Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the fi... | Define
$a=6-2\sqrt 3>0$
$b=3\sqrt 2-2>0$
$a-b = 8 - (2\sqrt 3 + 3\sqrt 2)$
$(2\sqrt 3 + 3\sqrt 2)^2 = 30+12\sqrt 6 = 6×(5+2\sqrt 6)
< 60 < 64$
because $6=2×3 < (5/2)^2$
$a-b > 8-8=0, a>b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
In a triangle $a:b:c =4:5:6$, then $3A+B$ equals to? In the above question $a,b,c$ are sides of triangle and $A,B,C$ are angles. The correct answer is $\pi$ but I am getting $\pi - C$.
| $$\cos{A}=\frac{5^2+6^2-4^2}{2\cdot5\cdot6}=\frac{3}{4}.$$
Thus, $$\sin{A}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}<\frac{1}{\sqrt2},$$
which says that $A<45^{\circ}$, which gives that $2A$ is an acute angle
and $$\sin2A=2\cdot\frac{\sqrt7}{4}\cdot\frac{3}{4}=\frac{3\sqrt7}{8}.$$
In another hand, $$\cos{C}=\frac{4^2+5^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Absolute value equation infinite solutions $$|3-x|+4x=5|2+x|-13$$
One of the solutions is $[3,\infty)$
I'm not familiar with interval solutions for absolute equations.
How to solve for this interval?
| The first thing to do is to notice that if $x$ is large and positive, then
$|3-x|$ is the same as $x-3$, so the equation becomes
$$
(x-3)+4x = 5(x+2) -13 \\
5x-3 = 5x -3
$$
and so any time $x$ is that large, the equation is automatically true. How large is "that large"?
Well, as long as $x\geq 3$ the essential proper... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$10(x + y) + 4z = 5xyz$. Find the min value of: $A = 2x + y + 2z$
PROBLEM: We have $x,y,z \in \mathbb{R}^+$ such as $10(x + y) + 4z = 5xyz$. Find the min value of:
$$A = 2x + y + 2z$$
My attempt:
I would find the min of:
$$A^2 = \frac{(2x+y+2z)^2(10x+10y+4z)}{5xyz}$$
and then let $z=1$, after that I used Lagrange m... | For $x=\frac{4}{\sqrt5}$, $b=\frac{6}{\sqrt5}$ and $c=\sqrt5$, we get a value $\frac{24}{\sqrt5}$.
We'll prove that this is a minimal value.
Indeed, let $x=\frac{4a}{\sqrt5}$, $y=\frac{6b}{\sqrt5}$ and $z=\sqrt5c$.
Hence, the condition gives $2a+3b+c=6abc$ and we need to prove that
$$4a+3b+5c\geq12$$ or
$$\frac{(4a+3b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
picking two same colored balls at the time
There are 5 white and 5 black balls in a bin. We take two balls at the time without replacement. Describe X = number of takes where both balls have the same color.
I realized that the minimal number of takes will be 0, and the maximum 5. I started describing for $X=0$:
$$P=5... | In every turn, you pick two balls. The color of the first ball does not matter, as long as the second ball has a different color. As such, we get:
$$P[X=0] = \frac{5}{9} \cdot \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1} = \frac{120}{945} = \frac{8}{63}$$
It is not possible to have $X = 1$, since i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tangents to the curve $y=x^3-3x^2-7x+6$ cut off on the negative semi axis $OX$ a line segment half that on the positive semi axis $OY$ is The coordinates of points at each of which the tangents to the curve $y=x^3-3x^2-7x+6$ cut off on the negative semi axis $OX$ a line segment half that on the positive semi axis $OY$ ... | Let the point of tangency be $(x_1,y_1)$
Then the equation of tangent is $\displaystyle\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$.
When $y=0$, $\displaystyle x=x_1+\frac{-y_1}{3x_1^2-6x_1-7}$.
When $x=0$, $\displaystyle y=y_1-x_1(3x_1^2-6x_1-7)$.
\begin{align}
-2\left(x_1+\frac{-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required... | HINT:
The equation of any straight line passing through $(1/2,2)$ is $$\dfrac{y-2}{x-1/2}=m\iff y=mx+2-m/2$$ where $2m$ is the gradient.
Let us find the intersection of this line with the given parabola.
$$mx+m-\dfrac12=2-\dfrac{x^2}2\iff x^2+2mx+2m-5=0$$ which is a quadratic equation in $x$ whose each of the two roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm st... | From where you were stuck,
$$\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)} = \frac{\cos^2x(1 - \sin^2x) + \sin^2x}{\cos x(\sin x \cos x+1)} = \frac{(\cos^2x + \sin^2x) - \cos^2x\sin^2x}{\cos x(\sin x \cos x+1)} = \frac{1 - \cos^2x\sin^2x}{\cos x(\sin x \cos x+1)} = \frac{(1+\sin x \cos x)(1-\sin x \cos x)}{\cos x(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| \begin{align*}
x^3+x&=5x^2\\
x+\frac{1}{x}&=5\\
\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2&=7\\
\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)&=\sqrt{7}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where $d$ is a fixed point of $u_n$ Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where:
*
*$d$ is a fixed point of the recurrence relation $u_n=au_{n-1}+b$;
*$A$ is a constant.
My attempt:
I have tried evaluating the recurrence relation and I've come ... | We have
$$\frac{u_n}{a^n}=\frac{u_{n-1}}{a^{n-1}}+\frac{b}{a^n}$$
So,
\begin{align}
\sum_{k=1}^n\frac{u_k}{a^k}&=\sum_{k=1}^n\frac{u_{k-1}}{a^{k-1}}+\sum_{k=1}^n\frac{b}{a^k}\\
\frac{u_n}{a^n}&=u_0+\frac{\frac{b}{a}\left(1-\left(\frac{1}{a}\right)^n\right)}{1-\frac{1}{a}}\\
&=u_0+\frac{b\left(1-\left(\frac{1}{a}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A simple method to obtain the inverse of Vandermonde matrix For any general $n$, is there any easy way of finding out the inverse of a square matrix of the form:
$$A=van(a_1,a_2,\cdots , a_n)=\pmatrix{ 1 \quad a_1 \quad a_1^2 \ \dots \ a_1^{n-1} \\ 1 \quad a_2 \quad a_2^2 \ \dots \ a_2^{n-1} \\\vdots \qquad \vdots \qq... | When I look at your matrix, I see a matrix that transforms a polynomial in the standard basis to $p(a_0), p(a_1)\cdots p(a_n)$
$\pmatrix{1 \quad a_0 \quad a_0^2 \ \dots \ a_0^n \\ 1 \quad a_1 \quad a_1^2 \ \dots \ a_1^n \\ 1 \quad a_2 \quad a_2^2 \ \dots \ a_2^n \\ \vdots \\ 1 \quad a_n \quad a_n^2 \ \dots \ a_n^n}\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circ... | First of all
$$
\int_{|z|=1}\overline{z}\,dz=\int_{|z|=1}\frac{dz}{z}=2\pi i,
$$
since $z\overline{z}=1$. Meanwhile, for $k>1$
$$
\int_{|z|=1}\overline{z}^k\,dz=\int_{|z|=1}\frac{dz}{z^k}=0.
$$
Now, when $|z|=1$, we have $\overline{z}=z^{-1}$ and hence
$$
|1+z+\cdots+z^{2n}|^2=(1+z+\cdots+z^{2n})(1+\overline{z}+\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Not getting the right answer to $t^2 y'' -2y = 3t^2-1$, given $y_1(t) =t^2$ and $y_2(t) = \dfrac 1t$ I am trying to find a particular solution of $t^2 y'' -2y = 3t^2-1$, given that $y_1(t) =t^2$ and $y_2(t) = \dfrac 1t$ are solutions to the corresponding homogenous equation. I rewrote the equation as $y'' - \dfrac {2}{... |
The following line is incorrect:
$$\displaystyle \int \dfrac {y_2(t) g(t)}{W(y_1, y_2)(t)} dt = \int \dfrac {\frac 1t \left(3-\frac {1}{t^2} \right)}{-3} dt = - \dfrac 13 \int \left(\dfrac 3t - \dfrac {1}{t^3}\right) dt = -\dfrac 13 \left( \ln t - \dfrac {1}{2t^2}\right)$$
This is simply because:
$$\int \frac{3}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining the image of a function
Determine the image of the following function without using calculus:
$$f\left(x\right)=\ \frac{x}{2\left|x\right|+1}$$
My attempt:
We know $\left| x \right| < 2\left| x \right| + 1$, then, $\frac{\left| x \right|}{2\left| x \right| + 1} < 1$ which implies $\left| f\left( x \rig... | As discussed in the comment, we also need to show that $(-\frac{1}{2},\frac{1}{2}) \subseteq r(f)$. For each $y \in (-\frac{1}{2},\frac{1}{2})$, let
$$g(y) = \frac{y}{1-2|y|}$$
Check that $f(g(y)) = y$, so that $(-\frac{1}{2},\frac{1}{2}) \subseteq r(f)$. It follows from Walton's hint that $r(f) \subseteq (-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability of rolling X successes with differently sized dice I'm trying out different systems for a board game, and my (weak) probability skills are failing me.
The basic mechanic is as follows: you roll a number of six-sided dice (d6). Any die showing 5 or more is a success.
A specific bonus can "upgrade" part of yo... | I jump directly to the second question:
same roll with bonus: roll $4d6$ and $3d10$; any die showing 5 or more
is one success.
4d6: The probability that one die show a 5 or a 6 is $\frac26=\frac13$. Now you can use the binomial distribution to calculate that x dice out of $4$ have an success.
$$P(X=x)=\binom{4}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=... | The formula
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-x^2}{1+x^2}}
$$
is wrong. You were probably thinking to
$$
\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}
$$
but there's a better one that doesn't have any sign ambiguity:
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
If $x=\arcsin(5/13)$, then $\cos x>0$, so
$$
\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Frobenius Method general solution not providing an expected solution I'm trying to solve the following ODE:
$$3xy''+(3x+1)y'+y=0$$
So I started by assuming a solution(and its derivates) with the form:
$$
\tag{1}y = \sum_{n=0}^{\infty } a_{n} x^{n + r}$$
$$y' = \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}$$
$$y'' = \su... | For solution $r_1$: $$a_n=\frac{(-1)^n}{n!}a_0$$$$y_1=a_0(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots)\ =\ a_0e^{-x}$$
For solution $r_2$: $$b_n=\frac{(-3)^n}{(5)(8)(11)\cdots(3n+2)}b_0$$
$$y_2=b_0(1-\frac{3}{5}x+\frac{9}{40}x^2-\frac{27}{440}x^3+\cdots)$$
The general solution is $\ y_1+y_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $... | Computing Tangents
Using that $1+\color{#090}{\cos\left(\frac{2\pi}5\right)}+\color{#C00}{\cos\left(\frac{4\pi}5\right)}+\color{#C00}{\cos\left(\frac{6\pi}5\right)}+\color{#090}{\cos\left(\frac{8\pi}5\right)}=0$,
$$
\begin{align}
0
&=\color{#090}{\cos\left(\frac{2\pi}5\right)}\color{#C00}{+\cos\left(\frac{4\pi}5\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Calculation with function For as long as I go around it I do not get to anything
If $\phi \left ( f(x)-1\right )=2x+5$
and
$\phi (x)=2f(x+1)+1$
find $f(4)$
| (Too long for a comment.) Let $\,\varphi(x)=\phi(x-1)\,$, then the equations can be written as:
$$
\begin{align}
\varphi(f(x)) &= 2x+5 \tag{1} \\
\varphi(x) &= 2 f(x) + 1 \tag{2}
\end{align}
$$
Substituting $x \mapsto f(x)$ in $(2)$ and equating to $(1)$ gives:
$$\varphi(f(x)) = 2 f(f(x)) + 1=2x+5 \quad\implies\quad f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to express a matrix as sum of two square zero matrices I have a real square matrix $M$ that I'd like to express as $M=A+B$ such that
$A^2=0$,$B^2=0$. $M$ has an additional property that $M^2$ is a scalar matrix :
($M^2=s^2I$); and it's dimension is a power of 2 : $dim(M)=2^n,n>0$; Any suggestions?
| This can always be done for $M$ of even size $2n$, i.e. $M$ is a $2n \times 2n$ matrix.
Set
$P = \begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix}, \tag{1}$
where $I_n$ is the $n \times n$ identity matrix. Then
$P^2 = \begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix} = 0; \tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))... | $$\frac{\sin{A}}{1+\cos{A}}+\frac{1+\cos{A}}{\sin{A}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}}+\frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}=$$
$$=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}+\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}=\frac{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\sin\frac{A}{2}\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of t... | Neither of your two integrals in the last step are divergent. The first one is
$$
I_1 = \int_0^\infty\frac{1 + \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2}}\,dx = \int_0^\infty \frac{x^2 + 1}{x^4 + 2x^2 + 1} \, dx = \int_0^\infty \frac{dx}{x^2 + 1} = \left[\arctan x\right]_0^\infty = \frac{\pi}{2}.
$$
So that's nice and con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
| You should post your attempts at the question.
Try rearranging some things to fit them together.
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
$$a+b+c=6$$
$$a+c=6-b$$
$$ab+ac+bc=11$$
$$b(a+c)+ac=11$$
$$b(a+c)+ac=11=b(6-b)+ac=11$$
$$ac=\frac{6}{b}$$
$$b(6-b)+ac=11=b(6-b)+\frac{6}{b}=11$$
now you have an equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to ... | Using AM-GM:
$$a^2+b^2+2(a+b)\ge 2ab+2(a+b)=4+2(a+b).$$
If $a,b>0,$ then $a+b \to min$ subject to $ab=2$. Using AM-GM: $a+b\ge 2\sqrt{ab}=2\sqrt{2}.$ Hence: $4+4\sqrt{2}$ is min.
If $a,b<0,$ then $a+b \to max$ subject to $ab=2.$ Maximize $f(a)=a+\frac2a$ to get $a=b=-\sqrt{2}$. Hence: $4-4\sqrt{2}$ is min.
| {
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"source": "stackexchange",
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Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:
If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if an... | $$\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} = 2a$$
$$\Rightarrow \left(\sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2}\right)^2 = 4a^2$$
$$\Rightarrow (x+c)^2 + y^2 + (x-c^2) + y^2 - 4a^2 = 2\sqrt{\left((x+c)^2+y^2\right)\left((x-c)^2 + y^2\right)}$$
$$\Rightarrow 2x^2 + 2c^2 + 2y^2 - 4a^2 = 2\sqrt{\left((x+c)^2+y^2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of all integers $n$ such that $n^2 + 2n +2$ divides $n^3 + 4n^2 + 4n -14$? How to proceed in this problem? I divided $n^3 + 4n^2 + 4n -14$ by $n^2 + 2n +2$ and equated the remainder to zero . I got $n=5$ as answer but the answer is wrong.
| It must help:
$$\frac{n^3+4n^2+4n-14}{n^2+2n+2}=n+2-\frac{2n+18}{n^2+2n+2}.$$
Thus, or
$$\left|\frac{2n+18}{n^2+2n+2}\right|\geq1,$$ which gives
$$-4\leq n\leq4$$
I got $n\in\{0,1,4,-2,-4\}$.
Also, we have $n=-9$.
| {
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If $\frac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$. If $\dfrac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$.
My Attempt:
$$\dfrac {\sin A+\tan A}{\cos A}=9$$
$$\dfrac {\sin A+ \dfrac {\sin A}{\cos A}}{\cos A}=9$$
$$\dfrac {\sin A.\cos A+\sin A}{\cos^2 A}=9$$
$$\dfrac {\sin A(1+\cos A)}{\cos^... | By the Fundamental Theorem of Trigonometry we have:
$$\tag {*} \sin^2x+\cos^2x=1 \Rightarrow \cos x=\sqrt {1-\sin^2x}$$
Now:
$$\frac {\sin A + \tan A}{\cos A}=9$$
$$\frac {\sin A + \frac {\sin A}{cos A}}{\sqrt {1-\sin^2A}}=9$$
$$\frac {\sin A + \frac {\sin A}{\sqrt {1-\sin^2A}}}{\sqrt {1-\sin^2A}}=9$$
Let $\sin A=x$. T... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A limit involving $\sum_{k=1}^{n}\frac{2^k}{k}$ I would like to show that
$\lim_{n \to \infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} = 2$
I have seen proofs that $\sum_{k=1}^n \frac{2^k}{k} \sim \frac{2^{n+1}}{n}$ using the Euler-Maclaurin method but instead I tried with the squeeze theorem.
For one side I showed th... | It is enough to apply the Stolz-Cesàro theorem. By setting $A_n=\sum_{k=1}^{n}\frac{2^k}{k}$ and $B_n=\frac{2^n}{n}$ we have
$$ \frac{A_{n}-A_{n-1}}{B_n-B_{n-1}} = \frac{\frac{2^n}{n}}{\frac{2^n}{n}-\frac{2^{n-1}}{n-1}} = \frac{1}{1-\frac{1}{2}\cdot\frac{n}{n-1}}\tag{1}$$
hence
$$ \lim_{n\to +\infty}\frac{A_{n}-A_{n-1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Closed form of function composition Given that $f(x)=\dfrac{x+6}{x+2}$, find $f^{n}(x)$ where $f^{n}(x)$ indicates the $n$th iteration of the function.
I first tried to find a pattern but there didn't seem to be an obvious one:
$$f(x) = \dfrac{x+6}{x+2}$$
$$f^2(x) = \dfrac{7x + 18}{3x + 10}$$
$$f^3(x) = \dfrac{25x + 7... | If $A=\left(\begin{smallmatrix}1&6\\1&2\end{smallmatrix}\right)$ and if $A^n=\left(\begin{smallmatrix}a_n&b_n\\c_n&d_n\end{smallmatrix}\right)$, then$$f(z)=\frac{a_nz+b_n}{c_nz+d_n}.$$But$$A=\begin{pmatrix}2&1\\-3&1\end{pmatrix}\begin{pmatrix}4&0\\0&-1\end{pmatrix}\begin{pmatrix}\frac15&-\frac15\\\frac35&\frac25\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving an inequality with AM-GM
$X+Y>N$ and $$X+Y+2\sqrt{(X+13)(Y+26)}>-14$$ Find $N$.
Got really confused... Should I use AM-GM which $$2\sqrt{(X+13)(Y+26)}<(X+13)+(Y+26)$$ (since we want to minimize $M+N$) we maximize $$2\sqrt{(X+13)(Y+26)}$$? Got really confused. Please help!
| Since $2(a^2+b^2)\geq(a+b)^2$, we obtain:
$$2(x+13+y+26)\geq\left(\sqrt{x+13}+\sqrt{y+26}\right)^2>25.$$
Thus, $x+y>-26.5$.
For $x\rightarrow-\frac{27}{4}$ and $y\rightarrow-\frac{79}{4}$ say that our estimation was exact.
Thus, a best $N=-26.5$.
| {
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Riccati D.E., vertical asymptotes
For the D.E.
$$y'=x^2+y^2$$
show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:
$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^... | The usual trick to get a better manageable equation out of this Riccati equation is to substitute $y=-\frac{u'}{u}$ which results in the linear ODE of second order
$$
u''+x^2u=0,\quad u(0)=1,\, u'(0)=0
$$
While this still does not lead to a symbolic solution without involving (very) special functions (Convert $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Getting the RHS into summation form Some time ago, I wrote down this identity$$\frac 4\pi=1+\left(\frac 12\right)^2\frac 1{1!\times2}+\left(\frac 12\times\frac 32\right)^2\frac 1{2!\times2\times3}+\ldots$$And being the idiot I was, I didn't write down the RHS into a compact sum.
Question: How do you write the RHS with... | A straight forward use of hypergeometric functions is as follows:
Since it has been established that
$$a_{n} = \left(\frac{(2n)!}{2^{2n} \, n!}\right)^2\frac{1}{n!(n+1)!}$$
then
\begin{align}
a_{n} &= \left(\frac{(2n)!}{2^{2n} \, n!}\right)^2\frac{1}{n!(n+1)!} \\
&= \frac{1}{\Gamma(2) \, n! \, (2)_{n}} \, \left( \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^... | It is well-known from high school the roots of the quadratic $a^2+a+1=0$ are the non-real cube roots of unity, hence $a^3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate the integral $\int_{\pi/4}^{3\pi/4} x \cdot \cot(x) \cdot \csc(x) \mathop{dx}$ $$ \int_{\pi/4}^{3\pi/4} x \cdot \cot(x) \cdot \csc(x) \mathop{dx}
$$
Some working
Evaluate the indefinite integral first
Let $u = \csc(x)$ , then $\frac{du}{- \csc(x) \cdot \cot(x)} = dx$, and
$\csc^{-1} u = x$
This gives
\begin{... | HINT:
Integrate by parts with $u=x$ and $v=-\csc(x)$. This yields
$$\int_{\pi/4}^{3\pi/4} x\cot(x)\csc(x)\,dx= \left.\left (-x\csc(x)\right)\right|_{\pi/4}^{3\pi/4}+\int_{\pi/4}^{3\pi/4}\csc(x)\,dx$$
Can you finish?
| {
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | Hint. Note after the rationalization you should get $a=\frac{x+ \sqrt {x^2-4}}{2}$ which is a solution of the quadratic equation $z^2-xz+1=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\... | Take the original pythagorean identity, and divide by $\sin^2 x$ to get one of the identites, and divide by $\cos^2 x$ for the other.
| {
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How to integrate: $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}dx$ I want to evaluate $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}$. The lecture only provided me with a formula for $\int_{-\infty}^{\infty}dx \frac{A}{B}$ where $A,B$ are polynomials and $B$ does not have real zeros. Unfortunately, in the given case $B$ has a zero at... | $$\frac{1}{x^3+x^2+x+1}=\frac{1}{(x+1)(x^2+1)}=\frac{1}{2}\left(\frac{1}{1+x}-\frac{x}{x^2+1}+\frac{1}{x^2+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$
Find $$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$$
$$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)
= \lim\limits_{x \rightarrow \infty} \frac{e^{-\frac{1}{x}}-1}{x^{-0.5}}
= \lim\limits_{x \rightarrow \i... | I always prefer Taylor expansion to L'Hospital rule, as it is clearer and more generalized. For this one, write:
\begin{align}
\sqrt{x}(e^{-1/x} - 1) & = \sqrt{x}\left(1 - \frac{1}{x} + o\left(\frac{1}{x}\right) - 1\right) = -\frac{1}{\sqrt{x}} + o\left(\frac{1}
{\sqrt{x}}\right) \\
& \to 0 \quad \text{as } x \to \inf... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simultaneous equation with curve and a line $$2x+5y=1;$$
$$x^2+2xy=2y^2+1.$$
The answer is $(3,-1)$, but I keep overcomplicating the solution with fractions and I just want the easiest solution possible.
| Lets try like this
$$2x+5y=1\\x^2+2xy-2y^2=1$$Now square the first equation $$(2x+5y)^2=1\\x^2+2xy-2y^2=1$$Multiply the second by $4$ $$4x^2+20xy+25y^2=1\\4x^2+8xy-8y^2=4$$
Then first minus the second
$$12xy+33y^2=-3\\4xy+11y^2=-1\\y(4x+11y)=-1\\y(2(2x+5y)+y)=-1\\y(2+y)=-1\\y^2+2y+1=0$$
From there $y=-1$ then easily $2... | {
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Find the limit $ \lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} $ Question
$$
\lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x}
$$
I'm not sure how to go about this limit, I've tried to apply L'Hopital's rule
(as shown).
It seems that the form is going to be forever indeterminate? Unless I... | Let $y = \displaystyle \frac{1}{2} - x$. As $x \to \displaystyle \Big(\frac{1}{2}\Big)^{-}$, $y \to 0^{+}$. Then:
$$ \lim \limits_{x \to \Big( \frac{1}{2} \Big)^{-}} \frac{\ln(1-2x)}{\tan(\pi x)} = \lim \limits_{x \to 0^{+}} \frac{\ln(2y)}{\tan\Big( \frac{\pi}{2} - \pi y \Big)} $$
Because $\displaystyle \tan\Big( \frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$
I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph... | First, let us calculate few of $f$'s derivatives:
\begin{align}
f'(x) &= 15x^4+8x^3+3x^2+4x-1,\\
f''(x) &= 2(30x^3+12x^2+3x+2),\\
f'''(x) &= 6(30x^2+8x+1) > 0,\ \forall x\in\mathbb R.
\end{align}
Thus, $f'$ is convex, and $f''$ strictly increasing. Since $\lim_{x\pm\infty} f''(x) = \pm\infty$, $f''$ has exactly one roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363898",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | Let
$$y=\arctan\left(\frac{1+x}{4+x}\right).$$
Then
$$x=\frac{1}{2}\left[3\left(\frac{1+\tan y}{1-\tan y}\right)-5\right].$$
As $x \to \infty, y \to \frac{\pi}{4}.$
So your limit is
\begin{align*}
\lim_{ x\to \infty} \left( \arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x & = \lim_{y \to \pi/4}\left(y-\frac... | {
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Let $x$, $y$ and $z$ be natural numbers satisfying $x^2 + y^2 + 1 = xyz$. Prove that $z = 3$.
Let $x$, $y$ and $z$ be natural numbers satisfying $x^2 + y^2 + 1 = xyz$. Prove that $z = 3$.
I have managed to show that $z$ is a multiple of $3$ by looking at it modulo $3$ but not sure how to show $z=3$. Any tips?
| One could go about this in the following way: If $x = y$, then the equation becomes
\begin{align*}
2x^2 + 1 &= x^2 z
\end{align*}
where the left-hand side is congruent to $1$ (mod $x$) and the right-hand side is congruent to $0$. It follows immediately that $x = 1$, and consequently that $z = 3$.
Now, suppose that $y >... | {
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Evaluate the integral $\int\limits_{-\infty}^\infty \frac{\cos(x)}{x^2+1}dx$. Evaluate the integral $\displaystyle\int\limits_{-\infty}^\infty \frac{\cos(x)}{x^2+1} dx$.
Hint: $\cos(x) = \Re(\exp(ix))$
Hi, I am confused that if I need to use the Residue Theorem in order to solve this, and I am not sure where I should ... | Another approach: A combination of Feynman's Trick and Laplace Transforms:
\begin{equation}
J = \int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + 1}\:dx
\end{equation}
Here let:
\begin{equation}
I(t) = \int_{-\infty}^{\infty}\frac{\cos(xt)}{x^2 + 1}\:dx
\end{equation}
We see $I(1) = J$ and $I(0) = \pi$. Here we take the Lapl... | {
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"timestamp": "2023-03-29T00:00:00",
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If $(1-x^3)^n=\sum_{r=0}^{n} a_r x^r (1-x)^{3n-2r}$, then prove that $a_r=\binom{n}{r} 3^r$. If $$(1-x^3)^n=\sum_{r=0}^{n} a_r x^r (1-x)^{3n-2r}$$ then prove that $$a_r=\binom{n}{r} 3^r.$$
I have factored $1-x^3=(1-x)(1+x+x^2)$ and cancelled out one factor $(1-x)^n$ from both sides after which we get
$$(1+x+x^2)^n=\su... | I begin with the answer :
$\sum_0^n \binom{n}{r} 3^r x^r (1-x)^{3n-2r} =
\sum_0^n \binom{n}{r} (3x)^r ((1-x)^2)^{n-r} (1-x)^n$
which is $(1-x)^n ((3x) + (1-x)^2)^n = (1-x)^n (1+x+x^2)^n$
I tried to make something like $\binom{n}{r} A^r B^{n-r}$
| {
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Minimum of the given expression
For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$
I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its mini... | $$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2 = \left\| \,\, \begin{bmatrix} 1 & -1\\ 1 & 1\\ 2 & -3\end{bmatrix} \begin{bmatrix} a\\ b\end{bmatrix} - \begin{bmatrix} 0\\ 2\\ 0\end{bmatrix} \,\, \right\|_2^2$$
This is a least-squares problem. Since the matrix has full column rank, the minimum is
$$\left\| \,\, \begin{bmatrix} 1 & ... | {
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"answer_id": 1
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Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:
If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$
My solution:
$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number
$\implies ... | Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,
$$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$
Now $$(\cot A-\cot B)^2+(\cot B-\cot C)^2+(\cot C-\cot A)^2\ge0$$
$$\iff\cot^2A+\cot^2B+\cot^2C\ge\cot A\cot B+\cot B\cot C+\cot C\cot A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Need help integrating $\int e^x(\frac{x+2}{x+4})^2 dx $ I have simplified the problem a bit using integration by parts, with $u = e^x(x+2)^2$ and $v = 1/(x+4)^2$ but I'm then stuck with how to integrate this:
$$\int\frac{e^x(x^2+4x+8)}{x+4}dx. $$
I've considered substituting $t = e^x$, but this doesn't seem to make the... | Since
\begin{align*}
\left(\frac{x+2}{x+4}\right)^2 &= \frac{x^2+4x+4}{(x+4)^2} \\
&= \frac{4}{(x+4)^2}+\frac{x(x+4)}{(x+4)^2} \\
&= \frac{4}{(x+4)^2}+\frac{x}{x+4} \\
&=f'(x)+f(x), \\
\end{align*}
let
$$
\color{blue}{f(x)=\frac{x}{x+4}}.
$$
Now for $C$ constant, since
$$
\dfrac{d}{dx}(e^x f(x)+C) = (e^x f(x)+C)'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find out $ k $ in this problem If $ k $ is an non negative integer, such that $ 24^k $ divides $ 13! $, what is $ k $ then?
Explanation needed, as I am new to factorials.
| Hint:
$$24^k=(2^33)^k=2^{3k}3^k$$
\begin{align}
13! &= (13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)\\
&=2^2(3)(2)(3^2)(2^3)(2)(3)(2^2)(3)(2)M \\
&= 2^{10}3^5M
\end{align}
where $M$ is not divisible by $2$ or $3$.
Edit:
For $24^k$ to divide $13!$, we need $3k \leq 10$ (inspect power of $2$) and $k \leq 5$ (inspect power of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$
We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrigh... | An alternative proof to that presented by @Rigel and one that does not require any results from calculus/differentiation is as follows:
Let $\mathbb{u}\equiv\left(\sqrt{a},\sqrt{b},\sqrt{c}\right)$ and $\mathbb{v}\equiv\left(\sqrt{a+\frac{3}{a}},\sqrt{b+\frac{3}{b}},\sqrt{c+\frac{3}{c}}\right)$. From these definitions ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Product $\prod _{n=1}^{\infty } \frac{1}{2} \sqrt{\frac{4 (\pi n-x) (\pi n+x)}{\pi ^2 n^2-4 x^2}}$ $$\prod _{n=1}^{\infty } \frac{1}{2} \sqrt{\frac{4 (\pi n-x) (\pi n+x)}{\pi ^2 n^2-4 x^2}}=\frac{\sqrt{2}}{\sqrt[4]{1+e^{-2 i x}} \sqrt[4]{1+e^{2 i x}}}$$
Mathematica Could not find the solution it is possible to show?... | Hint. One may recall the Weierstrass product of the sine function
$$
\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right),
\qquad x\in(0,\pi),
$$ then one may write, for appropriate values of $x$,
$$
\prod _{n=1}^{\infty } \frac{1}{2} \sqrt{\frac{4 (\pi n-x) (\pi n+x)}{\pi ^2 n^2-4 x^2}}=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Probability of choosing a of b numbers in exactly c guesses (This is my first question, and I have not formally learned much probability, so forgive me if this question is too easy.)
Background information: while watching a UK gameshow called The Code, I began wondering how many questions a contestant would need to ans... | Note that this is can be modeled by a hypergeometric distribution for $a-1$ successes out of the first $c-1$ guesses, along with a $\frac{1}{b-c+1}$ chance of getting the last number on the next guess. Plugging in to the formula should give us $\frac{\binom{a}{a-1}\binom{b-a}{c-a}}{\binom{b}{c-1}} \times \frac{1}{b-c+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Formula for consecutive residue of primitive modulo n. \begin{align*}
3^0 \equiv 1\mod 7\\
3^1 \equiv 3\mod 7\\
3^2 \equiv 2\mod 7\\
3^3 \equiv 6\mod 7\\
3^4 \equiv 4\mod 7\\
3^5 \equiv 5\mod 7\\
3^6 \equiv 1\mod 7\\
3^7 \equiv 3\mod 7\\
\end{align*}
Now just focusing on 1, 3, 2, 6, 4, 5, 1....
How to devise a formula ... | You just need to multiply by $3$ and take the residue mod $7$. This is because $3^j$ mod $7$ is the same as $3^{j-1}\cdot 3$ mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I ... | From the angle addition formula, we have
$$\begin{align}
\cos(2x)&=\cos(x+x)\\\\
&=\cos(x)\cos(x)-\sin(x)\sin(x)\\\\
&=\cos^2(x)-\sin^2(x)\\\\
&=\left(1-\sin^2(x)\right)-\sin^2(x)\\\\
&=1-2\sin^2(x)\tag 1
\end{align}$$
Solving $(1)$ for $\sin^2(x)$ yields
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
If
$$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of $$\cos2\theta+\cos2\phi$$
My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2... | Let $s= \cos^2(\theta)+\cos^2(\phi)$, we want $2(s-1)$. Square the first equation and use Pythagorus
\begin{eqnarray*}
a^2+s-2 =2 \sin( \theta) \sin (\phi) \\
(a^2+s-2)^2 = 4(1- \cos^2( \theta))(1- \cos^2 (\phi)) \\
(a^2-2)^2-4+2a^2 s+s^2 =4 \cos^2( \theta) \cos^2 (\phi)
\end{eqnarray*}
From the second equation
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is there a mistake in my evaluation of $\int\frac{3x+1}{\sqrt{5x^2+1}}dx$? I want to evaluate the following integral:
$$\int\frac{3x+1}{\sqrt{5x^2+1}}dx$$
The answer given in my textbook is $\frac{3}{5}\sqrt{5x^2+1}+\frac{1}{\sqrt{5}}\ln(x\sqrt{5}+\sqrt{5x^2+1})$.
I think the author has excess factor $\sqrt{5}$ in l... | hint for another approach
Put $$x=\frac {\sinh (t)}{\sqrt {5}} $$
then it becomes
$$\frac {1}{\sqrt {5}}\int \frac { \frac {3\sinh (t)}{\sqrt {5}}+1}{\cosh (t)}\cosh (t)dt $$
$$=\frac {3}{5}\cosh (t)+\frac {t}{\sqrt {5}}+C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expectation of a die roll summed I know this problem involves conditional probability, but I'm confused as to how to tackle it.
Assume a die is rolled over and over, where the total is summed. If the die's roll is $\geq 3$ the game stops and the summed total is read out. What is the expectation of the total? What is th... | Another way of looking at it:
If you roll a 3,4,5, or 6 on the first roll (1/6 chance each), then the expected total will be that number (3, 4, 5, or 6).
If you roll a 1 or a 2, (also 1/6 chance each), then the expected roll is 1 or 2, PLUS the total expected roll.
So the total expected roll $T$ should be:
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Let $f(x)=(x+a)(x+b)(x+c) +(x-a)(x-b)(x-c)$. Find all factors of $f(x)$ Let $f(x)=(x+a)(x+b)(x+c) +(x-a)(x-b)(x-c)$. Find all factors of $f(x)$. Hence find the remainder of $2020 \times 2025\times 2027$ when divided by 2017.
When $x=0$ , $f(0)=0$. Therefore $(x-0)$ is a factor.
Then $f(x)\equiv x(Ax^2+Bx+C) \equiv (x+... | HINT: prove the $f(x)$ has the form $$f(x)=2\,x \left( ab+ac+bc+{x}^{2} \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of series of fractions I am trying to find the $f$ formula that returns the sum of the series created by fractions that have constant nominator and shifting by one denominator.
Here are some examples:
$$f(3) = \frac{3}{1} + \frac{3}{2} + \frac{3}{3} = 5.5$$
or
$$f(4) = \frac{4}{1} + \frac{4}{2} + \frac{4}{3} + \fr... | There is a way to approximate quite well the value
$$k H_k\approx \frac{1}{2}+k \left(\gamma -\log \frac{1}{k}\right)$$
where $\gamma\approx 0.577216$ is Euler constant and $\log$ are natural logarithms
Here is a table which shows
$$
\begin{array}{r|r|r}
k & \textit{actual value} & \textit{approximation}\\
\hline
10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | Noting
$$ 1+\sin\theta=1+\cos(\frac{\pi}{2}-\theta)=2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}), \cos\theta=\sin (\frac{\pi}{2}-\theta)=2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})$$
one has
\begin{eqnarray}
z&=&\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
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Matrices in group theory
Let
$$ S = \left\{ \left(\begin{matrix}
a & a \\
a & a \\
\end{matrix}\right) \ \middle| \ a \in \mathbb{R} \right\}. $$
and $\cdot$ be a usual matrix multiplication operation.
Then is $(S,\cdot)$ a group or semigroup or monoid ?
| We know that matrix multiplication is associative. We need to check that $S$ is also closed under this operation. Indeed,
$$
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
\begin{pmatrix}
b & b \\ b & b
\end{pmatrix}
=
\begin{pmatrix}
2ab & 2ab \\ 2ab & 2ab
\end{pmatrix}
$$
shows that it is closed under multiplication. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\o... | Here is a proof that uses the fact, once in $(4)$ and twice in $(5)$, that
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{1}
$$
A geometric proof of $(1)$ can be found in this answer.
Answer Using the Duplication Formulas for $\boldsymbol{\sin(x)}$ and $\boldsymbol{\cos(x)}$
Using the duplication formulas for $\sin(x)$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 5
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Qusetion about Gamma function ※I'm not good at English, so please edit my question※
I read a question ,
and tried to prove $\Gamma(x) \geq x^{3}$ For all real number $x \geq 6$ using gamma function's definition and its derivative , but I couldn't prove inequality $\Gamma(x) \geq x^{3}$
This is my attempt.
For $ n \in ... | 1) Statement is false as stated:
$$\Gamma\left(6\right) = 5!=120$$
$$6^{3} = 216$$
2) Proposition: $\Gamma\left(x\right)\geq x^{3}$
for all real $x\geq7$.
Proof:
$$\frac{\Gamma\left(x\right)}{x^{3}}=\frac{\left(x-1\right)\Gamma\left(x-1\right)}{x^{3}}=...=\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the remainder of $N= 10^{10}+10^{100}+10^{1000}+\cdots+10^{10000000}$ divided by $7$ $$N= 10^{10}+10^{100}+10^{1000}+\cdots+10^{10000000}$$. What is the remainder when N is divided by 7?
$$N=(10^{10}+10^{100}+10^{1000}+\cdots+10^{10000000})/7$$
$$Rem[3^{10}+3^{100}+\cdots+3^{10000000}]/7$$
Now I did not underst... | Here I will use notation common for modular arithmetic.
Given $A$ and $a$ integers and $n$ a natural number different from $0$, the following are equivalent:
*
*$(A-a)$ is a multiple of $n$
*$(a-A)$ is a multiple of $n$
*$n$ divides evenly into $(A-a)$
*the remainder of $(A-a)$ when divided by $n$ is zero
*$A-a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Taylor series of $\tan x - \tan (\sin x)$ has all coefficients positive. Why? It's well known that $x > \sin x$ for $x> 0$. The Taylor series of $ x - \sin x$ is also well known, and the coefficients are alternating. However, it appears that the Taylor coefficients of the function $\tan x - \tan (\sin x)$ are all posit... | The coefficients of the Taylor series at the origin of $\tan x$ are non-negative since $f(x)=\tan(x)$ is a solution of the differential equation $$f''(x) = 2\,f(x)\left(1+f(x)^2\right)\tag{1} $$
leading to
$$ f'''(x) = 2\,f'(x)\left(1+3\,f(x)^2\right)=2\left(1+f(x)^2\right)\left(1+3\,f(x)^2\right)\tag{2} $$
$$ f^{(4)}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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How to find the area inside $x^2-xy+y^2=1$?
Find the area inside $x^2-xy+y^2=1$. Tip: $x^2-xy+y^2=\big(x-\frac{y}{2}\big)^2+\big(\frac{\sqrt 3y}{2}\big)^2$.
I thought change of variables would be appropriate here.
$$
u=x-\frac{y}{2}\\
v=\frac{\sqrt 3y}{2}
$$
Then we'd get $u^2+v^2=1$. This is a circle so its area is ... | The answer is correct. An easier approach is to realize the shape is an ellipse--the area of which is $\pi a b$. Setting $ x=y$ will lead to a $a = \sqrt{2}$, setting $ x=-y$ will lead to $b = \sqrt{2}/\sqrt{3}$, so $A = 2\pi/\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$
By using the definition of a limit only, prove that
$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$
We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$
I have simplified $\left|\dfr... | No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $\left|\dfrac {1}{3x+1}\right|$ will be unbounded.
If you want to put an upper bound on $\left|\dfrac {1}{3x+1}\right|$, then you are going to need something like $3x + 1 > \frac 13$, which is equivalent to
$x > -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | By observing that
$$\left(\frac12\pm\frac{\sqrt 5}{2}\right)^3=\frac18\pm\frac{3\sqrt 5}8+\frac{15}{8}\pm\frac{5\sqrt 5}{8}=2\pm\sqrt 5$$
It follows
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\tfrac12+\tfrac{\sqrt 5}{2}+\tfrac12-\tfrac{\sqrt 5}2=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
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Find all $z$ in the form $a + ib$ such that $\sin z = 2$.
I need help on checking on my proof, especially on the part where i
convert it using logarithm identities.
$\begin{aligned}
\sin z = 2 & \Rightarrow \dfrac{e^{iz} - e^{-iz}}{2i} = 2\\
&\Rightarrow e^{iz} -e^{-iz} = 4i \\
&\Rightarrow e^{2iz} - 1 - 4ie^{... | Yes, your proof is correct. A few comments:
Note that because $(2 + \sqrt{3})(2 - \sqrt{3}) = 1$, we have $$-i \ln (2 \pm \sqrt{3}) = i \ln (2 \mp \sqrt{3}).$$
Another method of solution would be to use the angle addition identity to write $$ \sin (a+bi) = \sin a \cos bi + \cos a \sin bi = \sin a \cosh b + i \cos a \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Determinant of a $3 \times 3$ matrix involving a variable
I'm trying to find the determinant of
$$A=\begin{bmatrix}x& x^2 &x^3 \\x^8 & x^9 & x^4\\x^7 & x^6 & x^5 \end{bmatrix}$$
I'm using the cofactor formula,
$det(A)=x(x^{14}-x^{10})+x^2(x^{11}-x^{13})+x^3(x^{14}-x^{16})$
$=x^{15}-x^{11}+x^{13}-x^{15}+x^{17}-x^{... | By Sarrus rule,
\begin{align*}
\det(A)&=x^{1+9+5}+x^{2+4+7}+x^{3+8+6}-x^{7+9+3}-x^{6+4+1}-x^{5+8+2}\\
&=x^{15}+x^{13}+x^{17}-x^{19}-x^{11}-x^{15}\\
&=x^{11}(x^2+x^6-x^8-1) \\
&=-x^{11}(x^6-1)(x^2-1)\\
&=-x^{11}(x^2+x+1)(x^2-x+1)(x-1)^2(x+1)^2.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Linear [12,9,4]-code over F11 Concerning linear error-correcting codes over finite fields of odd characteristic.
Over $\mathbb F_{13}$ I can use the following Vandermonde-style parity check matrix:
$$H=\begin{pmatrix}1&1&1&1&1&1&1&1&1&1&1&1\\1& 2 &3 &4 &5& 6 &7& 8 &9& 10&11 &12\\1^2& 2^2 &3^2 &4^2 &5^2& 6^2 &7^2& 8^2 &... | This is possible.
Consider the check matrix
$$H=\begin{pmatrix}1&1&1&1&1&1&1&1&1&1&1&0\\0&1& 2 &3 &4 &5& 6 &7& 8 &9& 10&0\\0^2&1^2& 2^2 &3^2 &4^2 &5^2& 6^2 &7^2& 8^2 &9^2& 10^2&1\end{pmatrix}.$$
We need to prove that any three columns are linearly independent over $\Bbb{F}_{11}$. If the triple of columns does not incl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$ I need to calculate:
$$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$$
I get $0/0$ and can then use l'hopital's rule to find the limit, I can do this but someone asked me how I can do this without using l'hopital's rule.... | $$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}\times \frac{\sqrt{2x^2 - ax} +a}{\sqrt{2x^2 - ax} +a}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{2x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a}{1}=\\
\lim_{x \rightarrow a} \frac{x^2 + ax - a^2-a^2}{x^2+x^2-ax-a^2}\times \frac{\sqrt{2x^2 - ax} +a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final an... | Writing
$$
\frac{4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4},
$$
you saw that $A=1$, $B=-1$, and $C=0$.
So you have
$$
\int \frac{4}{x(x^2+4)}dx = \int \frac{dx}{x} - \int \frac{x}{x^2+4} dx.
$$
To integrate the second term on the right, do a $u$-substitution:
let $u=x^2+4$. Then $du=2xdx$.
So
$$
\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Taylor expansion of $\cos^2(\frac{iz}{2})$
Expand $\cos^2(\frac{iz}{2})$ around $a=0$
We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}... | We know $ e^{z}= \sum_{n=0}^{\infty} \frac{z^n}{n!}$ and $ \cos z = \frac{e^{iz}+e^{-iz}}{2}$
so $\cos^2 \frac{iz}{2} = \frac{e^z + e^{-z}}{4} + \frac{1}{2}= 1 + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z^{2n}}{2n!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$ $$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$
I am trying to use residue.
We first need to find the singularities, $x^2+4=0\iff x=\pm 2i$
Just $2i$ is in the positive part of $i$ so we take the limit
$lim_{z\to 2i}\frac{1}{(z^2+4)^5}$ but the limit is $0$
| set $x= 2t $ and then $t^2 = s$
Then
$$ \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 4 )^5}=\frac{2}{4^5}\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1 )^5}=\frac{1}{4^4}\int_{0}^{\infty} \frac{dt}{(t^2 + 1 )^5} =\frac{1}{2.4^4}\int_{0}^{\infty} \frac{s^{-1/2}ds}{(s + 1 )^5} $$
But
$$ \int_{0}^{\infty} \frac{s^{-1/2}ds}{(s ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Write complex number using rational numbers, square root and arithmetic operations I am asked to write the number
$$z=e^{2\pi i/5}=\mathrm{cis}\left(2\pi/5\right)$$ using only numbers from $\mathbb{Q}$, arithmetic operations and the square root operation.
I thought about using $i=\sqrt{-1}$, but I'm not sure if I can d... | Let $\theta = 2\pi/5$. We have $z^5 = 1$ and $z \neq 1$, so $z^4 + z^3 + z^2 + z + 1 = 0$.
Since $z^4 = \frac1{z} = z^{-1}$, $z^3 = z^{-2}$, we can rewrite:
$$z+ z^{-1} + z^2 + z^{-2} + 1 =0$$
Now $z+z^{-1} = 2\cos\theta$ and $z^2 + z^{-2} = 2 \cos 2\theta = 2(2 \cos^2 \theta -1)$. Putting $r = \cos \theta$, we have:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Polynomial $x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ is reducible over $\mathbb{Q}$? Clearly, there are no roots, but how can I find factors of higher degree?
| A good way to see this, if you don't know about geometric series or about cyclotomic polynomials, is to factor by grouping. Start by splitting the nine terms into three groups of three:
$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1\\
= (x^8+x^7+x^6)+(x^5+x^4+x^3)+(x^2+x+1)$$
Now, factor out a GCF from each group:
$$=x^6(x^2+x+1)+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Complex number quadratic Is the following equation
$$z^2 + z^* + \frac14 = 0$$
where $z$ is a complex number and $z^*$ is its conjugate
completely separate from ordinary quadratic equations? i.e. can I use the discriminant, quadratic formula etc. If not what, what type of equation is this? Can z* be treated independe... | $$z=a+ib\\z^*=a-ib\\\text{while $a,b\in\mathbb{R}$}\\z^2+z^*+\frac{1}{4}=\\(a+ib)^2+(a-ib)+\frac{1}{4}=\\a^2+2iba-b^2+a-ib=-\frac{1}{4}\in\mathbb{R}\\\therefore a^2+2iba-b^2+a-ib\in\mathbb{R}$$
so we can create 2 equations:
$$\begin{cases}
2iba-ib=0
\\[2ex]
a^2-b^2+a=-\frac{1}{4}
\end{cases}$$
we can solve this:$$ib(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding the quadratic equation from its given roots.
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$
Now, using Vieta's formula,
For new equation,
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Su... | $\alpha+\frac{1}{\beta}+\beta+\frac{1}{\alpha}=\frac{(\alpha+\beta)(\alpha\beta+1)}{\alpha\beta}=\frac{(\frac{-b}{a})(\frac{c}{a}+1)}{\frac{c}{a}}=-\frac{(b(c+a))}{ac}$
and
$(\alpha+\frac{1}{\beta})(\beta+\frac{1}{\alpha})=\frac{(\alpha\beta+1)^2}{\alpha\beta}=\frac{(\frac{c}{a}+1)^2}{\frac{c}{a}}=\frac{(c+a)^2}{ac}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the... | Updated Solution
Here's a neater solution!
$$\begin{align}
\sum_{k=0}^n \frac 1{2^{n+k}}\binom {n+k}n
&=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k 2^{n-k}\\
&=\frac 1{2^{2n}}\sum_{k=0}^n \binom {k+n}k \sum_{j=0}^{n-k}\binom {n-k}j\\
&=\frac 1{2^{2n}}\sum_{l=0}^n \binom {2n-l}{n-l}\sum_{j=0}^l \binom lj
&&(l=n-k)\\
&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
proof that the pattern exists for all n Explain the pattern
$$(\sqrt2-1)^1= \sqrt2 - \sqrt1$$
$$(\sqrt2-1)^2 = \sqrt9 - \sqrt8$$
$$ (\sqrt2-1)^3 = \sqrt{50} - \sqrt{49} $$
that is $(\sqrt2-1)^n$ is equal to difference of two consecutive numbers one of which are squares.
| Suppose $(\sqrt 2+1)^n=a\sqrt 2+b$ where $a$ and $b$ are integers. Then $(1-\sqrt 2)^n=b-a\sqrt 2$
We now compute $(\sqrt 2 +1)^n(1-\sqrt 2)^n$ in two ways. First, since $(1+\sqrt 2)(1-\sqrt 2)=-1$, the product is $(-1)^n$. Second it is $(b+a\sqrt 2)(b-a\sqrt 2)=b^2-2a^2$ so that $$b^2-2a^2=(-1)^n$$
So $(1-\sqrt 2)^n=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a cle... | HINT: we have $$\frac{1}{\sqrt{x^2+1}}-\frac{x^2}{\sqrt{x^2+1}^3}=\frac{x^2+1-x^2}{\sqrt{x^2+1}^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
In order to determine the coefficient of $x^{28}$ we obtain
\begin{align*}
\color{blue}{[x^{28}]}&\color{blue}{(1+x+\cdots+x^{27})(1+x+\cdots+x^{14})^2}\\
&=[x^{28}]\frac{1-x^{28}}{1-x}\cdot\frac{(1-x^{15})^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Evaluate a limit using l'Hospital rule Evaluate $$\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}$$
I tried to apply l'Hospital rule in order to get the limit to be equal to
$$\lim_{x\to0}\frac{e^x-1}{2(e^x-\frac{x^2}{2}-x-1)^{-\frac{1}{3}}(e^x-x-1)}$$
but the new denominator has an indeterminate fo... | Using L'Hospital's rule. Note the limit can be transformed for convenience as follows:
$$L=\frac13 \sqrt[3]{\lim_{x\to 0} \frac{(e^x-x-1)^3}{(e^x-\frac{x^2}{2}-x-1)^2}}=(L'H)=$$
$$\frac13 \sqrt[3]{\lim_{x\to 0} \frac{3(e^x-x-1)^2(e^x-1)}{2(e^x-\frac{x^2}{2}-x-1)(e^x-x-1)}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding ALL rational solutions to a diophantine equation I came accross a problem in an excerpt of the book "1001 problems in classical number theory" that asks to determine all rational solutions to the equation :
$$ x^3 + y^3 = x^2 + y^2$$
Apart from the trivial solutions $(1,0)$ $(1,1)$, $(0,1)$ $(0,0)$, I couldn't... | I get that all solutions are the intersection of the curve (over the reals) with lines through $(1,1)$ with slope a negative square, that is
$$ p^2 x + q^2 y = p^2 + q^2 $$
with intgers $p,q > 0$ and, might as well, $\gcd(p,q) = 1.$
ADDED: I went through it more carefully, for each positive pair we get two intersectio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\lef... | For all $ n \in \mathbb{N}$, we know that $0 < \frac{1}{n^2} \leq 1$, hence $\left\lceil \frac{1}{n^2}\right\rceil=1.$
$$\left\lceil \frac{4n^2+1}{n^2} \right\rceil=\left\lceil 4 + \frac{1}{n^2} \right\rceil= 4 + \left\lceil \frac{1}{n^2} \right\rceil=4+1=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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If $\sin{x}=t, \quad x\in(\frac{3\pi}{2},2\pi),$ what is $\tan{x}?$ So, according to the interval in which $x$ lies, the value of the sine is negative, and positive for the cosine. Using trig identity I get that
$$\cos{x}=\pm\sqrt{1-\sin^2{x}}=\pm\sqrt{1-t^2}.$$
Knowing the signs of sine and cosine, I pick $\sqrt{1-t^2... | You correctly found that
$$\tan x = \frac{t}{\sqrt{1 - t^2}}$$
Observe that since $x \in (\frac{3\pi}{2}, 2\pi)$, $t = \sin x < 0$. Hence, $t = -|t|$. Thus,
$$\tan x = \frac{t}{\sqrt{1 - t^2}} = -\frac{|t|}{\sqrt{1 - t^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n -... | \begin{align}
\lim_{n \rightarrow \infty} \dfrac{\sqrt{n^{2} + 2}}{4n + 1} = \lim_{n \rightarrow \infty} \dfrac{\frac{1}{n}\sqrt{n^{2} + 2}}{\frac{1}{n}(4n + 1)} = \lim_{n \rightarrow \infty} \dfrac{\sqrt{1 + \frac{2}{n^{2}}}}{\left(4 + \frac{1}{n} \right)} = \dfrac{1}{4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
De Moivre's Theorem to calculate the fourth roots of 8 I was given this question:
Use de Moivre's theorem to derive a formula for the $4^{th}$ roots of 8.
As far as my understanding of this theorem goes, it is only applicaple to complex numbers. How am I supposed to use it for 8?
My initial thought was use 8 to make $z... | Note that $z = 8+0i$ is equivalent to $z = 8(\cos(2k\pi) + i\sin(2k\pi))$
And then for the fourth roots:
$$z^{\frac{1}{4}}=8^\frac{1}{4}\left(\cos\left(\frac{k\pi}{2}\right) + i\sin\left(\frac{k\pi}{2}\right)\right)$$
For $k=0$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(0) + i\sin(0)) = 8^\frac{1}{4}$
For $k=1$: $z^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove the following norm inequality of a vector Prove that $\forall x=(x_1,x_2)\in \mathbb R^2$ satisfying $34x_1^2-24x_1x_2+41x_2^2=25$ we get the following inequality for its norm: $1/\sqrt 2\leq \|x\| \leq1$.
My attempt at solving it: obviously, the given equality references a quadratic form, let's denote it $q$, th... | Your form can be represented as a symmetric matrix:
$$A = \begin{bmatrix}
34 & -12 \\
-12 & 41 \\
\end{bmatrix}$$
$A$ diagonalizes to $\begin{bmatrix}
50 & 0 \\
0 & 25 \\
\end{bmatrix}$ in the basis $\left\{\frac{1}{5}\begin{pmatrix} -3 \\ 4 \end{pmatrix},\frac{1}{5}\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$ I was looking for the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$
My work:
Dividing the $\frac{8t^3 +13}{(t+2)(4t^2+1)}$, I get $2 + \frac{-16t^2 -2t + 9}{4t^3 + 8t^2 + t + 2}$ or $2 + \frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$
Using the partial fraction dec... | you should properly pose x=2t but then dx/dt=2
$I= \int 2 dt + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{6}{ \color{red}{4t^2 +1 }} dt$
$I= \int 2 dt + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{3}{ \color{red}{x^2 +1 }} dx$
$I= \int 2 dt + \int \frac{-3}{t+2} dt + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.
For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \vare... | You are on the right track.
Also you can do this:
Let $\epsilon >0$
If $\delta<1$ then $1<x<3$ thus $|x+2|<5$
\begin{align}|\sqrt{x^2+5}-3|&=\frac{(\sqrt{x^2+5}-3)(\sqrt{x^2+5}+3)}{\sqrt{x^2+5}+3}\\&=\frac{|x^2-4|}{\sqrt{x^2+5}+3} \\&\leq \frac{|x+2||x-2|}{3}\\&\leq \frac{5}{3}|x-2|\end{align}
Now take $\delta \leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $F_m = 2^{2^m} + 1$ is prime (with $m \geq 1$), does it follow that $3 \mid \frac{F_m + 1}{2}$? If $$F_m = 2^{2^m} + 1$$ is prime with $m \geq 1$, does it follow that $$3 \mid \frac{F_m + 1}{2}?$$
I have verified this to be true for $1 \leq m \leq 4$.
| $F_m=2^{2^m}+1$ are also known as Fermat numbers. From
$$2 \equiv -1 \pmod{3} \Rightarrow 2^{2^m-1} \equiv (-1)^{2^m-1} \equiv -1 \pmod{3}$$
or
$$2^{2^m-1} +1 = \frac{2^{2^m} +2}{2} \equiv 0 \pmod{3}$$
or
$$\frac{F_m +1}{2}=\frac{2^{2^m} +2}{2} \equiv 0 \pmod{3}$$
Thus, $F_m$ should not necessarily be prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the domain of Irrational function, which contains Log and Trigonometric functions I'm trying to find the domain of the following function:
$$f(x)=\Biggl( {\frac{\displaystyle\log(\arctan(x))}{\displaystyle \sqrt{\sin(x)-\frac{1}{2}} + \sin(x)}}\Biggr)^{1/3} $$
I reasoned this way:
$$
\left\{
\begin{array}{... | Since $\sqrt{\sin x-\frac{1}{2}}\ge0$, if $\sin x\ge\frac{1}{2}$, then it is guaranteed that $\sqrt{\sin x-\frac{1}{2}}+\sin x\ge\frac{1}{2}$, and therefore is nonzero. Thus, we only need $x>0$ and $\sin x\ge\frac{1}{2}$; the other inequality is redundant. This gives us $$\left\{x:2k\pi+\frac{\pi}{6}\le x\le2k\pi+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequalities involving geometry but I can't post a picture yet How do I show that
$$ \frac 12 \left(\frac 1 {3^2}+\frac 1{4^2}+ \frac 1{5^2}+\dots\right) < \frac 1 {3^2} + \frac 1{5^2} + \frac1{7^2} +\dots \quad ?$$
| Alternatively: Note that the RHS:
$$\frac{\pi ^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots =\\
\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots \right)+\frac{1}{2^2}\left(\underbrace{1+\frac{1}{2^2}+\frac{1}{3^2}\cdots}_{\frac{\pi^2}{6}} \right) \Rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$ where $n\in\mathbb{N}-\{0\}$ with induction? So the question is as above is shown:
How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$?
$n=1: 8^2 - 5^2 = 39$ is a multiple of $13$.
I.H.: Suppose $8^{2^n} - 5^{2^n}$ is divisible by $13$ is true $\forall n \in \ma... | More generally, $8^{m} - 5^{m}$ is divisible by $13$ when $m$ is even.
Indeed, $8^{m} - 5^{m} = (13-5)^m-5^m = 13a+5^m-5^m=13a$ by the binomial theorem.
Here is a proof by induction:
$8^{m} - 5^{m}=13a$
$8^{2} - 5^{2}=13b$
$8^{m+2} - 5^{m+2} = 8^m 8^2 - 5^m 5^2 = (13a+5^m)8^2 - 5^m 5^2 = 13c + 5^m 8^2 - 5^m 5^2 = 13c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine whether $A (2, 2, 3)$, $B(4, 0, 7)$, $C (6, 3, 1)$ and $D (2, −3, 11)$ are in the same plane.
a) Compute a suitable volume to determine whether $A (2, 2, 3)$, $B
(4, 0, 7)$, $C (6, 3, 1)$ and $D (2, −3, 11)$ are in the same plane.
b) Find the distance between the line $L$ through $A$, $B$ and the
line ... | Alternatively: a) find the vectors:
$$\vec{AD}\{0,-5,8\}; \ \ \ \vec{BD}\{-2,-3,4\}; \ \ \ \vec{CD}\{-4,-6,10\}.$$
The volume of the tetrahedron $ABCD$:
$$V=\pm \frac16 \begin{vmatrix} 0 & -5 & 8 \\ -2 & -3 & 4 \\ -4 & -6 & 10 \end{vmatrix}=\pm \frac16 \cdot (-20)=\frac{10}{3}.$$
It implies the four points are not copl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$ 3^{2^n }- 1 $ is divisible by $ 2^{n+2} $ Prove that if n is a positive integer, then $ \ \large 3^{2^n }- 1 $ is divisible by $ \ \large 2^{n+2} $ .
Answer:
For $ n=1 \ $ we have
$ \large 3^{2^1}-1=9-1=8 \ \ an d \ \ 2^{1+2}=8 $
So the statement hold for n=1.
For $ n=2 $ we have
$ \large 3^{2^2}-1=81-1=80 \ \... | The Carmichael function $\lambda(2^{n+2})=2^n$ for positive $n$. Thus for any odd number $k$, $k^{2^n}\equiv 1 \bmod 2^{n+2}$ and thus $2^{n+2}$ divides $k^{2^n}- 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Why must an ellipsoid's R.H.S. be $1$?
Why must the ellipsoid equation or ellipse equation have $1$ as the R.H.S?
The standard equation of an ellipse is:
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$
And similar one for the ellipsoid:
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{... | I assume you're asking about the standard equation for an ellipsoid:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
It is designed this way for convenience, so that the numbers $a,b,c$ represent half the lengths of the principal axes.
Let's suppose you want the right hand side to be $r$ instead of $1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$... | \begin{eqnarray*}
x+ \frac{1}{1- \cfrac{1}{z}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{1}{1- \cfrac{1}{\frac{7}3-\frac{1}{x}}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{1}{1- \cfrac{3x}{7x-3}} =4
\end{eqnarray*}
\begin{eqnarray*}
x+ \frac{7x-3}{4x-3} =4
\end{eqnarray*}
& after a little algebra $(2x-3)^2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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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.