theta$$
| Thank you to @GReyes for his help. He pointed out that the correct formula for $\| \cdot \|_{L^2}$ should've been
$$\| f \|_{L^2} = (2 \pi)^{-1} \int_{- \pi}^{\pi} |f(\theta)|^2 \mathrm{d} \theta .$$
With this fix, the solution to the problem is as follows:
\begin{align*}
\hat{f}(k) & = (2 \pi)^{-1} \int_{- \pi}^{ \pi} f(\theta) e^{- i k \theta} \mathrm{d} \theta \\
& = (2 \pi)^{-1} \int_{0}^{\pi} e^{- i k \theta} \mathrm{d} \theta .
\end{align*}
If $k = 0$, then $e^{- i k \theta} = 1$ for all $\theta$, so we know that $\hat{f}(0) = \frac{1}{2}$. But if $k \neq 0$, then
\begin{align*}
\hat{f}(k) & = (2 \pi)^{-1} \int_{0}^{\pi} e^{- i k \theta} \mathrm{d} \theta \\
& = (2 \pi)^{-1} \left[ \frac{e^{- i k \theta}}{- i k} \right]_{\theta = 0}^{\pi} \\
& = (2 \pi)^{-1} \frac{1}{- i k} \left( e^{- i k \pi} - 1 \right) \\
& = \frac{i}{2 \pi k} \left( (-1)^{k} - 1 \right) \\
& = \begin{cases}
0 & \textrm{$k \neq 0$ even} \\
- \frac{i}{\pi k} & \textrm{$k$ odd}
\end{cases}
\end{align*}
In summary,
$$\hat{f}(k) = \begin{cases}
\frac{1}{2} & k = 0 \\
0 & \textrm{$k \neq 0$ even} \\
- \frac{i}{\pi k} & \textrm{$k$ odd}
\end{cases}$$
Now we use Parseval's theorem. We can see that $\| f \|_{L^2}^2 = (2 \pi)^{-1} \int_{- \pi}^{\pi} |f(\theta)|^2 \mathrm{d} \theta = \frac{1}{2}$. Now we're going to use the fact that $\| f \|_{L^2}^2 = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2$ to find $\sum_{k = 1}^{\infty} k^{-2}$. Note that if we can show that $\sum_{\textrm{$k \geq 1$ odd}} k^{-2} = \pi^2 / 8$, the remainder of the proof will follow from the text's argument (see (5.4.21), page 194).
By Parseval's theorem, we have that
\begin{align*}
\| f \|_{L^2}^2 & = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2 \\
= \frac{1}{2} & = \frac{1}{4} + \sum_{\textrm{$k \in \mathbb{Z}$ odd}} \frac{1}{\pi^2 k^2} \\
& = \frac{1}{4} + \frac{2}{\pi^2} \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\
\Rightarrow \frac{1}{4} & = \frac{2}{\pi^2} \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\
\Rightarrow \frac{\pi^2}{8} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} .
\end{align*}
Finally, following the text's lead, we observe that
\begin{align*}
\sum_{k = 1}^{\infty} k^{-2} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \sum_{\textrm{$k \geq 1$ even}} k^{-2} \\
& = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \sum_{\ell = 1}^{\infty} (2 \ell)^{-2} \\
& = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \frac{1}{4} \sum_{\ell = 1}^{\infty} \ell^{-2} \\
\Rightarrow \frac{3}{4} \sum_{k = 1}^{\infty} k^{-2} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\
& = \frac{\pi^2}{8} \\
\Rightarrow \sum_{k = 1}^{\infty} k^{-2} & = \frac{4}{3} \frac{\pi^2}{8} \\
& = \frac{\pi^2}{6} .
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Explanation of an integration trick for $\int \frac{a_1 \cos x + b_1 \sin x}{a\cos x + b\sin x}dx$ I'm not sure how to formulate my question correctly. Basically it comes from solving the integral:
$$
\int \frac{a_1 \cos x + b_1 \sin x}{a\cos x + b\sin x}dx\\
a^2 + b^2 \ne 0
$$
I haven't been able to solve the integral without the trick I've found after a while. This trick suggests to rewrite:
$$
a_1 \cos x + b_1 \sin x = \frac{a_1a + b_1b}{a^2 + b^2}(a\cos x + b\sin x) + \frac{a_1b - ab_1}{a^2 + b^2}(b\cos x - a\sin x)\tag{1}
$$
After using this trick the integral becomes almost elementary:
$$
I = \int \frac{a_1a + b_1b}{a^2 + b^2} dx + \int \frac{a_1b - ab_1}{a^2 + b^2}\frac{b\cos x - a\sin x}{a\cos x + b\sin x} dx
$$
The first part is trivial, for the second one substitute $u = a\sin x + b\cos x$.
My question is how on earth one could arrive at $(1)$. Is that some sort of well-known expression that I just missed?
Going from RHS to LHS in $(1)$ is easy, but how do I make it the other way round?
Thank you!
| Any linear combination of sine waves with the same period and different phase shifts can be written as a single sine wave with that same period and a suitable phase shift.
\begin{align}
& A\cos(x+\varphi) + B\cos(x+ \psi) \\[8pt]
= {} & A\big(\cos x\cos\varphi - \sin x \sin\varphi\big) \\
& {} + B\big(\cos x\cos\psi - \sin x \sin \psi\big) \\[8pt]
= {} & C\cos x + D\sin x
\end{align}
where
\begin{align}
C & = A\cos\varphi + B \cos\psi \\[8pt]
D & = -A\sin\varphi - B\sin\psi
\end{align}
and then
\begin{align}
& C\cos x + D\sin x \\[8pt]
= {} & \sqrt{C^2+D^2} \left( \frac C {\sqrt{C^2+D^2}} \cos x + \frac D {\sqrt{C^2+D^2}} \sin x\right) \\[8pt]
= {} & \sqrt{C^2+D^2} \big( E\cos x + F\sin x\big).
\end{align}
We now have $E^2+F^2=1$ so $E= \cos\chi$ and $F=\sin\chi$ for some angle $\chi.$ Thus we have
\begin{align}
& E\cos x + F\sin x \\[8pt]
= {} & \cos\chi\cos x + \sin\chi\sin x \tag 1 \\[8pt]
= {} & \cos(x-\chi).
\end{align}
Line $(1)$ above is what you have in the problem you're facing.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} \equiv 9^{5} \bmod 11 = 9^{4} 9 \bmod 11, \:\: 9^{2} \equiv 4 \bmod 11 $$
$$ \implies 9^{5} \equiv 144 \bmod 11 \implies 20^{5} \equiv 1 \bmod 11 $$
Then the proof is done.
Now I will prove:
$$ 20^{15} \equiv 1 \bmod 31 $$
Notice $20^{2} \equiv 28 \bmod 31$, so
$$20 \times (20^{2})^{7} \equiv 20 \times (28)^{7} \bmod 31 \equiv 20 \times (-3)^{7} \bmod 31 \equiv -60 \times 16 \bmod 31\equiv 32 \bmod 31 $$
then the proof is done.
Also, in similar way to prove the $20^{15} \equiv 1 \bmod 61$.
Are there shorter or more efficient proof?
| $$20\equiv3^2\pmod{11}$$
$20^{15}\equiv(3^2)^{15}\equiv(3^{10})^3\equiv1^3$ by Fermat's Little Theorem
$$20=2^2\cdot5,\implies20^{15}\equiv2^{30}5^{15}$$
By Fermat's Little Theorem
$$2^{30}\equiv1\pmod{31},5^3\equiv1\pmod{31}\implies5^{15}=(5^3)^5\equiv1^5$$
Again $5^3\equiv3\pmod{61},2^6\equiv3$
$$20^{15}\equiv(2^6)^5(5^3)^5\equiv3^5\cdot3^5\pmod{61}$$
Finally
$3^5=243\equiv-1\pmod{61}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so there are two values
1 and 3.
Then $$2^x=1$$
$$x=0$$
And $$2^x=3$$
$$x=\log_2 3$$
But the answer doesn’t seem to consider the $\log_2 3$ as a viable root, and the answer is 1. Why is that the case?
|
Then there are two cases
$$a-1=a^2-2a-5$$
$$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
You made a small mistake in solving these.
The first equation is $a^2 - 3a - 4 = 0$, so $(a - 4)(a + 1) = 0$, so $a = 4$ or $a = -1$.
Looks like you solved the second equation correctly. You should have
$$
a = -1, 4, -2, 3
$$
But the answer doesn’t seem to consider the $\log_2 3$ as a viable root
Once we have found the values of $a$ that work, we're not done yet!
We have to plug in the values to the original equation to see if they work. The problem is that $|a - 1| = a - 1$ only if $a \ge 1$, and $|a - 1| = -(a - 1)$ only if $a \le 1$, so we might have introduced some "extraneous" answers that are not valid.
As you observed, since $a = 2^x$, $a$ has to be positive; so we are left with just $a = 3$ and $a = 4$ as possibilities.
Next, we have to plug in each to the original equation to see if they work.
$a = 4$ works, but $a = 3$ doesn't. So the only answer is
$$
a = 4 \implies 2^x = 2 \implies \boxed{x = 2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the complete Taylor expansion of $\frac{1}{1+z^2}$ around $z=0$. For an exercise, I need to find the complete Taylor expansion for $(1+z^2)^{-1}$ around $z=0$. I have tried decomposing first $(1+z^2)^{-1}$ into partial fractions. Since $1+z^2=0$ gives $z=\pm i$, the partial fractions are:
$$\frac{1}{1+z^2} = \frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)} = \frac{i}{2} \Big{(}\frac{1}{z+i} - \frac{1}{z-i}\Big{)}$$
Therefore, my idea was to find the much easier Taylor expansion of both fractions, and add them together. I have worked out each Taylor expansion, since the $n$-th derivative of each of the fractions can be easily found:
$$\frac{1}{z+i}=\sum_{n=0}^\infty -i^{n+1}z^n, \hspace{25px} \frac{1}{z-i}=\sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}$$
Typing each sum in WolframAlpha returns the original fraction, so I think they are okay. Now, I replace both sums into the partial fractions:
$$\frac{1}{1+z^2} = \frac{i}{2} \Big{(}\sum_{n=0}^\infty -i^{n+1}z^n - \sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}\Big{)}$$
But I always get a sum that does not return the correct Taylor expansion for $1/(1+z^2)$. Where am I getting this wrong?
| Your approach is all correct. Note that your sum can be rewritten as $$\frac{1}{2}\sum_{n=0}^\infty i^n ((-1)^n+1)z^n.$$
For an odd index value $n$, the term equals zero, and for even - it equals $2(-1)^kz^{2k}$, $k\in \mathbb{Z}$.
Therefore
$$\frac{1}{2}\sum_{n=0}^\infty i^n ((-1)^n+1)z^n=\frac{1}{2} ( 2-2z^2+2z^4-2z^6+\ldots)=1-z^2+z^4-z^6+\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Disproving $ 0^0 $ by binomial theorem For fun, is something like this true?
Let
\begin{equation} \nonumber
\begin{split}
k &= 0^0 = (a-a)^{a-a} = \frac{(a-a)^{a}}{(a-a)^{a}} \\
&= \frac{\binom{a}{0}a^a(-a)^0 + \binom{a}{1}a^{a-1}(-a)^1 + ... +
\binom{a}{a-1}a^{1}(-a)^{a-1} + \binom{a}{a}a^{0}(-a)^{a}}{\binom{a}{0}a^a(-a)^0 + \binom{a}{1}a^{a-1}(-a)^1 + ... +
\binom{a}{a-1}a^{1}(-a)^{a-1} + \binom{a}{a}a^{0}(-a)^{a}}. \\
\end{split}
\end{equation}
Furthermore, let $a$ be any odd number, then
\begin{equation} \nonumber
\begin{split}
k &= \frac{\binom{a}{0}a^a - \binom{a}{1}a^{a} + ... + \binom{a}{a-1}(a)^{a} - \binom{a}{a}(a)^{a}} {\binom{a}{0}a^a - \binom{a}{1}a^{a} + ... + \binom{a}{a-1}(a)^{a} - \binom{a}{a}(a)^{a}} \\
\end{split}
\end{equation}
since $a$ is odd there is an even number of terms in the polynomials, thus
\begin{equation} \nonumber
\begin{split}
k = \frac{0}{0}. \\
\end{split}
\end{equation}
| You are not disproving anything. $0^0$ is a mathematical expression with no agreed-upon value. The most common possibilities are $k=1$ or leaving the expression undefined, with justifications existing for each, depending on context.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of Complex Trigonometric Expression
Minimum value of of $\displaystyle f(\theta) = \frac{a}{\cos \theta}+\frac{b}{\sin \theta}+\sqrt{\frac{a^2}{\cos^2 \theta}+\frac{b^2}{\sin^2 \theta}}.$
Where $\displaystyle a,b>0, \theta \in \bigg(0,\frac{\pi}{2}\bigg).$
what i try
$$f(\theta)=\frac{2(a\sin \theta +b\cos \theta)+2\sqrt{\bigg(a\sin \theta+b\cos \theta\bigg)^2-a\sin 2 \theta}}{\sin 2 \theta }$$
How do i minimize is Help me please
| \begin{align*}
f'(\theta) &= a \tan \theta \sec \theta + b \tan \theta \sec \theta \hfill \\
&\quad {}+ \frac{2 a^2 \tan \theta \sec^2 \theta + 2 b^2 \tan \theta \sec ^2\theta }{2 \sqrt{a^2 \sec ^2 \theta + b^2 \sec^2 \theta}} \\
&= \tan \theta \sec \theta \left(\cos \theta \sqrt{\left(a^2+b^2\right) \sec^2 \theta }+a+b\right) \\
&= \tan \theta \sec \theta \left( a + b + \sqrt{a^2 + b^2} \right) \text{.}
\end{align*}
$\tan \theta \neq 0$ for $\theta \in (0, \pi/2)$. $\sec \theta \neq 0$ for any $\theta \in \Bbb{R}$. Since $a>0, b>0$, the parenthesized expression is always positive. Therefore, this function has no critical points.
So we check the endpoints to see which one would be the location of the minimum if it were a permissible point. $f(0) = a + b + \sqrt{a^2 + b^2}$ and $\lim_{\theta \rightarrow \pi/2^-} f(\theta) = \infty$ (requiring the use of $a>0, b>0$).
So for $\theta \in (0,\pi/2)$, $f$ has no minimum value, but it takes values arbitrarily close to $a+b+\sqrt{a^2 + b^2}$, a lower bound for values of $f$, as $\theta$ approaches $0$.
| {
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Why does $\left(x \cdot \tan\left(\frac{1}{x}\right)-1\right)^{-1}$ asymptotically approach $3x^2 - 6/5$? I noticed that $\lim_{x \to \infty}\tan\left(\frac{1}{x}\right)*x = 1$ and I was wondering how fast it approaches $1$. I looked at $\frac{1}{\tan\left(\frac{1}{x}\right)*x-1}$ and found that this grows slower than $x^3$, so to find what polynomial degree it grows as fast as, I plugged it into $\lim_{x \to \infty} \frac{\ln\left(f\left(x\right)\right)}{\ln\left(x\right)} = \lim_{x \to \infty} \frac{\ln\left(\frac{1}{\tan\left(\frac{1}{x}\right)*x-1}\right)}{\ln\left(x\right)}=2$ to find that it grows around as fast as $x^2$. Then I tried plugging $\lim_{x \to \infty}\left(\tan\left(\frac{1}{x}\right)*x-1\right)*x^2$ into Wolfram|Alpha and it produced $1/3$. It was not able to provide any steps. How is this limit calculated? Using this, the next question I come upon is how $\lim_{x\to\infty}3x^2-\frac{1}{\tan\left(\frac{1}{x}\right)*x-1} = 6/5$ is calculated. Why does $\left(\tan\left(\frac{1}{x}\right)*x-1\right)^{-1}$ asymptotically approach $3x^2 - 6/5$?
| We have that by Taylor's series
$$\tan\left(\frac{1}{x}\right)=\frac1x+\frac1{3x^3}+o\left(\frac1{x^3}\right)$$
and therefore
$$\left(\tan\left(\frac{1}{x}\right)\cdot x-1\right)\cdot x^2=\left(1+\frac1{3x^2}+o\left(\frac1{x^2}\right)-1\right)\cdot x^2=\frac 13+o(1) \to \frac 13$$
and since
$$\tan\left(\frac{1}{x}\right)=\frac1x+\frac1{3x^3}+\frac2{15x^5}+o\left(\frac1{x^5}\right)$$
$$\frac{1}{\tan\left(\frac{1}{x}\right)\cdot x-1}=\left(\frac1{3x^2}+\frac2{15x^4}+o\left(\frac1{x^4}\right)\right)^{-1}=$$
$$=3x^2\left(1+\frac2{5x^2}+o\left(\frac1{x^2}\right)\right)^{-1}=3x^2-\frac6{5}+o\left(1\right)$$
then
$$3x^2-\frac{1}{\tan\left(\frac{1}{x}\right)\cdot x-1}=\frac65+o(1) \to \frac65$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Any clues on how to do this modular arithmetic proof? Assume:
*
*$2x^3 - 8x^2 + 8y^3 - 12y^2 -10 \equiv 0 \mod 10$.
*$2y^3 - 8y^2 + 8z^3 - 12z^2 -10 \equiv 0 \mod 10$.
WTP:
*
*$2x^3 - 8x^2 + 8z^3 - 12z^2 -10 \equiv 0 \mod 10$.
I'm not sure where to start. How should I go about this?
| Reducing $\pmod {10}$ gives:
$2x^3 +2x^2 - 2y^3 - 2y^2 \equiv 0 \pmod {10}$
So:
$2x^3 +2x^2 \equiv 2y^3 + 2y^2 \pmod {10}$
Similarly from eqn. 2 we get:
$2z^3 +2z^2 \equiv 2y^3 + 2y^2 \pmod {10}$
The equivalence relation is transitive, therefore:
$2z^3 +2z^2 \equiv 2x^3 + 2x^2 \pmod {10}$
and so eqn. 3 is true.
| {
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Number in binary as a product I agree that any binary number that consists of $n$ ones (and no zeros) has as its decimal equivalent the number $2^n - 1$. However, the author of the book I'm reading next makes the following claim, which I don't quite see. He says that as a consequence of the fact above, it follows that any binary number that has the form of starting with $n$ ones and ending with $n - 1$ zeros (with nothing between them) has as its decimal equivalent the number $2^{n - 1}(2^n - 1)$. For example, the decimal number 496 in binary is 111110000, which consists of five 1s followed by four 0s, so $496 = 2^4 \times (2^5 - 1) = 16 \times 31$. But why is this true, i.e. why does it work?
| Since your binary number starts with $n$ ones followed by $n-1$ zeros, the number has $n+(n-1) = 2n-1$ binary digits. Therefore its decimal value is
$$
\begin{align}
& 0 \times 2^{0} + 0 \times 2^{1} + \cdots + 0 \times 2^{n-2} + 1 \times 2^{n-1} + 1 \times 2^{n} + 1 \times 2^{n+1} + \cdots + 1 \times 2^{2n-3} + 1 \times 2^{2n-2} \\
&= 2^{n-1} \left( 1 + 2 + 2^2 + \cdots + 2^{n-2 } + 2^{n-1} \right) \\
&= 2^{n-1} \frac{1 - 2^n }{1 - 2 } \\
&= 2^{n-1} \frac{2^n - 1 }{2-1} \\
&= 2^{n-1} \left( 2^n-1 \right).
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{2x^8}{8!}+...$
$e^x-\sin(x) = 1+\dfrac{x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{2x^7}{7!}+\dfrac{x^8}{8!}+...$
$e^x-\cos(x) = x+x^2+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{2x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$
Due to the missing term and the irregularity, I am unable to write the general formula for these series. I wish to find the general formula with compact sigma notation. Can you help with this?
| Formal power series can be added and then their area of convergence is limited by the most restricted one.
So one easy way would be to
*
*find separately expansions for $\{e^{x},\cos(x),\sin(x)\}$
*add (or subtract) them to each other
*put them under the same $\sum$ and then
*try to simplify the expression you get using algebra.
| {
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Solve system solutions The number of actual system solutions
$ \begin{cases}
a^2=b+2\\
b^2=c+2 \\
c^2=a+2\\
\end{cases}$
is equal to:
Solution:
$\cos 2\theta=2\cos^2\theta-1\implies 2\cos 2\theta=(2\cos\theta)^2-2$.
Using this results in all $8$ solutions to the system.
$(2,2,2)$, $(-1,-1,-1)$, and cyclic permutations of $\left(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7}\right)$ and $\left(2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right)$
How can i solve this in a simpler way
(Without using trigonometry)
| $$a^2=b+2 \tag1$$
$$b^2=c+2 \tag2$$
$$c^2=a+2 \tag3$$
By successive eliminations
$$(1) \implies b=a^2-2$$
$$(2) \implies c=a^4-4a^2-2$$
$$(3) \implies a^8-8 a^6+20 a^4-16 a^2-a+2=0\tag 4$$
$(4)$ can be factorized as
$$(a-2) (a+1) \left(a^3-3 a+1\right) \left(a^3+a^2-2 a-1\right)=0$$ Each cubic equation has three real roots that you can express using ... the trigonometric method !
| {
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For rational numbers $a, b$, what is the range of $b$ such that $\lceil a + b \rfloor = \lceil a \rfloor$ holds? For rational numbers $a, b$, what is the range of $b$ such that $\lceil a + b \rfloor = \lceil a \rfloor$ holds?
Clearly, b=0 gives us the result.
What are the lower and upper bounds of $b$?
$\lceil \cdot \rfloor$ : is a rounding function that rounds a rational number to the nearest integer
The range of $a$ is $[-\frac{x}{2}, \frac{x}{2})$ for some positive integer $x$.
| For positive $a$, $\lceil a \rfloor = \lfloor a + \frac 12 \rfloor$, being the floor function. If $a = a_q+a_r$ where $a_q$ is the whole number part and $a_r$ is the fractional part, $\lceil a \rfloor = a_q + \lfloor a_r + \frac 12 \rfloor$. Similarly, if $b$ (also positive) is $b_q + b_r$ then
\begin{align}
\lceil a + b \rfloor & = \big \lfloor a_q+a_r + b_q + b_r + \frac 12 \big \rfloor \\
& = a_q + b_q + \big \lfloor a_r + b_r + \frac 12 \big \rfloor
\end{align}
If this is to equal $\lceil a \rfloor$ then we must have $b_q = 0$ and we want to find when $\lfloor a_r + b_r + \frac 12 \rfloor = \lfloor a_r + \frac 12 \rfloor$.
This is best broken into parts: $a_r < \frac 12$ and $a_r \ge \frac 12$. For the first of these we find that $b_r \in [0, \frac 12 - a_r)$, while for the second $b_r \in [0, \frac 32 - a_r)$.
For negative $a$, it is easier to think of $a_q< 0$ and $a_r \ge 0$, thus $-5.6 = -6 + 0.4$. The above arguments are true and we still split it into two: when $a_r< \frac 12, b\in [0, \frac 12 - a_r)$ and when $a_r \ge \frac 12, b \in [0, \frac 32 - a_r)$.
| {
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Find G. C. F. of $(8n^3 + 8n, 2n+1)$ I'm stuck with this problem, I divided $8n^3 + 8n$ by $2n+1$ and obtained $5$, so now my G. C. F is $\gcd(2n+1, -5)$.
What's next? I can't divide $2n+1$ by $-5$.
| Well, you figured out that $\gcd(8n^3 +8n, 2n+1) = \gcd(2n+1, -5)$.
And as $\gcd(\pm a, \pm b) = \gcd(a b)$ we know $\gcd(8n^3 + 8n, 2n+1) = \gcd(2n+1,5)$.
As $5$ is prime then $\gcd(2n+1, 5)$ is either $1$ or $5$.
It is $5$ if $5|2n+1$. ANd it is $1$ if $5\not\mid 2n+1$.
And $5|2n+1 \iff$
$2n+1 \equiv 0 \pmod 5 \iff$
$2n \equiv -1 \pmod 5 \iff$
$n \equiv 2 \pmod 5$.
So the answer is:
$\gcd(8n^3+8n, 2n+1) =\begin{cases} 5 &\text{if } n\equiv 2 \pmod 5\\1&\text{if } n\not\equiv 2 \pmod 5\end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving system of Congruences with Two Variables (x and y) I know a bit about the Chinese Remainder theorem but what do I do in the case I was asked to solve a system of congruences such as this with two variables:
$3x + y = 7$ (mod 8)
$4x + 3y = 1$ (mod 8)
| The first equation is in a form suggesting substitution method as you have $y$ with coefficient $1$:
*
*$y \equiv 7-3x \pmod 8$
Plug this into the second equation and solve for $x$:
$$4x + 3(7-3x) \equiv 4x +21 -9x \stackrel{21=16+5}{\equiv} -5x +5 \equiv 1 \pmod 8$$
Now, note that the $5^2 \equiv (24+1) \equiv 1 \pmod 8$. Using this you get
$$\Leftrightarrow 5x \equiv 4 \pmod 8 \stackrel{\cdot 5}{\Leftrightarrow} \boxed{x =} 20 \equiv \boxed{4 \pmod 8}$$
Plugging this back into the equation $y \equiv 7-3x\pmod 8$ gives
$$\boxed{y=} 7-3\cdot 4 \equiv -5 \equiv \boxed{3 \pmod 8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits