Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to solve Quadratic Equations with an Unknown C and other variables. For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?
And also : If $x^2 +px + q$ is a... | A: $3x^2-11x +r=0$
The codition is that discreminet $\Delta\geqslant 0$, so we must have:
$\Delta= 11^2-12 r\geqslant0$
Or: $r\leqslant 10$
Now $121$ is odd and $12r$ is even and we have to find odd perfect squares less than $121$ which are $81, 49, 25$ and $9$. Only $49$ and $25$ give integers for $r$:
$r= 121-49=72... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find all natural $n$ numbers such that Find all natural $n$ numbers such that
$15(n!)^2+1$ is divisible by $2n-3$.
My try: First I assumed $2n-3$ is not prime number. Let $a$ be divisor of $2n-3$. It's clear that $a<n-1$, so $15(n!)^2$ is divisible by $a$. Which means $15(n!)^2+1$ is not divisble by $a$. But it is give... | You do need Wilson's Theorem, but observe:
\begin{align}n &\equiv -(n-3) \pmod {2n-3}\\
n-1 &\equiv -(n-2) \pmod {2n-3}\\
&\ \vdots\\
1 &\equiv -(2n-4)\pmod {2n-3}\end{align}
This gives:
\begin{align}(n!)^2&\equiv n!(-1)^n(2n-4)(2n-5)\dots(n-2)(n-3) \pmod {2n-3}\\&\equiv(-1)^n(2n-4)!(n)(n-1)(n-2)(n-3) \pmod {2n-3}\\\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Infinite product of $\sqrt{2}$ I am struggling to show the following relation
$$
\prod_{k=1}^\infty \left(1 - \frac{(-1)^k}{(2k-1)}\right) = \sqrt 2.
$$
I have tried to compute the sum
$$
\sum_{k=1}^\infty \log \left(1 - \frac{(-1)^k}{2k-1}\right),
$$
by using the expansion for $\log(1+x)$, however, I was not able to e... | $$\begin{align}\frac{2}{1}\frac{2}{3} &= 4^1\frac{2!^3}{4!1!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7} &= 4^2\frac{4!^3}{8!2!^2}\\\frac{2}{1}\frac{2}{3}\frac{6}{5}\frac{6}{7}\frac{10}{9}\frac{10}{11} &= 4^3\frac{6!^3}{12!3!^2}\\\ldots\end{align}$$
Use Stirlings approximation $n! \sim \sqrt{2\pi n}\left(\frac{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$ I tried integration by parts considering $dx=du$
$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.
My attempt....
$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{... | Let $x=\sinh^2t$. Then
$$\int\frac{dx}{\sqrt{x(x+1)(x+2)}}=\int\frac{2\sinh t\cosh t\,dt}{\sqrt{\sinh^2t\cosh^2t(\sinh^2t+2)}}=\sqrt2\int\frac{dt}{\sqrt{1+\frac12\sinh^2t}}
\\=\sqrt2\int\frac{dt}{\sqrt{1-\frac12\sin^2it}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving or disproving basic facts about sequences in Real Analysis I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is ... | $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.
(a) Consider the sequence $(x_n)$ given by $x_n = \sqrt{n+1}$ and the sequence $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solve $ y^2y’’ = y’, y(0) = 1, y’(0) = 1 $ Solve the ODE with the initial conditions:
$$
y^2y’’ = y’, y(0) = 1, y’(0) = 1
$$
I did the substitution:
$$
y’ = z
$$
$$
y’’ = z’ = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z \frac{dz}{dy}
$$
Putting in the ODE:
$$
y^2z \frac{dz}{dy} = z \Rightarrow y^2 \frac{dz}{dy} = ... | $y^2y''=y'$
Write as
$y''=\frac{y'}{y^2}$
Integrate both sides
$y'=-1/y +C$
$y'(0)=1,y(0)=1\to C=2$
$y'=-1/y+2$
$y=\frac{1}{2} \left(W\left(e^{4 x+1}\right)+1\right)$
Where $W(z)$ is Lambert $W$ function which solves $we^w=z$
The implicit form is
$x=\frac{1}{4} (2 y+\log (2 y-1)-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.
Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.
Assume $y \ge 3$ and $x \ge 1$.
By $\mod 9$, we get ... | examples to study
CW:Catalan Thue Ramanujan Nagell Tijdeman p^x - q^y = C
https://math.stackexchange.com/users/292972/gyumin-roh
Exponential Diophantine equation $7^y + 2 = 3^x$
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Elementary solution of exponential Diophantine equation $2^x - 3^y = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$\lim_{x\to 0}\frac{1}{x}−\frac{2}{e^{2x}−1}$ doesn't get solved as expected A question says $f:\Bbb R\setminus\{0\} \to \Bbb R$, $f(x)=\dfrac{1}{x}−\dfrac{2}{e^{2x}−1}$. The limit as the function approaches $x=0$ is logically $0$, as in;
$\lim \limits_{x \to 0}\dfrac{e^{2x}-1}{2x} = 1$ This function then becomes $\dfr... | You did an approximation error. When $x\to0$ it is $\frac{2}{e^{2x}-1}=\frac{2}{1+2x+\frac{1}{2}(2x)^2+\text{o}(x^2)-1}=\frac{1}{x+x^2+\text{o}(x^2)}$, so
$$\frac{1}{x}-\frac{2}{e^{2x}-1}=\frac{1}{x}-\frac{1}{x+x^2+\text{o}(x^2)}=\frac{1}{x}\left(1-\frac{1}{1+x+\text{o}(x)}\right)$$
Using the fact that $\frac{1}{1+\var... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do you solve the equation $3x^2+2\sin x=2$? This seems like a simple equation to solve, but I can't find a way to do it and wolfram alpha can't give me an exact answer.
| This is a transcendental equation; so, you will need numerical methods and then good starting guesses.
Consider that you look for the zero's of function $f(x)=3x^2+2\sin x-2$.
For example, using Taylor series built around $x=0$
$$f(x)=-2+2 x+3 x^2-\frac{x^3}{3}+O\left(x^5\right)$$ Ignoring the term in $x^3$, solving th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequal... | From Minkowski inequality: $\sqrt{x^2 + a^2} + \sqrt{y^2 + b^2} \ge \sqrt{(x + y)^2 + (a + b)^2}$
applied to the first two terms, with the notation of @Albus, we get:
$\sqrt{\sqrt{3}^2 +x^2} + \sqrt{\sqrt{3}^2 +y^2} \ge \sqrt{(2\sqrt{3})^2 + (x+ y)^2}$. Let's denote RHS of the inequality by $X$.
Then, apply Minkowski i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:
If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$
I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.
| Notice that:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right).$$
The minimum of the function $f(w) = w + \frac{1}{w}$ is attained for $w=1$. More specifically, $f(1)= 2.$ See addendum section below for further details.
Then:
$$\frac{(x^2+1)(y^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong? I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$
First of all I saw tha... | Clearly,
$$
\frac{1}{n+1}\le \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{n}
$$
for $k=1,2,\ldots,2n+1$. Hence
$$
\frac{2n+1}{n+1}<\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}<\frac{2n+1}{n}
$$
and
$$
\lim_{n\to\infty}\frac{2n+1}{n+1}=\lim_{n\to\infty}\frac{2n+1}{n}=2.
$$
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| If $x^2 + 5x+12 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (-8)(-6)-48 = 0.$$
If $x^2 + 5x-2 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = 6\cdot 8-48 = 0.$$
Therefore LHS and RHS have the same zeroes. Since RHS has no double roots and both sides are polynomials of degree $4$, we conclude LHS = RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \i... | $\sum_{k=1}^n k^{1/2}
\approx \int_0^n x^{1/2} dx
=\dfrac{n^{3/2}}{3/2}
=\dfrac23 n^{3/2}
$
and
$\sum_{k=1}^n k^{-1/2}
\approx \int_0^n x^{-1/2} dx
=\dfrac{n^{1/2}}{1/2}
=2n^{1/2}
$
so
$\begin{array}\\
s(n)
&=1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90... | Angle $ABC$ is $135°$ because it is half the concave angle $COA=270°$.
Thus angle $BAC=22.5°$ and $$BC=2r \sin 22.5°=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Trying to find the sum of cosines I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}... | You can use this :
$$S_n(x)= \mathcal{Re}\left( \frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)\\
= \mathcal{Re}\left(e^{ikx/2}\frac{e^{-i(k+1)x/2}-e^{i(k+1)x/2}}{e^{-ix/2}-e^{ix/2}} \right)\\
=\mathcal{Re}\left( e^{ikx/2}\frac{\sin((k+1)x/2)}{\sin{(x/2)}}\right)\\
= \cos\left(\frac{kx}{2}\right)\frac{\sin\left(\frac{(k+1)x}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
I'm having trouble solving this equation. I've tried squaring both sides and got this
$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$
after simplifying
$$19 - 6\sqrt{... | There is a formula to simplify nested radicals like that
$$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
It is effective only if $(a^2-b)$ is a perfect square. In your example both of them have this property
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
$$\sqrt {11 - \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding a linear transformation matrix for a set of vectors I am wondering if we can find a linear transformation matrix $A$ of size $3\times 3$ over the field of two elements $\mathbb{Z}_2$ i.e. a matrix $A$ of zeros and ones s.t.
$$A \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix... | There is no such $A$, invertible or not. For instance your equations lead to
$$
\begin{bmatrix} 1 \\ 0\\ 0\end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix} =A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} +
A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} =
A \begin{bmatrix} 1 \\ 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$ $\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}}$
How can I calculate this? Since there are $4^n$ and $\cos^2x$, I tried:
$$\sum_{n=1}^{\infty} {\frac{1}{4^n \cos^2 (\frac{\pi}{2^{n+2}})}} = 4\sum_{n=1}^{\infty}{\frac{\sin^2{\frac{\pi}{... | Notice
$$\begin{align}\frac{1}{\cos^2\frac{\theta}{2}}
&= \frac{2}{1+\cos\theta}
= 2\frac{1 - \cos\theta}{1-\cos^2\theta}
= \frac{4 - 2(1+\cos\theta)}{1-\cos^2\theta}\\
&= \frac{4}{\sin^2\theta} - \frac{2}{1-\cos\theta}
= \frac{4}{\sin^2\theta} - \frac{1}{\sin^2\frac{\theta}{2}}
\end{align}
$$
We have
$$\begin{align}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$... | Maybe your teacher want you to solve the equations one by one for
each column of $B$. For example let the first column of $Q$ be $q_{1}=\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}$ Then you have the equations $Aq_{1}=b_{1}$, that is
\begin{align*}
x_{1}+\frac{1}{3}x_{3} & =\frac{4}{3}\\
-\frac{x_{3}}{3} & =-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$ Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then
(A) $f$ is not invertible on (0,1)
(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}... | Just observe that there is no $x$ such that $f(x) =\frac{1}{b}$. If any such $x$ exists then it would imply that $b^2=1$ which is certainly not possible. This shows that $f$ is not surjective which implies that $f$ is not invertible.
Edit:
One more way to see that $f$ is not surjective is to observe that $-1 < f(x) < b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Use of definition of limit to prove $\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3\cdot \sqrt[3]{4} }$ I know that by definition I have to prove that $$\displaystyle\lim_{h→0} \dfrac{\sqrt[3]{2+h}- \sqrt[3] {2}} {h}= \dfrac {1}{ 3 \cdot \sqrt[3]{4} }⟺∀ϵ>0,∃δ>0,\, 0<|h−0|<δ⟹| \dfrac{\sq... | The limit $3\sqrt[3]4$ is given by the problem and it is easy to calculate. What the problem asks is to use the definition of limit to show that this limit is indeed $3\sqrt[3]4$. This presents some difficulty.
Let $\epsilon\gt0$ be given. We need to find out $\delta \gt0$ such that $|x|\le \delta\Rightarrow |3\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from the... | well if $2^{6k}4\equiv m \pmod {18}$ then $2^{2^{6k}4}\equiv 2^m \pmod {19}$.
$2^{6k}4 \equiv 0 \pmod 2$ and $\phi(9) = 6$ so $2^6 \equiv 1 \pmod 9$ so $2^{6k}4\equiv 4 \pmod 9$. So by Chinese remainder theorem there is one solution $\pmod {18}$ to $2^{6k}4\equiv 0 \pmod 2$ and $6^{6k}4\equiv 4 \pmod 9$.
And that uniq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule? I can evaluate the limit with L'Hospital's rule:
$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=... | Here's a trick to prove convergence of the continuous limit$$
\lim\limits_{x\rightarrow\infty} x \left(\sqrt[x]4 - 1\right)
$$
if you know also know how to integrate $2^x$.
Observe:\begin{eqnarray}
x \left(\sqrt[x]4 - 1\right) = x \left(\sqrt[x]2 - 1\right)\left(\sqrt[x]2 + 1\right) = 2x\left(\sqrt[2x]4 - 1\right)\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
All possible angles of triangle such that the feet of the angle bisectors determine a right angle
Triangle $ABC$ has angle bisectors $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Given that $\angle EDF = 90^\circ$, enter all possible values of $\angle BAC$ (in degrees), separated by commas.
I plan to use vec... | I entered $120^\circ$ and it was correct, but without much reason to it. Here is their solution.
As usual, let $a = BC,$ $b = AC,$ and $c = AB.$
By the Angle Bisector Theorem, $BD:DC = c:b,$ so
$\overrightarrow{D} = \frac{b}{b + c} \overrightarrow{B} + \frac{c}{b + c} \overrightarrow{C} = \frac{b \overrightarrow{B} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Can Trigonometric Substitution be used to derive the anti-derivative $\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$? My textbook (Calculus by James Stewart 8th edition) uses partial fraction decomposition to derive the anti-derivative
$$\int \frac{1}{x^2 -a^2}dx = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + ... | Use $\frac{x-a}{x+a}=\frac{1-\cos\theta}{1+\cos\theta}=\frac{(1-\cos\theta)^2}{\sin^2\theta}=(\csc\theta-\cot\theta)^2$ or $\left|\frac{x-a}{\sqrt{x^2-a^2}}\right|=\sqrt{\frac{x-a}{x+a}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(\frac{2}{3}+n)}{(1+\frac{3}{2})\ldots(\frac{3}{2}+n)}.$
Examine the convergence of the series:
$$\sum_{n=1}^{\infty}\frac{\frac{2}{3}(1+\frac{2}{3})\ldots(n+\frac{2}{3})}{(1+\frac{3}{2})\ldots(n+\frac{3}{2})}.$$
Attempt. Using ratio test for $a... | In terms of the Gamma and Beta functions we are discussing
$$ \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{5}{3}\right)/\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(n+\frac{5}{2}\right)/\Gamma\left(\frac{5}{2}\right)}=\underbrace{\frac{\Gamma(5/2)}{\Gamma(2/3)\Gamma(5/6)}}_{K}\sum_{n\geq 1}B(n+5/3,5/6) $$
or
$$ K \sum_{n\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$? I did
\begin{align}
& \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt]
= {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt]
= {} & \fra... | \begin{align}
i\times j & = k = -j\times i \\
j\times k & = i = -k\times j \\
k\times i & = j = - i\times k \\
i \times i & = j \times j = k\times k = 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$ I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.
$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$
$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1... | The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq i <j<k}}\frac{1}{5^i3^j2^k}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq i <j<k
\end{align*}
which has $1$ as lower limit and $k-1$ as upper limit. We have two indices $i$ and $j$, which means we can write it as double ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very ... | $$\sqrt{x^{2}-1}=\left(x+5\right)\frac{\sqrt{x+1}}{\sqrt{x-1}}$$
Both sides are only defined for $|x|>1$ or $x=-1$
Squaring both sides,
$$x^{2}-1=\left(x+5\right)^{2}\left(\frac{x+1}{x-1}\right)$$
We can factor the LHS using difference of squares,
$$\left(x+1\right)\left(x-1\right)=\left(x+5\right)^{2}\left(\frac{x+1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank y... | Note that $x>1$. To avoid squaring equations, factorize directly as follows
\begin{align}
0 &=x+\frac{x}{\sqrt{x^2-1}}-\frac{35}{12}\\
&= \left(x-\frac{35}{12} + \frac{25}{12x}\right)
+ \left( \frac{x}{\sqrt{x^2-1}} -\frac{25}{12x}\right)\\
&=\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
+ \frac{x^2-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \... | We can use
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$
Since both
$$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}$$
and
$$\int_{-1}^1 2\sqrt{1-x^2}dx$$
converge to $\pi$, you can either use
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty {2n \choose n} \left(\frac{x}{2}\right)^{2n}$$
or
$$2\sqrt{1-x^2} = \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substitute... | \begin{aligned}
& \int x^{3} \sqrt{x^{2}-9} d x \\
=& \int \frac{x^{3}\left(x^{2}-9\right)}{\sqrt{x^{2}-9}} d x \\
=& \int x^{2}\left(x^{2}-9\right) d\left(\sqrt{x^{2}-9}\right) \\
=& \int\left(y^{2}+9\right) y^{2} d y,\text { where } y=\sqrt{x^{2}-9} \\
=& \frac{y^{5}}{5}+3 y^{3}+c \\
=& \frac{y^{3}}{5} \cdot\left(y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$ Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Originally I thought to rephrase it in geometric terms, see... | Remark: @VIVID gave a nice solution. I rewrote it in elementary way (without using derivative, extrema). Note that while we used calculus to motivate our solution, we do not need to include any
calculus in the solution!
Letting $x = 0$, we have $|1 - q| \le \frac{\sqrt{2}-1}{2}$ or
$$1 - \frac{\sqrt{2}-1}{2} \le q \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Deriving Generating function for centered trinomial coefficients Let $c_n$ denote the $n$-th center trinomial coefficient (OEIS sequence here).
It appears they cannot be generated by a linear recurrence relation, so how should I go about to find the generating function $G(x)$ for the sequence?
$$G(x)=\sum_{n=0}^{∞}c_nx... | Here is a variation based upon G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. We start with the central trinomial coefficients:
\begin{align*}
[x^n](1+x+x^2)^n\qquad\qquad n\geq 0
\end{align*}
We consider the function
\begin{align*}
f(x)=1+x+x^2\tag{1}
\end{align*}
and der... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the vertex of $y = x^2 - 6x + 1$. My solution doesn't match the book. I'm studying on how to find the vertex of a Parabola and got stuck with this question.
Determine the vertex of $y = x^2 - 6x + 1$
I found the vertex is $V=(3,-8)$ (see work below), but my workbook showed it as $V=(3,10)$. Unfortunately, it ... | $4-36$ is indeed $-32$. Whenever you have $a - b$ and $b > a$, you should interpret it as $a - b = -1 \cdot (b - a)$. So your calculation seems to be right. And yes, the vertex of the parabola is at $(3, -8)$, this can be verified by differentiating and setting $y' = 0$.
$y' = 2x - 6 = 0 \implies x = 3$ and then $y(x=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin... | $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are both positive in the given domain, so their quotient is also positive, and $|x| = x$ when $x \in \mathbb R^+$. $\sin \alpha + \cos \alpha > 0$ as $\sin \alpha, \cos \alpha$ are positive in the domain. Hence we can focus our attention to just proving $\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}... | So, denoting $w_n = f_n -g _n$, you can write down the equation
\begin{align}
w_0 = f_0-g_0 = 4, \qquad &w_{n+1} = f_{n+1}-g_{n+1} = 3(f_n-g_n)+8n -12, \quad\text{i.e.}\\
w_0 = 4, \qquad &w_{n+1} = 3 w_n +8n -12.
\end{align}
This a liner difference equation for which you can have an explicit solution, in the same way t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Evaluating a triple summation ${\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k$
If $$a_i= \begin{cases} 1 && 1 \le i \le 10 \\ 2 && 11 \le i \le 20 \end{cases} $$
Find the value of $$\displaystyle {\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k $$
I could only evaluate the summation when $1 \le i < j \le... | Let
*
*$ A = \sum_{i < j < k } a_ia_ja_k$
*$ B = \sum_{i = j < k } a_ia_ja_k$
*$ C = \sum_{i < j = k } a_ia_ja_k$
*$ D = \sum_{i = j = k } a_ia_ja_k$
We're after $ A + C$.
Clearly,
*
*$ 30^3 = ( \sum a_i)^3 = D + 3C + 3B + 6A$.
*$ D = 10 ( 1^3 + 2^3 ) = 90$
*$C = 2^2 ( 10 \times 10\times 1 + \frac{10 \times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to get the value of a function dependant on $\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot \left(\frac{A}{2}\right)}$? The problem is as follows:
Find the value of:
$$R=\frac{\sec^2B-\cot A}{4+\csc^2 A}$$
Where $A$ is given by:
$\tan A = \frac{2+\cot \left(\frac{B}{2}\right)}{4+\cot
> \left(\frac{A}{2}\ri... | Let $\displaystyle t = \tan\frac{A}{2} \quad →
\tan\frac{B}{2} = \tan \left(45°\!-\frac{A}{2}\right) = \frac{1-t}{1+t}$
$\displaystyle \tan A = \frac{2+\cot\frac{B}{2}}{4+\cot\frac{A}{2}} \;\; →
\frac{2t}{1-t^2} = \frac{2+\frac{1+t}{1-t}}{4+\frac{1}{t}}\;\; → 1-t^2 = 6t
\;\; →\cot A = \frac{6t}{2t} = 3$
$\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with t... | Let the polynomial be denoted by $f(x)$.
$f'(x)=4(x^3+3x^2+2x-1)$ has only one real root $\approx 0.3247$. At that point $f''(x)=4(3x^2+6x+2)>0$. Thus $f(x)$ has minimum value at $\approx 0.3247$ and $f(0.3247)\approx 2.271$.
$\therefore f(x)>0 \;\forall x\in \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
theory of equation let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
( c+d)^{-1}+ (c+d)^{-1}$$
my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck
| If you combine your result with that of B. Goddard, you have
$$\frac{1}{a+b}+\frac{1}{c+d} = \frac{3}{(a+b)(c+d)}$$
Now suppose $a=1+w, b=1+w^2, c=1+w^3, d=1+w^4$ where $w$ is one of the imaginary roots of $w^5=1$, then
$$(a+b)(c+d) = (2+w+w^2)(2+w^3+w^4) \\= 4 + 2(w+w^2+w^3+w^4) + (w^4 + 1 + 1 + w)=4+w+w^4$$
Similarly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
kernel and image of a map help The question states:
a) Suppose that $U = V = M_{2,2} (\mathbb R)$ is the vector space of $2 × 2$ matrices over $\mathbb R $ Let $\phi$ be defined by $\phi(A) = A - A^T $ for each $A \in M_{2,2} \mathbb R$. Determine
$\ker \phi $ and state its dimension.
b) Determine $\operatorname{im}... | The first part is correct. It is easy to see that the kernel is the set of symmetric matrices, which for dimension $n$ has dimension $n(n+1)/2$.
Since the image is one dimensional, it suffices to find one nonzero element of it, and that will form a basis. For instance, we have that $\begin{pmatrix}1&1\\0&1\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty wei... | $$\left(\sqrt{x^2+12y}+\sqrt{y^2+12x}\right)^2=(x+y+10)^2$$
$$\implies x^2+12y+y^2+12x+2\left(\sqrt{(x^2+12y)(y^2+12x)}\right)=x^2 + y^2 + 100+ 2xy+20x+20 y $$
$$\implies \sqrt{(x^2+12y)(y^2+12x)}=xy+4(x+y)+50 $$
$$\implies(x^2+12y)(y^2+12x)=(xy+142)^2 $$
$$\implies x^3 - \dfrac{35 x y}{3}+ y^3 - \dfrac{5041}{3}=0$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
How to find $B=50-\overline{(n-1)n}$ when $n=\operatorname{Im}(z+w)$ and it requires the simplification of each complex? The problem is as follows:
First simplify the following complex numbers:
$z=3\left(\frac{1+i}{1-i}\right)+\left(\frac{1+i}{1-i}\right)^2+3\left(\frac{1+i}{1-i}\right)^3+\left(\frac{1+i}{1-i}\right)^4... | The complicated expressions are only there to confuse you. You've noticed that $\frac{1+i}{1-i}=i$, so
$$z=[3(i)+(-1)+3(-i)+(1)]_n+3(i)+(-1)$$
where the part in square brackets, which is zero, repeats $n$ times so it may be ignored. Similarly, every fraction in brackets in $w$ is of the form $\frac k{ik}=-i$ for some $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degre... | We use the identity $(u+v)^2=u^2+v^2.$ From $$x^4+x^2+x=0$$ we see that
$$x=x^4+x^2=(x^2)^2+x^2=(x^2+x)^2$$
And we can construct the root of an arbitrary polynomial
$$ax^3+bx^2+cx+d=x(ax^2+c)+bx^2+d=(x^2+x)^2(ax+c)^2+(bx+d)^2=((x^2+x)(ax+c)+(bx+d))^2=(ax^3+(a+c)x^2+(b+c)x+d)^2$$
So we see that every polynomial has exac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$
If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.
I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\fra... | Let $p = a + b + c = 1$, $q = ab + bc + ca$ and $r = abc$.
Fact 1: $q^2 \ge 3pr$.
(The proof is given at the end.)
Fact 2: $p^2 \ge 3q$.
(Proof: $p^2 - 3q = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0$.)
We split into four cases:
*
*$q \ge 0$: By Fact 2, we have $q \le \frac{1}{3}$. By Facts 1-2, we have $r\le \f... | {
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Looking for analytical proof that this function given as a power series is constant. Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$
Is there an analytical... | Too long for comments.
Assuming that $t$ is a positive integer, there is no problem to get explicit formulae up to $t=3$.
For $t \geq 4$ appear hypergeometric function and the result write
$$P_t(x)=\frac x {1-x} \big[Q_t(x)-1\big]$$
$$Q_4(x)=\,
_3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
S... | Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $a^2$ to find
$$0=a^2-a-1+\dfrac1a+\dfrac1{a^2}=\left(a-\dfrac1a\right)^2-\left(a-\dfrac1a\right)+1=0$$
$\implies a-\dfrac1a=-w$ where $w$ is a complex cube root of unity
$\implies a^2=1-wa$
$\implies... | {
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"source": "stackexchange",
"question_score": "2",
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Factorize polynomial of degree 4 given statements If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$
Such that$$cd=-8\\a>b\\c<d$$
What are $a,b,c$ and $d$?
Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be:
$$abcd=-8\\a+... | You know that $abcd = -8$ and $cd = -8.$ Therefore $ab= 1.$
From the coefficient of $x$ in the polynomial, you know that $abc + acd + abd + bcd = -26$ (note: not $+26$).
But since $ab=1$ and $cd=-8$, you can see that
$abc + acd + abd + bcd = c - 8a + d - 8b.$
Therefore
$$ -8a - 8b + c + d = -26. \tag1 $$
But you also k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different.... | Since $$0\lt \cot^{-1} (-\frac 34 ) \lt \pi$$
, $$0\lt \frac 12 \cot^{-1} (-\frac 34) \lt \frac{\pi}{2} $$ which means its sine must be positive, so Wolfram’s answer is wrong. $\frac{2}{\sqrt 5}$ is correct.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Sum of square roots inequality
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$
The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a t... | If you square both sides:
$(\sqrt a + \sqrt b + \sqrt c +\sqrt d)^2 =a + b+c+d + 2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})$
while $(2\sqrt {a+b+c+d})^2= 4(a+b+c+d)$ so it suffices to prove
$2(\sqrt{ab} + \sqrt{ac}+\sqrt{ad} + \sqrt{bc}+\sqrt{bd} + \sqrt{cd})\le 3(a+b+c+d)$.
If we apply AM/GM ... | {
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"source": "stackexchange",
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Integrate $\int \prod_{i=0}^n \frac{1}{(x+i)}dx$ I initially tried to find a pattern using partial fraction decomposition:
$$
n= 1 \, : \, \, \frac{1}{x(x+1)} = \frac{1}{x}+\frac{-1}{x+1}
$$
$$
n=2 \, : \, \, \frac{1}{x(x+1)(x+2)} = \frac{1/2}{x}+\frac{-1}{x+1}+\frac{1/2}{x+2}
$$
$$
n=3 \, : \, \, \frac{1}{x(x+1)(x+2)(... | Hint. If we have that
$$\prod_{i=0}^n \frac{1}{(x+i)}=\sum_{i=0}^n \frac{A_i}{(x+i)}.$$
then, for $k=0,\dots,n$,
$$\lim_{x\to -k}(x+k)\prod_{i=0}^n \frac{1}{(x+i)}=A_k.$$
The general formula of $A_k$ is related to a binomial coefficient.
Once that the partial fraction decomposition is ready, the integration is straight... | {
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Solve Diophantine Equation Show that $ x^3 - 6 = 25y^2 + 35y $ doesn't have any non-zero integer solution.
What I have tried
$$ 25y^2 + 35y - x^3 + 6 = 0 $$
solving for y we get
$$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$
so I need to show that
$$ 25 + 4x^3= z^2 $$
doesn't have a solution
or if it does I n... | Alternative solution
Suppose $x^3 = 25y^2 + 35y + 6 = (5y+6)(5y+1).$
If $(5y+6)$ and $(5y+1)$ have a common factor, then this common factor must divide $5$. If the common factor is $> 1$, then it must be $5$. But this would imply that $5$ is a factor of $x^3$, which would imply that $5 | x^3$ which would imply that $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equality of triple integrals over unit sphere $ \iiint_{\text{unit ball}} x e^{ax + by + cz} dV$ I have to calculate
$$ \iiint_{\text{unit ball}} x e^{ax + by + cz} \,dV,$$
where by "unit ball" I mean the region $x^2 + y^2 + z^2 \leq 1$.
I know how to calculate this (rotation matrix that takes $(a,b,c)$ to $(0,0,\sqrt{... | In the Cartesian coordinates $(u,v,w)$ with the unit vector $\hat w=\frac{1}{\sqrt{a^2+b^2+c^2}}(a,b,c)$,
$$\hat x= \cos\theta_u \hat u + \cos\theta_v \hat v + \cos\theta_w \hat w$$
where $\cos \theta_w =\frac{a}{\sqrt{a^2+b^2+c^2}}$. Then
\begin{align}
\iiint_{{r<1}} \frac xa e^{ax + by + cz} \,dV
& = \frac1a\iiint_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a, b, c, d)$? If $a+b+c+d=12$; $ 0≤a, b, c, d≤ 6$ and none of $a,b,c,d$ equals $3$, what is the number of possible solutions for $(a,b,c,d)$?
I need to solve it using combinatorics.
I've started with cal... | $a + b + c +d = 12 \ $ where $0 \leq a, b, c, d \leq 6$ and $a, b, c, d \ne 3$
Here is how you can solve using Principle of Inclusion-Exclusion, which is straightforward as well but I have taken more space and time to explain the steps.
A) Number of unrestricted solutions for non-negative numbers
$\displaystyle = {15 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Combinatorics That Looks Similar to Vandermonde's Identity How do I simplify:
$$\sum_{r = 0}^{\lfloor \frac{k}{2} \rfloor} \dbinom{k}{2r} \cdot \dbinom{n-k}{k-2r}?$$
Basically, the sum is: $\dbinom{k}{0} \cdot \dbinom{n-k}{k} + \dbinom{k}{2} \cdot \dbinom{n-k}{k-2} + \dbinom{k}{4} \cdot \dbinom{n-k}{k-4} \cdots.$
This ... | Note that
$$\sum_{r\ge 0} a_{2r} = \sum_{r\ge 0} \frac{1+(-1)^r}{2}a_r.$$
Taking $$a_r=\binom{k}{r}\binom{n-k}{k-r}$$ yields
\begin{align}
\sum_{r\ge0} \binom{k}{2r}\binom{n-k}{k-2r}
&= \sum_{r\ge0} \frac{1+(-1)^r}{2} \binom{k}{r}\binom{n-k}{k-r} \\
&= \frac{1}{2}\sum_{r\ge0} \binom{k}{r}\binom{n-k}{k-r} + \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013315",
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"source": "stackexchange",
"question_score": "3",
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Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin
First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 ... | The mistake lies in using $6x+8y-75=0$ further instead of $-12x-16y+150=0$
in the formula. Correct way is to write:
$$C(0,0) C(x,y)=(-12x-16y+150)^2.$$
to get the correct answer.
| {
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"source": "stackexchange",
"question_score": "1",
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If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem usefu... | hint: $x^2+y^2+z^2+t^2=(\frac{x^2}{3}+y^2)+(\frac{x^2}{3}+z^2)+(\frac{x^2}{3}+t^2)
=(\frac{x}{\sqrt{3}}-y)^2+(\frac{x}{\sqrt{3}}-z)^2+(\frac{x}{\sqrt{3}}-t)^2+\frac{2}{\sqrt{3}}\cdot x(y+z+t)\geqslant 0+1\cdot x(y+z+t)$
When would the equality hold?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
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"source": "stackexchange",
"question_score": "1",
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Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | Other answers have shown how you could continue with what you had. But just for interest, here's another proof, which doesn't use the difference, but the quotient:
Let's call the formula
$$ a_n = 4^{2^n} + 2^{2^n} + 1 $$
You already showed that $a_1 = 21$ is a multiple of $7$. (So is $a_0 = 4^1+2^1+1 = 7$, so we can ac... | {
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System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$ This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3... | Observe that your system is equivalent to $$\begin{cases}f(x)=y\\ f(y)=z\\f(z)=x\end{cases}$$ Where $$f(x):=3\cdot \left[\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x\right]$$ Let $g(x)=f(x)-x$. Adding up the three equations yields $\sum g(x)=0$, so at least one of $g(x), g(y), g(z)$ ... | {
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"url": "https://math.stackexchange.com/questions/4020476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve this limit $\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$? I need to find the limit of $$\lim_{x\to \infty}\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)$$
I did rooting like this:
$$\frac{\left(x-\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)*\left(x+\sqrt{\frac{4x^3+3x^2}{4x-3}}\right)}{x+\sqrt{\... | Putting the terms in the numerator over the common denominator one obtains:
$$\begin{align}\frac{x^2-\frac{4x^3+3x^2}{4x-3}}{x+\sqrt{\frac{4x^3+3x^2}{4x-3}}}&=\frac{-6x^2}{(4x-3)x\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}\\
&=\frac{-6}{(4-\frac3{x})\left(1+\sqrt{\frac{4x+3}{4x-3}}\right)}.
\end{align}$$
Can you get it fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Uniform Convergence of series expansion of $\frac{\log(1+x)}{x}$ We know that the series expansion of $\frac{\log(1+x)}{x}$ is
$$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}....$$
It is also known that the value of dilogarithm at 1 is
$$\int_{-1}^{1} \frac{\log(1+x)}{x}=2\sum_{k=1}^{\infty}\frac{1}{(2k... | For $x \in [-1+\delta,1-\delta]$ where $0 < \delta < 1$, we have
$$\left|(-1)^k \frac{x^k}{k+1}\right|\leqslant \frac{(1-\delta)^k}{k+1} \leqslant (1-\delta)^k$$
As the geometric series $\sum_{k \geqslant 0} (1-\delta)^k$ converges, it follows by the Weierstrass M-test that we have uniform convergence on $[-1+\delta,1+... | {
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"timestamp": "2023-03-29T00:00:00",
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Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}... | You could also evaluate the integral in this way:
$$\int \frac {1}{49x^2 + 25}dx=\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx$$
By knowing that $\int\frac{1}{x^2+a^2}dx=\frac1a\tan^{-1}(\frac xa)$ we have:
$$\frac1{49}\int\frac1{x^2+(\tfrac57)^2}dx=\frac1{49}\times\frac75\tan^{-1}\left(\frac{7x}{5}\right)+C=\frac1{35}\tan^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct re... | Note that $1-\cos\theta=2\sin^2\frac{\theta}2$:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}=\lim_{\theta\to0}\frac{2\sin^2\frac{\theta}2}{\theta \sin\theta}$$
Also we have $\sin u\sim u$ when $u\to0$ so:
$$\lim_{\theta\to0}\frac{2\times(\frac{\theta}2)^2}{\theta\times\theta}=\frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Selecting at least 4 questions of 12 and at least one of the first 3 questions must be taken?
There are 12 questions. Students must take at least 4 questions and at least one of the first 3 questions is taken too.
How many configurations are there?
My attempt
\begin{align}
\text{answer} &= \sum_{i=1}^3\left({3\choose... | Yes that is correct answer. Another way to approach it -
If there are $12$ questions, we have $2^{12}$ ways to select any number of questions including no selection $\big[{12 \choose 0} + {12 \choose 1} + ...+ {12 \choose 12}\big]$.
Out of that ${12 \choose 0} + {12 \choose 1} + {12 \choose 2} + {12 \choose 3} = 299$ a... | {
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"question_score": "2",
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Division in inequality I have the problem
$(x-3)*(x+3) \leq x*(3+x)$
and with the following steps
$(x-3)*(x+3) \leq x*(3+x)$
$x^2-9 \leq 3x+x^2$
$-9 \leq 3x$
$-3 \leq x$.
However, my first thought was not to go this way but rather divide by $(3+x)$, which yields
$x-3 \leq x$.
Obviously this does not work. My first thou... | Actually I think you can solve this problem by dividing by $x+3,$ provided that you handle the sign of this term correctly and also take care of the special case
$x + 3 = 0.$
Consider three cases: $x + 3 = 0,$ $x + 3 > 0$, $x + 3 < 0.$
Case $x + 3 = 0$:
In this case $(x-3)(x+3) = 0 = x(3+x),$ so it is certainly the cas... | {
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"source": "stackexchange",
"question_score": "1",
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Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$ I would like to ask how to integrate the following integration.
\begin{gather*}
\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}
I try to use integration by parts and I stuck with the following
\begin{gather*}
\frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\l... | Let $u=\log(x^2+1)$ and $dv=\frac{dx}{x^2}$.
Then $du = \frac{2xdx}{x^2+1}$ and $v=\frac{-1}{x}$.
Therefore
$$
\int_0^1 \frac{\log(x^2+1)}{x^2}dx = \left[\frac{-\log(x^2+1)}{x}\right]_0^1 - \int_0^1\frac{2dx}{x^2+1}
$$
The value of $\frac{\log(x^2+1)}{x}$ at 0 is undefined, but we can compute it with de l'Hôpital's ru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The way to deduce the following equations. Given the $5$ equations.
$a_1=a_2+a_5 \tag{1}$
$a_4=a_3+a_5 \tag{2}$
$b=a_1+a_2 \tag{3}$
$b=a_3+\frac{a_4}{2} \tag{4}$
$\frac{a_5}{2}+a_1=a_3 \tag{5}$
I want to deduce the following.
$a_1=\frac{9}{16}*b\tag{6}$
$a_2=\frac{7}{16}*b\tag{7}$
$a_3=\frac{5}{8}*b\tag{8}$
$a_4=\frac{... | Hint 1: As the comments have stated, one approach is to write the system as a matrix and then find the RREF. The matrix is given by $$A = \left(
\begin{array}{cccccc}
1 & -1 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & -1 & 1 & 0 \\
1 & 1 & 0 & 0 & 0 & b \\
0 & 0 & 1 & \frac{1}{2} & 0 & b \\
1 & 0 & -1 & 0 & \frac{1}{2} & 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$ If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
The solution goes as follows:
$a=\frac{x}{y+z}$, $b=\frac{y}{z+x}$, $c=\frac{z}{x+... | if $x,y,z > 0 $, say $a = \frac{x}{y+z} , b=\frac{y}{x+z} , c=\frac{z}{x+y} $
then we'll proof that $$(a+b+c)^2 \ge 4 - 14\cdot a \cdot b \cdot c $$
$(a+b+c)^2 = a^2+2.a.b+b^2+2.a.c+2.b.c+c^2 = a^2+b^2+c^2+2(a.b+b.c+a.c) $
as you said $a.b+a.c+b.c+2.a.b.c = 1$
$a^2+b^2+c^2 +2(1-2.a.b.c) \ge 4 - 14.a.b.c $
$ a^2+b^2+c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is a determinant in matrix ? NOT how to get it but WHAT is it in real life We know to get determinant in a matrix we use sarrus, laplace, x methods and so on
But question : WHAT is it, according to some pages is about permutations the products of element of the matrix , but WHAT IS IT ?
| \begin{align}
& \left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} ax+by \\ cx+dy \end{array} \right] \\[8pt]
& \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \mapsto \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the following integral converge or diverge and can it be solved? I have been stuck solving the following integral for some time now: $$\int_0^{2R} \frac{\sin^2\left(\frac{x}{4R}\pi\right)R^2}{\left(\sqrt{\left(d+Z\right)^2+\sin^2\left(\frac{x}{4R}\pi\right)}\right)^3}\,\mathrm{d}x$$
If it isn't possible to solve, ... | First note that your integral diverges for $d + Z = 0$. For $d + Z \neq 0$ we can write
\begin{align}
I (R,d,Z) &\equiv \int \limits_0^{2R} \frac{R^2 \sin^2 \left(\frac{\pi x}{4R}\right)}{\left[(d+Z)^2 + \sin^2 \left(\frac{\pi x}{4R}\right)\right]^{3/2}} \, \mathrm{d} x \stackrel{x = \frac{4 R t}{\pi}}{=} \frac{4 R^3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.
I solved it as follows:
$x+2\sqrt{x}-1\ge 0$
$(\sqrt{x}+1)^2-2\ge 0$
$(\sqrt{x}+1)^2\ge 2$
$... | Let's use Mathematica for an overview plot:
ReImPlot[Sqrt[x + 2 Sqrt[x] - 1] + Sqrt[x - 2 Sqrt[x] - 1],
{x, -1, 7}, PlotPoints -> 10^3,
GridLines -> {{0, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2]},
{0, 2 Sqrt[Sqrt[2] - 1], 2, 2 Sqrt[1 + Sqrt[2]]}},
PlotTheme -> "Scientific"]
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to factorize $P=(X^2-4X+1)^2+(3X-5)^2$ in $\mathbb R[X]$? First of all, I searched for roots. I knew that $\exists x\in\mathbb R,\, P(x)=0 \iff \exists x\in\mathbb R,\, x^2-4x+1 = 0 \text{ and } 3x-5 = 0$.
However, It is really easy to say that there is no real roots. So there is only complex non real roots.
And we... | Assume (hope) that $a,b,c,d\in\Bbb Z$. If $bd=26$ then either:
*
*$b=1,d=26$ or $b=26,d=1$ which gives $ac=0$ but this is absurd;
*$b=2,d=13$ which gives $13a+2c=-38$, equating with $a+c=-8$ gives $a=-2,c=-6$.
Indeed $(X^2-4X+1)^2+(3X-5)^2=(X^2-2X+2)(X^2-6X+13)$. This is the unique irreducible factorisation so we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Side of a square attached to two triangles Solve for $l$
I tried to solve this for $l$ with trigonometry, but I stopped at $l^2+4-2l\cdot \cos(x)=16, l^2 + 4 +2l\cdot \sin(x)=25$, where $x$ is the angle of the left triangle between 4 and $l$.
So I've tried with Heron's, and got stuck with some square roots...
Accordin... | Rearrange the two equations as
$$l^2-12=2l\cos x \quad , \quad l^2-21=-2l\sin x$$
On squaring both and adding, $x$ is eliminated,
$$(l^2-12)^2+(l^2-21)^2=(2l)^2$$
$$\Rightarrow 2(l^2)^2-70l^2+585=0$$
and a quadratic in $l^2$ is obtained. Its value can be computed using the quadratic formula.
To compute $l$, remember to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?
How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?
I thought that $a_n$ can be constructed in three ways:
*
*The strings that contain both three consecutive $0s$ and fo... | Well , first of all we should think about basic set theory . We know that all situations consist of $4$ subcases such that ( containing $3$ consecutive zeros and $4$ consecutive ones ) $\color{red}{\cup}$ ( containing $3$ consecutive zeros and do not contain $4$ consecutive ones ) $\color{red}{\cup}$ (do not contain $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Generating Function for Recursive Sequences (generalization to the binomial series) Let $S_k$ be a recursive sequence defined by
$S_k = (x+1)S_{k-1}+S_{k-2}$ with $k>1$
with initial conditions:
$S_0 = 2, S_1=x+1$.
Then the first few terms are
$2, x + 1, x^2 + 2x + 3, x^3 + 3x^2 + 6x + 4, x^4 + 4x^3 + 10x^2 + 12x + 7,\\... | The solutions to $X^2 -(x+1)X-1=0$ are $X_1= \frac{x+1+\sqrt{(x+1)^2+4}}{2}$ and $X_2= \frac{x+1-\sqrt{(x+1)^2+4}}{2}$. Then, with the given initial values, we have
$$S_k(x)= X_1^k+X_2^k.$$
Since by definition $$S_n(x):=\sum_{t\ge 0}x^{n-t}P_{t,n},$$
we have $$S_n(x^{-1})=\sum_{t\ge 0}x^{t-n}P_{t,n},$$
that is $$\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$ My question is to find the value of $$\lfloor (\sqrt{21}+\sqrt{29})^{1984}\rfloor \mod 100$$
Mathematica told me the result is 71 and $$\lfloor (\sqrt{21}+\sqrt{29})^{20k+4}\rfloor \mod 100$$ is always 71 for $k\in \mathbb N$, can anyone explain t... | Let $\alpha = (\sqrt{21}+\sqrt{29})^{2} = 50 + 2 \sqrt{609}$. Then $\alpha^n = x_n + y_n \sqrt{609}$ with $x_n, y_n \in \mathbb Z$.
Let $\beta = (\sqrt{21}-\sqrt{29})^{2} = 50 - 2 \sqrt{609}$. Then $\beta^n = x_n + y_n \sqrt{609}$ and so $\alpha^n+\beta^n = 2x_n \in \mathbb Z$.
Since $0 < \beta < 1$, we get $\lfloor \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Probability of Sum being $4$ with unfair dice You roll two six-sided dice. Die $1$ is fair. Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rollin... | There is a mistake. For the unfair die, the probability of an odd number is $\frac{2}{3}$ and of an even number is $\frac{1}{3}$. It also says that the probability of each odd number is same and of each even number is same.
So for the unfair die, probability of each odd number is $\frac{2}{9}$ and of each even number i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of an interception point as one parameter changes. Consider the functions
\begin{align*}
h_1(z)&=m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)\\
h_2(z)&=\frac{-M}{(a^2+z^2)^{3/2}}
\end{align*} where $M$, $m$ and $e$ are real parameters and $e<0$. I know that these two only have one interception point, $z_0$ (I chec... | It should actually be $-2e$, which assuming that $e<0$ will be positive. We can show this by considering solutions to $h_1(z) = h_2(z)$, since a point $z = z_0$ which satisfies this equation is an intercept. So, we have
\begin{align*}
h_1(z) &= h_{2}(z)\\[5pt]
\implies m\bigg(\frac{1}{z^3} + \frac{2e}{z^4}\bigg)&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$ Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting th... | It's $\,(\color{#0a0}{n^4\!+\!1})(\color{#c00}{n^4\!-\!1})^2$ so $\,n\,$ odd $\,\Rightarrow\begin{align}&\ \,2\mid \color{#0a0}{n^4\!+\!1}\\ &2^4\mid \color{#c00}{n^4\!-\!1}\end{align}\,$ since $\,\begin{align} n&\equiv \pm1\ \ \pmod{\!4}\\\Rightarrow \color{#c00}{n^4}&\equiv \color{#c00}{(\pm1)^4}\!\!\!\!\pmod{\!4^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | Using the result $\int_0^{\infty} \frac{\ln x}{x^2+1} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x &=\frac{1}{a} \int_0^{\infty} \frac{\ln a+\ln y}{y^2+1} d y \\
&=\frac{\ln a}{a} \int_0^{\infty} \frac{d y}{y^2+1}+\frac{1}{a} \int_0^{\infty} \frac{\ln y}{y^2+1} d y \\
&=\frac{\pi \ln a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be? If we let $x = 0$.
\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}
Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.
But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $... | $3 (x + 7) - y (2 x + 9)=21-9y$ for $x=0$ and
$3 (x + 7) - y (2 x + 9)=24 - 11 y$ for $x=1$
They are the same, so must be
$$21-9y=24-11y\to y=\frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \fr... | $$L=\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}$$
Lets try using our dear L'Hopital's rule to simplify the expression.
$$L=\lim_{x\to 0} \frac{2xe^{x^2} -2\sin x }{2x\sin (2x^2)}$$
We are very familiar with these functions! So lets expand them since they're easier to deal with now.
$$L=\lim_{x\to 0} \frac{x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$ I'm trying to solve the following sum for a project. Many thanks in advance for answering the question!
$$\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m+a)}{\Gamma(k+m+b)}(k+m)$$
where $m$ is an ... | Using this result for hypergeometric functions of unity argument as well as the property of the gamma function $x\Gamma(x)=\Gamma(x+1)$ we find:
$$
\begin{align}
S
&=\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m
+a)}{\Gamma(k+m+b)}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(m
+a)_k}{(m+b)_k}(k+m)\\
&=\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$ With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?
$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$
Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count ... | We have two multiplicity patterns possible for numbers given the digits available:
$(2,1,1,1) \approx wwxyz$ (A) and $(2,2,1)\approx wwxxy$ (B).
Pattern A can be filled in $\binom 31 \binom 33 = 3$ ways and permuted in $\binom{5}{2,1,1,1}$ $ = 5!/2! = 60$ ways
Pattern B can be filled in $\binom 32 \binom 21 = 6$ ways a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
matrices such that $A^2B+BA^2=2ABA$ Let be $A$, $B$ two matrices $3 \times 3$ with complex entries.
Prove that if $$A^2B+BA^2=2ABA$$
THEN
$$B^2A+AB^2=2BAB$$
I tried it and do not know how to continue.
If $A$ is invertible then
$$AB^2+A^{-1}BA^2B=2BAB$$
so I have to prove that
$$B^2A=A^{-1}BA^2B$$
How to continue and al... | The statement in question is equivalent to $[A,[A,B]]=0\rightarrow[[A,B],B]=0$ and it is false. Here is a counterexample taken from the last section of Irving Kaplansky, Jacobson's Lemma Revisited, Journal of Algebra, 62, 473-476 (1980):
\begin{aligned}
A&=\pmatrix{1&1&0\\ 0&1&0\\ 0&0&1},\quad B=\pmatrix{0&0&0\\ 0&0&1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$ The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have ... | By Taylor, we have
$$\csc \left(\frac{\pi n}{6 (n-2)}\right)=2-\frac{2 \pi }{\sqrt{3} n}+\frac{\frac{7 \pi ^2}{9}-\frac{4 \pi
}{\sqrt{3}}}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=4-\frac{8 \pi }{\sqrt{3} n}+\frac{8 \pi \left(5 \pi -6 \sqrt{3}\right)}{9
n^2}+O\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Congruency application I want to prove whether the following is true using congruencies: If $n$ is odd and $3\not | n$, then $n^2\equiv 1\pmod{24}$.
I tried a direct proof.
Let $n=2k+1$ for an integer $k$ such that $3\not| n$, or $3c=n+r$ such that $r=1$ or $r=2$ for an integer $c$. Then in the first case, $3c=n+1$ imp... | If $n$ is odd and not divisible by $3$. Then $n \equiv \pm 1 \pmod{6}$.
$$n = 6k \pm 1$$
$$n^2 = 36k^2 \pm 12k + 1=12k(3k \pm 1) +1$$
We consider two cases if $k$ is even, then $12k(3k \pm 1)$ is divisible by $24$.
If $k$ is odd, then $3k\pm 1$ is even, hence $12k(3k\pm 1)$ is again divisible by $24$.
Hence $n^2 \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to resolve this integral $J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$? $$J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$$
I'm already try to do this step in below:
$$\overset{split}{=} \int{\frac{2}{2+\cos{x}}dx}+\int{\frac{d(\cos{x}+2)}{2+\cos{x}}} = \int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx} + \ln({2+\cos{x}})$$
| We need to find $$J:=\int \frac{2-\sin(x)}{2+\cos(x)}\operatorname{dx}.$$
The answer is $$\boxed{J=\frac{4}{\sqrt{3}}\arctan\left(\frac{\tan(x/2)}{\sqrt{3}} \right)+\ln(\cos(x)+2)+C}$$
For solve $J$, we can use the substitution recommended by @Prime Mover.
Let, $\color{blue}{u=\tan(x/2)}$ so we have that $$\operatornam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proving that $3^n \gt n^4 \ \forall n \gt 8$ Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:
Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$
Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$
Inductive step ($P(n) \implies P(n+1)$): $\ 3^{n+... |
Inductive step ($P(n) \implies P(n+1):) \ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$
You need to PROVE $ 3^{n+1} \gt \ (n+1)^4 =n^4 +4n^3+6n^2+4n+1$. You can NOT start by assumenting that.
But you start by assuming $3^n > n^3$.
So $3^{n+1} = 3\cdot 3^n > 3n^4$
So how does $3n^4$ compare with $(n^4 + 4n^3 + 6n^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $P=ab+bc+3ca+\dfrac{3}{a+b+c}$ Let $a,b,c$ be non-negative real numbers such that $a^2+b^2+c^2=3$. Find the minimum value of $$P=ab+bc+3ca+\dfrac{3}{a+b+c}.$$
This is an asymmetric inequality. It is hard for me to find when the equation holds. I guess when it occurs if $a=c=0$ and $b=\sqrt{3}$... | As $a,c \ge 0$ then
$$P \ge S := ab+bc+ca + \frac{3}{a+b+c}$$
We minimize $S$
As $a^2+b^2 +c^2 = 3$ then
$$S =\frac{1}{2} (a+b+c)^2+\frac{3}{a+b+c} -\frac{3}{2}$$
Denote $x = a+b+c$. Because $a^2+b^2 +c^2 = 3$ then $\sqrt{3} \le x \le 3$
$$S = f(x) := \frac{1}{2}x^2+\frac{3}{x}-\frac{3}{2}$$
As $f'(x) = x-\frac{3}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $x^x+y^y\ge x^2+y^2$ for $x,y>0$ and $x+y\le 2$. We may prove the inequality for $x,y\in (0,1]$.
Note that, for $0<x\le 1$, it holds that
\begin{align*}
x^x&=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\cdots\\
&\ge1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\\
&\ge x^2.
\end{align*}
Similarily, for $y \in (0,1]$, it holds that $$... | Here is a sketch. As you have shown, it suffices to show that
$$x^x+y^y\ge x^2+y^2$$
if one of $x,y$ is bigger than 1. Assume $x\in (1,2]$, then $y\le 2-x\in (0,1).$ One can check that the function $x^x-x^2$ is decreasing on $(0,1)$ and thus
$$y^y-y^2\ge (2-x)^{2-x}-(2-x)^2.$$
Thus it suffices to show that
$$x^x-x^2+(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove the identity $\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}$ Prove the identity $$\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}=2\sqrt{2}.$$
We have $$\left(\sqrt{17+2\sqrt{30}}-\sqrt{17-2\sqrt{30}}\right)^2=17+2\sqrt{30}-2\sqrt{17+2\sqrt{30}}\cdot\sqrt{17-2\sqrt{30}}+17-2\sqrt{30}=34-2\sqrt{(17)^2-(2\sqrt{30})... | both of your original real numbers are roots of
$$ x^4 - 34 x^2 + 169 $$
Standard bit for quartic with no cubic term and no linear,
$$ (x^2 - 13)^2 - 8 x^2 = x^4 - 34 x^2 + 169 $$
$$ (x^2 - 13)^2 - (x \sqrt 8)^2 = x^4 - 34 x^2 + 169 $$
This becomes ( because a difference of squares)
$$ (x^2 - x \sqrt 8 - 13)(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2... | Using the Quadratic Formula is more general but here is an alternative method:
We have $abx^2-(a^2+b^2)x+ab=0 \iff x^2-\frac{a^2+b^2}{ab}x+1=0 \iff x^2-(\frac{a}{b}+\frac{b}{a})x+1=0$.
Let $r_1$ and $r_2$ be the (possibly equal, possibly complex) roots of the quadratic function $x^2-(\frac{a}{b}+\frac{b}{a})x+1$. Then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| COMMENT.-For $x=y$ one has $2x^3=44x^2$ so $x=0$ and $x=22$. Besides we have
$(x+y)^3-3xy(x+y)=(x+y)^2+40xy\iff S^3-3SP=S^2+40P$ where $S$ is sum and $P$ is product. Note that if $(x,y)$ is solution so is $(y,x)$.
We have $S|40P$, and $P=\dfrac{S^3-S^2}{3S+40}$. We consider first $S|40$ so $S=1,2,4,5,8,10,20,40$.
$$S=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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MGF of standard normal raised to a power Let $Z \sim \mathcal N(0,1)$ be a standard normal variable. What is the moment generating function $M_{Z^d}(t)$ for the variable $Z^d$ where $d$ is a positive integer?
For $d = 1$, we have $M_Z(t) = e^{t^2/2}$, and for $d = 2$, thinking of the variable as Chi-Squared with $1$ de... | It is worth noting that the moment-generating function $M_{Z^d}(t)$ of $Z^d$ corresponds to the moment-generating function of the Chi-square distribution for $d=2$. Furthermore, for all other $d>2$ where $d$ is even, it is only finite if $t \leq 0$, whereas it is infinite for $t>0$. This is because
\begin{align}
\mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings? 3 different families have 3 children each. In how many ways we can put them into 3 tents, such that in every tent, there are at least 2 siblings?
Finding the total different... | I'm assuming that the tents are indistinguishable and can contain $\leq3$ children.
First put the children family-wise into the tents, so that we have the three tents
$$(A_1,A_2,A_3),\quad(B_1,B_2,B_3),\quad(C_1,C_2,C_3)\ .$$
No we can make certain permutations, moving $\leq1$ children from each tent:
*
*Leave it as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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When we express $\sin x - \cos x$ as $A \sin (x+c)$, how many solutions are there for $c \in [0, 2\pi)$? This is a problem from problem set 1 of MIT OCW 18.01SC:
express $\sin x - \cos x$ in the form $A \sin (x+c)$.
Their solution is $\sqrt{2} \sin (x - \frac{\pi}{4})$
I found two solutions (for $c \in [0, 2\pi]$):
S... | As you wrote, $$A\sin(x+c)=A\sin x\cos c+A\cos x\sin c=\sin x-\cos x$$ leads, by identification, to the system
$$\begin{cases}A\cos c=1,\\A\sin c=-1.\end{cases}$$
This is a Cartesian-to-polar transformation, which can be represented graphically. The point is uniquely determined, and if you assume a positive modulus, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $\lim \sqrt{1-\frac{1}{n}} = 1$ by definition Is my attempt correct?
Proof. Let $\varepsilon > 0$. Take $N > \frac{1}{\varepsilon}$ and let $n \geq N$. Then
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lve... | As the function values are smaller than $1$, we have to solve the inequation
$$1-\sqrt{1-\frac1n}<\epsilon$$ (we can assume $\epsilon\le1$ because for larger $\epsilon$ the inequation is always true), or
$$1-\epsilon<\sqrt{1-\frac1n},$$
$$(1-\epsilon)^2<1-\frac1n,$$
$$\frac1n<(2-\epsilon)\epsilon,$$
$$n>\frac1{(2-\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.