Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
generalized binomial coefficient I know that the coefficient $$-\frac{1}{2} \choose k$$ can be simplified by multiplying both the nominator and the denominator by $$2^k$$ and then represented as $$ (-\frac{1}{4})^k {2k\choose k}$$
But what if the upper index is $$\frac{1}{3}$$it seems that our trick do not work anymore... |
There is no nice generalisation from $\binom{-\frac{1}{2}}{k}$ to $\binom{\pm \frac{1}{n}}{k}, 0\leq k \leq n$ available.
We obtain for $n=2$:
\begin{align*}
\binom{-\frac{1}{2}}{k}&=\frac{1}{k!}
\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{1}{2}-(k-1)\right)\\
&=\frac{1}{k!}\frac{(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$? Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$?
*
*$x^8+x^4+1$
*$x^4+1$
*$x^4-x^2+1$
*$x^5-x^4+x^3-x^2+x-1$
Is the answer $(1)$ correct? I am not sure which one is.
| By repeatedly using the simple factorizations:
$$x^{2n}-1=(x^n+1)(x^n-1)$$
and
$$x^{3n}-1=(x^{2n}+x^n+1)(x^n-1)$$
and
$$x^{3n}+1=(x^{2n}-x^n+1)(x^n+1)$$
We get:
$$x^{12}-1=(x^4-x^2+1)(x^2+1)(x^2-x+1)(x+1)(x^2+x+1)(x-1)$$
Ruling out every choice except for number 3:
$$x^4-x^2+1$$
Call this $p(x)$. There are many ways to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Laurent series in various regions I have the question: "Find the Laurent series which represents the function
$$
f(z) = (z^2 - 1)/(z + 2)(z + 3)\
$$
in the regions
(i) $\mid z\mid < 2\ $
(ii) $ 2 < \mid z\mid < 3\ $
(iii) $\mid z\mid > 3\ $"
I can rewrite this function into partial fractions as:
$$
f(z) = 5/(z+2) - 1... | (i) $|z| < 2$
$$\frac5{z+2}-\frac{10}{z+3} = \sum_{n=0}^\infty (-1)^n \left(\frac52\left(\frac{z}2\right)^n-\frac{10}3\left(\frac{z}3\right)^n\right)$$
(ii) $2<|z|<3$
\begin{align*}
& \frac{5}{z+2} - \frac{10}{z+3} \\
=& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{3} \frac{1}{1+\frac{z}{3}} \\
=& -\frac{10}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing an infinite trigonometric sum $\sum \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$ Let $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$
I'm trying to compute this sum by understanding it as an integral kernel. This question comes from Dym and Mckean Fourier Series and Integrals Ex... | Marty Cohen is correct in his approach to this $G(x,y)$ but there is the sign error in the trig expressions.
Defined on the interval $0 < t < P$, it is fairly easy to demonstrate that the function:
\begin{equation}
h(t)=\pi^2 \left[ \left( \frac{t}{P} \right)^2 -\left( \frac{t}{P} \right) + \frac{1}{6} \right]
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many rectangles are there on an $8 \times 8$ checkerboard?
How many rectangles are there on an $8 \times 8$ checkerboard?
\begin{array}{|r|r|r|r|r|r|r|r|}
\hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \... | First, the easy way to count them is to notice that each rectangle is completely determined by its top and bottom edges and its left and right edges. Pick any two of the nine horizontal lines and any two of the nine vertical lines, and you’ve picked out a rectangle. Conversely, each rectangle determines two horizontal ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Positive Integer solutions to $y = \frac{x z}{-x - z + x z}$ I'm trying to find positive integer solutions to the following diophantine equation:
$$y = \frac{x z}{-x - z + x z}$$
The first thing I did was split the fraction as follows:
$$ \frac{x z}{-x - z + x z} = 1 + \frac{x+z}{x z-x-z}$$
This doesn't seem to help an... | If $-x-z+xz=0$, then $(x-1)(z-1)=1$, so $(x,z)=(2,2)$, which doesn't give a solution.
If $-x-z+xz\neq 0$, then the equation is equivalent to $xy+yz+zx=xyz$, i.e. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Let wlog $x\ge y\ge z\ge 1$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\le \frac{3}{z}$, so $z\le 3$.
If $z=1$, then... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with a residue calculation I have a problem with calculating a residue. I want to calculate Res($f,\frac{1}{2}$), where
$$ f(z) = \frac{z^6 +1}{z^3(2z-1)(z-2)} = \frac{z^6 +1}{2z^5 -5z^4+2z^3}.$$
$$ $$
One method is to view $f(z)$ as $\frac{p(z)}{q(z)}$, where both $p$ and $q$ are analytic; here the residue o... | $\text{Res}(f,\frac 12)=\lim_{z \to \frac 12}(z-\frac 12)f(z)=\lim_{z\to \frac 12} \frac {(z-\frac 12)(z^6+1)}{(z^3)(z-2)(2z-1)} =\lim_{z\to \frac 12}\frac {(z-\frac 12)(z^6+1)}{2(z^3)(z-2)(z-\frac 12)}=\lim_{z\to \frac 12} \frac {z^6+1}{2(z^3)(z-2)}=-\frac {65}{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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minimum $x^2 + y^2$ on $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1 $ ellipse Given $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
T... | Another, maybe not as elegant, way is to solve for $y$, getting $$y=-5\pm\frac54\sqrt{16-\left( x -12\right)^2}$$ and then, since $$y_+(x):=\left(-5+\frac54\sqrt{16-\left( x -12\right)^2}\right)^2\le\left(-5-\frac54\sqrt{16-\left( x -12\right)^2}\right)^2=:y_-(x), $$ find the zero $x_0$ of the derivative of $x^2+(y_+(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
$$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to pr... | Say $$n \equiv 0,\pm1,\pm2 \pmod 5$$ or,
$$n^2 \equiv 0,1,4 \pmod 5$$ or,
$$n^6 \equiv 0,1,64 \pmod 5$$ or,
$$n^6 \equiv 0,1,-1 \pmod 5$$
Therefore, $5$ divides at least one of $n^6,n^6-1$ or $n^6+1$, that is, $5$ divides $n^6(n^6-1)(n^6+1)$ .
And
$n^{19}-n^7=n\cdot n^6(n^{12}-1)=n\cdot n^6(n^6-1)(n^6+1)$
Hence give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How does the Schwarz inequality prove the following? I don't quite see how the Schwarz inequality proves that, for
$$f_n(x) = \frac{x}{1+nx^2}$$
we have
$$|f_n(x)| \leq \frac{|x|}{2\sqrt{n}|x|} = \frac{1}{2\sqrt{n}}.$$
| Consider the vectors $a = (1,x\sqrt{n})$ and $b = (x\sqrt{n},1)$. By the Cauchy-Schwarz Inequality:
$$|a \cdot b| \le \|a\| \cdot \|b\|$$
$$|1 \cdot x\sqrt{n} + x\sqrt{n} \cdot 1| \le \sqrt{(1)^1+(x\sqrt{n})^2} \cdot \sqrt{(1)^1+(x\sqrt{n})^2}$$
$$|2x\sqrt{n}| \le 1+(x\sqrt{n})^2$$
$$2\sqrt{n}|x| \le 1+nx^2$$
Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$ I'd like to get an asymptotic expansion of $\int_0^1 \frac{x^n}{1+x} \, dx$ at order two in $\frac{1}{n}$.
I'm able to prove that $$\lim\limits_{n \to \infty} n \int_0^1 \frac{x^n}{1+x} \, dx = \frac{1}{2}$$ which provides an asymptotic expansion at or... | Method 1.
An elementary approach.
You may just integrate by parts twice,
$$
\begin{align}
I_n&=\int_0^1\frac{x^n}{1+x}\:dx
\\&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{1+x}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\frac{1}{n+1}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.
It looks like AM-GM should be used here but the square roots make it difficult. So maybe Cauchy-Schwarz works?
| Note that (by AM-GM),
$\frac{x^2+(y^2+z^2)}{2}\geq\sqrt{x^2(y^2+z^2)}=x\sqrt{y^2+z^2}$.
and similarly,
$\frac{y^2+(x^2+z^2)}{2}\geq\sqrt{y^2(x^2+z^2)}=y\sqrt{x^2+z^2}$.
Summing them up gives:
$x^2+y^2+z^2\geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sum $1^k+2^k+\cdots+n^k$ where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$.
Prove that the sum $$1^k+2^k+\cdots+n^k$$
where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$.
Question
In the solution to this problem it splits it up into two case... | Using Proof of $a^n+b^n$ divisible by a+b when n is odd,
$$r^k+(n-r)^k$$ is divisible by $r+n-r=n$ as $k$ is odd
$$\implies\sum_{r=1}^n(r^k+(n-r)^k)$$ will be divisible by $n$
Similarly,
$$\sum_{r=1}^n(r^k+(n+1-r)^k)$$ will be divisible by $r+n+1-r=n+1$
$$\implies\sum_{r=1}^n(r^k+(n+1-r)^k)=2\sum_{r=1}^n r^k$$ will b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$... | To prove:
$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq \frac{9}{a+b+c}$,
let $x=a+b, y=b+c, z=c+a$, then we have:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Rightarrow 3+(\frac{y}{x}+\frac{x}{y}) +(\frac{y}{z}+\frac{z}{y})+ (\frac{z}{x}+\frac{x}{z}) \geq 3+2+2+2=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Algebra: Rational roots of a polynomial of degree $4$ Consider the following polynomial with real coefficients: $x^4(t^5 - 1) - x^3(1+4t^5) + x^2(6t^5 -1) - x(1+4t^5) + (t^5 -1 ) = 0$, where $x$ and $t$ are both rational and $t$ is fixed.
By simple algebra, we find that the sum of the roots is $\dfrac{1+4t^5}{1-t^5}$ a... | Let $x$ be a rational solution to the equation. We can easily check that $x \not = 1$ and $t \not = 0$. Then
$$\begin{array}{c c c c}
&x^4 (t^5-1) - x^3 (4t^5+1) + x^2 (6t^5-1) - x (4t^5+1) + (t^5-1)&=&0 \\
\Leftrightarrow&t^5 (x^4 - 4x^3 + 6x^2 - 4x + 1) - (x^4 + x^3 + x^2 +x + 1) &=&0\\
\Leftrightarrow&t^5 (x-1)^4&=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591131",
"timestamp": "2023-03-29T00:00:00",
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Help with $\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$ I need help with differential equation $$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
I know that $y’=\frac{dy}{dx}$, but i have no idea what to do next.
| $$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
We know that $$y’=\frac{dy}{dx}$$
$$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) \frac{dy}{dx}=0
-\tan(x)\sin^2(y)=\cos^2(x)\cot(y)\frac{dy}{dx}
-\frac{\tan(x) dx}{\cos^2(x)}=\frac{\cot(y) dy}{\sin^2(y)} $$
We also know that $$ \tan(x)= \frac{\sin(x)}{\cos(x)} $$ and $$ \cot(x)=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591858",
"timestamp": "2023-03-29T00:00:00",
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | Suppose it's true for $n\ge1$: then
$$
5^n+2\cdot3^{n-1}+1=8k
$$
for some integer $k$; in particular, $5^n=8k-2\cdot3^{n-1}-1$. Then
\begin{align}
5^{n+1}+2\cdot3^{n}+1
&=5(8k-2\cdot3^{n-1}-1)+2\cdot3^{n}+1 \\[3px]
&=40k-10\cdot 3^{n-1}-5+6\cdot 3^{n-1}+1\\[3px]
&=40k-4\cdot 3^{n-1}-4
\end{align}
Thus you just need to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.
For every positive integer $n$ set $H_n = \dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$. Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.
Attempt
I will prove this result by induction. The base case holds since for $n = 2$ we have $(1+2)... | Notice that $$n+1 = \frac{2}{1}\frac{3}{2} \cdots \frac{n}{n-1}\frac{n+1}{n} $$
Hence by AM-GM we get
$$ \left (n+1\right )^{\frac{1}{n}} < \frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n} $$
But observe that
$$\frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n}= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?
| $$\begin{align}
\frac{2}{3}
&>
-4y-\frac{25}{3} \\
3\left(\frac{2}{3}\right)
&>
3\left(-4y-\frac{25}{3}\right) \\
3·\frac{2}{3}
&>3·\left(-4y\right)-3·\frac{25}{3} \\
\require{cancel} \cancel{3}·\frac{2}{\require{cancel} \cancel{3}}
&>
3·\left(-4y\right) -\require{cancel} \cancel{3} ·\frac{25}{\require{cancel} \cancel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | $$\int \frac{1}{2x^2+x+3}\, dx=\frac{1}{\sqrt{2}}\int \frac{d\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}\, $$
Let $\sqrt{2}x+\frac{1}{2\sqrt{2}}=\sqrt{\frac{23}{8}}\tan u$.
$$=\frac{1}{\sqrt{2}}\int \frac{\sqrt{\frac{23}{8}}\frac{1}{\cos^2 u}}{\frac{23}{8}(\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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find the tangent to the sphere obtain the equations of tangent to sphere
$$x^2+ y^2+z^2+6x-2z+1 = 0$$
which pass through the line
$$3 (16-x) = 3z=2y+30$$
Now I know if the plane is $$lx +my+n z=p$$
then $$-I/3 +m/2+n/3=0$$
also $(16,-15,0)$ is a point on the plane
I know that there is $2$ answers , but how to proceed
| The sphere has for equation $$S \equiv (x+3)^2-9+y^2+(z-1)^2-1+1=(x+3)^2+y^2+(z-1)^2-9=0$$ which means that its center is $C= (-3,0,1)$ and its radius is equal to $r=3$. The line has for equations $$L \equiv \begin{cases}
x+z=16\\
2y-3z=-30
\end{cases}$$ It goes through the point $P_L=(16,-15,0)$ and has $v_L=(-2,3,2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Integrate $\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$ How do I go about integrating:
$$\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$$
The common trigonometric substitutions don't seem to work here.
I think it requires to take some power of $x$ outside the square root but I am not able to solve fu... | The answer is (after a request from OP) updated with more details
I suggest you to set
$$
u=\sqrt{\frac{1-x^2}{1+x^2}}.
$$
Then
$$
x^2=\frac{1-u^2}{1+u^2}
$$
and so
$$
2x\,dx=-\frac{4u}{(1+u^2)^2}\,du.
$$
Thus
$$
\begin{aligned}
\int \frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,dx&=\int\frac{1}{2x^2}\sqrt{\frac{1-x^2}{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
To solve differential equation ($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$ The ODE is
($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$
I have tried everything like integrating factor,it is not homogenous and not linear differential equation..What should be done now?
| Re-arranging your differential equation, we have
$$\frac{dx}{dy}=xy^3+x^2y^7$$
$$\frac{dx}{dy}-xy^3=x^2y^7$$
$$\frac{1}{x^2}\cdot \frac{dx}{dy}-\frac{1}{x}\cdot y^3=y^7$$
$$-\frac{d}{dy}\left(\frac{1}{x}\right)-y^3\cdot \left(\frac{1}{x}\right) =y^7$$
Put $u=\frac{1}{x}$
You get $$\frac{du}{dy}+uy^3=y^7$$
The integrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \... | $A$ can be thought as $e^{-i\theta}$. So that for the natural homeomorphism $\phi:\mathbb{C}\rightarrow \mathbb{R}^2:x+iy \mapsto (x,y)$, $(\phi^{-1}\circ A\circ \phi )(z)=e^{-i\theta } z$. Hence, $(\phi^{-1} \circ A^{2010} \circ \phi)(z)= e^{- 2010 i \theta} z = e^{-60\pi i}z=z$. Hence, the answer is (B).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
| Note that, for each fixed $k$, $\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)=\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)$. Then:
\begin{eqnarray*}
\sum_{k=1}^n\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)&=&\sum_{k=1}^n\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
probability of coloring a $3\times 3$ table with two colors such that no $2\times 2$ square exists imagine we have a $3\times3$ table ( or $3\times3$ square ) and we want to color each place of that $9$ places with two colors ( red and blue ).
find the probability that no $2\times2$ square exists after coloring place... | There are $2^9$ ways to color the square. We now want to count how many ways there are to make a $2\times 2$ square of the same color.
Pick any $2\times 2$ square. There are $2^5$ ways to color the squares not inside the square we chose. The square we chose can be colored in two ways, so we get $2^6$ colors that resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | If you want to use the Maclaurin series for $\cos$ in order to obtain the first few terms of the Maclaurin series for $\cos^6$ proceed as follows: Start with
$$\cos x=1-{x^2\over2}+{x^4\over24}+?x^6=1-{x^2\over2}\left(1-{x^2\over12}+?x^4\right)\ .$$Using the binomial theorem you then obtain
$$\cos^6 x=1-{6\choose 1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Combinatorial proof that $\frac{({10!})!}{{10!}^{9!}}$ is an integer I need help to prove that the quantity of this division :
$\dfrac{({10!})!}{{10!}^{9!}}$
is an integer number, using combinatorial proof
| Here's a simple non-combinatorial proof. Still eager to see a truly combinatorial one. The denominator contains only powers of $2$, $3$, $5$, and $7$:
$$
(10!)^{9!}=(10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2)^{9!}=\left(7\cdot 5^2 \cdot 3^4 \cdot 2^7\right)^{9!}=7^{9!}5^{2\cdot 9!}3^{4\cdot 9!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Incorrect cancellation of fraction with correct answer If you cancel (incorrectly) the 6 in $\frac{26}{65}$, you get the (correct) equation $\frac{26}{65} = \frac{2}{5}$. What other fraction exhibits this property?
I tried considering only the simple case where the fraction is of the form $\frac{10a+n}{10n+b}$ (as in t... | We'll solve $$\frac{10 a + n}{10 n + b} = \frac{a}{b},$$ or equivalently $n (10 a - b) = 9 a b$; to interpret the l.h.s. as a ratio of $2$-digit numbers, we restrict to solutions for which $1 \leq a, b, n \leq 9$.
Reducing modulo $9$ reduces $n(10 a - b) = 9 ab$ to $$n (a - b) \equiv 0 \pmod 9.$$ In particular $n$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
geometric meaning of complex cubic polynomial coefficients A complex cubic polynomial arranged as $f(z)=z^3-3az^2+3b^2z-c^3$ with coefficients $a,b,c\in\mathbb{C}$ can be represented as the product of (unknown) factors $f(z)=(z-p)(z-q)(z-r)=z^3-3\left(\frac{p+q+r}{3}\right)z^2+3\left(\sqrt \frac{pq+qr+rp}{3}\right)^2z-... | Vieta's Formulae
\begin{align}
f(z) &= z^3-3az^2+3b^2z-c^3 \\
&= (z-p)(z-q)(z-r) \\
a &= \frac{p+q+r}{3} \\
b^2 &= \frac{pq+qr+rp}{3} \\
c^3 &= pqr
\end{align}
Foci of Steiner ellipse
\begin{align}
f'(\lambda) &=0 \\
\lambda &= a\pm \sqrt{a^2-b^2} \\
&= \frac{p+q+r \pm \sqrt{(p+q+r)^2-3(pq+qr+rp)}}{3}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x \to 1} \frac{\ln(x^3)}{x^2-1}$ My first idea was to use the definition of derivative (but it's giving me the wrong answer).
$\displaystyle \lim_{x \to 1}\frac{\ln(x^3)}{x^2-1} =\lim_{t \to 1}\frac{\ln(t^{3/2})}{t-1} = \frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)}{t} =\frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)-\ln(1)}{t... | Your error is that
$$\frac{d}{dx} \ln x \big|_{x = 1} = \frac 1 x \big|_{x = 1} = 1$$
Here's a slightly different way that doesn't involve making a substitution before applying the definition of the derivative:
\begin{align*}
\lim_{x \to 1} \frac{\ln(x^3)}{x^2 - 1} &= \lim_{x \to 1} \frac{3}{x + 1} \frac{\ln x}{x - 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Euler's theorem for congruences Using Euler's theorem, how to compute:
a) $3^{1000} \pmod{2500}$
b) $7^{1001} \pmod{2500}$
c) $101^{21600} \pmod{81000}$
| First note that $(3,2500)=1$ and then compute $\phi(2500)=5^3 \cdot 2 \cdot 4=1000$ .
Now Euler's theorem tells us that :
$$3^{\phi(2500)} \equiv 1 \pmod{2500}$$ or :
$$3^{1000} \equiv 1 \pmod{2500}$$
In the same way $7^{1000} \equiv 1 \pmod{2500}$ so :
$$7^{1001} \equiv 7 \pmod{2500}$$
Finally : $$\phi(81000)=3^3 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $P$ be permutation matrix then $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}} * \\ 0 \\ \end{array}} \right)$
*
*Let $A\in M_n$ is nonnegative(all $a_{ij}\ge0$).
*$x\in C^n$ be eigenvector of $A$ with $r ≥ 1$ positive entries and $n − r$ zero entries .
*There is $P\in M_n$(permutation matrix) su... | No, this won't hold for general matrices $A$.
For instance, you can take $n=2$, and $x = \left( {\begin{array}{*{20}{c}}
0 \\
1 \\
\end{array}} \right)$. Then the permutation matrix you are looking for is $P = \left( {\begin{array}{*{c}{c}}
0 & 1 \\
1 & 0\\
\end{array}} \right)$, and $Px = \left( {\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqr... | Intersecting the dual conics is one way to go.
The dual conics are $$a^2X^2-b^2Y^2=1\quad (I)$$ and $$-b^2X^2+a^2Y^2=1\quad(II).$$ $b^2(I)+a^2(II)$ gives $(-b^4+a^4)Y^2=b^2+a^2$ or $(a^2-b^2)Y^2=1$. Substituting back to get the corresponding $X$s, we get the four intersection points to be $(X,Y)=(\frac1{\sqrt{a^2-b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An application of Bernoulli's Inequality - relevant to an expression for $e^{x}$ I saw on another post the following proposition.
\begin{equation*}
1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{n}}{n!}
≤ \left( 1 − \frac{x}{n} \right)^{−n}
\end{equation*}
for every real number $0 < x < n$. The member posting this propos... | Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} < \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$ with $x > 0$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit $\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$ I need to calculate limit:
$\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$
I was thinking of using the formula for $\sin \alpha - \sin \beta$, but what can be the next step?
| $$
\sin\sqrt{x+1}-\sin\sqrt{x}=
2\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}
\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}
$$
Now
$$
\lim_{x\to\infty}\frac{\sqrt{x+1}-\sqrt{x}}{2}=0
$$
and
$$
\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}
$$
is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Best Fit Line with 3d Points Okay, I need to develop an alorithm to take a collection of 3d points with x,y,and z components and find a line of best fit. I found a commonly referenced item from Geometric Tools but there doesn't seem to be a lot of information to get someone not already familiar with the method going. ... | We start with a sequence of $m$ measurements $\left\{ x_{k}, y_{k}, z_{k}, f_{k} \right\}_{k=1}^{m}$, and the trial function
$$
f(x,y,z) = a_{0} + a_{1} x + a_{2} y + a_{3} z.
$$
The linear system is
$$
\begin{align}
\mathbf{A} a &= f \\
%
\left[ \begin{array}{cccc}
1 & x_{1} & y_{1} & z_{1} \\
1 & x_{2} & y_{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing th... | Put $$a=10^9=729\cdot1371742+82\equiv 1+9^2\ (mod\space 9^3)$$
$$N=111111111=729\cdot 152415+576\equiv 576 \ (mod\space 9^3)$$
Let $M$ be the number so $$M=N(a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$$
Furthermore $$a\equiv 1+9^2\ (mod\space 9^3)$$ $$a^2\equiv 1+2\cdot9^2\ (mod\space 9^3)$$ $$a^3\equiv 1+3\cdot9^2\ (mod\space 9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
If $\frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Then prove that $ab\leq 0$
If $\displaystyle \frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Prove that $ab\leq 0$
$\bf{My\; Try::}$ I am Trying To solve it Using Inequality.
Using $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\f... | Trivially, $\sec^8 \theta \ge 1$ and $\tan^8 \theta \ge 0$ for all $\theta \in \mathbb{R}$ for which $\sec \theta$ and $\tan \theta$ are defined.
If $a > 0$ and $b > 0$. Then, $\dfrac{\sec^8 \theta}{a}+\dfrac{\tan^8 \theta}{b} \ge \dfrac{1}{a}+\dfrac{0}{b} = \dfrac{1}{a} > \dfrac{1}{a+b}$, a contradiction.
If $a < 0$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate complex determinant $$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(a + 1)}^2}}&{{{(a + 2)}^2}}&{{{(a + 3)}^2}}\\{{b^2}}&{{{(b + 1)}^2}}&{{{(b + 2)}^2}}&{{{(b + 3)}^2}}\\{{c^2}}&{{{(c + 1)}^2}}&{{{(c + 2)}^2}}&{{{(c + 3)}^2}}\\{{d^2}}&{{{(d + 1)}^2}}&{{{(d + 2)}^2}}&{{{(d + 3)}^2}}\end{array}} \right|
$$
It's ve... | By the fourth column (resp. the third column, the second column) subtract the third column (resp. the second column, the first column), you get
$\left|\begin{array}{cccc}
a^2 & 2a+1 & 2a+3 & 2a+5 \\
b^2 & 2b+1 & 2b+3 & 2b+5 \\
c^2 & 2c+1 & 2c+3 & 2c+5 \\
d^2 & 2d+1 & 2d+3 & 2d+5 \\
\end{array} \right|=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_{-\pi}^{\pi} (e^{ix} + e^{-ix})^n dx $ In an exercise following identity is used:
$$ \int_{-\pi}^{\pi} (e^{ix} + e^{-ix})^n dx = \begin{cases} 0, \hspace{2.1cm} n = 2m+1 \\
2\pi {2m \choose m}, \hspace{1cm} n=2m. \end{cases}, $$
Does anybody know how to prove this result or has some ideas to do so? ... | Notice, $e^{ix}+e^{-ix}=2\cos x$, hence one should have
$$\int_{-\pi}^{\pi}(e^{ix}+e^{-ix})^n\ dx=\int_{-\pi}^{\pi}(2\cos x)^n\ dx$$
$$=2^n\int_{-\pi}^{\pi}\cos^n x\ dx$$
since, $\cos(-x)=\cos x$,
$$=2\cdot 2^{n}\int_{0}^{\pi}\cos^n x\ dx$$
$$=2^{n+1}\int_{0}^{\pi}\cos^n (\pi-x)\ dx$$
Case-1: if $n=2m+1$ then $\cos^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $0 \leq ab + ac + bc - abc \leq 2.$
Let $a,b,$ and $c$ be nonnegative real numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$0 \leq ab + ac + bc - abc \leq 2.$$
I tried using rearrangement to get $a^2+b^2+c^2+abc = 4 \geq ab+bc+ac+abc$. Then I just need to show that $0\leq ab + ac + bc - abc$ and $4-2a... | I have a diferent solution for this question using a trigonometric substitution.Let:
$\\ \\ \displaystyle u=a\sqrt{\frac{2a+bc}{(2b+ac)(2c+ab)}};v=b\sqrt{\frac{2b+ac}{(2a+bc)(2c+ab)}};w=c\sqrt{\frac{2c+ab}{(2a+bc)(2b+ac)}} \\ \\$
This implies that:
$\\ \displaystyle a=2\sqrt{\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Find all integer solutions to $a+b+c|ab+bc+ca|abc$ As you can see from the title, I am trying to find all integer solutions $(a,b,c)$ to $$(a+b+c) \ \lvert\ (ab+bc+ca) \ \lvert\ abc$$ (that is, $a+b+c$ divides $ab+bc+ca$, and $ab+bc+ca$ divides $abc$). Unfortunately, I could not find anything on this problem (although ... | $a,b,c$ are roots of $X^3-(a+b+c)X^2+(ab+ac+bc)X=abc$.
Putting $$\begin{cases}a+b+c=A\\ab+ac+bc=mA\\abc=n(ab+ac+bc)\end{cases}\qquad (*)$$
one gets $$X^3-AX^2+mAX=mnA$$ it follows
$$\begin{cases}a^3-Aa^2+mAa=mnA\\b^3-Ab^2+mAb=mnA\\c^3-Ac^2+mAc=mnA\end{cases}$$
Hence $a^3\equiv b^3\equiv c^3\pmod A$ so that
$a^3+b^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Finding the infinite series: $3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots$? I'm trying to find the infinite sum that is defined by:
$$
3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots
$$
However, I do ... | For $|r|<1$ we have $$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}$$
Then, by taking derivatives respect to $r$
$$\sum_{k=0}^{\infty}kr^{k-1}=\frac{1}{(1-r)^2}\quad\implies\quad \frac{1}{r}+2+\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}$$
Then
$$\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}-\frac{1}{r}-2$$
Thus
$$\sum_{k=3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integers represented by $x^2 + 3 y^2$ vs. integers represented by $x^2 + x y + y^2$. How does one show that the quadratic forms $x^2 + 3 y^2$ and $x^2 + x y + y^2$ represent the same set of integers?
I think it relates to a classical result of Euler about primes of form $6k+1$. In fact a positive integer $n$ is of t... | Hint
$$x^2+xy+y^2=(x+\frac{1}{2}y)^2+3 (\frac{y}{2})^2=(\frac{1}{2}x+y)^2+3 (\frac{x}{2})^2= (\frac{x-y}{2})^2+3(\frac{x+y}{2})^2$$
Split it in three cases by parity.
Backwards you need to do the back substitutions, i.e. denote the brackets on the RHS by $x',y'$ and solve for $x, y$. Then you get relations of the form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let a,b be rationals and x irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$. I'm trying to solve the following problems:
*
*Let $a$,$b$ be rationals and $x$ irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$
*Let $x$,$y$ be rationals such that $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}... | 1. There are $p,q\in\mathbb{Z}$, with $gcd(p,q)=1$ such that $\frac{x+a}{x+b}=\frac{p}{q}$. Then $$(x+a)q=(x+b)p\Leftrightarrow x(q-p)=bp-aq\Leftrightarrow x=\frac{bp-aq}{q-p}\in\mathbb{Q}\Leftrightarrow x=0.$$ This is absurd. Therefore $p=q$ and $x+a=x+b$, so $a=b$.
2. Let $r\in\mathbb{Q}$ such that $\frac{x²+x+\sqrt{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequalit... | For the other side, we note that
\begin{align*}
(a+b+c)^2 &\ge 3 (ab + bc + ca)\\
&= 9.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even. We say an integer $p>1$ is prime when its only positive divisors are $1$ and $p$. Let $a$ and $b$ be natural number not both $1$. Prove that if $a^4+4^b$ is prime, then $a$ is odd and $b$ is even.
I'm also given a hint:
Consider the expression $(x^2-2xy+2y^2)(x^... | I think that hint is terrible and only the geniuses in the class could get it right away. I put $(x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$ into Wolfram Alpha and it gave me $x^4 + 4y^4$ as an "alternate form."
But the problem says "$4^b$", not "$4y^4$." If you're a precocious genius you can right away see that if $b$ is od... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding $(a+\sqrt b)^n+(a-\sqrt b)^n$ where $n$ is natural For the expression $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$ where $n \in \mathbb{N}$, and $a,b, \in \mathbb{Q}$, the radical is always ends up cancelled, and the result is always in $\mathbb{Q}$. Is there any way that this could be reexpressed wit... | Since
$$\begin{bmatrix}
A & B \\
1 & A
\end{bmatrix} = \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix} \begin{bmatrix}
A - \sqrt B & 0 \\
0 & A + \sqrt B
\end{bmatrix} \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix}^{-1}
$$
It follows that
$$\begin{bmatrix}
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Laurent series of f(z) Can you check my calculations?
Find Laurent series for: $f(z)=\frac{6z+8}{(2z+3)(4z+5)}=\frac{1}{4z+5}+\frac{1}{2z+3}$
when a)$\frac{5}{4}<|z|<\frac{3}{2}$
b)$|z|>\frac{3}{2}$
a)$\sum_{n=0}^\infty \frac{(-2)^nz^n}{3^{n+1}} + \sum_{n=1}^\infty \frac{(-4)^nz^{-n}}{5^{n+1}}$
b)$\sum_{n=0}^\infty \fr... | Note: Regarding your question, the series we take is important. Here I provide a complete Laurent expansion around $z=0$. From this you should be able to check our results (which slightly differ).
The function
\begin{align*}
f(z)&=\frac{1}{4}\frac{1}{z+\frac{5}{4}}+\frac{1}{2}\frac{1}{z+\frac{3}{2}}\\
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is $x^2+x+1$ divisible by $101$, if $x\in\mathbb Z$? Prove $x^2+x+1$ isn't divisible by $101$, for any $x\in\mathbb Z$?
I think the way of solving the problem it by using "Fermat's Little Theorem".
| Oh, well. If $$ x^2 + x + 1 \equiv 0 \pmod {101}, $$ then multiplying by $4$ gives
$$ (2x+1)^2 + 3 \equiv 0 \pmod {101}, $$ and
$$ (2x+1)^2 \equiv -3 \pmod {101}. $$
However, Legendre
$$ (-3|101) = (-1|101) (3|101) = (3|101) = (101|3) = -1 $$
NOTE: for any odd prime $p,$ we always get $(-3|p) = (p|3),$ all that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solving $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^{k}}{\sum_{i=0}^{k-1}2015^i\sum_{l=0}^k2015^l}=$ $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^{k}}{\sum_{i=0}^{k-1}2015^i\sum_{l=0}^k2015^l}=$$
$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^k}{\frac{1-2015^k}{1-2015}\cdot\frac{1-2015^{k+1}}{1-2015}}=$$
$$\lim_{n\to\infty}... | More general, $\frac{x^k}{(1-x^{k})(1-x^{k+1})}=\frac{1}{x-1}\left(\frac{1}{x^k-1}-\frac{1}{x^{k+1}-1}\right)$.
Thus
$\begin{eqnarray}
\sum_{k=1}^n\frac{x^k}{(1-x^{k})(1-x^{k+1})}&=&\frac{1}{x-1}\sum_{k=1}^n\left(\frac{1}{x^k-1}-\frac{1}{x^{k+1}-1}\right)=\frac{x(x^n-1)}{(x-1)^2(x^{n+1}-1)}
\end{eqnarray}$
Therefore
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\int_{0}^{\pi}\frac{r}{1-2r\cos x+r^2}dx=\int_{0}^{\pi}\frac{\cos x}{1-2r\cos x+r^2}dx$ Prove for every $r\in(-1,1)$: $$\int_{0}^{\pi}\frac{r}{1-2r\cos x+r^2}\,dx=\int_{0}^{\pi}\frac{\cos x}{1-2r\cos x+r^2}\,dx$$
I tried proving that $$\int_{0}^{\pi}\frac{\cos x-r}{1-2r\cos x+r^2}dx=0$$ using variable subst... | By symmetry this is the same as proving that
$$\int_0^{2\pi} \frac{r}{1-2r \cos x + r^2} \; dx
= \int_0^{2\pi} \frac{\cos x}{1-2r \cos x + r^2} \; dx$$
for $r\in (-1,1).$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence
$\frac{dz}{iz} = dx$ to obtain for the first integral
$$\int_{|z|=1}
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where ... | Standart way of such:
$$J = \int\dfrac{dx}{(x^2+a^2)^n} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)}\int x\,d\, \dfrac1{(x^2+a^2)^{n-1}}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ $a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).
Here's my attempt:
$\frac{2ab}{a+b}\leq\sqrt {ab}$
$\frac{2a^... | This is just a rearrangement of the famous AM-GM inequality!
It states that $\frac{a+b}{2} \geq \sqrt{ab}$
for $a,b \geq 0$.
The proof is clear from the picture!
Now multiply by $\sqrt{ab}$ on both sides and rearrange with ease to get the desired: $\frac{2ab}{a+b}\leq\sqrt {ab}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Say for what values of $a \in \mathbb {R} $ this matrix system has solutions
Let $a \in \mathbb{R}$ and
$$
A_a =
\begin{pmatrix}
1 & a & 1 \\
a & 2 & 3 \\
2 & 3 & 4
\end{pmatrix}.
$$
*
*Say for each values of $a \in \mathbb{R}$ the system:
$$
A_a
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
... | Writing $A_a$ short as $A$, and the inhomogenous system $A x = b$ as augmented matrix $[A|b]$ we get these transformations:
$$
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
a & 2 & 3 & 0 \\
2 & 3 & 4 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
a-3 & 2-3a & 0 & -3 \\
-2 & 3-4a & 0 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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I need some help with Geometry. Is this a correct answer to this problem? Good day,
I have a question regarding geometry. I don't know whether my answer is correct because the answer in my book uses a totally different method for solving this particular problem.
Here's the problem:
Given is a triangle, ABC, in which th... | Note: My figure is misleading. ABC in figure is not isosceles.
$$2x+2y = 180^\circ$$
$$x+y = ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$... | The confusion you are having comes from the concept of finding the solutions to $z^2 = 36$. Definitely $z = \pm 6$ are solutions because $(-6)^2 = 36 = 6^2$, but this is not the same thing as $y = \sqrt{36}$. Otherwise, we get nonsense like $-6 = 6$ which isn't true.
If we plug the answer $x = -6$ back into the origina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the s... | Hint For the second:
$$\int \frac{x+1}{x^2+x+1}dx$$
Rewrite as $$\int\bigg( \frac{2x+1}{2(x^2+x+1)}+\frac{1}{2(x^2+x+1)} \bigg)dx$$
$$=\underbrace{\frac 1 2 \int\frac{2x+1}{x^2+x+1}dx}_{:=I}+\underbrace{\frac 1 2 \int \frac{1}{x^2+x+1}dx}_{:=J}$$
For $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$
For $J$ complete the sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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How to prove that $k^3+3k^2+2k$ is always divisible by $3$? How can I prove that the following polynomial expression is divisible by 3 for all integers $k$?
$$k^3 + 3k^2 + 2k$$
| Just to be different. If $k \in \mathbb Z$ then $k = 3m + i$ where $i = $ either $0, 1,$ or $-1$ and $m \in \mathbb Z$.
So
\begin{align*}
k^3 + 3k^2 + 2k
&=(3m + i)^3 + 2(3m + i)+ 3k^2 \\
&= 3^3m^3 + 3 \cdot 3^2m^2 \cdot i + 3 \cdot 3m \cdot i^2 + i^3 + 2 \cdot 3m + 2i + 3k^2 \\
&= i^3 + 2i + 3\big[3^2m^3 + 3^2m^2i + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 14,
"answer_id": 3
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Calculate the area enclosed by the curve and line Calculate the are enclosed by ${y = 2x - 1}$ and ${y= x^2 + 6x + 2}$
First of all I combine the equations into:
${x^2 + 4x + 3 = 0}$
${(x + 3)(x + 1), x = -3, x = -1}$
They intersect at ${(-3 -7) (-1, -3)}$
I would say that ${y = 2x -1}$ is the top equation so to wor... | $(\frac13-2+3)-(9-18+9)=\frac43$ when there are two equation, there are two roots. You should use
$$\int_{α}^{β}a(x-α)(x-β)dx=-\frac{a}6(β-α)^3$$
for example
$$\frac16(-1-(-3)^2)=\frac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$ How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$
without using induction? Note that clearly $n\neq 0$
Thanks for any help!!
| This inequality does not hold since for any $n\in \mathbb N$ you have
$$\frac{1}{2}\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}.$$
To see this, note first that any term above is greater than $\frac{1}{n+n}$, so that your sum is greater than $n\cdot \frac{1}{2n}=\frac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $x^2 \equiv -x\pmod{2015}$
Problem: Find all integer solutions of $x^2 \equiv -x \pmod{2015}$.
I proceeded this way: first, I realized that $2015 = 5 \times 13 \times 31$. I rewrote $x^2 \equiv -x$ as $x^2 + x \equiv 0$.
Then, for each factor $d$, I have that $d\ |\ x(x+1)$, therefore $d\ |\ x$ or $d\ |\ x + ... | There's this general trick for solving quadratic congruences: if $\gcd(4a,n)=1$, then:
$$ax^2+bx+c\equiv 0\pmod{n}\stackrel{\cdot 4a}\iff (2ax+b)^2\equiv b^2-4ac\pmod{n}$$
In your case, $2015=5\cdot 13\cdot 31$ and:
$$x^2\equiv -x\pmod{2015}\iff (2x+1)^2\equiv 1\pmod{2015}$$
$$\iff \begin{cases}(2x+1)^2\equiv 1\pmod{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What to do if the modulus is not coprime in the Chinese remainder theorem? Chinese remainder theorem dictates that there is a unique solution if the congruence have coprime modulus.
However, what if they are not coprime, and you can't simplify further?
E.g. If I have to solve the following 5 congruence equations
$x=1 \... | \begin{align}
x &\equiv 1 \pmod 2 \\
x &\equiv 1 \pmod 3 \\
x &\equiv 1 \pmod 4 \\
x &\equiv 1 \pmod 5 \\
x &\equiv 1 \pmod 6 \\
\end{align}
Note that
$x \equiv 1 \pmod 6 \implies
\left\{
\begin{array}{l}
x\equiv 1 \pmod 2 \\
x \equiv 1 \pmod 3
\end{array}
\right.$
Replace $x \equiv 1 \pmod 6$ in yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
The equation represents the line of intersection of two planes. Using augmented matrix
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
2 & 3 & 4 & 6
\end{bmatrix}$$
$R_2\rightarrow ... | A line passes through two points.
You can get the first point intersecting your line with the plane $z=0$. $p=(3,0,0)$.
Simply project another random point of your line, say $(0,6,-3)$, in the plane $z=0$, you get $q=(0,6,0)$.
The line you are looking for is thus $z=0$, $y=-2x+6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$, Prove that $x+y+z$ $=0$
Question If $$ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} $$ Prove that $x+y+z$ $=0$
I've attmempted this question by cross multiplying so that
$$ x(c-a)(a-b) = y(b-c)(a-b) = (b-c)(c-a)z $$
but that did not work
I also tried splitting ... | Using $k$-method, let $\displaystyle \frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}=k$, then $\displaystyle \sum_{xyz} x=\sum_{abc} (b-c)k=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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integrate $\int \sin^{4}x\cos^{2}x$
$$\int \sin^{4}x\cos^{2}xdx$$
$$\int \sin^{4}x\cos^{2}xdx=\int (\sin x \cos x)^{2}\sin^2xdx=\int \left(\frac{\sin^{2}2x}{2}\right)\left(\frac{1}{2}-\frac{\cos2x}{2}\right)dx=\int \left(\frac{\sin^{2}2x}{4}-\frac{\sin^{2}2x\cos2x}{4}\right)dx=\frac{1}{4}\int ({\sin^{2}2x}-{\sin^{2}2... | $$\int\sin^4(x)\cos^2(x)\space\text{d}x=$$
$$\int\sin^4(x)\left(1-\sin^2(x)\right)\space\text{d}x=$$
$$\int\left(\sin^4(x)-\sin^6(x)\right)\space\text{d}x=$$
$$\int\sin^4(x)\space\text{d}x-\int\sin^6(x)\space\text{d}x=$$
You've to use twice the reduction formula:
$$\int\sin^m(x)\space\text{d}x=-\frac{\cos(x)\sin^{m-1}... | {
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"url": "https://math.stackexchange.com/questions/1649563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{x\cos x}{\sin^2x}dx$
$$\int \frac{x\cos x}{\sin^2x}dx$$
$$\int \frac{x\cos x}{\sin^2x}dx=\int \frac{x\cos x}{1-\cos^2x}dx=\int \frac{x\cos x}{(1-\cos x)(1+\cos x)}dx$$
How can I find the two fractions? if there are at all?
| Since you wanted to solve using partial fractions,
$$\int\frac{x\cos x}{(1+\cos x)(1-\cos x)}dx=\frac{1}{2}\int\frac{x(1+\cos x)-x(1-\cos x)}{(1+\cos x)(1-\cos x)}dx$$
$$=\frac{1}{2}\int\frac{x}{1-\cos x}dx- \frac{1}{2}\int{\frac{x}{1+\cos x}}dx$$
$$= \frac{1}{2}\int \frac{x}{2\sin^2\frac{x}{2}}dx- \frac{1}{2}\int \fra... | {
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"answer_count": 7,
"answer_id": 5
} |
Integrate $\int \frac{x^7}{(1-x^4)^2}dx$
$$\int \frac{x^7}{(1-x^4)^2}dx$$
I have tried to simplify the expression, to use U substation, any idea where to start from?
| $$\int\frac{x^7}{\left(1-x^4\right)^2}\space\text{d}x=$$
$$\int\frac{x^7}{\left(x^4-1\right)^2}\space\text{d}x=$$
Substitute $u=x^4$ and $\text{d}u=4x^3\space\text{d}x$:
$$\frac{1}{4}\int\frac{u}{\left(u-1\right)^2}\space\text{d}u=$$
$$\frac{1}{4}\int\left[\frac{1}{u-1}+\frac{1}{(u-1)^2}\right]\space\text{d}u=$$
$$\f... | {
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Where did my simplification go wrong? Sum and difference formula simplification I'm struggling with the following: We are to use the sum and difference formulas to find the exact value of the expression. The problem is simplification has been tough. As a last resort I decided to use Symbolab to find the answer and step... | \begin{align*}
\frac{\sqrt{6}}{4}- \left(-\frac{\sqrt{2}}{4}\right) & = \frac{1}{4}(\sqrt6+\sqrt2)\\
& = \frac{\sqrt{2}}{4}(\sqrt{3}+1)\\
& = \frac{\sqrt{2}}{4}\sqrt{(\sqrt3+1)^2}\\
& = \frac{\sqrt{2}}{4}\sqrt{4+2\sqrt3}\\
& = \... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
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Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\r... | $$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
As $x\rightarrow-3^+$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightarrow0$ from the left side on the number line.
As $x\rightarrow-3^-$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightar... | {
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"answer_id": 2
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Schurs Lemma for Upper Triangular matrix Schurs Lemma Says : For any $A\in M_n(\mathbb{C})$ with eigen values $\lambda_1,\dots,\lambda_n$ There exists an unitary matrix $U$ such that $$UAU^*=T$$ where $T$ is an upper triangular matrix with diagonal $\lambda_1,\dots,\lambda_n$
If In particular $A$ is itself an upper tri... | To answer your first question: no, $A$ and $T$ are not equal in general (since Schur decomposition is not unique--even in the case where each $\lambda_1,\dots,\lambda_n$ are distinct). Take
$$A = \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix},$$
we can Schur factor $A$ as follows:
$$\begin{pmatrix} 1 ... | {
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"answer_id": 1
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Determinant equality issue Hi I am studying for an exam tomorrow and I have a question, How do I prove that the two determinants are equal ? is there a short way ?
$2abc\left|\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}
\end{array}\right|$
$
\left|\begin{array}{ccc}
b+c & c+a & a+b\\
b^{2}+c^{2} & c... | Use
$$abc\left|\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}
\end{array}\right| = \left|\begin{array}{ccc}
a & b & c\\
a^2 & b^2 & c^2\\
a^{3} & b^{3} & c^{3}
\end{array}\right| \quad\text{and}\quad
\left|\begin{array}{ccc}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{array}\right| = 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658043",
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"source": "stackexchange",
"question_score": "2",
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My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{... | Divide both sides by $x^2$ to get $$1-\dfrac4x+\dfrac6{x^2}=0\iff\dfrac4x-\dfrac6{x^2}=1$$
$$1-\dfrac4{3x}+\dfrac2{x^2}=1-\dfrac13\left(\dfrac4x-\dfrac6{x^2}\right)=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_... | If we write $$\frac{\sqrt{16-x^2}}{x} = \frac{x \sqrt{16-x^2}}{x^2},$$ then the substitution $u^2 = 16-x^2$, $x^2 = 16-u^2$, $x \, dx = -u \, du$ immediately yields $$\int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx = \int_{u=\sqrt{12}}^0 \frac{-u^2}{16-u^2} \, du = \int_{u=0}^{\sqrt{12}} \frac{16}{16-u^2} - 1 \, dy.$$ Then... | {
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Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.
$\bullet~$ I can ... | $$x_n-x_{n+1}=x_n-\frac12\left(x_n+\frac 2{x_n}\right)=\frac{2x_n^2-x_n^2-2}{2x_n^2}=\frac{x_n^2-2}{2x_n^2}.$$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$ For every $a,b,c>0$ such that $abc=1$ prove the inequality
$$\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$$
My work so far:
$abc=1 \Rightarrow \frac 13 (a+b+c)\ge 1$
Then $$c^2+a+b \le ... | Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, by Muirhead $\sum\limits_{cyc}\frac{a^2+b^2}{c^2+a+b}=\sum\limits_{cyc}\frac{x^6+y^6}{z^6+xyz(x^3+y^3)}\geq\sum\limits_{cyc}\frac{x^6+y^6}{z^6+z(x^5+y^5)}=\frac{\sum\limits_{cyc}(x^7y+x^7z)}{xyz(x^5+y^5+z^5)}\geq2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665254",
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"source": "stackexchange",
"question_score": "3",
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If $x^2=y+z$, $y^2=x+z$ and $z^2=x+y$, prove If $x^2=y+z$, $y^2=x+z$ and $z^2=x+y$,
Prove that
$$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$$.
My attempt:
$$L.H.S=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$$
$$=\frac{(y+1)(z+1)+(x+1)(z+1)+(x+1)(y+1)}{(x+1)(y+1)(z+1)}$$
$$=\frac{x^2+y^2+z^2+yz+xz+xy+3}{(x+1)(y+1)(z+1)}... | Note that
$$x^2 + x = y^2 + y = z^2 + z = x + y + z$$
$$x(x + 1) = y(y + 1) = z(z + 1) = x + y + z$$
so
$$\begin{align}\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} &= \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z}\\
&= \frac{x + y + z}{x + y + z}\\&= 1\end{align}$$
| {
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Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$
Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$
My attempt using the LHS
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$$
$$ \frac{2\sin(a)+\frac{1}{\cos(a)}}{1+\frac{\sin(a)}{\cos(a)}} $$
$$ \frac{\frac{2\sin(a)+1}{... | Let $s=\sin(a)$, $c=\cos(a)$, $k=\sec(a) = 1/c$, $t=\tan(a)=s/c \;$ to eliminate visual clutter.
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)} = \frac{2s+k}{1+t} = \frac{2s+1/c}{1+s/c}=\frac{2sc+1}{c+s}=\frac{(c+s)^2}{c+s}= c+s \\= \frac{1+s/c}{1/c}=\frac{1+t}k = \frac{1+\tan(a)}{\sec(a)}$$
| {
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"source": "stackexchange",
"question_score": "1",
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What function does this Taylor Series represent? What is the function $f$ who's Taylor series is $1 - \frac{x}{4} + \frac{x^2}{7} - \frac{x^3}{10} + \cdots$ ?
I need to find the value of the series $ \sum^{\infty}_{n = 0}a_n = 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots$ by finding $\lim_{x\rightarrow1^-} \s... | If
$f(x)
=\sum_{k=0}^{\infty} a_k x^k
$,
then
$f_{n, j}(x)
=\sum_{k=0}^{\infty} a_{j+kn} x^{j+kn}
=\frac1{n}\sum_{i=0}^{n-1} w^{ij} f(w^i x)
$
where
$w = e^{2\pi i /n}
$.
This is known as
multisection of series.
Here is one of many
available discussions:
http://mathworld.wolfram.com/SeriesMultisection.html
Since
$f(x)
... | {
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"source": "stackexchange",
"question_score": "4",
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why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \... | Hint summation of$n$ odd integers which are positive is $n^2$ which can be expresed as $\sum (2n-1)=n^2$ where $n\in N$
| {
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"source": "stackexchange",
"question_score": "2",
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Show that $\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$ If $n$ is a positive integer, show that
$\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$
This is how I tried to solve it:
the given inequality is equivalent to
$\binom{n}{1}...\binom{n}{n-1}\leq ({\frac{2^n-2... | As noted by @LeGrandDODOM from AM-GM that $\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{\binom{n}{1}+...+\binom{n}{n-1}}{n-1}$.
However, note that $\binom{n}{1}+...+\binom{n}{n-1}=2^n-2$ which follows from the binomial theorem.
$\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{2^n-2}{n-1}$.
This gives u... | {
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Integral Calculus: Plane Areas in Rectangular Coordinates Find the area bounded by the given curves:
$y^2+2x-2y-3=0$ and the $y$-axis
(using horizontal & vertical strip)
| Finding the points:
$x = 0$ so
$$ y^2+2(0)-2y-3=0 $$
$$ y^2-2y-3=0 $$
$$ (y-3)(y+1)=0 $$
Thus points $P$ are $(0,3),(0,-1)$.
By completing the square
$y^2-2y+1=-2x+3+1$.
$$ (y-1)^2=-2(2x-2) $$
$ y=1, x=2 ,
V = (2,1) $
The parabola opens left.
| {
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"source": "stackexchange",
"question_score": "2",
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Is $x^3+x^2+x+1$ divisible by $x^3+5x^2+x$ in $\mathbb{Z}_5\left[x\right]$? (for simplicity, denote $a(x)=x^3+x^2+x+1$ and $b(x)=x^3+5x^2+x$)
My assignment asks to show that $b(x)$ does not divide $a(x)$ in $\mathbb{Z}\left[x\right]$, but rather that it does divide it in $\mathbb{Z}_5\left[x\right]$.
Long division show... | I'm not sure where you had a leak, but:
$$x^3+x^2+x+1=(x+1)(x^2+1)$$
and this much is true in the integers $\;\Bbb Z\;$ , so it is also true in $\;\Bbb Z_5\;$ . It comes from noticing that $\;-1=4\pmod5\;$ is a root of the leftmost polynomial. Observe further that since $\;5=1\pmod4\;,\;\;-1\;$ is a square in this last... | {
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Finding the inverse of a matrix given an equation So I've been given this equation:
$A\begin{bmatrix}
2&3&1&5\\
1&0&3&1\\
0&2&-3&2\\
0&2&3&1
\end{bmatrix} = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$
I'm supposed to find the inverse of A using this but I'm not really sure where to start. Any i... | Bernard's answer above gives a lot of intuition about what's really going on, but in case you are not yet familiar with some of the terminology here is a more brutish answer.
The inverse of the right hand matrix can be calculated by examination. Notice that there is only one non-zero entry per column and row of the m... | {
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integral for range in x axis Calculate the area in the shaded region:
${f(x) = x^3 -2x + 7}$
${{{\int_{-1}^2}} f(x) dx = x^4 - x^2 + 7x}$
$= {[(2^4) -(2^2) + 14] - [(-1)^4 - (-1)^2 + 7(-1)]}$
$= [16 - 4 + 14] - [- 7] = 19$
But the answer in the book is $21{3\over 4}$.
| As @Sean said ,you forgot to divide $x^4$ by $4$
Also if you want to find the integral of ${f(x) = x^3 -2x + 7}$
So the result will be $\require{cancel} \cancel{\int_{-1}^{2}} \frac{x^4}{\color{blue}4}-x^2+7x+C$
$$\int_{-1}^{2}\left(x^3-2x+7\right)dx=\left(\frac{x^4}{\color{blue}4}-x^2+7x\right)\bigg|_{-1}^{2}=4-4+14-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682024",
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"source": "stackexchange",
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Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with ste... |
$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
You need to use the chain rule here,
$$ y'=\frac{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{(x^2+1)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x^2)}{(x^2+1)^2}\frac{2}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{2(1-x^2)}{(1+x^2)^2\left(1+\frac{4x^2}{(1+x^2)^2}\right)}=\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that if a and b have the same sign then |a| + |b| = |a + b| We start with a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result :
$|a| + |b| = a+b$
$(|a| + |b|)^2 =... | You can shorten your approach and deal with the negative case at the same time, by writing
$$
(|a|+|b|)^2=a^2+2|a||b| + b^2\stackrel{(1)}=a^2+2ab+b^2=(a+b)^2\tag{*}
$$
where (1) uses the fact that $|a||b|=ab$ if $a$ and $b$ have the same sign. Now take square roots of both ends of (*).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ Find the equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and passes through the point... | Can't comment:
To get to the result that J. Wu and Dr. Sonnard Graubner gets, you can assume that everything is equal to $t$ or seperately
$$t=\frac{x-1}{2}; t=\frac{y-2}{3};t=\frac{z-3}{4}$$
and solve for $x,y,$ and $z$ in each seperate equations
| {
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"url": "https://math.stackexchange.com/questions/1683817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find solutions for $a$ and $b$ where $9 \equiv 4a+b \pmod {26} $ and $10 \equiv 19a+b \pmod {26}$? $$9 \equiv 4a+b \pmod {26}$$
$$10 \equiv 19a+b \pmod {26}$$
How can I solve the following system?
| Notice that substracting the two equations gives us that $15a \equiv 1 \pmod {26}$.
This gives us that $a \equiv 7 \pmod {26}$.
Since $4 \times 7+b \equiv 9 \equiv 35 \equiv 5 \times 7 \pmod {26}$, this implies that $b \equiv 7 \pmod {26}$.
The answer is $a \equiv b \equiv 7 \pmod {26}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the $\epsilon$-$\delta$ definition of the limit, evaluate $ \lim_{x \to a} \frac{x+2}{x^2-5}$
Using the $\epsilon$-$\delta$ definition of the limit, evaluate $$ \lim_{x \to a} f(x)$$ where $$f(x) = \frac{x+2}{x^2-5}$$
Attempt
We need to show that $\forall \epsilon, \exists \delta$ such that $$0 < |x-a| < \delta... | To solve these types of problems, you usually start with what you want to show and work backwards to try to figure out a $\delta$ that will work.
So let $\epsilon > 0$ be arbitrary. At the end of the day, we want $\left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon$. Let's try to manipulate the left hand ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$
If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$, then find the value of $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots$
Firstly how is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$?
Secondly, I thoug... | If you already know or take as given the first result, then
$$\frac{\pi^4}{90}=\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}\implies$$
$$\sum_{n=1}^\infty\frac1{(2n-1)^4}=\left(1-\frac1{16}\right)\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Determining whether an orthogonal matrix represents a rotation or reflection The exercise asks us to determine whether the given orthogonal matrix represents a rotation or a reflection. If it is a rotation, give the angle of rotation; if it is a reflection, give the line of reflection.
$$
A = \begin{bmatrix}
-\f... | for the line of reflection look for a mirror lines that stays fixed under the transformation. for example, in the case of $A = \begin{bmatrix}
-\frac{3}{5} & -\frac{4}{5}\\[0.3em]
-\frac{4}{5} & \frac{3}{5}\\[0.3em]
\end{bmatrix}$ try the line $x = \pmatrix{5\\5a}.$ we want $Ax = x$ that is
$$\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\int_{0}^{\infty }\frac{\ln( 1+x^{4} )}{1+x^{2}}{d}x~,~\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x$ How to evaluate these two integrals below
$$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$
$$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$... | Using the identity $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the power $4$ of $x$ in the numerator makes the life easier.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x
= & \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x \\
= & \oper... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove: $\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$
Prove: $$\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$$
for $\: -\frac{\pi}{2} < x < \frac{\pi}{2}$
I have thought of three things so far that can be useful.
The Cauchy–Schwarz inequality
$$
\int_Efg\,\mathrm{d}x\le\left(... | Without loss of generality, let $0\le x < \frac{\pi}{2}$. Define
$$ f(x)=\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right). $$
Then $f(0)=0,f'(0)=0$ and
\begin{eqnarray}
f''(x)=\frac{1}{8\cos^2x}(4 \cos (2 x)-\sin x \ln (\frac{1+\sin
x}{1-\sin x})-\sin (3 x) \ln(\frac{1+\sin x}{1-\sin
x})+12).
\end{eqnarray}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Coordinates of the center of curvature of $y^2 = 4px$ I'm trying to find the coordinates of the center of curvature at any point of
$$
y^2 = 4px
$$
I start with the first and second derivatives
$$
y' = \sqrt{\frac{p}{x}} \\
y'' = - \frac{1}{2} \sqrt{\frac{p}{x^3}} \\
$$
I then use the curvature formula
$$
K = \frac{|y'... | Given $m = -\sqrt{\frac{x}{p}}$, let $\theta$ be the angle of your radius of curvature
$$
\cos \theta = \frac{\sqrt{p}}{\sqrt{x+p}} \\
\sin \theta = \frac{-\sqrt{x}}{\sqrt{x+p}} \\
$$
So the coordinates of the center of curvature are
$$
x = \rho \cos \theta + x = \frac{2(x+p)^{3/2}}{\sqrt{p}} \frac{\sqrt{p}}{\sqrt{x+p}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that $C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$, for every $n \in N$ EDIT: I forgot to mention $C = 0.5uu^T + 0.33vv^T$ and now if I use it, I solve it easily.
Given: $Cx = \frac{1}{2}au + \frac{1}{3}bv$, $x \in R^2$, $u,v$ are orthonormal vectors in $R^2$, $x = au + bv$ and $a,b \in R$, $C$ is a matrix $2x2$. N... | Think of $C$ as a linear transformation on $\mathbb R^2$.
Then, the matrix of $C$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2 & 0 \\ 0 &1/3}$.
Therefore, the matrix of $C^n$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2^n & 0 \\ 0 &1/3^n}$.
Concretely, if $Cx=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Quaternion exponential problem I have problem with Euler´s form of quaternion. My quaternion $q=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j,$ so $q^2=-1$, because $$q^2=(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)=-\frac{1}{2}+\frac{1}{2}k-\frac{1}{2}k-\frac{1}{2}=-1.$$ Thus I can writ... | For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$ the exponential is defined as:
$$
e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$
(see :Exponential Function of Quaternion - Derivation)
and in general, since quaternions are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Point P such that perimeter is least Given two points $A(-2,0)$ and $B(0,4)$ then find coordinate of point $P$ lying on the line $2x-3y=9$ so that perimeter of triangle $APB$ is least.
Doing it by traditional calculus is making calculations very complicated. Is there smart geometrical way to solve this?
| Well, it was easy to say what might help, but a bit less simple to find a way to avoid unpleasant radicals, which are essentially the same ones you would get using path-minimization in calculus. Here is a geometric argument that dodges them.
The big problem with a direct approach is that the lengths of $ \ AP \ $ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of $\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$ if $a+b+c=1$
Let $a,b,c\ge 0$, and $a+b+c=1$. Find the minimum value of
$$\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$$
I think the minimum value is $\sqrt{2} + 2$? when $a=1,b=c=0$.
Of course, I can't prove it. Can anyone help?
| Let
$$
U=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z\ge 0\}\\
S=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z>0\}.
$$
Define $f,g:U\to \mathbb{R}$ such that
\begin{align}
f(a,b,c)&=\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}\\
g(a,b,c)&=a+b+c-1,
\end{align}
then
\begin{align}
\nabla f&=\left(\frac{1}{2\sqrt{a+1}},\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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.