Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$ I saw this twitter post that reads:
Find all the pairs of positive integers $(x,y)$ satisfying $$ x^2 + y^2 - 5xy + 5 = 0 . $$
I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's int... | It's $(5y-2x)^2-21y^2=-20$, which is a Pell type equation.
I got that by completing the square:
$x^2-5xy+y^2=-5\implies \left(x-\frac52y\right)^2-\frac{21}4y^2=-5\implies (5y-2x)^2-21y^2=-20$.
So it's $X^2-21Y^2=-20$, with $X=5y-2x$ and $Y=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
weird finding function from the graph
The graph is on the interval [-3,3], defined on R
This function maintain the pattern on all of R
How to find the $f(x)$?
See how the graph goes parallel at certain intervals. I am thinking adding one more variables to indicate one of the direction. However, I cannot merge the two ... | Using the absolute value function together with the floor function, you can write
$$
f(x)=1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
$$
Verification:
With $f$ as defined above, we get
\begin{align*}
f(x+2)
&=
1-\left|(x+2)-\left(2\left\lfloor\frac{(x+2)+1}{2}\right\rfloor\right)\right|
\\[4p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Strong inductive proof for recursive sequence given Explicit formula I am stuck on this
Show that the recursively defined sequence $a_n=2·a_{n-1}+3·a_{n-2}, a_1=-1, a_2=13$ has the explicit formula $〖a_n=3^n+4(-1)^n〗$
I can prove the base cases no problem but I keep getting stuck on the transformation
so I have $a_{k+1... | You want to show:
$$3^{n}+4(-1)^{n} = 2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}]$$
Simplifying from the left we have:
$$2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}] = 2[3^{n-1}] + 8[(-1)^{n-1}] + [3^{n-1}] + 12[(-1)^{n-2}] =$$
$$=3[3^{n-1}] - 8[(-1)^{n-2}] + 12[(-1)^{n-2}] = 3^{n} + 4(-1)^{n-2} = 3^{n} + 4(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3... | $$
y=\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
$$
=\frac{x (x^4 + x^3 + x^2 + 1) (x^6 - x^5 + 2 x^4 + x + 1)}{(x + 1) (x^2 + 1) (x^2 - x + 1) (x^4 + 1)}$$
$$
=\frac{8 + 5 2^{1/5} + 7 2^{2/5} + 3 2^{3/5} + 6 2^{4/5}}{3 + 2 2^{1/5} + 3 2^{2/5} + 2^{3/5} + 3 2^{4/5}}$$
$$x=\sqrt[\large 5]{2}\ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Moment of inertia of sphere of changing density I need to calculate moment of inertia of a sphere. The density of sphere changing $\rho=cz^2$ All the three components $I_x,~I_y,~I_z~.$
I did the assumption $dm=\rho dV$ then I found $c$ as $=\frac{15m}{4\pi r^5}~.$ Then called
$$I=\int\rho dv z^2=c\int r^4 \cos^4\theta ... | $I_x = \int dm \ (y^2+z^2), I_y = \int dm \ (x^2+z^2), I_z = \int dm \ (x^2+y^2)$
As $\rho = cz^2, dm = \rho dV = cz^2 dV$
Using cylindrical coordinates,
$x^2+y^2 = r^2, r^2 + z^2 = R^2 \ $ where $R$ is the radius of the sphere.
$dm = \rho dV = c z^2 \ r \ dr \ dz \ d\theta \ $
$\displaystyle I_z = c \int_0^{2\pi} \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\det( J_f^T J_f)$ $f:R^n \rightarrow R^{n+1}, f(x) =\frac{(x_1,\dots,x_n,1) }{(1+x_1^2+\dots+x_n^2)^{\frac{1}{2}}} $
$J_f$ is the Jacobi matrix of f
I want to calculate $\det( J_f^T J_f)$ but:
$J_f^t J_f = \left( \begin{array}{cc} |(x,1)|^2 - x_1^2 & -x_1 x_2 & \dots & -x_1 x_n \\ \vdots & & & \vdots \\- x_1 x^n & -... | Let me use $n = 3$ to make this easier to write and picture
The trick here is that the matrix
$$\begin{pmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} x & y & z \end{pmatrix}$$
is rank 1.
So we have
$$J_f^\top J_f = \frac{1}{c^2}\left( c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$ I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$
We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.
But I have a question: what should I do if the problem is for 4 variables, eg:
Let $a;b;... | We have
If $x,\,y,\,z$ are non-negative real numers satisfy $x+y+z=3,$ then
$$xy^2+yz^2+zx^2\leqslant 4.$$
Now, suppose $b = \max \{a,\,b,\,c,\,d\}.$ From this inequality we put $x=a,\,y=b,\,z=c+d,$ we get
$$a^2(c+d)+ab^2+b(c+d)^2 \leqslant 4.$$
But
$$a^2(c+d)+ab^2+b(c+d)^2$$
$$= ab^2+bc^2+cd^2+da^2 + c(a^2+2bd)+d^2(b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the roots of a cubic polynomial? All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what if you can't do that?
For example, how would you find the roots of the following equati... | While there is a rather nice test for whether a cubic polynomial has a triple zero by inspecting its coefficients (it will even tell us directly the value of that root), there is not anywhere near as simple a method for testing for a double zero by using the coefficients. (It can be managed, but the calculations do no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Smallest possible area for triangle I'm solving the following question:
The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?
This is my attempt at solving the problem:
Let $a$ be the le... | You have made a small error. You wrote that
$$m = \dfrac{b-0}{0-a}$$ but then said that this is equal to
$$ \dfrac{-a}{b}$$
This is clearly wrong. In fact,
$$m=-\frac{b}{a}$$
The Solution
We have $y=-\frac{b}{a}x+b=\frac{ab-bx}{a}$. Substituting this into $2x^2+y^2=1$ yields
$$\begin{align}
&2x^2+\left(\frac{ab-bx}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{... | Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have
$$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3} =1-\frac{3}{xy} \tag{1}$$
But we have also
$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$
From (1) and (2), we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unkno... | Your method is not in vain. We have:
$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$
$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$
hence $x + y = 1, xy = -6$. Now:
$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$
$$691\times 1 = ax^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2... | Let $r > 0$ and $0 \le \theta < 2\pi$, and $x = r \cos \theta$, $y = 2r \sin \theta$, so that $$4x^2 + y^2 = 4r^2,$$ hence $$\begin{align}f(x,y) &= \frac{8x-3y}{\sqrt{4x^2 + y^2}} \\ &= \frac{8r \cos \theta - 6r \sin \theta}{2r} \\ &= 4 \cos \theta - 3 \sin \theta \\
&= 5 \left(\frac{4}{5} \cos \theta - \frac{3}{5} \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$ If $r$ and $s$ are roots of $x^3-cx+d=0$, then
$$r^3-cr+d=s^3-cs+d , $$
for $r \neq s$. Solving the equation for $c$ gives me
$$ c=r^2+rs+s^2$$
Now to find an expression for $d$
$$ r^3-cr+d=0 $$
$$ d=r(c-r^2) ... | Let $t$ be the $3$rd root of the equation $x^3 - cx + d = 0$
So $$t^3 - ct + d = 0 \tag{1}$$
Also $$r + s + t = 0 \implies t = -(r+s) \tag{2}$$
Substitute the value of $t$ from $(2)$ into $(1)$ to get
$$-(r+s)^3 + c(r+s) + d = 0 \equiv (r+s)^3 - c(r+s) - d = 0$$
Hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Intersection of perpendicular chords: is this true for a sphere? It is true that in a circle with radius $R$, if the intersection of any two perpendicular chords divides one chord into lengths $a$ and $b$ and divides the other chord into lengths $c$ and $d$, then
$$a^2 + b^2 + c^2 + d^2 = 4R^2.$$
Does this equation hol... | For circles:
From the circle center draw perpendicular lines to the chords. Let the intersection of the lines with the chords be $X$ and $Y$, respectively. Let $|OX|=x$, $|OY|=y$.
We have (see figure):
$$\begin{align}
a&=\sqrt{R^2-x^2}+y,\\
b&=\sqrt{R^2-x^2}-y,\\
c&=\sqrt{R^2-y^2}-x,\\
d&=\sqrt{R^2-y^2}+x,
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
With or without squeeze theorem I am doing a proof involving finding the sum of a series $\sum_{x\geq1} \frac{x}{2^x} - \frac{x+1}{2^{x+1}}$ and I need to show that $\lim_{x\to\infty} \frac{x+1}{2^x}$ converges to $0$ without using L'Hopital's theorem.
My attempt:
$\lim_{x\to\infty} \frac{x+1}{2^x} = \lim_{x\to\infty} ... | Since you are in the context of a series, $x$ is natural.
So, you can apply the binomial formula to $2^x = (1+1)^x=\sum_{n=0}^x\binom xn$:
$$0\leq \frac{x}{2^x} = \frac{x}{(1+1)^x} \stackrel{x\geq 2}{\leq} \frac{x}{\binom x2} = \frac 2{x-1}\stackrel{x\to \infty}{\longrightarrow}0$$
Similarly for the case $\frac 1{2^x}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the general solution to the ODE: I was asked to find the general solution to these ODEs.
*
*$$\frac{dy}{dx}=\frac{(-8x+3y-31)}{(-3x+y-11)}$$
I tried by rearranging to the form $M(x,y)+N(x,y)\frac{dy}{dx}=0$ and let $y=vx$, but what I got was $$(v^2x-6vx-11v+8x+31)dx+x(-3x+vx-11)dv=0$$
and I could not separate $x... | We have
$$
\frac{dy}{dx} = \frac{-8x+3y-31}{-3x+y-11}.
$$
One way to solve it is to introduce time variable $t$, obtaining a system
$$
\left\{
\begin{align*}
&\dot{y} = -8x+3y-31 \\
&\dot{x} = -3x+y-11
\end{align*}
\right. ,
$$
which is linear. A matrix of this system
$$
A =
\begin{pmatrix}
-8 & 3\\
-3 & 1
\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$
Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$... | You don't need $\beta \rightarrow 0$, you just need that for sufficiently large $n$ $$\beta < 1$$ Then $$|b_{n+M}-1|<\beta^n|b_M-1| \rightarrow 0$$
You can check that
$$ b_{n+1} = \sqrt{\frac{b_nn^2 +2n+1}{(n+1)^2}}{} = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1}$$
that is
$$ b_{n+1}-1 = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this us... | $$x+2y=8 \implies x = 8 - 2y\tag{1}$$
Let $$f(x, y) = x+y+\frac{3}{x}+\frac{9}{2y} \tag{2}$$
Using $(1)$ in $(2)$,
$$f(x, y) = 8 - 2y +y+\frac{3}{8 - 2y}+\frac{9}{2y} = 8 - y +\frac{3}{8 - 2y}+\frac{9}{2y} \tag{3}$$
Differentiate $(3)$ wrt $y$, set it equal to $0$ and solve for $y$
$$\frac{-9}{2y^2} + \frac{3}{2(4-y)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Show that $\triangle ABC$ satisfying $B^2=AC$ and $2b=a+c$ is equilateral
A $\triangle ABC$ satisfies the conditions below:
$$B^2=AC \qquad 2b=a+c$$
Show that it's an equilateral one, where $a=|BC|$, $b=|AC|$, $c=|AB|$ and $B=\angle ABC$, $A=\angle BAC$, $C=\angle ACB$.
What I have done:
When trying to solve these qu... | I am assuming $A\ge C$ without loss of generality. Therefore $A=Br$ and $C=\frac Br$ for some $r\ge1$, by the first equation. From the second equation, note that
\begin{align}
&2b=a+c\\
\implies &\dfrac{2b}{2R} = \dfrac{a}{2R}+\dfrac{c}{2R}\\
\implies &2\sin B = \sin A+\sin C&&(\text{By sine rule})\\
\implies &4\cos\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Show that $G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$ The problem:
Using the inequalities
$\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{\left(\frac{1}{12n}-\frac{1}{360n^{3}}\right)} < n! < \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n}}$
show that, for a constant $... | I put $q=1-p$. First, I shall use the inequalities
$$
\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}}
$$
which hold for all positive integer $n$. We obtain
$$
\binom{n}{pn} > \frac{{\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n }}{{\sqrt {2\pi pn}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of $(x^3 - 3x^2 + 2x)$ as $x \to 1$ using epsilon delta definition I need to prove the following limit using epsilon delta method. I have come up this on my own just to practice skills.
$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$
So, I need to come up with some $\delta >0$ such that given $0 < \lvert x - ... | Other methode which easier
$f(x) =x^3 - 3x^2 +2x$
Suppose $|f(x)-l|<\epsilon $
Let's show $|x-1|<\alpha$
$|f(x)-l|=|f(x)-0|=|x^3 - 3x^2 +2x|<\epsilon $
$|x^3 - 3x^2 +2x|=|x^2-2x|×|x-1|<\epsilon $
Suppose $x\in [\frac{1}{2}, \frac{3} {2}]$
$\Rightarrow $$ \frac{1} {4}\leq x^2 \leq \frac{9}{4}$
And
$-3\leq-2x\leq - 1$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Entropy of extractions A box contains $3$ white and $6$ black balls. We draw $2$ balls consequentially without replacement. Find the entropy of first and second extractions and the entropy for both of them.
| For the first case, when $2$ balls are drawn sequentially, the corresponding pmf is
$$P(B_1B_2)=\frac{6}{9}.\frac{5}{8}=\frac{5}{12}$$
$$P(W_1B_2)=\frac{3}{9}.\frac{6}{8}=\frac{1}{4}$$
$$P(B_1W_2)=\frac{6}{9}.\frac{3}{8}=\frac{1}{4}$$
$$P(W_1W_2)=\frac{3}{9}.\frac{2}{8}=\frac{1}{12}$$
Hence, the entropy is $$H(P) = -\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality proof $\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$ Can someone give me an idea about how to prove the inequality?
Mention that $a_0, a_1, a_2, ..., a_{2n}, a_{2n+1}$ is an AP and $0<a_0<a_1<...<a_{2n}<a_{2n+1}$
$$\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1... | For simplicity let us assume that the common difference of the arithmetic progression is equal to one (the same calculation works in the general case with small modifications).
You correctly noticed that the middle expression can be written as
$$
\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+\cdots+\frac{1}{a_{2n-1}a_{2n}} \\
= \f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quaternion sandwich simplification gives me a term too many Context:
I'm reading Foundations of Game Engine Development by Eric Lengyel. On the topic of rotations with quaternions it is explained that a vector v can be rotated by being "sandwiched" by quaternions: v' = $qvq^{-1}$. When the quaternion is a unit-quaterni... | Your mistake comes at the first step. Remember quaternion multiplication is $(a+\mathbf{b})(c+\mathbf{d})=(ac-\mathbf{b}\cdot\mathbf{d})+(c\mathbf{b}+a\mathbf{d}+\mathbf{b}\times\mathbf{d})$ so you should get
\begin{align*}\require{cancel}
&(\mathbf{b} \times \mathbf{v} + c\mathbf{v} - \mathbf{b} \cdot \mathbf{v})(-\m... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 1,
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Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+... | You can show $5^n\ge 3^n+4^n$ for $n\ge2$ faster: $5^2\ge 3^2+4^2$ and in the induction step
$$ 5^{n+1}=5\cdot 5^n\ge 5\cdot3^n+5\cdot4^n>3^{n+1}+4^{n+1}.$$
An alternative idea for negative $x$ s also by size comparison with $2^{-n}$ being quite big
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Supremum of $x^2y^2(x^2+y^2)$ Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is
*
*$3n^2$
*$2n^6$
*$4n^6$
*$n^6$
My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express ... | By the AM-GM inequality, we have
$$xy \leqslant \left(\frac{x+y}{2}\right) ^2 =n^2,$$
and
$$xy(x^2+y^2) \leqslant \frac{1}{2} \cdot \left(\frac{2xy+(x^2+y^2)}{2}\right)^2=\frac{1}{2} \cdot \left(\frac{(x+y)^2}{2}\right)^2 = 2n^4.$$
Therefore
$$x^2y^2(x^2+y^2) \leqslant 2n^6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the general solution of a differential equation using two methods. Can somebody please help me in finding the general solution of $2x^3dy=y(y^2+3x^2)dx$ using another method?
My attempt for the first method:
$2x^3y^{-3}dy-3x^2y^{-2}dx=dx$
Let $v=y^{-2}$; $dv=-2y^{-3}dy$
The equation now becomes:
$-x^3dv-3x^2vdx... | \begin{gather*}
\frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\
\\
Let\ y=tx\\
\frac{dy}{dx} =t+x\frac{dt}{dx}\\
t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\
x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\
\\
Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solution of a trigonometric linear equation with the method of the added angle This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals.
The equation is:
$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$
Sur... | Putting together the two comments we have that the initial equation is equivalent to
$$\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$$
Leading to the solutions
$$ x = \arctan \frac12 - \arcsin \frac{\sqrt 5}5 + 2k \pi = 2k \pi$$
and
$$x = \pi + \arcsin \frac{\sqrt 5}5 + \arctan \frac12 + 2k \pi=\underbrace{\pi + 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4045774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Radius of Convergence of $(1-x)^{1/4}$ This is my first post so I am very new to MathJax formatting - i apologize in advance for the messy formatting
This equation becomes $$\sum_{n=2}^\infty \frac{(4n-5) x^n}{(4^n)n!}$$
The textbook says the ratio test would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(4n-5)(x^{n+... | The generalized binomial series is
$$ (1 \ + \ x)^p \ \ = \ \ \sum_{n=0}^\infty \ \left( \begin{array}{c} p \\ n \end{array} \right) \ x^n \ \ , $$
which becomes $$ (1 \ - \ x)^p \ \ = \ \ \sum_{n=0}^\infty \ (-1)^n \ \left( \begin{array}{c} p \\ n \end{array} \right) \ x^n \ \ $$
for the base $ \ (1 \ - \ x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$? $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$
$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$
$3x+5=x-1$
$x=-3$
But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$
So I tried to do other things... | The domain is $x\ge3$ due to the power functions involved, and it may easier to examine the equation with $x=t+3$, which leads to
$$(t^2+6t)^{t+2}[( t^2+6t)^{2(t+6)} -1]=0$$
The first factor produces the solution $t=0$, or $x=3$, and the second factor reduces to $t^2+6t=1$,
which yields $t=\sqrt{10}-3$, or $x=\sqrt{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Looking for general approach to find number of $(x,y,z)$ so that $x+y=z$ How many $(x,y,z)$ exist so that $x+y=z$ holds for $x<y<z\le10$ ($x,y,z\in \mathbb{Z}^+$ ) ?
$1)15\quad\quad\quad\quad\quad\quad2)20\quad\quad\quad\quad\quad\quad3)40\quad\quad\quad\quad\quad\quad4)45\quad\quad\quad\quad\quad\quad5)50\quad\quad\qu... | One of the ways would be to first look at the number of positive integer solutions of $x+y \leq 10$. Taking $x = a + 1, y = b +1$, it is equivalent to,
Number of non-negative solutions of $a + b + c = 8$ and that is $\displaystyle {10 \choose 2} = 45$. We discard solutions where $a = b = 0$ (as $x+y \gt 2)$ and we dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?
| Well $a^3 + b^3 = (a+b)(a^2 -ab + b^2)=1$
The only factors of $1$ are either $1\cdot 1$ or $-1 \cdot (-1)$.
So we must have $a+b = a^2 - ab + b^2 =\pm 1$.
And that just won't work unless $ab=0$.
It would follow that $b= \pm 1 -a$
And therefore $a^2 - a(\pm 1 - a) + (\pm 1 - a) = \pm 1$ so
$a^2+a^2 \mp a -a \pm 1 = \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Show that $a^2 - 2021b^2 = 13$ has no integer solutions
Show that $a^2 - 2021b^2 = 13$ has no integer solutions.
I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A s... | A slightly different alternative that does not require separate consideration of zero residues:
Because any odd number $n$ is a product of only odd primes, quadratic reciprocity implies $(p|n)=(n|p)$ for any $4m+1$ prime $p$. Thus $(13|2021)=(2021|13)=(6|13)=-1$, and then we can't have $a^2\equiv13\bmod 2021$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005) Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.
Source : Vietnamese TST 2005
I'm sure that the only solutions are the zero function, the identity, and the negative... | Using hgmath's comment, here is a way to write $n^3$ as a sum of five cubes of integers with absolute values $<|n|$ for $n=2k$ and $k\ge 8$. We write
$$(2k)^3=(2k-4)^3+(k+7)^3-(k-9)^3-10^3-2^3.$$
For this to work, we need to have $f(0)$, $f(1)$, $f(2)$, ..., $f(15)$. I already got $f(0)$ until $f(12)$.
For $f(13)$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the example give here we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\m... | The intuitive answer to why
$$\frac{x^3}{3!}+ \frac{x^4}{4!} + \cdots = \mathcal{O}(x^3)$$
as $x \to 0$ is that $x^3$ is the term that goes to $0$ at the lowest rate. All higher-order terms go to $0$ faster, so the expression will "at worst" approach $0$ like $x^3$. More precisely, the reason in that
$$|x^3| \geq C|x^4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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how to prove this trig inequality: $\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)$ The following inequality has come up in my work. I'm trying to prove that it is true:
$$
\begin{equation}
\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)
\end{equa... | Replacing $(x, z)$ by $(-x, -z)$ or $(y, z)$ by $(-y, -z)$ does not change the inequality, therefore we can assume that
$$
0 \le x, y \le \pi \, , \, -1 \le z \le 1 \, .
$$
Case 1: $x+y \le \pi$. The solution for this case is inspired by Proving that $\cosh^{-1} (\cosh(x)\cosh(y) ) \geq \sqrt{x^2 + y^2}$ and the CAT(... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | There is a much simpler way to solve this.
You know how to compute the geometric series
$$\frac{1}{3} + \frac{1}{3^2}+ \frac{1}{3^3}+ \frac{1}{3^4}+ \cdots = \frac{1}{2}$$
you also know how to compute the geometric series
$$\frac{1}{9} + \frac{1}{9^2}+ \frac{1}{9^3}+ \frac{1}{9^4}+ \cdots= \frac{1}{8}$$
But this can be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | Contour Integration Approach
Consider the keyhole contour, $\gamma$, that goes from $0$ to $\infty$ just above the positive real axis, circles the complex plane counterclockwise, then returns to $0$ just below the positive real axis.
Now compute the integral
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
Existence of a particular function Given positive integers $n,d,\ell\ge1$ such that $n\ge 1+(d-1)\ell$, I must find a a function
$$
a=a(n,d,\ell),
$$
such that
\begin{align*}
0< a(n,d,\ell)&<1,\\
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&\le a(n,d,\ell),
\end{align*}
for all $n,d,\ell\ge1,\ n\ge1+(d-1)\ell.$
I've been a... | Consider the function $a(n, d, l) = \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)}$ where $b(c, l) = \Gamma(3+ \frac {c \cdot \ell} {\ell-1})$. This function satisifies the constraint since:
\begin{align*}
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&=\frac{n}{n-(d-1)(\ell-1)} \cdot \frac {1} {b(n-1-(d-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | If
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n}$$
then
$$\log T = \mathop {\lim }\limits_{n \to \infty } n\log{\left( {1 + \frac{H_n}{{{n^2}}}} \right)}\le \mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}.$$
It can be easily ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$ Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,
$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$
A teacher grouped elements as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cd... | If I read well you are evaluating the sum
$$ \sum_{k=0}^{p-1}2^k + \sum_{k=0}^{p-1}2^k(2^p-1) = 2^p\sum_{k=0}^{p-1}2^k = 2^p\frac{1-2^p}{1-2} = 2^p(2^p-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$Q(x)=f(x^3)+xg(x^3)$ and $Q(x)$ is divisible by $x^2+x+1$ Let $Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$ and $Q(x)$ is divisible by $x^2+x+1$, then find the value of $Q(1)$
My approach is as follow
$Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$
$Q\left( x \right)... | Hint: if $x^2+x+1|Q(x)$, then
$$x^3-1\ \big| (x-1)Q(x)=(x^2-x)g(x^3)+(x-1)f(x^3).$$
See if you can use this, combined with the property that
$$f(x)-g(x)\ \big|\ h(f(x))-h(g(x))$$
for any polynomials $f,g,h$, to find $f(1)$ and $g(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the matrix of the linear transformation $T$ Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatri... | The last equation should be $$\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}2\\3\end{pmatrix}.$$
The right hand side of the equation is $D^*$, not $D$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$ Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$
Initially I tried working out how ... | I think the "too-complicated" solution you cite would be more understandable if it were phrased constructively, for example:
For a given integer $m$ with $3\nmid m$, there is a smallest $k>0$ for which $3^{k}\equiv 1\bmod{m}$ - the multiplicative order of $3 \bmod m$. Then for any $j>0$, $3^j\equiv 1 \bmod m$ $\iff$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to separate real and imaginary parts of $\arccos(x+iy)$? I know how to separate the real and imaginary parts of $\ln z$ and $\arctan z$:
$$\begin{align}\ln(x+iy)&=\frac{1}{2}\ln(x^2+y^2)+i\arctan\frac{y}{x}\\
\arctan(x+iy)&=\frac{1}{2}\arctan\frac{2x}{1-x^2-y^2}+\frac{i}{2}\operatorname{artanh}\frac{2y}{1+x^2+y^2}\... | Let
$$
\arccos(x+iy)=u+iv.\tag1
$$
Then
$$\begin{align}
x+iy&=\cos(u+iv)=\cos u\cosh v-i\sin u\sinh v\\
\\
&\implies
\begin{cases}
x=\hphantom-\! \cos u\,\!\cosh v\\
y=-\sin u\,\sinh v
\end{cases}.
\end{align}$$
Eliminating $u$ one obtains:
$$\begin{align}
&\left(\frac x{\cosh v}\right)^2+\left(\frac y{\sinh v}\right)^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right)... | $$y=\frac{x^{2x}}{(x+1)^{x+1}}\implies \log(y)=2x \log(x)-(x+1)\log(x+1)$$
$$\log(y)=2x \log(x)-(x+1)\left(\log(x)+\log \left(1+\frac{1}{x}\right)\right)$$
$$\log(y)=(x-1) \log (x)-(x+1)\log \left(1+\frac{1}{x}\right)$$
Now, by Taylor series
$$(x+1)\log \left(1+\frac{1}{x}\right)=(x+1)\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. than... | Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots
While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get... | {
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"answer_count": 4,
"answer_id": 3
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Show that $O_2(\mathbb{R})$ contains only rotational and reflective symmetries. Show that $O_2(\mathbb{R})$ contains only rotational and reflective matrices.
I know that rotational and reflective symmetries are part of $O_2(\mathbb{R})$. I want to show that there is no other matrix that satisfies $O_2(\mathbb{R})$. Thi... | By definition $A \in O_2(\mathbb{R})$ iff $AA^T = A^TA = I$.
Let
\begin{align}
A =
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\end{align}
then we have that
\begin{align}
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
a & c\\
b & d
\end{pmatrix}
=
\begin{pmatrix}
a^2+b^2 & ac+bd\\
ac+bd & c^2+d^2
\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Something maybe related to stochastic process We have $5$ floors and we are on the ground(treat the ground as floor$0$).
The probability of jumping from the ground to floor$1$ is $1$.
For $i\in\{1,2,3,4\}$,
the probability of jumping from the $i$-th floor to the $i+1$-th floor is $\frac{1}{3^i}$,
the probability of sta... | Labelling floors $0 - 5$ as $a - f$, and using first-step-analysis, we have
$\displaylines{a = 1+b\\
b= 1+c/3 + (1/3)(b+a)\\
c= 1+d/9 + (4/9)(c+b)\\
d=1+e/27 +(13/27)(d+c)\\
e = 1 +(40/81)(e+d)}$
and wolfram confirms your answer of $\;\fbox{14543}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos4\theta=\cos3\theta$ to find the roots of $8x^3+4x^2-4x-1=0$. Find all the solution to the equation
$\cos 4\theta=\cos 3\theta$. Hence or otherwise, show that the roots of the equation
$$8x^3+4x^2-4x-1=0$$ are $$\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$$
Therefore prove that
$$\sec \frac... | $\cos(x) = \cos(y) \Leftrightarrow x = 2n\pi \pm y, n\in \Bbb Z$
So,
$4\theta = 2n\pi \pm3\theta \Rightarrow \begin{cases}\theta = 2n\pi\\\text{ or} \\\theta = \dfrac{2n\pi}{7}\end{cases}$
Now using compound angle formula,
$\begin{align}&\cos4\theta = \cos3\theta \\\Rightarrow &2\cos^2(2\theta )-1 = 4\cos^3\theta-3\co... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of a symmetric function on a square In a paper of Frits Beukers "A note on the irrationality of $\zeta(2)$ and $\zeta(3)$", he says
It is a matter of straightforward computation to show that
$$\frac{y(1-y)x(1-x)}{1-xy} \le \left\{\frac{\sqrt{5}-1}{2}\right\}^5$$ for all $0\le x\le1, 0\le y \le 1$.
I not... | $$\frac{y(1-y)x(1-x)}{1-xy} = \frac{xy(1-(x+y)+xy)}{1-xy} \le \frac{xy(1-2\sqrt{xy}+xy)}{1-xy}= \frac{xy(1-\sqrt{xy})}{1+\sqrt{xy}}$$
Denote $t = \sqrt{xy}$ then $0 \le t\le 1$.
The function $f(t) = \frac{t^2(1-t)}{1+t}$ reaches its maximum in $t \in [0,1]$ when $t = \frac{\sqrt{5}-1}{2}$ (it suffices to calculate $f'(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that odd partitions and distinct partitions are equal I am working through The Theory of Partitions by George Andrews (I have the first paperback edition, published in 1998).
Corollary 1.2 is a standard result that shows that the number of partitions of $n$ into odd parts is the same as the number of partitions... | See if this helps?
\begin{align*}
\prod_{n = 1}^\infty \frac{1 - q^{2 n }}{1 - q^n} &= \frac{1 - q^{2}}{1 - q}\cdot\frac{1 - q^{4 }}{1 - q^2}\cdot\frac{1 - q^{6 }}{1 - q^3}\cdot\dots\\
&= \left[\frac{1}{1-q}\cdot\frac{1}{1-q^3}\cdot\frac{1}{1-q^5}\dots\right]\left[\frac{1 - q^{2}}{1 - q^2}\cdot\frac{1 - q^{4 }}{1 - q^4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $\langle 2,1+\sqrt{-5}\rangle \langle3,1+\sqrt{-5} \rangle = \langle1-\sqrt{-5} \rangle$ In $\mathbb Z[\sqrt{-5}]$ I compute
$\langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle = \langle 6,2+2\sqrt{-5},3+3\sqrt{-5},-4+2\sqrt{-5} \rangle $
So there must be $a+b\sqrt{-5}\in\mathbb Z[\sqrt{-5}]$ that $... | As you've showed, \begin{align} \langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle &= \langle 6,\color{blue}{2+2\sqrt{-5}},\color{blue}{3+3\sqrt{-5}},-4+2\sqrt{-5} \rangle \\ \tag{1}&= \langle \color{blue}{6},1+\sqrt{-5}, \color{blue}{-4+2\sqrt{-5}} \rangle \\ \tag{2}&=\langle 6, \color{blue}{1+\sqrt{-5}},\co... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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On bounds for the quantity $n^2 / D(n^2)$ when $q^k n^2$ is an odd perfect number with special prime $q$ and $k > 1$ Let $N = q^k n^2$ be an odd perfect number given in Eulerian form (i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$).
In this post, I would like to compute t... | Yes.
We have
$$\frac{n^2}{D(n^2)}=\frac{\sigma(q^k)}{2s(q^k)}=\frac{q^{k+1}-1}{2(q-1)\bigg(\dfrac{q^{k+1}-1}{q-1}-q^k\bigg)}=\frac{q^{k+1}-1}{2q^k-2}$$
Let $f(k):=\dfrac{q^{k+1}-1}{2q^k-2}$. Then, we get $f'(k)=-\dfrac{(q-1)q^k\log q}{2(q^k-1)^2}$ which is negative.
Since $f(k)$ is decreasing, we obtain
$$\lim_{k\to\in... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1) \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ t... | A slightly worse estimate may be obtained as follows. If $|\arg z|<\pi$ then
$$
\log (1 + z) = \sum\limits_{k = 1}^N {( - 1)^{k + 1} \frac{{z^k }}{k}} + ( - 1)^N z^{N + 1} \int_0^1 {\frac{{t^N }}{{1 + zt}}dt}
$$
for any $N\geq 0$. Substituting $z = - e^i$, taking the imaginary part of each side and rearranging, we f... | {
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"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$ \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\f... | $$x=\frac{t^2}{1-2t}$$
$$dx=\frac{2t(1-2t)+2t^2}{(1-2t)^2}dt$$
$$=\frac{2(t-t^2)}{(1-2t)^2}dt$$
$$\sqrt{x^2+x}=x+t=\frac{t-t^2}{1-2t}$$
$$\frac{x^3dx}{\sqrt{x^2+x}}=$$
$$\frac{t^6}{(1-2t)^3}\frac{1-2t}{t-t^2}\frac{2(t-t^2)dt}{(1-2t)^2}$$
$$=\frac{2t^6dt}{(1-2t)^4}$$
To make the integration easier, put $$u=2t$$
to get
$... | {
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"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Repeated eigenvalue = 0 I have this quite straightforward system:
$$ \displaystyle
\begin{array}{l}
\frac{d x}{d t}=\left[\begin{array}{cc}
3 & 9 \\
-1 & -3
\end{array}\right] x \\
\text { with } x(0)=\left[\begin{array}{c}
2 \\
4
\end{array}\right] .
\end{array}
$$
My reasoning:
Using the traditional eyeball-method, o... | Your solution is correct up until you plug the constants into your general solution:
$$-\frac53\begin{bmatrix}3 \\ -1\end{bmatrix} + 14\begin{bmatrix}\frac12 + 3t \\ \frac16 - t\end{bmatrix} = \begin{bmatrix}-5 \\ \frac53\end{bmatrix} + \begin{bmatrix}7+42t \\ \frac73 - 14t\end{bmatrix} = \begin{bmatrix}2 + 42t \\ 4 - ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing an equality of complex numbers Let $e_k \in \mathbb{C}$ such that
$$e_1+e_2+e_3 =0$$ and
$$g =4(e_1 e_2 + e_2 e_3 + e_1 e_3)$$
Prove that $$
g^2=16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2)
$$
What I got :
$$
g^2 =16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2+2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2)
$$
However, I fail... | You can factor out $2e_{1}e_{2}e_{3}$ from $$2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2},$$ which gives you $$2e_{1}e_{2}e_{3}(e_{1} + e_{2} + e_{3}).$$ Because $e_{1} + e_{2} + e_{3} = 0$, then $2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Calculate the value of the expression for polynomial roots The roots of the polynomial equation: $ x ^ 3 - 3x -1 = 0 $ are: $ a, b $ and $ c $. Based on this, calculate the value of:$$ L = \frac {1} {a ^ 2-2a + 2} + \frac {1} {b ^ 2-2b + 2} + \frac {1} {c ^ 2-2c + 2} $$
Attemp:
$L=\frac{1}{(a-1)^2+1}+\frac{1}{(b-1)^2+1... | This page gives a nice summary of common polynomial root transforms, which we can employ here.
First, let's find a polynomial that has roots that are $a-1, b-1, c-1$. This is just $$Q(x) = P(x+1) = (x+1)^3 -3(x+1) -1 = x^3 +3x^2 -3.$$
Next, we find the polynomial that has roots $(a-1)^2, (b-1)^2, (c-1)^2.$ This is
$$R(... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solve the inequality $\log_3(2^{x+1}-8)\le x$ Find x
$$\log_3(2^{x+1}-8)≤x$$
I found that $x>2$
$$\log_3(2^{x+1}-8)\le x\log_33$$
$$(2^{x+1}-8)\le 3^x$$
$$(2^{x+1}-8)> 0$$
$$(2^{x+1})>8$$
$$2^{x+1}>2^3$$
$${x+1}>3$$
$${x}>2$$
| We have
$$\log_3(2^{x+1}-8)≤x$$
$$\log_3(2^{x+1}-8)≤\log_33^x$$
$$\begin{align}&\begin{cases} 2^{x+1}-3^x-8≤0 \\x+1>3 \end{cases} \\\\ \iff &\begin{cases}3^x-2^{x+1}≥-8 \\ x>2\end{cases}\end{align}$$
Then, note that $f(x)$ is a strictly increasing function, $\forall x>2.$
This means,
$$f(x)=3^x-2^{x+1}>1≥-8, \forall ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n... | I have no idea what you are doing, but the generating function of $(k+1)(n-k+1)$ is the derivative of the generating function of $(n-k+1).$ In other words,
if $$A_n(x) = \sum_{i=1}^{n+1} (n-k+1) x^k,$$ then
$$A^\prime_n = \sum_{i=0}^n (k+1)(n-k+1),$$ and your sum is just equal to $A_n^\prime(1).$
Now, $$A_n(x) = n \sum... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Use laws of logarithms to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ I am to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ using the laws of logarithms. The solution is provided as $\ln(2)+\ln(b)+\frac{\ln(b+1)-\ln(b-1)}{2}$ whereas I got $\ln(2)+\ln(b)+\frac{1}{2}\ln(b+1)-\frac{1}{2}\ln(b-1)$
My working:
$$... | $$\frac{1}{2}A-\frac{1}{2}B=\frac{A-B}{2}$$
where $A=\ln(b+1)$, $B=\ln(b-1)$.
In short, you arrived at the given answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
wit... | $$\cos \beta = \frac{|AB|}{|BC|} = \frac{\sqrt 5}{5},\quad \sin \beta = \frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}$$
Given $|AB|: |BC| = \sqrt5:5$, we assume $$|AB| = \sqrt5k, |BC| = 5k $$ where $k$ is a constant. Now using $$\frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}, \quad |AH| = 2k$$
$$|AH| + |BH| + |HC| = |AH| + |BC| = 7k... | {
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"url": "https://math.stackexchange.com/questions/4127013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $... | There is a slightly more general result:
Claim 1: Let $a$ and $b$ be 2 positive nonintegral reals satisfying $a+b=n$ for some natural number $n$. Then $$\lfloor a \rfloor + \lfloor b\rfloor =n-1.$$
To see this, note that as $a+b$ is integral, it follows that $(a-\lfloor a \rfloor) + (b-\lfloor b \rfloor)$ must also b... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding all tangent lines to circle $x^2+y^2=9$ having exactly one common point with parabola $y=x^2+3$
I have to find all tangent lines to the circle $x^2+y^2=9$, which have exactly one common point with the parabola $y=x^2+3$.
I drew it and I saw that $(0,3)$ is a common point circle and parabola. I found that thre... | At any given point on the circle, there is a unique tangent line. At the point $(0,3)$ you have identified 3. Only one is tangent.
The other lines:
Since this is tagged pre-calculus, I will not use calculus in this answer.
Suppose $x_1,y_1$ lies on the circle.
Slope of the tangent line though that point is negative ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\... | The first rule is that, when you use Taylor's expansion, you use it at the same order for all terms. The second rule is that, when you have a product of expansions, or a composition, you expand it with the usual rules for polynomials, and truncate the result at the chosen order.
I'll start it for the first example, for... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \t... | You can do the following: $\tan (50) = \tan (70 - 20)= \frac{\tan (70) - \tan (20)} {1+ \tan( 70). \tan (20)}.$ The bottom is $2,$ so you are done.
In general, if $ A +B = \pi/2,$ then $\tan A = \tan B + 2\tan(A-B)$, where $A$ is the bigger one.
Edit: If you can also start from the right/left side. Essentially do the s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Proof of Least upper bound of $\frac{x}{x+1}$ The questions states:
Prove that there is a unique positive real number $m$ that has the following two properties.
*
*For every positive real number $x$, $\frac{x}{x+1}<m$.
*If $y$ is any positive real number with the property that for every positive real number $x$ , $\... | $$ \dfrac{x}{x+1} < \dfrac{x+1}{x+1} \implies \dfrac{x}{x+1} < 1 $$
So 1 is an upper bound. Let $\epsilon < 1$ We will show that there is an $x$ for which
$\dfrac{x}{x+1} = 1-\epsilon$
\begin{align}
\dfrac{x}{x+1} &= 1-\epsilon \\
x(1-\epsilon) + 1(1-\epsilon) &= x \\
-\epsilon x &= -(1-\epsilon) \\
x &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Denesting Cardano's Formula For a depressed cubic equation $x^3 + px + q =0$ having exactly one real root, Cardano's formula gives the real root as $$\sqrt[3]{-\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$ Suppose we know that the cubic equation ... | If $n$ is your one real integer solution, then $$x^3+px+q=(x-n)(x^2+nx+b)$$ where $-nb=q, b-n^2=p$ and $n^2<4b,$ that last because there are real roots to $x^2+nx+b$ otherwise.
Substitute $p=b-n^2, q=nb$ into your formula, then see if you can eliminate the cube roots in the general case. You should be able to.
In parti... | {
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"source": "stackexchange",
"question_score": "2",
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Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \i... | Note $\cos(x-n) = \cos x\cos n + \sin x \sin n$ and rewrite the integrand as
\begin{align}
\frac{1}{{\cos(x-1)\cos(x-2)\cos(x-3)}}
={}\frac{\csc1\csc2\csc3\sec^3 x}{(\tan x+\cot 1) (\tan x+\cot 2) (\tan x+\cot 3)}
\end{align}
Substitute $t=\tan x $ and perform the partial fractionization
\begin{align}
&\int \frac{1}{\c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$ Given a,b,c are positive real numbers and n is positive integer number.
Prove that
$$\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$$
I tried expanding it, using equivalence transformation, using inequalities AM-GM, Holder.., but no... | Your conjecture does not hold for $n > 1$. Let $a = b > 0$ and $c \to 0$ then, for $n > 1$,
$$
\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \to 2 \le \frac{3}{\sqrt[n]{2}}
$$
Indeed, $2$ is a tight lower limit for $n > 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\sum_{0 \le r We know that $$\sum_{0 \le r <s \le n} A_r A_s=\frac{1}{2}\left [(\sum_{t=0}^{n} A_t)^2-\sum_{t=0}^{n} A^2_t\right]$$
Nest using $A_r={n \choose r}x^r$, we can get
$$S(x)=\sum_{0 \le r <s \le n} {n \choose r}{n \choose s}x^{r+s}=\frac{1}{2}\left[(\sum_{t=0}^{n} {n \choose r} x^r)^2=\sum_{t=0}^{n}... | $$ \begin{align} \tag{1} S &\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose r}{n \choose s} \\
&\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose n-r}{n \choose n-s} \\
&\ \tag{2} = \sum_{0 \le r <s \le n}~ (n-r+n-s){n \choose n-r}{n \choose n-s} \\
\end{align}$$
Add $(1)$ and $(2)$
$$ 2S= \sum_{0 \le r <s \le n}~ 2n{n \choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.... | If the right hand side is negative then the inequality trivially holds. Otherwise $a,b,c$ are the sides of a triangle, and by Heron's formula for the area $S$ of a triangle the inequality reduces to:
$$
a^2+b^2 \ge 4 S
$$
This follows from $S = \frac{1}{2} ab \sin C \le \frac{1}{2} ab$ and the means inequalities $ab \l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sequence $ a_0=1,a_1=3$ and $a_{n+2}=4a_{n+1}-a_n $. Prove that $2017|a^2_{2017}-4a_{2017}+3$ I need a solution verification for next problem.
Let sequence $a_0,a_1,...$ be such that:
$ a_0=1,a_1=3$ and $a_{n+2}=4a_{n+1}-a_n $. Prove that $$2017\mid a^2_{2017}-4a_{2017}+3$$
By solving this recurrence we get $$a_n = ... | Your proof works.
I wondered why they chose $$f(m)=m^2-4m+3=(m-1)(m-3),$$ I see when $p=5,7$ that $a_p\equiv 1\pmod p.$
I wonder if this is true for all primes. To prove that, you’d want to to show that if $\left(\frac 3p\right)=-1$ then $a_p\equiv 1\pmod p.$
Proof as outlined by commenter Derive Foiler.
If $\left(\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In the following question find the ratio of $a:b:c$
If the equations $x^2+2x+3=0$ and $ax^2+bx+c=0$; $\enspace$ $a,b,c \in \mathbb{R}$, have a common root, then $a:b:c$ is ?
This is my approach:
let the common root be $\alpha$. Then
$$\alpha^2+2\alpha+3=0------(1)$$
$$a\alpha^2+b\alpha+c=0------(2)$$.
Now the above ... | Here is a direct answer to your question. You should have continued with a Gauss decomposition of the quadratic expression:
$$9a^2+3b^2+c^2-6ab-2ac-2bc=0\tag{1}$$
into:
$$(3a-b-c/3)^2+2(b-2c/3)^2=0 \ \ \iff \ \ \begin{cases}3a-b-c/3=0 & Eq. (2)\\b-2c/3=0& Eq. (3)\end{cases}$$
From (3), one deduces that $b=2c/3$ ; plugg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why the PIs using different methods are different? What am I doing wrong? I have a PDE
$$
\text { Solve }\left(D^{2}+3 D D^{\prime}+2 D^{\prime 2}\right) z=x+y .
$$
where $D = \frac{\partial }{\partial x}$, $D^{\prime} = \frac{\partial }{ \partial y}, D^2 = \frac{\partial^2}{\partial x^2}, DD^{\prime}= \frac{\partial^2... | $$
\frac{(x+y)^3}{36} = \frac{(y-x)^3}{12} - \frac{(y-2x)^3}{18} + \frac{x^2 y}{2} -\frac{x^3}{3}
$$
So the two answers are equivalent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Harmonic-analysis Fourier series proof - general fourier series. Let $f(x) \in L^2[-\pi,\pi]$ such that $f(x)$ suffice: $f(x) \sim \frac{a_0}{\sqrt{2}} + \sum_{n=1}^{\infty}a_n\cos(nx)+b_n\cos(nx) $
$f(x)$ is also even function. Prove:
$
\int_{-\pi}^{\pi} f^2(x)\cos^2(x)\, dx = \frac{1}{2}
\int_{-\pi}^{\pi} f^2(x)\ \if... | Since $f$ is even, we have
$$f(x) = \frac{a_0}{\sqrt2}+\sum_{n=1}^\infty a_n \cos (nx)$$
\begin{align}
f(x)\cos x &= \frac{a_0}{\sqrt2}\cos x + \sum_{n=1}^\infty a_n \cos (nx) \cos (x)\\
&=\frac{a_0}{\sqrt2}\cos x + \sum_{n=1}^\infty \frac{a_n}{2}\left[ \cos(n+1)x + \cos (n-1) x \right] \\
&= \frac{a_0}{\sqrt2}\cos x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\iiint x+y+z$ over the region inside $x^2+y^2+z^2 \le 1$ in the fist octant I feel this should be an easy question, but I seem to be struggling with it. So, I started by finding my bounds of integration. In this case, I get $0 \le x \le 1$, $0 \le y \le \sqrt{1-x^2}$ and $0 \le z \le \sqrt{1-x^2-y^2}$. Then, u... | If $E$ is the region $x^2 + y^2 + z^2 \leq 1$ in first octant. Please note that due to symmetry,
$$I = \iiint_E (x+y+z) \ dV = 3 \iiint_E z \ dV$$
In spherical coordinates,
$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
So $x^2 + y^2 + z^2 \leq 1 \implies \rho \leq 1$
As we are in first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Reversing the divisibility of 2 out of 3 formula I was looking into a couple of exercises:
First exercise:
Prove that if $x \mid (x^2 + 1)$ the $x = 1$ or $x = -1$
My solution was that if $x \mid (x^2 + 1)$ this means that $x | x^2$ and $x | 1$. The only number that is a divisor of $1$ is 1/-1. Hence $x = 1$ or $x = ... | As mentioned by shoteyes in the comments, if $x \mid (a + b)$ and either $x \mid a$ or $x \mid b$, then $x$ divides both $a$ and $b$.
In the first exercise, since it is known that $x \mid x^2$ and it is given that $x \mid (x^2 + 1)$, $x \mid 1$. Therefore, the only values of $x$ for $x \mid 1$ to be true are $-1$ and $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit ... | The number has to be divisible by $3$, so the sum of the digits must also have that property. Since the sum of the six digits is $15$, the missing digit has to be either $0$ or $3$. (This was your mistake -- you thought that either 1 or 4 needed to be unused).
*
*If the missing digit is $0$, then the units digit ca... | {
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Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} ... | The first thing I would try is $v = y - 1$
$$
v' = \left(1+\frac{v}{2x}\right)^2
$$
Let $v = wx$
$$
w + xw' = \left(1+\frac{w}{2}\right)^2
$$
which is basically
$$
xw' = g(w)
$$
now see if you can solve this ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Dual Basis
In $\mathbb{R^{3}}$, let be $v_{1}=\begin{bmatrix} 1 \\2 \\0 \end{bmatrix}$i $v_{2}=\begin{bmatrix} 0 \\3 \\1 \end{bmatrix}$ and $v_{3}=\begin{bmatrix} 0 \\0 \\1 \end{bmatrix}$. If f is a linear functional on $\mathbb{R}^{3}$ such that $f(v_1)=-1$,$f(v_2)=2$ and $f(v_3)=4$ and if $v=\begin{bmatrix} ... | Let $f=(p,q,r)_{1\times 3}$, then $f(v_1)=-1 \implies p+2q=-1, f(v_2)=3q+r=2, f(v_3)=4 \implies r=4 \implies p=-2/3, q=1/3,r=4$. Finally
$$f(v)=(-2/3,1/3,4)\begin{pmatrix} x\\ y\\ z \end{pmatrix}=-2x/3+y/3+4z.$$
We can write $v_1=(1,2,0), v_2=(0,3,1), v_3=(0,0,3)$, then the dual basis is given by cross products:
$$u_1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$
let $\overrightarrow{a} = 2 \hat {i}+ \hat {j}-2 \hat {k}$ and $\overrightarrow{b}= \hat {i} + \hat{j}$. $\:$ if $\overrightarrow{c}$ is a vector such that $\overrightarrow{a} \cdot \overrightarrow{c}+2|\overrightarrow{c}|=0$ an... | The first approach gives you a result for $(\vec a \times \vec b) \times \vec c \ $ as it does not test the compatibility of the given data.
Just simply applying,
$|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|=|(\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of ... | Hint
For $$\int_0^\infty\dfrac{\ln x}{x^2+bx+b^2}dx$$
set $x=by$
then we can easily solve $$\int_0^\infty\dfrac{dy}{y^2+y+1}$$ using trigonometric substitution
Set $\dfrac1y=u$ in $$J=\int_0^\infty\dfrac{\ln y}{y^2+y+1}dy$$ to find $$J=-J$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I evaluate $\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$ I've been trying to find and prove that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\pi \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\frac{\pi }{2}\left(\ln \left(\frac{1+\... | $$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\int _0^1\frac{1}{\sqrt{1-x^2}}\left(\int_0^1\frac{x^2\ln(y)}{1+x^2 y}\mathrm{d}y\right)\mathrm{d}x$$
$$=\int _0^1\ln(y)\left(\int_0^1\frac{x^2}{(1+x^2y)\sqrt{1-x^2} }\mathrm{d}x\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}... | {
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"timestamp": "2023-03-29T00:00:00",
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Showing that that x^2 ≡ (−1)^k (mod p). I'm stumped on using my observations to properly solve this problem. The problem is as follows:
Suppose p is an odd prime expressible as $p = 2k − 1$ and let $x$ be the product of all positive odd integers less than p. Show that $x^2 ≡ (−1)^k $(mod p).
Here are my thoughts on the... | Recall that $(p-1)! \equiv -1 \pmod{p}.$ Note that
$$(p-1)! = x \cdot (2 \cdot 4 \cdots (p-1)) = x \cdot 2^{\frac{p-1}{2}} \cdot \left(1 \cdot 2 \cdots \frac{p-1}{2}\right).$$
So,
$$-1 \equiv x \cdot 2^{\frac{p-1}{2}} \cdot \left(\frac{p-1}{2}\right)! \pmod{p}.$$
Squaring this, we get that
$$1 \equiv x^2 \cdot 2^{p-1} ... | {
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"timestamp": "2023-03-29T00:00:00",
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How can one evaluate $\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$ using only elementary means? How can one evaluate $$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$$ using only elementary means?
By using only elementary means I mean methods not exceeding the basic methods taught in a first cour... | One way:
Let $3x^2-x^2+x-4=A(x^2-3x-2)(2x-3)+B(x^2-3x+4)+C(2x-3)+D.$
Find $A,B,C,D$ by comparing the coefficients of various powers of $x$. then
$$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=A\int_{0}^{1}\sqrt{x^2-3x+2} ~(2x-3)~dx+B\int_{0}^{1} \sqrt{x^2-3x+2} ~dx+ C\int_{0}^{1}\frac{(2x-3)dx}{\sqrt{x^2-3x+2}}+D\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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mathematical induction proof aren't solving i am trying to solve for a mathematical induction for a while now but i seem to be getting the incorrect answer
$$
\frac{1}{10}+\frac{1}{40}\:+\frac{1}{88}+_...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{6n+4}
$$
Equation
i am unable to solve the left hand side... | HINT: Write the $n^{\text{th}}$ term.
$$T_n = \frac{1}{(3n-1)(3n+2)} = \frac{1}{3}\frac{(3n+2) - (3n-1)}{(3n-1)(3n+2)} = \frac{1}{3}\frac{1}{3n-1} - \frac{1}{3}\frac{1}{3n+2}$$
Can you identify what happens on adding successive terms?
Using Mathematical Induction, first of all, verify the base case of $n=1$ which you ... | {
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Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1... | I found another way to solve it, by Taylor series for $a^x$, we have:
\begin{align*}
a^x\geq 1+x.
\end{align*}
As long as $a>1$. Now if $x = y-1$, we have:
\begin{align*}
a^{y-1}&\geq 1+(y-1)=y\\
a^{y-1}&\geq y\\
a^{y}&\geq ya
\end{align*}
Thus, if $y = n+1$,
\begin{align*}
n+a^{n+1}\geq a^{n+1}\geq (n+1)a.
\end{align*... | {
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Finding the maximum and minimum values of $f(x)=|x-1|+|x^2-2x|$ for $x \in [0,2]$ Can someone help me solve this problem?
If a function $f(x)$ in the domain $x \in [0,2]$ is
$$f(x)=|x-1|+|x^2-2x|$$
then the minimum value is [A] and the maximum one is [B].
I've tried a lot of things:
As I said earlier, I tried to set ... | I think you are looking for a hint rather than a full solution.So this is how I would go about it.
We start by getting rid of modulus symbol
Case 1: $0\leq x<1$
$f(x)=-(x-1)+x(-(x-2))=-x^2+x+1=\frac{5}{4}-(x-\frac{1}{2})^2$
Case 2: $1\leq x\leq 2$
$f(x)=(x-1)+x(-(x-2))=-x^2+3x-1=\frac{5}{4}-(x-\frac{3}{2})^2$
| {
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"url": "https://math.stackexchange.com/questions/4187387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up. Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up.
The only way I can think of is: ... | This question is very good place to use exponential generating functions .
It is said that all letter will be used ,so
the generating function for $A$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $B$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating functio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$ Let $a,b,c>0$ satisfy $abc=1 $
Prove that: $$(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$$
I have tried:
Let $p=a+b+c;q=ab+bc+ca;r=abc$, the problem is: $$pq-r\ge r+p+q+1 \;\;\text{or}\;\; pq-p-q\ge3 $$
I have $pq\ge 9r$, so we need to prove $$9-p-q\ge3$$ or $$p+q\le 6$$ I have $p\... | Alternative solution:
Since we are given that $abc=1$, we can assume that $a=\frac x y, b=\frac y z, c=\frac z x$.
We have to prove that $$({x^2} + yz)({y^2} + zx)({z^2} + xy) \ge xyz(x+ y)(y + z)(z+ x)$$ Notice that we have
$ ({x^2} + yz)({y^2} + zx) - xy(z + x)(z + y) = z(x + y){(x - y)^2} \ge 0 $
$ \Rightarrow ({x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I evaluate the following limit $\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$? Evaluate this limit :
$$\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$$
I tried to simplify the function :
$$x^2e^{\frac{1}{x}}-x^2e^{\frac{1}{x+1}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x... | I think you're already on the right path, try to use the classical equivalents :
$$e^x\overset{x\to 0}\sim x+1$$
Using your results :
When $x$ tends to $0$ you'll have :
\begin{align}
\lim_{x\to 0} \frac{e^x}{x^2} -\frac{e^{\frac{x}{x+1}}}{x^2}&= \lim_{x\to 0} \frac{x+1}{x^2}-\frac{\frac{x}{x+1}+1}{x^2}\\
&=\lim_{x\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$ are roots of the equation $x^2 - bx + 3 = 0$. What is the value of $b$?
If roots of the equation $x^2 - bx + 3 = 0$ are $\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$, then what is the value of $b$ ?
$1)\pm2\sqrt6\qquad\qquad2)\pm2\sqrt3... | Let $\alpha$ and $\beta $ be the roots of the equation with $\beta>\alpha$.
$${1\over\alpha^2}=1-\frac1{r^2}$$
$${1\over\beta^2}=1+\frac1{r^2}$$
$$\implies \frac{1}{\alpha^2}+\frac{1}{\beta^2}=2$$
$$\implies \alpha^2+\beta^2=18$$
$$\implies \alpha^2+\beta^2+2\alpha\beta=24$$
$$\implies b^2=24$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:
Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.
given in book is as follows:
$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}... | First, the quadratic formula deals with polynomials. If you introduce some unspecified function $f$, you muddy the waters, since $f$ needn't be a polynomial. For example, if $f(x)=\sin x$, then $x^2+f(x)$ is not quadratic in $x$.
I understand that you saw the quadratic formula derived, it was assumed that $a,b,c$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality... | Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}\frac{1}{a^2+b^2+ab}-\frac{9}{(a+b+c)^2}=\sum_{cyc}\left(\frac{1}{a^2+b^2+ab}-\frac{3}{(a+b+c)^2}\right)=$$
$$=\frac{1}{(a+b+c)^2}\sum_{cyc}\frac{c^2-2a^2-2b^2-ab+2ac+2bc}{a^2+b^2+ab}=$$
$$=\frac{1}{2(a+b+c)^2}\sum_{cyc}\frac{(c-a)(c+b+4a)-(b-c)(a+c+4b)}{a^2+b^2+ab}=$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 4
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Prove that $ \det(A^4 + A^2 B^2 + 2A^2 + I) \geq 0 $ Problem:
Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$
My Approach:
If $ A $ invertible, then
$$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \impl... | This answer is not really different from Anurag A's, but I think it is clearer to present the answer in the way below.
Let $A=\pmatrix{0&A_2\\ 0&A_4}$ where $A_4$ is invertible. The condition $AB=-BA$ implies that $B=\pmatrix{B_1&B_2\\ 0&B_4}$ where $A_4$ and $B_4$ anti-commute. Hence it suffices to prove the inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What is the sum of the maximum and minimum values BC can take? $\small AB$, $\small BD$ and $\small BC$ are are integer measurements
If $\small AB + BD = k$, find the maximum and minimum values that the $\small BC$ side can assume and then add the values found. (Answer: $k$)
My progress:
$\small \triangle ADB: |AB - BD... | $AB=a, BD=b, BC=c$
$\triangle ABC: a\sin 3\theta=c\sin \theta \Rightarrow$
$a(3\sin\theta-4\sin^3\theta)=c\sin\theta\Rightarrow$
$a(3-4\sin^2\theta)=c\Rightarrow$
$\sin^2\theta=\frac{3a-c}{4a}\Rightarrow$
$c<3a$
$\triangle BCD: c\sin\angle BCD=b\sin 2\theta \Rightarrow$
$\sin\angle BCD=\frac{b}{c}\sin 2\theta$
$0 < \an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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General Formula for an arbitrary Rotation of Sphere I am currently reading F. Klein's book "Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree" and in Part 1, Chapter 2, we desire to deduce a general formula for an arbitrary rotation of sphere. The idea is as follows: first we use stereograph... | Letting $A:=\frac{\xi+i\eta}{1+\zeta}$ and $B:=\frac{\xi+i\eta}{1-\zeta}$ we obtain from $(*)$:
\begin{align*}
e^{-i\frac{\alpha}{2}}\ \frac{z^{\prime}+A}{z^{\prime}-B}&=e^{i\frac{\alpha}{2}}\ \frac{z+A}{z-B}\\
e^{-i\frac{\alpha}{2}}\left(z^{\prime}+A\right)\left(z-B\right)&=e^{i\frac{\alpha}{2}}(z+A)\left(z^{\prime}-B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | Here is a one-liner:
$$2^{2k+1}+5^{2k+1}+56\equiv2(1+3)^k+5(1+24)^k-7\equiv2(1+3k)+5(1+24k)-7\equiv126k$$
where we used binomial theorem in middle step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 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.