Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
line through $P(l,3)$ intersect ellipse at $A$ and $D$ and axis at $B$ and $D$. then min. of $|l|$ A line through $P(l,3)$ meets the ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1$ at $A$ and $D$ and meets the $x$ axis and $y$ axis at $B$ and $C$ so that $PA.PD =PB.PC.$ find minimum value of $|l|$
equation of l... | Eliminating $y$ from $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $y-3=m(x-l)$ gives
$$(16m^2+9)x^2+16(-2m^2l+6m)x+16(m^2l^2-6ml)=0$$
Letting $\alpha,\beta$ be the $x$-coordinate of $A,D$ respectively, we get
$$\alpha+\beta=-\frac{16(-2m^2l+6m)}{16m^2+9},\quad \alpha\beta=\frac{16(m^2l^2-6ml)}{16m^2+9}\tag1$$
Since we have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
group 12 boys and 8 girls with restrictions
There're 12 boys and 8 girls in a class. The teacher wants to randomly split them into 3 groups: 5 kids in group A, 11 kids in group B and 4 kids in group C.
1) What is the probability that John and Peter will not be in the same group?
2) What is the probability that in eac... | The total number of ways to split them into groups is:
$$\frac{(12+8)!}{5!\times11!\times4!}=21162960$$
Question #$1$:
The number of combinations with John and Peter in the 1st group is:
$$\frac{(12+8-2)!}{(5-2)!\times11!\times4!}=1113840$$
The number of combinations with John and Peter in the 2nd group is:
$$\frac{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the quadratic equation from highest point on parabola? If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:
$$f(x) = ax^2 + bx + c$$
| Vertex form of a quadratic is: $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex. So, in our case the vertex is $(2,0)$ so $h=2$ and $k=0$. So we have, $f(x)=a(x-2)^2$. Now, find $a$ by using the point $(-1,-6$). So we have, $-6=a(-1-2)^2$ which gives us $-6=9a$ so $a=-\frac{2}{3}$. So our equation is $f(x)=-\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$
Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$
where $a,b,c>0$
My attempt:
I couldn't proceed after that. Please help me in this regard, thanks!
| We need to prove that
$$(a+b+c)\ln\frac{ab+ac+bc}{a+b+c}\geq\frac{1}{2}\sum\limits_{cyc}a\ln{bc}$$ or
$$\ln\left(\frac{ab+ac+bc}{a+b+c}\right)^2\geq\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}.$$
But by Jensen $$\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}\leq\ln\sum\limits_{cyc}\frac{abc}{a+b+c}$$
Thus, it remains to prove tha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of $\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}$? Not so thrilling... An exercise of one of my daughters.
How to evaluate
$$\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}?$$ I made several substitution namely:
*
*Factorisation of $x^2+x+1$
*Then use of $\sinh t$
*Then substitution by $e^u$
*To get a rational fracti... | $$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} & \stackrel{x\mapsto\frac{1}{x}}{=}\int_0^1 \frac{d x}{\sqrt{1+x+x^2}} \\
&=\int_0^1 \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}
\end{aligned}
$$
Letting $x+\frac{1}{2} =\frac {\sqrt 3}{ 2 }\sinh \theta$ transforms the int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle i... | A formula for determining the quadrant containing an angle $\theta$ when $\theta$ is not an integer multiple of $\frac{\pi}{2}$ (the quadrantal angles) is
\begin{equation}
Q\left(\theta\right)=\left(\left\lfloor\frac{2\theta}{\pi} \right\rfloor\mod{4}\right)+1
\end{equation}
For example,
\begin{eqnarray}
Q\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
P(X > Y - 1/2), given X and Y are uniform(0,1) I'm trying to calculate this, using the law of total probability. My understanding is that the answer should be 1/2.
$\displaystyle P\left(X > Y - \frac{1}{2}\right) = \int_{0}^{1} P\left(X > Y - \frac{1}{2} \middle| Y = y\right) \cdot f_Y(y) \cdot dy$
Solving the first te... | For $0<y<1$ we have
$$P\left(X>y-\frac{1}{2}\right)=
\begin{cases}
1 & \text{ if } y\le \frac{1}{2}\\
\frac{3}{2}-y & \text{ if } y>\frac{1}{2}
\end{cases}
$$
So
$$P\left(X>Y-\frac{1}{2}\right)=\int_0^{1/2}1\,dy+\int_{1/2}^1 \left(\frac{3}{2}-y\right) dy=\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\sum\limits_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,...$ It is n-th Fibonacci.
Prove this sum
$$\sum_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$$
I try:
We know that $\sum_\limits{k=1}^{n}F_n^2=F_nF_{n+1}$
$$\sum_{k=1}^{n}(n^2+1+k^2+2n-2nk-2k)F_k^2=\sum_{k=1}... | As noted in a comment, this sequence is the convolution of the two sequences $a_n=F_n^2$ and $b_n=(n+1)^2$, so we can multiply the two (standard) generating functions.
First, the generating function for $a_n$ — that is, $\displaystyle\sum_{n=0}^\infty F_n^2 x^n$ — is $\dfrac{x(1-x)}{(1+x)(1-3x+x^2)}$ (see, for instance... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Pro... | By Jensen $\sum\limits_{cyc}\sqrt{x+y-x^2}\geq2\sqrt{\sum\limits_{cyc}(2x-x^2)}=2\sqrt{2-\sum\limits_{cyc}x^2}\leq2\sqrt{2-\frac{1}{4}}=\sqrt7<4\sqrt2-3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any he... | $$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x + m\right)=10$$
$$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }\left(x + \dfrac{2x^2+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }(x +\ell x)=0 \implies \ell = -1$$
$$\lim_{x \to\infty }\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How does this separation of the variable from the fraction result in $3\cdot \frac{1}{x}$ & not $\frac{3}{x}$ I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3... | As you stated we have
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x}$
Now we know that $\frac{3}{1}=3$ (dividing by 1 does not change a number).
Hence, we have
$ \frac{3}{1} \cdot \frac{1}{x}=3\cdot \frac{1}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show $y^2,z^2,u^2$ are roots of $(1)$ Euler's solution for the general quartic are as follows:
From the depressed quartic, $x^4+px^2+qx+r=0$, assume $x=y+z+u$ and it may be shown that $u^2,y^2,z^2$ are roots of the cubic$$t^3+\dfrac q2t^2+\dfrac {q^2-4s}{16}t-\dfrac {r^2}{64}=0\tag1$$
$$\vdots$$
However, how w... | That's Decartes–Euler solution
\begin{align*}
x &= u+v+w \\[5pt]
x^2 &= u^2+v^2+w^2 + 2(uv+vw+wu) \\[5pt]
x^4 &= (u^2 + v^2 + w^2)^2 + 4(u^2+v^2+w^2)(uv+vw+wu) \\
&\quad +4(u^2 v^2+v^2 w^2+w^2 u^2)+8uvw(u+v+w) \\[5pt]
0 &= x^4+px^2+qx+r \\[5pt]
0 &= (u^2+v^2+w^2)^2+(uv+vw+wu)[\color{blue}{4(u^2+v^2+w^2)+2p}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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The decay of the series given by $x_n=x_{n-1}+\cos x_{n-1}$. Let $x_n=x_{n-1}+\cos x_{n-1}$, $x_1=1$. It is easy to see that $x_n\to \dfrac{\pi}{2}$. However, how can we show that $n^n(x_n-\dfrac{\pi}{2})\to 0\ (n\to\infty)$?
I find Stolz formula hard to use.
| We can manipulate the recursion as follows:
\begin{align}
x_n &= x_{n-1}+\cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \sin\left(\dfrac{\pi}{2}-x_{n-1}\right)
\end{align}
Since $|\theta-\sin \theta| \le \dfrac{1}{6}|\theta|^3$ for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that :$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$ How do I show that
Let $n\ge0$. Then,
$$I:=\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$$
Here, $m!!$ denotes the product of all positive integers $i \in \left\{1,2,\ldots,m\right\}$ that have the same parity as $m$.
My try:
Using Binomial theorem
$$(1... | Let $z=x^{2}$
\begin{align}
\int\limits_{0}^{1} (1-x^{2})^{n} dx &= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{n} z^{-1/2} dz \\
&= \frac{1}{2} \mathrm{B}(1/2,n+1) \\
&= \frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)} \\
\tag{1}
&= \frac{\sqrt{\pi}n!}{(2n+1)\Gamma(n+1/2)} \\
\tag{2}
&= \frac{\sqrt{\pi}n!}{(2n+1)} \frac{2^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be ... | $$\cos{C}=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{a^2+b^2}{4ab}\ge\dfrac{1}{2}$$
so
$$C\le \dfrac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
| Starting with $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$, take $x=b\sinh\theta$:
$$f(x)=2(a-b\sinh\theta)(b\sinh\theta+b\cosh\theta)=2b(a-b\sinh\theta)\exp\theta$$
Now, let $\gamma=\exp\theta$:
$$f(x)=2b\gamma\left(a-b\frac{1}{2}\left(\gamma-\frac{1}{\gamma}\right)\right)=-b^2\left(\gamma^2-2\frac{a}{b}\gamma-1\right)$$
Now, comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Diophantine system of equations I'm trying to know if there is an efficient way to find the smallest (i.e lexicographically) triplet $(a,b,c)$ of integers verifying $$a^2+b^2+c^2 = x$$ $$a^3+b^3+c^3=y$$ $$a^4+b^4 + c^4 = z$$ if $(x,y,z)$ is known.
We assume that a solution exists for that triplet $(x,y,z)$.
Originally... | If $a,b,c$ are integers such that
$$a^2 + b^2 + c^2 = x$$
$$a^3 + b^3 + c^3 = y$$
$$a^4 + b^4 + c^4 = z$$
then the following divisibility conditions must hold:
$$
\left(
a + b + c
\right)
\mid
\left(
x^2 - 2z
\right)
$$
$$
\left(
4\left(ab + bc + ca\right)
\right)
\mid
\left(
x^4 + 6x^2z - 16xy^2 + 9z^2
\right)
$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alte... | Hint: for $0<x<1$, the inequality
$$
(1-x)^n<\frac{1}{(1+x)^n}
$$
is equivalent to
$$
(1-x^2)^n<1
$$
Or, mimicking the proof of Bernoulli’s inequality, we have to prove that
$$
(1-x)^{-n}>1+nx
$$
The statement is true for $n=1$, because it is $1-x^2<1$. Suppose it holds for $n$. Then
$$
(1-x)^{-n-1}=(1-x)^{-n}(1-x)^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Infinitely nested radical formulas for $\pi$ Has anyone come by the infinitely nested radical formula:
$$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty} 2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2... | From here:
If there exists an infinite iterative sequence $A=\{a_1, a_2,\cdots a_n\}$, and a sequence $B=\{c_1, c_2,\cdots c_n\}$ obtainable from $A$ such that for all $b, a_n =\cos b$, $a_{n+1} =\cos b/2, c_{n+1}=\sin b/2$. Then $$a_{n+1} =\sqrt{\frac{1+a_n}{2}} \text{ and } c_n =\sqrt{\frac{1-a_n}{2}}$$
If $a_1 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^... | When $x\rightarrow-\infty$, you don't have $0$ as denominator, but $+\infty$, because when $x \rightarrow-\infty, \sqrt{x^2 -2x} \sim |x|$ and, as $x < 0$, $|x| = -x$, so you have $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
| Let $x < 0$ be a real root of $x^4+2ax^3+x^2+2ax+1=0$. Since there are two different
negative real roots we can assume that $x \ne -1$.
You already found out that $t=x+ \frac{1}{x}$ satisfies $t^2+2at-1=0$. Then
$$
-t = \lvert x \rvert + \frac{1}{\lvert x \rvert} > 2
$$
from the AM-GM inequality, with strict inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don'... | If $c/d$ is positive, you will have $|-c/d|=c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=\dfrac{c}{d}(bc-ad),
$$
or
$$
ac+\frac{bc^2}{d}=\dfrac{bc^2}{d}-ac,
$$
or
$$ac=0.$$
Now if $c/d$ is negative, you will have $|-c/d|=-c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=-\dfrac{c}{d}(bc-ad),
$$
or
$$
ac+\frac{bc^2}{d}=-\dfrac{bc^2}{d}+ac,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$ Reshetnikov gave the remarkable evaluation,
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\... | I think you should start by the following general form
$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k}
\cdots ,\frac{k-1}{k},\frac{k+1}{k};\left(
\frac{m(1-m^k)}{f_k}\right)^k \right) = \frac{1}{1-m^k}$$
Where
$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$
Put $k=4$
$${}_{4}F_{3}\left(\frac{1}{5} ,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
Factorization of quartic polynomial. I want to know other ways of factorization to get quadratic factors in this polynomial:
$$x^4+2x^3+3x^2+2x-3$$
Thanks in advance for your suggestions.
The original polynomial is $$x^6-2x^3-4x^2+8x-3$$ where the two found factors are $(x+1)$ and $(x-1)$ by synthetic division.
| Note that $$(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$$which means that your polynomial is equal to
$$
(x^2 + x + 1)^2 - 4 \\
= (x^2 + x + 1)^2 - 2^2\\
= (x^2 + x + 1 - 2)(x^2 + x + 1 + 2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$ I figured that
\begin{align*}7^{x+1} &= 21^{x}\\
7^{x+1} &= 3^{x} \times 7^{x} \\
7 &= 3^{x}
\end{align*}
but I can't go further at the moment.
| You have : $7^{\frac{1}{x}} = 3$ and $3^{2x} = 7^2$. Thus $7^{\frac{1}{x}} + 3^{2x} = 52$
| {
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"source": "stackexchange",
"question_score": "2",
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Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\df... | Although this is clearly a faulty approach, take the first 15 formable numbers.
These are 1,2,3,10,11,12,13,20,21,22,23,30,31,32,33.
The decimal reciprocals are roughly equal to $1+0.5+0.35+0.1+0.09+0.08+0.08+0.05+0.05+0.04+0.04+0.03+0.03+0.03+0.03$, more or less by a very small amount.
This is slightly less than 2.5... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi... | You can try this method:
$$x+y+z=0 \implies x+y=-z\tag1$$
And $$\sin{x}+\sin{y}+\sin{z}=0$$
$$\implies \sin{x}+\sin{y}=-\sin{z}\tag2$$
And $$\cos{x}+\cos{y}+\cos{z}=0$$
$$\cos{x}+\cos{y}=-\cos{z}\tag3$$
So $$(2)^2+(3)^2 \implies 2+2\cos(x-y)=1$$
$$\implies \cos(x-y)=-\frac12\tag4$$
Similarly we get $$\cos(y-z)=-\frac12... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How to prove this sum $\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$ Let $n\ge 2$ postive integer, show that
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$
I have done this works
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\bin... | It is a convolution identity. You may consider that:
$$ \sum_{k\geq 1}\frac{x^k}{k}\binom{2k-2}{k-1} = \frac{1-\sqrt{1-4x}}{2}\tag{1}$$
holds as a consequence of the generating function of Catalan numbers. If you square both sides of $(1)$ and consider the coefficient of $x^n$, you get:
$$ \sum_{k=1}^{n-1}\frac{1}{k(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Determine whether the following sequence is increasing or decreasing $\frac{(n-8)^2}{(1-n)^2}, n\geq 2$ Determine whether the following sequence is increasing or decreasing:
$$\frac{(n-8)^2}{(1-n)^2}, n\geq 2$$
So the first few terms are: $36,\frac{25}{4},\frac{16}{9},...$ so let's assume the sequence is decreasing.
... | You can rewrite your expression using polynomial division:
$$\frac{(n-8)^2}{(n-1)^2} = \left(\frac{n-8}{n-1} \right)^2 = \left(1-\frac{7}{n-1} \right)^2$$
since $\frac{7}{n-1}$ is a positive but decreasing sequence, $1-\frac{7}{n-1}$ is a increasing sequence. But since $1 < \frac{7}{n-1}$ for $n<8$ we need to consider ... | {
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"url": "https://math.stackexchange.com/questions/2086431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How do we show that $\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2$ I was going through the the top votes questions and I saw this quite of interesting post posted by @Sangchul Lee. I was just messing around with it and found a slightest Variation of Ahmed's inte... | Using your approach $$u=x(x^2+2)^{1/2}\implies x=\sqrt{\sqrt{u^2+1}-1}\implies dx=\frac{u}{2 \sqrt{u^2+1} \sqrt{\sqrt{u^2+1}-1}}\,du$$ and, after simplifications $$\int{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int\frac{\tan ^{-1}(u)}{2 u^2+2}\,du=\frac{1}{4} \tan ^{-1}(u)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Cute limit: $\lim\limits_{x \to 0^{+}}\frac{\sin(x)^x − x^ {\sin(x)}}{\tan(x) ^x − x^{\tan (x)}}$ What's the fastest way to calculate the below limit (ideally, without resorting to Taylor expansions)?
$$\lim_{x \to 0^{+}}\frac{\sin(x)^x
− x^ {\sin(x)}}{\tan(x)
^x − x^{\tan (x)}}$$
| To expand on my comment, I will use the following two limits apart from the standard lmits: $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6},\,\lim_{x \to 0}\frac{x - \tan x}{x^{3}} = -\frac{1}{3}$$ Both the above are easily obtained either via Taylor series or via L'Hospital's Rule.
We have then
\begin{align}
L ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How can I solve the following sequence Let we have the following sequence
$$y_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\ \sqrt{n^2+n}}$$ find the limit of the sequence $y_n$ decide whether it is increasing or decreasing
| Hint
Prove that $$\frac{n}{\sqrt{n^2+n}}\leq \sum\frac{1}{\sqrt{n^2+k}}\leq\frac{n}{\sqrt{n^2}}$$
$$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{n}{\sqrt{n^2}}$$
Then use squeeze lemma
It is a increasing sequence
Knowing that $$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{1}{n}$$
We have that $y_n\leq\frac{1}{n}$ and $\frac{n+1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ We have to solve for $x$ and $y$:
$$\left \{
\begin{align*}
\cos (2x+3y) &= \frac{1}{2} \\
\cos (3x+2y) &= \frac{\sqrt{3}}{2}
\end{align*}
\right.$$
I got
$$\left \{
\begin{align*}
2x+3y &= 2n\pi \pm\frac{\pi}{3} \\
3x+2y &= 2m\pi \... | To get rid of $y$,
the first equation
is multiplied by 2
and the second by 3
giving
$\pm 2\pi/3\pm\pi/2$.
To get rid of $x$,the first equation is multiplied by 3
and the second by 2
giving
$\pm \pi\pm\pi/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| Hint 1:
For $x\in [0,\frac{\pi}{2}]$:
$$\tan \frac{x}{2} = \frac{\sin x}{1+\cos x}$$
$$\cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}$$
$$\sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}}$$
Hint 2:
$$7^{\circ}30' = 15^{\circ}\cdot \frac{1}{2} =30^{\circ}\cdot \frac{1}{2}\cdot \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?
| \begin{align*}
x\sqrt{y^{2}+1}+y\sqrt{x^{2}+1}&=\left(x,y\right)\cdot\left(\sqrt{y^{2}+1},\sqrt{x^{2}+1}\right)\le\sqrt{(x^{2}+y^{2})(x^{2}+y^{2}+2)}\\
&\le\frac{2x^{2}+2y^{2}+2}{2}=x^{2}+y^{2}+1
\end{align*}
where I have used Cauchy-Schwarz followed by the arithmetic-geometric inequality. Note that equality cannot occ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Inequality solution with absolute value I have the following inequality:
$$ \left\rvert x + \frac{1}{x} \right\lvert \ge 2$$
can I say that the distance of the expression inside the absolute value from $0$ is greater or equal to 2 (from the positive side of the real axis) and smaller or equal to $-2$ (from the negative... | This line is incorrect:
$$\rvert x + \frac{1}{x} \lvert \ge 2 \implies -2\ge x + \frac{1}{x} \ge 2$$
The absolute value means the value inside could have been positive or negative so it actually becomes:
$$\pm\left(x+\frac{1}{x}\right)\ge2$$
When you go to move the negative sign to the other side you must split this in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If function $f$ satisfies given conditions, then find $\frac{1}{3} f(1,98)-f(1,99)$ Let $f$ be a function defined on $\{(m,n):$ $m$ and $n$ are positive integers $\}$ satisfying:
$$1. f(m,m+1)=\frac{1}{3}$$, for all m
*$$f(m,n)=f(m,k)+f(k,n)-2f(m,k) \cdot f(k,n)$$ for all $k$ such that $m<k<n$, then find the value of... | First note the factor of $\frac{1}{3}$ and try to exploit it with the first defining equation:
$$\begin{align}S &= \frac{1}{3}f(1, 98) - f(1, 99)\\
&= f(98, 99)f(1, 98) - f(1, 99)\\
&= f(1, 98)f(98, 99) - f(1, 99)\end{align}$$
Let $m = 1, n = 99, k = 98$. Then,
$$\begin{align}S &=\frac{f(1,98) + f(98, 99) - f(1, 99)}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find the value of $\alpha + \beta + \alpha\beta$ from the given data Suppose the quadratic polynomial
$P(x) = ax^2 + bx +c $
has positive coefficients $ a, b, c $ in an arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha $ and $\beta $, then $\alpha + \beta + \alpha\beta $ equals?
1) $3$
2) $... | Any quadratic (or for that matter polynomial) can be written in terms of its roots:
$$P(x)=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta$$
So $\alpha+\beta+\alpha\beta=-\frac{b}{a}+\frac{c}{a}=\frac{-b+c}{a}$
We also know $a,b,c$ form an AP so lets rewritten them as $a,a+d,a+2d$ where $d$ is the common differ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are ... | As noted in the comments, you mean to write "$b^2$ divides $a^2$" in your third line. I wouldn't jump immediately from that to concluding though. So $b^2|a^2$ so there exists a $k$ such that $b^2k=a^2$.
Lemma: Let $x$ and $y$ be natural numbers. If $xy$ is a perfect square, then it is not the case that exactly one of ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two sm... | Considering the integral $$I=\int{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx$$ using partial fraction decomposition, we have $$I=\int\left(\frac{a^5 b^3}{\left(a^4-b^4\right) \left(a^2 x^2+b^2\right)}-\frac{a^3
b^5}{\left(a^4-b^4\right) \left(a^2+b^2 x^2\right)}\right)\,dx$$ from which $$I=\frac{a^3 b^3 \left(a^2 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt... | $$
|z + 1 - i\sqrt{3}| = |z - 1 + i\sqrt{3}| \\
\implies |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|
$$
If I take $z_0 = 1 - i\sqrt{3}$, then the equation
$$
|z - (-z_0)| = |z - z_0|
$$
describes the perpendicular bisector of the straight line joining the points $P(z_0)$ and $Q(-z_0)$. What is clear is that the requ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
determinant of 4x4-matrix occuring in Zarhin's trick What's the easiest/fastest way of calculating the determinant of $$\begin{pmatrix}a & b & c & d\\ -b & a & -d & c\\ -c & d & a & -b \\ -d & -c & b & a\end{pmatrix}$$? The result is $(a^2+b^2+c^2+d^2)^2$. This determinant occurs in Zarhin's trick (if $A$ is an Abelian... | Write $X$ for the matrix in question. Then
$$
X \cdot X^T = (a^2+b^2+c^2+d^2) I.
$$
This implies that
$$
\det(X) =f(a,b,c,d)= \pm (a^2+b^2+c^2+d^2)^2 .
$$
Now as
$$
f(a,0,0,0)=a^4
$$
for all $a \in \mathbf R,$ we have that
$$
\det(X)=(a^2+b^2+c^2+d^2)^2
$$
for all $a,b,c,d \in \mathbf R.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For all triangle prove that $\sum\limits_{cyc}\frac{a}{(a+b)^2}\geq\frac{9}{4(a+b+c)}$
Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that:
$$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$
The Buffalo way kills it, but I am looking for a nice proof for this nice inequa... | After using Ravi subtitution, we need to prove
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{9}{8(x+y+z)}.$$
By the Cauchy-Schwarz inequality we get
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{\left[\displaystyle \sum (x+y)(42x+3y+55z) \right]^2}{\displaystyle \sum (2y+z+x)^2(x+y)(42x+3y+55z)^2}.$$
Therefore, we will... | {
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"url": "https://math.stackexchange.com/questions/2109187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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integral calculus problem involving infinity problem
The value of the integral $\displaystyle \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx$ is?
The value of the integral $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy$
I couldn't figure out the trick... | Here are solutions.
Solution of 1. Apply the Fubini's theorem to interchange the order of integration. The domain of integration is specified by the inequality $0 \leq x \leq y$, which tells you that
$$ \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx
= \int_{0}^{\infty}\int_{0}^{y} \frac{e^{-y}}{y} \, dxdy
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
2nd order linear differential equation with piece wise continuous in homogeneous term. $y^{ii}=f(x)$ where
$f(x)=\begin{cases}
0 ,& \text{ if } x \in [0,1/2]\\
1, & \text{ if } x \in (1/2,1]
\end{cases} $\
subject to $y'(0)=0,y(1)=1$
We can solve this differential equation by finite difference method.But how can we fin... | $y^{''}=f(x)$ where
$f(x)=\begin{cases}
0 ,& \text{ if } x \in [0,\frac{1}{2}]\\
1, & \text{ if } x \in (\frac{1}{2},1]
\end{cases} $\
subject to $y^{'}(0)=0,y(1)=1$
The function must be continuous and differentiable. In $[0,\frac{1}{2}]$ we have $ y^{'} $ is constant: $ y = a + rx $. Then $ y^{'}(0)=0 $ gives: $ y=a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Sum calculation with binomial coefficients Knowing that $$\sum_{k} \binom{a+b}{a+k} \binom{b+c}{b+k} \binom{c + a}{c + k} (-1)^k = \frac{(a + b + c)!}{a!b!c!}$$ where $a,b,c ∈ \mathbb Z $ solve the following sum
$$\sum_{k} \binom{2a}{a+k} \binom{2b}{b+k} \binom{2c}{c + k} (-1)^k $$
| We have
$$
\sum_k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}(-1)^k
=\sum_k\frac{(a+b)!(b+c)!(c+a)!(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!} \\
=(a+b)!(b+c)!(c+a)!\sum_k\frac{(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!}
$$
and
$$
\sum_k\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}(-1)^k
=\sum_k\frac{(2a)!(2b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convexity of $x\left(1+\frac1x\right)^x,\ x\ge 0$ This may turn out to be really simple but I do not see a quick way to the proof. How would one show $\displaystyle x\Big(1+\frac1x\Big)^x,\ x\ge 0$ is convex?
I derived the second derivative. It has a negative term. I suppose I could combine certain terms to make the ne... | Let $f(x)=x\left(1+\frac{1}{x}\right)^x$, where $x>0$.
Hence,
$$f''(x)=\frac{\left(1+\frac{1}{x}\right)^x\left(x(x+1)^2\ln^2\left(1+\frac{1}{x}\right)+2(x+1)\ln\left(1+\frac{1}{x}\right)-x-3\right)}{(x+1)^2}.$$
Thus, we need to prove that
$$\ln\left(1+\frac{1}{x}\right)>\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$$
Let $g(x)=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
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Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$ Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's
$$ {(... | we have $$\frac{1}{a^3(b+c)}=\frac{bc}{a^2(b+c)}=\frac{(bc)^2}{a(b+c)}$$ then we get
$$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)}\geq \frac{3}{2}$$ by $AM-GM$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$... | If $2 \arcsin(x) = \arcsin(3x/4)$, we apply $\sin$ to both sides, and use the double-angle formula and the fact that $\cos(\arcsin(x)) = \sqrt{1-x^2}$ to get
$$ 2 x \sqrt{1-x^2} = 3x/4 $$
thus either $x=0$ or $\sqrt{1-x^2} = 3/8$. Squaring both sides of the latter,
$ 1 - x^2 = 9/64$, so $x = \pm \sqrt{55}/8$. Note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Limit of a sequence by Cauchy second test The sequence is
$$ \left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{1/n} $$
as $n$ goes to infinity. By Cauchy second test it's pretty clear that it's limit will be equal to $1+\frac{n}{n}$ which is $2$. ... | we can write $$\left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right] = n!\binom{2n}{n}\cdot \frac{1}{n^n}$$
Using $$\binom{2n}{n}\leq \binom{2n}{0}+\binom{2n}{1}+\cdots \cdots \binom{2n}{2n}=2^{2n} = 4^n$$
And Using Power mean Inequality $$\frac{\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$ How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$
Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$.
If I checked from the graph, t... | Hint: $\sin 2x -\sin Ax = 2\sin {\frac{2-A}{2}x} \cos {\frac{2+A}{2}x}$
Solution: if this hint is used, $$\lim_{x\to 0}{ \frac{2\sin \big( {\frac{2-A}{2}x}\big) \cos \big( {\frac{2+A}{2} x}\big) } {x(1+x^2)}}$$
$$2 \frac{2-A}{2} \lim_{x\to 0} {\frac{\cos \big( {\frac{2+A}{2}x}\big) }{1+x^2}}=2-A$$
And this is equal to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$... | This proof is at least unlucky written down.
Normally, we start with the claim for $n$ (Here $1+2+\cdots n=\frac{n(n+1)}{2}$) and then proof the claim for $n+1$.
The correct way is to start with $1+2+\cdots n+n+1=\frac{n(n+1)}{2}+n+1$ (we assume that the formula is correct for $n$) and show that this is equal to $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Using induction to show that if $x_n^2 \geq 2$, then $x_{n+1}^2 \geq 2$ Let $x_1=2$. For all $n \geq 1$, define
$$x_{n+1}=\frac{1}{2} \left( x_n + \frac{2}{x_n} \right) $$
Prove by induction that $x_n^2 \geq 2$ for any $n \geq 1$.
I have a way to solve the question, which is to use AM-GM inequaity to conclude in the i... | Hint: Use induction to show $x_n > 0$ always and then
$$x_{n+1}=\frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \geq \sqrt{2} \Leftrightarrow \\x_n + \frac{2}{x_n} \geq 2\sqrt{2} \Leftrightarrow x_n^2 + 2 \geq 2x_n\sqrt{2} \Leftrightarrow \\ x_n^2 - 2x_n\sqrt{2} + 2 \geq 0 \Leftrightarrow \left(x_n - \sqrt{2} \right)^2 \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt... | Here's your equations:
$x+y^2 = y^3$
$y+x^2 = x^3$
I could substitute
$x=y^3-y^2$
and get a degree 9 equation
in $y$,
but I'll try something else.
Looking at these,
I notice that if I subtract them,
I get something
in which everything
is divisible by
$x-y$.
Subtracting the second from the first,
I get
$(x-y)+(y^2-x^2)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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Problem of Complex Numbers in Geometry using Roots of Unity Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
$x^2 + px + q = 0$
for some integers $... | From the development of the geometric series with reason $\omega$ we get $1 + \omega + \omega^2 + \dots + \omega^6 = (\omega^7 - 1) / (\omega - 1) = 0$, from which we deduce $1 + \alpha + \beta = 0$.
Using the same development again, $\alpha \beta = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does the discriminant tell us how many zeroes a quadratic equation has? The quadratic formula states that:
$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The part we're interested in is $b^2 - 4ac$ this is called the discriminant.
I know from school that we can use the discriminant to figure out how many zeroes a quad... | If the equation is $ax^2+bx+c=0$, with $a\ne0$, it is equivalent to
$$
4a^2x^2+4abx+4ac=0
$$
that can also be rewritten as
$$
4a^2x^2+4abx+b^2=b^2-4ac
$$
or, recognizing the square on the left-hand side,
$$
(2ax+b)^2=b^2-4ac
$$
Now, if $b^2-4ac<0$, we cannot find a real number $x$ such that $(2ax+b)^2=b^2-4ac$, because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 9,
"answer_id": 3
} |
Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \fra... | Try to multiply the given sum by $\frac{3}{2}=1+\frac{1}{2}$:
$$\begin{eqnarray*}\left(1+\frac{1}{2}\right)\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^{n+1}}\\&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=2}^{101}\frac{(-1)^{n}(n-1)}{2^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How can I show that modul of this complex number equal to 1? Let $z$ be a complex number such that
$$(1+2 i) \left| z\right| -\frac{\sqrt{10}}{z}+2-i=0.$$
Prove that $\left| z\right|=1$.
I tried
Put $z = a + bi$ $(a,b \in \mathbb{R})$, we have
$$2-\frac{\sqrt{10}a}{a^2+b^2}+\sqrt{a^2+b^2}+i \left(-1+\frac{\sqrt{10}b}{a... | Hint: note that $1+2i= i(2-i)$ then collect and rewrite:
$$
(2-i)(1+i|z|)= \frac{\sqrt{10}}{z}
$$
Note that $|2-i|^2=5$ and take the square of the modulus on both sides:
$$
5(1+|z|^2) = \frac{10}{|z|^2}
$$
The latter is a simple quadratic in $|z|^2$ which gives $|z|=1$ in the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form Question
find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form
my steps:
So i first find the magnitude of the root to get $|2-2i|^{\frac{1}{3}}$
and the argument of the root is $\frac{\frac{-\pi}{4}}{3}$
thus the first root must be $|2-2i|e... | Cube roots are equally distributed around a circle centered at the pole and having radius equal to the cube root of the modulus (absolute value).
In the case of $z^3=2-2\,i$ the modulus is, as you stated, $\vert 2-2\,i\vert=2\sqrt{2}$. So the modulus of the roots is $\sqrt[3]{2\sqrt{2}}=\sqrt{2}$.The argument for $2-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2... | There is a mistake in the line below:
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
Actually you should be integrating the difference of the $2$ curves within that limit, i.e. $(4x-x^2-1)-(x+1)=3x-x^2-2$, since each of them represents the area under the curve and bounded by the x-axis.
The magnitude of the enclosed ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Which integration techniques I should use for $\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$
$$\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$$
I can simplify it to:
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}$$
but I can't go from here.
| 1.$\displaystyle\int{\frac{1}{\sqrt{x^2+3}}}dx=\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx $
Substitute $\tan u=\frac{x}{\sqrt3}\rightarrow dx=\sqrt3\sec^2udu$, Then
$$\begin{align}\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx&=\int\frac{\sec^2u}{\sqrt{\tan^2u+1}}du\\&=\int \sec udu\\&=\ln(\vert \sec u+\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$ Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$.
I computed the fourier transform of $e^{-a|x|}$, which is $\frac{a}{\pi (a^2+w^2)}$
I'm not sure how to continue from here,... | Actually:
$$ \mathcal{F} \left [ e^{- a |x|} \right ] (\omega) = \frac{ 2a}{a^2 + \omega^2} $$
As for the integral, you could use one of Plancherel's formulae:
$$ \int_{-\infty}^{+\infty} f(t) \overline{g (t)} \ \mathrm dt = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{f} (\omega) \overline{ \widehat{g} (\omega)} \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What are the solutions to $a^2+ab+b^2$ $=$ $3^n$? What are all known integer solutions ($a, b, n$) to $a^2+ab+b^2$ $=$ $3^n$ besides ($1, 1, 1$) and ($-2, 1, 1$)? Do any others even exist?
This question comes from the identity that ($a^3±b^3$)/($a±b$) $=$ $0, 1$ $\pmod 3$. If ($a^3±b^3$)/($a±b$) $=$ $0$ $\pmod 3$, then... | The set of solutions to $$ x^2 + xy + y^2 = 1 $$ in integers is finite (6).
x = 1, y = 0 target 1
x = -1, y = 0 target 1
x = 1, y = -1 target 1
x = -1, y = 1 target 1
x = 0, y = 1 target 1
x = 0, y = -1 target 1
The set of solutions to $$ x^2 + xy + y^2 = 3 $$ in integers is finite(6).
x = 2, y = -1 target 3
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the prob... | $$\begin{align*}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac7h\cdot \left( \frac1{\sqrt{x+h}} -\frac1{\sqrt{x}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \cdot \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding the solution of equation I have to find out the integral solution of a equation $ax+by=c$ such that $x \geq 0$ and $y \geq 0$ and value of $(x+y)$ is minimum.
I know if $c \equiv 0 \pmod{\gcd(a,b)}$ then it's always possible. How to find the values of $x$ and $y$?
| From the method of solving a linear Diophantene Equation we have that the solutions are $x = x_0 \pm n\cdot\frac{b}{(a,b)}$ and $y = y_0 \mp n \cdot \frac{a}{(a,b)}$, where the pair $x_0, y_0$ is a solution of the equation.
Now assume that $x+y \le x_0 + y_0$. If $x = x_0 + n\cdot\frac{b}{(a,b)}$, then $y = y_0 - n\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove in this trig problem i have to prove this
$$\frac{\cos 3x}{\sin 2x \sin 4x}+\frac{\cos 5x}{\sin 4x \sin 6x}+\frac{\cos 7x}{\sin 6x \sin 8x}+\frac{\cos 9x}{\sin 8x \sin 10x} = \frac{1}{2}\csc x(\csc 2x - \csc 10x)$$
i tried taking lcm but does not leads to anything. i believe i have to write numerator as so... | Hint: Observe
\begin{align}
\frac{\cos 3x}{\sin 2 x \sin 4x } = \frac{1}{2\sin x}\left(\frac{1}{\sin 2x}-\frac{1}{\sin 4x} \right)
\end{align}
since
\begin{align}
\frac{1}{\sin 2x}-\frac{1}{\sin 4x}=&\ \frac{\sin 4x-\sin 2x}{\sin 2x \sin 4x}\\
=&\ \frac{\sin 3x\cos x+\sin x\cos 3x-\sin 3x \cos x+\sin x\cos 3x}{\sin 2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.
For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$
I've managed to prove this via brut-force and Muirhead's inequality (Very unsatis... | I found this identity: $$\text{LHS-RHS} = \frac{1}{6} \sum\, \left( a-b \right) ^{2}\Big[ \left( a+b+c \right) ^{2}{c}^{2}+2\,
\left( ab+ca+bc \right) ^{2}\Big] \geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What is the $\prod \frac{1}{n-1}$ I am trying to compute the probability of none the events occurs where the probability for each event is $Pr[A_i]=\frac{1}{n-1}$ for all i and these events are independent.
What is the $\prod_{i=3}^{n} \frac{1}{n-1}$ when n >= 3
I know that the Pr(none event occur) = 1 - Pr(at least ... | *
*$\prod_{1}^{n} \frac{1}{n-1} = (\frac{1}{n-1})^{n}$
*$\Pr($none event occur$) = 1-(\frac{1}{n-1})^{n}$
For any $n\geq 3$, we have
$log(\frac{8}{7}) < n.log(n-1) \Rightarrow -log(\frac{8}{7}) > -n.log(n-1) \Rightarrow log(\frac{8}{7})^{-1} > n.log(n-1)^{-1} \Rightarrow log(\frac{7}{8}) > n.log(\frac{1}{n-1}) \Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Creating a non-recursive formula I'm a bit stuck on how to convert a non-recursive formula.
Assuming:
f(0) = 0
f(1) = 0
f(n) = f(n - 2) + $2^{n-1}\ if\ n \ge 1$
I can see that:
f(2) = 0 + $2^{2-1}$
f(3) = 0 + $2^{3-1}$
f(4) = (0 + $2^{2-1})\ +\ 2^{4-1}$
f(5) = (0 + $2^{3-1})\ +\ 2^{5-1}$
f(6) = ((0 + $2^... | See, that you create geometric series, so
$$f(n)=\begin{cases}2\frac{1-4^{\frac{n}{2}}}{1-4} &, n\in 2\mathbb{Z}\\
4\frac{1-4^{\frac{n-1}{2}}}{1-4} &, n\not\in 2\mathbb{Z}\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the following without the calculator
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$
Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step... | We know that $$4x^4-16x^3-17x^2+27x-7 =(4x^2-20x+7)(x^2+x-1) $$ $$=(x- \frac{5+3\sqrt {2}}{2})(x-\frac {5-3\sqrt {2}}{2})(x-\frac {-1 +\sqrt {5}}{2})(x+\frac {-1-\sqrt {5}}{2}) $$
Observe that $\frac {5+3\sqrt {2}}{2} $ is a root of this polynomial. Also observe that the above polynomial is $4$ less than the required p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
|
"Can the function $f(x)$ be considered as $x^2 + 3$ and go about solving the limit using the Limit laws?"
No, since we have only that $|f(x)-3|\le x^2\implies 3-x^2\le f(x)\le 3+x^2$.
But we can proceed by using $\color{blue}{f(x)-3=O(x^2)}$, where we are using the ("Big O notation").
Then, we can evaluate the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
| The complex numbers are isomorphic to the set of matrices of the form
$$
a+bi \sim
\begin{pmatrix} a & -b \\ b & a \end{pmatrix}
=a I + b J,
\text{ where }
J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
J^2=-I
$$
Your first matrix corresponds to $z=\sqrt3 +i$.
Now $\dfrac{iz}{2}=\dfrac{-1+\sqrt3}{2}$ is a third root ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I d... | Substitute $$y=x-1$$ to get $$\frac{\ln(x^3)}{x-1}=\frac{3\ln(y+1)}{y}$$
and now use $$\ln(y+1)=y+O(y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $0 \le \frac{1+\cos\theta}{2+\sin\theta} \le \frac{4}{3}$ for all real $\theta$. I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer.
| Let $\dfrac{1+\cos\theta}{2+\sin\theta}=y$
$$\iff2y-1=\cos\theta-y\sin\theta=\sqrt{y^2+1}\cos\left(\theta+\arccos\dfrac1{\sqrt{y^2+1}}\right)$$
$$-\sqrt{y^2+1}\le2y-1\le\sqrt{y^2+1}$$
$$\implies(2y-1)^2\le y^2+1\iff0\ge3y^2-4y=3y\left(y-\dfrac43\right)$$
$$\implies0\le y\le\dfrac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For $a,b,c>0$. Minimize $P=a+b+c$ For $a,b,c>0$ and $\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1$, minimize $$P=a+b+c$$
| By Cauchy Schwarz:
$$
1\times P=\left(\frac{2}{a}+\frac{5}{b}+\frac{3}{c}\right)(a+b+c)\geq(\sqrt{2}+\sqrt{5}+\sqrt{3})^2
$$
so $\min P=(\sqrt{2}+\sqrt{5}+\sqrt{3})^2$ when
$$
\frac{2}{a^2}=\frac{5}{b^2}=\frac{3}{c^2}\quad\text{and}\quad\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1;
$$
i.e.
$$
a=2+\sqrt{6}+\sqrt{10},\quad b=5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
... | Note that we have $$a^{2}>(b-c)^{2}\;,b^{2}>(a-c)^{2}\;,c^{2}>(b-a)^{2}$$
As $a,b,c$ are sides of a trinagle. Adding these, we get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$
Which is equivalent to $$a^2+b^2+c^2<2ab+2bc+2ca$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Producing circles with $r^2=x^2+y^2$ $r^2=x^2+y^2\ +\pi $ produces graph with two circles :
$r = 2$
This graph was produced using Desmos, is this correct or a quirk of the graph software ?
Ive noticed $\forall n \epsilon N [r^2=x^2+y^2+n\pi]$ appears to produce same circle, so $r^2=x^2+y^2\ +3\pi $ produces same circl... | The equation $2^2=x^2+y^2+\pi$ is equivalent to $x^2+y^2=4-\pi=0.8584\ldots$, which produces a circle of radius $\sqrt{4-\pi}=\sqrt{0.8584\ldots} = 0.9265\ldots$, which is the red circle in the top figure.
The equation $2^2=x^2+y^2+3\pi$ is equivalent to $x^2+y^2=4-3\pi=-5.4247\ldots$. Since that's a negative number, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to prove the two identities in number theory are equivalent? Let
\begin{align}
Li_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}.
\end{align}
There are two different forms of Abel identities for polylogarithms:
1.
\begin{align}
& Li_2(-x) + \log x \log y \\
& + Li_2(-y) + \log ( \frac{1+y}{x} ) \log y \\
& + Li_2(-... | My comment is a bit too long therefore I write it as an answer.
Thanks for the link, it makes the question much clearer. With $(6.63)$ follows directly $2.$ by the substitution $x\to\frac{x}{1-y}$ and $y\to\frac{y}{1-x}$ . The script uses the symbol-technique $S$ to transform the terms of $(6.63)$ into other terms gett... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | $(n^3 - 1)n^3(n^3 + 1) =n^3(n^6 -1)$
Since $\phi(7)=\phi(9)=6$, we get that $7$ and $9$ always divide $n^3(n^6 -1)$, by Euler–Fermat.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Finding null space of matrix. I need to make sure I'm understanding this correctly.
I skipped a few steps to reduce typing, but let me know if I need to clarify something.
Question asks:
Find $N(A)$ for $A$ = \begin{bmatrix}
-3 & 6 & -1 & 1 & -7 \\
1 & -2 & 2 & 3 & -1\\
2 & -4 & 5 & 8 & -4 \\
... | As mentioned in the comments, provided your arithmetic is accurate, this is the correct response. The idea behind the null space of a matrix is that it is precisely those vectors in the domain being sent to the $\mathbf{0}$ vector in the codomain. So, what you have (correctly) done, is determined the solution set of $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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How to find all solutions for $x^3=6x+6$ Could anyone help me to find how to find all solution for $x^3=6x+6$?
| Set $x=ay$; the equation becomes $a^3y^3-6ay=6$. We want that
$$
\frac{a^3}{6a}=\frac{4}{3}
$$
so we can take $a=2\sqrt{2}$. Then we have
$$
16\sqrt{2}y^3-12\sqrt{2}y=6
$$
that becomes
$$
4y^3-3y=\frac{3}{4}\sqrt{2}>1
$$
Now set $y=\cosh z$, so the equation becomes
$$
4\cosh^3z-3\cosh z=\frac{3}{4}\sqrt{2}
$$
that is,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve this trigonometric equation Plese help to solve this equation:
$$ \sin x=2\sin20^{\circ}\sin\left(170^{\circ}-x\right)$$
I tried to convert this equation:
$$\sin x=2\sin20^{\circ}\left(\sin170^{\circ}\cos x-\cos170^{\circ}\sin x\right)$$
$$\sin x\left(1+2\sin20^{\circ}cos170^{\circ}\right)=2\sin20^{\circ}\... | $\sin170^\circ=\sin(90+80)^\circ=\cos80^\circ(?)$
$\cos170^\circ=\cos(90+80)^\circ==-\sin80^\circ(?)$
$$\implies\frac{2\sin(20^{\circ})\sin(170^{\circ})}{1+2\sin(20^{\circ})\cos(170^{\circ})}=\dfrac{2\sin20^\circ\cos80^\circ}{1-2\sin20^\circ\sin80^\circ}$$
Now, $1-2\sin20^\circ\sin80^\circ=1-(\cos60^\circ+\cos80^\circ)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$ without L'hopital
Question:
Let
\begin{align}
&S(t):=\int_{\pi/4}^t (\sin t-\sin\left(\frac\pi4\right))dt\\
&T(t):=\frac{\left(\sin t-\sin\left(\frac\pi4\right)\right)\left(t-\frac\pi4\right)}2\\
\end{align}
Using $$\lim_{t\to0}\... | Based on your calculations, the given limit and the sum to product formulae,
\begin{align}
& \lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)} \\
=&\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac
t2}{\left(\sin\left(t+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Limit of a sequence of fractions I have that:
$$S_n=\frac{b\cdot S_{n-1}}{b+S_{n-1}}+a$$
where $n\in\mathbb{N}^+$ and $S_1=a+b$.
So, for $S_2$ we will get:
$$S_2=\frac{b\cdot S_{2-1}}{b+S_{2-1}}+a=\frac{b\cdot S_1}{b+S_1}+a=\frac{b\cdot(a+b)}{b+(a+b)}+a$$
And for $S_3$:
$$S_3=\frac{b\cdot S_{3-1}}{b+S_{3-1}}+a=\frac{b\... | Assume that $S_n = \frac{p_n}{q_n}$ is associated with $(p_n,q_n)\in\mathbb{R}^2$. The recurrence
$$ S_{n} = \frac{(a+b)S_{n-1}+ab}{S_{n-1}+b}\tag{1} $$
can be written in the following form
$$ \begin{pmatrix}p_n \\ q_n \end{pmatrix} = \begin{pmatrix} a+b & ab \\ 1 & b\end{pmatrix} \begin{pmatrix}p_{n-1} \\ q_{n-1} \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $CE=2AD$ I already have a proof, but if you can please give another:
Let $ABC$ be an isosceles triangle with $AB=AC$ and point $D$ be on segment $AB$. The line perpendicular to $AB$ which passes $D$ intersects $BC$ (extended) and $AC$ at $E$ and $F$ respectively. $C$ is on segment $BE$, between $B$ and $E$.... | $\dfrac{CF}{FA}\cdot\dfrac{EF}{FD} = 2$ and $\dfrac{DF}{FE}\cdot \dfrac{EC}{CB}\cdot \dfrac{BA}{AD} = 1.$ This implies that $\dfrac{CE}{AD} = 2\dfrac{BC}{BA}\cdot\dfrac{AF}{FC}$. But $\dfrac{AF}{FC} = \dfrac{AD}{DB}\cdot \dfrac{BE}{CE} = \dfrac{BE}{BD}\cdot\dfrac{AD}{CE} = \dfrac{2AB}{BC}\cdot \dfrac{AD}{CE}$. Multiply... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence m... | Consider $(a+b+c)^2$
Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$
Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.
and since squares of odd is odd and evens is even $a+b+c$ has to be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 1
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Factorising Non-Monic Quadratic Equations I am trying to find a way to factorise a non-monic quadratic using a system of linear equations. I know there exist various algorithms which can help with this, however I'm not fond of such methods.
I first started by creating a general equation with 4 unkown variables excludi... | You can solve it in 2 ways:
FIRST WAY (longer version):
$6x^2-19x+15=0$
$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}=\frac{-(-19)\pm\sqrt{(-19)^2-4\cdot 6\cdot 15}}{2\cdot 6}$
$x=\frac{19\pm\sqrt{361-360}}{12}=\frac{19\pm\sqrt{1}}{12}=\frac{19\pm 1}{12}$
$x_1=\frac{19+1}{12}=\frac{20}{12}=\frac{5}{3} \quad \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Equation of circles tangent to $x$-axis with a given radius and point Find the equations of the circles which are tangent to the x-axis, with radius of 5 units and passing through the point $(0,8)$.
I know how to formulate an equation with the given radius of 5 and then substitute the point $(0,8)$, but then I get stuc... | The equation of circle is $(x-a)^2+(y-b)^2=r^2$ centered ar point $(a,b)$.
The equation of line is $y=mx+c$. The x axis is the line $y=0$.
All circles having tangent to x axis implies it have point $y=0$ and only one solution for x.
$(x-a)^2+b^2=5^2$
$x^2-2ax+a^2+b^2-25=0$
We want solutions for x.
We thus require the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Critical points of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ Domain: $0 \le x<\pi$ and $0 \le y<\pi$
After setting the gradient of $f(x,y)$ equal to zero, I obtained the following system:
$\cos(x)=\cos(x+y),$ $\cos(y)=\cos(x+y)$
I am not sure how to solve this system but what I tried was:
On the given domain, the first equat... | Setting the gradient to zero yields
$$
\begin{cases}
\cos x - \cos(x+y)=0 \\[6px]
\cos y - \cos(x+y)=0
\end{cases}
$$
Therefore $\cos x=\cos y$. If we are interested in the whole plane, this means
$$
y=x+2k\pi
\qquad\text{or}\qquad
y=-x+2k\pi
$$
With the second set of solutions we get $x+y=2k\pi$, so $\cos(x+y)=1$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Co-ordinates of the vertices an icosahedron relative to its centroid.
This is a picture of an icosahedron. I need to know the coordinates of the vertices of the icosahedron relative to it's centroid in order to programme a projection of one on a three dimensional plane. (By the way, it has twelve vertices, so it is go... | PolyhedronData["Icosahedron", "VertexCoordinates"]
$$
\left(
\begin{array}{ccc}
0 & 0 & -\frac{5}{\sqrt{50-10 \sqrt{5}}} \\
0 & 0 & \frac{5}{\sqrt{50-10 \sqrt{5}}} \\
-\sqrt{\frac{2}{5-\sqrt{5}}} & 0 & -\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
\sqrt{\frac{2}{5-\sqrt{5}}} & 0 & \frac{1}{\sqrt{10-2 \sqrt{5}}} \\
\frac{1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Can't calculate the integral $\int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)\,dx$ Let $I(a,b):= \int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)\,dx$
Calculate $I(a,b)$.
My attempt:
Define function $F(a,b,x)$ as following $\frac{\partial f}{\partial a}F(a,b,x)=\ln(a^2\cos^2x+b^2\sin^2x)$
Then $I(a,b):=
\int_0^{\pi/2}\ln(a^2\cos^2x+... | Hint. One may set
$$
f(s):=\int_0^{\pi/2}\ln(s+\sin^2 x)dx, \qquad s\geq0.
$$
Then differentiating under the integral sign with respect to $s$ you get
$$
\begin{align}
f'(s)&=\int_0^{\pi/2}\frac1{s+\sin^2 x}dx\\\\
&=\int_0^{\infty}\frac1{s+\dfrac{t^2}{t^2+1}}\dfrac{dt}{t^2+1}\quad (t=\tan x)\\\\
&=\int_0^{\infty}\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to integrate fractions of polynomials with complex roots $$\int\frac{4x^3-2x^2+60x+10}{x^4+30x^2+125} dx $$
The denominator has no rational roots. It can be factored as $(x^2+5)(x^2+25)$ which gives roots of $\pm i \sqrt{5}$ and $\pm 5i$.
How can these complex roots be used to integrate the function?
| First split up the integral into two parts and then use the natural logarithm and arctan.
$$\begin{align*} P = &\frac{4x^3-2x^2+60x+10}{x^4+30x^2+125} \\
= &\frac{4x^3-2x^2+60x+10}{(x^2+25)(x^2+5)} \\
= &\frac{2x-3}{x^2+25} + \frac{2x+1}{x^2+5}\end{align*}$$
So $\int P \,\mathrm{d}x = \int \frac{2x-3}{x^2+25}\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Envelope of isoperimetric ellipses Find envelope of ellipses $ \frac{x^2}{a^2} + \frac{y}{b^2}=1 $ which have $a,b$ and its associated eccentricity $e$ variable while holding its perimeter
$$ p= 4 a E(e) $$
as constant.
Expected to be asteroid-like, passing through
$$
\begin{pmatrix} 0 \\ \frac{\pi a}{2} \end{pmatrix}... | \begin{align*}
p &= 4a E(k) \\
a(k) &= \frac{p}{4E(k)} \\
F(x,y,k) &=
x^2+\frac{y^2}{1-k^2}-\frac{p^2}{16E^2(k)} \\
\frac{\partial F}{\partial k} &=
\frac{2ky^2}{(1-k^2)^2}-
\frac{p^2}{8E^3(k)} \left[ \frac{K(k)-E(k)}{k} \right]
\end{align*}
The envelope is given by $$F=\dfrac{\partial F}{\partial k}=0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Maximum/Minimum Find the Max/Min values:
$f(x)=x^3-x^2-8x+1$ on $[-2,2]$
$$\begin{align}f'(x)&=3x^2-2x-8\\&=3x^2+4x-6x-8\\&=x(3x+4)-2(3x+4)\\&=(x-2)(3x+4)\end{align}$$
Thus, $x=-2,0.75,2$
Now sub these into the original function
$f(2)=2^3-2^2-8(2)+1=-11$
$f(0.75)=0.75^3-0.75^2-8(0.75)+1=-5.14$
$f(-2)=(-2)^3-(-2)^2-8(-2... | You've got it right so far- we wish to solve $5x^4+1=0$, and this will give us the location of the maximum and minimum of the function.
But consider, if we simplify $5x^4+1=0$, we get $x^4=-1/5$... This obviously has no real solutions, and so there is no point at which a maximum or minimum can exist.
However, we are gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Largest square that can fit into region. Let $R$ be the region bounded by the lines $y=cx+b$ and $y=cx+d$. Find the sides of largest square, with sides parallel to the $x$ and $y$ axis in $R$.
Here is what I know. The distance between the two lines is $\frac{|b-d|}{ \sqrt{c^2+1}}$. Without the requirement, this is the ... | Suppose without loss of generality that $b> d$, so that line $l_1: y=cx+b$ lies above line $l_2:y=cx+d$. Moreover, in this answer we will assume that $c>0$, but the solution can be easily adapted to the case $c<0$. The case $c=0$ is trivial.
Choose any point on $l_1$, say $A=(0,b)$. Our square will be $ABCD$. We will f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limits at infinity, in terms of another function Suppose:
$$\lim_{x\to\infty}\; f(x) + 2x^2 =1$$
Determine if possible:
$$\lim_{x\to\infty}\; \frac{f(x)}{x^2+1}$$
I determined the answer to be $-2$, through substituting for $f(x)$:
$$\frac { 1 - 2x^2 }{x^2 + 1}$$
and I solved for the infinite limit.
however I feel li... | I think it's okay, but you're skipping a lot of steps that may not work in other problems. To be thorough (and to see exactly what's going on), I'd write it like so:
\begin{align}
\frac{f(x)}{x^2+1} &= \frac{f(x)+2x^2-2x^2}{x^2+1} \\
&= \frac{f(x)+2x^2}{x^2+1} - \frac{2x^2}{x^2+1}
\end{align}
By the laws of limits, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sixth degree polynomial problem If the graph of $$y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2$$ always lies above the line $y = bx + c$, except for $3$ points where the curve intersects the line.
What is the largest value of $x$ for which the line intersects the curve?
*
*A) 4
*B) 5
*C) 6
*D) 7
Through general idea ... | As there are three double roots, $$x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2-bx-c$$ is a perfect square, and we can evaluate its square root.
Looking at the first two terms,
$$x^6-10x^5\leftrightarrow(x^3-5x)^2=x^6-10x^5\cdots$$
Next
$$x^6-10x^5+29x^4\leftrightarrow(x^3-5x^2+px)^2=x^6-10x^5+(2p+25)x^4\cdots$$ so that $p=2$.
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to see if this functions are linearly independent?
Is this set linearly independent? $A=\left\{ \arctan(x), \arctan(2x), \arctan\big(\frac{3x}{1-2x^2}\big)\right\}$
I've tried using the Wronskian
$$W(A)= \left| \begin{array}{ccc}
\arctan(x) & \arctan(2x) & \arctan\big(\frac{3x}{1-2x^2}\big) \\
\frac{1}{1+x^2} & \... | Hint:
$$\arctan\frac{\tan a+\tan b}{1-\tan a\tan b}=a+b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Make $\$100$ by taking $\$1$, $\$5$, and $\$10$... but we can take only $21$ notes I am weak in mathematics, but I need to know if this is possible. I will take $\$100$ from my friend, but he will give me only $21$ notes of $\$1$, $\$5$, $\$10$. I need to tell him the numbers of each notes that will make $100$ dollars.... | Expanding the answer by @user3558 , we can show, that these two solutions are the only solutions.
We have:
$$\begin{cases}x,y,z \geq 0 \\
x,y,z \in \mathbb{Z} \\
z=t\\
y=19+\frac{3-9t}{4}\\
x=2-\frac{3-5t}{4}
\end{cases}$$
We want $y$ to be integer, so if $z$ is an integer, then we have
$3-9t \equiv 0 \mod 4$
$t \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. Prove that $abcd > a + b + c + d + 8$. Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. How to prove that $abcd > a + b + c + d + 8$? I've tried using AM-GM inequality but with no outcome. ... | It suffices to check the case $a, b, c, d>0$. Indeed, $abcd>0$, and so the problem becomes strictly stronger upon the transformation $(a, b, c, d)\to (|a|, |b|, |c|, |d|)$ if any of the variables are negative.
Now let $\frac{a+b+c+d}{4}=t$. Then $t^4\ge abcd>a^2+b^2+c^2+d^2\ge 4t^2$ due to AM-GM and Cauchy Schwarz, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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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.