Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Area of ellipse which is not in standard form By graphing device i understand that $x^2+xy+y^2=1$ is ellipse. By some geometry i find area of above ellipse which comes out $\pi$ (is it right?), but it was easy case. Is there any quick method or standard formula to calculate it or we have to convert it into standard form always and then calculate?
| If the conic $ Q(x,y) = ax^2 + by^2 + c + 2hxy + 2fy + 2gx$ is an ellipse, then if we translate the orgin of our coordinate system to the centre of $ Q$ to get,
$$ {R(X,Y) = Q(X + u, Y + v)} = aX^2 + b Y^2 + 2hXY + c^\prime$$
where $(u, v)$ is the centre of the conic given. We need to find $(u, v)$ and $c^\prime$ (practically we only need $c'$).
Anyway $(u, v)$ is the solution of $$\begin{cases}\dfrac{\partial Q}{\partial x} = 0 \\ \dfrac{\partial Q}{\partial y} = 0 \end{cases}$$
and $c^\prime$ is given by $ c^\prime =\dfrac{\Delta}{\delta}$ where $$\delta = \begin{vmatrix} a & h \\ h & b\end{vmatrix}.$$ and $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}.$$
The problem reduces to finding the area of the curve $R(X, Y) = 0$, which is equivalent to finding the area of $ S (X, Y) = \dfrac{-a}{c^\prime}X^2 + \dfrac{-b}{c^\prime} Y^2 + 2\dfrac{-h}{c^\prime}XY - 1$.
Let $\alpha = \dfrac{-a}{c^\prime}
, \beta = \dfrac{-b}{c^\prime}$ and $\gamma = \dfrac{-h}{c^\prime}$, then we have $ S(X, Y) = \alpha X^2 + \beta Y^2 + 2\gamma XY - 1$.
To write $ S$ in standard form we need to eliminate $ XY$ term. If $ \gamma = h \ne 0$ then that means the ellipse is rotated by some angle $\theta$ given by the relation, $$\tan (2\theta) = \dfrac{2\gamma}{ \alpha - \beta} = \dfrac{2h}{a - b}.$$
So to remove $XY$ term we rotate $S$ by $-\theta$, $${X \choose Y} = \left[\begin{matrix}\cos\theta & -\sin \theta \\ \sin \theta& \cos \theta \end{matrix}\right] {X^\prime \choose Y^\prime}.$$
We will then get,
$$ S(X^\prime, Y^\prime) = \dfrac{X{{^\prime}^2} }{\alpha'^2} + \dfrac{Y{{^\prime}^2} }{\beta'^2} - 1 = 0$$
The area of which is given by $\pi \alpha' \beta'$.
For example, Let $Q(x,y) = x^2 + 2y^2 + 2xy + 4x + 4y + 1$,
Here we have $c^\prime = -3$ and $\theta = \dfrac12 \tan^{-1} (-2)$.
First translating the origin we get, $S(X, Y) = \dfrac{1}{3} X^2 + \dfrac23 Y^2 + 2\dfrac13XY - 1$.
Now we rotate the cooridinate system by $\dfrac12 \tan^{-1} (2)$ to get the equation of the ellipse as,
$$\dfrac{X'^2}{\dfrac6{3- \sqrt{5}}} + \dfrac{Y'^2}{\dfrac6{3+ \sqrt{5}}} = 1 $$
So the area is $3\pi$.
Not the easiest method but the simplest method for sure.
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $p$ is an odd prime and there exists some integer $x$ such that $x^2 \equiv -1 \pmod{p}$. Prove that $p \equiv 1 \pmod{4}$. Suppose $p$ is an odd prime and there exists some integer $x$ such that $x^2 \equiv -1 \pmod{p}$. Prove that $p \equiv 1 \pmod{4}$. Use Fermat's Theorem.
| Let's call the order of $x$ to be the smallest positive integer power, $k$ so that $x^k \equiv 1 \mod p$. We'll denote the order of $x$ as $|x|$.... (assuming some a power exists; which it won't if $x$ is not relatively prime to $p$.)
Clearly $x^{k|x| + j; 0 \le j < |x|} = (x^{|x|})^k*x^j \equiv 1^k*x^j\equiv x^j \mod p$ and furthermore $x^{m} \equiv 1 \mod p \iff m \equiv 0 \mod |x|$
.....
Okay we know 1) $x^2 \equiv -1\mod p$.
And 2) Fermat's Theorem: $x^{p-1} \equiv 1 \mod p$.
For 1) It is easy to verify $|x| = 4$. [$*$].
So 2) implies $p-1 \equiv 0 \mod 4$.
======
[$*$]If the order of $x$ were $1$ then $x\equiv 1\mod p \implies x^2 \equiv 1\not \equiv -1 \mod p$.
If the order of $x$ were $3$ then $1 \equiv x^3 = x*x^2 \equiv -x \mod p\implies x\equiv -1 \mod p \implies x^2 \equiv 1 \mod p$.
And, of course $x^4 \equiv 1 \mod p$.
[Note; this assumes $p > 2$ so $1 \not \equiv -1 \mod p$]
| {
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Minimum and maximum value related to the sides of the quadrilateral.
If $a$, $b$, $c$ and $d$ are the sides of the quadrilateral then find the minimum value of
$$\frac{a^2+b^2+c^2}{d^2}.$$
I have tied by the inequality $a+b+c>d$, but it doesn't work.
| The maximum does not exist: $a=b=c\rightarrow+\infty$.
The minimum does not exist.
Indeed, by C-S $$\frac{a^2+b^2+c^2}{d^2}=\frac{(1+1+1)(a^2+b^2+c^2)}{3d^2}\geq\frac{(a+b+c)^2}{3d^2}>\frac{d^2}{3d^2}=\frac{1}{3}.$$
Since for $a=b=c\rightarrow\frac{d}{3}$ we have $\frac{a^2+b^2+c^2}{d^2}\rightarrow\frac{1}{3}$, we obtain that $\frac{1}{3}$ is an infimum of our expression.
| {
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Prove the inequality $a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$ for positive $a, b, c$ I faced problem proving this inequality for positive $a$, $b$, $c$:
$a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$
I tried to simplify it and I got that:
$bc^3 + a^3 c + a b^3 \geq a b^2 c + a^2 bc + a b c^2$
Then I tried to prove it using MMI, but it gave me nothing.
Please just give me a hint, maybe I am missing something important.
| We need to prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c,$$
which is true by Rearrangement because
$(a^2,b^2,c^2)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered,
which gives
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{a^2}{a}+\frac{b^2}{b}+\frac{c^2}{c}=a+b+c.$$
C-S also works:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{(a+b+c)^2}{a+b+c}=a+b+c.$$
AM-GM:
$$\sum_{cyc}\left(\frac{a^2}{b}+b\right)\geq2\sum_{cyc}\left(\frac{a^2}{b}\cdot b\right)=2\sum_{cyc}a.$$
SS:
Let $c=\min\{a,b,c\}$.
Thus,
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-a-b-c=$$
$$=\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}-c+\frac{c^2}{a}-\frac{b^2}{a}=$$
$$=\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}\geq0.$$
| {
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Why the remainders of the division of a number by 2 results in the number in binary? For example 14, if you divide be 2 it equals 7 and the remainder = 0, then 7/2 = 3 with r = 1 then 3/2 = 1 and r = 1 then 1/2 = 0 with r = 1. If we take the r's and put them togheter 1110 is 14 in binary. I don't understand the intuition behind this, why does this work?
| This is because of Horner's scheme for evaluating polynomials:
\begin{align}&a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}\dots+a_1x+a_0\\
={}(\dots(((&a_nx+a_{n-1})x+a_{n-2})x+\dots+a_1)x+a_0.
\end{align}
The successive divisions by $2$ yield
\begin{align}
14&=7\cdot 2+0\\&=(3\cdot2+1)\cdot 2+0=3\cdot2^2+1\cdot 2+0\\
&=\color{red}{((1\cdot2+1)\cdot2+1)\cdot 2+0}=(1\cdot2+1)\cdot2^2+1\cdot 2+0 \\ &=1\cdot2^3+1\cdot2^2+1\cdot 2+0
\end{align}
| {
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Computing $\int \frac{1}{\sqrt{25y^2-30y-7}} \,dy$ $$\int \frac{1}{\sqrt{25y^2-30y-7}} \, dy$$
I start by completing the square: so that $(5y-3)^2-16=25y^2-30y-7$.
Then I let $u=5y-3$, and then let $ u=4 \sec \theta$.
This leads to the equation becoming $\frac{1}{4\tan\theta}$ which is the same as $\frac{1}{4}\cot \theta$.
Taking the integral of this is $\frac{1}{4}\ln\left|\sin\theta\right|$.
Substituting back I get $\frac{1}{4}\ln\left| \sin \left( \operatorname{arcsec} \frac{5y-3} 4 \right)\right|$.
I drew a triangle and came up with this: $\frac{1}{4}\ln\left|\frac{\sqrt{25y^2-30y-7}}{5y-3}\right|$. Please, where did I go wrong?
| HINT: set $$\sqrt{25y^2-30y-7}=5y+t$$ then you will get
$$y=-\frac{7+t^2}{10t+30}$$
$$dy=-1/10\,{\frac { \left( t+7 \right) \left( t-1 \right) }{ \left( t+3
\right) ^{2}}}dt
$$
| {
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Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}{3-1}\right)\\ = 2\times 3^n - 4(3^n - 1) \\ = -2\times 3^n +4.$$
Method 2:
Solving the homogenous equation using $x_n = r^n$,
$$r^n - 3r^{n-1} = 0 \\ \implies r = 3$$
So the homogenous solution is $h_n = a\times 3^n$ for some $a\in\mathbb{R}$.
With the initial condition, $h_n = 2\times 3^n$.
Also, by guessing the particular solution with $x_n = C$,
$$C - 3C = -8 \implies C = 4$$.
So, the final solution is
$$x_n = 2\times 3^n + 4.$$
I'm unsure why the two methods differ. It looks obvious that the first one gives the correct result.
| Safety way
Set
$v_n =x_n-a$ such that $v_n$ satisfies $$ v_{n+1} =3v_n\Longleftrightarrow x_{n+1}-a =3(x_n-a)$\Longleftrightarrow 3x_{n}-8-a =3(x_n-a)\Longleftrightarrow a=4$$
Then
$$v_n =3^nv_0\Longleftrightarrow x_n-4 = 3^n(x_0-4)\Longleftrightarrow \color{red}{x_n = -2\cdot3^n +4}$$
| {
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Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty odd integer.]
What I got:
Using the hint. $4n^2, n^4-4, n^4+4$ is a Phytagorean triple if $x^2+y^2=z^2$. Replacing and solving the equation it is clear that, $(4n^2)^2+(n^4-4)^2=(n^4+4)^2$ where $n$ is odd is indeed a Pythagorean triple with $x=4n^2=(2n^2)^2$ a perfect square.
Now I have to prove that $gcd(4n^2, n^4-4, n^4+4)=1$.
But I'm stuck here. I tried $gcd(n^4-4, n^4+4)=1$ without success. Any ideas?
| If $n$ is odd, $\gcd (n^4-4, n^4+4)=\gcd (n^4-4,8)=1$ by the Euclidean algorithm.
| {
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minimum value of a two variable problem Suppose that $x$ and $y$ are in $(−2, 2)$ and $xy = −1$. The minimum value of
$\frac4{4-x^2}+\frac9{9-y^2}$ is ?
So I have tried to manipulate this equation but what I got is a complex equation with square roots in it...
| $x, y \in (−2, 2)$ and $xy = −1$.
Let $3x+2y=k$. Then $9x^2-12+4y^2=k^2$. That is, $9x^2+4y^2=k^2+12$.
\begin{align}
f(x,y)
&= \frac4{(4-x^2)}+\frac9{(9-y^2)} \\
&= \dfrac{72 - (9x^2+4y^2)}{36-(9x^2+4y^2)+1} \\
&= \dfrac{72 - (k^2+12)}{36-(k^2+12)+1} \\
&= \dfrac{60 - k^2}{25-k^2} \\
&= 1 + \dfrac{35}{25-k^2}
\end{align}
This attains its smallest value, $\frac{12}{5}$, when $k=0$, for example, when $x = \frac 23$ and $y=-1$.
| {
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Prove this inequality $1\le \sum _{cyc}\frac{a}{1+bc}\le \frac{3\sqrt{3}}{4}$ Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that
$$1\le\dfrac{a}{1+bc}+\dfrac{b}{1+ca}+\dfrac{c}{1+ab}\le\dfrac{3\sqrt{3}}{4}$$
*) $LHS\le \frac{3\sqrt{3}}{4} \Rightarrow LHS^2\le \frac{27}{16}$
$\Leftrightarrow \left(1+1+1\right)\left(\sum _{cyc}\frac{a^2}{\left(ab+1\right)^2}\right)\le \frac{27}{16}\Leftrightarrow \sum _{cyc}\frac{a^2}{\left(1+ab\right)^2}\le \frac{9}{16}$
We have:
$\frac{a^2}{\left(ab+1\right)^2}=\frac{a^2}{\left(ab+a^2+b^2+c^2\right)^2}\le \frac{a^2}{\left(ab+bc+ab+ac\right)^2}\le \frac{1}{16}\left(\frac{a}{ab+bc}+\frac{a}{ac+ab}\right)^2$
$\Rightarrow \sum _{cyc}\frac{1}{16}\left(\frac{a}{ab+bc}+\frac{a}{ac+ab}\right)^2\le \frac{9}{16}$
$$\Leftrightarrow \left(\frac{b}{bc+ab}+\frac{b}{bc+ac}\right)^2+\left(\frac{c}{ac+ab}+\frac{c}{ac+bc}\right)^2+\left(\frac{a}{ab+bc}+\frac{a}{ac+ab}\right)^2\le 9=\sqrt{3}^2+\sqrt{3}^2+\sqrt{3}^2$$
It looks easier than before but i can't continue. Help me
*)$LHS\ge 1$I haven't got idea about this problem
| Firstly, the left inequality.
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{1+bc}=\sum_{cyc}\frac{a^2}{a+abc}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+abc)}\geq$$
$$\geq\frac{(a+b+c)^2}{a+b+c+\frac{(a+b+c)(ab+ac+bc)}{3}}=\frac{3(a+b+c)}{3+ab+ac+bc}.$$
Thus, it remains to prove that
$$3(a+b+c)\geq3+ab+ac+bc$$ or
$$9(a+b+c)^2(a^2+b^2+c^2)\geq(3(a^2+b^2+c^2)+ab+ac+bc)^2.$$
Let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, $k\geq1$ and we need to prove that
$$9(k+2)k\geq(3k+1)^2$$ or
$$12k-1\geq0,$$
which is obvious.
Your right inequality is wrong.
Try $c\rightarrow0^+$ and $a=b\rightarrow\frac{1}{\sqrt2}.$
| {
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Arithmetic of complex numbers
If $z = \cos x + i\sin x$ , show that
$$\frac{2}{1+z} =1-i\tan\left(\frac{x}{2}\right),$$
I get up to:
$$\frac{1+\cos x -i\sin x}{1+\cos x}.$$
| Note that
\begin{align*}\frac{2}{1+z}&=\frac{2(1+\overline{z})}{|1+z|^2}=\frac{2(1+\cos x - i\sin x)}{(1+\cos x)^2 + (\sin x)^2}\\
&=\frac{2(1+\cos x - i\sin x)}{2(1+\cos x)}=1-i\cdot\frac{\sin x}{1+\cos x}
\end{align*}
and recall the tangent half-angle formula.
| {
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Solving Recurrence Relation of $T(n)=4T(n-2)+2$ Question
Solve Recurrence Relation of $T(n)=4T(n-2)+2$
Base case-: $T(1)=1,T(2)=2$
My Approach/solution
$$T(n)=4T(n-2)+2$$
$$T(n-2)=4T(n-4)+2 \tag{1}$$
$$T(n-4)=4T(n-6)+2 \tag{2}$$
Using $(1)$ and $(2)$ in my equation
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2}\cdot T(n-2\cdot 2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{2}\cdot(4T(n-6)+2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{3}\cdot T(n-2\cdot 3)+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
\vdots \\
&=4^{k}\cdot T(n-2\cdot k)+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}
\end{align*}$$
Substituing $T(n-2\cdot k)$ by $2$, i.e $T(2)=2$
$$n-2\cdot k=2 \Rightarrow k=\frac{n-2}{2}$$
So our equation will look like
$$\begin{align*}
T(n)&=2\cdot 4^{k}+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
T(n)&=2\cdot \left(4^{0}+4^{1}+4^{2}+...+4^{k-1}+4^{k}\right)\\
T(n)&=2\cdot \left(4^{0}\cdot \frac{(4^{k+1}-1)}{4-1}\right)
\end{align*}$$
$k=\frac{n-2}{2}$
$$\begin{align*}
T(n)&=2\cdot \left(\frac{(4^{k+1}-1)}{4-1}\right)\\
T(n)&=2\cdot \frac{2^{n}-1}{3}
\end{align*}$$
Is it correct? Also if it is correct, can anyone hint me another approach as it is bit lengthy.
Thanks!
| Using generating functions technique we have
$$f(x)=\sum\limits_{n=0}T(n)\cdot x^n=T_0+1\cdot x+2\cdot x^2+\sum\limits_{n=3}T(n)\cdot x^n=\\
T_0+1\cdot x+2\cdot x^2+\sum\limits_{n=3}\left(4T(n-2)+2\right)\cdot x^n=\\
T_0+4\sum\limits_{n=3}T(n-2)\cdot x^n + x + 2\cdot x^2 +\sum\limits_{n=3}2\cdot x^n=\\
T_0+4x^2\sum\limits_{n=3}T(n-2)\cdot x^{n-2}-2-x+\sum\limits_{n=0}2\cdot x^n=\\
T_0+4x^2\sum\limits_{n=1}T(n)\cdot x^{n}-2-x+\sum\limits_{n=0}2\cdot x^n=\\
T_0-4T_0x^2-2-x+4x^2\sum\limits_{n=0}T(n)\cdot x^{n} +\sum\limits_{n=0}2\cdot x^n=\\
T_0-4T_0x^2-2-x+4x^2f(x)+ \frac{2}{1-x}$$
or
$$f(x)=T_0-\frac{2+x}{1-4x^2}+\frac{2}{(1-x)(1-4x^2)}=\\
T_0-\left( \frac{5}{4(1-2x)}+\frac{3}{4(1+2x)} \right)+\left(-\frac{2}{3(1-x)}+\frac{2}{1-2x}+\frac{2}{3(1+2x)} \right)=\\
T_0-\frac{2}{3(1-x)}+\frac{3}{4(1-2x)}-\frac{1}{12(1+2x)}=\\
T_0-\frac{2}{3}\sum\limits_{n=0}x^n+\frac{3}{4}\sum\limits_{n=0}(2x)^n-\frac{1}{12}\sum\limits_{n=0}(-2x)^n=\\
T_0+\frac{2}{3}\sum\limits_{n=0}\left(\frac{3}{4}\cdot 2^n -\frac{1}{12}\cdot (-2)^n-\frac{2}{3}\right)x^n$$
or
$$T(n)=\frac{3}{4}\cdot 2^n -\frac{1}{12}\cdot (-2)^n-\frac{2}{3}, n\geq 1$$
or
$$T(n)=\frac{1}{12}\left(9\cdot 2^n + (-1)^{n+1}\cdot 2^n-8\right)$$
as presented by Dr. Sonnhard Graubner, only with a complete proof this time. The result is easy to validate $T(1)=1, T(2)=2, T(3)=6$.
Some of the shortcuts are explained here.
More learning materials here.
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prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y^2}<\frac{x^2 \sqrt{x^2+y^2} -y^2\sqrt{x^2+y^2}}{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2-y^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2-y^2) }{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2) }{x^2+y^2}=\sqrt{x^2+y^2}<\delta=\epsilon$$
I use that $x^2<x^2+y^2$ twice, can I do it? or some identity to use when you have $x^2-y^2$
| If the question is correct:
$f(x,y) = \frac {xy^2 - x^2y}{x^2 + y}$
Suppose $y = -x^2 + \zeta$
where $\zeta$ is something very small.
$f(x,y) \approx \frac{x^2}{\zeta}$
There exists an $(x,y)$ in a neighborhood of $(0,0)$ such that $f(x,y)$ is large.
The limit does not exist.
If it is
$f(x,y) = $$\frac {xy^2 - x^2y}{x^2 + y^2}\\
\frac {xy(x-y)}{x^2 + y^2}$
$2xy < x^2+y^2 $ by the AM-GM inequality
$f(x,y) < \frac 12 (x-y)$
and $|f(x,y)| < \frac 12 (|x| + |y|)$
| {
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Counting with combinations vs manually Lets say there is an urn containing 10 balls, 5 are red and 5 are black. 3 are drawn without replacement. I am finding the probability that I get 3 red balls.
I understand that I can count this in two different ways:
1.) (5/10)(4/9)(3/8)
2.) 5C3/10C3
and that when I expand and simplify the second expression, it reduces to the first one. My question is that how do I intuitively recognize this?
I realize that by expanding nCr=n!/(n-r)!r!, I am counting all the possible arrangements of n objects and then dividing by (n-r)! and r! because I don't want to count the different permutations both divisions. How can I intuitively know NOT to count this same situation with 2.) when the balls are drawn WITH replacement?
|
Find the probability that three red balls are drawn when three balls are drawn without replacement from an urn containing five red and five black balls.
You correctly found the answer in two ways. In the first, you counted ordered selections. In the second, you counted unordered selections. However, this is not the best example to illustrate this, so let's consider a different example.
Find the probability that three balls of different colors are selected when three balls are drawn without replacement from a bag containing four blue, three green, and two red balls.
Ordered selections
There are $9$ ways to select the first ball, $8$ ways to select the second, and $7$ ways to select the third, so there are $9 \cdot 8 \cdot 7$ selections in our sample space.
Let $b$ denote blue, $g$ denote green, and $r$ denote red. There are $3! = 6$ favorable cases: $$(b, g, r), (b, r, g), (g, b, r), (g, r, b), (r, b, g), (r, g, b)$$ Hence, the probability of selecting three different colors is
\begin{align*}
p & = p((b, g, r)) + p((b, r, g)) + p((g, b, r)) + p((g, r, b)) + p((r, b, g)) + p((r, g, b))\\
& = \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7} + \frac{4}{9} \cdot \frac{2}{8} \cdot \frac{3}{7} + \frac{3}{9} \cdot \frac{4}{8} \cdot \frac{2}{7} + \frac{3}{9} \cdot \frac{2}{8} \cdot \frac{4}{7} + \frac{2}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} + \frac{2}{9} \cdot \frac{3}{8} \cdot \frac{4}{7}\\
& = 6 \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}\\
& = \frac{2}{7}
\end{align*}
Unordered selections
There are $$\binom{9}{3}$$ ways to select three of the nine available balls.
The favorable cases consist of choosing one of the four blue balls, one of the three green balls, and one of the two red balls. Hence, the number of favorable cases is
$$\binom{4}{1}\binom{3}{1}\binom{2}{1}$$
Therefore, the probability of selecting three balls of different colors is
$$p = \frac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{2}{1}}{\dbinom{9}{3}} = \frac{24}{84} = \frac{2}{7}$$
The reason that we obtain the same answer for ordered and unordered selections is that the order in which we select the elements of the subset does not matter, just which elements we select.
How do I know not to count the same situation with combinations if the balls are drawn with replacement?
The binomial coefficient
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
counts the number of ways of selecting a subset of $k$ elements from a set with $n$ elements.
If the balls are chosen without replacement and $k > 1$, we are not choosing a subset of $k$ balls since all the balls are available for each selection. In particular, the same ball may be selected more than once.
Find the probability that three balls of different colors when three balls are drawn with replacement from a bag containing four blue, three green, and two red balls.
Thre are $9$ ways to select the first ball, $9$ ways to select the second ball, and $9$ ways to select the third ball. Hence, there are $9^3$ selections in our sample space.
The favorable cases are the same as in the preceding example. However, the probability of selecting three different colors is
\begin{align*}
p & = p((b, g, r)) + p((b, r, g)) + p((g, b, r)) + p((g, r, b)) + p((r, b, g)) + p((r, g, b))\\
& = \frac{4}{9} \cdot \frac{3}{9} \cdot \frac{2}{9} + \frac{4}{9} \cdot \frac{2}{9} \cdot \frac{3}{9} + \frac{3}{9} \cdot \frac{4}{9} \cdot \frac{2}{9} + \frac{3}{9} \cdot \frac{2}{9} \cdot \frac{4}{9} + \frac{2}{9} \cdot \frac{4}{9} \cdot \frac{3}{9} + \frac{2}{9} \cdot \frac{3}{9} \cdot \frac{4}{9}\\
& = 6 \cdot \frac{4}{9} \cdot \frac{3}{9} \cdot \frac{2}{9}\\
& = \frac{16}{81}
\end{align*}
Notice that the order of selection matters, so using combinations would not be appropriate.
| {
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minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0 \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well. This one gives me an upper bound of $3$.
$$ x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq
3\, \big( x^2 + y^2 + z^2 \big)^2 = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2 + z^2 = 1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.
They might not be extraneous we could set spherical coordinates:
$$ (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
&=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big) \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi - \sin^2 2\varphi
$$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi - \sin^2 2\varphi
= \cos^2 2\varphi - (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1 \geq - 1$$
This is very similar to what I obtained before.
| A different approach. As you noted, being on or within the unit sphere just implies $x^2+y^2\leq 1$, and we can then ignore $z$.
$$F = x^4 - 6x^2 y^2 + y^4$$
$$F = (x^2-y^2)^2 - 4x^2 y^2$$
$$F = (x^2-y^2+2xy)(x^2-y^2+2xy)$$
Now substitute $x = r \cos\theta, y = r \sin\theta$. We know $0 \leq r\leq 1$.
$$F = (r^2\cos^2\theta-r^2\sin^2\theta+2r^2\cos\theta\sin\theta)
(r^2\cos^2\theta-r^2\sin^2\theta-2r^2\cos\theta\sin\theta)$$
$$F = r^4(\cos^2\theta-\sin^2\theta+2\cos\theta\sin\theta)
(\cos^2\theta-\sin^2\theta+2\cos\theta\sin\theta)$$
$$F = r^4[\cos(2\theta)+\sin(2\theta)][\cos(2\theta)-\sin(2\theta)]$$
$$F = r^4[\cos^2(2\theta) - \sin^2(2\theta)]$$
$$F = r^4\cos(4\theta)$$
Therefore
$$-1 \leq F \leq 1$$
with minimum $-1$ at
$$\theta = \frac{\pi}4, \frac{3\pi}4, \frac{5\pi}4, \frac{7\pi}4 \Rightarrow x= \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}}$$
and maximum $1$ at
$$\theta = 0, \frac{\pi}2, \pi, \frac{3\pi}2 \Rightarrow x = \pm 1, y = 0 \mbox{ or } x = 0, y = \pm 1.$$
| {
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If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
| Let' s try like this: Suppose $a,b,c$ are the roots of the polynomial $x^3-mx-n$.
By Vieta's formula's we have:
$$a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca)= 0-2m = 2m$$
Since $x^3 =mx+n \;\;\;\;(*)$ we have:
$$x^5 = x^2(mx+n) = mx^3+nx^2= m^2x+mn+nx^2$$
so
$$a^5+b^5+c^5 = m^2(a+b+c)+3mn+n(a^2+b^2+c^2) =3mn+n(a^2+b^2+c^2)=5mn$$
Also because of $(*)$ we have $$a^3+b^3+c^3 = m(a+b+c) +3n = 3n$$
and we are done.
| {
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How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"? How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"?
My attempt: Let $F(n, k)$ be the number of bit strings of length $n$ that contain exactly $k$ blocks of $10.$ Note that for $k \neq0, $ $F(0, k) = F(1, k)= 0.$
Consider a bit string of length $n$ where $ n \geq 2.$
Every bit string either begins with:
1) $10$,
or
2) $01$, $00$ or $11.$
In the former case, we get $F(n-2, k-1)$ since the remaining $n-2$ bits must contain $k-1$ blocks of $10$ in exactly $F(n-2, k-1)$ ways.
In the latter case, in either of the $3$ situations we get $F(n-2, k)$ since the remaining $n-2$ bits must contains $k$ blocks of $10$ in $F(n-2, k)$ ways.
Therefore for $n \geq 2, k \geq 1$ we have:
$F(n, k) = F(n-2, k-1)+ 3F(n-2, k)$.
For $k \geq 1$ let $A_k(x) = \displaystyle \sum_{n \geq2}F(n,k)x^n$ (and let $A_0(x)=1$) then multiplying the above recurrence by $x^n$ and summing over all $n \geq 2$ we get:
$A_k(x) = x^2 A_{k-1}(x) +3x^2A_{k}(x) $
$\Rightarrow A_k(x) = \dfrac{x^2}{1-3x^2} A_{k-1}(x)$
$\Rightarrow A_{k}(x) = \dfrac{x^{2k}}{(1-3x^2)^k} $
$\Rightarrow F(n, k) = [x^n] A_k(x)$
$\Rightarrow F(n, k) = [x^n] \dfrac{x^{2k}}{(1-3x^2)^k}$
$\Rightarrow F(n, k) = [x^{n-2k}] \displaystyle \sum_{r \geq 0} \binom{k+r-1}{r} 3^r x^{2r}$
which clearly gives a wrong answer for odd $n$. Could somebody point out where in the above calculation have I gone wrong? Thanks in advance.
| Essentially this amounts to finding the places where the string switches from a run of $1$'s to a run of $0$'s. Note that there are $n-1$ places where such a switch may occur (between each pair of numbers). In any event, there must be $k$ places where the string switches from $1$ to $0$:
*
*If the string starts and ends with $0$, then (since it starts with $0$) there must also be $k$ places where it switches from $0$ to $1$. There are $\binom{n-1}{2k}$ ways of making these choices.
*If the string starts with $1$ and ends with $0$, then there must (since it ends in $0$) also be $k-1$ places where it switches from $0$ to $1$; there are $\binom{n-1}{2k-1}$ ways of making these choices.
*If the string starts with $0$ and ends with $1$, then there must (since it starts with $0$) also be $k+1$ places where it switches from $0$ to $1$; there are $\binom{n-1}{2k+1}$ ways of making these choices.
*Finally, if the string starts and ends with $1$, there must (since it ends with $1$) also be $k$ places where it switches from $0$ to $1$. There are $\binom{n-1}{2k}$ ways of making these choices.
So finally, the total number of such strings is
$$\binom{n-1}{2k} + \binom{n-1}{2k-1} + \binom{n-1}{2k+1} + \binom{n-1}{2k}
= \binom{n}{2k} + \binom{n}{2k+1} = \binom{n+1}{2k+1}.$$
| {
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Question on uniqueness of projections in linear algebra Suppose A and B are projection matrices and suppose that for some vector x in the column space of B, BAx=x. Can I say that Ax=x?
Since x is in the column space of B, I know that Bx=x it projects itself in its own column space. I guess my question is, is x the only vector that can give the outcome of x when premultiplied by the projection matrix B?
| Let $A = \begin{pmatrix}0&1\\0&1\end{pmatrix},$ $B = \begin{pmatrix}0&0\\0&1\end{pmatrix},$ and $x = \begin{pmatrix}0\\1\end{pmatrix}.$ Then $A^2 = A$ and $B^2 = B,$ so $A$ and $B$ are indeed projection matrices. On the other hand,
$$
\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}= \begin{pmatrix}0\\1\end{pmatrix},
$$
so that $BAx = x,$ but $Ax\neq x.$
| {
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Prove $\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$ So I have this rather simple trigonometric identity that, for the life of me, I cannot solve. I have worked on it for about 2 hours and still can't get it.
Show that
$$\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$$
Here's what I've done so far:
\begin{align}
{2-\csc^2 A \over \csc^2 A+2\cot A} & = {2 - ({1 \over \sin A })^2 \over ({1 \over \sin A })^2 + 2({1 \over \tan A})} \\
&= {{2\sin^2 A \over \sin^2 A} - {1\over \sin^2 A} \over {1\over \sin^2 A} + 2({\cos A \over \sin A})} \\
& = {{2\sin^2 A -1 \over \sin^2 A} \over {1\over sin^2 A}+2({\sin A \cos A \over \sin^2 A})}\\
& = {2 \sin ^2 A -1 \over \sin^2 A} \times {\sin^2 A \over 2\sin A \cos A +1} \\
& = {2\sin^2 A -1 \over 2\sin A \cos A + 1}
\end{align}
Lots of fractions are involved so I fear I may have made a mistake somewhere.
If anyone has any tips on proving these trigonometric identities, could they please add them in their answer? I've been told just to keep trying; though I believe there must be some 'troubleshooting' method to finish the problem.
| =$\frac{2 sin^2(A)-1}{1+2sin(A)cos(A)}$
=$\frac{2sin^2(A)-(cos^2(A)+sin^2(A))}{(cos^2(A)+sin^2(A))+2sin(A)cos(A)}$
=$\frac{sin^2(A)-cos^2(A)}{(sin(A)+cos(A))^2}$
=$\frac{(sin(A)+cos(A))(cos(A)+sin(A))}{(sin(A)+cos(A))^2}$
=$\frac{sin(A)-cos(A)}{(sin(A)+cos(A))}$
| {
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Inverse of Quadratics with Horizontal Transformations This might sound like a beginner question for a lot of the community, I apologize, but, I really need help understanding it. I was trying to find the inverse of the quadratic $ f(x)= (\frac 12x +2)^2 +4$. Using WolframAlpha, I was able to find that the inverse function was $ f(x) = ± 2 \sqrt{x - 4}-4$
My steps are the following;
$y=(\frac 12x +2)^2 +4$
$x=(\frac 12y +2)^2 +4$
$x-4=(\frac 12y +2)^2$
$ \sqrt{x-4} =\frac 12y +2$
$ \sqrt{x-4} -2 =\frac 12y$
$ 2\sqrt{x-4} -2 =y$
I don't understand why I am getting a different inverse equation than the original.
| $y=(\frac 12x +2)^2 +4$
$x=(\frac 12y +2)^2 +4$
$x-4=(\frac 12y +2)^2$
Everything is right up to here.
$\sqrt{x^2} = |x|$, so you need a $\pm$ in front of the square root, because you need to account for both positive and negative values of $y$ for a given $x$ while solving for the inverse.
$\pm \sqrt{x-4} =\frac 12y +2$
$\pm \sqrt{x-4} -2 =\frac 12y$
You need to multiply both sides by $2$ (above, you only multiplied $\sqrt{x-4}$ by $2$, but not the $-2$).
Final answer: $y = \pm 2\sqrt{x-4} - 4 $
| {
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Conjugate of Quaternion
The conjugate of $$\alpha=\left[ {\begin{array}{cc}
a+bi & c+di \\
-c+di & a-bi \\
\end{array} } \right]$$ is$$\overline{\alpha}=\left[ {\begin{array}{cc}
a-bi & -c-di \\
c-di & a+bi \\
\end{array} } \right]$$
The norm of $\alpha$ is $a^2+b^2+c^2+d^2$ and is written $\lvert\lvert\alpha\rvert\rvert$. Show directly that $$\overline{\alpha}\alpha=\alpha\overline{\alpha}=\left[ {\begin{array}{cc}
t & 0 \\
0 & t \\
\end{array} } \right]$$ where $t= \lvert\lvert\alpha\rvert\rvert$.
Conclude that the multiplicative inverse of $\alpha$ is $(1/t)\overline{\alpha}$.
So to show directly I should do matrix multiplication, so
$\overline{\alpha}\alpha=\left[ {\begin{array}{cc}
a-bi & -c-di \\
c-di & a-bi \\
\end{array} } \right]\cdot \left[ {\begin{array}{cc}
a+bi & c+di \\
-c+di & a-bi \\
\end{array} } \right] = \left[ {\begin{array}{cc}
(a - bi) (a + bi) + (c - di) (c + di) & (a + bi) (-c - di) + (a + bi) (c + di) \\
(a - bi) (c - di) + (a - bi) (di - c) & (a - bi) (a + bi) + (-c - di) (di - c) \\
\end{array} } \right] = \left[ {\begin{array}{cc}
a^2-(bi)^2+c^2-(di)^2 & 0 \\
0 & a^2-(bi)^2+c^2-(di)^2 \\
\end{array} } \right] = \left[ {\begin{array}{cc}
a^2-(bibi)+c^2-(didi) & 0 \\
0 & a^2-(bibi)+c^2-(didi) \\
\end{array} } \right]$.
EDIT: Since complex number commute I am able to conclude that $\overline{\alpha}\alpha=\alpha\overline{\alpha}= \left[ {\begin{array}{cc}
t & 0 \\
0 & t \\
\end{array} } \right]$.
| You do not need to bother with matrix multiplication being commutative. Once you have figured out the matrix product, multiplication of the complex numbers within each component of that product is still commutative. What doesn't always commute is multiplication of one matrix by another, which you are already done with.
So, in your product, $(bi)^2$ in the upper left corner is just a product of complex numbers, not any matrices, and that commutes, so it equals $b^2i^2=-b^2$. Similarly for $(di)^2$ in the same component and the imaginary number squares in the lower right corner.
| {
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Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.
| From this result: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
Then $\dfrac{1^4+2^4+...+n^4+(n+1)^4}{1^4+2^4+...+n^4}=1+\dfrac{(n+1)^4}{O(n^5)}=1+O(\frac 1n)\to 1$ and so is the inverse of that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recurrence induction finding a constant $b$ such that $f(n) \leq bn$
Pretend induction is just a weird way our teacher uses induction when it comes to finding constant. Same as induction
Base Case:
let $n = 1$
$f(n) = 6(1) = 6$ and $bn = b$. Therefore we need $b \geq 6$ (*)
let $n = 2$
$f(n) = 6(2) = 6$ and $bn = 2b$. Therefore we need $b \geq 6$ (*)
$\vdots$
as wanted
IND STEP: Let $n \geq 9$. Suppose $f(j) \leq bj $ whenever $0 \leq j < n$. [IH]
WTP: $f(n) \leq bn$
$f(n) = 2f(\lfloor \frac{n}{9} \rfloor) + 3f(\lceil \frac{n}{4} \rceil) + 5n$ [def of f; $n \geq 9$]
$\leq 2b(\frac{n}{9}) + 3b \frac{n+3}{4} + 5n$ [IH, and $\frac{n}{9} \geq \lfloor \frac{n}{9} \rfloor$ and $\frac{n+3}{4} \geq \lceil \frac{n}{4} \rceil $]
$\leq 2b(\frac{n}{9}) + 3b \frac{2n}{4} + 5n$ [$\frac{2n}{4} \geq \frac{n+3}{4}$ for $n \geq 9$]
$= n \left(b \frac{2}{9} + b \frac{6}{4} + 5\right)$ [n common factor]
$= n \left( \frac{31}{18}b + 5\right)$ [Basic Arithmetic]
Thus we need $= n \left( \frac{31}{18}b + 5\right) \leq bn$
$5 \leq b \frac{-13}{18}$
$-\frac{90}{13} \leq b$
I'm lost. I don't know where I went wrong.
| I think the problem is that you are using $\frac{n+3}{4}\leq\frac{2n}4$ which isn't strong enough (if the problem had $\lfloor 2n/4\rfloor$ instead of $\lceil n/4\rceil$ no such $b$ would exist).
Instead, keep $\frac{n+3}{4}$. That means that $$f(n)\leq\bigg(\frac{35}{36}b+5+\frac{3b}{2n}\bigg)n.$$
We $n$ to be quite large to have a hope here: since this bound needs to be at most $bn$ we must have $\frac{35}{36}+\frac{3}{2n}<1$. This is true only for $n>216$.
I think what you need to do here is bound $f(n)$ up to some suitable $n_0>216$. You can do this in stages; for $n\leq 32$, $\lceil n/4\rceil<9$ which gives you some bound like $13n$ in that region. Then you get a worse bound in the region $32<n\leq 128$, and so on.
This means you will get one bound $b_1$ for $n\leq 256$ and another value $b_2$ such that for all $b\geq b_2$ the inductive step works for all $n>256$. Then that means you can prove $\max(b_1,b_2)$ will work by induction, with the "base case" being all $n\leq 256$.
| {
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Basis for Linear Transformation with Matrix Multiplication Let $V$ be the vector space of $ 2 \times 2 $ real matrices.
Find a basis for the kernel of the linear transformation $T :V \rightarrow V$ given by $T(A)= XA $.
$X$, being the matrix below.
\begin{bmatrix}1&1\\1&1\end{bmatrix}
The matrix $X$ isn't invertible, so I don't know what the kernel should be... And when I do, how do I find the basis for it?
*Sorry about the formatting, would appreciate an edit!
| You are looking for the set of all vectors $x \in V$ such that $T(x) = 0$, which is the definition of the null space (kernel). So then you want to look for the elements in $V$ such that
$$T(X) = 0 $$
First we observe the transformation
$$T\left(\begin{pmatrix} a&b\\c&d\end{pmatrix}\right) = \begin{pmatrix} 1&1\\1&1\end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix} = \begin{pmatrix} a+c&b+d\\a+c&b+d\end{pmatrix} $$
And in this case we are looking for it to be the zero matrix:
$$T\left(\begin{pmatrix} a&b\\c&d\end{pmatrix}\right) = \begin{pmatrix} 0&0\\0&0\end{pmatrix} \iff a+c =0 \ \ \ \text{and } \ \ b+d=0$$
So, because $c=-a$ and $d = -b$, we can write it as
$$\begin{pmatrix} a&b\\-a&-b\end{pmatrix} = a\begin{pmatrix} 1&0\\-1&0\end{pmatrix} + b\begin{pmatrix} 0&1\\0&-1\end{pmatrix}$$
And now we can write a basis $\beta $ for the kernel of $T$ as follows:
$$\beta = \Bigg\{ \begin{pmatrix} 1&0\\-1&0\end{pmatrix},\begin{pmatrix} 0&1\\0&-1\end{pmatrix} \Bigg\}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality with cosine and sine Let $A:=f^2+g^2$, where $f,g$ are functions of $x$ such that
$$f'=(c-1)(f\cos(x)\sin(x)+g\sin^2(x)),$$
$$g'=-(c-1)(f\cos^2(x)+g\cos(x)\sin(x)),$$
for some constant $c$.
(Note: $f'\cos(x)+g'\sin(x)=0$)
How do I show that $A'\leq4|c-1|A?$
I see that
\begin{align}
A'&=2ff'+2gg'\\
&=2(c-1)\left(f^2\cos(x)\sin(x)+fg\sin^2(x)-fg\cos^2(x)-g^2\cos(x)\sin(x)\right)\\
&=2(c-1)(f\cos(x)+g\sin(x))(f\sin(x)-g\cos(x)).
\end{align}
But how does the desired inequality follow?
| Cauchy-Schwartz' inequality tells you that
$$
|a f + b g| \le \sqrt{a^2 + b^2}\sqrt{f^2+ g^2}
$$
Starting from your result, we obtain
$$
A^\prime\le |A^\prime|\le 2 |c-1|1.\sqrt{f^2+g^2}1.\sqrt{f^2+g^2} = 2 |c-1|A
$$
| {
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"source": "stackexchange",
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Prove the intersection between $z^2=x^2+y^2$ and $x+y+2z=2$ is an ellipse. I want to prove that the intersection between the cone $z^2=x^2+y^2$ and the plane $x+y+2z=2$ is an elispe in this plane.
My work:
I suppose to prove it I have to see that the equation $x+y+2\sqrt{x^2+y^2}=2$ can be rewritten as $\frac{x^2}{a}+\frac{y^2}{b}=c$. To do so I have tried to completing the square without any results.
| You cannot put it in this form, because the ellipse is not always centered at the origin, or having its major axis on either the $y$ or $x$ axes.
$$\frac{x+y-2}{2}=z$$
$$\left(\frac{x+y-2}{2}\right)^2=x^2+y^2$$
$$x^2+xy-2x+xy+y^2-2y-2x-2y+4=4x^2+4y^2$$
$$3x^2+3y^2-2xy+4x+4y=4$$
This conic is an ellipse. You can use orthogonal diagonalisation to transform it into standard form.
$$3x^2-2xy+3y^2+4x+4y-4=0$$
$$a=3,\ b=-1,\ c=3, \ d=4$$
$$X=
\begin{pmatrix}
x\\
y\
\end{pmatrix}
$$
$$ A= \begin{pmatrix}
3 & -1\\
-1 & 3\
\
\end{pmatrix}
$$
$$v=
\begin{pmatrix}
4\\
4\
\\
\end{pmatrix}
$$
$$X^TAX+v^TX=4$$
Find the eigenvalues and eigenvectors of $A$, use the normalized eigenvectors as the columns of a rotation matrix $P$ with determinant $1$. It becomes relatively straightforward after that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$ How do I evaluate $\int_{0}^{+\infty} \cos(x^2)\cos(x)dx$?
I don't know what to do. Should I use a contour integration?
| Let $I$ be the desired integral. Then, by the product-to-sum trigonometric identity,
\begin{align*}
2I&=\int_{0}^{+\infty} \cos (x-x^2)\,dx+\int_{0}^{+\infty} \cos (x^2+x)\,dx\\
&=\int_{0}^{+\infty} \cos ((x-1/2)^2-1/4)\,dx+\int_{0}^{+\infty} \cos ((x+1/2)^2-1/4)\,dx\\
&=\cos(1/4)\int_{0}^{+\infty} \cos ((x-1/2)^2)\,dx+\sin(1/4)\int_{0}^{+\infty} \sin((x-1/2)^2)\,dx\\
&\quad+\cos(1/4)\int_{0}^{+\infty} \cos ((x+1/2)^2)\,dx+\sin(1/4)\int_{0}^{+\infty} \sin ((x+1/2)^2)\,dx\\
&=\cos(1/4)\int_{-1/2}^{+\infty} \cos (x^2)\,dx+\sin(1/4)\int_{-1/2}^{+\infty} \sin(x^2)\,dx\\
&\quad+\cos(1/4)\int_{1/2}^{+\infty} \cos (x^2)\,dx+\sin(1/4)\int_{1/2}^{+\infty} \sin (x^2)\,dx.
\end{align*}
By the Fresnel integrals,
$$\int_{0}^{+\infty} \cos (x^2)\,dx=\int_{0}^{+\infty} \sin (x^2)\,dx=\sqrt{\pi/8},$$
we obtain
\begin{align*}
\int_{-1/2}^{+\infty} \cos (x^2)\,dx+\int_{1/2}^{+\infty} \cos (x^2)\,dx&=\int_{-1/2}^{0} \cos (x^2)\,dx +\sqrt{\pi/8}\\
&\quad+\sqrt{\pi/8}-\int_{0}^{1/2} \cos (x^2)\,dx=2\sqrt{\pi/8}
\end{align*}
and
\begin{align*}
\int_{-1/2}^{+\infty} \sin (x^2)\,dx+\int_{1/2}^{+\infty} \sin (x^2)\,dx&=\int_{-1/2}^{0} \sin(x^2)\,dx +\sqrt{\pi/8}\\
&\quad+\sqrt{\pi/8}-\int_{0}^{1/2} \sin(x^2)\,dx=2\sqrt{\pi/8}
\end{align*}
Hence
$$I=\sqrt{\pi/8}\left(\cos(1/4)+\sin(1/4)\right)=
\frac{\sqrt{\pi}}{2}\cos\left(\frac{\pi-1}{4}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that series is convergent and find its sum $$\sum\nolimits\arccos\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}, n\in\mathbb{N}^{*}$$
I need help proving that this series is convergent and calculating its sum.
What I've done so far:
$$\lim_{n\to\infty}\arccos\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}=0$$
The result shows that the series may be convergent, but I don't know how to continue. WolframAlpha shows that the series is convergent (by the comparison test), but I have no idea to what other convergent series could I compare it.
Thank you!
| To know the answer, first you must know some results:
the propose (1) is sum of angle of cosine:
$$\arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1^2\right)\left(1-x_2^2\right)}\ \right)\quad, x_1<x_2$$
hence
$$ \arccos\left(x_1\right)-\arccos\left(x_2\right)=\arccos\left(x_1\cdot x_2+\sqrt{\left(1-x_1\right)\left(1+x_1\right)\left(1-x_2\right)\left(1+x_2\right)}\right)\tag{1}$$
So compare the expression (1) with expression (2):
$$\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)\tag{2}$$
hence can be easily deduced:
$$x_1=\frac{n}{2n+1}\ \ ,\ \ \ x_2=\frac{n+1}{2n+3}\tag{3}$$
where $x_1<x_2$.
so using (3) in (1) obtain:
$$S(k)=\sum_{n=1}^k\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right) $$
$$ S(k)=\sum_{n=1}^k\left(\arccos\left(\frac{n}{2n+1}\right)-\arccos\left(\frac{n+1}{2n+3}\right)\right)$$
$$ S(k)=\sum _{n=1}^{k }\arccos\left(\frac{n}{2n+1}\right)-\sum _{n=1}^{k }\arccos\left(\frac{n+1}{2n+3}\right)$$
it is easy to see that this is a telescoping series.
So.
$$S(k)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{k+1}{2k+1}\right)$$
hence using limits I have:
$$\lim_{k\to\infty}S(k)=\lim_{k\to\infty}\left(\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$
$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\lim_{k\to\infty}\left(\arccos\left(\frac{k+1}{2k+1}\right)\ \right)$$
$$\lim_{k\to\infty}S(k)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\lim_{k\to\infty}\frac{k+1}{2k+1}\right)\ $$
$$\lim_{k\to\infty}S(k)= \arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$
Finally:
$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1}{3}\right)-\arccos\left(\frac{1}{2}\right)$$
that which is the same:
$$\sum_{n=1}^{\infty }\arccos\left(\frac{n(n+1)+\sqrt{(n+1)(n+2)(3n+1)(3n+4)}}{(2n+1)(2n+3)}\ \right)=\arccos\left(\frac{1+2\sqrt{6}}{6}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Tossing a biased coin We toss a biased coin $k$ times with the probability of tossing a head being $1/m$. We know that in the first two attempts there was at least one tail. What the probability that we tossed exactly $n$ heads?
Well, the tosses are independent so getting at least one tail in the first two attempts is the sum of $2\cdot (\tfrac{m-1}{m})^1 \cdot (\tfrac{1}{m})^2 = \tfrac{2m-2}{m^3}$ and $1\cdot (\tfrac{m-1}{m})^2 \cdot (\tfrac{1}{m})^0 = \tfrac{(m-1)^2}{m^2}$, which is $\tfrac{(m-1)(m^2-m+2)}{m^3}$. Thus getting at least on head in the first two attempts is $1-\tfrac{(m-1)(m^2-m+2)}{m^3}$.
Now we are left with $k-2$ tosses so we want to hit $n-1$ heads. The chance is $${k-2 \choose n-1}\tfrac{1}{m^{n-1}}\cdot (\tfrac{m-1}{m})^{k-2 - n +1}.$$
So the answer should be
$$1-\tfrac{(m-1)(m^2-m+2)}{m^3} + {k-2 \choose n-1}\tfrac{1}{m^{n-1}}\cdot (\tfrac{m-1}{m})^{k-1 - n}$$
Is this right?
| The probability that we tossed exactly n heads, given there was exactly 1 tail in the first two tries, is $Q_{1}={{k-2}\choose{n-1}}(\frac{m-1}{m})^{n-1}(\frac{1}{m})^{k-n-1}$, and the probability that we tossed exactly n heads, given there was exactly 2 tails in the first two tries, is $Q_{2}={{k-2}\choose{n}}(\frac{m-1}{m})^{n}(\frac{1}{m})^{k-n-2}$.
The probability that we had exactly $1$ tail in the first two tries, given that we know there was at least $1$ tail, is the conditional probability given by $P_{1}=\frac{P((exactly \ one \ tail)\cap(at \ least \ one \ tail))}{P(at \ least \ one \ tail)}=\frac{P(exactly \ one \ tail) }{P(at \ least \ one \ tail)}=\frac{2(\frac{1}{m})(\frac{m-1}{m})}{2(\frac{1}{m})(\frac{m-1}{m})+(\frac{m-1}{m})^{2}}$.
Similarly, the probability that we had exactly $2$ tails in the first two tries, given that we know there was at least $1$ tail, is given by $P_{2}=\frac{(\frac{m-1}{m})^{2}}{2(\frac{1}{m})(\frac{m-1}{m})+(\frac{m-1}{m})^{2}}$.
The total probability is $P_{1}Q_{1}+P_{2}Q_{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve system $\cos^2x+\cos^2y+\cos^2z=1,$ $\cos x+\cos y+\cos z=1,$ $x+y+z=\pi$ I want to solve the system:
$$\cos^2(x)+\cos^2(y)+\cos^2(z)=1,$$
$$\cos(x)+\cos(y)+\cos(z)=1,$$
$$x+y+z=\pi.$$
I tried to prove that only one of cosines can be not a zero, but I just prove that one or three cosines can not be zero.
I get, that $$\cos(x)~\cos(y)+\cos(x)~\cos(z)+\cos(y)~\cos(z)=0$$
Also after substituting
$$\cos(x) = 1 - \cos(y)-\cos(z),$$
we get $$\cos^2(y) + \cos^2(z)=\cos(y)+\cos(z) - \cos(y)\cos(z)$$
| Since $z=\pi-x-y$, you have $\cos z=\cos(x+y)$; then
\begin{align}
\cos^2x&+\cos^2y+\cos^2z
\\[6px]
&=\cos^2x+\cos^2y+\cos^2x\cos^2y-2\cos x\cos y\sin x\sin y+\sin^2x\sin^2y
\\[6px]
&=\cos^2x+\cos^2y+\cos^2x\cos^2y-2\cos x\cos y\sin x\sin y
\\
&\qquad+1-\cos^2x-\cos^2y+\cos^2x\cos^2y\\[6px]
&=1+2\cos x\cos y(\cos x\cos y-\sin x\sin y)\\[6px]
&=1+2\cos x\cos y\cos z
\end{align}
which is an interesting identity of its own.
In your case, we get $\cos x\cos y\cos z=0$.
Assume $\cos x=0$; then $x=\pi/2+m\pi$ and $\cos y+\cos z=1$. By squaring we get also $\cos y\cos z=0$, so $y=\pi/2+n\pi$ or $z=\pi/2+n\pi$.
In the first case, $\cos z=1$, so $z=2k\pi$ and we also need
$$
\frac{\pi}{2}+m\pi+\frac{\pi}{2}+n\pi+2k\pi=\pi
$$
so $m+n+2k=0$. Similarly in the second case and for $\cos y=0$ or $\cos z=0$.
If you also have the limitation that the angles are those of a (possibly degenerate) triangle, so in $[0,\pi]$, you get that two of the angles are $\pi/2$ and the other one is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ Find Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ in $\left [0 \:\: 2\pi\right]$
The equation can be written as:
$$\sin (2x)+1-2 \sin^2x+\sin x+\cos x=1$$ $\implies$
$$\sin x+\cos x=2\sin^2 x-2 \sin x\cos x$$ $\implies$
$$\sin x+\cos x=2\sin x\left(\sin x-\cos x\right)$$
$\implies$
$$\frac{\sin x+\cos x}{\sin x-\cos x}=2\sin x$$ $\implies$
$$\frac{1+\tan x}{1-\tan x}=-2\sin x$$
$$\tan \left(\frac{\pi}{4}+x\right)=-2\sin x$$
Now i have drawn the graphs of $\tan \left(\frac{\pi}{4}+x\right)$ and $-2\sin x$ and observed there are two solutions.
is there any other way?
| I think your reasoning with graphs is not so right because if so,
why not to draw the graph of $f(x)=\sin2x+\cos2x+\sin{x}+\cos{x}-1$?
By the Claude's hint from your equation
$$\frac{1+\tan{x}}{1-\tan{x}}=-2\sin{x}$$ after substitution $\tan\frac{x}{2}=t$ we obtain
$$\frac{1+\frac{2t}{1-t^2}}{1-\frac{2t}{1-t^2}}=-2\cdot\frac{2t}{1+t^2}$$ or
$$t^4+2t^3+8t^2-6t-1=0.$$
Now, let $$f(t)=t^4+2t^3+8t^2-6t-1.$$
Thus, $$f''(t)=12t^2+12t+16>0,$$ which says that $f$ is a convex function.
Hence, a graph of $f$ and the $t$-axis have two common points maximum.
But $f(0)<0$, which says that $f$ has two roots exactly and since the period of $\tan$ is $\pi$, we get that the starting equation has two roots exactly on $[0,2\pi]$.
Done!
| {
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Proof of an inequality by induction Prove using induction that
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}$$
for all positive whole numbers $n$.
I began by showing that it is true for $n=1$
I then assumed that it is true for $n=p$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} = \sum_{k=1}^p \frac{1}{k^2} \le 2-\frac{1}{p}$$
I now want to show that it is true for $n=p+1$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2}= \sum_{k=1}^{p+1} \frac{1}{k^2} $$
If I add $\frac{1}{(p+1)^2}$ to $\sum_{k=1}^{p} \frac{1}{k^2}$, I will then get
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2} \le 2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
If this is true then
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
or
$$-\frac{1}{(p+1)}=-\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{1}{(p+1)} -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)-(p+1)^2 + p}{p(p+1)^2}$$
$$0= \frac{(p^2 + p) - (p^2 + 2p + 1) + p}{p(p+1)^2}$$
$$0= \frac{-1}{p(p+1)^2}$$
This is invalid. I am not sure where I have made a mistake but I think it is
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
It then must be that
$$-\frac{1}{(p+1)} < -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0 \le \frac{1}{(p+1)} - \frac{1}{p} + \frac{1}{(p+1)^2}$$
$$ 0< \frac{-1}{p(p+1)^2}$$
which is true for all positive whole numbers $p$.
I am pretty sure it is proved now but I would be happy if someone can confirm this.
| I think the following induction a bit of better.
$$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1+1-\frac{1}{n}=2-\frac{1}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving the integral :$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2 x}}dx$ Solving the integral :
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=?$$
My try:
$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}\cdot \frac{1-\sqrt{\sin 2x}}{1-\sqrt{\sin 2x}}dx$$
$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{1-\sin 2x}dx$$
$$(\sin x-\cos x)^2=1-\sin 2x$$
So :
$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{(\sin x-\cos x)^2}dx$$
Now what?
$$$$
| The given integral equals
$$ \frac{1}{2}\int_{0}^{\pi}\frac{\sin\tfrac{x}{2}}{1+\sqrt{\sin x}}\,dx =\frac{1}{2\sqrt{2}}\int_{0}^{\pi}\frac{\sqrt{1-\cos x}}{1+\sqrt{\sin x}}\\=\frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1-\cos x}+\sqrt{1+\cos x}}{1+\sqrt{\sin x}}\,dx\\
=\frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sqrt{\sin x}}\,dx$$
or
$$ \frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}(1+\sqrt{x})}\,dx = \frac{1}{2}\int_{0}^{1}\frac{dx}{(1+\sqrt{x})\sqrt{1-x}}=\int_{0}^{1}\frac{x\,dx}{(1+x)\sqrt{1-x^2}}$$
which equals $\frac{\pi}{2}-\int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}$. On the other hand
$$ \int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}=\int_{0}^{\pi/2}\frac{d\theta}{1+\sin\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos\theta}=\int_{0}^{\pi/2}\frac{d\theta}{2\cos^2\tfrac{\theta}{2}}=1, $$
hence:
$$\boxed{\int_{0}^{\pi/2}\frac{\sin x}{1+\sqrt{\sin(2x)}}\,dx = \color{blue}{\frac{\pi}{2}-1},} $$
no particular elliptic integral like $K\left(\frac{1}{2}\right)$ or $E\left(\frac{1}{2}\right)$ is really involved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On the summation $\sum \limits_{n=1}^{\infty} \arctan \left ( \frac{1}{n^3+n^2+n+1} \right )$ Here is a problem that I ran into. I seriously doubt if there is a closed form but you never know.
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^\infty \arctan \left ( \frac 1 {n^3+n^2+n+1} \right) $$
I searched in vain to attack it using telescopic summation but I failed miserably. Then I remembered the following technique. Since ${\rm Im} \log (1+ix) = \arctan x$ we can express the sum as follows
\begin{align*}
\sum_{n=1}^\infty \arctan \left ( \frac{1}{n^3+n^2+n+1} \right ) &= \sum_{n=1}^\infty \arctan \left [ \frac{1}{(n+1)(n^2+1)} \right ] \\
&= \sum_{n=1}^\infty \operatorname{Im} \left [ \log \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] \\
&= \operatorname{Im} \log \left [ \prod_{n=1}^\infty \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ]
\end{align*}
I tried to combine it with the famous Euler product
$$ \frac{\sin \pi z}{\pi z} = \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) \tag{1} $$
but I see no connection. So, is there a possible way to evaluate it?
| We have
$$S = \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^3 + n^2 + n + 1}\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \left(1 + \frac{i}{n^3+n^2+n+1}\right)\right)\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \frac{n^3+n^2+n+1+i}{n^3+n^2+n+1}\right)\right) \\ = \Im\left(\ln\left(\frac{\pi\operatorname{csch}(\pi)}{1+i}\prod_{k=1}^{3}\frac{1}{\Gamma(r_k)}\right)\right)$$
where $r_k$ is the $k$th root of $x^3-x^2+x-1-i$ (order doesn't matter since the product iterates over all of them). The last step comes from employing equation 19 on here (which comes directly from the Weierstrass factorization product formula for $\Gamma(x)$), and the fact that $\Gamma(i)\Gamma(-i) = \pi\operatorname{csch}(\pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
What would the determinant of the following matrix be? " If $$\det\begin{pmatrix}a&1&d\\ b&1&e\\ c&1&f\end{pmatrix}=4$$
and $$\det
\begin{pmatrix}a&1&d\\ b&2&e\\ c&3&f\end{pmatrix}=3$$
What is $$\det \begin{pmatrix}a&-1&d\\ b&-3&e\\ c&-5&f\end{pmatrix}?$$"
This does not seem to fit into any of the regular changes in the values of matrices. A row does not seem to be multiplied. It does not seem like a row is being added to another. So how would I find the required determinant?
| $\det \begin{pmatrix}a&-1&d\\ b&-3&e\\ c&-5&f\end{pmatrix}=\det\begin{pmatrix}a&1-2\times1&d\\ b&1-2\times 2&e\\ c&1-2\times 3&f\end{pmatrix}=\det \begin{pmatrix}a&1&d\\ b&1&e\\ c&1&f\end{pmatrix}-2\det \begin{pmatrix}a&1&d\\ b&2&e\\ c&3&f\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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How to find the number of solutions for $x_1+2x_2+5x_3=10$? This is a question taken from Discrete mathematics by Kenneth Rosen:
Find the number of ways to make change for \$100 using \$10, \$20 and \$50 bills.
My approach:
Let number of \$10 notes be $x_1$.
Let number of \$20 notes be $x_2$.
Let number of \$50 notes be $x_3$.
Then the number of ways is equal to
number of solutions of $10x_1+20x_2+50x_3=100$ and so $x_1+2x_2+5x_3=10$.
Now I don't know how to find the number of solutions to this equation.
|
first variant
When looking for non-negative integral solutions $x_1,x_2,x_3$ we notice that the possible solutions of $x_3$ in
\begin{align*}
x_1+2x_2+5x_3=10\tag{1}
\end{align*}
are $x_3\in\{0,1,2\}$ since $0\leq 5x_3\leq 10$.
Setting these three values for $x_3$ we consider instead of (1) the three equations
\begin{align*}
x_1+2x_2&=10\\
x_1+2x_2&=5\\
x_1+2x_2&=0\\
\end{align*}
The first equation has $6$ admissible values $x_2\in\{0,1,2,3,4,5\}$, the second equation has $3$ admissible values $x_2\in\{0,1,2\}$ and the third equation has one admissible value $x_2\in\{0\}$. The value of $x_1$ is then uniquely determined.
The number of admissible solutions of (1) is
\begin{align*}
\color{blue}{6+3+1=10}
\end{align*}
second variant
We follow the comment from @JackDAurizio and use generating functions to find the number of admissible solutions.
Values of $x_3$ represent zero or more multiples of $5$ which can be encoded as
\begin{align*}
1+x^5+x^{10}+\cdots=\frac{1}{1-x^5}
\end{align*}
We argue similarly when considerung values of $x_1$ and $x_2$. Since the right hand side of (1) is $10$ we look for the coefficient of $x^{10}$ in
\begin{align*}
\frac{1}{1-x^5}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}
\end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series $A(x)$. This way we can write e.g.
\begin{align*}
[x^n]A(x)=[x^n]\sum_{j=0}^\infty a_jx^j=a_n
\end{align*}
We obtain
\begin{align*}
\color{blue}{[x^{10}]}&\color{blue}{\frac{1}{1-x^5}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}}\\
&=[x^{10}](1+x^5+x^{10})\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}\tag{2}\\
&=\left([x^{10}]+[x^5]+[x^0]\right)\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}\tag{3}\\
&=[x^{10}](1+x^2+x^4+x^6+x^8+x^{10})\cdot\frac{1}{1-x}\\
&\qquad +[x^5](1+x^2+x^4)\frac{1}{1-x}+[x^0]\frac{1}{1-x}\tag{4}\\
&=6+3+1\tag{5}\\
&\color{blue}{=10}
\end{align*}
showing the number of solutions is $10$.
Comment:
*
*In (2) we expand $\frac{1}{1-x^5}$ up to $x^{10}$ since other terms do not contribute to $[x^{10}]$.
*In (3) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (4) we expand $\frac{1}{1-x^2}$ similarly as we did in (2).
*In (5) we notice that we could work as we did in (3) and since $\frac{1}{1-x}=1+x+x^2+\cdots$ each term has a contribution of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $ T(n) = 2 T(\frac{n}{8}) + \Theta(\sqrt[3]{n}) $ I'm tasked to give the best possible asymptotic bounds for the following recurrence:
$$ T(n) = 2 T\left( \frac n 8 \right) + \Theta(\sqrt[3]{n}) $$
I got the following using the iterative method:
\begin{align} T(n) & = 2 T \left( \frac n 8 \right) + \Theta(\sqrt[3]{n}) \\
& = 2^2 T \left( \frac n {8^2} \right) + \Theta(\sqrt[3]{n}) \\
& \,\,\,\vdots \\
& = 2^k + \Theta(\sqrt[3]{n})
\end{align}
By setting $ k = \log_8 (n) $ we get:
$$ = 2 ^{\frac{\log_2(n)}{\log_2(8)}} + \Theta(\sqrt[3]{n}) $$
$$ = \sqrt[3]{2^{\log_2(n)}} + \Theta(\sqrt[3]{n}) $$
$$ = \Theta(\sqrt[3]{n}) $$
Is this correct ?
Thanks for any answers.
| The strictest proof goes like this. Following the definition of $\Theta(\sqrt[3]{n})$, there exists two constants $C_1$ and $C_2$, such that for all sufficiently large $n\geq n_0$, where $n_0$ is a constant (may depend on $C_1,C_2$)
$$C_1\sqrt[3]{n}\leq T(n)-2T(\frac{n}{8})\leq C_2\sqrt[3]{n}.$$
Then multiply everybody by $2$ and replace $n$ by $n/8$ to get
$$2C_1\sqrt[3]{\frac{n}{8}}\leq 2T(\frac{n}{8})-4T(\frac{n}{8^2})\leq 2C_2\sqrt[3]{\frac{n}{8}}.$$
Notice that $8=2^3$ is the critical case. So we have
$$C_1\sqrt[3]{n}\leq 2T(\frac{n}{8})-4T(\frac{n}{8^2})\leq C_2\sqrt[3]{n},$$
$$C_1\sqrt[3]{n}\leq 4T(\frac{n}{8^2})-8T(\frac{n}{8^3})\leq C_2\sqrt[3]{n},$$
$$\cdots\,\cdots\;\cdots\,\cdots\;\cdots\,\cdots\;\cdots\,\cdots\;$$
$$C_1\sqrt[3]{n}\leq 2^kT(\frac{n}{8^k})-2^{k+1}T(\frac{n}{8^{k+1}})\leq C_2\sqrt[3]{n}.$$
We can do this all the way until $\frac{n}{8^k}\geq n_0$ is unsatisfied, i.e., $k\leq\log_8\frac{n}{n_0}=\frac{1}{3}\,\log_2\frac{n}{n_0}$. Adding the inequalities up, we get
$$kC_1\sqrt[3]{n}\leq T(n)-2^{k+1}T(\frac{n}{8^{k+1}})\leq kC_2\sqrt[3]{n}.$$
Therefore, the complexity is given by
\begin{align}
T(n)&=\Theta(2^{k+1}T(\frac{n_0}{8}))+\Theta(k\sqrt[3]{n})\\
&=\Theta(\sqrt[3]{n})+\Theta(\sqrt[3]{n}\log_2n)=\Theta(\sqrt[3]{n}\log_2n).
\end{align}
So you missed a factor of $\,\log_2n\,$ by assuming $\,\Theta(\sqrt[3]{n})+\cdots+\Theta(\sqrt[3]{n})=\Theta(\sqrt[3]{n})$ ($\times$). We need to be careful when adding infinite terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Linearization of a Differential Equation Can someone please help me to linearize this system, which is given by the differential equation shown in the picture below.
All variables are expressed as deviations from initial values
(0, for all variable states (x, x)).
My task is to find the model described as the one shown in the attached picture. Find the matrix A, and display the linearized model in the matrix form just like in the picture.
Find the picture here
|
$$m \ddot x+2c(x^2-1) \dot x+kx=0. \tag{1}$$
We define $y:=\dot x$ and plugging this into (1) + some algebra yields
\begin{align} m \dot y+2c(x^2-1) y+kx=0 \\ \Longleftrightarrow \dot y=-\frac{2c}{m}(x^2-1)y-\frac{k}{m}x. \end{align}
To get back into the notation of your picture, we set $\mathbf{X}:=(x,y)^T$ and get the system
$$\dot{\mathbf{X}}=f(\mathbf{X})=\begin{pmatrix}y \\ -\frac{2c}{m}(x^2-1)y-\frac{k}{m}x \end{pmatrix}. $$
Now, as seen in the picture, the matrix $A$ is the Jacobian of $f$ evaluated at some point $\mathbf{X}_0:=(x_0,y_0)^T$, i.e.
$$A=\begin{pmatrix} 0 & 1 \\ -\frac{k}{m}-\frac{4c}{m} xy & -\frac{2c}{m}(x^2-1) \end{pmatrix}\Bigg|_{(x_0,y_0)}=\begin{pmatrix} 0 & 1 \\ -\frac{k}{m}-\frac{4c}{m} x_0y_0 & -\frac{2c}{m}(x_0^2-1) \end{pmatrix}$$
and the linearization of the ODE at some point $\mathbf{X}_0$ is given by
$$\dot{\mathbf{X}}=A\mathbf{X}.$$
For example around the origin we have
$$\dot{\mathbf{X}}=\begin{pmatrix} 0 & 1 \\ -\frac{k}{m} & \frac{2c}{m} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} y \\ -\frac{k}{m}x+\frac{2c}{m}y \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$.
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
| Here I use the formulas
$$\cos(x+y) = \cos x \cos y - \sin x \sin y$$
$$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$
So $$\cos 2\theta = \cos(\theta+\theta) = \cos^2 \theta -\sin^2 \theta = 2\cos^2 \theta - 1.$$
Similarly
\begin{align}
\cos 3\theta = \cos (2\theta+\theta) & = \cos 2\theta\cos \theta-\sin 2\theta \sin \theta \\
& = 2\cos^3 \theta-\cos \theta-2\sin^2 \theta \cos \theta \\
& = 4\cos^3 \theta -3\cos \theta.
\end{align}
Now if $\cos 3\theta = \cos 2\theta$ you get
$$4\cos^3 \theta - 2\cos^2 \theta - 3\cos \theta + 1 = 0$$
which is what you claim putting $z = \cos \theta$. It's easy to see that $z = 1$ is a solution, so
$$4z^3-2z^2-3z+1 = (z-1)(4z^2+2z-1) = \bigg(z-1\bigg)\bigg(z+\frac{1+\sqrt{5}}{4}\bigg)\bigg(z-\frac{-1+\sqrt{5}}{4}\bigg). $$
Now $\theta = \frac{2\pi}{5}$ satisfies $\cos 3\theta = \cos 2\theta$, and $\cos \frac{2\pi}{5}$ must be positive (because $\frac{2\pi}{5} < \frac{\pi}{2}$) and different from $1$ (because $\frac{2\pi}{5} \neq 0$). Hence
$$\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}.$$
| {
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"url": "https://math.stackexchange.com/questions/2505537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $x^2 \equiv 3$ (mod $q$) (where $q$ is an odd prime) have infinite solutions? Not sure how to prove/disprove this. One thought I had for proving this was doing an indirect proof, assuming there are only finitely many solutions $x_1,x_2,...,x_n$ and perhaps:
1) constructing a new solution using these solutions
or
2) taking the largest of the solutions and show an even larger solution exists
| There is an ambiguity in the question. I address the version of the question "For how many primes $q$ does $x^2 \cong 3 \pmod{q}$ have a solution?" (I ignore the other interpretation because either we recognize that $\mathbb{Z} / q \mathbb{Z}$ is a finite set of residue classes, so there are only finitely many classes which are solutions, or we realize that the solution classes each contain infinitely many integers. Either way, the answer depends on the question I choose to address.)
You as whether $3$ is a quadratic residue modulo $q$. It is known that $3$ is a quadratic residue modulo $q$ if $q \cong 1$ or $q \cong 11 \pmod{12}$ (excepting $q = 1$, which is not prime). We use the Legendre symbol to represent quadratic character:
$$ \left( \frac{3}{q} \right) = \begin{cases} 0, \text{if $q$ divides $3$} \\+1, \text{if $3$ is a quadratic residue modulo $q$ and $q$ does not divide $3$}\\ -1, \text{if $3$ is a quadratic nonresidue modulo $q$ and $q$ does not divide $3$} \end{cases} \text{.} $$
This may be shown using quadratic reciprocity. By quadratic reciprocity,\begin{align*}
\left( \frac{3}{q} \right) \left( \frac{q}{3} \right) &= (-1)^{\frac{3-1}{2} \cdot \frac{q-1}{2}} \\
&= (-1)^{\frac{q-1}{2}} \text{.} \\
\end{align*}
*
*If $q \cong 0 \pmod{3}$, then $q$ is not prime and we exclude it from consideration, since the question concerns odd primes, $q$.
*If $q \cong 1 \cong 1^2 \pmod{3}$, then $\left( \frac{q}{3} \right) = +1$. So
$$ \left( \frac{3}{q} \right) \left( \frac{q}{3} \right) = \left( \frac{3}{q} \right) = (-1)^{\frac{q-1}{2}} \text{.} $$ This last is $+1$ if and only if $q \cong 1 \pmod{4}$. Putting the two congruences for $q$ together using the Chinese Remainder Theorem, $q \cong 1 \pmod{12}$.
*If $q \cong 2 \pmod{3}$, then $\left( \frac{q}{3} \right) = -1$. So
$$ \left( \frac{3}{q} \right) \left( \frac{q}{3} \right) = -\left( \frac{3}{q} \right) = (-1)^{\frac{q-1}{2}} \text{.} $$ Then $\left( \frac{3}{q} \right) = +1$ if and only if $q \cong 3 \pmod{4}$ and combining congruences gives $q \cong 11 \pmod{12}$.
We have hereby demonstrated the claim that $3$ is a quadratic residue modulo $q$ if $q \cong 1$ or $q \cong 11 \pmod{12}$.
Edit:
Occurred to me that there is one more, perhaps not obvious, fact to use. Dirichlet's result on the distribution of primes among residue classes (also known as Dirichlet's theorem on arithmetic progressions) shows that there are infinitely many primes congruent to $1 \pmod{12}$ and congruent to $11 \pmod{12}$. Therefore, there are infinitely many such primes $q$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?
| Here is one approach. Note from the second equation you have
$$
3a^2 = 5/b - 13b^2
$$
which we can substitute into the first to get
$$
18 = a\left(a^2+39b^2\right)
= a\left(\frac{5}{3b} - \frac{13b^2}{3}+39b^2\right)
$$
so
$$
3a^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2
$$
which implies an equation in $b$,
$$
\frac{5}{b} - 13b^2 = \left(\frac{18\cdot 3}{\frac{5}{3b} - \frac{13b^2}{3}+39b^2} \right)^2,
$$
since both sides are equal to $3a^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the first same-proportion rectangle(s) whose both sides are also whole numbers? Let's elaborate the issue with an example:
I have an arbitrary rectangle. Let's choose the rectangle to be $5.6 \times 4.2$. The width-to-height ratio is $\frac 4 3$. The rectangle is in between the two whole-numbered rectangles of $4 \times 3$ and $8 \times 6$, both having the same ratio.
So, given an arbitrary rectangle, how do I find (if any) the closest* rectangle(s) whose sides are both whole numbers and also has (have) the same ratio?
* By closest, I mean the rectangle(s) whose sides have the same ratio, and both sides have a whole-numbered length that are the first possible whole numbers that form such a ratio.
| General algorithm
Let's say we start from a rectangle with side lengths $w$ and $h$ (in your case $w=5.6$ and $h=4.2$). Now you try to express $w/h$ as a rational number $p/q$ with $\gcd(p,q)=1$. In your case we have
$$\frac{5.6}{4.2}=\frac{56}{42}=\frac{\color{red}2\cdot 2\cdot 2\cdot \color{blue}7}{\color{red}2\cdot 3\cdot \color{blue}7}=\frac{2\cdot 2}{3}=\frac 43.$$
So we have $p=4$ and $q=3$. Define $\alpha=w/p$. In your case we find $\alpha=5.6/4=1.4$. Then compute
$$\check w
= p\cdot \lfloor \alpha\rfloor
= 4\cdot \lfloor1.4\rfloor
= 4\cdot1
= 4,
\qquad
\check h
= q\cdot \lfloor \alpha\rfloor
= 3\cdot \lfloor1.4\rfloor
= 3\cdot1
= 3,
$$
$$\hat w
= p\cdot \lceil\alpha\rceil
= 4\cdot \lceil 1.4\rceil
= 4\cdot 2
= 8,
\qquad
\hat h
= q\cdot \lceil \alpha\rceil
= 3\cdot \lceil 1.4\rceil
= 3\cdot 2
= 6.
$$
and so we found your nearest rectangles $(\check w,\check h)=(4,3)$ and $(\hat w, \hat h)=(8,6)$. And this works in general.
| {
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"url": "https://math.stackexchange.com/questions/2513697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding the convolution of two chi squared variables? If I have variables $X_i^2$ that are chi squared distributed with $1$ degree of freedom: $$f_{X_i^2}=\frac 1 {\sqrt{2\pi}}\frac 1 {\sqrt x}e^{-\frac x 2}$$
Then I want to derive the distribution of $Z^2=X_1^2+X_2^2$ where these $X's$ are i.i.d.
We do this by convolution:
$$f_{Z^2}=\int_0^\infty \frac 1 {\sqrt{2\pi}}\frac 1 {\sqrt x}e^{-\frac x 2}\frac 1 {\sqrt{2\pi}}\frac 1 {\sqrt {y-x}}e^{-\frac {y-x} 2}dx=\frac 1 {\sqrt{2\pi}^2}e^{-\frac {y} 2}\int_0^\infty \frac 1 {\sqrt x}\frac 1 {\sqrt {y-x}}dx$$
I think I can rewrite the latter integral as $$\int_0^\infty \frac {2 \sqrt x \sqrt y} {\sqrt x\sqrt {y-x}}d\left(\frac {\sqrt x} {\sqrt y}\right)=2\int_0^\infty \frac 1 {\sqrt {1-\frac x y}}d\left(\frac {\sqrt x} {\sqrt y}\right)$$
However, I don't know how to proceed from this to get the desired result.
| Let $u=\sqrt{x} \implies \frac{du}{dx}=\frac1{2\sqrt{x}} \implies 2\, du = \frac1{\sqrt{x}}\, dx$. Also $u^2=x$ implies $y-x=y-u^2$. So your integral is
$$\int_0^{\sqrt{y}}\frac{dx}{\sqrt{x(y-x)}}=\int_0^{\sqrt{y}}\frac{2}{\sqrt{y-x}}\cdot \frac1{2\sqrt{x}}\, dx=\int_0^{\sqrt{y}}\frac{2}{\sqrt{y-u^2}}\, du$$ as $0\le x\le \sqrt{y}$. This evaluates to $\left[2\sin^{-1}\left(\frac u{\sqrt{y}}\right)\right]_0^{\sqrt{y}}=2\left(\frac\pi2\right)=\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2516718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Maximum area of a triangle defined by a point $P$ inside another triangle (Try using ceva's theorem) Let $P$ be a point inside an acute triangle $\triangle ABC$ and let $D$, $E$, and
$F$ be the points of intersection of the lines $AP$, $BP$ and $CP$ with sides $BC$, $CA$ and $AB$, respectively. Determine $P$ so that the area of the triangle $\triangle DEF$ is maximal.
| Let $\frac{AF}{FB}=x$, $\frac{BD}{DC}=y$ and $\frac{CE}{EA}=z$.
Thus, by Cheva's theorem $xyz=1$ and
$$\frac{S_{\Delta AFE}}{S_{\Delta ABC}}=\frac{AF\cdot AE}{AB\cdot AC}=\frac{x}{(1+x)(1+z)},$$
$$\frac{S_{\Delta BFD}}{S_{\Delta ABC}}=\frac{BD\cdot BF}{BA\cdot BC}=\frac{y}{(1+y)(1+x)},$$
$$\frac{S_{\Delta CDE}}{S_{\Delta ABC}}=\frac{CD\cdot CE}{CB\cdot CA}=\frac{z}{(1+z)(1+y)}$$ and by AM-GM
$$\frac{S_{\Delta FDE}}{S_{\Delta ABC}}=1-\sum_{cyc}\frac{x}{(1+x)(1+z)}=$$
$$=\frac{\prod\limits_{cyc}(1+x)-\sum\limits_{cyc}(x+xy)}{\prod\limits_{cyc}(1+x)}=\frac{2}{\prod\limits_{cyc}(1+x)}\leq\frac{2}{8\sqrt{xyz}}=\frac{1}{4}.$$
The equality occurs for $x=y=z=1$, which gives that $P$ is is an intersect point of medians of the triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify solutions for $y'' + 4y = 2 \tan x$? The original equation is $$y'' + 4y = 2 \tan x$$
What I did so far:
$$\lambda^2+4 = 0$$
$$\lambda_1 = -2i \quad \lambda_2 = 2i$$
$$y(x) = C_1\cos2x+C_2\sin2x$$
Write down the system:
$$\begin{cases} C_1'\cos2x + C_2'\sin2x = 0 \\
2C'_2\cos2 x - 2C_1'\sin2x = 2\tan x \end{cases}$$
Applying Krammer's formulas
$$\Delta = \begin{vmatrix}\cos2x & \sin2x \\ 2\cos2x & -2\sin2x \end{vmatrix} = -2\cos4x$$
$$\Delta_2 = \begin{vmatrix} 0 & \sin2x \\ 2\tan x & -2\sin2x\end{vmatrix} = -2\tan x\sin2x = -4\sin^2x$$
$$\Delta_3 = \begin{vmatrix}\cos2x & 0 \\ 2\cos2x & 2\tan x \end{vmatrix} = -2\cos2x\tan x$$
but when I tried to find solutions I got different expressions which I do not know how to integrate, what is the best way to simplify solutions here to get better integrals?
| There is an error in your calculation of $\Delta$.
$$ \Delta = \begin{vmatrix}\cos(2x) &\sin(2x) \\ -2\sin(2x) &2\cos(2x)\end{vmatrix} = 2\cos^2(2x)+2\sin^2(2x) = 2$$
$$ C'_1 = \frac{\Delta_2}{\Delta} = -2\sin^2 x$$
$$ C_1 = -2\int sin^2 x \; dx = \frac{1}{2}\sin(2x) -x$$
$$ C'_2 = \frac{\Delta_3}{\Delta} = -\cos(2x)\tan (x)$$
\begin{align}
C_2
&= -\int \cos(2x)\tan (x) \; dx
=-\int \frac{(2\cos^2 x -1)\sin x}{\cos x} \, dx \\
&= \int \frac{\sin x}{\cos x} - \int 2\cos x \sin x\, dx
= -ln|\cos x| + \cos^2 x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding the roots of simultaneous equations like these
I am unable to find the answer to this question. It would also be helpful if you post what areas of mathematics this relates to.
| Expand $(a+b+c)^3$, giving
$$a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^3+3c^2b+3cb^2+6abc\\
=3(a^2+b^2+c^2)(a+b+c)-2(a^3+b^3+c^3)+6abc.$$
$$7^3=3\cdot35\cdot7-2\cdot151+6abc$$
$$abc=15.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this partial differential equation ? (PDE) (UPDATED) How can I solve this PDE?
$$ \frac{{∂}^2z}{∂x^2} - \frac{∂^2z}{∂x∂y} - 2\frac{∂^2z}{∂y^2} +6\frac{∂z}{∂x}- 9\frac{∂z}{∂y} +5z = e^{2x +y} + e^{x+y} $$
| The given equation is $(D^2-DD'-2D'^2+6D-9D'+5)z=e^{2x+y}+e^{x+y}$
The auxillary equation is obtained by setting $D=m, D'=1$ since there are no common factors. This gives $m^2-m-2+6m-9+5=0\implies m^2+5m-6=0$. Call the roots $c_1$ and $c_2$. So the homogeneous (complementary) solution is $z_c(x,y)=\phi_1(y+c_1x)+\phi_2(y+c_2x)$.
The particular integral is $$\dfrac{1}{(D^2-DD'-2D'^2+6D-9D'+5)}(e^{2x+y})=\dfrac{1}{4-2-2+12-9+5}e^{2x+y}=\dfrac{1}{8}e^{2x+y}$$
For the 2nd term the denominator becomes $0$. In that case differentiate $F(D,D')$ wrt $D$ and multiply by $x$. This means, we have
$$\dfrac{1}{(2D-D'+6)}(x\cdot e^{x+y})=\dfrac{x}{7}e^{x+y}$$
So the final solution is $z(x,y)=\phi_1(y+c_1x)+\phi_2(y+c_2x)+\dfrac{1}{8}e^{2x+y}+\dfrac{1}{7}x\cdot e^{x+y}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Fourier series of $f(x) = 1$ for $x\in[-\pi,0]$, $f(x) = -1$ for $x\in[0,\pi]$. I am attempting to solve a Fourier series problem where we have the question defined by a piecewise function:
$$f(x) = \begin{cases} 1 & -\pi \leq x \leq 0 \\ -1 & 0 \leq x \leq \pi \end{cases}$$
I can solve it out where I calculate that $A_{n} = 0$, and $A_{0} = 1$, and I then integrate
$$B_{n} = \frac{2}{\pi} \, (1-\cos(\pi \,n))$$
At this point when trying to solve for an equation, it doesn't match up to any of the example solutions online I can find.
| Just for the Fourier series:
\begin{align}
f(x) &= A_{0} + \sum_{n=1}^{\infty} \left( A_{n} \, \cos(n x) + B_{n} \, \sin(n x) \right) \\
A_{0} &= \frac{1}{2 \pi} \, \int_{-\pi}^{\pi} f(x) \, dx \\
A_{n} &= \frac{1}{\pi} \, \int_{-\pi}^{\pi} f(x) \, \cos(n x) \, dx \\
B_{n} &= \frac{1}{\pi} \, \int_{-\pi}^{\pi} f(x) \, \sin(n x) \, dx.
\end{align}
With
$$f(x) = \begin{cases} 1 & -\pi \leq x \leq 0 \\ -1 & 0 \leq x \leq \pi \end{cases}$$
then
\begin{align}
A_{0} &= \frac{1}{2 \pi} \, \left[ \int_{-\pi}^{0} (1) \, dx + \int_{0}^{\pi} (-1) \, dx \right] = 0 \\
A_{n} &= 0 \\
B_{n} &= \frac{2}{\pi \, n} \, (1 - \cos(n \pi)) = \frac{2 \, (1 - (-1)^{n})}{n \pi }
\end{align}
and
$$f(x) = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1 - (-1)^{n}}{n} \, \sin(n x) = \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{\sin(2n+1)x}{2n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculation of $\int \sin^2 x\frac{\sin x-\cos x}{\sin x+\cos x}dx$
Calculation of $\displaystyle \int\frac{\sin^2 x\cos x}{
\sin x+\cos x}dx$ and $\displaystyle \int^{\pi}_{0}\frac{1}{1-2a\cos x+a^2}dx, ,0<a<1$
$\bf{Attempt}$ For (a) $\displaystyle \frac{1}{2}\int\frac{\sin^2 x\bigg[(\sin x+\cos x)+(\sin x-\cos x)\bigg]}{\sin x+\cos x}dx $
$\displaystyle =\frac{1}{4}\int (1+\cos 2x)dx+\frac{1}{2}\int \sin^2 x\frac{\sin x-\cos x}{\sin x+\cos x}dx$
could some help me how to solve $\displaystyle \int \sin^2 x\frac{\sin x-\cos x}{\sin x+\cos x}dx$
For (b) Can we solve it any geometrical way like taking unit circle . thanks
| For (a),
\begin{eqnarray}
&&\int\frac{\sin^2 x\cos x}{\sin x+\cos x}dx\\
&=&\int \frac{\sin^2x(\cos^2x-\sin x\cos x)}{\cos^2x-\sin^2x}dx\\
&=&\frac14\int \frac{(1-\cos(2x))(1+\cos(2x)-\sin(2x))}{\cos(2x)}dx\\
&=&\frac14\int\frac{1-\cos^2(2x)-(1-\cos(2x))\sin(2x)}{\cos(2x)}dx\\
&=&\frac14\int(\sec(2x)-\cos(2x)-\tan(2x)+\sin(2x))dx\\
&=&\frac14(\frac12\ln(\frac{\cos x+\sin x}{\cos x-\sin x})-\frac12\sin(2x)+\frac12\ln(\cos (2x))-\frac12\cos(2x))+C\\
&=&\frac18(\ln(\frac{\cos x+\sin x}{\cos x-\sin x})-\sin(2x)+\ln(\cos(2 x))-\cos(2x))+C\\
&=&\frac18(2\ln(\cos x+\sin x)-\sin(2x)-\cos(2x))+C.
\end{eqnarray}
For (b), let $t=\tan\frac{x}{2}$ and then
\begin{eqnarray}
&&\int^{\pi}_{0}\frac{1}{1-2a\cos x+a^2}dx\\
&=&\int^{\pi}_{0}\frac{1}{1-2a\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}+a^2}dx\\
&=&\int_0^\infty\frac{1}{1-2a\frac{1-t^2}{1+t^2}+a^2}\frac{2}{1+t^2}dt\\
&=&\int_0^\infty\frac{1}{(1+a)^2t^2+(1-a)^2}dt\\
&=&\frac{2}{(1+a)^2}\int_0^\infty\frac1{t^2+(\frac{(1-a}{1+a})^2}dt\\
&=&\frac{2}{(1+a)^2}\frac{1+a}{1-a}\arctan\frac{t}{\frac{1-a}{1+a}}\bigg|_{0}^\infty\\
&=&\frac{\pi}{1-a^2}.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$L^1$ norm of derivative of Dirichlet kernel Let $D_n(x)$ denote the $n$-th Dirichlet kernel. We all know that $D_n$ has $L^1$ norm $O(\log n)$ (over a full period). But I'm wondering about the $L^1$ norm of its derivative. Is it $O(n)$ or $O(n \log n)$? Would anyone know a proof? Thanks.
Edit: Here's a proof that it is at least $O(n \log n)$ (I wonder if one could do better):
An application of Abel's summation formula shows that
$$
-iD_n'(x) := \sum_{|m| \leq n}me^{imx} = nD_n(x) - 1 - \int_1^n D_{\lfloor t \rfloor}(x)dt
$$
and so
$$
|D_n'(x)| \leq n|D_n(x)| + 1 + \int_1^n |D_{\lfloor t \rfloor}(x)|dt.
$$
Now integrate the above over $x \in [0,2\pi]$ to get $O(n \log n)$.
| By symmetry, it suffices to look at the interval $[0,\pi]$. From
$$D_n(t) = \frac{\sin \bigl(\bigl(n+\frac{1}{2}\bigr)t\bigr)}{\sin \frac{t}{2}}$$
(with the appropriate interpretation if $t = 0$) we can read off the zeros of $D_n$, they are $z_k = \frac{k\pi}{n+\frac{1}{2}}$ for $1 \leqslant k \leqslant n$. Between $z_k$ and $z_{k+1}$, $D_n$ has constant sign, and its modulus attains a maximum $M_k$, so
$$\int_{z_k}^{z_{k+1}} \lvert D_n'(t)\rvert\,dt = 2M_k.$$
Since the denominator is strictly increasing on $[0,\pi]$ and the modulus of the numerator is bounded by $1$, with that bound attained at $w_k = \frac{k+\frac{1}{2}}{n+\frac{1}{2}}\pi$ and nowhere else in $[z_k,z_{k+1}]$, we have
$$\frac{1}{\sin w_k} < M_k < \frac{1}{\sin z_k}.$$
On $\bigl(0,\frac{\pi}{2}\bigr)$ we have
$$0 < \frac{1}{\sin x} - \frac{1}{x} < \sin x < x.$$
The first inequality follows from $\sin x < x$ there, and the second from
\begin{align}
&&x\cos x < x &< \tan x \\
&\implies& \cos^2 x = 1 - \sin^2 x &< \frac{\sin x}{x} \\
&\implies& 1 - \frac{\sin x}{x} &< \sin^2 x \\
&\implies& \frac{1}{\sin x} - \frac{1}{x} &< \sin x.
\end{align}
Thus
$$2\sum_{k = 1}^{n-1} \frac{1}{w_k} < \int_{z_1}^{z_n} \lvert D_n'(t)\rvert\,dt < 2\sum_{k = 1}^{n-1} \biggl(\frac{1}{z_k} + z_k\biggr).$$
Now we can "explicitly" compute those sums,
\begin{align}
\sum_{k = 1}^{n-1} \frac{1}{w_k} &= \frac{2n+1}{\pi} \sum_{k = 1}^{n-1} \frac{1}{2k+1} = \frac{2n+1}{\pi}\biggl(-1 + H_{2n}- \frac{1}{2}H_n\biggr) \\
&= \frac{2n+1}{\pi}\biggl(\frac{1}{2}\log n + \log 2 + \frac{1}{2}\gamma - 1 + O(n^{-1})\biggr), \\
\sum_{k = 1}^{n-1} z_k &= \frac{\pi}{n+\frac{1}{2}}\frac{n(n-1)}{2} = \frac{\pi n(n-1)}{2n+1}, \\
\sum_{k = 1}^{n-1} \frac{1}{z_k} &= \frac{n+\frac{1}{2}}{\pi} H_{n-1} = \frac{n+\frac{1}{2}}{\pi}\bigl(\log n + \gamma + O(n^{-1})\bigr)
\end{align}
and obtain
$$\frac{2n+1}{\pi}\bigl(\log n + \gamma - 2(1-\log 2) + O(n^{-1})\bigr) < \int_{z_1}^{z_n} \lvert D_n'(t)\rvert\,dt < \frac{2n+1}{\pi}\bigl(\log n + \gamma + O(n^{-1})\bigr).$$
With $D_n(0) = 2n+1$ and $D_n(\pi) = (-1)^n$ we have
$$\int_0^{z_1} \lvert D_n'(t)\rvert\,dt = 2n+1\quad\text{and}\quad \int_{z_n}^{\pi} \lvert D_n'(t)\rvert\,dt = 1,$$
so
$$\lVert D_n'\rVert_{L^1([-\pi,\pi])} = 2\int_0^{\pi} \lvert D_n'(t)\rvert\,dt = \frac{4}{\pi}n\log n + O(n).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to put a complex number into the form a+ib if it has an exponent So I'm trying to put this into the form a+ib
$(1 + i)^{1000}$
(Hint: Use the polar form of the number).
I know the polar form without the exponent would be
$√2(cos(π/4)+isin(π/4))$
Do i just throw the exponent on at the end? I'm not sure how this helps get it into a+ib form
| Apropos of the answers from other posters applying DeMoivre's Theorem, the problem can be thought of in terms of the geometrical interpretation of multiplication between complex numbers. When we multiply a number $ \ a + bi \ = \ \sqrt{a^2 + b^2}·cis(\theta) \ \ $ by a second number $ \ c + d \ = \ \sqrt{c^2 + d^2}·cis(\phi) \ \ , \ $ with $ \ \theta \ = \ \arctan\left(\frac{b}{a} \right) \ $ and $ \ \phi \ = \ \arctan\left(\frac{d}{c} \right) \ $ (adjusted to the appropriate quadrants), the product $$ \sqrt{a^2 + b^2} \ · \ \sqrt{c^2 + d^2} \ · \ cis(\theta + \phi) $$
can be regarded as the "re-scaling" (dilation) of the length $ \ \sqrt{a^2 + b^2} \ $ of a vector representing the first number by the factor $ \ \sqrt{c^2 + d^2} \ $ and a counter-clockwise rotation of the direction $ \ \theta \ $ of that vector by an angle $ \ \phi \ $ (or vice versa, of course, with the first number acting on the "vector" of the second number).
So for $ \ (1 + i)^2 \ = \ (1 + i)·(1 + i) \ \ , \ $ we can treat this as a vector of length $ \ \sqrt2 \ $ pointing in the direction $ \ \frac{\pi}{4} \ $ having its length multiplied by $ \ \sqrt2 \ $ and its direction rotated counter-clockwise by $ \ \frac{\pi}{4} \ \ , \ $ producing a "product vector" with length $ \ 2 \ $ and direction $ \ \frac{\pi}{2} \ \ , \ $ thus
$ \ 2 \ cis \left(\frac{\pi}{2} \right) \ \ $ or $ \ 0 + 2i \ \ . $ Repeated multiplication for $ \ (1 + i)^n \ $ creates a succession of vectors, the "tips" of which lie on an equiangular (or "logarithmic") spiral, with each new vector being $ \ \sqrt2 \ $ times longer than its predecessor and pointed $ \ \frac{\pi}{4} \ $ radians counterclockwise relative to that predecessor. We conclude that $ \ (1 + i)^{1000} \ \ $ corresponds to a vector with length of $ \ (\sqrt2)^{1000} \ = \ 2^{500} \ $ and direction $ \ \frac{1000·\pi}{4} \ = \ 250 \pi \ \equiv \ 0 \ \ , \ $ which is to say $ \ (1 + i)^{1000} \ = \ 2^{500} \ cis \ 0 \ = \ 2^{500} + 0·i \ \ . $
$$ \ \ $$
Alternatively, we can examine the implied "binomial-power". We calculate $ \ ( 1 + 1 )^n \ $ by summing the terms $ \ \binom{n}{k}·1^k·1^{n - k} \ $ using the rows of the (Yang Hui/Pingala/Khayyam/Tartaglia/...) Pascal triangle,
$ 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \ = \ 1 \ = \ 2^0 $
$ 1 \ \ + \ \ 1 \quad \quad \quad \quad \quad \quad \ \ \ \ = \ 2 \ = \ 2^1 $
$ 1 \ \ + \ \ 2 \ \ + \ \ 1 \quad \quad \quad \quad \ \ = \ 4 \ = \ 2^2 $
$ 1 \ \ + \ \ 3 \ \ + \ \ 3 \ \ + \ \ 1 \quad \quad = \ 8 \ = \ 2^3 \ \ , \ $ etc.
as a way to demonstrate that $ \ \sum_{k=0}^{n} \ \binom{n}{k} \ = \ 2^n \ \ . $ If we do something similar for the terms $ \ \binom{n}{k}·1^k·i^{n - k} \ $ of $ \ (1 + i)^n \ \ , \ $ we have (for the "even" rows)
$ \require{cancel} \cdots $
$ \cancel{1} \ \ + \ \ 2·i \ \ + \ \ \cancel{1·i^2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ = \ 2i $
$ 1 \ \ + \ \ \cancel{4·i} \ \ + \ \ 6·i^2 \ \ + \ \ \cancel{4·i^3} \ \ + \ \ 1·i^4 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ = \ -4 \ = \ (2i)^2 $
$ \cancel{1} \ \ + \ \ 6·i \ \ + \ \ \cancel{15·i^2} \ \ + \ \ 20·i^3 \ \ + \ \ \cancel{15·i^4} \ \ + \ \ 6·i^5 \ \ + \ \ \cancel{1·i^6} \quad = \ -8i \ = \ (2i)^3 $
$ 1 \ \ + \ \ \cancel{8·i} \ \ + \ \ 28·i^2 \ \ + \ \ \cancel{56·i^3} \ \ + \ \ 70·i^4 \ \ + \ \ \cancel{56·i^5} \ \ + \ \ 28·i^6 \ \ + \ \ \cancel{8·i^7} \ \ + \ \ 1·i^8 \ = \ 16 \ = \ (2i)^4 \ \ , \ \text{etc.} $
So the binomial theorem establishes that $ \ (1 + i)^{2n} \ = \ (2i)^n \ \ , \ $ which Michael Rozenberg applies in his answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I found a correlation between n^x and x! can anyone explain it? This is the example to numbers to the 3rd.
Here is the same thing just with numbers to the 4th
Well I was messing around with numbers I made this discovery, with the correlation between n^x and x!. What is the reason for this? Is it easily explainable?
In the sheets every column is found from the differences of two numbers in sequence in the last column
| The binomial theorem says
$$
(a+b)^x = a^x + xa^{x-1}b + \frac{x(x-1)}2 a^{x-2} b^2 + \frac{x(x-1)(x-2)} 6 a^{x-3} b^3 + \cdots.
$$
The differences in the column to the right of the one that lists values of $a^x$ are
$$
(a+1)^x - a^x = \left( a^x + x a^{x-1} + \frac{x(x-1)} 2a^{x-2} + \cdots \right) - a^x.
$$
Since $a^x$ cancels, you have $xa^{x-1}$ as the highest-degree term in the resulting polynomial function of $a.$
Then for the next row, for the same reason you get $x(x-1)x^{x-2}$ as the highest-degree term.
And next you get $x(x-1)(x-2)a^{x-3},$ and so on until finally you have $x(x-1)(x-2)\cdots 1 a^{x-x}.$
| {
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About factoring and simplifying the following algebra expression... today a problem like the following was sent to me by a friend: $$\dfrac{(a^3+a+10)}{(a^3+3a^2+a-2)}\cdot\dfrac{(a^2-2a+5)}{(a^4-3a^2+1)}$$ find the simplest form of the expression above. The answer as the key suggests turns out to be $a^2-a-1$.
I've tried to factor out $(a+1)$ but there were too many factors. It is possible to do it using the polynomial division. But it will take too long. There is a general problem about the fact that these questions take too long. Is there any way to simplify only using the factors of the variables?
What are your suggestions?
| Probably there is a typo. The reducible form occurs if the operation is a division, not a multiplication. Accordingly, inverting the numerator and the denominator of the second fraction, the expression becomes
$$\dfrac{(a^3+a+10)}{(a^3+3a^2+a-2)}\cdot\dfrac{( a^4-3a^2+1 )}{( a^2-2a+5 )}$$
and taking into account that
$$a^3+a+10=(a+2)(a^2-2a+5)$$
$$a^3+3a^2+a-2=(a+2)(a^2+a-1)$$
$$a^4-3a^2+1 =(a^2-a+1)(a^2+a-1)$$
we get that the whole expression reduces to $a^2-a+1\,\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$
My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?
| Hint:
$$4f(x)-5\sqrt3=\sqrt{16-(x^2-5)^2}+\sqrt3(x^2-5)$$
Now set $x^2-5=4\cos t,0\le t\le\dfrac\pi2$
We can prove $$a\cos t+b\sin t\le\sqrt{a^2+ b^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does there exist a closed form for the sinc function series $\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}}$? Here I want to get the closed form solution of the following summation
$$
\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}} \qquad(1)
$$
Or the more general form ($x$ be an arbitrary real number, and $a\geq0$ is a constant):
$$
f_a(x) = \sum_{n=1}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(2)
$$
I tried the numeircal simulations before I post the question. I truncated the first $1,000,000$ terms of equation (1) and it turned $0.781233190560320$.
Anyone can help me?
In fact, I made it the reduced case when $a=0$. It can be proved by Fourier series:
$$
f_0(x)=\sum_{n=1}^\infty \frac{\sin nx}{n} = \left\{ \matrix{\dfrac{\pi-x}{2}, 0<x<2\pi\\0, x=0,2\pi} \right.
$$
And the function is periodical:
$$
f_0(x) = f_0(x+2\pi)
$$
Edit: How about this one?
$$
g_a(x) = \sum_{n=1}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(3)
$$
We get the "diverging wave solution" in physics when we combine the equation (2) and (3):
$$
h_a(x)=g_a(x)+\text{i}f_a(x)=\sum_{n=1}^\infty \frac{\exp\left(\text{i}x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(4)
$$
Edit #2:
I tested the solution solved by Random Variable (see the most ranked answer and thousands thanks to it!) compared with the truncating results:
$$
\sum_{n=1}^N\frac{\sin \left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}, N = 1,000,000, a=1
$$
Here is the solution by @Random Variable:
the solution of equation (2) as the following:
$$
\sum_{n={1}}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}
=
\frac{\pi}{2} J_0(ax) -\frac{\sin(ax)}{2a}, a>0, 0<x<2\pi\qquad(2*)
$$
where $J_0(ax)$ is the Bessel function of the first kind of order zero.
Here is the comparison:
It can be found that the both agree well when $0 <x<2\pi$,but differ in other domain. So, how about the solution beyond $(0,2\pi)$?
Edit #3:
Inspired by Random Variable's answer, I found the solution of equation (3) as the following:
$$
\sum_{n={1}}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}
=
-\frac{\pi}{2} Y_0(ax) -\frac{\cos(ax)}{2a}, a>0, 0<x<2\pi\qquad(3*)
$$
where $Y_0(ax)$ is the Bessel function of the second kind of order zero.
Here is the comparison:
Note that equation (3) is divergent when $x=0$.
Possible relating QUESTIONS:
Does there exist a closed form for the non-integer shifted sinc-function series: $\frac{\sin(n+a)x}{(n+a)x}$?
| COMMENT to users: Achille-Hui and Random-Variable
How about this ?:
$$\sum _{n=0}^{\infty } \frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\sum _{n=0}^{\infty } \mathcal{L}_x\left[\frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\right](s)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } \frac{1}{a^2+n^2+s^2}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}+\frac{\pi \sqrt{-a^2-s^2} \cot \left(\pi \sqrt{-a^2-s^2}\right)}{2 \left(a^2+s^2\right)}\right](x)=\frac{1}{2}
\pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}\right](x)+\mathcal{L}_s^{-1}\left[-\frac{\pi \cot \left(\pi \sqrt{-a^2-s^2}\right)}{2
\sqrt{-a^2-s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
for $a>0$ and $s>0$
$$\frac{\sin (a x)}{2 a}+\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2
a}$$
$$\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)$$
$$\mathcal{L}_x\left[\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)\right](s)=\mathcal{L}_x\left[\frac{1}{2} \pi
J_0(a x)\right](s)$$
$$\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\neq \frac{\pi }{2 \sqrt{a^2+s^2}}$$
and then:
$$\sum _{n=0}^{\infty } \frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\neq \frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$
EDITED:
Comparison a numeric InverseLaplaceTransform and Bessel function:
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)}$ I want to solve the following problem:
Find $B \in \mathbb{R}$ such that
$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} \in \left] \frac{1}{2}, 2 \right[ \quad.$$
My attempt is as follows:
$$ \left( \frac{2}{\sqrt{x^2 + 4x} - x} \right) ^{B(x+2)} = \frac{2^{B(x+2)}}{2^{B(x+2)\log_{2} \left( \sqrt{x² + 4x} - x \right)}} = 2^{B(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)}.$$
Thus,
$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} = 2^{B \lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)} $$
and I'm left with
$$\lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right) = \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}}$$
that I can compute using L'Hôpital's rule and really boring computations, arriving at
$$\tag{*} \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}} = \frac{1}{\log(2)}.$$
From here I'd compute a valid interval for $B$.
Question:
Computing (*) is a horrible hassle. Is there an easier way to solve this problem?
| $$\frac{2}{\sqrt{n^2+4n}-n}=1+\frac{\sqrt{n^2+4n}-n}{2n}$$
$$=1+\frac{a_n}{n}$$ where $a_n\to 1$
It follows that
$$(1+\frac{a_n}{n})^{n+2}\to e.$$ Does this help ?
| {
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The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
If $a=2$,what is the solution set of the equation?
Options
*
*{$3,6$}
*{$2$}
*{$3$}
*{$6$} Correct Answer
I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equation, you should get:
For 3:
$\sqrt {3-2} = 3-4$
Since $\sqrt {1} = ±1$
$±1 = -1$
For 6:
$\sqrt {6-2} = 6-4$
Since $\sqrt {4} = ±2$
$±2 = 2$
For the answer, they(SAT) evaluated $\sqrt {1}$ as $\sqrt {1} = -1$ and $\sqrt {4}$ as $\sqrt {4} = 2$ Why is it that $\sqrt {1}$ is equal to $-1$ and not $1$ and why $\sqrt {4}$ is equal to $2$ and not $-2$ Why isn't the solution set {$3,6$} a correct answer?
| The standard way to solve equations with square roots is to use this rule:
$$ \sqrt A=B\iff (A=B^2\quad\textbf{and}\quad B\ge 0). $$
So here you obtain
$$\sqrt{x-2}=x-4\iff x-2=x^2-8x+16\;\text{and}\; x\ge 4\iff(x-3)(x-6)=0\;\text{and}\; x\ge 4,$$
which shows there's only one root: $6$.
| {
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Number of seven digit numbers without repetition of digits divisible by $3$ Find number of seven digit numbers divisible by $3$ with
$1.$ Repetition
$2.$ Without Repetition
For Part $1.$ The least seven digit number divisible by $3$ is $1000002$ and highest seven digit number is $9999999$
So total is $3000000$
For Part $2$ The least such number is $1023456$ and the next is $1023459$
Now sum of the digits of $1023456$ is $21$.
So if we include $7$ by removing one of the digits in $1023456$ then the digits that can be removed is $1$ or $4$.
If we include $8$ then digits that can be removed is $2$ or $5$.
but we get more cases here, any better way to approach?
| $(1)$ Number of possible digit combinations without $0$ digit: $\binom{9}{7}=36 $
$(2)$ Number of possible digit combinations including $0$ digit: $\binom{9}{6}=84 $
So there are $36 \cdot 7!+84 \cdot 6 \cdot 6!=544320$ numbers without digit repetition.
Now we have to find the numbers divisible by $3$.
We can seperate the cases:
$(1.a)$ $1$ digit divisible by $3$, $6$ digits not divisible by $3$: $\binom{3}{1}\binom{6}{6}=3\cdot1=3$
Out of the $\binom{6}{6}=1$ case
$$(1,2,4,5,7,8)$$
$1$ is divisible by $3$, so there are $\binom{3}{1}\cdot 1=3$ numbers divisible by $3$.
$(1.b)$ $2$ digits divisible by $3$, $5$ digits not divisible by $3$: $\binom{3}{2}\binom{6}{5}=3\cdot6=18$
Out of the $\binom{6}{5}=6$ cases
$$(1,2,4,5,7),(1,2,4,5,8),(1,2,4,7,8),(1,2,5,7,8),(1,4,5,7,8),(2,4,5,7,8)$$
$0$ is divisible by $3$, so there are $\binom{3}{2}\cdot 0=0$ numbers divisible by $3$.
$(1.c)$ $3$ digits divisible by $3$, $4$ digits not divisible by $3$: $\binom{3}{3}\binom{6}{4}=1\cdot15=15$
Out of the $\binom{6}{4}=15$ cases
$$(1,2,4,5),...,(4,5,7,8)$$
$9$ is divisible by $3$, so there are $\binom{3}{3}\cdot 9=9$ numbers divisible by $3$.
So in the $(1)$ case there are $(3+0+9)\cdot 7!=12\cdot 7!=60480$ numbers divisible by $3$.
Similarly we can find that in the $(2)$ case there are $=30\cdot 6\cdot 6!=129600$ numbers divisible by $3$.
Maybe it's not the best approach and for some numbers we have to test the 3-divisibility, but we got the answer: $60480+129600=190080$
| {
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How to factorize polynomials to the 5th degree? I have the polynomial:
$$2x^5-x^4+10x^3-5x^2+8x-4$$
and I know that the final result is:
$$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$
But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a common factor between the expressions but in this case their are none.
| The question slightly wrong in my opinion. Please try multiplying the factors or the eqn. and lets see if you will get back the original question.
The correct equation is $P(x): 2x^5-x^4-10x^3+5x^2+8x-4$
Group and Factorize: $(2x^5-x^4) (-10x^3+5x^2) (8x-4) =0$
$x^4(2x-1) -5x^2(2x-1) 4(2x-1)$
Factor out $(2x-1)$ then we have $x^4-5x^2+4$
Our first factor of the equation is $(2x-1)$
This looks like a quadratic form but our first coefficient has 4th degree
look let's expand $x^4-5x^2+4$ and we have $(x^2-4)$ and $(x^2-1)$. $-4$ and $-1$ when added will be $-4-1 = -5$ and when multiply $-4x-1 = 4. x^2 x x^2 = x^4$
so its just splitting them and distributing them on $-4$ and $-1$ which are the product of the constant and sum of the middle term.
Now with $(x^2-4)$ and $(x^2-1)$ we can have $(x-2) (x+2)$ and $(x-1) (x+1)$.
so our final factors of the polynomial are $(2x-1) (x-2) (x+2) (x-1) (x+1)$
Therefore; the zeros of the polynomial are $1/2, 2, -2, 1, -1$.
| {
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Using Euclidean Algorithm in RSA? it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification.
it leverages multiplication and subtraction, which humans are fairly good at, to make fractions like 15996751/3870378 reducible. Also useful in scaling equations down to their simplest integer representation in one step, given with extra integers, GCD(C,GCD(A,B)) is equivalent to GCD(A,B,C). It's been around for over two thousand years and mathematicians have bent it to many purposes , but just the first bit should be justification.
ind a and b?
| Here is the continued fraction way of writing this. It is, of course, equivalent to the "back-substitution" way one often sees. The convergents can bbe written in a single row of fractions, as below.
$$ \gcd( 1680, 71 ) = ??? $$
$$ \frac{ 1680 }{ 71 } = 23 + \frac{ 47 }{ 71 } $$
$$ \frac{ 71 }{ 47 } = 1 + \frac{ 24 }{ 47 } $$
$$ \frac{ 47 }{ 24 } = 1 + \frac{ 23 }{ 24 } $$
$$ \frac{ 24 }{ 23 } = 1 + \frac{ 1 }{ 23 } $$
$$ \frac{ 23 }{ 1 } = 23 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccc}
& & 23 & & 1 & & 1 & & 1 & & 23 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 23 }{ 1 } & & \frac{ 24 }{ 1 } & & \frac{ 47 }{ 2 } & & \frac{ 71 }{ 3 } & & \frac{ 1680 }{ 71 }
\end{array}
$$
$$ $$
$$
\begin{array}{ccc}
\frac{ 1 }{ 0 } & \mbox{digit} & 23 \\
\frac{ 23 }{ 1 } & \mbox{digit} & 1 \\
\frac{ 24 }{ 1 } & \mbox{digit} & 1 \\
\frac{ 47 }{ 2 } & \mbox{digit} & 1 \\
\frac{ 71 }{ 3 } & \mbox{digit} & 23 \\
\frac{ 1680 }{ 71 } & \mbox{digit} & 0 \\
\end{array}
$$
$$ 1680 \cdot 3 - 71 \cdot 71 = -1 $$
| {
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Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit
I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
| If $\gcd(a,10) = 1$ then $a^{\phi(10)} = a^4 \equiv 1\mod 10$. So $a^{4n+1} = (a^4)^n*a\equiv 1^na \equiv a \mod 10$.
If $\gcd(a,10) \ne 1$ then $a = 0, 5$ or $2^kb; \gcd(b,10) =1$.
$0^k = 0\equiv 0 \mod 10$ (duh) and $5^k \equiv 5 \mod 10$ (do I really need to show that?)
$6^2 = 36\equiv 6 \mod 10$ so by induction $6^k \equiv 6 \mod 10$.
So if $a = 2^k*b$ then $$a^{4n+1} = a *(a^{4n})=a*((2^4)^{kn}b^4)\equiv 16^{kn}a\equiv 6^{kn}a$$$$ \equiv 6a\equiv a(5+1) \equiv 5a + a \equiv 10*\frac a2 + a \equiv a\mod 10$$.
| {
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Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?
| Observe that
$$g(x) = 16x^4 + x^2+1 = 16(x^2 + \frac 1{32})^2 + \frac{63}{64}$$
Take the fourth root to get
$$\sqrt[4]{g} \sim 2 \sqrt{x^2 + \frac 1{32}}$$
as $x \to \infty$.
We still have to multiply by $x$ and then subtract $2x^2$. We see that
$$x \cdot 2 \sqrt{x^2 + \frac 1{32}} = 2 \sqrt{x^4 + \frac {x^2}{32}} = 2 \sqrt{(x^2 + \frac 1{64})^2 -\frac{1}{4096}}$$
And $$2 \sqrt{(x^2 + \frac 1{64})^2 -\frac{1}{4096}} \sim 2(x^2 + \frac 1{64}) = 2x^2 + \frac 1{32}$$ as $x \to \infty$.
Finally we subtract the $2x^2$ to get $$2x^2 + \frac 1{32}- 2x^2 = \frac 1{32}$$
Hence $$\lim_{x\to \infty} x \sqrt[4]{16x^4 + x^2+1}-2x^2 = \frac 1{32}$$ which can be verified using L'Hôpital's Rule.
But your follow-up question was important, so I'll answer it as if the limit were infinity.
When we write $$\lim_{x\to \infty}f(x)=\infty$$ we are saying that for any real number $M$, there is a real number $\delta$ (which will depend on $M$) such that $f(x)\gt M$ for all $x$ such that $0\lt |x| \lt \delta$. It is important to realize, however, that the limit does not exist. For the limit to exist, it needs to be a number, and $\infty$ is not a number. The expression "$\lim_{x\to \infty}f(x)=\infty$" is just a concise way to express the idea that $f$ increases without bound.
| {
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Help with this determinant I need help to calculate this determinant:
$$\begin{vmatrix}
1^k & 2^k & 3^k & \cdots & n^k \\
2^k & 3^k & 4^k & \cdots & (n+1)^k\\
3^k & 4^k & 5^k & \cdots & (n+2)^k\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k
\end{vmatrix}$$
Where $2\leq n$ and $0\leq k \leq n-2.$
I did the $2\times 2$, $3\times 3$ and $4\times 4$ cases but I couldn't find a pattern to follow.
I did those cases by making zeros in the first column (except the $1^k$ in the first row) and then using the Laplace expansion.
The $2\times 2$ case equals: $3^k -4^k$, the $3\times 3$ case equals: $15^k -16^k -20^k -27^k +24^k +24^k$
| Another proof.
This involves some tricks on manipulating polynomials.
$\mathit{Proof}.\blacktriangleleft$
Consider
$$
f(z) =
\begin{vmatrix}
(z+1)^k & (z+2)^k & (z+3)^k & \cdots & (z+n)^k\\
2^k & 3^k & 4^k & \cdots & (n+1)^k\\
3^k & 4^k & 5^k & \cdots & (n+2)^k \\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k
\end{vmatrix}.
$$
Then the computational definition determinant, i.e.
$$
\det(\boldsymbol A) = \sum_{\sigma \in \mathfrak S_n} \mathrm{sgn}(\sigma) \prod_1^n a_{\sigma(j), j}
$$
implies that $f(z)$ is a polynomial of degree $k$. The goal is to compute $f(0)$. Note that $f(1) = f(2) = \cdots = f(n-1) = 0$, since the $1^{\mathrm {st}}$ row coincides with the $ (j+1 )^{\mathrm {st}}$ row, and a determinant with duplicated rows is $0$. Therefore $f(z)$ is a real polynomial with $n-1$ zeros while its degree is $k \leqslant n-2$. Thus the polynomial is just the zero polynomial, and the original determinant is $f(0) = 0$. $\blacktriangleright$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone explain this proof to me for $2n < 2^n - 1? \,\, n >= 3$ Prove $2n < 2^n - 1;n \geq 3$
Prove for $n = 3$
$2\cdot 3 < 2^6 - 1$
$6 < 7$
Prove for $n \mapsto n + 1$
$$2(n + 1) = 2n + 2$$
$$< 2^n - 1) + 2$$
$$= 2^n + 1$$
$$< 2^n + 2^n - 1$$
$$ = 2^{n + 1} - 1$$
Can someone explain this proof to me? I get lost on the third step: $2^n + 1$ and I don't understand how this shows that $2n < 2^n − 1$
| Your inductive hypothesis is that $2n < 2^n - 1$.
If the inductive hypothesis holds, then by adding 2 to both sides, you find that $$(2n) + 2 < (2^n - 1) + 2$$
On the left hand side, $2n + 2$ is equal to $2(n+1)$.
On the right hand side, $(2^n -1) + 2 = 2^n + 2 - 1 < 2^{n} + 2^{n} - 1 = 2^{n+1} - 1$.
And putting them together proves what you wanted to show:
$$2(n+1) < 2^{n+1} - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the indefinite integral $\int\frac{dx}{(x^2+2x+5)^2}$ I need help with the indefinite integral
\begin{align}
& \int\frac{dx}{(x^2+2x+5)^2} \\[10pt]
= {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt]
= {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt]
= {} & \frac 1 8 \int \frac{ds}{(s^2+1)^2} = \text{?} & & \frac u 2 = s, \quad 2\,ds=du
\end{align}
or maybe there is an easier way ?
Any ideas ? thanks !
| it is $$x^2+2t+5=(x+1)^2+4$$ we Substitute $$x+1=t$$ then we have $$dx=dt$$
and $$\int\frac{1}{(t^2+4)^2}dt$$ and then we Substitute $$t=2\tan(s)$$ with $$dt=2\sec^2(s)ds$$ and we get $$(t^2+4)^2=16\sec^4(s)$$ and our integral is $$2\int \frac{\cos^2(s)}{16}ds$$ and in the last step note that $$\cos^2(s)=\frac{1}{2}\cos(2s)+\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Binomial to the power of five, equality proof I want to find out when the equality $(x+y)^5=x^5+y^5$ for real numbers $y$ and $x$ holds.
Expanding this binomial yields $(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4$
Factoring the right side of the above equality gives $x^5+y^5+5xy(x^3+2x^2y+2xy^2+y^3)$
If $(x+y)^5=x^5+y^5$, then $5xy(x^3+2x^2y+2xy^2+y^3)=0$ which implies that
$xy=0$ or $x^3+2x^2+2xy^2+y^3=0$
if $xy\neq0$ then $x^3+2x^2y+2xy^2+y^3=0$ and thus $(x+y)^3=xy(x+y)$
This is where I'm stuck. How can the last equality help me to figure out when the original equality holds? I know that $xy\neq0$ and I can somehow see that showing that $x+y=0$ seems to be an appropriate step at this moment, but I'm now sure.
Thanks
| It can be helpful with equations like your last one there to gather terms on one side, and factor. Thus: $$(x+y)^3-xy(x+y)=0\\\implies (x+y)((x+y)^2-xy)=0\\ \implies x+y=0\,\,\,\text{ or }\,\,\, (x+y)^2-xy=0$$
What can you do with that second equation?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x) &= (x^2c) \cdot \int \frac{x^3}{x^2} dx \\
&= (x^2c) \cdot \frac{x^2}{2}
\end{align}
but this solution is wrong. Where is the mistake?
| $$y'=\frac { 2 }{ x } y+x^{ 3 }\\ y'-\frac { 2 }{ x } y=0\\ \frac { dy }{ dx } =\frac { 2y }{ x } \\ \int { \frac { dy }{ y } } =2\int { \frac { dx }{ x } } \\ \ln { y } =2\ln { x } +C\\ \ln { y } =\ln { C{ x }^{ 2 } } \\ y=C{ x }^{ 2 }\\ y=C\left( x \right) { x }^{ 2 }\\ { y }'={ C }'\left( x \right) { x }^{ 2 }+2xC\left( x \right) \\ { C }'\left( x \right) { x }^{ 2 }+2xC\left( x \right) =\frac { 2 }{ x } C{ x }^{ 2 }+{ x }^{ 3 }\\ { C }'\left( x \right) { x }^{ 2 }={ x }^{ 3 }\\ { C }'\left( x \right) ={ x }\\ C\left( x \right) =\frac { { x }^{ 2 } }{ 2 } +C\\ y={ x }^{ 2 }\left( \frac { { x }^{ 2 } }{ 2 } +C \right) \\ \\ $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $|f(y)-f(x)|\leq f(|y-x|)$ for $f(x)=-x\log_2(x)$ I need to prove (or disprove) the following claim:
for $0 \leq x \leq y \leq 1$:
$$|f(y)-f(x)|\leq f(y-x)$$
where $f(x) = -x \log_2(x)$ if $x \in (0,1)$ and $f(x)=0$ elsewhere.
I managed to prove that: $f(y)-f(x) \leq f(y-x)$ using Lagrange's mean value theorem (MVT), here is my proof:
Note that $f'$ is decreasing for $x \geq 0$, also$f'(0) \rightarrow \infty$ when $x \rightarrow 0^+$.
Let $z = y-x$.
If $x \geq z$, then using MVT:
\begin{equation*} \begin{aligned}
f(y)-f(x) &= z f'(c_1) && \text{for } c_1 \in (x, y) \\
f(0+z)-f(0) &= z f'(c_2) && \text{for } c_2 \in (0, z)
\end{aligned} \end{equation*}
Since $f'$ is decreasing and $z \leq x$ we have that:
$$ f(y)-f(x) = z f'(c_1) \leq z f'(c_2) = f(0+z)-f(0) = f(z) $$
If $x < z$, again, using MVT:
\begin{equation*} \begin{aligned}
f(x)-f(0) &= x f'(c_1) && \text{for } c_1 \in (0, x) \\
f(y)-f(z) &= x f'(c_2) && \text{for } c_2 \in (z, y)
\end{aligned} \end{equation*}
Since $f'$ is decreasing and $x < z$ we have that: $x f'(c_2) \leq x f'(c_1)$ which implies:
\begin{align*}
f(y)-f(x)
&= (f(y)-f(z)) + (f(z)-f(x)) \\
&= x f'(c_2) + (f(z)-f(x)) \\
&\leq x f'(c_1) + (f(z)-f(x)) \\
&= (f(x)-f(0)) + (f(z)-f(x)) \\
&= f(z)
\end{align*}
Now, I'm stuck here... Any idea of how to prove or disprove the claim with absolute value?
| The claim is false as stated. Indeed, consider the case when $y=1$ and $x=\frac{1}{4}$. Then
$$
f(y) =f(1)= 0 \qquad\text{and}\qquad
f(x) =f\left(\frac{1}{4}\right)=\frac{1}{4}\log(4)
$$
and
$$
f(y-x) = f\left(\frac{3}{4}\right) = \frac{3}{4}\log\left(\frac{4}{3}\right) = \frac{1}{4}\log\left(\frac{4^3}{3^3}\right).
$$
Note that $4>\frac{4^3}{3^3}$ and thus
\begin{align*}
\left\lvert f(y)-f(x)\right\rvert &= \frac{1}{4}\log(4) \\
&> \frac{1}{4}\log\left(\frac{4^3}{3^3}\right)\\
& = f(y-x).
\end{align*}
Other counterexamples with $y\neq1$ can be generated and checked numerically. For example, if we choose $x=0.2$ and $y=0.99$, then $f(y−x)−|f(y)−f(x)|≈−0.125$.
The claim is true if we remove the absolte value. Note that $f$ is concave, since $f''(x)=-\frac{1}{x}$. The claim that $f(y)-f(x)\leq f(y-x)$ follows from concavity of $f$, since
$$
f(0)+f(y)\leq f(x) +f(y-x)
$$
and $f(0)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Let $(a_n)_n$ the sequence verifying $\begin{cases} a_0=2\\ a_1=1\\a_{n+2}=a_{n+1}-2a_n\end{cases}$
The characteristic equation of this linear recurrence relation is $x^2=x-2$
Whose roots are $\dfrac{1\pm i\sqrt{7}}2$.
Thus $a_n=\alpha r^n+\beta {\bar r}^n$ and given the initial conditions then $\alpha=\beta=1$.
So $a_n=\left(\dfrac{1+i\sqrt{7}}2\right)^n+\left(\dfrac{1-i\sqrt{7}}2\right)^n$
We are asked to calculate $a_4$ ?
*
*$a_2=a_1-2a_0=1-4=-3$
*$a_3=a_2-2a_1=-3-2=-5$
*$a_4=a_3-2a_2=-5+6=1$
Use the integer sequence to calculate any other power you desire...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?
| Say $z=x+y\geq 0$ then $$ z^2 = 14^2(zx-x^2)-14^2\cdot 48$$ so $z= 14t$ and thus we have $$t^2= 14tx-x^2-48$$ or $$t^2-14tx +49x^2 =48(x^2-1)$$ so $$(t-7x)^2= 48(x^2-1)$$ and now we have $t-7x = 12s$ for some integer $s$. So we have $$ 3s^2= x^2-1$$ If $3|x-1$ then $$x-1 =3u^2\;\;\;{\rm and} \;\;\;x+1 = v^2$$ where $u,v$ are relatively prime. In this case take mod 3 and we get $v^2 =_3 2$ which is impossible.
If $3|x+1$ then $$x+1 =3u^2\;\;\;{\rm and} \;\;\;x-1 = v^2$$ where $u,v$ are relatively prime. Then we get Pell's equation $3u^2-v^2=2$ which has infinitely solutions...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Calculate the limit: $\lim_{x\to+\infty}(\frac{x^2 -x +1}{x^2})^{\frac{-3x^3}{2x^2-1}}$ without de l'Hôpital rule I was wondering how can I calculate the limit:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}$$
without de l'Hôpital rule.
I tried to reconduct the limit at the well known one:
$$\lim_{x\to+\infty}\left(1+\frac1x\right)^x = e$$
Now I'm concentrating only on the part of the limit with still the Indeterminate Form, I reached this form elevating $e$ to the neperian logarithm of the function, trying to get rid of the $1^\infty$ I.F.
but, at the end of the day, I could only obtain:
$$\begin{align}\lim_{x\to+\infty}\ln\left(\frac{1}{x^2}(1+x^2-x)\right) &= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} \ln\left(1 + \frac{1}{\frac{1}{x^2-x}}\right) \\&= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} (x^2-x) \ln\left(\left(1 + \frac{1}{\frac{1}{x^2-x}}\right)^{\frac{1}{x^2-x}}\right)\end{align}$$
But then defining
$$t= \frac{1}{x^2-x}$$
the limit
$$\lim_{t\to 0} \ln\left(\left(1 + \frac{1}{t}\right)^{t}\right)$$
goes no more to
$$ \ln(e)$$
because now $$t \to 0$$
Can you please give me some help?
|
I thought it might be instructive to present an approach that uses only the squeeze theorem and elementary inequalities obtained using pre-calculus analysis. To that end we proced.
In THIS ANSWER
, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1\tag1$$
Using $(1)$, it is straightforward to see that
$$-\frac{x-1}{x^2-x+1}\le \log\left(1-\frac{x-1}{x^2}\right)\le -\frac{x-1}{x^2}\tag 2$$
Multiplying $(2)$ by $\displaystyle -\frac{3x^3}{2x^2-1}$ and inverting the inequality signs accordingly reveals
$$\begin{align}
\underbrace{\left(\frac{3x^3}{2x^2-1}\right)\,\left(\frac{x-1}{x^2}\right)}_{\displaystyle=\frac{3-3/x}{2-1/x^2}}\le -\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\le \underbrace{ \left(\frac{3x^3}{2x^2-1}\right)\,\left(\frac{x-1}{x^2-x+1}\right)}_{\displaystyle \frac{3-3/x}{(2-1/x^2)(1-1/x+1/x^2)}}\tag 3
\end{align}$$
Applying the squeeze theorem to $(3)$, we find that
$$\lim_{x\to \infty }\left(-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\right)=\frac32$$
from which we find by virtue of the continuity of the exponential function that
$$\begin{align}
\lim_{x\to \infty }\left(\frac{x^2-x+1}{x^2}\right)^{-\frac{3x^3}{2x^2-1}}&=\lim_{x\to \infty }e^{-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)}\\\\
&=e^{\lim_{x\to \infty}\left(-\frac{3x^3}{2x^2-1}\log\left(1-\frac{x-1}{x^2}\right)\right)}\\\\
&=e^{3/2}
\end{align}$$
And we are done!
Main Tools Used: The Squeeze Theorem, Inequalities in $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^3 + 2x^2 + 5x + 2\cos x = 0$
$x^3 + 2x^2 + 5x + 2\cos x = 0$
How do I find the number of solutions of this equation (in $[0, 2\pi]$) without a graph?
Attempt:
The equation simplifies to $x(x^2 + 2x + 5)=- 2\cos x $
Minima of the quadratic occurs at $x= -1$ and it's value is $4$
Minima of $-2\cos x$ is $-2$
| $$y_1=-x^3-5x$$ and $$y_2=2x^2+2\cos x$$
for $x\ge0, y_1<0, y_2>0$
for $-1<x<0$ both function decrease and have only one intersection
for $x\le-1$ there are no intersection $$-x^3-5x<2x^2+2\cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does $\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$ hold for all n values? I was looking at formula $(81)$ on here which is shown below
$$\sum_{k=0}^{n}{2n\choose k}2^kF_k=F_{3n}$$.
$F_n$ is the $n^{th}$ Fibonacci number
I wrote out the sum and just alternate the signs and found out that it has a simple answer in the form of $5^n$
$$\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$$.
Here notice that it is only work for even terms.
Does this formula $$\sum_{k=0}^{2n}(-1)^k{2n\choose k}2^kF_{k+1}=5^n$$
works for all n values or it just accidentally for some n values?
Examples
for $n=1,2$ and $3$
$$1-4+8=5$$
$$1-8+48-96+80=5^2$$
$$1-12+120-480+1200-1536+832=5^3$$
| Yes, the identity holds and is easily proven with induction on $n$. We first recall the familiar identity $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1},$$ which is of course simply Pascal's triangle. If we apply this twice, we get $$\binom{n}{k-1} + 2\binom{n}{k} + \binom{n}{k+1} = \binom{n+2}{k+1},$$ which is equivalent to $$\binom{2n}{k-1} + 2\binom{2n}{k} + \binom{2n}{k+1} = \binom{2(n+1)}{k+1}.$$ Now define $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+1}.$$ Then recalling the convention that $\binom{n}{k} = 0$ when $k < 0$ or $k > n$,
$$\begin{align*} S(n+1) &= \sum_{k=0}^{2(n+1)} (-2)^k \binom{2(n+1)}{k} F_{k+1} \\
&= \sum_{k=0}^{2(n+1)} (-2)^k \binom{2n}{k-2}F_{k+1} + 2 \sum_{k=0}^{2(n+1)} (-2)^k \binom{2n}{k-1}F_{k+1} + S(n) \\
&= S(n) + 4\sum_{k=0}^{2(n+1)} (-2)^{k-2} \binom{2n}{k-2} (2F_{k-1} + F_{k-2}) - 4 \sum_{k=0}^{2(n+1)} (-2)^{k-1} \binom{2n}{k-1} (F_k + F_{k-1}) \\
&= S(n) + 8 S(n) + 4 \sum_{k=0}^{2(n+1)} (-2)^{k-2} \binom{2n}{k-2} F_{k-2} - 4S(n) - 4 \sum_{k=0}^{2(n+1)} (-2)^{k-1} \binom{2n}{k-1} F_{k-1} \\
&= S(n) + 8 S(n) - 4 S(n) \\
&= 5 S(n). \\
\end{align*}$$
It is important to understand that while all of the sums appear to be taken over $\{0, \ldots, 2(n+1)\}$, in actuality some terms are zero because the corresponding binomial coefficient in the summand is zero.
Consequently, we have $S(n+1) = 5S(n)$ which with the initial condition $S(0) = (-2)^0 \binom{0}{0} F_1 = 1$, we get $S(n) = 5^n$ as claimed.
The initial condition part of the above proof leads to an interesting observation for the astute reader: note that the identity only needs the fact that in the initial case, the Fibonacci number is $1$, and $F_n = F_{n-1} + F_{n-2}$. Therefore, if we use the usual definition $F_0 = 0, F_1 = 1$, then the identity can be written with $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+1}$$ or $$S(n) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+\color{red}{2}}.$$ In fact, it is a special case of the more general identity $$S(n,m) = \sum_{k=0}^{2n} (-2)^k \binom{2n}{k} F_{k+m} = F_m 5^n.$$ Another nice corollary is that when $m = 0$, we immediately get $S(n,0) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many numbers can you get by multiplying two or more distinct members of the set $\{1,2,3,5,11\}$ together? Apparently the answer is 15, but I got 26.
My process:
The set $\{1,2,3,5,11\}$ has five numbers.
Ways to choose two members: $_5C_2$
Ways to choose three members: $_5C_3$
Ways to choose four members: $_5C_4$
Ways to choose five members: $_5C_5$
Total: $_5C_2 + {}_5C_3 + {}_5C_4 + {}_5C_5 = 26$
Is there a detail I have misunderstood in this question?
| Note that, because two or more members can be multiplied, multiplying by $1$ will only make a difference if it is one of two numbers. Thus, multiplying by $1$ adds four potential numbers.
Now, we only need to consider the number of combinations that can be made from $2$, $3$, $5$, and $11$.
Choosing two from this set offers six possiblities: $2 \cdot 3$, $2 \cdot 5$, $2 \cdot 11$, $3 \cdot 5$, $3 \cdot 11$, and $5 \cdot 11$.
Choosing three offers four possibilities: $2 \cdot 3 \cdot 5$, $2 \cdot 3 \cdot 11$, $2 \cdot 5 \cdot 11$, and $3 \cdot 5 \cdot 11$.
Finally, there is one possibility with four chosen: $2 \cdot 3 \cdot 5 \cdot 11$. Thus, there are $4 + 6 + 4 + 1 = \boxed{15}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring 2 in $\mathbb{Q}(\sqrt{-7})$ $7$ is a Heegner number, so all numbers in $\mathbb{Q}(\sqrt{-7})$ have a unique factorization.
I'm told that:
*
*$2$ is not prime in $\mathbb{Q}(\sqrt{-7})$,
*$3$ is not prime in $\mathbb{Q}(\sqrt{-11})$,
*$5$ is not prime in $\mathbb{Q}(\sqrt{-19})$,
*$11$ is not prime in $\mathbb{Q}(\sqrt{-43})$,
*$17$ is not prime in $\mathbb{Q}(\sqrt{-67})$,
*$41$ is not prime in $\mathbb{Q}(\sqrt{-163})$.
For the first of these, I've tried finding prime numbers $(a+b (1+\sqrt{-7})/2)$ and $(c+d (1+\sqrt{-7})/2)$, with product $2$, but I haven't had luck so far.
| Your mileage may vary...
I find these problems much easier if instead of trying to solve $(a - b \theta)(a + b \theta)$ (where $\theta = \frac{1 + \sqrt d}{2}) = p$, I try to solve $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = p.$$ Then $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = \frac{a^2 + (-d)b^2}{4}$$ and $a^2 + (-d)b^2 = 4p$.
So then, for $d = -7$, $p = 2$, I solve $a^2 + 7b^2 = 8$. The answer then becomes obvious: $a = 1$, $b = 1$ also.
With the thetas, the whole thing is kind of confusing.
$$2 = \frac{1 + 7}{4} = \left( \frac{1 + \sqrt{-7}}{2} \right) \left( \frac{1 - \sqrt{-7}}{2} \right) = (1 - \theta) \theta,$$
(adjust $\theta$ as you go on to each of these)
$$3 = \frac{1 + 11}{4} = \left( \frac{1 + \sqrt{-11}}{2} \right) \left( \frac{1 - \sqrt{-11}}{2} \right) = (1 - \theta) \theta,$$
$$5 = \frac{1 + 19}{4} = \left( \frac{1 + \sqrt{-19}}{2} \right) \left( \frac{1 - \sqrt{-19}}{2} \right) = (1 - \theta) \theta,$$
$$11 = \frac{1 + 43}{4} = \left( \frac{1 + \sqrt{-43}}{2} \right) \left( \frac{1 - \sqrt{-43}}{2} \right) = (1 - \theta) \theta,$$
which is to say, for your $(a + b \theta)(c + d \theta)$, we have $a = 1$, $b = -1$, $c = 0$, $d = 1$. The thetas obscure your approach to $4p$. But, like I said, your mileage might vary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solution of the Pell equation $3a^2-7b^2=3$.
| You can always find solutions $(a,b)$ that give smaller values of $S$.
Consider solutions to the equation $x^2-21y^2=1$, $(x_1,y_1)=(55,12),(x_2,y_2)=(6049,1320)$ solutions from $(x_3,y_3)$ onward can be generated using solutions to the recursive solution $$x_{k+1}=x_1x_k+21y_1y_k\\y_{k+1}=x_1y_k+y_1x_k$$
Solutions that satisfy $3a^2-7b^2=3$ are then given by $(a_i,b_i)=(x_i,3y_i)$
We can also write $(a,b)$ as $(\sec t, \sqrt\frac{3}
{7}\tan t$. Subtitute this in the expression for $S$.
$$S=(\sqrt{3}\sec t-\sqrt{7}\sqrt\frac{3}
{7}\tan t)(\sec t+\sqrt\frac{3}
{7}\tan t)\\ =\sqrt3(\sec t-\tan t)(\sec t+\sqrt\frac{3}
{7}\tan t)$$
Now at the global minimum we must have $dS/dt = 0$. We compute the derivative and we get $$\frac{dS}{dt}=\frac{\sqrt3}7(\sqrt{21}-1)\sec t(\sec t - \tan t)^2=0 \\ \implies \sec t - \tan t=0.$$
The solutions to this lie at $t=\frac{\pi}2+2n\pi$. At these points $(a,b)=(\infty, \infty)$, hence there is no upper bound on integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Contest Inequality - Is it AM GM? To prove:
$$(a^2+2)(b^2+2)(c^2+2)\ge9(ab+bc+ca)$$
where $a,b,c$ are positive real numbers.
When does equality hold?
I thought equality would hold when $a = b = c$, but that doesn't seem to fit the statement.
To prove the inequality, I tried substituting $a = \tan A, b=\tan B$ and $c=\tan C$ in the hope of using some trigonometric manipulation.
Also, my observational skills suggest that the LHS is the obvious cube of the GM of $3$ numbers; but that didn't help either.
The above inequality is same as proving $(a^2+b^2+c^2+6)^3≥81(ab+bc+ca)$
I know that $a^2+b^2+c^2\ge ab+bc+ca$, but I'm not able to take it from here.
Is this approach fine, or is there a shorter method for the problem? I doubt is $AM\ge GM$ is the most elegant way of arriving at the required statement.
Please help.
| By C-S $$\left(\frac{(a+b)^2}{2}+1\right)(2+c^2)\geq(a+b+c)^2$$ and since
$$(a^2+2)(b^2+2)\geq3\left(\frac{(a+b)^2}{2}+1\right)$$ it's $$(a-b)^2+2(ab-1)^2\geq0,$$ we obtain:
$$(a^2+2)(b^2+2)(c^2+2)\geq3(a+b+c)^2\geq9(ab+ac+bc).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Different answers on $\frac{1}{2\pi}\int_0^{2\pi} \frac{1}{e^{ix}+c}\ dx$ where $c\in\mathbb{C}$ is fixed Upon the substitution $z=e^{ix}$, I get that this is equal to
$$\frac{1}{2\pi i} \oint_{C} \frac{dz}{z(z+c)},$$
where the integral is over the unit circle. I then split this up into
$$\frac{1}{2c\pi i}\left(\oint_{C} \frac{dz}{z}-\oint_{C}\frac{1}{z+c}\right)dz.$$
From here, I get that the first integral is $2\pi i$ and the second integral is $2\pi i$ if $c$ is inside the unit circle and $0$ if $c$ is outside. From this question, it seems as if the second integral should be $\pi i$ if $c$ is on the unit circle itself. Thus, I get that this integral should be
$$f(c)=\frac{1}{2\pi}\int_0^{2\pi} \frac{1}{e^{ix}+c}\ dx = \begin{cases} 0 & \mathrm{if}\ |c|<1 \\ \frac{1}{2c} & \mathrm{if} \ |c|=1 \\ \frac{1}{c} & \mathrm{otherwise.}\end{cases}$$
However, upon experimentation with various values of $c$ on the Online Integral Calculator, it seems as if
$$f(c)=\begin{cases}0 & \mathrm{if}\ c=0 \\ \frac{3}{2} & \mathrm{if} \ c=1 \\ \frac{1}{c} & \mathrm{otherwise.}\end{cases}$$
What's going on?
| If $c=0$, then
$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{e^{ix}+c}dx=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ix}dx = 0.
$$
If $c\ne 0$, then
\begin{align}
\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{e^{ix}+c}&=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\frac{1}{ie^{ix}}d(e^{ix})}{e^{ix}+c} \\
&= \frac{1}{2\pi i}\int_{|z|=1} \frac{dz}{z(z+c)} \\
&=\frac{1}{2\pi ic}\int_{|z|=1}\frac{1}{z}-\frac{1}{z+c}dz.
\end{align}
If $|c| < 1$ the above is $0$. If $c > 1$, the above is $\frac{1}{c}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
limit of a trigonometric function $ \lim\limits_{x\to\pi/3} \frac{1 - 2\cos (x)}{\sin (3x)} $ compute the limit of
$$ \lim_{x\to \frac{\pi }{3} } \frac{1 - 2 \cos (x)}{\sin (3x)} $$
I would like to not do a translation with the change of variable $ t = x - \frac{\pi }{3} $
| $$\frac{1 - 2 \cos x}{\sin (3x)}=\frac{1- 2 \cos (x)}{-4\sin^3 x+3\sin x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4\sin^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4+4\cos^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{4\cos^2 x-1}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{(2\cos x-1)(2\cos x+1)}=\frac{1}{\sin x}\frac{-1}{(2\cos x+1)}\to\frac{2}{\sqrt3}\frac{-1}{2}=\frac{-1}{\sqrt3}=-\frac{\sqrt3}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$
$$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$
and the 2nd way as -
$$(\sin A + \cos A)^2 = 2$$
$$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$
$$\Rightarrow 2\sin A\cos A=1$$
$$\Rightarrow \sin A\cos A=\frac12$$
As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.
I know few trigonometry identities as per my textbook, those are
*
*$\sin^2 A + \cos^2 A = 1$
*$1 + \cot^2 A = \csc^2 A$
*$\tan^2A + 1 = \sec^2 A$
| $\cos\frac {\pi}4=\sin\frac {\pi}4=\frac1{\sqrt{2}}$. Hence what you have got is $$\cos\frac {\pi}4\sin A+\sin\frac {\pi}4\cos A=1$$ or $\sin\Big(A+\frac {\pi}4\Big)=1$. Can you solve from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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In a square ABCD with side 14cm, 2 quadrants were made with centres A & B respectively and AB as radius. Find area of region I and II https://photos.app.goo.gl/5ibXDN5u6s0yo6KB3
I could do the following:
II + III = $\frac{1}{4}$ × $π$ × $14^2$ = $49π$ = $154cm^2$
II + IV = $\frac{1}{4}$ × $π$ × $14^2$ = $49π = 154 cm^2$
Area of square = $I + II + III + IV$ = $14^2 = 196 cm^2$
$(I + II + III +।V ) - ((II + III) + (II + IV)) = 196 - (154 + 154) = -112$
$\mapsto$ I - II = -112
$\mapsto$ II - I = 112
Am I going on the correct path?
|
Start by constructing $\overline {AE}$ and $\overline {EB}$. $E$ is the intersection of the arcs created by $ABD$ and $BAC$.
Notice that $\overline{AE} = \overline{EB} = \overline{AB} = 14$ because they are all radii of the circle. The points $A$, $B$ and $E$ form an equilateral triangle $\triangle ABE$. Hence, we know $\angle EAB = \angle ABE = \angle BEA = \frac{\pi}{3}$rad. You will need to be confident with radians rather than degrees, as we are working with circular measure.
\begin{align}
A_{\text{segment}} & = \frac{1}{2}r^2(\theta-\sin \theta) \\
\text{Segment} ~ EB & = \frac {1}{2} \cdot 14^2 ( \frac{\pi}{3}-\sin {\frac{\pi}{3}})\\
& = 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2})
\end{align}
Segment $EB$ is congruent to segment $AE$, so both are $98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2})$ cm$^2$.
The area of II can be found by the following formula:
\begin{align}
\text{II} & = \text{Quadrant $ABC$} - \text{Sector $EBC$} + \text{Segment $EB$} \\
& = \frac{1}{4}\pi r^2 - \frac{1}{2}r^2 \theta + \text{Segment $EB$}\\
& = \frac{1}{4} \pi \cdot 14^2 - \frac{1}{2} \cdot 14^2 \cdot \frac{\pi}{6} + 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2}) \\
& = 49\pi - \frac{49}{3}\pi + 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2}) \\
& = 98(\frac{\pi}{3}-\frac{\sqrt3}{2})+\frac{98}{3}\pi \\
& = 98(\frac{2}{3}\pi-\frac{\sqrt3}{2}) \\
& \approx 120.380
\end{align}
As you see, it's not a very nice answer, but it is an exact answer for the area of II.
As for the area of I:
\begin{align}
\text{I} & = ABCD - \text{Quadrant $ABD$} - \text{Quadrant $CBA$} + \text{II} \\
& = 14^2 - 49\pi - 49\pi + 98(\frac{\pi}{3}-\frac{\sqrt3}{2})+\frac{98}{3}\pi \\
& = 98(\frac{\pi}{3}-\frac{\sqrt3}{2})-\frac{196}{3}\pi + 196\\
& \approx 8.504
\end{align}
So to summarise, $\text{I} = 8.504 ~\text{cm}^2$ and $\text{II} = 120.380 ~\text{cm}^2$ which sound about right.
| {
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"url": "https://math.stackexchange.com/questions/2586363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something.
Hints and answers appreciated. Sorry if this is a duplicate.
| You certainly forgot an absolute value in your computations.
Here is a computation for $x<0$:
\begin{align}
f(x)&= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=-\frac 1{\Bigl(x^3-\sqrt{x^6+1}\Bigr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)}\\
&= -\frac 1{\biggl(x^3-|x^3|\sqrt{1+\dfrac1{x^6}}\mkern2mu\biggr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)}
\sim_{-\infty} \frac{-1}{2x^3\cdot2x^2}=-\frac1{4x^5}\to 0^+.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 + 3 + 5 + \dots &= x\\
1 + 1 + 2 + 3 + \dots &= x\\
1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)}
\end{align}$$
We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is $x-1$.
But $x-1=2x$ which implies that $x=-1$.
So $1+1+2+3+5+8+\dots=x$ which means...
$1+1+2+3+5+8+13+21+\dots=-1$
Is this right or wrong? Can someone please tell? Thanks...
| The problem is that the series you're trying to sum is divergent. You cannot manipulate divergent series by rules you can use with absolutely convergent series! Otherwise, by following the same "method" as yours, I can also claim the ridiculous statement that $1+2+3+\cdots=0$ as follows:
$$\begin{align}
1 + 2 + 3 + 4 + 5 + \dots &= x\ \fbox1\\
1 + 2 + 3 + 4 + \dots &= x\\
1 + 3 + 5 + 7 + 9 + \dots &= 2x\ \fbox2
\end{align}$$
Also multiplying $\fbox1$ by 2 gives
$$2+4+6+8+10+\dots=2x\ \fbox3$$
So $\fbox1=\fbox2+\fbox3$ $\implies$ $x=2x+2x=4x$ $\implies$ $x=0$ as $1\ne4$.
Obviously this is totally absurd. This is because I'm manipulating a divergent series by rules only allowed to be used with absolutely convergent series. If a series is divergent, it diverges – trying to make anything out of it is just pointless.
| {
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"url": "https://math.stackexchange.com/questions/2591315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Some infinite series (just for fun!) I have a few infinite series problems that I think MSE might enjoy, whose answers I already know:
$$\sum_{n=0}^\infty \frac{2^{n-2^n}}{1+2^{-2^n}}=\text{?}$$
$$\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}=\text{?}$$
$$\sum_{n=0}^\infty \frac{3^n(1+2\cdot3^{-3^n})}{2\cosh(3^n\ln 3)+1}=\text{?}$$
After these all get answered, I will post explanations of how I came up with answers for them. Have fun!
| The tag telescoping is a great hint:
\begin{align*}
A_n &:= \frac{2^n}{2^{2^n}-1}
&\Rightarrow & \qquad A_n - A_{n+1} = \frac{2^n}{2^{2^n}+1} \\
B_n &:= \frac{4^n \cdot 4^{2^n}}{(4^{2^n}-1)^2}
&\Rightarrow & \qquad B_n - B_{n+1} = \frac{4^n \cdot 4^{2^n}}{(4^{2^n}+1)^2} \\
C_n &:= \frac{3^n}{3^{3^n}-1}
&\Rightarrow & \qquad C_n - C_{n+1} = \frac{3^n(3^{3^n}+2)}{3^{2\cdot 3^n}+3^{3^n} + 1}
\end{align*}
From these, we obtain
\begin{align*}
\sum_{n=0}^{\infty} \frac{2^n}{2^{2^n}+1}
&= \lim_{N\to\infty} (A_0 - A_{N+1}) = 1 \\
\sum_{n=0}^{\infty} \frac{4^n \cdot 4^{2^n}}{(4^{2^n}+1)^2}
&= \lim_{N\to\infty} (B_0 - B_{N+1}) = \frac{4}{9} \\
\sum_{n=0}^{\infty} \frac{3^n(3^{3^n}+2)}{3^{2\cdot 3^n}+3^{3^n} + 1}
&= \lim_{N\to\infty} (C_0 - C_{N+1}) = \frac{1}{2}
\end{align*}
For a generalization, let $a \geq 2$ be an integer and consider
$$ P_n = \frac{a^n}{X^{a^n}-1}. $$
Then it is not hard to check that
$$
P_n - P_{n+1}
= \frac{a^n \sum_{k=0}^{a-2} (a-1-k) X^{k \cdot a^n}}{\sum_{k=0}^{a-1} X^{k \cdot a^n}}
$$
So if $|X| > 1$, we obtain
$$ \sum_{n=0}^{\infty} \frac{a^n \sum_{k=0}^{a-2} (a-1-k) X^{k \cdot a^n}}{\sum_{k=0}^{a-1} X^{k \cdot a^n}} = P_0 = \frac{1}{X-1}. \tag{*}$$
Then
*
*The 1st series corresponds to $a = 2$ and $X = 2$ applied to $\text{(*)}$.
*The 2nd series corresponds to $a = 2$ and $X = 4$ applied to the derivative of $\text{(*)}$ w.r.t. $X$.
*The 3rd series corresponds to $a = 3$ and $X = 3$ applied to $\text{(*)}$.
| {
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"url": "https://math.stackexchange.com/questions/2592425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find all ordered pairs such that $x + y^2 = 2$ and $y + x^2 = 2$ How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?
A solution/explanation would be greatly appreciated.
If it helps, the solutions given in the textbook are
x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; x= (1-root 5)/2, y=(1+root 5)/2
| If we leave $y$ alone in the second equation, we have $y = 2-x^2$. Then putting it in the first equation, we get $x+(2-x^2) = 2 \implies x^4-4x^2+x+2 = 0$. Now, notice that $x=1$ satisfies this equation, therefore we can say that the polynomial $x^4-4x^2+x+2$ can be factorized as
$$x^4-4x^2+x+2 = (x-1)(Ax^3+Bx^2+Cx+D)$$
Here, we can find $A$ and $D$ easily because there is only one term to consider and $A = 1$, $D = -2$. Notice that after that, we can conclude that $B-1 = 0$ and $C-B = -4$ by polynomial equality. Then we have $B = 1$ and $C = -3$, so we have
$$x^4-4x^2+x+2 = (x-1)(x^3+x^2-3x-2) = (x-1)(x+2)(x^2-x-1)$$
From here, you can find all $x$ and corresponding $y$'s by using the equations given (Don't forget to consider $x^2-x-1 = 0$ because it has two roots as well).
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration using only two rules and by parts. I need to integrate a few fractions of form $\frac{p(x)}{q(x)}$ where ${p(x)}$ is a polynomial of degree 1 or and $q(x)$ being a polynomial of degree 2 only using integration by parts and following two rules:
$$\int \frac{f'(x)}{f(x)}\, dx = \ln| f(x)|+C$$
$$\int \frac{a}{a^2+x^2}\, dx = \arctan\left(\frac{x}{a}\right)+C$$
Would someone be able to integrate $\frac{3x-1}{x^2-2x+5}$ as an example for me using only these three ways? Thanks in advance.
| $$I=\int\frac{3x-1}{x^2-2x+5}$$
$$\int\frac{ 2x-2}{(x-1)^2+4}+\int\frac{ x+1}{(x-1)^2+4}$$
$$\int\frac{( 2x-2)dx}{(x-1)^2+4}+\int\frac{ (z+2)dz}{z^2+4}$$
$$I=\frac 3 2\int\frac{ (2z)dz}{z^2+4}+\int\frac{ 2dz}{z^2+4}$$
Where $z=x-1$
| {
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"url": "https://math.stackexchange.com/questions/2594630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int^{1}_{0}x^{x}dx$ The answer can be in the form of two defined constants:
$$A = \frac{1}{1^1}+\frac{1}{2^2} + \frac{1}{3^3} + \cdots$$
$$B = \frac{1}{2^2} + \frac{1}{4^4} + \frac{1}{6^6} + \cdots $$
I would highly appreciate it if you included your thought process in the answer itself.
As pointed out by zz20s
, the answer to this integral can be seen on the wikipedia page: Sophomore's dream and the page accompanies a proof too
| I have written a detailed explanation to find the integral of a general expression of the form $f(x) =x^{cx^a} $ from $0$ to $1$ here.
$$\int_{0}^{1} x^{cx^a}\, dx = 1- \frac{c}{(a+1)^2} + \frac{c^2}{(2a+1)^3} - \frac{c}{(3a+1)^4}+\ldots$$
On substituting $c =1$ and $a=1$, we get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this substitution to solve an integral correct? To solve the integral:
$\int\frac {x}{\sqrt{4-x^2}}\ dx$
I used the substitution:
$u = \sqrt{4-x^2}$
hence:
$\frac{du}{dx} = -\frac {x}{\sqrt{4-x^2}}$
$dx = -\frac {\sqrt{4-x^2}}{x}\ du$
So the integral becomes:
$\int\frac {x}{u}(-\frac {\sqrt{4-x^2}}{x})\ du$
$= \int\frac {x}{u}(-\frac {u}{x})\ du$
$= \int -1\ du$
$= -u+c$
$= -\sqrt{ 4-x^2}+c$
However the solution I have seen is:
Use the substitution $x = 2\sin(t)$, so $dx = 2\cos(t)\ dt$
So the integral becomes:
$\int \frac {2\sin(t)(2\cos(t))}{\sqrt{\cos(t)^2}}\ dt$
After some rearranging the integral becomes:
$\int 2\sin(t)\ dt$
Which gives:
$-2\cos(t)+c$
Then using $\sin(t) = \frac x2$:
$\cos(t)^2 = 1 - \sin(t)^2$
$\cos(t)^2 = 1 - \frac {x^2}{4} = \frac{4-x^2}{4}$
$\cos(t) = \frac{\sqrt{4-x^2}}{2}$
Which then gives the solution as:
$= -\sqrt{ 4-x^2}+c$
I feel like the method I used was more efficient as it didn't require the use of trig functions. I'm fairly confident my method is correct so why would the solution use trig functions? Was I just lucky that I decided $u = \sqrt{ 4-x^2}$ was a good substitution?
| The two have more in common than they first seem to because $u=\sqrt{4-4\sin^2 t}=\cos t,\,du=-\frac{x}{2}dt$. A trigonometric substitution is so useful in such a variety of problems it's become a standard part of the toolkit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$ . Prove that $ABC$ is the equilateral triangle. For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$. Prove that $ABC$ is the equilateral triangle.
| Since $\measuredangle GBH=\measuredangle GAH$, we see that $ABHG$ is cyclic.
Thus, by Ptolemy $$AB\cdot GH+BH\cdot AG=AH\cdot BG$$ or in the standard notation
$$\frac{c^2}{2}+\frac{ab}{4}=\frac{1}{4}\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)},$$
which after squaring of the both sides gives
$$(a-b)^2(a+b-c)(a+b+c)=0$$ or
$$a=b.$$
Now, by law of cosines for $|delta AHC$ we obtain
$$\cos30^{\circ}=\frac{AH^2+AC^2-HC^2}{2AH\cdot AC}$$ or
$$\frac{\sqrt3}{2}=\frac{\frac{1}{4}(2b^2+2c^2-a^2)+b^2-\frac{a^2}{4}}{2\cdot\frac{1}{2}\sqrt{2b^2+2c^2-a^2}b}$$ or
$$\sqrt3b\sqrt{2b^2+2c^2-a^2}=3b^2+c^2-a^2$$ or
$$3b^2(a^2-b^2)+(a^2-c^2)^2=0,$$ which gives $$a=c$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$ I wish to solve the equation
$$(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$$
I tried factorizing until I reached
$(\sqrt{2}+\sqrt{3})^x + (\sqrt{2}-\sqrt{3})^x = 10$
But from there I don't know what to do any help would be welcome
Thanks in advance
| There are two solutions: $x=2$ and $x=-2$.
We easily see that $x=2$ is a solution. There are no other solutions $x>0$ because the left-hand side is an increasing function on ${\mathbb R}^+$.
Indeed, noticing that $1/(5+2\sqrt{6})=5-2\sqrt{6}$, we then find that
$$
f(x) = (5+2√6)^{\frac{x}{2}} + ( 5-2√6)^{\frac{x}{2}}
$$
is an even function $(f(x)=a^x+a^{-x}=2\cosh(x\log a))$. So we have the second solution $x=-2$, and no other solutions for $x<0$ because $f(x)$ is decreasing on ${\mathbb R}^-$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$. I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$. I am stuck here. How do I simplify this? Or is it that my method is wrong?
| We want to find the integer part of
$\sum_{x=1}^{50} f(x)$
where
$f(x) = \sqrt{2x}-\sqrt{2x-1}$.
For decreasing $f(x)$,
$$\int_a^{b+1} f(x) \; dx < \sum_{x=a}^b f(x) < \int_{a-1}^b f(x) \; dx$$
(It may help to make a sketch to see this.)
It's easier to work with the sum from $x=2$ to $50$ instead of $x=1$ to 50 because $f(x)$ is not defined when $x=0$. We can adjust for the missing $x=1$ term later.
By elementary calculus,
$$\int f(x) \; dx = \frac{1}{3} (2x)^{3/2} - \frac{1}{3} (2x-1)^{3/2} + C$$
so
$$\int_2^{51} f(x) \; dx < \sum_{x=2}^{50} f(x) < \int_{1}^{50} f(x) \; dx$$
yields
$$4.10 < \sum_{x=2}^{50} f(x) < 4.38$$
Now to adjust for the missing $x=1$ term. Since $$\sum_{x=2}^{50} f(x) + \sqrt{2} - \sqrt{1} = \sum_{x=1}^{50} f(x)$$
we have
$$4.10 + \sqrt{2} - \sqrt{1} < \sum_{x=1}^{50} f(x) < 4.38 + \sqrt{2} - \sqrt{1}$$
which yields
$$4.51 < \sum_{x=1}^{50} f(x) < 4.79$$
so the integer part of the sum is $\boxed{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Arithmetic and harmonic mean of two numbers. What is the maximum value of arithmetic mean of two integer number if their harmonic mean is 40?
$$\frac{1}{x}+\frac{1}{y} = \frac{2}{40}$$
$$xy=20x+20y$$
$$xy−20x−20y+20⋅20=400$$
$$(x−20)(y−20)=400$$
| $$\frac{1}{x}+\frac{1}{y} = \frac{2}{40}$$
$$xy=20x+20y$$
$$xy−20x−20y+20⋅20=400$$
$$(x−20)(y−20)=400$$
$400=1\cdot 400 =2\cdot 200 =4\cdot 100=5\cdot 80 = 8\cdot 50= 10\cdot 40=16\cdot 25=20\cdot 20$
$$ x+y = \{21+420; \;22+220;\;24+120;\;25+100;\;28+70;\;30+60;\;36+45;\;40+40\}
$$
the maximum value of summ is 441 and arithmetic mean is 220.5
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of simultaneous equations involving integral part (floor) Solve the following system of simultaneous equations:
$$
\left \{
\begin{matrix}
x^2+[y]=10 \\
y^2+[x]=13
\end{matrix}
\right .
$$
where square brackets $[...]$ denote integral part of a number (a.k.a. floor)
This is what I've come to:
Case I. $y\ge 0$.
First equation yields: $|x|\le\sqrt{10}$ (|..| denotes absolute value), therefore $-4\le [x] \le 3$. Substitution into the second equation gives: $10\le y^2 \le 17$, so $3\le [y] \le 4$.
Then, from the first equation it can be evaluated that $x=\pm\sqrt{7}$ for
$[y]=3$ and $x=\pm\sqrt{6}$ for $[y]=4$.
If $x>0$ (i.e $x=+\sqrt{6}$ or $x=+\sqrt{7}$), then $[x]=2$, so (from 2-nd equation) $y=\sqrt{11}$ and $[y]=3$. This gives a solution: ($\sqrt{7}$,$\sqrt{11}$).
If $x<0$ (i.e $x=-\sqrt{6}$ or $x=-\sqrt{7}$), then $[x]=-3$, so $y^2=16$ and $[y]=y=4$. This gives another solution: ($-\sqrt{6}$, $4$),
Case II. $y \lt 0$, $x \ge 0$.
From second equation $-\sqrt{13} \le y \lt 0$, therefore $-4 \le y \le -1$, so from the first equation $\sqrt{11} \le x \le \sqrt{14}$ and $[x]=3$. Then from second equation $y=-\sqrt{10}$, so $[y]=-4$ and $x=\sqrt{14}$. Hence the third solution: ($\sqrt{14}$,$-\sqrt{10}$).
Case III. $x \lt 0$, $y \lt 0$.
Any ideas?
| Hint: $\;\lfloor x \rfloor \le x \lt \lfloor x \rfloor + 1\,$ and the same for $\,y\,$, then:
$$
\left \{
\begin{matrix}
10 \le x^2+y \lt 11 \\
13 \le y^2+x \lt 14
\end{matrix}
\right .
\;\;\implies\;\; 23 \;\le\; x^2+y+y^2+x \,=\, (x+1/2)^2+(y+1/2)^2 - 1/2 \;\lt\; 25
$$
It follows that $|x+1/2|, |y+1/2| \lt \sqrt{25+1/2}\,$ which gives useful bounds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the infinitesimal order of the function in $x0= 0$ Problem: infinitesimal order. If you can explain me how you solve this
function:
$$(1+3x)^x -1 -x\ln(1+3x)$$
Mine resolution: $$(1+3x)^x$$ = 1 (taylor series)
$$xln(2+3x)$$ = $3x^2$
so $$lim \frac {1-1-3x^2}{x^n}$$= -3 (this is not equal to 0 so the infinitesimal order is 2? Is this the correct solution?)
| Note that
$$x\ln(1+3x)=x\left(3x-\frac{9}{2}x^2+9x^3+o(x^3)\right)=3x^2-\frac{9}{2}x^3+9x^4+o(x^4)$$
$$(1+3x)^x=e^{x\ln(1+3x)}=e^{3x^2-\frac{9}{2}x^3+9x^4+o(x^3)}=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{(3x^2-\frac{9}{2}x^3+9x^4+o(x^4))^2}{2}+o(x^4)=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{9}{2}x^4+o(x^4)$$
thus
$$(1+3x)^x -1 -x\ln(1+3x)=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{9}{2}x^4-1-3x^2+\frac{9}{2}x^3-9x^4+o(x^4)=\frac{9}{2}x^4+o(x^4)$$
| {
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"url": "https://math.stackexchange.com/questions/2606439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while and could not figure it out. Can anyone give me some hints? Thank you.
| With a bit of linear algebra:
An equivalent problem is
$$\max_{(a,b,c)\ne 0} \frac{(3a+28b+35c)(20a+23b+33c)}{a^2 + b^2 + c^2}$$
or, equivalently
$$\max_{a^2 + b^2 + c^2 =1} (3a+28b+35c)(20a+23b+33c)$$
The symmetric quadratic form $$(a,b,c) \mapsto (3a+28b+35c)(20a+23b+33c)=\\=60 a^2 + 629 a b + 799 a c + 644 b^2 + 1729 b c + 1155 c^2$$ is given by the symmetric matrix
$$\left( \begin{matrix} 60 & 629/2& 799/2 \\629/2 & 644 & 1729/2 \\ 799/2 &1729/2 &1155 \end{matrix}\right)$$
with eigenvalues $\frac{3877}{2}$, $-\frac{159}{2}$ and $0$. By the theory, the above maximum is $\frac{3877}{2}=1938.5$. So the best estimate for the problem is $a^2+b^2+c^2 \ge \frac{1}{1938.5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x+\frac{1}{x}$ is an integer Is the following Proof Correct? In particular please comment on the correctness of the given formulas.
Theorem. Given that $x$ is a real number, $x\neq 0$, and $x + \frac{1}{x}$ is an integer. For all $n\ge 1$, $x^n+\frac{1}{x^n}$ is an integer.
Proof. We construct the proof by recourse to Strong-Induction. Assume for an arbitrary $n\in\mathbf{Z^+}$ that $x^k+\frac{1}{x^k}$ is an integer for any positive integer $k$ strictly less than $n$. Now Consider the following cases.
Case-1: If $n$ is even then for some $l\in\mathbf{Z^+}$, $n = 2l$ thus the Binomial-Theorem implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l-1}\binom{n}{2j}\left(x^{2j}+\frac{1}{x^{2j}}\right)-\binom{n}{l} = x^n+\frac{1}{x^n}$$
from the inductive hypothesis we know that $x+\frac{1}{x}\in\mathbf{Z}$ which implies that $(x+\frac{1}{x})^n \in\mathbf{Z}$ in addition it also follows from the inductive hypothesis that $x^{2j}+\frac{1}{x^{2j}}\in\mathbf{Z}$ for $j\in\{1,2,3,...,l-1\}$ moreover we also that $\binom{n}{r}\in\mathbf{Z^+}$ is always a positive integer implying that $x^n+\frac{1}{x^n}$ is an integer.
Case-2: If $n$ is odd then for some $l\in\mathbf{Z^+}$, $n = 2l+1$ thus the Binomial-Theorem implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l}\binom{n}{2j-1}\left(x^{2j-1}+\frac{1}{x^{2j-1}}\right) = x^n+\frac{1}{x^n}$$
and by using the same reasoning as in the previous case we can deduce that $x^n+\frac{1}{x^n}$ is an integer.
| I think it is easier to see that
$$
x^{n+2}+\frac{1}{x^{n+2}}=\left(x+\frac{1}{x}\right)\left(x^{n+1}+\frac{1}{x^{n+1}}\right)-\left(x^n+\frac{1}{x^n}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2609537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume of Region Bounded by $y = \sqrt{x}$ and $y = 18-x^2$ I haven't a clue why I'm having so much issues solving this problem. The two curves are $y=\sqrt{x}$ and $y=18-x^2, x \ge 1$, and rotated about the $y$-axis.
If I'm not mistaken, I need to use the shell method. In that case, the volume integration formula is $\int 2\pi x f(x)dx$.
So my thickness is $dx$, my height is $f(x)$, I believe my range is from $1$ to $\sqrt{18}$ ($1$ because $x \ge 1$ and $\sqrt{18}$ because $0 = 18-x^2$).
So I integrate $2\pi \cdot x \cdot x^{1/2} dx$ and get $(2/5)x^{5/2}$ from $1$ to $\sqrt{18}$, but that isn't even close to the answer.
I've been working on this problem far too long and am getting brain fog.
| Note that the graph of $y = \sqrt{x}$ and $y = 18-x^2$ shows that these two curves meet at $x = 4$ and $y = 18-4^2 = \sqrt{4} = 2$. Write out explicitly $f$ in the question body.
$$f(x) = 18-x^2-\sqrt{x} \quad \forall\,x \in [1,4]$$
\begin{align}
\text{Required volume of revolution}
&= 2\pi \int_1^4 x f(x) dx \\
&= 2\pi \int_1^4 (18x - x^{3/2} - x^3) dx \\
&= 2\pi \left[9x^2 - \frac{x^{5/2}}{5/2} - \frac{x^4}{4} \right]^4_1 \\
&= 2\pi \left[ 9(4^2 - 1^2) - \frac{64}{5} + \frac25 - \frac{4^4 - 1^4}{4} \right] \\
&= \frac{1177\pi}{10}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Maclaurin series questions How can I find the Maclaurin series for $\ln(1+x^2+x)$? I'm not sure if I should try writing 1+x+x^2 in some other way or..Also if I want to find $h^{(3k)}(0)$ (the derivative of order 3k) where should I start?
| Take $f(x)=\ln(x^2+x+1)$ therefore $f'(x)=\frac{2x+1}{1+x+x^2}=\frac{1+x-2x^2}{1-x^3}$. Also we have:$$\frac{1}{1-x^3}=\sum_{n=0}^{\infty}x^{3n}$$then we have$$f'(x)=(1+x-2x^2)\sum_{n=0}^{\infty}x^{3n}=\sum_{n=0}^{\infty}x^{3n}+\sum_{n=0}^{\infty}x^{3n+1}+\sum_{n=0}^{\infty}-2x^{3n+2}=\sum_{n=0}^{\infty}a_nx^n$$where $a_n=1$ for $n=3k$ or $3k+1$ and $a_n=-2$ for $n=3k+2$. Therefore we have:$$f(x)=C+\sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}$$and $C=0$ obviously. Therefore we obtain:$$f(x)=\sum_{n=1}^{\infty}\frac{a_{n-1}}{n}x^{n}$$with the same definition of $a_n$ or equivalently:$$f(x)=\sum_{n=0}^{\infty}\frac{1}{3n+1}x^{3n+1}+\sum_{n=0}^{\infty}\frac{1}{3n+2}x^{3n+2}-2\sum_{n=1}^{\infty}\frac{1}{3n}x^{3n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we prove the same cardinality of the sets $\mathbb{N}$ and $\mathbb{N^2}$ this way? I tried to prove that the sets $\mathbb{N}$ and $\mathbb{N^2}$ have the same cardinality and I concluded the following:
Consider the function $f:\mathbb{N}\rightarrow \mathbb{N^2}$ that achieves the mapping:
$ \begin{cases} 1 \rightarrow (0,0) \end{cases} \\
\begin{cases} 2 \rightarrow (1,0) \\
3 \rightarrow (1,1) \\
4 \rightarrow (0,1) \end{cases} \\
\begin{cases} 5 \rightarrow (2,0) \\
6 \rightarrow (2,1) \\
7 \rightarrow (2,2) \\
8 \rightarrow (1,2) \\
9 \rightarrow (0,2) \end{cases} \\
\quad \vdots$
This way $\forall k \in \mathbb{N}\cup\{0\}$ we construct the function:
$f(n) = \begin{cases} \begin{aligned}
&(\sqrt{n-1},0) \\
&(\sqrt{n-1},1) \\ \vdots \\
&(\sqrt{n-1},\sqrt{n-1}-1) \\
&(\sqrt{n-1},\sqrt{n-1}) \\
&(\sqrt{n-1}-1,\sqrt{n-1}) \\ \vdots \\
&(1,\sqrt{n-1}) \\
&(0,\sqrt{n-1})
\end{aligned}
&&
\begin{aligned}
&, n=k^2+1 \\
&, n=k^2+2 \\ \\
&, n=k^2+k \\
&, n=k^2+k+1 \\
&, n=k^2+(k+1)+1 \\ \\
&, n=k^2+(2k-1)+1 \\
&, n=k^2+2k+1=(k+1)^2
\end{aligned} \end{cases}$
Is the above mapping correct? If not then why? If yes then how can I prove rigorously that this function is bijective?
| With $0 \in \mathbb N$ we trace out a path:
$(0,0) \to$
$(1,0) \to (1,1) \to (0,1) \to $
$(2,0) \to (2,1) \to (2,2) \to (1,2) \to (0,2) \to $
$(3,0) \to (3,1) \to (3,2) \to (3,3) \to (2,3) \to (1,3) \to (0,3) \to $
$\text{etc.} \qquad \text{Figure 1}$
Here is the corresponding mapping:
$$
\pi(m,n) = \left\{\begin{array}{lr}
n^2+2n-m, & \text{for } m \le n\\
m^2+n, & \text{for } m \gt n
\end{array}\right\}
$$
As a check, apply $\pi$ to Figure 1:
$\pi(0,0) = 0$
$\pi(1,0)=1 \; \pi (1,1)=2 \; \pi (0,1)=3$
$\pi(2,0)=4 \; \pi (2,1)=5 \; \pi (2,2)=6 \; \pi (1,2)=7 \; \pi (0,2)=8 $
$\pi(3,0)=9 \; \pi (3,1)=10 \; \pi (3,2)=11 \; \pi (3,3)=12 \; \pi (2,3) =13 \; \pi (1,3) =14 \; \pi (0,3)=15 $
$\text{etc.} \qquad \pi\text{(Figure 1)}$
Proposition: If $k \in \mathbb N$ then the mapping $\pi$ restricted to
$\quad \{(m,n) \, | \, max(m,n) \le k$}
is $\text{1:1}$ and onto the initial segment $[0,\,{(k+1)}^2-1]$.
Proof
Hint: Use induction on $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Proof verification: If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$ Would someone be willing to verify the following proof?
Theorem: Suppose $a, b \in \mathbb{Z}; n \in \mathbb{N}$. If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$.
Proof:
$a \equiv b (mod$ $n) \rightarrow xn = a - b; x \in \mathbb{Z}$
$\rightarrow a = xn + b$
Then, $a^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) + b^3$.
This gives us $a^3 - b^3 = (xn)^3 + 3b(xn)^2 + 3b^2(xn) = n(x^3n^2 + 3bx^2n + 3b^2x)$
$n | n(x^3n^2 + 3bx^2n + 3b^2x) \rightarrow n | (a^3 - b^3)$
Therefore $a^3 \equiv b^3 (mod$ $n)$.
| The posted proof is correct. It might be even easier, however, to first prove the more general:
$$
a \equiv b \pmod{n} \quad\text{and}\quad a' \equiv b' \pmod{n} \quad \implies \quad a \cdot a' \equiv b \cdot b' \pmod{n}\quad
$$
The proof goes the same: $a\cdot a' = (xn+b)\cdot(x'n+b')=b\cdot b'+n\cdot (x\cdot b'+ x' \cdot b+n\cdot x\cdot x')\,$.
Then, using the proposition with $a'=a, b'=b$ gives $a \equiv b \implies a^2 \equiv b^2 \implies a^3 \equiv b^3 \pmod{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.