Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$.
| Consider the diagram
where the number of dots in column $1$ is no greater than the number of dots in column $2$, which is no greater than the number of dots in column $3$, which is equal to the number of dots in column $4$ (columns $3$ and $4$ are identical).
The diagram represents a solution to
$$
a\le b\le c\quad\text{and}\quad c\le a+b\quad\text{and}\quad a+b+c=n\tag1
$$
where $a$ is the number of dots in column $2$, $b$ is the number of dots in column $3$, $c$ is the sum of the number of dots in columns $1$ and $4$, and $n$ is the number of dots in all columns. All solutions to $(1)$ can be uniquely represented this way.
We can count the number of solutions to $(1)$ by looking at the diagrams sideways. Each diagram represents a term in the expansion of
$$
\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\tag2
$$
The diagram above represents the $x^{16}$ term from
$$
\underbrace{\color{#AAA}{\overbrace{\left(1+\color{#000}{x^2}+x^4+x^6+\cdots\right)}^{\large\frac1{1-x^2}}}}_\text{$1$ rows with $2$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^3+\color{#000}{x^6}+x^9+\cdots\right)}^{\large\frac1{1-x^3}}}}_\text{$2$ rows with $3$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^4+\color{#000}{x^8}+x^{12}+\cdots\right)}^{\large\frac1{1-x^4}}}}_\text{$2$ rows with $4$ dots}\tag3
$$
The diagram
represents the $x^{16}$ term from
$$
\underbrace{\color{#AAA}{\overbrace{\left(1+x^2+\color{#000}{x^4}+x^6+\cdots\right)}^{\large\frac1{1-x^2}}}}_\text{$2$ rows with $2$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(\color{#000}{1}+x^3+x^6+x^9+\cdots\right)}^{\large\frac1{1-x^3}}}}_\text{$0$ rows with $3$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^4+x^8+\color{#000}{x^{12}}+\cdots\right)}^{\large\frac1{1-x^4}}}}_\text{$3$ rows with $4$ dots}\tag4
$$
Thus, the number of such diagrams with $n$ dots is
$$
\left[x^n\right]\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\tag5
$$
That is, the generating function for the number of solutions to $(1)$ is given by $(2)$.
Each solution to $(1)$ can be mapped uniquely to a solution of
$$
a\le b\le c\quad\text{and}\quad c\lt a+b\quad\text{and}\quad a+b+c=n+3\tag6
$$
by adding $1$ to $a$, $b$, and $c$. Thus, the number of solutions to $(6)$ is given by $(5)$.
Therefore, the number of solutions to
$$
a\le b\le c\quad\text{and}\quad c\lt a+b\quad\text{and}\quad a+b+c=n\tag7
$$
is
$$
\left[x^{n-3}\right]\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}=\left[x^n\right]\frac{x}{1-x^2}\frac{x}{1-x^3}\frac{x}{1-x^4}\tag8
$$
Thus, the generating function for the number of solutions to $(7)$ is
$$
\bbox[5px,border:2px solid #C0A000]{\frac{x^3}{\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)}}\tag9
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$ If $ a $ and $ b $ are real numbers that satisfy the equation $ 17 (a ^ 2 + b ^ 2) -30ab-16 = 0 $, the maximum value of the expression$$\sqrt{16a^2+4b^2-16ab-12a+6b+9}$$goes:
a) $1$
b) $3$
c) $5$
d) $7$
e) $9$
Can anyone give a hint?
Attempt: I tried Variation, Jacobian, Hessian matrices.
| Hint
$$16a^2+4b^2-16ab-12a+6b+9=(4a-2b)^2-2(4a-2b)\cdot(3/2)+9=(4a-2b-3/2)^2+9+(3/2)^2$$
Let $4a-2b=2c\implies b=2a-c$
From the given condition,
$$0=17a^2+17(2a-c)^2-30a(2a-c)-16=a^2(17+68-60)+a(60c-68c)+17c^2-16=0$$
which is a quadratic equation in $a$
As $a$ is real, the discriminant must be $\ge0$
Use this fact to find the range of values of $c$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f(z)= z + \frac{1}{z}$ and $z$-points move on the circle $|z|=R$, then what is the locus of $f(z)$ points Here's what I did to graph
$$f(z) = z + \frac{1}{z}$$
where $|z|=R$. Given $|z|=R$ (all points being in a circle of radius R), I can get y depending on x and vice-versa, so $y=\pm \sqrt{R^2-x^2}$.
After some rearrangements
$$f(x,y) = x\frac{R^2+1}{R} + i*y\frac{R^2-1}{R^2}$$
From where I got the real and imaginary parts of the graph.
Real: $$u(x) = x\frac{R^2+1}{R^2}$$
And 2 imaginary part functions:
$$v_1(x) = \sqrt{R^2-x^2} \frac{R^2-1}{R^2}$$
$$v_2(x) = -\sqrt{R^2-x^2} \frac{R^2-1}{R^2}$$
Also some helpful inequations I deduced: $$\frac{R^2+1}{R^2} > 1$$
$$\frac{R^2-1}{R^2} < 1$$
$$|x|\leq R$$
$$|y|\leq R$$
By calculating a few values of the functions I got some points on the Argand plane, and a graph might look like an ellipse centered at the origin, but I'm not sure about that.
The question is: what is a graph of that function and how I can obtain it analytically?
| Continue with what you obtained and let $f(x,y)=u+iv$,
$$u+iv= x\frac{R^2+1}{R^2} + iy\frac{R^2-1}{R^2}$$
Then, equate both real and imaginary parts,
$$x=\frac{R^2}{R^2+1}u,\>\>\>\>\>y=\frac{R^2}{R^2-1}v\tag 1$$
Given that $|z|=R$, a circle of radius $R$ center at the origin, we have
$$x^2+y^2 = R^2$$
Substitute (1) into above equation to obtain
$$\frac{u^2}{\frac{(R^2+1)^2}{R^2} }+ \frac{v^2}{\frac{(R^2-1)^2}{R^2} }=1$$
which represents an ellipse centered at origin, with the major and the minor axes of $\frac{R^2+1}{R}$ and $\frac{R^2-1}{R}$, respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Find $a, b$ so polynomial is divisible
Find values $a,b\in\Bbb{R}$ so the polynomial $$P(x)=6x^4-7x^3+ax^2+3x+2$$
is divisible by the polynomial
$$Q(x)=x^2-x+b$$
So what I know and how I do these problems most of the time, is since:
$$P(x)=Q(x)D(x)$$
I would know that by plugging the roots of $Q(x)$ in $P(x)$ should give me enough equations for me to solve this.
So I tried finding the roots of $Q(x)$:
$$x^2-x+b=0\\x_{1/2}={1\pm\sqrt{1-4b}\over2}$$
Okay so this $b$ value is giving me a headache here. The only thing I gathered from this is that (probably) $b\le0$. I tried now plugging this in $P(x)$, as I know that
$P(x_1)=0$ and $P(x_2)=0$
The exponents on everything made this a real pain and I'm pretty certain that it shouldn't be done this way. I'm stuck.
| Simply dividing works:
$$
\require{enclose}
\begin{array}{l}
\phantom{x^2-x+b\,\,\,}6x^2-x+(a-6b-1)\\[-4pt]
x^2-x+b\enclose{longdiv}{6x^4-7x^3+\phantom{6}ax^2+3x+2}\\[-4pt]
\phantom{x^2-x+b\,\,\,}\underline{6x^4-6x^3+6bx^2}\phantom{.00\,00\,00}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad{-x^3+(a-6b)x^2}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad\underline{-x^3+x^2-bx}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad(a-6b-1)x^2+(b+3)x\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4}\quad\underline{(a-6b-1)x^2-(a-6b-1)x+b(a-6b-1)}\\[-4pt]
\phantom{x^2-x+b\,\,\,6x^4(a-6b-1)x^2\quad\quad}(a-5b+2)x+2+b+6b^2-ab
\end{array}
$$
So we need $a=5b-2$ and $6b^2-ab+b+2=b^2+3b+2=0$.
Thus, $(a,b)\in\{(-7,-1),(-12,-2)\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble finding limits at $-\infty$. I'm being asked to evaluate $\lim\limits_{x\rightarrow\infty}f(x)$ and $\lim\limits_{x\rightarrow -\infty}f(x)$ for the following funtion: $f(x)=\frac{4x^3+2}{x^3+\sqrt{4x^6+3}}$.
For the first limit I proceed as follows:
$\begin{align*}
\lim\limits_{x\rightarrow\infty}f(x)&=\lim\limits_{x\rightarrow\infty}\frac{4x^3+2}{x^3+\sqrt{4x^6+3}}\\&=\lim\limits_{x\rightarrow\infty}\frac{(4+\frac{2}{x^3})}{(1+\sqrt{4+\frac{3}{x^6})}}\\&=\frac{4}{3}
\end{align*}$
This is the horizontal asymptote as $x\rightarrow\infty$, however, when I work through it again for $-\infty$, I get the same answer, but I know that as $x\rightarrow -\infty$, $f(x)$ approaches $-4$. Obviously there is some hole in my understanding. Any clarification would be greatly appreiciated! Thank you.
| Hint
Setting limit to $0^+$ often helps me in limit
Set $-\dfrac1x=h\implies h\to0^+$
$$\sqrt{4x^6+3}=\sqrt{\dfrac{4+3h^6}{h^6}}=\dfrac{\sqrt{4+3h^6}}{|h^3|}$$
As $h>0,|h^3|=|h|^3=(+h)^3$
But $x^3=\left(-\dfrac1h\right)^3=-\dfrac1{h^3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
$\lim\limits_{x \to 0}\frac{2x^6+6x^3}{4x^5+3x^3}$ Find $$\lim\limits_{x \to 0}\frac{2x^6+6x^3}{4x^5+3x^3} .$$
The answer is $2$. Can someone explain to me how to arrive at that?
| Hint: If $x\neq0$ then
$$\frac{2x^6+6x^3}{4x^5+3x^3}=\frac{2x^3+6}{4x^2+3}.$$
| {
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"url": "https://math.stackexchange.com/questions/3515309",
"timestamp": "2023-03-29T00:00:00",
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Evaluating the limit $ \lim_{x \to 0} \frac{x}{p} \lfloor \frac{q}{x} \rfloor$
$$\lim_{x \to 0} \frac{x}{p} \left\lfloor \frac{q}{x} \right\rfloor =\;\;\; ?$$
My attempt:
$\frac{q}{x}-1 < \lfloor \frac{q}{x} \rfloor \le \frac {q}{x} $ . Then I consider four cases. In case 1 I took x and p both positive and then I multiplied above by $\frac{x}{p} $ and took limit and applied squeeze theorem. In other cases I did the same. My answer is $ \frac{q}{p} $.
Is the solution correct?
| Your solution is correct.
Another way to solve it without squeeze theorem: write
$$
\frac{q}{x} = n + r,
$$
where $n \in \mathbb{Z}$ and $0 \le r < 1$. As $x \to 0$, $n \to \infty$ (or $-\infty$ if $q$ is negative) and $r$ varies between $0$ and $1$. Solving for $x$ in the above we have
$$
x = \frac{q}{n + r}.
$$
Now, in the original expression we get
\begin{align*}
\frac{x}{p} \left\lfloor \frac{q}{x} \right\rfloor
&= \frac{1}{p} \left(\frac{q}{n + r} \right) \lfloor n + r \rfloor \\
&= \frac{1}{p} \left(\frac{q}{n + r} \right) \cdot n \\
&= \frac{q}{p} \cdot \frac{n}{n + r}.
\end{align*}
As $x \to 0$, as we observed earlier $n \to \infty$ and $r$ varies between $0$ and $1$. But as $n \to \infty$, $\frac{n}{n + r} \to 1$ uniformly in $r$ -- meaning that it converges to $1$ regardless of how $r$ varies. So the limit of the above as $x \to 0$ becomes
$$
\frac{q}{p} \cdot 1 = \frac{q}{p}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Find all positive integers $n $ such that $n^2+n+7$ devided by 13. An idea please Find all positive integers $n $ such that $n^2+n+7$ devided by 13.
An idea please
| Since $13$ is a prime number, then $\mathbb Z_{13}$ is a field. Hence the quadratic equation will work.
$$n^2+n+7 \equiv 0 \pmod {13}$$
Note that $2 \cdot 7 \equiv 1 \pmod {13}$. So $\dfrac 12 \equiv 7 \pmod{13}$.
\begin{align}
n &\equiv \dfrac{-(1) \pm \sqrt{(1)^2-4(1)(7)}}{2(1)} \pmod{13} \\
n &\equiv 7 \cdot \left(-1 \pm \sqrt{1-28} \right) \pmod{13} \\
n &\equiv 7 \cdot \left(-1 \pm \sqrt{-27} \right) \pmod{13} \\
n &\equiv 7 \cdot \left(-1 \pm \sqrt{12} \right) \pmod{13} \\
n &\equiv 7 \cdot \left(-1 \pm 5 \right) \pmod{13}
&\text{(Note $5^2 \equiv 25 \equiv 12 \pmod{13}$.)} \\
n &\in \{2, 10\}
\end{align}
So $n^2+n+7$ is divisible by $13$ when $n$ is equivalent to $2$ or $10$ modulo $13$.
With some fooling around, you can do this without using the quadratic formula.
$$n^2 + n + 7 \equiv n^2-12n+20 \equiv (n-2)(n-10) \pmod{13}$$
so $n \in \{2,10\}$
or
$$n^2 + n + 7 \equiv n^2+n-6 \equiv (n+3)(n-2) \pmod{13}$$
so $n \in \{2,-3\} \equiv \{2,10\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta+\frac{\pi}{2}\cot\theta\csc\theta$
Prove $$
\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta+\frac{\pi}{2}\cot\theta\csc\theta
$$
$$
\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=-\frac{\sin\theta}{\cos\theta}\int_0^\theta\frac{-\cos\theta\cos x+1-1}{(1-\cos\theta\cos x)^2}dx\\
=-\tan\theta\int_0^\theta\bigg[\frac{1}{1-\cos\theta\cos x}-\frac{1}{(1-\cos\theta\cos x)^2}\bigg]dx\\
$$
How do I solve it ?. Can I use Leibniz rule here ?
Thanks @Peter Foreman
$$
\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=-\tan\theta\int_0^\theta\Bigg[\frac{1}{1-\cos\theta.\dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}-\frac{1}{\bigg(1-\cos\theta.\dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\bigg)^2}\Bigg]dx\\
=-\tan\theta\int_0^\theta\Bigg[\frac{\sec^2\frac{x}{2}}{1-\cos\theta+\tan^2\frac{x}{2}[1+\cos\theta]}+\frac{(1+\tan^2\frac{x}{2})\sec^2\frac{x}{2}}{\bigg(1-\cos\theta+\tan^2\frac{x}{2}[1+\cos\theta]\bigg)^2}\Bigg]dx
$$
Set $t=\tan\frac{x}{2}\implies dt=\frac{1}{2}\sec^2\frac{x}{2}dx$
$$
I_1=-\tan\theta\int_0^{\tan\frac{\theta}{2}}\frac{2dt}{1+t^2-\cos\theta[1-t^2]}
$$
| Note that
$$I(\theta) = \int_0^\theta\frac{1}{1-\cos\theta\cos x}dx=\frac\pi2 \csc\theta $$
and take the derivative with respect to $\theta$ to obtain
$$\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta-I'(\theta)
= \csc^2\theta+\frac{\pi}{2}\cot\theta\csc\theta $$
| {
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Let $a,b ∈N$ with $\gcd(a,b) = 1$. Find the number of ordered pairs $(a,b)$ such that: $\frac {a}{ b} + \frac {201b}{10201a} ∈ N$.
Let $a,b \in \mathbb{N}$ two coprime numbers, so $\gcd(a,b) = 1$.
Find the number of ordered pairs $(a,b)$ such that: $\frac {a}{ b} + \frac {201b}{10201a} \in \mathbb{N}$.
I got $10201ab | (10201a^2 +201b^2)$. How to proceed further?
Edit: As pointed out in the comment, I noticed $101^2=10201$. Therefore it can also be written as: $101^2ab|101(101a^2+2b^2)-b^2$
| Right.. so $\frac {201b^2}{10201a} = kb -a$ (for some natural $k$) is an integer.
But $a$ and $b$ are relatively prime so $a|201 = 3*67$.
So 4 cases: $a=1, 3, 67,$ or $201$.
In each case if we let $a'=\frac {201}3$ we get $\frac ab + \frac {a'b}{101^2} = k$ and $a + \frac {a'b^2}{101^2} = kb$ and as $\gcd(201, 101) = 1$ we have $\gcd(201,a)=1$ and $b = 101m$ is a multiple of $101$.
So we have $a + a'm^2 = 101km$ which means $m|a$.
If $m=a$ we get $a + 201*a = 202a = 101ka$ and $k= 2$.
And we get
$\frac {1,3,67,201}{101,303,6767,20301} + \frac {201*(101,303,6767,20301)}{10201*(1,3,67,201)}= 2$.
The only things to check are if $m|a$ but $m < a$. But if $m < a$ then $a'm\le 67$ and $\frac am+a'm = 101k$ will never be possible.
| {
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Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$. This is the final step of one old competition problem, and in textbook it simply says "reader can prove this", but I found this a bit nontrivial.
Show that $\sin^220^\circ\sin40^\circ = \sin10^\circ \sin30^\circ \sin60^\circ$.
I tried a lot of things and formula on this, but never succeeded.
Can someone help?
| $LHS=\sin^2 20^\circ \sin 40^\circ =\sin 20^\circ\cdot\frac{1}{2}(\cos20^\circ -\cos60^\circ)$
$=\frac{1}{2}(\sin 20^\circ \cos20^ \circ-\frac{1}{2}\sin 20^\circ)=\frac{1}{4}(\sin 40^\circ-\sin 20^\circ)$
$ =\frac{1}{4}\cdot2\sin 10^\circ\cos 30^\circ=\frac14\sqrt3\sin 10^\circ. $
$RHS=\sin10^\circ \sin30^\circ \sin60^\circ=\frac14\sqrt3\sin 10^\circ.$
$LHS=RHS.$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$$
Let's see the term on R.H.S with coefficient as $b_n$
$$T_{n+1}=b_n(x-3)^n$$
Comparing the coefficients of $x^0$
$$a_0-2a_1+(-2)^2a_2+(-2)^3a_3+\cdots a_{2n}(-2)^{2n}=b_0-3b_1+(-3)^2b_2+(-3)^3b_3+\cdots+b_n(-3)^n+\cdots b_{2n}(-3)^{2n}$$
Comparing the coefficients of $x^1$
$$a_1+2a_2(-2)^1+3a_3(-2)^2\cdots 2n\cdot a_{2n}(-2)^{2n-1}=b_1+2a_2(-3)^1+3a_3(-3)^2\cdots nb_n(-3)^{n-1} \cdots (2n)a_{2n}(-3)^{2n-1}$$
Now I am not getting any way to find $b_n$ , its so mixed up in these expressions, not able to get any breakthrough. Any inputs?
| Hint Derivate $n$ times and plug in $r=3$.
Let $f(x)=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$. Then
$$f^{(n)}(3)=n! b_n$$
Since
$$f(x)=a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}$$
you also have
$$f^{(n)}{3}=n!a_n+\frac{(n+1)!}{1!}a_{n+1}+\frac{(n+2)!}{2!}a_{n+2}+...+\frac{(2n)!}{n!}a_{n+1}$$
Now make those equal and plug in $a_k=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 \ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 $ Let real $a,b,c,d > 0$. Show that
\begin{align}
\frac{1}{64}(a^{15} + b^{15} + c^{15}+ d^{15})^4 &\ge \sum_{\rm {cyc}} (a b^2 c^4 d^8 )^4 \\
&= (a b^2 c^4 d^8 )^4 +(b c^2 d^4 a^8 )^4 +(c d^2 a^4 b^8 )^4 +(d a^2 b^4 c^8 )^4
\end{align}
This is obviously homogeneous. Equality appears at $a=b=c=d$, I haven't found other equality points. Applying AM-GM inequality to the LHS is too crude because it doesn't take into account the rising exponentials $(1,2,4,8)$ on the RHS. Indeed, AM-GM leaves to prove
\begin{align}
4 &\ge \sum_{\rm {cyc}} \frac{c\cdot d^{17}}{b^7 \cdot a^{11}}
\end{align}
which can be made to fail easily, e.g. by setting $a=b=c =0.01 \cdot d$.
| The answer to the corresponding problem (see Michael's comment) has been provided by me at math.stackexchange.com/a/3566860/317854
Hence the problem is solved.
| {
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"url": "https://math.stackexchange.com/questions/3527230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible to solve an equation such as $x^2+5^x - 10 = 0$ without using graphical methods? I tried using logarithms to find the answer to $x^2+5^x-10=0$ but I didn't have any luck. Is there a way of solving the above equation algebraically, or do you have to use a graphical method?
| $x^2+5^x-10=0$
$x= ± \sqrt{10-5^x}$
$10-5^x=1, 4, 9$
$10-5^x=1$ ⇒ $x=\frac{log (9)}{log (5)}=1.36$
$10-5^x=4$ ⇒ $x=\frac{log (6)}{log (5)}=1.11$
$10-5^x=9$ ⇒ $x=\frac{log (1)}{log (5)}=0$
Non of these results satisfy the equation.But we can use try and error method.We can see that x must be between $1.36$ and $1.11$, we try $x=1.3$:
$1.3^2=1.69=1-5^x$ ⇒ $x=\frac{log 8.31}{log 5}≈1.3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?
Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$
Can I now say that $\ln\left(\frac{1+x}{1-x}\right)=2\left(\sum^{\infty}_{n=0}\int x^n\right)^2$ ? This would get: $2\sum^{\infty}_{n=0}\frac{x^{2(n+1)}}{(n+1)^2}$.
I'm assuming the answer is wrong because the answer key did not agree with me. Did I mess up somewhere in this problem?
The answer key only doubles the $x$ in this step: $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$ so instead of $2\int\frac1{(1-x)^2}\ dx$ they get just $2\int\frac1{1-x^2}\ dx$. Why is this correct? (or is it incorrect?)
| $$\frac{d}{dx} \ln \left(\frac{1+x}{1-x} \right) = \frac{d}{dx} [\ln (1+x) -\ln (1-x)] = \frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{(1+x)(1-x)} = \frac{2}{1-x^2}.$$
$$\ln \left(\frac{1+x}{1-x} \right) = 2\int \frac{dx}{1-x^2} = 2\int dx [1+x^2 +x^4 + x^6 + \cdots] = 2[x+x^3/3+x^5/5+x^7/7 \cdots]. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve system of trigonometric equations $24^2=l^2+45^2-90 l\cos\alpha$, $51^2=l^2+45^2-90 l \cos(\frac{\pi}{3}-\alpha)$ Recently, I have found this problem:
If $\alpha\in (0,\pi/3)$, then find the solutions of the trigonometric system:
$$\left\{\begin{matrix}
24^2=l^2+45^2-90 l\cos\alpha
\\ 51^2=l^2+45^2-90 l \cos(\frac{\pi}{3}-\alpha)
\end{matrix}\right.$$
I have definitely no idea of how to proceed: I think that maybe the sum-substraction method can be useful, but when I reach: $$24^2-51^2=2\cdot l \cdot \left ( \cos\left ( \frac{\pi}{3}-\alpha \right )-\cos\alpha\right )$$ I am stuck. Any idea of how to go further?
| Rearrange the two equations as
$$
l^2+45^2-24^2 =90l\cos\alpha\tag 1$$
$$l^2+45^2-51^2=90l \cos(\frac{\pi}{3}-\alpha)\tag 2
$$
(1) - (2)
$$51^2 - 24^2 = 90l\left[ \cos\alpha - \cos(\frac{\pi}{3}-\alpha)\right]
=90l \cos(\frac\pi3+\alpha)\tag 3$$
(2) - (3)
$$l^2+45^2 + 24^2 - 2\cdot 51^2
=90\sqrt3 l\sin\alpha \tag 4$$
$\frac13 (4)^2 + (1)^2$ to get the quadratic equation in $l^2$,
$$\frac13(l^2+45^2 + 24^2 - 2\cdot 51^2)^2 + (l^2+45^2-24^2)^2 = 90^2l^2$$
Solve for $l^2=9(289\pm 120\sqrt3)$ and then plug into (1) to identify the valid solutions below
$$l=3\sqrt{289+ 120\sqrt3}, \>\>\>\>\>\alpha=\cos^{-1}\left(\frac l{90}+\frac{161}{10l}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Subsets of $\{1,2 \dots n\}$ with no consecutive integers How many subsets with cardinality k of $\{1, 2, \dots n\}$ contain no consecutive integers? I know that there are $F_{n+2}$ subsets of $\{1, 2, \dots n\}$ with no consecutive integers, but I do not know how to go about finding the number for a given $k$.
| We select the first value of the set:
$$\frac{z}{1-z}$$
followed by $k-1$ differences that are at least two:
$$\frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}$$
and we conclude by collecting the count for all subsets with maximum
element $\le n:$
$$[z^n] \frac{1}{1-z} \times \frac{z}{1-z}
\left(\frac{z^2}{1-z}\right)^{k-1}.$$
This is
$$[z^n] \frac{z^{2k-1}}{(1-z)^{k+1}}
= [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}}
= {n+1-2k+k\choose k}$$
or equivalently
$$\bbox[5px,border:2px solid #00A000]{
{n+1-k\choose k}.}$$
We get for the total
$$\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} {n+1-k\choose k}
= \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}}
\\ = [z^{n+1}] \frac{1}{1-z}
\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} z^{2k} \frac{1}{(1-z)^{k}}.$$
Here the coefficient extractor enforces the range and we may continue
with
$$[z^{n+1}] \frac{1}{1-z}
\sum_{k\ge 0} z^{2k} \frac{1}{(1-z)^{k}}
\\ = [z^{n+1}] \frac{1}{1-z} \frac{1}{1-z^2/(1-z)}
\\ = [z^{n+1}] \frac{1}{1-z-z^2}
= [z^{n+2}] \frac{z}{1-z-z^2} = F_{n+2}.$$
The above construction works for $k\ge 1.$ For $k=0$ we get the empty set, for a total count of one. Note however that ${n+1\choose 0} = 1$ so the formula holds there as well.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Evaluate $\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$ $$\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$$\lim_{x\to 0}\frac{xe^x+x-2e^x+2}{x^3}$$
Limit is in the form "$\frac{0}{0}$", using L'hopital once:
$$\lim_{x\to 0}\frac{e^x+xe^x+1-2e^x}{3x^2}=\lim_{x\to 0}\frac{-e^x+xe^x+1}{3x^2}$$
Again the limit is of the form "$\frac{0}{0}$", using L'hopital again:
$$\lim_{x\to 0}\frac{-e^x+e^x+xe^x}{6x}=\lim_{x\to 0}\frac{xe^x}{6x}$$
Once again the limit is of the form "$\frac{0}{0}$", using L'hopital one more time:
$$\lim_{x\to 0}\frac{e^x+xe^x}{6}=\frac{1}{6}$$
Is there a way to solve it without L'hopital?
| $$
\frac{x(e^x+1)-2(e^x-1)}{x^3}\approx\frac{x(2+x+x^2/2)-2x-x^2-x^3/3}{x^3}=\frac{x^3/2-x^3/3}{x^3}=\frac{1}{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14
Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.
Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function.
Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation.
In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!
$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$
$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$
Time do the chain rule!!!
$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$
Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$
$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$
multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get
$0 = (16x+1)$
And thus $x = \frac{-1}{16}$
Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..
$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$
But $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
| Given
$$f(x) = \sqrt{8x^2+x-3}$$
the horizontal tangent lines of $f(x)$ occur all at points in the domain of $f(x)$ where $f'(x)=0$. You have correctly found the derivative
$$f'(x)=\frac{16x+1}{2\sqrt{8x^2+x-3}}$$
which is a rational function. The zeroes of a rational function are where the numerator is zero. Solving $16x+1=0$, we find that $x=-1/16.$
In order to have a horizontal tangent line at $x=-1/16$, $f(x)$ must be defined at $x=-1/16$.
The domain of $f(x)=\sqrt{8x^2+x-3}$ is where $8x^2+x-3\ge0$. Solving this inequality, we find the domain of $f(x)$ as
$$x\ge \frac{1}{16}\left(\sqrt{97}-1\right)$$
$$x\le \frac{1}{16}\left(-1-\sqrt{97}\right)$$
therefore since $$\frac{-1-\sqrt{97}}{16} < \frac{-1}{16}$$
we see that $x=-1/16$ is not in the domain of $f(x)$. So, $f(x)$ doesn't have any horizontal tangent lines.
| {
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"url": "https://math.stackexchange.com/questions/3537941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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After simplifying $\frac{2^{2017}+1}{3\cdot 2^{2017}}$ to $\frac{n}{m}$ where n and m are coprime, find the remainder when $m+n$ is divided by 1000. After simplifying $\frac{2^{2017}+1}{3\cdot 2^{2017}}$ to $\frac{n}{m}$ where $n$ and $m$ are coprime, find the remainder when $m+n$ is divided by $1000$.
Since the top is odd and the bottom is even, the only thing that can cancel is the $3$. Therefore, I'm looking for $2^{2017}+\frac{2^{2017}+1}{3}$ (mod $1000$). However, I don't know how to find this value. Could someone give me a hint?
| We can write $3m=2^{2017}+1$ and $3n=3\cdot2^{2017}$
Now $3(m+n)=1+2^{2017}(1+3)=1+2^{2019}$
Now as $\phi(125)=100$ and $(2,125)=1$
Like How to find last two digits of $2^{2016}$
$2^{2016}\equiv2^{16}\equiv(2^8)^2\equiv6^2\pmod{125}$
$\implies2^{2016+3}\equiv2^36^2\pmod{125\cdot2^3}$
$\implies3(m+n)\equiv1+36\cdot8\pmod{1000}$
$\implies m+n\equiv289\cdot3^{-1}\equiv(96\cdot3+1)(3^{-1})\equiv96-333+1000$
as $3^{-1}\equiv1\equiv-333\equiv-333+1000$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have that $x + y + z = xyz \implies \dfrac{1}{yz} + \dfrac{1}{zx} + \dfrac{1}{xy} = 1$ and $$\left(\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}\right) \cdot \left(\frac{1}{x - 1} + \frac{1}{y - 1} + \frac{1}{z - 1}\right)$$
$$ \ge \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)^2 \ge 3 \cdot \left(\frac{1}{yz} + \frac{1}{zx} + \frac{1}{xy}\right) = 3$$
Now we need to find the maximum value of $\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1}$.
Let $\dfrac{1}{x} = a, \dfrac{1}{y} = b, \dfrac{1}{z} = c$ which implies that $a, b, c \in (0, 1)$.
It could be observed that $ab + bc + ca = 1$ and $$\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1} = \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c}$$
$$ = 3 - \left(\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}\right) \le 3 - \frac{9}{3 - (a + b + c)} = \frac{-3(a + b + c)}{3 - (a + b + c)}$$
Then I'm stuck.
| Two solution, one is mine, one is other people.
Solution 1:
Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$ then $0<a,b,c<1$ and $ab+bc+ca=1$
Need to prove:$$\sum \frac{b^2-ab^2}{a} \geqq \sqrt{3}-1$$
Or $$(\sum \frac{b^2}{a})^2 (ab+bc+ca) \geqq \Big[(\sqrt{3}-1)(ab+bc+ca) +(a^2+b^2+c^2)\Big]^2 $$
Assume $c=\min\{a,b,c\}$ then let $a=c+u,b=c+v (u,v\geqq 0)$ you get something is obvious, but ugly ;)
Equality holds when $x=y=z=\sqrt{3}$
Solution 2:(of other people)
Let $$P= \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
This solution has been solved by me and Nguyễn Việt Lâm, see here: https://hoc24.vn/hoi-dap/question/972598.html
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$ Let $x,y\in \mathbb{R}$ such that $$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$. Find the minimize value of $x$.
[Edit by Michael Rozenberg] I tried to use AM-GM to retire the radical but failed:
$$27=\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\geq2\sqrt[4]{\left(\frac{x+y}{2}\right)^3\left(\frac{x-y}{2}\right)^3}=$$
$$=2\sqrt[4]{\frac{(x^2-y^2)^3}{64}}=\sqrt[4]{\frac{(x^2-y^2)^3}{4}}.$$
Help me
| Hint
$a+b=27,a,b\ge0$
$a^{2/3}+b^{2/3}=a^{2/3}+(27-a)^{2/3}=f(a)$
Use
http://mathworld.wolfram.com/SecondDerivativeTest.html
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Probability of collecting all 5 buying 7 chocolates I'm struggling to find where I made a mistake on the way to solving the following problem.
Problem description: A grocery sells chocolates bars. There are $5$ kinds of stickers. Each chocolate is sold with $1$ sticker of one of these $5$ kinds. The probability to find any kind of stickers in any chocolate bar is the same.
What's the probability of collecting all $5$ types of stickers if you buy $7$ chocolates at once?
My solution: since the order of chocolate bars that I've bought is irrelevant, the total number of 7-chocolate bar sets that I can buy equals
$$\left(\binom{5}{7}\right)=\binom{7+5-1}{7}=\binom{11}{7}=330$$
Now I'm solving the reverse problem: let's calculate the probability of failing to collect all $5$ kinds of stickers. To do that I need to calculate the number of 7-chocolate bar sets that have less than $5$ kinds of stickers, which means that I need sets with only $1$ kind of stickers, $2$, $3$ and $4$. Since, again, the order of chocolate bars doesn't matter, the number of such sets is
$$\left(\binom{4}{7}\right)=\binom{7+4-1}{7}=\binom{10}{7}=120$$
Finally, my answer to the initial problem should be
$P = 1 - \frac{120}{330}= 0.6363636364$
The answer key says it's $\approx 0.215$
Where's the flaw in my solution?
I appreciate any help.
| The flaw in your solution is that the $330$ cases you enumerated are not equally likely to appear. For instance, there is only one way for all seven chocolate bars to show the first label. However, if we list the stickers in the order in which we look at the chocolate bars we collect, there are $\binom{7}{2}\binom{5}{2}3!$ ways to obtain a collection with two stickers of the first type, two stickers of the second type, and one of each of the other types.
There are five possible stickers for each of the seven chocolate bars, so there are $5^7$ ways to distribute stickers to chocolate bars.
For the favorable cases, we must subtract those distributions in which not all five kinds of stickers appear in the collection of purchased chocolate bars.
There are $\binom{5}{k}$ ways to exclude $k$ of the $5$ stickers and $(5 - k)^7$ ways to distribute stickers of the remaining $5 - k$ kinds to the chocolate bars. Hence, by the Inclusion-Exclusion Principle, the number of ways the stickers may be distributed to the seven chocolate bars so that all five kinds of stickers appear is
$$\sum_{k = 0}^{5} (-1)^k\binom{5}{k}(5 - k)^7 = 5^7 - \binom{5}{1}4^7 + \binom{5}{2}3^7 - \binom{5}{3}2^7 + \binom{5}{4}1^7 - \binom{5}{5}0^7$$
Thus, the probability that all five kinds of stickers appear on the collection of seven chocolate bars is
$$\frac{1}{5^7}\left[5^7 - \binom{5}{1}4^7 + \binom{5}{2}3^7 - \binom{5}{3}2^7 + \binom{5}{4}1^7 - \binom{5}{5}0^7\right]$$
| {
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"url": "https://math.stackexchange.com/questions/3545136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$ I want to prove that for $a,b,c>0$ we have
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=
\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$$
My first attempt: By Cauchy-Schwarz we have $$\left(\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}\right)^2\le3\sum_{cyc}\frac{a^2}{a^2+3b^2+3c^2}$$ so we only need to prove that the right-hand side is always less than $\frac{9}{7}$, but this is false. Failed
Second attempt: By Cauchy-Schwarz
$$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}=\sum_{cyc} \frac 1{\sqrt{1+3\frac{b^2}{a^2}+3\frac{c^2}{a^2}}}\le\sum_{cyc} \frac{\sqrt 7}{1+3\frac{b}{a}+3\frac{c}a}$$
so it remains to prove that $$\sum_{cyc} \frac{a}{1+3b+3c}\le\frac37$$ but this is wrong for example for $a=1,b=1,c=2$. Failed
Third attempt: Let $S=3(a^2+b^2+c^2)$. We need to prove $$\sum_{cyc} \frac{a}{\sqrt{S-2a^2}}\le \frac37.$$ But $x\mapsto \frac{x}{\sqrt{S
-2x^2}}$ is convex so Jensen has the wrong direction...
| Alternative proof:
Use the argument in https://artofproblemsolving.com/community/c6h1822770p12198977 or https://artofproblemsolving.com/community/c6h548438p3180154.
Let $x, y, z > 0$ such that
$\frac{7a^2}{a^2 + 3b^2 + 3c^2} = x^2, \
\frac{7b^2}{b^2 + 3c^2 + 3a^2} = y^2, \
\frac{7c^2}{c^2 + 3a^2 + 3b^2} = z^2$.
It is not hard to obtain
$$F(x, y, z) = 4 x^2 y^2 z^2+8 x^2 y^2+8 x^2 z^2+8 y^2 z^2+7 x^2+7 y^2+7 z^2-49 = 0.$$
It suffices to prove that if $x, y, z > 0$ and $F(x,y,z) = 0$, then $x+y+z \le 3$.
It suffices to prove that if $x, y, z > 0$ and $x + y + z > 3$, then $F(x, y, z) > 0$.
Note that $F(\alpha x, \alpha y, \alpha z) > F(x, y, z)$ for any $\alpha > 1$
and $x, y, z > 0$.
Thus, it suffices to prove that if $x, y, z > 0$ and $x+y+z = 3$,
then $F(x, y, z) \ge 0$.
We use pqr method. Let $p = x + y + z = 3$, $q = xy + yz + zx$ and $r = xyz$.
From $p^2 \ge 3q$, we have $q \le 3$. Let $q = 3(1-u^2)$ for $0 \le u\le 1$.
We have
\begin{align}
(x-y)^2(y-z)^2(z-x)^2 &= -4p^3r+p^2q^2+18pqr-4q^3-27r^2\\
& = 108u^6 - 27(3u^2+r-1)^2
\end{align}
which results in $108u^6 - 27(3u^2+r-1)^2 \ge 0$ and hence $r \le (2u+1)(1-u)^2 \le 3$.
Thus, we have
\begin{align}
F(x, y, z) &= 7p^2-16pr+8q^2+4r^2-14q-49 \\
&= 4(6-r)^2+72u^4-102u^2-100\\
&\ge 4(6 - (2u+1)(1-u)^2)^2 + 72u^4-102u^2-100\\
&= 2u^2(2u^2-4u+9)(2u-1)^2\\
&\ge 0.
\end{align}
We are done.
| {
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"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $m_a\geq \dfrac{b^2+c^2}{4R}$ Let triangle ABC, $m_a$ the lenght of the median from A, $b,c$ the lenghts of the segments AC and AB respectively and R the circumradius. Prove that:
$m_a\geq \dfrac{b^2+c^2}{4R}$.
I found this in a book and the hint was to denote M the midpoint of BC and $A_2$ the second intersection of the median and the circumcircle of the triangle. Then by power of a point we have that $AM \cdot MA_2=a^2/4$. Also $AM+MA_2 \leq 2R$. And then they said this to imply the conclusion.
Please help me understand. Thank you in advance.
| In the standard notation we need to prove that
$$\frac{1}{2}\sqrt{2b^2+2c^2-a^2}\geq\frac{b^2+c^2}{\frac{abc}{S}}$$ or
$$2abc\sqrt{2b^2+2c^2-a^2}\geq(b^2+c^2)\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ or
$$(b^2-c^2)^2(b^2+c^2-a^2)^2\geq0.$$
| {
"language": "en",
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"answer_count": 2,
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Using partial fraction to power that less than 1 I need help to check this answer since, unfortunately, the worksheet doesn't attach the key.
Integrate:
$$\int \frac{1}{\sqrt{x} - \sqrt[3]{x}}dx$$
So, I tried use a substitution that allows the variables to become polynomials. I rewrote the expression to
$$\int \frac{1}{x^\frac{5}{6}\left(x^\frac{-2}{6}- x^\frac{-3}{6}\right)} \, dx$$
when I set $$u = x^\frac{1}{6} \text{ and } \frac{dx}{x^\frac{5}{6}}=6 \, du$$ then
$$\int \frac{6 \, du}{\frac{1}{u^2}-\frac{1}{u^3}} = \int \frac{6 u^3 \, du}{u-1}.$$
This becomes
$$6 \int \left( u^2 + u + 1 + \frac{1}{u-1} \right) \, du = 6 \left(\frac{u^3}{3} + \frac{u^2}{2} + u + \ln\lvert u-1\rvert \right) + c.$$
and last part is to substitute back $u = x^\frac{1}{6}$.
I don't know whether this right or not. This is first time I am dealing with partial factions that have less than one power. Please correct it if this work is wrong. Thanks in advance.
| This is correct. As a check: \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}u} & 6 \left( \frac{u^3}{3} + \frac{u^2}{2} + u + \ln |u-1| \right) \\
&= 6 \left( u^2 + u + 1 + \frac{1}{u-1} \right) \\
&= \frac{6u^3}{u-1} \text{,}
\end{align*}
as expected.
It can be a little easier to start with $\mathrm{lcm}(2,3) = 6$, so
$$ \frac{1}{x^{1/2} - x^{1/3}} = \frac{1}{x^{3/6} - x^{2/6}} $$
and $u = x^{1/6}$, so both $x^{5/6} = u^5$ and $\mathrm{d}u = \frac{1}{6x^{5/6}} \,\mathrm{d}x$, and we conclude $\mathrm{d}x = 6u^5 \,\mathrm{d}u$. Substituting into the integral leaves
$$ \int \frac{6 u^5 \,\mathrm{d}u}{u^3 - u^2} = \int \frac{6 u^3 \,\mathrm{d}u}{u - 1} \text{,} $$
and continue as you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549513",
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"source": "stackexchange",
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} |
Find the derivative of $\frac{5-\frac{1}{x^2}}{x+3}$ Test 2 Review #4
Find the derivative of $\frac{5-\frac{1}{x^2}}{x+3}$
$Solution$:
The first step to calculating this derivative is to rewrite it as so:
$\frac{5-x^{-2}}{x+3}$
Now we will apply the quotient rule:
$\frac{d}{dx}(\frac{5-x^{-2}}{x+3})$
$=(\frac{(x+3)(\frac{d}{dx}(5-x^{-2}))-(5-x^{-2})(\frac{d}{dx}(x+3))}{(x+3)^2})$
$=(\frac{(x+3)((\frac{d}{dx}5-\frac{d}{dx}x^{-2}))-(5-x^{-2})((\frac{d}{dx}x+\frac{d}{dx}3))}{(x+3)^2})$
$=(\frac{(x+3)((0-(-2x^{-3}))-(5-x^{-2})((1)+0))}{(x+3)^2})$
$=(\frac{(x+3)(2x^{-3})-(5-x^{-2})}{(x+3)^2})$
$=(\frac{(2x^{-2}+6x^{-3})-5+x^{-2})}{(x+3)^2})$
$=(\frac{2x^{-2}+6x^{-3}-5+x^{-2})}{(x+3)^2})$
$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})$
This is an acceptable answer. We could multiply the top and bottom by $x^3$:
$=(\frac{3x^{-2}+6x^{-3}-5}{(x+3)^2})(\frac{x^3}{x^3})$
$=(\frac{3x+6-5x^3}{x^3(x+3)^2})$
| I like the following way.
$$\left(\frac{5-\frac{1}{x^2}}{x+3}\right)'=\frac{\frac{2}{x^3}}{x+3}+\left(5-\frac{1}{x^2}\right)\left(-\frac{1}{(x+3)^2}\right)=$$
$$=\frac{2(x+3)-5x^3+x}{x^3(x+3)^2}=\frac{3x+6-5x^3}{x^3(x+3)^2},$$ which gives the same result.
| {
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Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers) Answers obtained from two online integral calculators:
$$\begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\arcsin\left(\dfrac1{\sqrt2}\sqrt{1 - x}\right) + C \\
\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= 2\sin^{-1}\left(\dfrac{\sqrt{x + 1}}{\sqrt2}\right) - \sqrt{1 - x^2} + \text{ constant}
\end{align}$$
Answers from online calculator shown above, and my answers shown in the link:
Update:
I realized that the substitution for $\theta$ was supposed to be $\arcsin$ not $\arccos$, so the answer would have been the same as the right hand side.
But I also noticed that using the initial substitution to plug $x$ back in the final answer will not always give the correct answer because in a similar question:
$$\int \frac{\sqrt{x^2-1}}x dx$$
has a trig-substitution of $x = \sec\theta$, and the answer in terms of $\theta$ would be: $\tan \theta - \theta + C$. Then the final answer in terms of $x$ should be : $\sqrt{x^2-1} - \operatorname{arcsec}(x) + C$.
But online integral calculators give the answer: $\sqrt{x^2-1} - \arctan(\sqrt{x^2-1}) + C$, which doesn't match the original substitution of:
$$x = \sec\theta \to \theta = \operatorname{arcsec}(x)$$
Anyone know why the calculator gives that answer which doesn't match the original trig-substitution of $x = \sec \theta \to \theta = \operatorname{arcsec}(x)$?
| Let
\begin{align}
I &= \int \frac{\sqrt{1+x}}{\sqrt{1-x}}\;dx
= \int \frac{1+x}{\sqrt{1-x^2}}\;dx
\end{align}
Left side:
Let $x = \sin\theta$, $dx = \cos\theta\;d\theta$.
\begin{align}
I &= \int \frac{1+\sin\theta}{\sqrt{\cos^2\theta}}\cos\theta\;d\theta \\
&= \int \left(1+\sin\theta \right)d\theta \\
&= \theta - \cos\theta + c \\
&= \arccos\left(\sqrt{1-x^2}\right) - \sqrt{1-x^2} + c
\end{align}
Right side:
\begin{align}
I &= \int \frac{1}{\sqrt{1-x^2}}\;dx + \int \frac{x}{\sqrt{1-x^2}}\;dx \\
u &= 1 - x^2,\;\; -\frac{1}{2}du = x\,dx \\ \implies
I &= \arcsin x - \frac{1}{2}\int u^{-1/2}\;du \\
&= \arcsin x - \sqrt{u} + c \\
&= \arcsin x - \sqrt{1 - x^2} + c
\end{align}
The answers would be the exact same, if $\arccos\left(\sqrt{1-x^2}\right) = \arcsin x$. And therein lies the difference. On the "Left side", the substitution you originally made was $x = \sin\theta$. So when you replace $\theta$ you should substitute $\theta = \arcsin x$.
By the rules of trig substitution, they should be equivalent. But canonically, the arcsin function has a range of $-\pi/2$ to $\pi/2$, while the arccos function has a range of $0$ to $\pi$. So when you use $\arccos\left(\sqrt{1-x^2}\right)$, as-is you are losing the case where $-1 < x < 0$. The integral has a kink in it, but that's not what you want, seeing as how the function being integrated is continuous and differentiable at $x=0$.
You could shift arccos by an appropriate amount and use that solution, but I think it would be easier to go with arcsin here.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Absolute value of an exponential sum Consider the following sequence
$$
x_n=\Bigl|\sum_{t=1}^n\exp\Bigl\{2\pi i\Bigl[\frac1d-\frac {\lfloor n/d\rfloor}n\Bigr]t\Bigr\}\Bigr|
$$
with some fixed positive integer $d$ such that $d<n$. I am trying to establish a lower bound for $x_n$ for all large values of $n$.
Is it true that $x_n\ge cn$ with some absolute positive constant $c$ for large values of $n$?
If $n$ is an integer multiple of $d$, then $x_n=n$ since $1/d-\lfloor n/d\rfloor/n=0$. In the general case, we have that
$$
\frac1d-\frac{\lfloor n/d\rfloor}n
=\frac{n-d\lfloor n/d\rfloor}d\cdot\frac1n
$$
and
$$
0\le\frac{n-d\lfloor n/d\rfloor}{d}<1
$$
so it seems that $x_n$ should be close to $n$ for large values of $n$, but I have no idea if it is possible to show this rigorously.
Any help is much appreciated!
| (I found my error and corrected it.
I also made the proof simpler.)
$x_n=\Bigl|\sum_{t=1}^n\exp\Bigl\{2\pi i\Bigl[\frac1d-\frac {\lfloor n/d\rfloor}n\Bigr]t\Bigr\}\Bigr|
$.
I will show that
$\begin{array}\\
x_n
&=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\
&\ge n\frac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})|}\\
&\ge n(1-\frac16(\pi\left\{\frac{n}{d}\right\})^2)\\
\end{array}
$
Let
$a
=2\pi(\dfrac1d-\frac {\lfloor \frac{n}{d}\rfloor}{n})
=\dfrac{2\pi}{n}(\dfrac{n}{d}-\lfloor \dfrac{n}{d}\rfloor)
=\dfrac{2\pi}{n}\left\{\dfrac{n}{d}\right\}
$.
Then
$\begin{array}\\
x_n
&=\Bigl|\sum_{t=1}^ne^{ i at}\Bigr|\\
&=\Bigl|e^{ i a}\sum_{t=0}^{n-1}e^{ i at}\Bigr|\\
&=\Bigl|e^{ i a}\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr|\\
&=|e^{ i a}|\Bigl|\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr|\\
&=\Bigl|\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr|
\qquad\text{since } |ab| = |a|\,|b|\\
&=\dfrac{|2\sin( n a/2)|}{2|\sin( a/2)|}
\qquad\text{since } |1-e^{ix}| = 2\sin(x/2)
\text{ (see below)}\\
&=\dfrac{|\sin( n a/2)|}{|\sin( a/2)|}\\
&=\dfrac{|\sin( n \frac{2\pi}{n}\left\{\frac{n}{d}\right\}/2)|}{|\sin(\frac{2\pi}{n}\left\{\frac{n}{d}\right\}/2)|}\\
&=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\
&=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}{|(\frac{\pi}{n}\left\{\frac{n}{d}\right\})\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\
&=n\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})\sin(\frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\
&\ge n\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})|}
\qquad \text{since } \sin(x) \le x\\
&\ge n(1-\frac16(\pi\left\{\frac{n}{d}\right\})^2)
\qquad \text{since } \sin(x) \ge x-x^3/6\\
\end{array}
$
Auxiliary results.
$\begin{array}\\
|1-e^{ix}|
&=|1-(\cos(x)+i\sin(x))|\\
&=|1-\cos(x)-i\sin(x)|\\
&=\sqrt{(1-\cos(x))^2+\sin^2(x)}\\
&=\sqrt{1-2\cos(x)+\cos(x)^2+\sin^2(x)}\\
&=\sqrt{2-2\cos(x)}\\
&=\sqrt{2}\sqrt{1-\cos(x)}\\
&=2\sqrt{(1-\cos(x))/2}\\
&=2|\sin(x/2)|\\
|1+e^{ix}|
&=|1+(\cos(x)+i\sin(x))|\\
&=|1-\cos(x)-i\sin(x)|\\
&=\sqrt{(1+\cos(x))^2+\sin^2(x)}\\
&=\sqrt{1+2\cos(x)+\cos(x)^2+\sin^2(x)}\\
&=\sqrt{2+2\cos(x)}\\
&=\sqrt{2}\sqrt{1+\cos(x)}\\
&=2\sqrt{(1+\cos(x))/2}\\
&=2|\cos(x/2)|\\
(a+ib)(c+id)
&=(ac-bd)+i(ad+bc)\\
|(a+ib)(c+id)|
&=\sqrt{(ac-bd)^2+(ad+bc)^2}\\
&=\sqrt{a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2}\\
&=\sqrt{a^2c^2+b^2d^2+a^2d^2+b^2c^2}\\
|(a+ib)||(c+id)|
&=\sqrt{a^2+b^2}\sqrt{c^2+d^2}\\
&=\sqrt{a^2c^2+a^2d^2+b^2c^2+b^2d^2}\\
&=|(a+ib)(c+id)|\\
\end{array}
$
| {
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Which are the $\max$ and $\min$'s of $\frac{\sin{2t}}{2}$, $t \in [0, 2\pi]$? I have to find the max and mins of:
$f(x,y) = xy $ in set $ x²+y²<=1$
After derivating and finding $(0,0)$ as saddle point by hessian...
and testing the border:
$x = cost$, $y = sentf(cost,sent) = 1/2.sin(2t)..$
Wich are the max and mins of $(1/2)*sin(2t)$, $0<=t<=2pi$?
I know that critical points are $t = \frac{\pi}{4}$ and $t = \frac{3\pi}{4}$
But the book shows $t = \frac{5\pi}{4}$ and $t = \frac{7\pi}{4}$ critical points too.Why? Someone can help?
Thanks!
| As indicated in Maximilian Janisch's question comment, since $\sin(x)$ is periodic with period $2\pi$, you have that $\sin(2t)$ is periodic with period $\frac{2\pi}{2} = \pi$.
Thus, the values of $\sin(2t)$ for $t = \frac{\pi}{4}$ and $t = \frac{5\pi}{4} = \frac{\pi}{4} + \pi$ are the same, with both giving $\sin(2t) = 1$, so $\frac{\sin(2t)}{2} = \frac{1}{2}$ is a maximum at both points. Also, those for $t = \frac{3\pi}{4}$ and $t = \frac{7\pi}{4} = \frac{3\pi}{4} + \pi$ are the same, with both giving $\sin(2t) = -1$, so $\frac{\sin(2t)}{2} = -\frac{1}{2}$ is a minimum at both points. Also, finally note that $\frac{5\pi}{4} \in [0,2\pi]$ and $\frac{7\pi}{4} \in [0,2\pi]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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why is $e^{\frac{\pi i}{3}}$ a root of $f(x) = x^{2}-x+1$? I am not sure how this comes about since, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$
| $(z^a)^b \ne (z^b)^a$ if $a,b$ aren't both integers.
Indeed $z^{\frac 1n}$ is not well (or usefull) defined at all. There are $n$ values of $w$ so that $w^3 = z$ so if we decide arbitrarily that one is correct and the other $n-1$ aren't we can do anything because if we have $w^n = z$ we can't play God and assume that one of them is right and the others aren't.
$e^{\frac {2\pi i}n} := \cos \frac {2\pi}n + i\sin \frac {2\pi}n$.
It is NOT true that $e^{\frac{2\pi i}n} = (e^{2\pi i})^{\frac 13}$. And although $e^{2\pi i} =1$. And $1^{\frac 13}$ is not well defined and it doesn't make any sense to say "$1^{\frac 13} =1$" when talking of complex numbers.
$f(e^{\frac {2\pi i}3}) = (e^{\frac {2\pi i}3})^2 -e^{\frac {2\pi i}3} + 1=$
$e^{\frac {4\pi i}3}-e^{\frac {2\pi i}3} + 1=$
$(\cos \frac {4\pi}3 + i\sin \frac {4\pi}3) - (\cos \frac {2\pi}3 + i\sin \frac {2\pi}3) + 1=$
$(-\frac 12-i\frac {\sqrt 3}2)-(-\frac 12+i\frac {\sqrt 3}2) + 1 = $
$(-\frac 12 -\frac 12) + (-\frac {\sqrt 3}2 + \frac {\sqrt 3}2)i + 1=$
$-1 + 0i + 1 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561215",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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if $x$ is odd, show that $x^3+x$ has a remainder 2 when divided by 4 I did part of this question but am stuck and don't know how to continue
I let $x= 2k +1$
Also noticed that $x^3+x = x(x^2+1)$
therefore
$4m+2 = 2k+1((2k+1)^2+1)$
I simplified this and ended up with
$4m+2 = 8k^3+12k^2+8k+2$
I don't know how to continue from and prove that $x^3+x$ has remainder 2 when divided by 4
| You didn't do your math right Let $x= 2k + 1$ so $x(x^2 + 1) = (2k+1)((2k+1)^2 + 1)=$
$(2k+1)((2k+1)^2 + 1) = 8k^3 + 12k^2 + 8k +2$ so you had a typo.
But it's easier to do
$(2k+1)^3 + (2k + 1)=$
$(2k)^3 + 3(2k)^2 + 3(2k)+ 1 + (2k + 1)=$
$8k^3 + 12k^2 + 6k + 1 + 2k + 1=$
$8k^3 + 12k^2 + 8k + 2$.
So $4|8, 12, 8$ the $4|8k^3 + 12k^2 + 8k$ and the remainder of $8k^3 + 12k^2 + 8k + 2$ is $2$.
....
Also ... this may be overly abstract: If we let $x = 2k + 1 = m + 1$ where $m=2k$ is an even number, then $x^3 +x = (m+1)^3 + (m+1) = a_3m^3 + a_2m^2 + a_1m + a_0$, a polynomial. For $j\ge 2; m^j = (2k)^j = 2^j*k^j=4*2^{j-2}{k^j}$ we have $4|a_jm^j$.
So the remainder when divided by $4$ will be the same as the remainder of $a_1m + a_0$.
By binomial theorem: $(m+1)^3 = \sum_{i=0}^3 {3\choose i}m^i$ so for $(m+1)^3 + (m+ 1)$ we have $a_1 = {3\choose 1} + 1$ and $a_0 = {3\choose 0} + 1$.
${3\choose 1} = 3$ and $3+1=4$ so $4|a_jm$ and the remainder when divided by $4$ will be the same as the remainder of $a_0 = {3\choose 0} + 1$ when divided by $4$.
${3\choose 0} = 1$ and 1+1 = 2
| {
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What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$ What does the following expression converge to?
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$$
(It looks like this problem)
$\displaystyle\sum_{k=0}^n a_{2k+1}+a_{2k+2}-a_{k+1}$ ;
$\displaystyle\sum_{k=0}^n a_{k+1}$ does not converge.
| We can write the sum like this:
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\sum_{k=0}^n\left(\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)-\sum_{k=0}^n\left(\frac{1}{4k+4}+\frac{3}{2}\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{2k+2}\right)\right)=H_{4n+4}-\frac{1}{4}H_{n+1}-\frac{3}{2}H_{2n+2}+\frac{3}{4}H_{n+1}=H_{4n+4}+\frac{1}{2}H_{n+1}-\frac{3}{2}H_{2n+2}$$
Using $H_n = \ln n+\gamma +\mathcal{E}_n$ with $\mathcal{E}_n \to 0$, we can see that:
$$\lim_{n\to \infty}\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\lim_{n\to \infty} \left(\ln(4n+4)+\frac{1}{2}\ln(n+1)-\frac{3}{2}\ln(2n+2)\right)=\ln \sqrt{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i \ne a_j$ for $1\le i,j \le n$
Example: $2^6=64=7^2+3^2+2^2+1^2+1^2$ here $1^2$ repeat two times so this is not allowed.
My incomplete attempt for arithmetic squares
Edit: check related new post, Can a sum arithmetic square ever equal to power of two?
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
| No. 2 is a direct prime in the gaussian integers, which means that some power of it is twice another mumber. Thus 1+i squared is 2i, whereas 2+i squared is 3+4i.
Since the sum of two squares equate to non-direct numbers squared, it does not happen for direct numbers.
A similar thing is seen in eisenstein numbers, which leads to a^2+ab+b^2 being square. 3 divides instances, but you get free powers only for numbers comprised of primes 6n+1.
| {
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Find all integer solution for $x^2$ $+$ $y^2$ $=(x-y)^3$ The question is
Find all integer solution for $x^2$ $+$ $y^2$$=(x-y)^3$
For $x$ and $y $ belongs to integer
I tried this
$(x-y)^2=(x-y)^3-2xy$
Then let $x-y=a$
By solving it further and substituting $y =x-a$
We get $2x^2-2ax+a^2-a^3=0$
Now using quadratic formula
And little but of working on conditions i derived
$x= \dfrac{m^3+m^2+m+1}4 , \dfrac{-m^3+m^2-m+1}4$
And
$x-y=\dfrac{m^2+1}2$
Where $m$ belongs to all odd numbers
no doubt that this gives all solution but my doubt is
*
*Are there infinitly many solution .If not how to prove finitely many exists
*I want to see other elegant solutions because it took me lot of time and computations for getting the above result.
All answers would be appreciated.thanks in advance
Edit : my first querry has been solved but i would like to see some more answers.
| $x=a\ GCD(x,y)\ \ ,\ \ y=b\ GCD(x,y)$ where $GCD(a,b)=1$. Now continuing Your work.
$$
\begin{aligned}
(x-y)^{2}&=(x-y)^{3}-2xy\\
2xy&=(x-y)^{2}(x-y-1)\\
\\
2ab\ (GCD(x,y))^{2}&=(GCD(x,y))^{2}(a-b)^{2}((a-b)\ GCD(x,y)-1)\\
\\
2ab&=(a-b)^{2}((a-b)\ GCD(x,y)-1)
\end{aligned}
$$
Now, $(a-b)^{2}|2ab$ but since $GCD(a,b)=1$ then $(a-b)^{2}|2$. The only integer solution is $a-b=1$. Substitute $a=b+1$ and $a-b=1$ back to the previous equation.
$$
\begin{aligned}
2(b+1)b&=GCD(x,y)-1\\
GCD(x,y)&=2(b+1)b+1
\end{aligned}
$$
So the solutions are $x=2(b+1)^{2}b+b+1$ , $y=2(b+1)b^{2}+b$ for all integers $b$. There are infinitely many solutions.
Sorry i missed a possibility that at least one of them can be zero as well. Substitute back to original equation to obtain $(0,0)\ (0,-1)\ (1,0)$.
| {
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Find $a,b$ and $c$ if $(1+\sqrt[3]{2})^{-1}$ in the form of $a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$ Given that
$$ (1+\sqrt[3]{2})^{-1} =a+b\sqrt[3]{2}+c\sqrt[3]{2^2}$$
find the value of rationals $a,b,c.$
Solution I tried: I tried to rationalize it, but I'm not getting the answer:
$$\frac{1}{(1+\sqrt[3]{2})}\times \frac{(1-\sqrt[3]{2})}{(1-\sqrt[3]{2})}$$
$$\frac{(1-\sqrt[3]{2})}{1-2^{\frac{2}{3}}}$$
so doing so not getting answer; also, I tried to expand it, but we have condition that $(1+x)^n$ where $n$ is in fraction can be expandable only when $x < 1$, but here the cube root of $2$ is not less than $1.$
Thank you
| If you multiply both sides by $1+\sqrt[3]2$ you get $$1=(1+\sqrt[3]2)(a+b\sqrt[3]2+c\sqrt[3]4)\\
=a+2c+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$$
which we can resolve into three equations
$$1=a+2c\\0=a+b\\0=b+c$$
by equating the parts proportional to $1,\sqrt[3]2,\sqrt[3]4$
which give
$$b=-\frac 13\\
a=\frac 13\\c=\frac 13$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x= \frac{\pi}{10}$, what is the value of $\cos(8x) +\cos(4x)$?
If $x= \dfrac{\pi}{10}$, what is the value of $\cos(8x) +\cos(4x)$?
My try:
$5x=\dfrac{\pi}{2}$ and $10x=\pi \to \cos(8x)=-\cos(2x)\;$ &
$\;\cos(4x)=\sin(x) $ .
How can I complete?
| We can apply sum-to-product identities, i.e. $\quad \cos \alpha + \cos \beta = 2\cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$.
So, we have: $$\cos (8x) + \cos (4x) = 2\cos(6x)\cos(2x) = 2\cos \frac{6\pi}{10} \cos \frac{2\pi}{10} = 2 \cos \frac{3\pi}{5} \cos\frac{\pi}{5}.$$
Next, we know that holds (see https://www.math-only-math.com/exact-value-of-cos-36-degree.html): $$\cos \frac{\pi}{5} = \frac{\sqrt{5} + 1}{4} \quad \quad (\text{hint:} \quad \frac{\pi}{5}=36°).$$
Similarly, we know that holds:
$$\cos \frac{3\pi}{5} = \frac{1 - \sqrt{5}}{4}.$$
Finally, we have:
$$\cos (8x) + \cos (4x) = 2 \frac{1 - \sqrt{5}}{4} \frac{\sqrt{5} + 1}{4} = ... =-\frac{1}{2}.$$
| {
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Minimum $9a+25b+49c$ when $ab+bc+ca+abc=4$ If $a,b,c\ge 0$ such that $ab+bc+ca+abc=4$, find the minimum value of $9a+25b+49c$.
I know that $a+b+c\ge 3$, but I don't this is good to use here. So I tried with Lagrange multipliers:
$$L(x,y,z)=9a+25b+49c+\lambda(ab+bc+ca+abc-4)$$
With the partial derivative I found:
$$\frac{b+c+bc}{9}=\frac{a+c+ac}{25}=\frac{a+b+ab}{49}$$
and with $ab+bc+ca+abc=4$, I found minimum $59$ at $(3,1,1/7)$. My question is, can it be done with traditional ways? I tried to prove $ab+bc+ca+abc\le 4$ when $9a+25b+49c=59$ (with idea from this question: Minimum value when $abc+ab+4bc+9ca=144$), but I got lost after expanding.
| An alternative solution relies on the Cauchy–Bunyakovsky–Schwarz inequality (CBS).
The given constraint $\,ab+bc+ca+abc=4\,$ can be written as
$(a+1)(b+1)(c+1) = 2+(a+1)+(b+1)+(c+1)$, which in turn is equivalent to
$$\sum_{\text{cyc}}{1\over a+2} \:=\:1\tag{1}\,.$$
Thanks to (CBS) and benefitting from $(1)$ we have
$$\begin{align*}
(3+5+7)^2 & \:=\: \left(3\sqrt{a+2}\cdot\frac1{\sqrt{a+2}}+5\sqrt{b+2}\cdot\frac1{\sqrt{b+2}}+7\sqrt{c+2}\cdot\frac1{\sqrt{c+2}}\right)^2 \\[1ex]
& \:\leqslant\: 9(a+2)+25(b+2)+49(c+2) \\[1.5ex]
\iff\quad 59 & \:\leqslant\:9a+25b+49c
\end{align*}$$
Recall that (CBS) gives equality only if one argument is a scalar multiple of the other. This leads to the solution given in the OP.
| {
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Two projectiles at an initial height with a missing angle of projection
David and Goliath are testing what they learned in school, that the
distance travelled by launching a projectile is dependent on a range
of variables; the launch angle, height and velocity of projection.
As Goliath is significantly taller, David decides to throw a small
projectile while standing on a ladder so that he is throwing from the
same vertical height and same spot as Goliath.
Goliath launches first with an angle of 60$^{\circ}$ to the
horizontal. David will then launch the projectile next with the same
velocity of projection as Goliath.
If David wants to match Goliath’s attempt and achieve the same
horizontal distance, at what angle should he launch the projectile if
he can’t do it at an angle of 60$^{\circ}$?
(Use $ = −10ms^{-1}$ in your calculations and use this as the only
force acting upon the projectile.)
If David's projectile's displacement is given by $r_d = v t \cos \theta \hat{i} + (h + v t \sin \theta - 5t^2) \hat{j}$ then David's projectile hits the ground when $\frac{v \sin \theta \pm \sqrt{v^2 \sin^2\theta+20h}}{10} = 0$. Solving this, the time it takes David's projectile to reach the ground is $T_d = \frac{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}{10}$.
If Goliath's projectile's displacement is given by $r = \tfrac{v}{2} t \hat{i} + (h + \tfrac{\sqrt{3}v}{2}t - 5t^2) \hat{j}$ then Goliath's projectile hits the ground when $\frac{\sqrt{3}v \pm \sqrt{3v^2 + 80h}}{20} = 0$. Solving this, the time it takes Goliath's projectile to reach the ground is $T_g = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}$.
For the horizontal distance travelled to match, their terminal horizontal components must match. Hence,
\begin{align*}
v T_d \cos \theta &= \tfrac{v}{2} T_g\\
T_d \cos \theta &= \tfrac{1}{2} T_g\\
\cos \theta &= \frac{T_g}{2} \frac{1}{T_d}\\
&= \frac{\frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}}{2} \frac{10}{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}\\
\cos \theta &= \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) }\\
\end{align*}
I'm having a major mental block on how to obtain an expression for $\theta$ in terms of $h$ and $v$. For what its worth, wolframalpha gives $\theta = -\cos^{-1}\left(-\frac{\sqrt{200 h^2 + 70 h v^2 + 10 \sqrt{3} h v \sqrt{80 h + 3 v^2} + \sqrt{3} v^3 \sqrt{80 h + 3 v^2} + 3 v^4}}{\sqrt{800 h^2 + 80 h v^2 + 8 v^4}}\right)$. I seem to missing out some details in my calculations.
What's the simplest way to obtain the formula for $\theta$ in terms of $v$ and $h$?
| I got a very different answer.
I start with the following formula given here:
https://en.wikipedia.org/wiki/Range_of_a_projectile
$d = \frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right)$
Now we must have
$\frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right) = \frac{v^2 \sin 120^\circ}{g} = \frac{\sqrt{3}v^2}{2g}$
Solving,
$\theta = \frac{1}{2} \sin^{-1} \frac{3v}{\sqrt{6v^2 + 20y_0}}$
Please check the calculations.
| {
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At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$?
Problem:
At which values of the parameter $k$, there is no solution to the inequality $$(k+1)x^2-2kx+2k+2<0.$$
The solution in my textbook is as follows:
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
$\begin{cases} k+1>0 \\k^2-2(k+1)^2<0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-2k-2 <0 \end{cases} \Longrightarrow -1<k<+\infty.$
Answer: That is, there is no solution to the inequality of $ (k + 1) x ^ 2-2kx + 2k + 2 <0$ for the $ k $ `s that satisfy the $ -1 <k <+ \infty $ condition.
Firstly, I understand the question as follows:
At which values of the parameter $k$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0$ , for all $x\in\mathbb{R}.$
The last sentence is logically equivalent to :
At which values of the parameter $k$, the inequality $(k+1)x^2-2kx+2k+2\geq 0$ holds on for all $x\in\mathbb{R}.$
If I understand the question correctly, here is my solution:
It is obvious that, for $k=-1$ is not a solution.
$\color{black}{\large\text{Case} \thinspace 1:}$ $k+1>0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \geq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0, ∀ x\in\mathbb {R}$$
Then, applying $x=\dfrac{k}{k+1}$ we get, $\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$. We have,
$$\begin{cases} k+1>0 \\ \dfrac{k^2+4k+2}{(k+1)^2}\geq 0 \end{cases} \Longrightarrow k\geq -2+\sqrt2.$$
$\color{black}{\large\text{Case} \thinspace 2:}$ $k+1<0$
We have,
$$x^2-\dfrac{2k}{k+1}x+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2-\dfrac{k^2}{(k+1)^2}+\dfrac{2k+2}{k+1}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2 + \dfrac{2k+2}{k+1}-\dfrac{k^2}{(k+1)^2}\leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac {2(k+1)^2-k^2}{(k+1)^2} \leq 0 \\ \left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\leq 0, ∀ x\in\mathbb {R}$$
For sufficiently large $ x $ `s, we have $\left(x-\dfrac{k}{k+1} \right)^2+\dfrac{k^2+4k+2}{(k+1)^2}\geq 0$, which gives a contradiction.
Finally we deduce that, for all $x\in\mathbb{R}$ satisfying the condition $k\in\mathbb[\sqrt 2-2; +\infty)$, there is no solution to the inequality $(k+1)x^2-2kx+2k+2<0.$
It does not match the solution in my book. Probably, maybe I misunderstand the question or my solution is wrong. Or the book says it wrong.
Did I get the question right? If so, is my solution correct?
Thank you very much.
| There seems some wrong points in your 'textbook solution' you posted, but I don't know if it is your typo or the textbook. Your solution is correct.
Using the textbook solution, we can achieve same result.
$a=k+1, D=4\left[k^2-2(k+1)^2\right]$
If $k + 1 < 0$, we can be sure that for sufficiently large $x$, we always will have solution. we hence assume that $k+1 \geq 0$.
$$\begin{cases} k+1>0 \\k^2-2(k+1)^2 \leq 0 \end{cases} \Longrightarrow \begin{cases} k>-1 \\ -k^2-4k-2 \leq0 \end{cases} \Longrightarrow k \geq \sqrt{2}-2$$
Hence solution is $k \in [\sqrt{2}-2, \infty)$.
| {
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Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???
| Note that $abc = 1 \implies \sqrt[3]{a} \cdot \sqrt[3]{b} = c^{-\frac{1}{3}}$.
Now, cube both sides of $\sqrt[3]{a} + \sqrt[3]{b} = - \sqrt[3]{c}$ and re-arrange the terms to get:
$a + b + c = -3 \left(\sqrt[3]{a} \cdot \sqrt[3]{b} \right) \left(\sqrt[3]{a} + \sqrt[3]{b} \right) = -3 c^{-\frac{1}{3}} (-c^{\frac{1}{3}}) = 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$
My Attempt:
$c$ cannot be $5$ since $70 \times 5 > 343$.
It can't be $4$ either since if we put $c = 4$, we get $30a + 50b < 63$, which would mean there are no positive integer solutions for at least one of $a$ or $b$.
So, there are 3 options for $c$.
Proceeding similarly, we will get that $b = 1,2,3,4$ or there are $4$ options for b.
Similarly, there are $7$ options for $c$.
Since any value of $a$ can be paired with any value of $b$ and $c$, we get $3 \times 4 \times 7=84$ total triplets which is not the answer
| $(x^3+x^6+x^9+x^{12}+...x^{33})\cdot (x^5+x^{10}+x^{15}+x^{20}+x^{30}) \cdot (x^7+x^{14}+x^{21}+x^{28})$
Expand the above using geometric progression sum, and then find the sum of coefficients of $x^k$ where $k\leq 34$
| {
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Is my proof of the Fibonacci sequence correct? Been working on this for some time now but have no idea if it's correct! Any hints are appreciated.
Recall the Fibonacci sequence: $f_1 = 1$, $f_2 = 1$, and for $n \geq 1$, $f_{n+2} = f_{n+1} + f_n$. Prove
that $f_n > (\frac{5}{4})^n\ \forall \ n \geq 3$.
My answer:
"base case"
$[f_3 = 2 > (\frac{5}{4})^3\ = \frac{125}{64}\ correct$
$[f_4 = 3 > (\frac{5}{4})^4\ = \frac{625}{256}\ correct$
$assume\ f_k > (\frac{5}{4})^k\ for\ some\ k \geq 3$
$[and\ f_{k-1} > (\frac{5}{4})^{k-1}$
$then \ f_{k+1} = f_k + f_{k+1} > (\frac{5}{4})^k + (\frac{5}{4})^{k-1}$
$so \ f_{k+1} > (\frac{5}{4})^k + (\frac{5}{4})^{k-1}\ > (\frac{5}{4})^k (\frac{5}{4})^k = (\frac{5}{4})((\frac{5}{4})^k) = (\frac{5}{4})^{k+1}$
$f_{k+1} > (\frac{5}{4})^{k+1}$
$so \ f_n > (\frac{5}{4})^n \ \forall \ n \geq 3$
QED
| As J.W. Tanner mentioned, it's not true that $$\left(\frac{5}{4} \right)^k+ \left(\frac{5}{4} \right)^{k-1} > \left(\frac{5}{4} \right)^k\left(\frac{5}{4} \right)^k$$
(consider for example $k=3$ then $\frac{125}{64} + \frac{25}{16} < \frac{125} {64} \times\frac{125}{64}$
Hint: You can consider $$\left(\frac{5}{4} \right)^k+ \left(\frac{5}{4} \right)^{k-1} = \left(\frac{5}{4} \right)^{k-1} \left(\frac{5}{4} + 1\right) = \left(\frac{5}{4} \right)^{k-1} \left(\frac{9}{4} \right) > \left(\frac{5}{4} \right)^{k-1} \left(\frac{5}{4} \right)^{2}$$
and proceed from there.
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"answer_id": 0
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How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$$\lim_{x\to \pi/3} \left[\dfrac{\sin^2(x) - \sin^2\left(\dfrac{\pi}{3}\right)}{x^2 -\left(\dfrac{\pi}{3}\right)^2}\right]$$
| Considering
$$f=\frac{\sin ^2(x)-\sin ^2\left(\frac{\pi}{3}\right)}{x^2-\left(\frac{\pi}{3}\right)^2}$$ let $x=y+\frac{\pi}{3}$ and work for $y \to 0$. Thsi gives
$$f=\frac{3 \sqrt{3} \sin (y) \cos (y)-3 \sin ^2(y)}{6 y^2+4 \pi y}=\frac{3 \sqrt{3} \cos (y)-3 \sin (y)}{6 y+4 \pi }\times\frac{\sin(y)}y$$ Going to the limit
$$\lim_{y \to 0}\,f=\frac{3 \sqrt{3}}{4\pi}$$
If you want to go beyond the limit, use Taylor series
$$f=\frac{3 \sqrt{3}}{4 \pi }-\frac{3 \left(3 \sqrt{3}+2 \pi \right) y}{8 \pi
^2}+O\left(y^2\right)$$
USe it for $y=\frac \pi {12}$ the exact value is
$$f=\frac{4 \left(\sqrt{3}-1\right)}{\pi ^2}\approx 0.29669$$ while the above truncated series gives
$$f=\frac{21 \sqrt{3}}{32 \pi }-\frac{1}{16}\approx 0.29931$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If I know that $x\in[0,2]$, how can I find the set of values that $2x^2-2x$ can take? I know that:
$$x\in[0,2]$$
and I have to find the set of values that
$$2x^2-2x$$
can take. I tried factoring the above expression into:
$$2x(x-1)$$
in the attempt to form inequalities from the given information (the fact that $0\le x \le 2$) and then multiply the inequalities, but I get inequalities that I cannot multiply, they would give the wrong answer. What approach should I use?
| $2x^2 - 2x = 2(x-1)x$.
The hardest part is when $x-1 < 0$ and $x > 0$ i.e. when $0 < x < 1$
But note: If $0 < x < 1$ then $0 < 1-x$ and $0 < x$ and by AM-GM then $(1-x)x \le ....$ Oh, I can't do that in my head....
$\frac {x + (1-x)}2 \le \sqrt{x(1-x)}$ so
$\frac 12 \le \sqrt {x(1-x)}$ so
$\frac 14 \le x(1-x)$ and $x(1-x) = \frac 14$ when $x =1-x$ or in other words $x=\frac 12$.
So for $0< x < 1$ we have $2(x-1)x \ge -\frac 12$ with $2(x-1)x = -\frac 12$ when $x = \frac 12$.
For $1< x \le 2$ we have $x,x-1 > 0$ so $0=2*0*1 < 2(x-1)x\le 2*1*2=4$.
And at $x = 0$ we have $2x^2 - 2x = 0$ and at $x=2$ we have $2x^2 -2x = 4$.
And as $2x^2 -2x$ is continuous we know it doesn't "jump around" so it hits all points.
So the range of values is from $[-\frac 12, 4]$.
=====
Oh, d'oh.....
A basic study of parabolas
If $y= ax^2 + bx + c$ then the graph is a parabola. The $x$ intercepts are at $x = \frac {-b \pm {\sqrt b^2 -4ac}}2a; y=0$ and the axis of symmetry is at $x= \frac {-b}{2a}$ And the minimum/max value (if $a$ is positive/negative) is $c-\frac {b^2}{4a}$
So minimum value here where $a=2; b=-2; c=0$ is at $x =\frac{-2}{2*2}=\frac 12$ and is $y=0 - \frac {4}{8}=-\frac 12$.
And maximum is at the one of the endpoints... in this case at $x = 2; y=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving two inequalities in calculus I'm trying to prove this two statements:
$$1) \quad \forall\,x\geq 1, \quad \left|\arctan(x)-\frac{\pi}{4}-\frac{x-1}{2}\right|\leq \frac{(x-1)^2}{2}; \\
2)\quad\textrm{If}\;e <a<e^{2}\implies a^{\ln a}-e<4e^3(a-e)\qquad\quad ;
$$
For the second inequality, before trying to solve I saw some things in the statement.
If $a>e\implies \ln(a)>\ln(e)=1\implies a^{\ln a}-e=e^{(\ln a)^{2}}-e>e^{(\ln(e))^{2}}-e=0\implies a^{\ln a}-e>0.$
And... this is all I got. I don't know what to do.
For the first inequality, I've been searching for some similar questions. The answers always comes from a clever function definition and applying the mean value theorem. Is that the way to solve this problem?
I know that I didn't get so close, but any help would be appreciated.
| Here's a proof of the first.
Since
$\arctan(x)-\arctan(y)
=\arctan(\dfrac{x-y}{1+xy})
$
and
$\dfrac{\pi}{4}
=\arctan(1)
$,
$\arctan(x)-\frac{\pi}{4}
=\arctan(x)-\arctan(1)
=\arctan(\dfrac{x-1}{1+x})
$
the first inequality is
$\left|\arctan(\dfrac{x-1}{1+x})-\dfrac{x-1}{2}\right|\leq \dfrac{(x-1)^2}{2}
$.
Let
$y
= \dfrac{x-1}{1+x}
=1-\dfrac{2}{1+x}$
so
$0 \le y \le 1$.
$y+yx = x-1,\\
y+1 = x(1-y),\\
x = \dfrac{1+y}{1-y},\\
x-1
= \dfrac{1+y}{1-y}-1
=\dfrac{1+y-(1-y)}{1-y}
=\dfrac{2y}{1-y},
$
or
$\dfrac{x-1}{2}
=\dfrac{y}{1-y}
$.
This becomes
$\left|\arctan(y)-\dfrac{y}{1-y}\right|
\leq \dfrac{2y^2}{(1-y)^2}
$.
We have
$y
\ge \arctan(y)
\ge y-\dfrac{y^3}{3}
$
(to prove this, integrate
$1 \ge \dfrac1{1+t^2}
\ge 1-t^2$
from $0$ to $y$)
so
$\begin{array}\\
f(y)
&=\dfrac{y}{1-y}-\arctan(y)\\
&\ge \dfrac{y}{1-y}-y\\
&=\dfrac{y-y(1-y)}{1-y} \\
&=\dfrac{y^2}{1-y}\\
&\ge 0\\
\text{and}\\
g(y)
&=\dfrac{y}{1-y}- \dfrac{2y^2}{(1-y)^2}-\arctan(y)\\
&\le \dfrac{y}{1-y}- \dfrac{2y^2}{(1-y)^2}-(y-\dfrac{y^3}{3})\\
&= \dfrac{(y - 3) y^2 (y^2 + y + 1)}{3 (y - 1)^2}
\qquad\text{(according to Wolfy)}\\
&\le 0
\qquad\text{since } y < 3\\
\end{array}
$
Therefore
$0
\le \arctan(y)-\dfrac{y}{1-y}
\le \dfrac{2y^2}{(1-y)^2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Doubt on a recurrence equation I have the following recursive equation for a pmf $P(s)$:
$$P(s+1)=2 \lambda (1-\lambda) P(s)+ \lambda^2 \sum_{s'=0}^s P(s') P(s-s')$$
Here $s$ is a natural number and I have two initial conditions $P(0)=0$ and $P(1)=(1-\lambda)^2$.
I tried to solve it using a generating function $G(x)=\sum_{s=0}^\infty x^s P(s)$. Using it I get:
$$\frac{1}{x} G(x)=2 \lambda (1-\lambda) G(x) + \lambda^2 G^2(x)$$
but I think something is wrong (i.e. it does not satisfy $G(1)=1$). Any suggestions?
| Cintuing Simply Beautiful Art's work.
$G(x)=(1-\lambda)^2x+2\lambda(1-\lambda)xG(x)+\lambda^2xG(x)^2
$
so,
writing $c$ for $\lambda$,
$0
=c^2xG(x)^2+(2c(1-c)x-1)G(x)+(1-c)^2x
$.
$\begin{array}\\
D
&=(2c(1-c)x-1)^2-4c^2x(1-c)^2x\\
&=4c^2(1-c)^2x^2 -4c(1-c)x+1-4c^2x(1-c)^2x\\
&=-4c(1-c)x+1\\
&=1+(4c^2-4c)x\\
&=1+(4c^2-4c+1-1)x\\
&=1-x+(2c-1)^2x\\
\end{array}
$
so
$\begin{array}\\
G(x)
&=\dfrac{-(2c(1-c)x-1)\pm\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\
&=\dfrac{2c^2x-2c+1\pm\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\
&=1-\dfrac{2c-1\mp\sqrt{1-x+(2c-1)^2x}}{2c^2x}\\
G(1)
&=1-\dfrac{2c-1\mp\sqrt{(2c-1)^2}}{2c^2}\\
&=1-\dfrac{2c-1\mp(2c-1)}{2c^2}\\
&=1-\dfrac{0, 4c-2}{2c^2}\\
&=1, 1-\dfrac{3c-1}{c^2}\\
\end{array}
$
So $G(1) = 1$
for the positive square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all positive integer $a,b,c$ that $a^3+b^3+c^3$ can be divided by $a^2b,b^2c,c^2a$
Find all triplets of positive integers $(a,b,c)$ for which
$$a^3+b^3+c^3$$
is divisible by $a^2b$, $b^2c$ and $c^2a$.
I just found that $a=b=c$ satisfies the problem. Are there any other possible answers?
How to find all possible answers?
| Let $a$, $b$ and $c$ be positive integers for which $a^3+b^3+c^3$ is divisible by $a^2b$, $b^2c$ and $c^2a$. Then $\gcd(a,b)^3$ divides $a^2b$, $a^3$ and $b^3$, and hence also
$$(a^3+b^3+c^3)-a^3-b^3=c^3,$$
which shows that $\gcd(a,b)$ divides $c$, so $\gcd(a,b,c)=\gcd(a,b)$. By symmetry
$$\gcd(a,b,c)=\gcd(a,b)=\gcd(a,c)=\gcd(b,c),$$
and dividing $a$, $b$ and $c$ by this common factor yields another solution, so we may assume without loss of generality that $\gcd(a,b,c)=1$. Then it follows that $a^3+b^3+c^3$ is divisible by
$$\gcd(a^2b,b^2c,c^2a)=a^2b^2c^2.$$
Also without loss of generality $a\leq b\leq c$, and so
$$a^2b^2c^2\leq a^3+b^3+c^3\leq3c^3,$$
which shows that $a^2b^2\leq3c$. In particular, because $c^2$ divides $a^3+b^3+c^3$ we also have
$$c^2\leq a^3+b^3\leq2b^3,$$
and putting this together with $b^2\leq a^2b^2\leq3c$ we find that
$$c^4\leq4b^6\leq108c^3,$$
and hence that $c\leq108$. From here it is easy to verify that $(1,1,1)$ and $(1,2,3)$ are the only valid triplets (up to permutation and scaling) though without the help of a computer this may take some time.
To verify this manually, note that from $a^2b^2\leq3c$ and $a\leq b$ and $c\leq108$ we get
$$b\leq\frac{\sqrt{3c}}{a}\qquad\text{ and }\qquad a\leq\sqrt[4]{3c}<5.$$
For $a=4$ we have $b\leq\tfrac{\sqrt{3c}}{a}<5$ and so from $a\leq b$ we get $b=4$. This contradicts $\gcd(a,b)=1$.
For $a=3$ we have $b\leq\frac{\sqrt{3c}}{a}\leq6$. From $\gcd(a,b)=1$ it follows that $b\in\{4,5\}$ and so either
$$a^3+b^3=7\times13\qquad\text{ or }\qquad a^3+b^3=2^3\times19.$$
Because $c^2$ divides $a^3+b^3$ and $c\geq b$ we again reach a contradiction.
For $a=2$ we have $b\leq\frac{\sqrt{3c}}{a}\leq9$ . Then $b\in\{3,5,7,9\}$ and so the the factorizations
$$2^3+3^3=3\times7\qquad 2^3+5^3=7\times19\qquad 2^3+7^3=3^3\times13\qquad 2^3+9^3=11\times67,$$
again yield a contradiction with the facts that $c\geq b$ and $c^2$ divides $a^3+b^3$.
For $a=1$ we have $b\leq\frac{\sqrt{3c}}{a}\leq18$. Clearly $(a,b,c)=1$ is a solution, and for any other solution we have $c\geq2$ and $c>b$. We know that $c^2$ divides
$$a^3+b^3=1+b^3=(1+b)(b^2-b+1),$$
and computing the factorizations for all $b\leq18$ shows that this product is divisible by a square greater than $b^2$ only if $b=2$, and then $c=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Upper bound on summation involving fractional part Let $x\in[1,+\infty)\subset\mathbb{R}$. I would like to show that
$$\sum_{d=1}^{\lfloor x\rfloor}\left(\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\right)\leq x-1,$$
where $\lfloor\cdot\rfloor$ is the floor function. I understand that, since $$\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\leq1,$$
we have
$$\sum_{d=1}^{\lfloor x\rfloor}\left(\frac{x}{d}-\left\lfloor\frac{x}{d}\right\rfloor\right)\leq \sum_{d=1}^{\lfloor x\rfloor}1=\lfloor x\rfloor\leq x.$$
How can I come up with the finer bound? Any hints? Thank you in advance.
| We have
\begin{align*}
& \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \left\lfloor {\frac{x}{d}} \right\rfloor } \right)} = \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \left\lfloor {\frac{{\left\lfloor x \right\rfloor }}{d}} \right\rfloor } \right)} = \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{x}{d} - \frac{{\left\lfloor x \right\rfloor }}{d}} \right)} + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {\frac{{\left\lfloor x \right\rfloor }}{d} - \left\lfloor {\frac{{\left\lfloor x \right\rfloor }}{d}} \right\rfloor } \right)}
\\ & \le (x - \left\lfloor x \right\rfloor )\sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\frac{1}{d}} + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\left( {1 - \frac{1}{d}} \right)} = (x - \left\lfloor x \right\rfloor - 1)\sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } {\frac{1}{d}} + \sum\limits_{d = 1}^{\left\lfloor x \right\rfloor } 1
\\ & = x - 1 + (x - \left\lfloor x \right\rfloor - 1)\sum\limits_{d = 2}^{\left\lfloor x \right\rfloor } {\frac{1}{d}} \le x - 1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the Taylor polynomial for a function $f(x) = (1+x^2)^{1/3}$ I have to find the Taylor polynomial $T_2(x)$for a function $f$ of grade $2$ around $x_0 = 0$ given by
$$
f(x) = (1+x^2)^{\frac{1}{3}} \ \text{for} \ x \in \mathbb{R}
$$
I am very new to Taylor polynomials so I think what I have done is correct but I am just unsure whether or not they should give something 'pretty'. I know that they are approximations so they probably shouldn't but are you able to tell me whether or not my calculations are correct as I need this to be correct to be able to do some other questions which relies on me having the Taylor polynomial correct.
I have done the following:
\begin{align*}
T_2(x) & = \sum_{n=0}^2 \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n = f(x_0) + f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 \\
& = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2
\end{align*}
Thus
$$
f(0) = (1+0^2)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1
$$
and
$$
f'(0) = \frac{2}{3} \cdot \frac{1}{(1+0^2)^{\frac{2}{3}}} \cdot 0 = 0
$$
and lastly
\begin{align*}
f''(x) & = (f'(x))' = \bigg( \frac{2}{3} \cdot \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' = \frac{2}{3} \cdot \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' \\
& = \frac{2}{3} \cdot \Bigg( \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \bigg)' \cdot x + \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x' \Bigg) \\
& = \frac{2}{3} \cdot \Bigg( - \frac{1}{((1+x^2)^\frac{2}{3})^2} \cdot \frac{2}{3} \cdot (1+x^2)^{\frac{2}{3}-1} \cdot 2x \cdot x + \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot 1 \Bigg)\\
& = \frac{2}{3} \cdot \Bigg( - \frac{1}{(1+x^2)^\frac{4}{3}} \cdot \frac{4}{3} \cdot \frac{1}{(1+x^2)^\frac{1}{3}} \cdot x^2 + \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) \\
& = \frac{2}{3} \cdot \Bigg( - \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) = \frac{2}{3} \cdot \Bigg (- \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{3(1+x^2)}{3(1+x^2)^{\frac{5}{3}}} \Bigg) \\
& = \frac{2}{3} \cdot \Bigg ( \frac{x^2+3}{3(1+x^2)^{\frac{5}{3}}} \Bigg) = \frac{2(x^2+3)}{9(1+x^2)^{\frac{5}{3}}}
\end{align*}
which if we evaluate in $x_0 = 0$ gives
$$
f''(0) = \frac{6}{9^{\frac{5}{3}}}
$$
Thus
$$
T_2(x) = 1 + \frac{6}{2! \cdot 9^{\frac{5}{3}}}\cdot x^2
$$
I know that this is a lot of calculations but I hope some of you still wants to help me.
Thank you very much.
Regards
Mathias
| The simplest consists in using the standard binomial expansion at order $\mathbf 1$:
$$(1+u)^{1/3}=1+\tfrac 13u+o(u) $$
and substituting $u=x^2$, whence
$$\bigl(1+x^2\bigr)^{1/3}=\underbrace{1+\tfrac 13x^2}_{T_2(x)}+o\bigl(x^2\bigr).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I derive a general formula for $I(n,m)$?
To prove the first one, here's what I did:
$\int x^nln(x)^mdx=$
IBP: $ln(x)^m=u \Rightarrow \frac{mln(x)^{m-1}}{x}dx=du$
$x^ndx=dv \Rightarrow v=\frac{x^{n+1}}{n+1}$ $\Rightarrow$
$$ln(x)^m\frac{x^{n+1}}{n+1}-\int\frac{x^{n+1}}{n+1}\frac{mln(x)^{m-1}}{x}dx=ln(x)^m\frac{x^{n+1}}{n+1}-\frac{m}{n+1}\int x^nln(x)^{m-1}dx=ln(x)^m\frac{x^{n+1}}{n+1}-\frac{m}{n+1}I(n,m-1)$$
To show the second:
$I(n,1)=\frac{x^{n+1}}{n+1}ln(x)-\frac{1}{n+1}I(n,0)$
$I(n,0)=\int x^nln(x)^{0}= \int x^n= \frac{x^{n+1}}{n+1} \Rightarrow I(n,1)=\frac{x^{n+1}}{n+1}ln(x)-\frac{1}{n+1}\Biggl( \frac{x^{n+1}}{n+1}\Biggr)$
Now, for the third:
$I(0,1)=xln(x)-I(0,0)=x\Bigl(ln(x)-1\Bigr)$ since $I(0,0)=\int x^0ln(x)^0= \int 1= x$
$I(0,2)=xln(x)^2-I(0,1)=xln(x)^2-x(ln(x)-1)=x\Biggl(ln(x)^2-ln(x)+1\Biggr)$
$I(0,3)=xln(x)^3-x(ln(x)^2-ln(x)+1)= x\Biggl(ln(x)^3-ln(x)^2+ln(x)-1\Biggr)$
$\Rightarrow I(0,m)=x(ln(x)^m+(-1)^mln(x)^{m-1}+...+(-1)^m)$
Would this be correct?
Also how do I derive a general formula for $I(n,m)$?
| You start iterating the recurrence like this:
\begin{align*}
& I(n,m) = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{m}{{n + 1}}I(n,m - 1)
\\ &
= \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{m}{{n + 1}}\left[ {\frac{{x^{n + 1} }}{{n + 1}}\log ^{m - 1} x - \frac{{m - 1}}{{n + 1}}I(n,m - 2)} \right]
\\ &
= \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x + \frac{1}{{(n + 1)^2 }}m(m - 1)I(n,m - 2)
\\ & = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x \\ & \quad \, + \frac{1}{{(n + 1)^2 }}m(m - 1)\left[ {\frac{{x^{n + 1} }}{{n + 1}}\log ^{m - 2} x - \frac{{m - 2}}{{n + 1}}I(n,m - 3)} \right]
\\ & = \frac{{x^{n + 1} }}{{n + 1}}\log ^m x - \frac{{x^{n + 1} }}{{(n + 1)^2 }}m\log ^{m - 1} x + \frac{{x^{n + 1} }}{{(n + 1)^3 }}m(m - 1)\log ^{m - 2} x \\ &\quad\,- \frac{1}{{(n + 1)^2 }}m(m - 1)\frac{{m - 2}}{{n + 1}}I(n,m - 3),
\end{align*}
etc. From this, you can conjecture that
$$
I(n,m) = \frac{{x^{n + 1} }}{{n + 1}}\sum\limits_{k = 0}^m {\frac{{( - 1)^k }}{{(n + 1)^k }}\frac{{m!}}{{(m - k)!}}\log ^{m - k} x} ,
$$
and verify it by using the recurrence and induction.
| {
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"url": "https://math.stackexchange.com/questions/3593053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
= 9 + 3i + 3i + i^2 * (3 + i)
= 9 + 6i - 1 * (3 + i)
= (8 + 6i) * (3 + i)
= 24 + 8i + 18i + 6i^2
= 24 + 26i - 6
= 18 + 26i
Find the magnitude of OB = 3 + i and OC = 18 + 26i to determine ->OB and ->OC
|OB| = sqrt(3^2 + 1^2) = sqrt(10)
|OC| = sqrt(18^2 + 26^2) = 10 * sqrt(10)
Therefore,
->OB = 3 + i = sqrt(10) * cis(AOB)
->OC = 18 + 26i = 10 * sqrt(10) * cis(AOC)
Cube ->OB and show to show AOC = 3AOB
->OB^3 = (3 + i)^3 = (sqrt(10))^3 * cis^3(AOB)
= 18 + 26i = 10 * sqrt(10) * cis^3(AOB)
We found that (3 + i)^3 = 18 + 26i and by cubing ->OB we find the magnitude is the same as ->OC. Since this is the case, this shows that the angle AOC = 3AOB.
Would this be the correct way to solve this question?
| The method works. You might be able to do it a little easier.
To cube $3+i$ you can use the binomial formula:
$$ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3,$$
so
\begin{align}
(3+i)^3 &= 3^3 + 3\cdot 3^2 i + 3\cdot 3 i^2 + i^3 \\
&= 27 + 27 i + 9 (-1) + (-i) \\
&= 18 + 26 i.
\end{align}
You found that
$\DeclareMathOperator\cis{cis}|OB|^3 \cis(3\angle AOB) = |OC| \cis(\angle AOC)$,
and if you have studied the $r\cis\theta$ form sufficiently already,
this is enough to prove that $\cis(3\angle AOB) = \cis(\angle AOC)$.
The tricky part is to show that this really means $3\angle AOB = \angle AOC$
and not something like $3\angle AOB = \angle AOC+2\pi$ or
$3\angle AOB = \angle AOC - 2\pi.$
(Consider the fact that $(-3+2i)^3 = 9 + 46 i$; plot $D = (-3,2)$ and $E = (9,46)$
and consider what this says about the ratio of $\angle AOD$ to $\angle AOE$.)
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How do I prove that a non-differentiable point of a function is a local maximum? Determine global and local maximum/minimum points of :
$f(x,y) = \ln(1-x+y)+x-\sqrt{|y|}$
over its domain.
So I figured out that the only possible local maximum is (0,0), where $f$ is not differentiable. I think I should now show that there exist an $\epsilon >0$ such that $f(x,y)\le f(0,0)=0$ for $\sqrt{x^2+y^2} \lt \epsilon$ .
However, I don't know how to prove it. Any suggestions?
| We split up the problem and define
$$
f_1(x, y) = \ln (1-x+y) + x - \sqrt{y}\quad \quad y \geq 0, x \leq y+1
$$
and solve that for
$$
\nabla f_1(x, y) = (\frac{-1}{1-x+y} + 1, \frac{1}{1-x+y} -\frac{1}{2\sqrt{y}})= \mathbf{0},
$$
then
$$
\begin{align}
y &= x \\
\sqrt{y} &=\frac{1}{2}(1-x+y)
\end{align}
$$
which gives $x = \frac{1}{4}, y= \frac{1}{4}.$ We inspect the Hessian,
$$
\nabla^2 f_2(x, y) =
\begin{bmatrix}
\frac{1}{(1-x+y)^2} & \frac{-1}{(1-x+y)^2} \\
\frac{-1}{(1-x+y)^2} & \frac{-1}{(1-x+y)^2} + \frac{1}{4}y^{-3/2}
\end{bmatrix}
$$
at $(0.25, 0.25)$
$$
\nabla^2 f_2(x, y) =
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix}
$$
which has a positive $\lambda_1 = 1 + \sqrt{2}$ and a negative $\lambda_2 = 1 - \sqrt{2}$ eigenvalue. Thus this is a saddle point.
Moving on to the halfplane where $y\leq 0$, define
$$
f_2(x, y) = \ln(1-x+y) + x -\sqrt{-y} \quad \quad y \leq 0, x \leq 1 + y
$$
and solve for
$$
\nabla f_2(x, y) = (\frac{-1}{1-x+y} + 1, \frac{1}{1-x+y} +\frac{1}{2\sqrt{y}})= \mathbf{0}
$$
which implies
$$
\begin{align}
y &= x \\
-2\sqrt{y} &= 1-x+y
\end{align}
$$
Now $\sqrt{y} = -1/2$ which we are going to throw away as we're not interested in complex solutions. Thus our system permits only a single critical point, i.e. $(0.25, 0.25)$. Now we have examined our function at all points where it is differentiable.
But where have the maximums and minimums gone? We do some parametrizations to see what happens at infinity:
Let $y = 0, x = -t$ for $t\geq 0$, then
$$
g_1(t) = f(x(t), y(t)) = \ln(1 + t) - t
$$
which is unbounded below. Our functions permits no global minimum.
Let $x(t) = y(t) = t$ for $t \geq 0$ and observe that
$$
g_2(t) = f(x(t), y(t)) = 0 + t - \sqrt{t}
$$
is unbounded above. Our function permits no global maximum.
Moving on to the point your question originally concerned, $x=0,y=0$. What happens there? This is a point where we're gonna struggle in defining a critical point without understanding subdifferentials.
| {
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"source": "stackexchange",
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How to prove $\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx =\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$ If $v>0$ why
$$\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{1}{\sqrt{2\pi}} e^{-\frac{v^2}{2}}$$.
I saw it in
I tried to solve it by $t=\frac{1}{x^2}-1$
$$\frac{v}{\pi} \int_0^1 \frac{1}{x^2 \sqrt{1-x^2}} e^{\frac{-v^2}{2x^2}} dx
=\frac{v}{\pi}\int_0^{\infty} \frac{1}{\frac{1}{t+1} \sqrt{1-\frac{1}{t+1}}} e^{\frac{-v^2}{2}(t+1)} \, \frac{1}{2}(t+1)^{\frac{-3}{2}}dt $$
$$=\frac{v e^{\frac{-v^2}{2}} }{2\pi}\int_0^{\infty}\frac{1}{\sqrt{t}} e^{\frac{-v^2}{2}t} \, dt $$
Thanks in advance for any help you are able to provide.
| To arrive to a gaussian integral
$$I= \int \frac{e^{\frac{-v^2}{2x^2}}}{x^2 \sqrt{1-x^2}}\, dx$$
Let
$$x=\frac{1}{\sqrt{t^2+1}}\implies dx=-\frac{t}{\left(t^2+1\right)^{3/2}}\,dt$$ This makes
$$I=-\int e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}} \text{erf}\left(\frac{t
v}{\sqrt{2}}\right)}{v}$$
$$J=-\int_0^\infty e^{-\frac{1}{2} \left(t^2+1\right) v^2}\,dt=-\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{v^2}{2}}}{\sqrt{v^2}}$$
| {
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Evaluate $\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{\cos 2x}} \,\rm{d}x$
Evaluate $$\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{\cos 2x}} \,\rm{d}x$$
I was thinking to write $\cos 2x = \cos^2 x - \sin^2 x$ and then to factor out $\sin^2x$ to try to simplify the expression, but gets me nowhere. Any help will be appreciated!
| \begin{align}J&=\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{\cos 2x}} \,\rm{d}x\\
&=\int_0^\frac{\pi}{4}\frac{\sin x}{\cos x\sqrt{2\cos^2 x-1}} \,\rm{d}x\\
&\overset{t=\cos x}=\int_{\frac{1}{\sqrt{2}}}^1\frac{1}{x\sqrt{2t^2-1}}\,dt\\
&\overset{x=t^2}=\frac{1}{2}\int_{\frac{1}{2}}^1\frac{1}{x\sqrt{2x-1}}\,dx\\
&\overset{y=\sqrt{2x-1}}=\int_0^1 \frac{1}{1+y^2}\,dy\\
&=\arctan(1)-\arctan(0)\\
&=\boxed{\frac{\pi}{4}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find this matrix? a) Let $\mathbf{P}$ be the $2\times 2$ matrix that projects vectors onto $\mathbf{u} = \begin{pmatrix}2\\ -1 \end{pmatrix}$. That is,
$\mathbf{P} {v} = \operatorname{proj}_{u} ({v}) = \text{Projection of $\mathbf{v}$ onto $\mathbf{u}$}.$
Using the geometric meaning of the matrix and the picture, find
$\mathbf{P} \begin{pmatrix}2 \\ -1 \end{pmatrix} \text{ and } \mathbf{P} \begin{pmatrix}1 \\ 2 \end{pmatrix} .$
The answer to part a)
\begin{align*}
\mathbf{P} \begin{pmatrix}2 \\ -1 \end{pmatrix} = \boxed{\begin{pmatrix} 2 \\- 1\end{pmatrix}}, \mathbf{P} \begin{pmatrix}1 \\ 2 \end{pmatrix} = \boxed{\begin{pmatrix} 0 \\ 0\end{pmatrix}}.
\end{align*}
b) Now, let $\mathbf{P}$ be the $2\times 2$ matrix that projects vectors onto $\mathbf{u} = \begin{pmatrix}2\\ -1 \end{pmatrix}$. That is,
$\mathbf{P} {v} = \operatorname{proj}_{{u}} ({v}) = \text{Projection of $\mathbf{v}$ onto $\mathbf{u}$}.$Use your answers from part (a) to figure out $\mathbf{P}$.
I set $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and I have $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$ How should I continue?
| You can argue as follows. From (a), we have:
$$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
2\\
-1
\end{pmatrix} =
\begin{pmatrix}
2\\
-1
\end{pmatrix}
$$
$$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
1\\
2
\end{pmatrix} =
\begin{pmatrix}
0\\
0
\end{pmatrix}
$$
We obtain two linear systems of equations for $a, b, c, d$:
\begin{cases}
2a-b = 2\\
a+2b=0
\end{cases}
\begin{cases}
2c-d=-1\\
c+2d=0
\end{cases}
Now, you just need to solve them.
| {
"language": "en",
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"source": "stackexchange",
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Let $l_1$ and $l_2$ be the lengths of perpendicular chords of $y^2=4ax$ drawn through the vertex and ...
Let $l_1$ and $l_2$ be the lengths of perpendicular chords of $y^2=4ax$ drawn through the vertex and $\left (l_1l_2 \right)^{\frac 43}= 4a^2\lambda (l_1^{\frac 23} + L_2^{\frac 23})$. Find $\lambda$
Let the chord be PQ where $P(t_1)$ and $Q(t_2)$
Since they subtend a right angle at the vertex $t_1t_2=-4$
Also Let OP be $l_1$ and OQ be $l_2$
$$PQ^2= l_1^2 +l_2^2$$
$$a^2(t_1^2-t_2^2)^2+4a^2 (t_1-t_2)^2=l_1^2+l_2^2$$
$$a^2(t_1-t_2)^2
\left [(t_1+t_2)^2+4\right ]=l_1^2+l_2^2$$
$$a^2(t_1^2+t_2^2+8)\left [t_1^2+t_2^2-4 \right ]=l_1^2+l_2^2$$
I couldnt solve further. Proceeding calculations are lengthy enough to make me think I am doing it wrong. How should I correctly solve it?
| Take two generic perpendicular chords of $\mathcal C:y^2 = 4ax$ passing through the vertex $O=(0,0)$:
\begin{gather}
r:\ y=mx\\
s: \ y = -\frac{1}{m}x
\end{gather}
with $m> 0$. Our hope is that your condition is independent of $m$.
Find the intersection points between these lines and $\mathcal C$ in order to find $l_1$ and $l_2$. The points are:
\begin{gather}
P=\left(\frac{4a}{m^2},\frac{4a}{m}\right)\\
Q=\left(4am^2,-4am\right)
\end{gather}
And length are:
\begin{gather}
\overline{OP} = l_1 = \left[\left(\frac{4a}{m^2}\right)^2+\left(\frac{4a}{m}\right)^2\right]^{\frac{1}{2}} = \frac{4a}{m^2}\left(m^2+1\right)^{\frac{1}{2}}\\
\overline{OP} = l_2 = \left[\left(4am^2\right)^2+\left(4am\right)^2\right]^{\frac{1}{2}} = 4am\left(m^2+1\right)^{\frac{1}{2}}
\end{gather}
Now we are ready for the computation:
\begin{equation}
\left(l_1 l_2\right)^{\frac{4}{3}} = \left(\frac{4a}{m^2}\left(m^2+1\right)^{\frac{1}{2}}\cdot 4am\left(m^2+1\right)^{\frac{1}{2}}\right)^{\frac{4}{3}} = (4a)^{\frac{8}{3}}\left(\frac{m^2+1}{m}\right)^{\frac{4}{3}}
\end{equation}
And
\begin{equation}
4a^2\lambda\left(l_1^{\frac{2}{3}}+ l_2^{\frac{2}{3}}\right) = 4a^2\lambda\left(4a\right)^{\frac{2}{3}}\left(m^2+1\right)^{\frac{1}{3}}\left(\frac{1}{m^{\frac{4}{3}}} + m^{\frac{2}{3}}\right) = 4^{\frac{5}{3}}(a)^{\frac{8}{3}}\lambda\left(\frac{m^2+1}{m}\right)^{\frac{4}{3}}
\end{equation}
And imposing the equality, we simply obtain:
\begin{gather}
\lambda = 4
\end{gather}
| {
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"source": "stackexchange",
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Why is the constant term of $(1+x+y+xy)^n$ equal to $\frac{1}{2}\binom{2n}{n}$？
If we define this: for any $x,y$ such that
$x^2=y^2=1,xy\neq yx$, express in terms of $n$ the constant term of the expression $$f_{n}=(1+x+y+xy)^n\,.$$
I guess this result is $\dfrac{1}{2}\binom{2n}{n}$.
for $n=1$, we have
$f_{1}=1+x+y+xy$ the constant term is $1=\dfrac{1}{2}\binom{2}{1}$
for $n=2$, we have
$$f_{2}=(1+x+y+xy)(1+x+y+xy)=1+x+y+xy+x+x^2+xy+x^2y+y+yx+y^2+yxy+xy+xyx+xy^2+xyxy=1+x+y+xy+x+1+xy+y+y+yx+1+yxy+xy+xyx+x+xyxy=3+3x+3y+3xy+yx+yxy+xyx+xyxy$$ the term is $3=\dfrac{1}{2}\binom{4}{2}$
for $n=3$,we have $$f_{3}=f_{2}(1+x+y+xy)=(3+3x+3y+3xy+yx+yxy+xyx+xyxy)(1+x+y+xy)$$ the constant term is $$3+3x^2+3y^2+yxxy=3+3+3+yy=3+3+3+1=10=\dfrac{1}{2}\binom{6}{3}$$
I tink this problem very interesting,But maybe use induction to prove it?I can't it
| Let $R$ be algebra generated over $\mathbb{Z}$ by non-commuting variables $x$, $y$, subject to $x^2=y^2=1$. There is an algebra homomorphism $\varphi:R\to M_2(\mathbb{Z}[t,t^{-1}])$ given by
$$
x\mapsto \begin{bmatrix}0&1\\1&0\end{bmatrix},\hspace{10mm}y\mapsto \begin{bmatrix}0&t^{-1}\\t&0\end{bmatrix}.
$$
I claim that $\varphi$ is injective. To check this, note that there is a $\mathbb{Z}$-basis for $R$ consisting of $(xy)^n$ and $x(xy)^n$ for $n\in\mathbb{Z}$, and observe that
$$
\varphi((xy)^n)=\begin{bmatrix}t^{n}&0\\0&t^{-n}\end{bmatrix},\hspace{10mm}\varphi(x(xy)^n)=\begin{bmatrix}0&t^{-n}\\t^{n}&0\end{bmatrix}
$$
are $\mathbb{Z}$-linearly independent.
Now, we have
$$
\varphi(1+x+y+xy)=\begin{bmatrix}1+t&1+t^{-1}\\1+t&1+t^{-1}\end{bmatrix}=(1+t)\begin{bmatrix}1&t^{-1}\\1&t^{-1}\end{bmatrix}.
$$
It is easy to check by induction that
$$
\begin{bmatrix}1&t^{-1}\\1&t^{-1}\end{bmatrix}^n = (1+t^{-1})^{n-1}\begin{bmatrix}1&t^{-1}\\1&t^{-1}\end{bmatrix},
$$
so we get
$$
\varphi\big((1+x+y+xy)^n\big)=(1+t)^n (1+t^{-1})^{n-1}\begin{bmatrix}1&t^{-1}\\1&t^{-1}\end{bmatrix}=t^{-n+1}(1+t)^{2n-1}\begin{bmatrix}1&t^{-1}\\1&t^{-1}\end{bmatrix}.
$$
The constant coefficient of $(1+x+y+xy)^n$ is the coefficient of $1$ in the upper left entry of the matrix above, which is
$$
{2n-1\choose n-1}=\frac{1}{2}{2n\choose n}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3623355",
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"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
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Find the real factors of $X^{2n}-1$ Find the real factors of $x^{2n}-1$
I have tried to solve this way
let $x^{2n}=1=e^{2kπi}$
$$x=e^{\pm kπi/n}=a+ib, \text{ for } k=0,1,2,\ldots,(n-1)$$
From $x^{2n}=1,$ $-1$ and $1$ are roots, hence $(x+1)$ and $(x-1)$ are factors, other factors are given as product of conjugate factors
$$\Rightarrow \prod_{k=1}^{(n-1)}(x-(a+ib))(x-(a-ib))=\prod_{k=1}^{(n-1)}(x^2-2ax+a^2+b^2)$$
$$\therefore(x^{2n}-1)=((x+1)(x-1)\prod_{k=1}^{(n-1)}(x^2-2xa+a^2+b^2))$$
Then,according to my solution the factors are;
$$(x-1)(x+1)(x^2-2x\cos\frac{π}{n}+1)\cdots(x^2-2x\cos\frac{π(n-1)}{n}+1)$$
But the text book answer is
$$(x-1)(x+1)(x^2-2x\cos\frac{π}{n}+1)\cdots (x^2-2x\cos\frac{π(n-1)}{n}-1)$$
Please correct me wherever I have went wrong,your contribution is highly appreciated
| We can use the fact that $A^2-B^2=(A+B)(A-B)$, so:
$$x^{2n}-1=(x^n+1)(x^n-1)$$
We can factor this in $C$, using DeMoivre's law. In fact we want to find the roots of $1$ and $-1$. So, let:
$$z=r(\cos(\theta)+i\sin(\theta))$$
then we have:
$$z^n=1\leftrightarrow r^n(\cos(n\theta)+i\sin(n\theta))=1(\cos(0)+i\sin(0))$$
Clearly, $r=1$ and so:
$$\theta_k=\frac{2k\pi}{n}$$
Also, we have:
$$z^n=-1 \leftrightarrow r^n(\cos(n\theta)+i\sin(n\theta))=1(\cos(\pi)+i\sin(\pi))$$
And from here:
$$\theta_k=\frac{\pi}{n}+\frac{2k\pi}{n}$$
So, we can write:
$$x^{2n}-1=(x-1)(x+1)\prod_{k=0}^{n-1}(x-z_k)(x-w_k)$$
where:
$$z_k=cos\left(\frac{\pi}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\pi}{n}+\frac{2k\pi}{n}\right)$$
and:
$$w_k=\cos\left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right)$$
Now, in the real factorization there will be the factors with imaginary part equal to $0$. In other words when tge number $n$ can be written in the form:
$$n=\frac{2k}{j} \vee n=\frac{2k+1}{f}$$
where $f,j\in Z$.
In your answer the error is contained in the productory, in fact you proceed in group of two factor, so the productory must be rewritten in the form:
$$\prod_{k=1}^{n}$$
and then not considering the terms $(x-1)(x+1)$ because they are contained in the product itself. So, the answer given by your book is wrong.
| {
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"source": "stackexchange",
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Why are these two inequalities not the same even though they use the same equation? I just don't get it, like at all.
$U_{n}$ is an iteration defined on $\mathbb{N}$, BTW.
The question was:
$$\begin{align}
U_{n+1} &=
\frac{8U_n - 8}{U_{n} + 2} = 8 - \frac{24}{U_{n} + 2}\\
U_0 &= 3
\end{align}
$$
"Prove that $3 \leqslant U_n \leqslant 4$ by using mathematical induction"
Step 1: Test $n = 0$, yes, it's correct.
Step 2: Let's say the inequality is correct and test it for $n+1$
I used the first equation BTW. The result(for me) was:
$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$
Now, the exercise solved it by using the second equation, and its result was:
$\tfrac{16}{5} \leqslant U_{n+1} \leqslant 4 $
And since $16/5$ is bigger than 3 then it's correct and works.
How are there two different results and inequalities when you use each one of them?
Isn't $\tfrac{8U_{n} - 8}{U_{n} + 2}$ supposed to be equal to $8 - \tfrac{24}{U_{n} + 2}$ which means no matter which one you use, and there will be the same result?
I just don't get it. Which one is correct?
How are there two different results and inequalities when you use each one of them?
My solution:
An inequality for the numerator:
\begin{align}
3 &\leqslant U_n \leqslant 4\\
24 &\leqslant U_n \leqslant 32\\
16 &\leqslant U_n \leqslant 24
\end{align}
Now the equality for the dominator:
\begin{align}
3 &\leqslant U_n \leqslant 4\\
5 &\leqslant U_n + 2 \leqslant 6\\
\tfrac{1}{6} &\leqslant \tfrac{1}{U_{n} + 2} \leqslant \tfrac{1}{5}
\end{align}
By multiplying each side of the inequality then:
$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$
| I think I see what is going on. Let's focus on your left inequality. You basically do this:
$$
U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{U_n+2}\ge \frac{8\cdot3-8}{4+2} = \frac{16}{6}
$$
This is absolutely correct. However, we lose some tightness in the bound, because $\frac{8U_n-8}{U_n+2}$ is in fact increasing in $U_n$ (for $U_n>-2$). This is not obvious, but can be seen immediately in the rewritten form $8-\frac{24}{U_n+2}$. Therefore, you can replace $U_n$ with $3$ everywhere to get:
$$
U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{3+2} = \frac{16}{5}
$$
| {
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If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?
| The answer is B). See my solution.
By your work we need to prove that:
$$\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}\geq6,$$ which is impossible because it's wrong (you got a reversed inequality).
Another way.
$$\sum_{cyc}\frac{a}{b+c-a}-3=\sum_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum_{cyc}\frac{a-b-(c-a)}{b+c-a}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{c+a-b}\right)=\sum_{cyc}\frac{2(a-b)^2}{(b+c-a)(c+a-b)}\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deducing an inequality in sharpening results in an analytic number theory paper I am studying a research paper in analytic number theory
It involves an inequality which is to be used in sharpening a result. Unfortunately I am unable to see how to get the required inequality (See proposition $1$) Or read below
$a$ is odd integer $\geq$3 and $r$ is an integer lying between $1\le r<a/2$ . Author deduced $(12)$
French to English in 2nd line after $(12)$ is
and $s_0$ is the only root in $(0, 1)$ of polynomial $Q(s) =~\dots$
In particular $\delta(a) \geq~\dots$
Now, inequality to be deduced is $(13)$ of image using $$\phi_{r, a}\geq \frac{2^{r+1}} { r^{a-2r}}\text{ and }2r\leq{2r+1}\leq{2(r+1) }.$$
Taking $\log$ both sides in $\phi_{r, a} $ and putting it in $(12)$ I got RHS of $(12)$ equals
$$\frac{ (2a-3r-1) \log2 + (2r+1) \log(2r+1) +(a-2r) \log r } { a + (a-2r) \log2+(2r+1)\log(2r+1) } $$
which is not equal to $(13)$.
Can someone please tell how to derive it to $(13)$?
| Consider the numerator and denominator of RHS in (12). We have
\begin{align}
&(a-2r)\log(2) + (2r+1)\log(2r+1) - \log(\varphi_{r,a})\\
\ge\ & (a-r)\log(2) - r\log(2) + (2r+1)\log (2r) - \log(\varphi_{r,a})\\
=\ & (a-r)\log(2) + (r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a})
\end{align}
and
\begin{align}
&a + (a-2r)\log(2) + (2r+1)\log(2r+1)\\
\le\ & a + 1 + (a-2r)\log(2) + (2r+1)\log (2r+2)\\
=\ & (a+1) + (a+1)\log(2) + (2r+1)\log(r+1).
\end{align}
Thus, to prove (13), it suffices to prove that
$$(r+1)\log(2) + (2r+1)\log(r) - \log(\varphi_{r,a}) \ge (a+1)\log(r)$$
or
$$\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}.$$
Consider $Q(s) = rs^{a+2} - (r+1)s^{a+1} + (r+1)s - r$.
Note that $Q(0) < 0$, $Q(1)=0$, and
\begin{align}
Q(1 - \tfrac{1}{r}) = -2 (1-\tfrac{1}{r})^{a+1} - \tfrac{1}{r} < 0.
\end{align}
Thus, $s_0 > 1 - \frac{1}{r}$.
Now we are ready to prove that $\frac{2^{r+1}}{r^{a-2r}} \ge \varphi_{r,a}$.
We have to prove that
$$2^{r+1} \ge ((r+1)s_0 - r)^r(r+1-rs_0)^{r+1} (r - rs_0)^{a-2r}. \tag{1}$$
Since $(r+1)s_0 - r < r+1 - r = 1$ and $r+1 - rs_0 \le r + 1 - r(1-\tfrac{1}{r}) = 2$
and $r - rs_0 \le r - r(1 - \tfrac{1}{r}) = 1$, (1) is true.
We are done.
| {
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"source": "stackexchange",
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Find locus of $S$ denoting set of complex numbers $\frac{z+1}{z-3}$, where $z$ varies over set of $|z|=1$.
Question: Let $S$ denote the set of all complex numbers of the form $\frac{z+1}{z-3}$, where $z$ varies over the set of all complex numbers with $|z|=1.$ Find the locus of the points in set $S$.
My approach:
Let $z=x+iy$, with $x,y\in\mathbb{R}$ such that $|z|=1\implies |z|^2=1$. This implies that we must have $x^2+y^2=1$. Now $z+1=(x+1)+iy$ and $z-3=(x-3)+iy$.
Thus
\begin{align*}
\frac{z+1}{z-3}&=\frac{(z+1)(\overline{z-3})}{|z-3|^2}\\
&=\frac{(z+1)(\overline{z}-3)}{|z-3|^2}\\
&=\frac{x^2-2x-3+y^2-4iy}{(x-3)^2+y^2}\\
&=\frac{-2x-2-4iy}{10-6x}\\
&=\frac{-x-1-2iy}{5-3x}.
\end{align*}
Thus we have
$$
\Re\left(\frac{z+1}{z-3}\right)=\frac{x+1}{3x-5}
$$
and
$$
\Im\left(\frac{z+1}{z-3}\right)=\frac{2y}{3x-5},$$ and our task is to find a relationship between these two given that $|z|=1$.
For our ease let us have $\alpha=\frac{z+1}{z-3}\,\, \forall z$ satisfying $|z|=1$.
Now we obtain two useful information using the triangle inequality. We have
$$
|z+1|\le |z|+1=2
$$
and
$$
|z-3|\le |z|+3=4.
$$
From here we can conclude that $$|z+1|^2=(x+1)^2+y^2=2+2x\le 4 \iff 1+x\le 2\text{ and } |z-3|^2=5-3x\le 16.$$
Thus we have $$0\le 1+x\le 2 \text{ and } 0\le 5-3x\le 16\implies -1\le x\le 1.$$
Observe that even from $x^2+y^2=1$, we can directly conclude that $-1\le x,y\le 1$.
But this doesn't help much to find a relationship between $\Re(\alpha)$ and $\Im(\alpha)$.
How to proceed?
| Write $w = 1 +\frac{4}{z-3}$, then recognize that this transformation is the composition of 3 transformations: a shift to the left by 3, inversion with a scale factor of 4, then a shift to the right by 1. The tricky part for you is to show that applying inversion to a circle (that does not pass through the origin) yields another circle. Once you do that, you will find that the locus you are looking for is a circle with radius 0.5 centered on $z=-0.5$.
(BTW, this is an example of a Mobius transformation. The neat thing about such transformations is that they preserve the class of circles and lines. Since you start with a circle, you will either get another circle or a line. A line results when the original circle passes through the pole of the transformation. Knowing this, all you need to do is compute a few image points and you can deduce the image locus. For example, $-1\rightarrow 0$, $1\rightarrow -1$, and $i\rightarrow -0.2-0.4i$. You can confirm that $0, -1$ and $-0.2-0.4i$ lie on the aforementioned circle.)
| {
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Integer solutions to $b^3=(a \bmod b)\cdot(a-a\bmod b)$ Recently, I have found this problem:
Determine all the possible integers solutions $(a,b) \in N_0$ to the following equation with modulus:
$$b^3=(a \bmod b)\cdot(a-a\bmod b)$$
To solve this, I have thought to use the relation:
$$a \bmod b=a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor$$
Substituting, we have:
$$b^3=\left (a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\cdot\left(a-\left(a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\right)$$
Which is the same as:
$$b^3=\left (a-b\cdot\left \lfloor \frac{a}{b} \right \rfloor\right)\cdot b\cdot\left \lfloor \frac{a}{b} \right \rfloor$$
And so:
$$b^3=ab\cdot\left \lfloor \frac{a}{b} \right \rfloor-b^2\cdot\left \lfloor \frac{a}{b} \right \rfloor^2$$
Being $b\neq0$, we can divide and rearrange:
$$b^2=\left \lfloor \frac{a}{b} \right \rfloor\left(\frac{a}{b}-\left \lfloor \frac{a}{b} \right \rfloor\right)$$
Now, we can use the identity:
$$\frac{a}{b}-\left \lfloor \frac{a}{b} \right \rfloor=\left \{ \frac{a}{b}\right \}$$
And so:
$$b=\left \lfloor \frac{a}{b} \right \rfloor\cdot\left \{ \frac{a}{b}\right \}$$
But now, how can we go on? Any idea?
| Let $q,r$ be any positive integers such that $b^2=qr$ and $r<b<q$. Then $a=bq+r$ gives a solution: $(a\bmod b)=r$, $a-(a\bmod b)=bq+r-r=bq$, so $(a\bmod b)(a-(a\bmod b))=rbq=b^3$.
For example, let $b=10$, so $b^2=100=20\times5$, so take $q=20$, $r=5$. Then $a=205$, and you can check that $(a\bmod b)(a-(a\bmod b))=(5)(200)=1000=10^3$.
EDIT: Moreover, all solutions are of this form. Given any $a,b$, $b>0$, there exist unique $q,r$ with $a=bq+r$, $0\le r<b$. Then $(a\bmod b)(a-(a\bmod b))=rbq$, so we must have $rbq=b^3$, which is $b^2=qr$. Since $r<b$, we must have $q>b$.
| {
"language": "en",
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"source": "stackexchange",
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Is $2^{n+1} \leq a + b \leq 4^n$ when $4^n - 1 = ab \quad \text{; with } n,a,b \in \mathbb{N}$? $4^n - 1 = ab \quad \text{; with } n,a,b \in \mathbb{N} \text{ such that } a, b \mid 4^n - 1$
Is this inequation always true: $2^{n+1} \leq a + b \leq 4^n$ ?
I already know that:
$4^n - 1 = \left(2^n\right)^2 - 1 = \left(2^n - 1\right) \left(2^n + 1\right)$
If it would be:
$a = \left(2^n - 1\right)$ and $b = \left(2^n + 1\right)$
then $a + b = 2^{n+1}$.
If it would be:
$a = 1$ and $b = 4^n - 1$
then $a + b = 4^n$.
But what about the other cases?
| Trying to do a proof by checking certain cases, as you're trying, is not sufficient as there's an infinite # of them. Instead, you need to use a more general technique. First, though, I assume by $\mathbb{N}$ you mean integers which are positive, i.e., it doesn't include $0$, since otherwise your inequalities are not always necessarily true (e.g., you could get $4^{0} - 1 = 0 = ab$, so $a = b = 0$, but $2^{0 + 1} = 2 \le 0 + 0 = 0$ is not true).
As you stated, you're trying check if it's always true that
$$2^{n+1} \leq a + b \leq 4^n \tag{1}\label{eq1A}$$
given that
$$4^n - 1 = ab \implies 4^n = ab + 1 \tag{2}\label{eq2A}$$
Note $4^n - 1$ is odd, so $a$ and $b$ are odd integers. Also, due to the symmetry in $a$ and $b$ in the problem, assume WLOG that $a \ge b$. First, checking the right side of the inequality, you want to see if
$$a + b \le 4^n = ab + 1 \tag{3}\label{eq3A}$$
If $b = 1$, then \eqref{eq3A} becomes $a + 1 \le a + 1$, which is obviously true. Next, if $b \ge 3$, you then have
$$ab + 1 \ge 3a + 1 = 2a + a + 1 \ge 2a + b + 1 \gt a + b \tag{4}\label{eq4A}$$
This shows \eqref{eq3A} also holds in that case too. Update: As bjorn's comment below states, it's easier to see this since $a + b \le ab + 1$ is equivalent to $(a - 1)(b - 1) \ge 0$.
Next, note all odd perfect squares are congruent to $1$ modulo $4$, but since $4^n - 1 \equiv 3 \pmod 4$, this means you have that $a \neq b$. Since have assumed $a \ge b$, this means, using \eqref{eq2A}, you have
$$\begin{equation}\begin{aligned}
a - b & \ge 2 \\
(a - b)^2 & \ge 4 \\
a^2 - 2ab + b^2 & \ge 4 \\
a^2 + 2ab + b^2 & \ge 4ab + 4 \\
(a + b)^2 & \ge 4(ab + 1) \\
(a + b)^2 & \ge 4(4^n) \\
(a + b)^2 & \ge 2^{2(n+1)} \\
a + b & \ge 2^{n+1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows the left side of the inequality in \eqref{eq1A} is also always true.
| {
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Solve in integers the equation $(a-b)^3(a+b)^2 = c^2+2(a-b)+1$ $$(a-b)^3(a+b)^2 = c^2+2(a-b)+1$$
So it is evident that a solution to the equation $(a, b, c) = (0, 1, 0)$. The expression $c^2+1$ never divides $4$. Now let's assume that $c^2+1$ is even, then we can write it in the form $2k$ where $k$ is odd. Rewrite the equation as
$$ ((a-b)(a+b))^2 = \frac{c^2+1}{a-b}+2$$
Now if $a-b$ is even then we get a contradiction as the LHS would be even but the RHS would be odd. and if $a-b$ is odd, then LHS would be odd but the RHS would be even another contradiction. Therefore $c^2+1$ is odd and $a-b$ is odd. In fact some further analysis in $mod$ $4$ would yield that $a-b$ is of the form $4k-1$.
That is how far I got and right now I am stuck. Any ideas?
| If $p\mid a-b$ then $c^2\equiv -1\pmod p$, so $p=2$ or $p\equiv 1\pmod 4$. You already ruled out $p=2$, hence $a-b$ is $\pm$ a (possibly empty) product of primes $\equiv 1\pmod 4$. In case of $+$ sign, this give $a-b\equiv 1\pmod4$ overall, whereas you already know $a-b\equiv -1\pmod 4$.
It follows that $a-b$ is negative. Let $d=b-a>0$.
Then
$$1\le c^2+1=d(2-d^2(a+b)^2)$$
implies
$$ d^2(a+b)^2\le 2.$$
As $a+b=0$ is impossible when $a-b$ is odd, we conclude $d=1$ and $a+b=\pm1$, so either $a=0$, $b=1$, and then $c=0$ as already found; or $a=-1$, $b=0$, and again $c=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that sum of matrices equals zero
The matrix $A$ has size $3 \times 3$ and we know that for any column
vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
$A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.
So if
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
2a_{11} & a_{12} + a_{21} & a_{13} + a_{31}\\
a_{12} + a_{21} & 2a_{22} & a_{23} + a_{32}\\
a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.
| Since all vectors $v$ are orthogonal to $Av$ then $v^TAv=0$ and likewise $(Av)^Tv=v^TA^Tv = 0$ and so adding them together we get $v^TAv + v^TA^Tv = v^T(A+A^T)v=0$ and finally since $v$ was arbitrary we must have $A+A^T=0$.
| {
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If $\sec x + \csc x =p$ has four distinct solutions between $(0,2\pi)$, then which if the following is incorrect?
a) $p^2-8>0$
b) $p=\sqrt 2$
c) $p=-\sqrt 2$
d) $p=0$
My attempt
$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$
$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$
$$\sin x +\cos x \ge 2\sqrt {\sin x \cos x}$$
$$(\sqrt {\sin x}-\sqrt {\cos x})^2\ge 0$$
$$\sin x \ge \cos x$$
I realise that some of that squaring might have removed or added some roots, but I don’t know what else to do
From this result, the interval for $x$ is $[\frac{\pi}{4}, \pi]\cup [\frac{5\pi}{4}, \frac{3\pi}{2}]$
Don’t know what do next. Can I get some insight?
Another attempt
$$\sin x +\cos x =p \sin x \cos x$$
$$1+2\sin x\cos x =\frac 14 p^2 4\sin^2x \cos ^2x$$
$$1+\sin 2x =\frac 14 p\sin^2 2x$$
$$p\sin^22x-4\sin 2x -4=0$$
| $$\sin x+\cos x=p\cos x\sin x$$
Now if $y=\sin x+\cos x=\sqrt2\sin(\pi/4+x),y^2\le2$
$$2y=p(y^2-1)\iff py^2-2y-p=0$$
As the discriminant $>0,$ there will always be four real roots in$(0,2\pi)$
Observe that if $z=t$ is a solution of $$\sin z=a$$ so will be $z=\pi-t$
Now $t,\pi-t$ will coincide if $t=2m\pi+\pi-t\implies\cos2t=-1$
So, here $$1-y^2=-1\iff y=?$$
To coincide, we need $$p(2-1)=2\cdot\pm\sqrt2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a system of linear equations using Cramer's Rule I'm trying to solve the following Cramer's system
$$AX=B \rightarrow \left\{\begin{matrix}x + y + z = 1\\ ax + by + cz = m\\ a^{2}x + b^{2}y + c^{2}z = m^{2}\\
\end{matrix}\right.$$
I tried to find the determinant of the A matrix:
$$\begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^{2} & b^{2} & c^{2}\end{vmatrix}\underset{l_{2}-al_{1}, l_{3}-a^{2}l_{2}} {\rightarrow}\begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & b^{2}-a^{2} & c^{2}-a^{2}\end{vmatrix}\underset{l_{3}-(b+a)l_{2}} {\rightarrow}\begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & 0 & (c^{2}-a^{2}) - (c-a)\cdot(b+a) \end{vmatrix}$$
It seems like my determinant is wrong, because I'm getting the wrong solution to the system... Since the B matrix later substitutes a column in each of the 3 different matrices in the rule, the determinant obtained should be the same, i.e:
$$\det(A_{b_{1}}) = \begin{vmatrix}1 & 1 & 1\\ m & b & c\\ m^{2} & b^{2} & c^{2}\end{vmatrix}$$
We would end up swapping a for m.
I've tried re-doing the problem over and over, but I end up getting the same wrong solution and I need help...
| Note that the determinant
$$
d = \begin{vmatrix}1 & 1 & 1\\ 0 & b-a & c-a\\ 0 & b^{2}-a^{2} & c^{2}-a^{2}\end{vmatrix}
$$
can be easily found by cofactor expansion along the first column, you quickly get
$$
\begin{split}
d &= 1 \cdot \begin{vmatrix}b-a & c-a\\ b^2-a^2 & c^2-a^2\end{vmatrix} \\
&= (b-a)(c^2-a^2) - (b^2-a^2)(c-a) \\
&= (b-a)(c-a)[c+a - (b+a)] \\
&= (b-a)(c-a)(c-b).
\end{split}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:
Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?
| Not sure if this is helpful, but it may be fruitful to consider a sequence of functions $g_n(x)$ where $g_0(x) = x$ and
$$g_n(x) = [g_{n-1}(x)]^2 - x - n$$
In particular,
$$\lim_{n \to \infty} g_n(x_0) = 0.$$
I found this by setting
$$f_n(x) = 0$$
and moving all roots to the other side. For example, with $n = 2$,
$$\begin{align*}
0 &= \sqrt{x + \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}}, \\
-x &= \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}, \\
x^2 - x - 1 &= \sqrt{x+2 + \sqrt{x+3}}, \\
(x^2 - x - 1)^2 - x - 2 &= \sqrt{x+3}, \\
((x^2 - x - 1)^2 - x - 2)^2 - x - 3 &= 0.
\end{align*}$$
This makes it clear that if $a$ satisfies $f_n(a) = 0$, then $g_n(a) = 0$. I suppose then that the solution to the question is dependent on if the sequence $g_n$ defined by $g_0 = x$ and
$$g_n = g_{n-1}^2 - x - n$$
has an explicit form.
| {
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} |
If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.
If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $
A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$
Source: 4/12/2020, Competitive Exams Reasoning Sample Paper 3- Translation in Hindi, Kannada, Malayalam, Marathi, Punjabi, Sindhi, Sindhi, Tamil, Telgu - Examrace. Downloaded from examrace.com
I can only see the pitfall consisting in inferring that all 3 numbers must be equal to 0.
What I can conclude from the premise is that one of the 3 numbers is the additive inverse of the sum of the 2 others.
Admitting it is number $c$, we get
$$a+b+c = 0= (a+b) + \left( - (a+b) \right) \tag{1}$$
In that case
$$c^3 = [- (a+b)]^3 = - (a+b) (a+b)(a+b) = - ( a^3 +2a^2b+2ab^2+b^3) \tag{2}$$
So
$$\begin{align}
a^3+b^3+c^3 &= a^3+b^3 - ( a^3 +2a^2b+2ab^2+b^3) \tag{3} \\
&= a^3+b^3 - a^3 - 2a^2b- 2ab^2- b^3 \tag{4}\\
&= 2a^2b - 2ab^2 \tag{5} \\
&=2 ( a^2b - b^2a) \tag{6} \\
&= 2 ( a) (ab-b^2) \tag{7} \\
&= 2 ( a) (b) (a-b) \tag{8} \\
&= 2 ( a) (b) (- c) \quad\text{[ Since $c = -(a+b) = b - a = - (a-b) $]} \tag{9} \\
&= - 2 ( a) (b) (c) \tag{10}
\end{align}$$
However, this isn't one of the possible answers.
What did I miss? Was I wrong in supposing that I could take any number $a$, $b$, or $c$ to play the role of additive inverse of the sum of the two others?
| I believe you did the expansion of $(a+b)(a+b)(a+b)$ wrong. It should be $a^3 + 3ab^2 + 3a^2b + b^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{a_1} + \frac{2}{a_1+a_2} + \frac{3}{a_1+a_2+a_3}<2(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}) $ Prove that $$\frac{1}{a_1} + \frac{2}{a_1+a_2} + \frac{3}{a_1+a_2+a_3}<2(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}) $$ where $a_1, a_2, a_3 >0$.
From AM-HM I got that $\frac{2}{a_1+a_2}\le \frac{1}{2}(\frac{1}{a_1}+\frac{1}{a_2})$ and $\frac{3}{a_1+a_2+a_3}\le \frac{1}{3}(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3})$, but adding these is not enough.
| Further to my comment and your progress with AM-HM
$$\frac{1}{a_1}<\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}$$
$$\frac{2}{a_1+a_2}\leq \frac{1}{2}\left(\frac{1}{a_1}+\frac{1}{a_2}\right)<\frac{1}{2}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)$$
$$\frac{3}{a_1+a_2+a_3}\leq \frac{1}{3}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)$$
Add together
$$\frac{1}{a_1}+\frac{2}{a_1+a_2}+\frac{3}{a_1+a_2+a_3}<\color{red}{\left(1+\frac{1}{2}+\frac{1}{3}\right)}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)<\color{red}{2}\cdot ...$$
| {
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Prove that $\sum \tan \frac{n+1}{n^2}$ diverges Prove that
$$\sum\tan \frac{n+1}{n^2}$$
diverges.
I know that the solution should be using the limit comparison test with $1/n$.
Therefore, I think that in the limit:
$$
\lim_{n \to \infty} \frac{\tan \frac{n+1}{n^2}}{1/n}
$$
I should get a positive finite value... but, its not working, I don't know how to solve that limit...
I tried using L'Hospital, but its getting complicated:
$$
\frac{\frac{1}{\cos^2\frac{n+1}{n^2}}(\frac{n+1}{n^2})'}{-(1/n^2)}
$$
Help.
Thank you.
| Let $L=\lim_{n \rightarrow \infty} \frac{|\tan \frac{n+1}{n^2}|}{\frac{1}{n}}$
Then we have
$L=\lim_{n \rightarrow \infty} \frac{1}{|\cos\frac{n+1}{n^2}|} \frac{|\sin\frac{n+1}{n^2}|}{\frac{1}{n}}$
$=\lim_{n \rightarrow \infty} \frac{1}{|\cos\frac{n+1}{n^2}|}\frac{n+1}{n} \frac{|\sin\frac{n+1}{n^2}|}{\frac{n+1}{n}\frac{1}{n}}$
$=\bigg(|\frac{1}{\cos\lim_{n \rightarrow \infty}\frac{n+1}{n^2}}|\bigg)\bigg(\lim_{n \rightarrow \infty}\frac{n+1}{n}\bigg) \bigg(|\lim_{n \rightarrow \infty} \frac{\sin\frac{n+1}{n^2}}{\frac{n+1}{n^2}}| \bigg)$
$=(1)(1)\bigg(|\lim_{n \rightarrow \infty} \frac{\sin\frac{n+1}{n^2}}{\frac{n+1}{n^2}}| \bigg)$.
Let $x=\frac{n+1}{n^2}.$ Then $x \rightarrow 0$ as $n \rightarrow \infty$. It follows that
$L=|\lim_{x \rightarrow 0} \frac{\sin(x)}{x}|=1.$
This proves that $\sum |\tan \frac{n+1}{n^2}|$ converges. It follows that $\sum \tan \frac{n+1}{n^2}$ converges.
Please let me know if there is any clarification necessary.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ 4\sum^{30}_{n=1} n\;T(n)$, where $T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$
For $n$ measured in degrees, let
$$T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$$
Evaluate $$ 4\sum^{30}_{n=1} n \cdot T(n)$$
I have tried to use double-angle identities but got stuck with the coefficient $n$. I am new to trig, so I probably miss some advanced concepts.
| Like Ginger bread,
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos^2x+\cos^2y=1+\cos(x-y)\cos(x+y)$$
$$\cos x\cos y=\dfrac{\cos(x-y)+\cos(x+y)}2$$
If $\cos(x+y)=\dfrac12\iff x+y=360^\circ m\pm60^\circ,$
$$T(n)=1-\dfrac{\dfrac12}2=?$$
Here $x=30^\circ-n,y=30^\circ+n$
| {
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"url": "https://math.stackexchange.com/questions/3643516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A series of multiplication leads to $\frac{1}{2} = 2$ I'm presented with the equation $\frac{a+b}{a} = \frac{b}{a+b}$
Performing cross multiplication yields $a^2+2ab+b^2 = ab$
Subtracting $ab$ from both sides, we get $a^2+ab+b^2 = 0$
Multiplying both sides by (a-b) and simplifying:
$(a-b)(a^2+ab+b^2) = 0 * (a-b)$
$a^3 - b^3 = 0$
$a^3 = b^3$
$a = b$
Substituting into the original equation, we finally arrive at $2 = \frac{1}{2}$ ... something has obviously gone terribly wrong. Where did I mess up?
Also it is worth noting that the original problem explicitly states that $a ≠ b$
| Here
$$(a-b)(a^2+ab+b^2) = 0 (a-b)$$
When $ a=b$. both side becomes $0$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does a complex function plots a given circle line? If I have a complex function $w = \frac{1}{z}$ and I have to show how does it display a circle line: $x^2+y^2 +2x-4y+1 = 0$. I am not sure how to do it. Here is my try:
$z = x+yi \Rightarrow w = \frac{1}{x + yi}$ Then $u(x;y) = \frac{1}{x}$ and $v(x;y) = \frac{1}{y}$. Then I express x from the circle line: $x(x+2) = y^2 -4y +1 \Rightarrow x_1 = y^2 -4y +1 $ and $x_2 = y^2 -4y -1$. From there I get a system: $$\begin{matrix} u = y^2-4y +1 \\ v = \frac{1}{y} \end{matrix} \Rightarrow y = \frac{1}{v} \Rightarrow u = \frac{v^2}{v-1}$$. I know that I am doing something completely wrong, because $u$ is the real part of the function and not a circle line, so I don't know how to show how would this function display the circle line indicated above. Any help would be appreciated!
| I'll find the (complex) locus of $x^2+y^2+2x-4y+1=0$.
Notice we can complete the square:
$$(x+1)^2+(y-2)^2=4$$
Now this is a circle with radius 2 centered at $(-1,2)$. We can represent this as
$$|z-(-1+2i)|=2$$
$$|z+1-2i|=2$$
EDIT: To find the locus of $w=\frac{1}{z}$:
$$z=\frac{1}{w}$$
$$|\frac{1}{w}+1-2i|=2$$
$$|1+w-2wi|=2|w|$$
Let $w=u+vi$
$$|(1+u+2v)-(2u-v)i|=2|u+vi|$$
$$(1+u+2v)^2+(2u-v)^2=4(u^2+v^2)$$
$$1+u^2+4v^2+2u+4v+4uv+4u^2-4uv+v^2=4u^2+4v^2$$
$$1+u^2+2u+4v+v^2=0$$
$$(u+1)^2+(v+2)^2=4$$
Therefore, the locus of $w$ is a circle around $(-1,-2)$ with radius $2$. In other words, $|w+1+2i|=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x+y+2xy=83$, find the value of $x+y$.
Let $x$ and $y$ be integers. If $x+y+2xy=83$, find the value of $x+y$.
I tried to multiply both sides by $x+y-2xy$ but I could never manage to simplify it. Is there a better way to solve this question?
| Wolog assume $x \le y$ and $y-x = m\ge 0$ then
$2x + m + 2x(x+m) = 83$ and
$2x^2 + (2+2m)x + (m-83) = 0$
$x = \frac {-(2+2m) \pm\sqrt{4m^2+8m + 4-4(m-83)*2}}4=$
$ \frac {-(2+2m) \pm\sqrt{4m^2 + 668}}4=$
$\frac {-1-m\pm \sqrt{m^2 +167}}2\in \mathbb Z$
So $m^2 + 167 = k^2$ for some non-negative integer, $k$, so
$k^2 - m^2 = (k-m)(k+m) = 167$ but $167$ is prime so $k-m =1$ and $k+m=167$ so $m=83$ and $k = 84$
So $x = \frac {-1-83\pm84}2$
So $x = 0, -84$ and $y =83, -1$.
So $x+y = 83$ or $-85$.
| {
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"url": "https://math.stackexchange.com/questions/3648309",
"timestamp": "2023-03-29T00:00:00",
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Determining equations for the compactified version of the system using $X$ and $Y$ as dependent variables Can someone please solve the problem? I am not sure how to approach it. Thank you for your time.
Consider the system of equations $$x' = x+y-y^3$$ $$y' = -x+y+x^3.$$ I wish to determine equations for the compactified version of this system using $X$ and $Y$ as dependent variables. Desingularize the equation at infinity, i.e. for $Z=0,$ and how would the phase portrait look like?
| $\textbf{Solution:}$ Let $x = \frac{X}{Z}$ and $y= \frac{Y}{Z}$. So, $$X' = x'Z + xZ' = Z(x+y-y^3) + x\left(-\frac{(\frac{x(x+y-y^3) + y(-x+y+x^3)}{\sqrt{1+x^2+y^2}})}{(\sqrt{1+x^2+y^2})^2}\right)$$ $$=X+Y-\frac{Y^3}{Z^2} + \frac{X}{Z}\left(\frac{-\frac{X^2}{Z^2} - \frac{Y^2}{Z^2}+\frac{XY^3}{Z^4}-\frac{X^3Y}{Z^4}}{Z^{-3}}\right)$$ $$=X+Y - \frac{Y^3}{Z^2} + \frac{X}{Z}\left(-X^2Z-Y^2Z+ \frac{XY^3}{Z}-\frac{X^3Y}{Z}\right)$$ $$=X+Y - \frac{Y^3}{Z^2} - X^3 - XY^2+\frac{X^2Y^3}{Z^2}-\frac{X^4Y}{Z^2}$$ $$=\frac{1}{Z^2} (-Y^3 + X^2Y^3-X^4Y).$$ $$\text{Now, }Y'= y'Z + yZ' = Z(-x+y+x^2) + \frac{Y}{Z}(-X^2Z - Y^2Z + \frac{XY^2}{Z} - \frac{X^3Y}{Z})$$ $$=-X + Y + \frac{X^3}{Z^2} - X^2Y - Y^3 + \frac{XY^4 - X^3Y^2}{Z^2}$$ $$=\frac{1}{Z^2} (X^3 + XY^4 - Z^3Y^2).$$ Now, for $Z=0$ $$X=-Y^3 + X^2Y^3 - X^4Y = Y(-Y^2+X^2Y^2-X^4) \text{ and } Y= X^3 + XY^4 -X^3Y^2 = X(X^2+Y^4-X^2Y^2).$$ Observe $(0,\pm 1)$ is an equilibria point for $X,$ and $(\pm 1, 0)$ is an equilibria for $Y$. By comparing the right hand side equations with our given system, we arrive at $$Y^4 - X^4 -Y^2 + X^2= 0 \implies Y^4-X^4 = Y^2-X^2 \implies (X,Y) = (0,0).$$ However, by our established equilibrium points we have the following circle basically with points on the axis as $x=-1,1$ and at $y=-1,1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Establishing $\frac{ \sin mx}{\sin x}=(-4)^{(m-1)/2}\prod_{1\leq j\leq(m-1)/2}\left(\sin^2x-\sin^2\frac{2\pi j}{m}\right) $ for odd $m$
Let $m$ be an odd positive integer. Prove that
$$ \dfrac{ \sin (mx) }{\sin x } = (-4)^{\frac{m-1}{2}} \prod_{1 \leq j
\leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
j }{m } \right) \right) $$
Atempt to the proof
My idea is to use induction on $m$. The base case is $m=3$ and we obtain
$$ \dfrac{ \sin (3x) }{\sin x } = (-4) ( \sin^2 x - \sin^2 (2 \pi /3 ) ) $$
and this holds if one uses the well known $\sin (3x) = 3 \sin x - 4 \sin^3 x $ identity.
Now, if we assume the result is true for $m = 2k-1$, then we prove it holds for $m=2k+1$. We have
$$ \dfrac{ \sin (2k + 1) x }{\sin x } = \dfrac{ \sin [(2k-1 + 2 )x] }{\sin x } = \dfrac{ \sin[(2k-1)x ] \cos (2x) }{\sin x } + \dfrac{ \cos [(2k-1) x ] \sin 2x }{\sin x } $$
And this is equivalent to
$$ cos(2x) \cdot (-4)^{k-1} \prod_{1 \leq j
\leq k-1 }\left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
j }{m } \right) \right) + 2 \cos [(2k-1) x ] \cos x $$
Here I dont see any way to simplify it further. Am I on the right track?
| Note that $\sin (x-\frac{2\pi j }{m})=-\sin(x+\frac{(m-2j)\pi}{m})$ and $m-2j$ goes through the odd numbers $1,...m-2$ when $ 1\le j \le \frac{m-1}{2}$
By the paralelogram rule for sine $\sin^2 x- \sin^2 y=\sin(x-y)\sin(x+y)$ so we get that the RHS product
$P=\sin x \prod_{1 \leq j
\leq \frac{(m-1)}{2} } \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
j }{m } \right) \right)=(-1)^{\frac{m-1}{2}}\prod_{0 \leq j
\leq m-1}\sin (x+\frac{j\pi}{m})=$
$=(-1)^{\frac{m-1}{2}}2^{-(m-1)}\sin mx$ by the classic product formula, so we are done!
(the product formula is obtained by taking the imaginary part of both sides in $e^{2imx}-1=\Pi_{k=0,..m-1} {(e^{2ix}-e^{-\frac{2\pi ik}{m}})}$)
| {
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$K_{\frac{3}{2}}(z)=?$ where $K_{\nu}$ is the modified Bessel function The modified bessel function of the second kind is the function $K_n(z)$ which is one of the solutions to the modified Bessel differential equation.
Using Wolframalpha, we get $$K_{\frac{3}{2}}(z)= \sqrt{\frac{π}{2}} \frac{e^{-z} (1 + 1/z)}{\sqrt{z}} \qquad (*)$$
I'm looking for a reference (book or article) which I can find $(*)$.
Thank you in advance
| For large values of $z$, the asymptotics are given by
$$\sqrt{\frac{2z}{\pi }}\, e^z\, K_n(z)=1+\frac{4 n^2-1}{8 z}+\frac{16 n^4-40 n^2+9}{128 z^2}+\frac{64 n^6-560 n^4+1036
n^2-225}{3072 z^3}+O\left(\frac{1}{z^4}\right)$$
Now, notice that
$$16 n^4-40 n^2+9=16\left(n-\frac 32\right)\left(n-\frac 12\right)\left(n+\frac 12\right)\left(n+\frac 32\right)$$
$$64 n^6-560 n^4+1036
n^2-225=64\left(n-\frac 52\right)\left(n-\frac 32\right)\left(n-\frac 12\right)\left(n+\frac 12\right)\left(n+\frac 32\right)\left(n+\frac 52\right)$$ and for the particular case of $n=\frac 32$ the rhs reduces exactly to $1+\frac 1 z$. In fact, if $n=k+\frac 12$, you have simple and nice forms (as asymptotics - no $O(.)$)
$$\left(
\begin{array}{cc}
k & \sqrt{\frac{2z}{\pi }}\, e^z\, K_{k+\frac 12}(z) \\
0 & 1 \\
1 & 1+\frac{1}{z} \\
2 & 1+\frac{3}{z}+\frac{3}{z^2} \\
3 & 1+\frac{6}{z}+\frac{15}{z^2}+\frac{15}{z^3} \\
4 & 1+\frac{10}{z}+\frac{45}{z^2}+\frac{105}{z^3}+\frac{105}{z^4} \\
5 & 1+\frac{15}{z}+\frac{105}{z^2}+\frac{420}{z^3}+\frac{945}{z^4}+\frac{945}{z^5} \\
6 & 1+\frac{21}{z}+\frac{210}{z^2}+\frac{1260}{z^3}+\frac{4725}{z^4}+\frac{10395}{z^5}+\frac{10395}{z^6}
\end{array}
\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maclaurin series expansion for $\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$
Show that the expansion of
$$\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt{t}} \log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$
as a Maclaurin series in power of $t$ to obtain
$$\dfrac{4}{3}-4\sum_{n=1}^{\infty}\dfrac{t^n}{(4n^2-1)(2n+3)}$$
I am not sure how to do this. Below are some steps I have managed to get which gets more complicated and am not sure how to continue. Help is appreciated.
Expand into:
$$\dfrac{1}{2t}+\dfrac{1}{2} - \dfrac{(1-t)^2}{4t\sqrt{t}}\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)$$
$$=\dfrac{1}{2t}+\dfrac{1}{2}- \dfrac{\log\left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)}{4t\sqrt{t}}(-t+1)^2$$
Multiply by congugate:
$$=\dfrac{1}{2t}+\dfrac{1}{2}-\dfrac{\log\left(\dfrac{(1+\sqrt{t})^2}{-t+1}\right)}{4t\sqrt{t}}(-t+1)^2$$
| Considering $$A=\dfrac{1}{2t}\left(1+t-\dfrac{(1-t)^2}{2 \sqrt {t}}\log \left(\dfrac{1+\sqrt{t}}{1-\sqrt{t}}\right)\right)$$ just to make life easier, start making $t=x^2$ to get
$$A=\frac{2 x(1+x^2)-\left(1-x^2\right)^2 \log
\left(\frac{1+x}{1-x}\right)}{4 x^3}$$ and use first, as @metamorphy commented, the expansion of $\log\left(\frac{1+x}{1-x}\right)$ and expand. You should arrive at
$$\frac{4}{3}-\frac{4 x^2}{15}-\frac{4 x^4}{105}-\frac{4 x^6}{315}-\frac{4
x^8}{693}-\frac{4 x^{10}}{1287}+O\left(x^{12}\right)$$ Back to $t$
$$A=\frac 43-4\left(\frac{ t}{15}+\frac{t^2}{105}+\frac{t^3}{315}+\frac{
t^4}{693}+\frac{t^5}{1287} \right)++O\left(t^{6}\right)$$ The denominators seem to be what you need.
I must confess that I did not find a way to prove it without using a truncated series and finding a pattern for the coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $ \begin{array}{l}\sum _{n=0}^{\infty }\frac{\left(2k^2+2k+1\right)}{k^4+2k^3+k^2}\end{array}$ Prove that :$ \begin{array}{l}\sum _{n=0}^{\infty }\frac{\left(2k^2+2k+1\right)}{k^4+2k^3+k^2}\end{array}$ where $k=2n+1$ converges. Also find the value it converges to.
| HINT:
$$\frac{2 (2 n+1)^2+2 (2 n+1)+1}{(2 n+1)^4+2 (2 n+1)^3+(2 n+1)^2}=\frac{1}{(2n+1)^2}+\frac{1}{4(n+1)^2}\tag1$$
| {
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} |
Proving a function covers the unit sphere How can I show $f:\mathbb R^2\to\mathbb R^3$ defined as$$f(u,v)=\left(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1}\right)$$covers the whole unite sphere except the point $(0,0,1)$? It’s easy to show that the image of $f$ lies in $S^2$ but I can’t show it covers every point except a pole.
| Take any $(x,y,z)\in S^2$, so $x^2+y^2+z^2=1$. Let $(u,v)=(\frac{x}{1-z},\frac{y}{1-z})$. [This is the inverse of the stereographic projection map.]
Then $$u^2+v^2+1=\frac{x^2}{(1-z)^2}+\frac{y^2}{(1-z)^2}+1=\frac{x^2+y^2+(1-z)^2}{(1-z)^2}=\frac{2-2z}{(1-z)^2}=\frac{2}{1-z}$$ if $z\ne1$. We also have
$$u^2+v^2-1=\frac{x^2+y^2-(1-z)^2}{(1-z)^2}=\frac{x^2+y^2-z^2+2z-1}{(1-z)^2}=\frac{-2z^2+2z}{(1-z)^2}=\frac{2z(1-z)}{(1-z)^2}=\frac{2z}{1-z}$$ if $z\ne1$.
Therefore, if $z\ne1$, we have
$$f(u,v)=\left(\frac{2x}{1-z}\cdot\frac{1-z}{2},\frac{2y}{1-z}\cdot\frac{1-z}{2},\frac{1-z}{2}\cdot\frac{2z}{1-z}\right)=(x,y,z).$$
So every $(x,y,z)\in S^2$ except the north pole is in the image.
| {
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"timestamp": "2023-03-29T00:00:00",
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Deeper meaning and intuition behind $\frac{x}{1+x^2}$ having the same values for $x$ and for $\frac{1}{x}$ I was playing around with the function $f(x)=\frac{x}{1+x^2}$ and I noticed that $f(x)$ has the same values for $x$ and for $\frac{1}{x}$.
Is there any intuition or deeper meaning behind this?
Thanks in advance!
| Consider $g(x) = x + \frac{1}{x}$. Then $g(x) = g(\frac{1}{x})$, because $x + \frac{1}{x}$ = $\frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x$.
Your function $f(x) = \frac{1}{g(x)} = \frac{x}{x} \frac{1}{x + \frac{1}{x}} = \frac{x}{x(x + \frac{1}{x})} = \frac{x}{x^2 + 1}$
So some intuitions or deeper meanings behind this are: addition commutes and a reciprocal of a reciprocal is the original value (inverse of inverse). BTW note that $x\ne0$ for this to work - which we know it is, because we have already $\frac{1}{x}$.
| {
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$\mathbf{A}^T = p(\mathbf{A})$, prove that $\mathbf{A}$ is invertible Let $\mathbf{A}$ be a square matrix defined over a field $\mathbb{R}$. It is known that $\mathbf{A}^\text{T} = p(\mathbf{A})$, where $p(\mathbf{A})$ is a polynomial with a constant coefficient $a_0 \neq 0$.
*
*Prove that $\mathbf{A}$ is invertible.
*Is it true that for every operator $\mathbb{\phi}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ there exists some polynomial $p(x)$ and the basis for which the matrix $\phi$ satisfies the condition of $\mathbf{A}^\text{T} = p(\mathbf{A})$?
Solution
I.
By the definition of $\mathbf{A}$: $$\mathbf{A}^t = P(\mathbf{A}) = a_n \cdot \mathbf{A}^n + \cdots + a_1 \cdot \mathbf{A} + a_0 \cdot \mathbf{I}$$
Consider
$$\mathbf{A}^t \cdot \mathbf{A} = a_n \cdot \mathbf{A}^n \cdot \mathbf{A} + \cdots + a_1 \cdot \mathbf{A} \cdot \mathbf{A} + a_0 \cdot \mathbf{I} \cdot \mathbf{A} \\ = a_n \cdot \mathbf{A}^{n + 1} + \cdots + a_1 \cdot \mathbf{A}^2 + a_0 \cdot \mathbf{A}$$
$$\mathbf{A} \cdot \mathbf{A}^t = a_n \cdot \mathbf{A} \cdot \mathbf{A}^n + \cdots + a_1 \cdot \mathbf{A} \cdot \mathbf{A} + a_0 \cdot \mathbf{A} \cdot I \\ = a_n \cdot \mathbf{A}^{n + 1} + \cdots + a_1 \cdot \mathbf{A}^2 + a_0 \cdot \mathbf{A}$$
hence, $\mathbf{A}^t \cdot \mathbf{A} = \mathbf{A} \cdot \mathbf{A}^t$. Therefore, the matrix $\mathbf{A}$ is normal, which assumes $\text{ker}(\mathbf{A}) = \text{ker}(\mathbf{A^t})$. Assume that $\mathbf{A}$ is not invertible, then its kernel is non-trivial. Take non-zero $v$ from $\text{ker}(\mathbf{A})$:
$$\mathbf{A}^t \cdot v = a_n \cdot \mathbf{A}^n \cdot v + \cdots + a_1 \cdot \mathbf{A} \cdot v + a_0 \cdot \mathbf{I} \cdot v \\ \mathbf{A}^t \cdot v = 0 + \cdots + 0 + a_0 \cdot v = a_0 \cdot v$$
As per definition, $a_0 \neq 0$, so $a_0 \cdot v \neq 0$, and it follows that $\mathbf{A}^t \cdot v \neq 0$. However, the condition $\text{ker}(\mathbf{A})=\text{ker}(\mathbf{A^t})$ implies that $\mathbf{A}^t \cdot v = 0$. We have a contradiction, hence $\mathbf{A}$ is invertible.
II.
It was proven that $\mathbf{A}$ is normal if its transpose can be represented by a polynomial. But normality is independent of the basis chosen, so the second statement would mean that every operator is normal, which is not the case.
Any mistakes, improvements?
| For the first question, suppose $Ax=0$. Then
$$
0=(Ax)^Tx=x^TA^Tx=x^T\left(p(A)x\right)=a_0x^Tx=a_0\|x\|_2^2.
$$
Therefore $x$ must be zero and $A$ is invertible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Uniform convergence and boundedness of $\sum_{k=1}^n \frac{\sin kx}{k}$ Why is $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly convergent on $[\delta,\pi]$ for all $0<\delta<\pi$ from this nice property of Fourier series?
Furthermore, why is the partial sum $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly bounded for all $n$ and $x \in (0,\pi)$ as discussed in this question.
| Note that for $x > 0$ we have $|\sin (nx)| \leqslant |nx| = nx$ and
$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \sum_{k=1}^M\frac{|\sin (nx)|}{n} \leqslant \sum_{k=1}^M \frac{nx}{n} = Mx,$$
and since $M = \min(N, \lfloor \frac{\pi}{x}\rfloor)\leqslant \lfloor \frac{\pi}{x}\rfloor \leqslant \frac{\pi}{x}$ we get
$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \frac{\pi}{x} x = \pi$$
For the second sum in the linked question it was shown that
$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant \frac{1}{|\sin(x/2)|}\left(\frac{1}{N} + \frac{1}{M+1} + \sum_{n=M+1}^{N-1}\left(\frac{1}{n} - \frac{1}{n+1} \right) \right) = \frac{2}{(M+1)|\sin(x/2)|}$$
For $z \in (0,\pi/2)$ we have the inequality $\frac{2z}{\pi} \leqslant \sin z < z$ and, thus, for $x \in (0,\pi)$ we have $\sin(x/2) \geqslant x /\pi$. Hence,
$$\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}$$
In bounding this sum we can assume that $M < N$ (otherwise there are no terms) and, hence, $M = \lfloor \frac{\pi}{x} \rfloor$ implying that $M \leqslant \frac{\pi}{x} < M+1$ and $\frac{1}{(M+1)x} < \frac{1}{\pi}$.
Therefore,
$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}< \frac{2\pi}{\pi} = 2$$
| {
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Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$ Let $a,\,b,\,c$ are non-negative such that $ab+bc+ca>0.$ Prove that
$$\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}. \quad (1)$$
Note. Because
$$\sum \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)} = 2.$$
So, from $(1)$ we get know inequality of Darij Grinberg
$$\frac{a(b+c)}{b^2+bc+c^2}+\frac{b(c+a)}{c^2+ca+a^2}+\frac{c(a+b)}{a^2+ab+b^2} \geqslant 2.$$
My proof use sum of squares method.
| Thank @Michael Rozenberg :D We write inequality as
$$f(a) = 3(b+c)a^3-(5b^2+2bc+5c^2)a^2+(b+c)(b^2+bc+c^2)a+(b^2+bc+c^2)(b-c)^2 \geqslant 0.$$
But
$$\left(a+\frac{7b+7c}{4}\right) \cdot f(a)=\frac{3}{4}(b+c)(2a^2-b^2-c^2-ab+2bc-ca)^2+\frac{5bc(b+c)(b-c)^2}{4}$$
$$+\frac{17abc[(a-b)^2+(a-c)^2]}{4}+\frac{25a(b^2+c^2-ab-ca)^2}{8}+\left(\frac{a}{8}+b+c\right)(b-c)^2(b+c-a)^2.$$
So $f(a) \geqslant 0.$
| {
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Solve definite integral $\int_0^{2\pi} \frac{\cos^2x}{(1+b\cos x)^4} dx$ I am struggling with analytically solving the definite integral
$$\int_0^{2\pi} \frac{\cos^2x}{(1+b\cos x)^4} dx$$
I am more generally having issues with solving integrals of the form $\int_0^{2\pi}\frac{1}{(1+b\cos(x))^a} dx$, how can I solve these (both numerically and analytically. ($b<1$))
| Let
\begin{align}
I(b) &= \int_0^{\pi} \frac{1}{(1+b \cos x)^2} dx\\
&= \frac1{1-b^2}\int_0^{\pi}\left(
-d (\frac{b\sin x}{ 1+b \cos x} ) + \frac{1}{1+b \cos x} dx\right) \\
&= \frac1{1-b^2}\int_0^{\pi} \frac{1}{1+b \cos x} dx
= \frac\pi{(1-b^2)^{3/2}}
\end{align}
Then
$$\int_0^{2\pi} \frac{\cos^2 x}{(1+b \cos x)^4} dx
=\frac13\frac {d^2I(b)}{db^2}=\frac{\pi (1+4b^2)}{(1-b^2)^{7/2} }$$
——————————————-
Edit:
\begin{align}
& \int_0^{\pi} \frac{1}{1+b \cos x} dx \\
=&\int_0^{\pi} \frac{1}{1+b (2\cos^2\frac x2 -1)}dx
=\int_0^{\pi} \frac{2d(\tan \frac x2)}{(1-b )\tan^2\frac x2 +(1+b)}dx \\
=& \frac2{\sqrt{1-b^2}}\tan^{-1}\left(\sqrt{\frac{1-b}{1+b}}\tan\frac x2\right)_0^{\pi}= \frac\pi{\sqrt{1-b^2}}
\end{align}
| {
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A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of $\frac{a + b}{c}$.
A right triangle has legs $a$ and $b$ and hypotenuse $c.$ Find the largest possible value of
$$\frac{a + b}{c}.$$
I used the QM-AM inequality. For a set of numbers $\{a_1, a_2, \dots, a_n\}$ this inequality is,
$$\sqrt{\frac{a_1^2 + a_2^2+\dots+a_n^2}{n}}\geq\frac{a_1 + a_2 + \dots + a_n}{n}.$$
Since $(a,b,c)$ form a right triangle with $c$ being the hypotenuse, we have $a^2 + b^2 = c^2$. Using the QM-AM inequality on $a, b, $ and $c$, we have,
$$\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}.$$
Multiplying both sides by $9$ and plugging in $a^2 + b^2 =c^2$, we have
$$a+b+c\leq \sqrt{6c^2}.$$
Taking the $c$ out of the square root, we have,
$$a+b+c\leq c\sqrt6.$$
Subtracting $c$ from both sides, we have,
$$a+b\leq c\sqrt{6}-c.$$
Factoring $c$ out, we have $$a+b\leq(1-\sqrt{6})\cdot c.$$ I don't know where to go from here. I know for a fact that $1-\sqrt{6}$ isn't the answer, and this is probably because $a, b, c$ have to be positive, but I don't know how to solve for it. Any help? Thanks!
| Note $2ab\le a^2+b^2$ and
$$\frac{a+b}c = \sqrt{\frac{(a+b)^2}{a^2+b^2}}= \sqrt{1+\frac{2ab}{a^2+b^2} }\le \sqrt{1+1} =\sqrt2
$$
| {
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Trigonometry parametric equation $\sin(\sqrt {ax-x^2})=0 $ This is the hardest problem on Georgian (country) high school math exam.
Find all values for parameter a for which the sum of all the roots of the equation
$\sin(\sqrt {ax-x^2})=0 $
equal to 100.
| Let's assume for simplicity $a>0$, for $a<0$ we take $-x$ in place of $x$.
$y=ax-x^2\ge 0$ gives $0\le x\le a$ and $ax-x^2=c$ has exactly $2$ solutions for each $0\le c< y(\frac{a}{2})=\frac{a^2}{4}$ and exactly one for $c = y(\frac{a}{2})$, so does $u(x)=\sqrt{ax-x^2}=\sqrt{c}$.
On the other hand, $\sin u=0$ gives $u=\pi n,\,n\in\mathbb{Z}$ so we consider $2$ cases: $\sqrt{y(\frac{a}{2})}=\frac{a}{2}=\pi n$ and $a\ne\pi n$.
In the second case $\sin \sqrt{ax-x^2}=0$ for $0\le u\le \frac{a}{2}$ has $N+1$ pairs of roots $x = \frac{1}{2} \left(a \pm \sqrt{a^2 - 4 π^2 k^2}\right),\,0\le k\le N,\,k\in Z$ where $N=\lfloor\frac{a}{2\pi}\rfloor$, each pair sums up to $a$, so we have sum of roots $a\left(\lfloor\frac{a}{2\pi}\rfloor+1\right)$, in the first case we have one more root $\frac{a}{2}$, but $a$ has to be $2\pi n,\,n\in\mathbb{Z},\,n\ge 0$ so $\pi 2n(n+1)+\pi n$ has to be $=100$ for $n\in\mathbb{Z}$, which can't be as $\frac{100}{\pi}\notin \mathbb{Z}$, but has to be $=2n^2+3n\in\mathbb{Z}$.
So we consider $a\left(\lfloor\frac{a}{2\pi}\rfloor+1\right)=100$ and consider all factorizations of $100$ into product of two different positive integers: $(100,1),(50,2),(25,4),(20,5)$, checking only $a=25,\lfloor\frac{a}{2\pi}\rfloor+1=4$ fits.
For $a<0$ the roots count will be positive, but $a$ will be negative, so the sum can't be $=100$.
Answer: 25
| {
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Solving recurrence relation with initial conditions given I am trying to solve 2 recurrence equations:
a) $a_{n+1} = 2a_n + 2^n + 3, a(0)=4$
b) $a_{n+2} = a_n + 4n, a(0)=1, a(1)=1$
For (a), I don't know how to deal with $2^n$ part.
For (b), I am confused because $a(0)=a(1)$.
| $$a_{n+1}=2a_n+2^n+3,a(0)=4$$
Let's try it a bit:
$$a_{n+1}=2(2a_{n-1}+2^{n-1}+3)+2^n+3=4a_{n-1}+2\cdot2^{n}+(3+3\cdot2)$$
$$a_{n-1}=2a_{n-2}+2^{n-2}+3 \implies a_{n+1}=4(2a_{n-2}+2^{n-2}+3)+2\cdot2^{n}+9$$
$$\iff a_{n+1}=8a_{n-2}+3\cdot2^n+(3+3\cdot2+3\cdot4)$$
So, we can conjecture, (and we shall prove by induction): let $k$ be a positive integer,
$$a_{n+1}=2^{k+1}\cdot a_{n-k}+(k+1)\cdot2^{n}+3(1+2+\dots+2^k)$$
Now, with our base case $k=0$, we can have the inductive assumption:
$$P_m:a_{n+1}=2^{m+1}\cdot a_{n-m}+(m+1)\cdot2^{n}+3(1+2+\dots+2^m)$$
So, we shall prove
$$P_{m+1}:a_{n+1}=2^{m+2}\cdot a_{n-m-1}+(m+2)\cdot2^{n}+3(1+2+\dots+2^{m+1})$$
Now, by our original equation,
$$a_{n-m}=2a_{n-m-1}+2^{n-m-1}+3$$
So,
$$a_{n+1}=2^{m+1}\cdot (2a_{n-m-1}+2^{n-m-1}+3)+(m+1)\cdot2^{n}+3(1+2+\dots+2^m)$$
$$a_{n+1}=2^{m+2}a_{n-m-1}+2^{n}+3\cdot 2^{m+1}+(m+1)\cdot2^{n}+3(1+2+\dots+2^m)$$
$$=2^{m+2}a_{n-m-1}+(m+2)\cdot2^{n}+3(1+2+\dots+2^m+2^{m+1}) \ \Box.$$
So, now we've proved:
$$a_{n+1}=2^{k+1}\cdot a_{n-k}+(k+1)\cdot2^{n}+3(1+2+\dots+2^k)$$
$$=2^{k+1}\cdot (a_{n-k}+3)+(k+1)\cdot2^{n}-3 \tag{by geometric series formula}$$
Putting $k=n$ to relate it to $a_0$, we get
$$a_{n+1}=2^{n+1}\cdot (a_0+3) + (n+1) \cdot 2^{n}-3 =7 \cdot 2^{n+1}+ (n+1) \cdot 2^{n}-3 $$
$$\implies a_n = 7 \cdot 2^{n}+ n2^{n-1}-3 =2^{n-1}(n+14)-3 \ \Box.$$
| {
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Find all $(a , b , c)$ such that $\frac{(a - b)(b - c)(c - a) + 4}{2}$ is power of $2016$. Find all $(a , b , c)$ such that $\frac{(a - b)(b - c)(c - a) + 4}{2}$ is power of $2016$. (Power of $2016$ means the number can be written in the form $2016^n$ for some non-negative $n$.
I don’t know what to do. Should I factorize $2016$ and consider by mod $32$ , mod $9$ and mod $7$?
Can anyone give me some hint please. Thank you!
| Hints: First, let $x = b-a$ and $y = c-b$. Then, $(a-b)(b-c)(c-a) = xy(x+y)$.
We want $\dfrac{xy(x+y)+4}{2} = 2016^n$, i.e. $xy(x+y) = 2\cdot 2016^n-4$ for some integer $n \ge 0$.
If $n \ge 1$, then we need $xy(x+y) \equiv 5 \pmod{9}$. Is this possible? You have quite a few cases to check, but remember that none of $x$, $y$, and $x+y$ can be $0 \pmod{3}$.
If $n = 0$, then we need $xy(x+y) = -2$. This should be easy enough to solve.
| {
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What is the density of Z? if $f_{x,y}(x,y)= I_{(0,1)}(x) I_{(0,1)}(y) $ find the density of $Z$, where
$$Z= (X+Y) I_{(-\infty,1]} \hspace{.2cm} (X+Y) + (X+Y-1)I_{(1,\infty)} \hspace{.1cm} (X+Y) .$$
I know that $V=X+Y$ has the following distribution
$$f(v) = v I_{(0,1]} \hspace{.1cm}(v) + (2-v)I_{(1,2)} \hspace{.1cm} (v) $$
the solution puts it this way
$P[Z \leq z]= P[X+Y \leq z ; X+Y \leq 1] +P[X+Y-1 \leq z; X+Y>1] $
$=P[ X+Y \leq z ]+ P[ 1 < X+Y \leq z+1 ] = z I_{(0,1)}\hspace{.1cm} (z)$
so Z is uniformly distributed over (0,1).
but I don't understand this solution, could you explain to me, or if there is another solution.
| The solution omits some important steps and for that reason, I think it is sloppy. The main problem is that the result is incorrect: $$\Pr[Z \le z] = \begin{cases} 0, & z < 0 \\ z, & 0 \le z \le 1 \\ 1, & z > 1. \end{cases}.$$ Writing it as $z I_{(0,1)}(z)$ is plainly flawed since this would suggest the CDF is zero for $z \ge 1$.
As such, we cannot actually write $$\Pr[(X+Y \le z) \cap (X+Y \le 1)] = \Pr[X+Y \le z].$$ This is, I repeat, sloppy. The correct reasoning proceeds as follows:
$$\begin{align*}
\Pr[Z \le z]
&= \Pr[(Z \le z) \cap (X+Y \le 1)] + \Pr[(Z \le z) \cap (X+Y > 1)] \\
&= \Pr[(X+Y \le z) \cap (X+Y \le 1)] + \Pr[(X+Y-1 \le z) \cap (X+Y > 1)] \\
&= \Pr[X+Y \le \min(z, 1)] + \Pr[1 < X+Y \le z+1].
\end{align*}$$
Now, there are three cases to consider. In the first case, $z < 0$ results in both probabilities being zero since $\Pr[X + Y < 0] = 0$ and $\Pr[1 < X+Y < 1] = 0$. In the second case $z > 1$ gives $$\Pr[Z \le z] = \Pr[X+Y \le 1] + \Pr[1 < X+Y \le z+1] = \Pr[X+Y \le z + 1] = 1,$$ since $z > 1$ implies $z+1 > 2$ and because $X+Y \le 2$ on $[0,1]^2$, the claim follows. The only case remaining is $0 \le z \le 1$. In this situation, we have $$\Pr[X+Y \le z] = \frac{z^2}{2},$$ since the set of points $(X,Y) \in [0,1]^2$ satisfying $X+Y \le z$ comprise the triangle with vertices $(0,0)$, $(z,0)$, and $(0,z)$. We also have $$\Pr[1 < X+Y \le z+1] = 1 - \Pr[X+Y \le 1] - \Pr[X+Y > z+1] = 1 - \frac{1}{2} - \frac{(1-z)^2}{2}$$ using similar reasoning as above. This gives $$\Pr[Z \le z] = \frac{z^2}{2} + \frac{1}{2} - \frac{(1-z)^2}{2} = z$$ as desired. This completes the calculation.
| {
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Explain why $\{0,2,4,6\}\le\Bbb Z_8$, $\{1,4\}\le\Bbb Z_5^*$, and $\{1,5,7,11\}\le\Bbb Z_{12}^*.$ Could someone please explain the examples?
I can see that the first example is using modular addition, where 4+6=2. But I lack the maths background to understand the next two examples, I thought the * symbol mean multiply?
| It suffices to check for closure, an identity, and inverses, since associativity is inherited from each parent group; that they are each subsets is clear. I will demonstrate these properties by means of their Cayley tables.
For $\{0,2,4,6\}\le \Bbb Z_8$, we have
$$\begin{array}{c|cccc}
+_8 & 0 & 2 & 4 & 6\\
\hline
0 & 0 & 2 & 4 & 6 \\
2 & 2 & 4 & 6 & 0 \\
4 & 4 & 6 & 0 & 2 \\
6 & 6 & 0 & 2 & 4,
\end{array}$$
from which one can see that the identity is $0$ and $-2=6$, and $-4=4$.
For $\{1,4\}\le\Bbb Z_5^*$, we have
$$\begin{array}{c|cc}
\times_5 & 1 & 4 \\
\hline
1 & 1 & 4\\
4 & 4 & 1,
\end{array}$$
from which one can see that the identity is $1$ and $4^{-1}=4$.
For $\{1,5,7,11\}\le\Bbb Z_{12}^*$, we have
$$\begin{array}{c|cccc}
\times_{12} & 1 & 5 & 7 & 11\\
\hline
1 & 1 & 5 & 7 & 11 \\
5 & 5 & 1 & 11 & 7 \\
7 & 7 & 11 & 1 & 5 \\
11 & 11 & 7 & 5 & 1,
\end{array}$$
from which one can see that $1$ is the identity, $5^{-1}=5, 7^{-1}=7$, and $11^{-1}=11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate the volume of the solid determined by $S_1$ and $S_2$ I want to calculate the volume of the solid determined by this tho surfaces:
$$S_1=\{(x,y,z)\in\mathbb{R}:x^2+y^2+z^2=R^2\}$$
$$S_2=\{(x,y,z)\in\mathbb{R}:x^2+y^2=Rx\}$$
The solid is the intersection of a sphere of radius $R$ ($S_1$) and a cylinder of diameter $R$ (centered in $(R/2,0,0)$)($S_2$)
I guess i must change to spherical or cylindrical coordinates, and that's what i have problems with. I'm stucked in finding the new values of the variables. Also, which coordinate system will work better for this problem? Spherical or cylindrical? I will thank any help.
| Note that $S_2$ is cylinder $$
(x-\frac{R}{2})^2+ y^2 = (\frac{R}{2})^2 $$
We will use cylindrical coordinate : Here the curve $ z=0,\
(x-\frac{R}{2})^2+ y^2 = (\frac{R}{2})^2$ is parametrized by
$$x=r\cos\ \theta,\ y=r\sin\ \theta,\ r=R\cos\ \theta,\
-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$$
That cylinder has top and bottom roofs : $$z=\sqrt{R^2-x^2-y^2},\
z=-\sqrt{R^2-x^2-y^2}$$
Hence \begin{align*} dxdy &= rdrd\theta \\
V&=
2\int_0^{\frac{\pi}{2}}\int^{R\cos\ \theta}_0\ z \cdot r drd\theta
\\&=2\int_0^{\frac{\pi}{2}}\int^{R\cos\ \theta}_0\ \sqrt{R^2-r^2}
\cdot r drd\theta
\\&=2\int_0^{\frac{\pi}{2}}\int^{R^2\sin^2 \theta}_{R^2}\ \sqrt{T}
\frac{dT}{(-2)} d\theta
\\&= \int_0^{\frac{\pi}{2}} \ (\frac{-2}{3}) R^3 \{ \sin^3 \theta -1\}\
d\theta\\&= \frac{3\pi -4}{9} R^3
\end{align*}
since $\int\ \sin^3\theta = -\frac{1}{3}\cos\
\theta(\sin^2\theta+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I show that the zeroes of $f(z) = z^4 -2\cos(2\theta) z^2 +1$ are $e^{i\theta}$, $e^{-i\theta}$, $-e^{i\theta}$, $-e^{-i\theta}$? N.B. $z$ is a complex number.
This is what I have so far:
$$f(z) = z^{4} -2 \cos(2\theta) z^2 +1 =R^4 (\cos(4\theta)+i\sin(4\theta))-2R^2\cos(2\theta)(\cos(2\theta)+i\sin(2\theta)) +1 =0$$
$$ = R^4 (\cos(4\theta)+i\sin(4\theta)) -2R^2 \cos^2(2\theta) -2R^2i\cos(2\theta)\sin(2\theta)+1$$
$$=R^4 (\cos(4\theta)+i\sin(4\theta)) -R^2 (\cos(4\theta)+1)-R^2i\sin(4\theta)+1$$
$$=\cos(4\theta)(R^4 - R^2)+i\sin(4\theta)(R^4-R^2)-(R^2-1)$$
$$=(R^2-1)(R^2\cos4\theta +R^2 i \sin(4\theta)-1)$$
$$=(R^2-1)(R^2e^{i4\theta}-1)$$
and that's as far as I can get - if somebody would be able to point out where I've gone wrong, how I can get from here to the answer or a better way of doing this, anything would be much appreciated - thanks!
| You can view the equation $z^4-2\cos(2\theta)z^2+1=0$ as a quadratic in the variable $z^2$. Hence by the quadratic formula
$$z^2=\dfrac{2\cos(2\theta)\pm\sqrt{4\cos^2(2\theta)-4}}{2}=\cos(2\theta)\pm i\sin(2\theta)=e^{\pm2i\theta}.$$
Therefore, upon square rooting both sides (consider roots of unity) you obtain the values given above.
To see the expression for $z^2$ simplifies down sufficiently recall $\cos^2(x)+\sin^2(x)=1$ for all $x\in\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Values that make matrix positive definite Assume the symmetric matrix:
$$
M=
\begin{bmatrix}
\frac{\sigma\omega\pi^2}{4L^2} + g & 0 & -\frac{\sigma + c^2Qg}{2\sigma} \\
0 & a\mu & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right) \\
-\frac{\sigma + c^2Qg}{2\sigma} & -\frac{Q}{2}\left(\frac{c^2}\sigma + g - \mu \right)& \frac{c^2Q}{\sigma}
\end{bmatrix}
\
$$
with $\omega \in (0,1)$, $\sigma > 0$, $\mu > 0$.
The objective is to find some values of $Q,a > 0, g \in \mathbb{R}$ that make $M$ positive definite.
Attempt:
Using Sylvester's criterion, if the determinants:
\begin{align}
d_1 &= \left( \frac{\sigma\omega\pi^2}{4L^2} + g\right)a\mu \\
d_2 &= \left[ \frac{a\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} + g -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} + g \right) - a\mu \left(\frac{\sigma + c^2Qg}{2\sigma} \right)^2
\end{align}
are positive then $M$ is positive definite.
$d_1$ is easy to make it positive, for $g = 0$ and $a = 1$ for example. But then, I'm having trouble finding a value for $Q$ such that:
$$
d_2 = \left[ \frac{\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} -\mu \right)^2\right]\frac{\sigma\omega\pi^2}{4L^2} - \frac{\mu}{4} > 0
$$
Any thoughts? Is my initial choice of $g,a$ good?
| You may take $g=0$ and $a=Q^2$. Then $d_1$ is obviously positive and
\begin{align}
d_2
&= \left[ \frac{a\mu c^2Q}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} + g -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} + g \right) - a\mu \left(\frac{\sigma + c^2Qg}{2\sigma} \right)^2\\
&= \left[ \frac{\mu c^2Q^3}{\sigma}- \frac{Q^2}{4}\left(\frac{c^2}{\sigma} -\mu \right)^2\right]\left( \frac{\sigma \omega \pi^2}{4L^2} \right) - Q^2\mu \left(\frac{\sigma}{2\sigma} \right)^2\\
&= \frac{\mu c^2Q^3}{\sigma}\left( \frac{\sigma \omega \pi^2}{4L^2} \right) - Q^2\times\text{constant}\\
\end{align}
is positive when $Q>0$ is sufficiently large.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can this function be defined in a way to make it continuous at $x=0$? We have $$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}$$
If we want to "define" this function to be continuous at $x=0$, it's limit at $0$ must equal $f(0)$. So we should find this limit and assign it to be equal to $f(0)$, then the function is continuous at $0$. Since we are looking at the function when $x\to 0$, $x\neq 0$. Lets divide both sides by $x$.
$$f=\frac{x}{\vert x-1 \vert - \vert x +1 \vert}=\frac{1}{\frac{\vert x-1 \vert}{x}-\frac{\vert x+1\vert}{x}}=\frac{1}{\vert 1-\frac{1}{x}\vert - \vert 1+ \frac{1}{x}\vert }$$
We can use $\lim \phi(x)^{-1}=\frac{1}{\lim \phi(x)}$ here ( the limit $\neq$ 0, by hypothesis ). The inverse of the limit of $\phi(x)=\vert 1 - \frac{1}{x} \vert-\vert 1+\frac{1}{x}\vert$, when $x\to 0$. If $x<1$, we have that $$\frac{1}{x}>1\implies0>1-\frac{1}{x}\implies \Bigg\vert 1-\frac{1}{x}\Bigg\vert=-\Big(1-\frac{1}{x}\Big)$$
Now if $x>0$, we have that $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=-2$$
and if $x<0$, then $$\Bigg\vert 1 - \frac{1}{x} \Bigg\vert-\Bigg\vert 1+\frac{1}{x}\Bigg\vert=1-\frac{1}{x}-1-\frac{1}{x}=\frac{(-2)}{x}$$
The limit of $f$ when $x\to 0$ appears to be $\frac{-1}{2}$. Could anyone tell me what errors I made in the limit finding process?
| You get the correct result but the derivation is incorrect (or perhaps just unclear to me). You are performing the correct calculation for the $x>0$ case, but the $x<0$ should be more precise.
The second line should be:
\begin{equation}
\frac{1}{\frac{\left|x-1 \right|}{x} - \frac{\left|x+1 \right|}{x} } = \frac{sgn(x)}{\frac{\left|x-1 \right|}{|x|} - \frac{\left|x+1 \right|}{|x|} } = \frac{sgn(x)}{\left| 1 -\frac{1}{x} \right| - \left| 1 +\frac{1}{x} \right|},
\end{equation}
where $sgn(x)$ is the sign function, which is $1$, when $x>0$, and $-1$, when $x<0$.
When $x<0$ and close to zero $\left(1 + \frac{1}{x}\right)<0$. Hence
\begin{equation}
\left|1+ \frac{1}{x}\right| =- \left(1+ \frac{1}{x} \right).
\end{equation}
Thus in the case of $lim_{x\rightarrow0^-} f(x)$ we have
\begin{equation}
lim_{x\rightarrow0^-} \frac{sgn(x)}{\left|1 - \frac{1}{x} \right| - \left| 1 + \frac{1}{x} \right|}=lim_{x\rightarrow0^-} \frac{-1}{\left( 1 - \frac{1}{x} \right) + \left( 1 + \frac{1}{x} \right)} = -\frac{1}{2}.
\end{equation}
So the limits on both sides match, hence the function is continuous. You have missed two minus signs. One from the $sgn(x)$ at first, and another one when computing $\left| 1+ \frac{1}{x} \right|$ so your result ends up correct but only because you got lucky in getting an even number of minus signs wrong.
As a final comment, I will say that in order to see that $\left(1+ \frac{1}{x} \right)<0$ for small negative x, we can observe that $\lim_{x\rightarrow 0^-}\frac{1}{x} \rightarrow - \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse
Let the point $(h,k)$ lie on the given circle
The chord of the contact drawn to the given ellipse is
$$\frac{hx}{4}+\frac{ky}{9}-1=0$$
This line is coincident with the the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$y=mx\pm \sqrt{a^2m^2+b^2}$$
Then comparing the two equations
$$m=\frac{-9h}{4k}$$
And $$\frac{81}{k^2}=a^2m^2+b^2$$
$$\frac{81}{k^2}=\frac{81a^2h^2}{16k^2}+b^2$$
$$(81)(16)=81a^2h^2+16k^2b^2$$
How do I proceed from here? Simply substituting $h^2=100-k^2$ doesn’t give any details for $a$ and $b$
| Apply the same method that I did in this answer to a previous question of yours.
Using the usual parameterization of a circle, we get the one-parameter family of polar lines $$\frac52x\cos t+\frac{10}9y\sin t-1=0.$$ Equate coefficients with the generic line $\lambda x+\mu y+\tau=0$ and eliminate $t$ to obtain $$\frac4{25}\lambda^2+\frac{81}{100}\mu^2-\tau^2=0.$$ This conic is dual to $$\frac{25}4x^2+\frac{100}{81}y^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Cannot solve a differential equation using Laplace transformations $\dddot x + x = 2t$, $x(0)=0$, $\dot x(0)=1$, $\ddot x(0)=0$ and the answer should be $x=t^2-2+2\cos{t}+\sin{t}$
I start off with assuming $L(x)=X(p)$ then $$\dddot x=p^3X(p)-p^2X(0)-pX'(0)-X''(0)=p^3X-p$$
$$p^3X-p+X=\frac2{p^2}$$
$$X=\frac2{p^2(p^3+1)}+\frac{p}{p^3+1}=2\frac1{p^2}+\frac13\frac1{p+1}-\frac13\frac{p+1}{p^2-p+1}$$
$$\frac{p+1}{p^2-p+1}=\frac{p-\frac12+\frac12+1}{p^2-p+\frac14+\frac34}=\frac{p-\frac12}{(p-\frac12)^2+\frac34}+\sqrt{3}\frac{\frac{\sqrt3}{2}}{(p-\frac12)^2+\frac34}$$
$$X=2t +\frac13e^{-t}-\frac13e^{\frac{t}2}cos{\frac{\sqrt3}2t}-\frac{\sqrt3}3e^{\frac{t}2}sin{\frac{\sqrt3}2t}$$
| If $\dddot x+\dot x=2t, x(0)=\ddot x(0)=0, \dot x(0)=1$
$$p^3X-p+pX=\frac2{p^2}$$
$$X=\frac{p^3+2}{p^3(p^2+1)}=\frac{A}{p^3}+\frac{B}{p^2}+\frac{C}{p}+\frac{Dp+E}{p^2+1}$$
$$A=2, B=0, C=-2, D=2, E=1$$
$$X=2\frac1{p^2}-2\frac1{p}+\frac{2p+1}{p^2+1}=2\frac1{p^2}-2\frac1{p}+2\frac{p}{p^2+1}+\frac1{p^2+1}=2t^2-2+2cost+sint$$
which matches the answer I've been given. Thanks to @Lutz Lehmann for pointing it out
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.