Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$... | $$
\begin{align}
\int_0^\infty\frac{1+t^2}{(a^2+b^2t^2)^2}\,\mathrm{d}t
&=\frac1{a^3b}\int_0^\infty\frac{1+\frac{a^2}{b^2}t^2}{(1+t^2)^2}\,\mathrm{d}t\tag{1}\\
&=\frac1{2a^3b}\int_0^\infty\frac{t^{-1/2}+\frac{a^2}{b^2}t^{1/2}}{(1+t)^2}\,\mathrm{d}t\tag{2}\\
&=\frac1{2a^3b}\frac{\Gamma\left(\frac12\right)\Gamma\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
When does the function $a\cdot \sin(x)+b\cdot \cos(x)-x$ have exactly one real root with multiplicity $1$?
*
*When does the function $$f(x)=a\cdot \sin(x)+b\cdot \cos(x)-x$$ have exactly one real root ($a$ and $b$ are real numbers) ?
*When does $f(x)$ have a multiple root , in other words for which real numbers $a$ ... | Okay, so we have:
$$a\cdot\sin(x)+b\cdot\cos(x)=x$$
$$a\cdot\cos(x)-b\cdot\sin(x)=1$$
multiply bottom one by $\dfrac ab$
$$\dfrac{a^2}{b}\cdot\cos(x)-b\cdot\sin(x)=\dfrac ab$$
Add it to the top one:
$$(\dfrac{a^2+b^2}{b})\cos(x)=x+\dfrac{a}{b}$$
So:
$$\cos(x)=\dfrac{bx+a}{a^2+b^2}$$
Now multiply the bottom one by $-\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Elliptical Integral: Finding $I= \int \sqrt{1-k^2\sin^2(x)} \cos^2(x) dx $ Just a sidenote: The part where I am stuck on is at the very end , this is a long question so bear with me until I get to the last part!
Question 1: Two famous integrals are $F= \int \frac{dx}{\sqrt{1-k^2\sin^2(x)}}$ and $E= \int \sqrt{1-k^2\s... |
$$A:=k^2\int\frac{v^2}{\zeta}\;dx$$
$$
\zeta^2=1-k^2v\implies k^4v^2=(1-\zeta^2)^2\\
\begin{align}
\therefore k^2A&=\int\frac{1}{\zeta}\;dx-2\int{\zeta}\;dx+\int\zeta^3\;dx\\
&=F-2E+\int\zeta(1-k^2v)dx\\
&=F-E-k^2\int\zeta v\;dx\\
&=F-E-k^2\int\zeta(1-\cos^2x)\;dx\\
&=F-(1+k^2)E+k^2\int\zeta\cos^2x\;dx\\
&=F-(1+k^2)E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty ... | Rinse and repeat the first line below
$$\begin{align}
{1\over \color{blue}{r}-1}&={1\over \color{blue}{r}+1}+{2\over \color{blue}{r}^2-1}\\
&={1\over r+1}+2\left({1\over\color{blue}{r^2}+1}+{2\over (\color{blue}{r^2})^2-1}\right)\\
&={1\over r+1}+\frac{2}{r^2+1}+\frac{4}{\color{blue}{r^4}-1}\\
&={1\over r+1}+\frac{2}{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Given an $n \times n $ grid, how many squares exist? How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.
How can I go about counting the number of squares of each size?
| Let the vertices of the $n \times n$ grid by $\{(x,y)| 0\le x \le n; 0 \le y \le n\}$.
(Is that what an $n \times n$ grid is? A $1 \times 1$ has $4$ vertices and an $n \times n$ grid has $(n+1)^2$ vertices? Or is a $1 \times 1$ grid a single point? I'm assuming the former.)
A $k \times k$ square will have an lower... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$ without using Calculus techniques
Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$.
It is easy if we apply Calculus techniques (e.g., derivatives, Lagrange multipliers, etc). However, the problem is assigned for students who have not yet learned Calculus. Is ther... | Your constraint is a circle centered at $(-1,0)$ with radius $2$. And the level curves of $xy$ with positive level are hyperbolae in the first and third quadrant. Since the circle is centered left of the origin, visualizing this it should be clear that the maximum value of $xy$ happens when both $x$ and $y$ are negativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| Easy, download Photmath!
Answer = $$(x^2-3x+1)^2=4x^2-12x+9$$ $$(x^2-3x+1)^2=(2x-3)^2$$
$$(x^2-3x+1)^2-(2x-3)^2=0$$ $$x^2-5x+4=0$$ $$x^2-x-2=0$$
$$x=4$$
$$x=1$$
$$x=2$$
$$x=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
| $$5_bx^2-50_bx+118_b=5_b(x-4_b)(x-7_b)$$
From $118_b$, we can assume that $b > 8$.
This implies $5_b = 5,\; 4_b = 4,\; 7_b = 7,\;$ and $\; 118_b = b^2 + b + 8$.
Hence
\begin{align}
b^2 + b + 8 &= 5 \times 4 \times 7 \\
b^2 + b - 132 &= 0 \\
(b - 11)(b + 12) &= 0 \\
b &= 11
\end{align}
CHECK:
\begin{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Show that there are infinitely many pairs (a,b) such that both $x^2+ax+b$ and $x^2+2ax+b$ have integer roots The question is :
Show that there are infinitely many pairs (a,b) of relatively prime integers (not necessarily positive ) such that both quadratic equations $x^2+ax+b$ =0
and $x^2+2ax+b$ =0 have integer roots.
... | Wanting $a^2-b=c^2$ and $a^2-4b=d^2$, we can see that we want $4c^2-d^2=3a^2$. If $a$ is odd, we can solve $2c-d=3$, $2c+d=a^2$ getting:
$$c=\frac{a^2+3}{4}, d=\frac{a^2-3}{2}$$
Then use $3b=c^2-d^2$ to determine $b$ in terms of $a$, and then do a lot of arithmetic to get an explicit infinite family. There is another (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Probability of 4 consecutive heads in 10 coin tosses I am trying to compute the probability of having 4 (or more) consecutive heads in 10 coin tosses.
I tried using recursion but it led to a complicated expression so i think i did not quite manage.
I saw similar questions asked here that were solved with difficult ap... | Here we are looking for the number $f_N$ of binary strings of length $N$ which do not contain the substring $HHHH$. The probability $p$ is then $$p=1-\frac{f_{10}}{2^{10}}$$
The so-called Goulden-Jackson Cluster Method is a convenient technique to derive a generating function for problems of this kind.
We consider wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$ I am trying to formally evaluate the following limit:
$$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$$
Empirically, the limit seems to be converging to $1.5$, although I am not sure how to formally prove this. I had on... | $$\sqrt{x^2+2x+3}-(x+1) = \frac{2}{x+1+\sqrt{x^2+2x+3}}=O\left(\frac{1}{x}\right)\tag{1}$$
$$\sqrt{x^2-x-5}-\left(x-\frac{1}{2}\right) = -\frac{5+\frac{1}{4}}{\left(x-\frac{1}{2}\right)+\sqrt{x^2-x-5}}=O\left(\frac{1}{x}\right)\tag{2} $$
$$\lim_{x\to +\infty}\sqrt{x^2+2x+3}-\sqrt{x^2-x-5}=\lim_{x\to +\infty}(x+1)-\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplifying a radical with complex fractions So I understand to simplify this:
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}}
$$
I can just multiply
$$
\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \cdot\frac{2t^4}{2t^4}
$$
and get
$$
\frac{-3}{t\sqrt{\frac{1}{4t^6} - 1}}
$$
B... | $\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{\frac{1}{4t^6} - 1}} \frac{\sqrt {4t^6}}{\sqrt{4t^6}}$ first step will be to get rid of the fraction inside the radical. Note that ${\sqrt{4t^6}} = |2t^3|$
$\frac{\frac{-3}{2t^4}|2t^3|}{|\frac{1}{2t^3}|\sqrt{1 - 4t^6}} \frac{\sqrt {1-4t^6}}{\sqrt{1-4t^6}}$ Then we get the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$
Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$
Let $R$ be the region. I substitute $u=x^3-3xy^2,v=3x^2y-y^3$. $\det J(x,y)=9(x^2+y^2)^2.$ Also notice tha... | I think that your calculations are correct. The final result is given in terms of an hypergeometric function.
You may make use of
$$
\int \frac{\mathrm{d} u}{(a^2+u^2)^{1/3}} = \frac{u}{a^{2/3}} \, {}_2F_1 \left( \frac{1}{3}, \frac{1}{2}; \frac{3}{2}; -\frac{u^2}{a^2} \right) \, ,
$$
to find a closed form of your int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | *
*Take any sequence of consecutive square numbers. What do you know (or, if you don't know, try it: what do you notice) about the second differences?
*Each row of the matrix has four consecutive square numbers.* Note that column operations preserve the determinant in the same way that row operations do (this is obvi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 6
} |
Finding that a two variable limit does not exist trouble: $\lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4}$ I have
$$ \lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4} $$
I've tried setting $y= 0, x=0, y=mx, y=x$, and lots of other things but because of the $\sin$ I keep getting a limit of zero. How do yo... | As you say: if you let $x = 0$ or $y = 0$ your answer will be 0. If you let $y = x$, however, you get
\begin{equation*}
\lim_{x \rightarrow 0} \frac{x^2 \sin^2(x)}{x^4 + x^4}
= \lim_{x \rightarrow 0} \frac{\sin^2(x)}{2x^2}
= \lim_{x \rightarrow 0} \frac{\sin(x)}{x} \cdot \frac{\sin(x)}{x} \cdot \frac{1}{2} = \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$
Prove that $$\binom{2n}{n}>\frac{4n}{n+1}\forall \; n\geq 2, n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+........+\binom{n}{n}x^n$$
and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\bin... | We prove the stronger bound $\binom{2n}{n} \geq \frac{4^n}{2n+1}$, which is bigger than $\frac{4}{n+1}$ for all $n \geq 2$.
Note that the $2n+1$ binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have average
$$
\frac{\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{2n}}{2n+1} = \frac{2^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Nonlinear equations in 3 variables Solve for $x,y\ \mbox{and}\ z$:
$$
\left\{\begin{array}{rcr}
x + y + z & = & 2
\\[1mm]
\left(x + y\right)\left(y + z\right) + \left(y+z\right)\left(z+x\right) +
\left(z + x\right)\left(x + y\right) & = & 1
\\[1mm]
x^{2}\left(y + z\right) + y^{2}\left(x + z\right) + z^{2}\left(x ... | Considering the equations$$x+y+z-2=0\tag 1$$ $$(x+y) (x+z)+(x+z) (y+z)+(x+y) (y+z)-1=0 \tag 2$$ $$x^2 (y+z)+y^2 (x+z)+z^2 (x+y)+6=0\tag 3$$ first eliminate $z$ from $(1)$ $$z=-x-y+2\tag 4$$ Replace in $(2)$ to get $$\left(-x^2+2 x+3\right)+(2-x) y-y^2=0\tag 5$$ Solve the quadratic in $y$ and keep the largest root; this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \geq \frac{1}{3}$ for $x+y+z=1$ We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?... | We have
$$
x+y+z=1\\
(x+y+z)^2=1\\
x^2+y^2+z^2+2(xy+xz+yz)=1
$$
By the rearrangement inequality, $x^2+y^2+z^2\geq xy+xz+yz$. Inserting that gives you
$$
3(x^2+y^2+z^2)\geq1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$ Let $a,b,c>0$ prove or disprove
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
since
$$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$
other
$$(a+b+c)^2\le 3(a^2... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sqrt[3]{u(3u^2-2v^2)}\geq\sqrt[4]{3v^4-2uw^3}$$ or
$f(w^3)\geq0$, where $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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cubic equation has $3$ distinct roots $\alpha,\beta,\gamma$ equation $x^3-9x^2+24x+c=0$ has $3$ distinct roots $\alpha,\beta,\gamma$, then $\lfloor \alpha \rfloor+\lfloor \beta \rfloor +\lfloor \gamma \rfloor =$
and $\lfloor \alpha \rfloor = \alpha-\{\alpha\},\lfloor \beta \rfloor = \beta-\{\beta\},\lfloor \gamma \rflo... | We may assume that $\alpha\lt \beta\lt \gamma$.
We have to have
$$f(2)=20+c\gt 0\quad\text{and}\quad f(4)=16+c\lt 0$$
Note that we have
$$f(1)=16+c\lt 0\quad\text{and}\quad f(5)=20+c\gt 0$$
from which
$$1\lt\alpha\lt 2\implies 0\lt\{\alpha\}\lt 1$$
$$4\lt\gamma\lt 5\implies 0\lt\{\gamma\}\lt 1$$
follow.
Thus, since
$$0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime
I tried to show that the ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime in $\mathbb{C}[X,Y]$.
I noticed that $X^{2}-Y^{3}-(Y^{2}-X^{3})=(X-Y)(X^{2}+XY+Y^{2}+X+Y)$. Now, if $X-Y$ is in $(X^{2}-Y^{3},Y^{2}-X^{3})$ there exists $f(X,Y... | Here is a more geometric solution.
To show that $I=(x^2-y^3,y^2-x^3)$ is not prime, one way is to show that their set of common zeroes has cardinality $>1$.
This works because the irreducible components of their common zero set correspond to the radicals of the primary components in the decomposition of $I$.
So suppose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
To find number of distinct terms of binomial expansion I have binomial expansion as
$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$.
How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?
I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed... | $$\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}=\frac{1}{x^{30}}(x^3(x+1)+x+1)^{15}=$$
$$=\frac{1}{x^{30}}(x^3+1)^{15}(x+1)^{15}=\frac{1}{x^{30}}\sum_{i=0}^{15}\binom{15}{i}x^{3i}\sum_{j=0}^{15}\binom{15}{j}x^{j}=$$
$$=\sum_{i=0}^{15}\binom{15}{i}\sum_{j=0}^{15}\binom{15}{j}x^{3i+j-30}$$
Number of elements in set
$$\{3i+j-30:i,j\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$ Find the number of integer solution of $$\left\lfloor \dfrac{x}{100} \left\lfloor \dfrac{x}{100}\right\rfloor \right\rfloor=5$$
$\lfloor . \rfloor$ is Greatest Integer Function or floor function.
| $\lfloor\frac{{x}}{\mathrm{100}}\lfloor\frac{{x}}{\mathrm{100}}\rfloor\rfloor=\mathrm{5}\:\Rightarrow\lfloor\frac{{x}}{\mathrm{100}}\rfloor=\lfloor\sqrt{\mathrm{5}}\rfloor=\mathrm{2} \\ $
$\lfloor\frac{{x}}{\mathrm{100}}.\mathrm{2}\rfloor=\mathrm{5}\Rightarrow\frac{{x}}{\mathrm{50}}\in\left[\mathrm{5},\mathrm{6}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Angle of a triangle using trigonometry. If in a triangle ABC , cos A+ cos C= sin B then what is the measurement of angle A? I have tried to solve it using trigonometric identities but failed to solve it.
| Using Prosthaphaeresis Formula, $$\cos A+\cos C=2\cos\dfrac{A+C}2\cos\dfrac{A-C}2$$
Now $\cos\dfrac{A+C}2=\cos\dfrac{\pi- B}2=\sin\dfrac B2$
and $\sin B=\sin\left(2\cdot\dfrac B2\right)2=\sin\dfrac B2\cos\dfrac B2$
As $0<B<\pi,0<\dfrac B2\le\dfrac\pi2\implies\sin\dfrac B2>0$
So we have $$\cos\dfrac{A-C}2=\cos\dfrac B2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the coefficient of $x^{100}$ in the power series representing the function $\dots$ Find the coefficient of $x^{100}$ in the power series representing the function:
$$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$
Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdo... | A different conceptual interpretation: We have
$$
F(x) = x^6 (1+x+x^2+\cdots)^3
$$
We can interpret this as a generating function representing a fixed prefix of six balls, plus three (distinguishable) sacks of zero or more balls each. The coefficient of $x^{100}$ is then the number of ways to total $100$ balls. Facto... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$ I want to verify the following inequality:
Let $a, b$ be non negative number
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} +3 \ge 2\left(\frac{a}{b} + \frac{b}{a}\right)$$
I decided to analyse the sign of $$\frac{a^2}{b^2} + \frac{b^2}{a^2} +... | You actually can prove for +2 also like $\frac{a^2}{b^2}+1\geq 2\frac{a}{b}$ and
$\frac{b^2}{a^2}+1\geq 2\frac{b}{a}$ by AM-GM and by adding you get the desire result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | Let $f(x) = x^2 -ax + 1$ and let $g(x) = f(x+2) = x^2 - (a-4)x + (5-2a)$
Find all $x$ such that $|f(x)|\le\frac{1}{4}$ and $|g(x)|\le\frac{1}{4}$
If $f(x)$ and $g(x)$ are to share that property at the same point, that point must be the one point that they have in common.
\begin{align}
f(x) &= g(x) \\
x^2 -ax + ... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 7
} |
Solve $\frac{n(n+1)}{2}= 45$ I have this equation:
$\frac{n(n+1)}{2}= 45$
I started:
$$n^2+n= 90$$
What do I do next?
| $n^2+n=90$
We can write this as-
$n^2+n-90=0$
Now, it is in the quadratic equation form, .i.e., $ax^2+bx+c=0$
Factor out the resulting equation, break $n$ into two terms $10n-9n$
$$n^2+10n-9n-90=0$$
Take $n$ common from first two terms and $-9$ from last two terms-
$$n(n+10)-9(n+10)=0$$
$$(n-9)(9+10)=0$$
$n-9=0\;\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integration of $\frac{2x^2}{\sqrt{x^2-1}}\, dx$ Integration of $\dfrac{2x^2}{\sqrt{x^2-1}}\,\mathrm dx$
My try:
Let $u=\sqrt{x^2-1}$
Then $x=\sqrt{u^2+1}$
$x \,\mathrm dx =u\,\mathrm du$
$$\int \frac{2u(u^2+1)}{u\sqrt{u^2+1}} \, \mathrm du$$
$$=2\int \sqrt {u^2+1} \, \mathrm du$$
True ? and what about the last integra... | \begin{align}
x & = \sec\theta \\[5pt]
dx & = \sec\theta\tan\theta\, d\theta \\[5pt]
\sqrt{x^2-1} & = \tan\theta \\[5pt]
\frac{2x^2}{\sqrt{x^2-1}}\, dx & = \frac{2\sec^2\theta}{\tan\theta} \sec\theta\tan\theta\,d\theta = 2\sec^3\theta\,d\theta
\end{align}
The integral of secant cubed is well known to be challenging to ... | {
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"url": "https://math.stackexchange.com/questions/1974319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
| Note that
$$
\left(\sqrt{x-1}+\sqrt{5-x}\right)^2=4+2\sqrt{4-(x-3)^2}
$$
Since $4-(x-3)^2\le4$ and $4+2\sqrt{4-(x-3)^2}\ge4$, we get that
$$
2\le\sqrt{x-1}+\sqrt{5-x}\le2\sqrt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$ solve this following equation
$$\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$$
the equation obvious have a root $x=3$. But how to solve other roots?
| Obviously $x>0$, suppose that $\lfloor x \rfloor =k$ (clearly $k$ must be between $0$ and $5$), then the solutions are $\lfloor \frac{9}{x} \rfloor = 6-k\iff 6-k\leq \frac{9}{x} < 6-k+1\iff \frac{9}{6-k+1}< x \leq \frac{9}{6-k}$.
So the solution set is $\bigcup\limits_{k=0}^{5}( (\frac{9}{6-k+1},\frac{9}{6-k}]\cap [k,k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to put $\frac{1}{\cos\theta - j\sin \theta}$ in the form $a+jb$? ($j^2=-1$) I am new to complex numbers and am having trouble putting them in the form $a+jb$ (or $a+bi$) How would I go about putting this expression in the form $a+jb$?
$$\frac{1}{\cos\theta - j\sin \theta}$$
| $$\frac{1}{\cos \theta - j \sin \theta}=\frac{\cos \theta + j \sin \theta}{(\cos \theta - j \sin \theta)(\cos \theta + j \sin \theta)}=\frac{\cos \theta+j \sin \theta}{\cos^{2}\theta-j^2 \sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta - (-1)\sin^2 \theta}=\frac{\cos \theta + j \sin \theta}{\cos^2 \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Solving $xy''-y'+4x^3y=0$ I need to solve
$$xy''-y'+4x^3y=0$$
by reduction or order, knowing one solution is $y=\sin x^2$. That's what I did:
$y_2 = v(x)\sin x^2\implies \\y_2' = v'\sin x^2+2xv\cos x^2\implies\\y_2'' = v''\sin x^2+2xv'\cos x^2+2(-2vx^2\sin x^2+v'x\cos x^2+v\cos x^2)$
Then,
$$xy_2''-y_2'+4x^3y_2 = \\xv... | $y=\sin x^2 \int v(x) dx $
$y' =2x \cos x^2 \int v(x) dx +v(x)\sin x^2 $
$y''=2\cos x^2 \int v(x) dx -4x^2 \sin x^2 \int v(x) dx +2x v(x)\cos x^2 +v'(x)\sin x^2 +2xv(x) \cos x^2 $
so
$xy'' -y' +4x^3 y =x\left(2\cos x^2 \int v(x) dx -4x^2 \sin x^2 \int v(x) dx +2x v(x)\cos x^2 +v'(x)\sin x^2 +2xv(x) \cos x^2\right) - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ If $\frac{a}{b}$ < $\frac{c}{d}$ then
$$\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$$
I have been searching and trying and couldnt find a reliable proof.
One might do this like here which I think it is very wrong:
$\frac{a}{b}$ < $\frac{c}{d}$ the... | Hint. As regards $\frac{a}{b} < \frac{a+c}{b+d}$, show that
$$\frac{a+c}{b+d}-\frac{a}{b}=\frac{\frac{c}{d}-\frac{a}{b}}{\frac{b}{d}+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $ I've been working through the following integral and am stumped:
$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$
Given the questions in my class that have proceeded and followed t... | Integration by parts can be performed for the indefinite integral, using relations
$$\dfrac{x\cos x-\sin x}{x^2} = \left(\dfrac{\sin x}{x}\right)',\quad
\sin^3 z =\frac{3\sin z-\sin3z}4,\quad \cos^3z=\frac{3\cos
z+\cos3z}4.$$
One can get
\begin{align}
&\int \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989935",
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"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 3
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Why $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim\limits_{t\to 0}\frac{\sin t}{t}$? Why $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{t\to 0}\frac{\sin t}{t}$( and hence equals to $1$)?
Any rigorous reason? (i.e. not just say by letting $t=x^2+y^2$.)
| Using $x = t \cos \alpha $ and $y = t \sin \alpha$, then:
$$ \frac{ \sin (x^2 + y^2 ) }{x^2+y^2} = \frac{ \sin (t^2(\sin^2 \alpha + \cos^2 \alpha)) }{t^2(\sin^2 \alpha + \cos^2 \alpha)} = \frac{ \sin (t^2) }{t^2}$$
.
Note as $x,y \to 0,0$, then $t \to 0 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking ... | A slight detour from Ron Gordon's answer, using the square of the $\arcsin$ series:
$$\begin{align*}
I &= \int_0^1 \frac{\arcsin(x) \arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} \, dx \\[1ex]
&= \frac{\pi^3}{64} - \frac12 \int_0^1 \arcsin^2\left(\frac x{\sqrt2}\right) \, \frac{dx}{\sqrt{1-x^2}} \tag{1} \\[1ex]
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| Use this simple lemma
$(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+... | Let $\sin (a)=x,\sin (b)=y,\sin (c)=z $ thus continuing from your simplified version we have $$\frac {\sum ^{cyc} x\sqrt {(1-y^2)(1-z^2)}}{x+y+z}$$ then using Am-Gm for each sqyare root we have $$\frac {\sum^{cyc} x\sqrt{(1-y^2)(1-z^2)}}{x+y+z }\leq \frac {2 (x+y+z)-2 (xy+yz+xz)}{2 (x+y+z)} $$ thus its $1-\bf{something... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge I was wondering if my proof is correct, and if there are any better alternative proofs. Or maybe proof that use nice tricks i might need in the future.
$$1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} \ldots = \sum_... | Call $s_k$ the partial sums of the series. That is $s_k=\sum_{n=1}^ka_n$, where $a_n=\frac{1}{n}$ if $n$ is odd, and $a_n=-\frac{1}{n^2}$ if $n$ is even. As $\sum_{j=1}^\infty \frac{1}{j^2}=\frac{\pi^2}{6}$, we have the estimate
$$
s_k\geq-\frac{\pi^2}{6}+\sum_{n=1}^ka_{2n-1}=-\frac{\pi^2}{6}+\sum_{n=1}^k\frac{1}{2n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \frac{1}{1+x^4} dx$
Evaluate
$$\int_0^\infty \frac{1}{1+x^4} dx$$
It seems to me that the most natural way would be to use contour integral. The integral on the semi circle will disappear. The only thing I'm not sure about is how to find the residue. Four roots of $1+z^4 = 0$ can be found easily and th... | Let $x^{4}=z$
\begin{align}
\int\limits_{0}^{\infty} \frac{1}{1+x^{4}} dx
&= \frac{1}{4} \int\limits_{0}^{\infty} \frac{z^{-3/4}}{1+z} dz \\
&= \frac{1}{4} \mathrm{B}\left(\frac{1}{4}, \frac{3}{4}\right) \\
&= \frac{\Gamma(1/4)\Gamma(3/4)}{4\Gamma(1)} \\
&= \frac{\pi}{2\sqrt{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \... | Hint: if $\sin\alpha+\cos\alpha=1.2$ then
$$
(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+2\sin\alpha\cos\alpha=1.44.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and ... | We can use Telescoping series as well
$$\dfrac{4r-2}{3^r}=f(r+1)-f(r)$$ where $f(n)=\dfrac{a_0+a_1n+a_2n^2+\cdots}{3^n}$
So that $$\sum_{r=1}^n\dfrac{4r-2}{3^r}=\sum_{r=1}^n(f(r+1)-f(r))=f(n+1)-f(1)$$
$$4r-2=a_0+a_1(r+1)+a_2(r+1)^2+\cdots-3\left(a_0+a_1r+a_2r^2+\cdots\right)$$
As the coefficients of $r^m$ is zero for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | For the first Q.
Let $17=x_1.$ For each $n,$ if $|x_n|>1,$ take positive integer $K_n$ such that $$|K_nx_n|<41<|(1+K_n)x_n| .$$ Take $x_{n+1}$ with $41>|x_{n+1}|$ where: ...
(i). If $41-|K_nx_n|<|(1+K_n)x_n|-41$ then $x_{n+1}\equiv K_nx_n \pmod {41}$...
(ii). Otherwise $x_{n+1}\equiv (1+K_n)x_n \pmod {41}.$
(This is f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
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points on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is minimum
Find point on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is, minimum
$\bf{My\; Try::}$ Let $(x,y)$ be any point on the curve $x^2+2y^2=6\;,$ Then we have to
minimize $\displaystyle \left|\frac{x+y-7}{\sqrt{2}}\right|... | WLOG any point on $x^2+2y^2=6$ can be P$(\sqrt6\cos t,\sqrt3\sin t)$
Now the distance of $P$ from $x+y-7=0$ is $$\dfrac{|\sqrt6\cos t+\sqrt3\sin t-7|}{\sqrt{1^2+1^2}}=\dfrac{\left|3\cos\left(t-\arccos\dfrac{\sqrt6}3\right)-7\right|}{\sqrt2}$$
Now $-1\le\cos\left(t-\arccos\dfrac{\sqrt6}3\right)\le1$
$\implies-3-7\le\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $x^2/(x^2 + \alpha^2)^2$ I want to find the indefinite integral
$$\int\frac{x^2}{(x^2 + \alpha^2)^2}\,dx.$$
The answer (from Mathematica) is
$$-\frac{x}{2(x^2+\alpha^2)} + \frac{1}{2\alpha}\tan^{-1}\left(\frac{x}{\alpha}\right).$$
I can tell from the answer that there must be some integration by parts and ... | The usual method of trigonometrical substitution would suggest to put
\begin{align*}
x&=a\cdot \tan\theta\\
dx&=a\cdot \sec^2\theta\,d\theta
\end{align*}
and go on. We obtain
\begin{align*}
\int \dfrac{x^2}{(x^2+a^2)^2}\,dx&=\int \dfrac{a^2\tan^2\theta}{(a^2\tan^2\theta+a^2)^2}\,a\sec^2\theta\,d\theta\\
&=\int \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding the nullspace of a square matrix I wish to find the nullspace of $(A-2I)$, where $A$ is given below, as part of finding the kernel of a linear transformation.
And the answer provided is
I am able to obtain the vector on the left, but I am not too sure how the vector on the right was acquired. I do not requir... | You're computing the eigenspaces, the characteristic polynomial of the matrix is
$$
(3-X)(2-X)^2
$$
so the eigenvalues are $3$ and $2$. The eigenspace relative to $2$ is the null space of
$$
\begin{pmatrix}
3-2 & 0 & -1 \\
0 & 1-2 & 1 \\
0 & -1 & 3-2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & -1 \\
0 & -1 & 1 \\
0 & -1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\te... | We have $$p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$$ so, clearly, $n=1$ and $n\ge p$ are discarded hence one has $1\lt n\lt p$ and $(n,p)=1$ and $p$ neither divides $n^3$ nor $n-1$. Besides the possibility $p=n+1$ is easily discarded.
It follows $p$ divides $n^2+1$ and since $n^2+1$ neither divides $p-1$ nor $p+1$ then we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to direc... | It is easy to check that the equality holds if $bc=0$, $ca=0$ or $ab=0$.
Let $bc=x$, $ca=y$ and $ab=z$. Then the determinant in L.H.S. can be written as
\begin{align*}
\begin{vmatrix} -x & x+zx/y & x+xy/z \\ y+yz/x & -y & y+xy/z \\ z+yz/x & z+zx/y & -z \end{vmatrix}&=\frac{1}{xyz}\begin{vmatrix}-x^2 & zx+xy & zx+xy \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
what kind of test should I use to determine if this sequence converge or diverge? what kind of test should I use to determine if this sequence converge or diverge?
$$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right)$$
| $$\sum_{n=1}^\infty \left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right) = \left(\sin1-\sin \frac{1}{2}\right)+ \left(\sin\frac{1}{2}-\sin \frac{1}{3}\right)+\cdots+\left(\sin\frac{1}{n}-\sin \frac{1}{n+1}\right) = \sin1+\left(-\sin \frac{1}{2}+ \sin\frac{1}{2}\right)+\left(-\sin \frac{1}{3}+\cdots+\sin\frac{1}{n}\right)-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | Well done. And for those of us unaware of the box method, the following would have also worked: $$x^2+4x+3=x^2+4x+4-1=(x+2)^2-1=(x+2-1)(x+2+1)=(x+1)(x+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
| To denest the nested radical $\sqrt{3+2\sqrt{2}}$, we have a useful formula. Namely
Given a radical of the form $\sqrt{X\pm Y}$, we can rewrite it into$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag1$$
With $X>Y$ and $X,Y\in\mathbb{R}$.
In $\sqrt{3+2\sqrt{2}}$, we have $X=3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equation of a line making equal and positive intercepts Q. Find the equation of line passing through the pt of intersection of the lines- $3x-y=5$ and $x+3y=1$ and making equal and positive intercepts on both the axes.
A. To find the pt of intersection substitute $x=1-3y$ in $3x-y=5$. Thus, $y= \frac{-1}{5}$ and $x= \f... | Hint:
A line passing thorough the point $P=(\frac{8}{5},\frac{-1}{5})$ (and not parallel to the $y$ axis) has equation:
$$
y+\frac{1}{5}=m\left(x-\frac{8}{5}\right)
$$
and, if this line has equals positive intercepts with the $x$ and $y$ axis, than $m=-1$, so it has equation:
$$
y+\frac{1}{5}+\left(x-\frac{8}{5}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \fr... | I do not knwo how much this is valid or not.
When $x\to\infty$, $\tanh(x)\to 1$. So, it seems to me that $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}+x^2}}{x+1}-x$$ Now, $$\sqrt{x^4+x^{3}+x^2}=x^2\sqrt{1+\frac 1x+\frac 1 {x^2}}$$ Now, using $$\sqrt{1+\epsilon}=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Quadratic equation 4 need some guidance with a quadratic equation.
Suppose $x^2+20x-4000=0$
Here is what I have done so far;
Using the Quadratic Equation $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where from the above equation, $a=1$, $b=20$, $c=-4000$, we find
$$\begin{align*}x&=\frac{-20+\sqrt{(20)^2-(4)(1)(-4000)}}{2\times ... | $2\times 1=2$, not $1$. In the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| $x^8+x^6−4x^4+x^2+1$
$=(x^8−2x^4+1)+(x^4-2x^2+1)x^2$
$=(x^4-1)^2+(x^2-1)^2x^2$
$≥0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Primitive $6^{th}$ root of unity Why does any extension of $\mathbb{Q}$ that contains a root of $x^6+3$ have to contain a primitive $6^{th}$ root of unity? The roots of this equation are $\{\sqrt[6]{3}e^{\pi i/2},\sqrt[6]{3}e^{5\pi i/6},\sqrt[6]{3}e^{7\pi i/6},\sqrt[6]{3}e^{3\pi i/2},\sqrt[6]{3}e^{11\pi i/6},e^{13\pi i... | Let $\alpha$ be such a root, then $\beta=\alpha^3$ satisfies $\beta^2+3=0$, hence ${1+\beta\over 2}={1+\alpha^3\over 2}$ is a $6^{th}$ root of $1$.
You can even verify this directly:
$$(1+\beta)^6 = 1+6\beta+15\beta^2+20\beta^3+15\beta^4+6\beta^5+\beta^6$$
Now use that $\beta^2=-3$ to reduce this to:
$$(1+5\cdot(-3)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Choosing a basis for a subspace in $\mathbb{R}^{3\times 3}$ and finding a coordinate vector with respect to it We observe the subspace known as $U\subset \mathbb{R}^{3\times 3}$ that is spanned by the following vectors:
$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\e... | We have:
$$ (1) \qquad a \begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
+b \begin{bmatrix}
0 & -3 & 0 \\
0 & 2 & 0 \\
0 & -1 & 0 \\
\end{bmatrix}
+c \begin{bmatrix}
0 & 0 & 1 \\
0 & -2 & 0 \\
3 & 0 & 0 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Axes of Symmetry for a General Ellipse
Given a general ellipse
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?
It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellip... | The trigonometric manipulations are not messy.
After centering, the equation is, in polar coordinates,
$$\rho^2=-\frac{F'}{A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta}.$$
The denominator
$$A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta=\frac12\left((A-C)\cos\frac\theta2+B\sin\frac\theta2\right)$$ achieves its ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $x^7- 2ix^4 - ix^3 -2 = 0$ for $x$ I am having difficulty factorising the equation.
| $$x^7-ix^3-2ix^4-2=0\\(x^7-ix^3)-2(ix^4+1)=0\\
x^3(x^4-i)-2(ix^4+1)=0\\\frac{i}{i}x^3(x^4-i)-2(ix^4+1)=0\\
\frac{1}{i}x^3(ix^4-i^2)-2(ix^4+1)=0\\
\frac{1}{i}x^3(ix^4+1)-2(ix^4+1)=0\\
(\frac{x^3}{i}-2)(ix^4+1)=0\\
(\frac{x^3-2i}{i})(ix^4+1)=0\\ \to
(x^3-2i)(ix^4+1)=0\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| After proceeding with the polar coordinate, it suffices to evaluate the following: $$I = 4\int_{0}^{\frac{\pi}{2}}\dfrac{d\theta}{2-\sin^2(2\theta)}= 2\int_0^{\pi}\dfrac{dx}{2-\sin^2x}$$.
Now do a tangent substitution: $$\tan x = t\Rightarrow \sin^2x = \dfrac{t^2}{1+t^2},\,\, dx = \dfrac{dt}{1+t^2}$$
Finally, $$I = 4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$ I set up the triple integral as follows:
$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\s... | The region $E$ in cylindrical coordinate is given by
$$ E = \left\{(r,\theta,z)\colon 0\le r\le 1/\sqrt{2}, 0\le\theta\le 2\pi, r\le z\le \sqrt{1-r^2}\right\}. $$
To see this, observe that $z$ is bounded below by the given cone means that
$$ z\ge \sqrt{x^2+y^2} = r,$$
and $z$ is bounded above by the sphere of radius 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| $\frac{5}{x+3}$+$\frac{2}{2(x+3)}$=4
$\frac{(5*2+2)}{2(x+3)}$=4
12=8(x+3)
x=$\frac{-12}{8}$
x=$\frac{-3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
| $\displaystyle \int \cos^2 x+\dfrac{1}{\cos^2 x}\,dx$
$=\displaystyle \int \dfrac{1}{2}(1+\cos 2x)+\sec^2 x\,dx$
$=\dfrac{1}{2}x+\dfrac{1}{4}\sin 2x+\tan x+C$
So integrating between the limits
$\left[0,\dfrac{\pi}{3}\right]$
$=\dfrac{\pi}{6}+\dfrac{1}{4}\sin \dfrac{2\pi}{3}+\tan \dfrac{\pi}{3}-0$
$=\dfrac{\pi}{6}+\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that $ \lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2} $ $$\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2}$$
Attempt:
According to me, it has to be $1$, since th... | Let us consider $n=2$ first, for simplicity, and use $x^2$ instead of $\sin^2x$, as $x\to0$. As you suggested, since the function has the indeterminate form $\infty^0$ as $x\to0$, let us consider the logarithm of the original function
$$
x^2\log\left(1+2^{1/x^2}\right) = \frac{\log\left(1+2^{1/x^2}\right)}{1/x^2},
$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$? Let $a,b,c\in \mathbb{R}$ and $a,b,c \ne 0$ and $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.
Why does $a=b=c$?
| Let $a=cx$ and $b=cy$. The equation becomes $x^2y^2+y^2+x^2=xy(x+y+1)$, or
$$(y^2-y+1)x^2-(y^2+y)x+y^2=0$$
As a quadratic in $x$, the discriminant, which must be non-negative in order for $x$ to be real, is
$$(y^2+y)^2-4(y^2-y+1)y^2=-3y^2(y-1)^2$$
So if $y\not=0$, we must have $y=1$, which in turn implies $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Probability that the sum 3 integers is divisible by 3
I found the sample space, but i was having difficulty finding the favorable outcomes.
But how do i find numbers that add up to be divisible by 3?
| Al integers are of the form $m=3i_m + k_m$ where $i_m$ is an integer and $k_m$ is either $0,1$ or $2$. As the the $3n$ numbers are consecutive the number of integers where $k_m$ is $0, 1$ or $2$ are equal.
The ways in which $a+b+c$ is divisible by three is that: i) all three have $k_a = k_b = k_c$ or that ii) $\{k_a,k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Reflexion of an element on $P_1(\Bbb R)$ Let $p, q \in P_3(\Bbb R)$. Define the inner product by
$$\langle p(x), q(x)\rangle = \int_0^1 p(x) q(x) \, dx $$
What is the reflection of $q(x) = x ^ 2$ on the subspace $P_1(\Bbb R)$?
My result is $R(q)(x) = \frac{2}{3} + \frac{3x}{2} - x^2 $
But the answer in the book is $ R... | You have to start with an orthogonal base for $P_1(\Bbb R)$ in order to calculate the reflection.
*
*We take $1$ as the first element of our orthogonal base for $P_1(\Bbb R)$.
*
*$\lVert1\rVert^2 = \langle1,1\rangle = 1$ is obvious.
*We apply the Gram-Schmidt process on $x$ and $x^2$ to obtain the second and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Mathematical Induction with Exponents: $1 + \frac12 + \frac14 + \dots + \frac1{2^{n}} = 2 - \frac1{2^{n}}$ Prove $1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}$ for all positive integers $n$.
My approach was to add $\frac{1}{2^{n + 1}}$ to both sides for the induction step. However, I got ... | Hint
if we add $\frac{1}{2^{n+1}}$ to the right side, we get
$$2-\frac{1}{2^n}+\frac{1}{2^{n+1}}$$
$$=2-\frac{2}{2^{n+1}}+\frac{1}{2^{n+1}}$$
$$2-\frac{1}{2^{n+1}}$$
qed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$ for $k\le n$ and $3k>n+1$ How to prove the following inequality:
$$(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$$
where $k\le n$ and $3k>n+1$ for integers $n,k$.
| Since $3k-1>k$, we obtain $k\geq1$.
In another hand, $k\leq n\leq3k-2$.
Let $f(n)=(n-k+3)(n-k+2)(n-k+1)-k(n+1)(n+2)$.
$f(3k-2)=-k(k-1)(k-2)\leq0$
and $f(k)=-k^3-3k^2-2k+6\leq0$.
$f'(x)=3x^2-(8k-12)x+3k^2-15k+11$,
which says that our inequality is proven for $3k^2-15k+11\geq0$
because there is also $x_1<0$ for which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle i... | Notice this. Quite simple approach
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Residue ${{(z+1)^2}\sin(1/z)}\over{(3z+1)^2}$
Calculate residue at $z=0$ of $\dfrac{{(z+1)^2\sin(1/z)}}{(3z+1)^2}$
It is essential singularity.So I am trying to get Laurent series expansion.Which in this case is hard because requires combing two expansions.
| If you are familiar with the residue at infinity, you can reduce your problem to a simpler one. Set
$$ f(z) = \left( \frac{z+1}{3z+1} \right)^2 \sin \left( \frac{1}{z} \right). $$
The function $f$ has isolated singularities at $z = 0, -\frac{1}{3}, \infty$. Let us calculate the residue at infinity first. We have
$$ -\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the derivative of $\sqrt{x^2+9}$ using the definition of a derivative? I had this question on a not-so-recent math test. It asked to find the derivative of $\sqrt{x^2+9}$ with the definition of a derivative. Using the chain rule, I can figure out that the derivative is $\frac x{\sqrt{x^2+9}}$, but how c... | So we have $f(x) = \sqrt{x^2 +9}$.
Then using the definition of the derivative, we have $f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} =\lim_{h \to 0} \frac{\sqrt{(x+h)^2 +9} - \sqrt{x^2+9}}{h} * \frac{\sqrt{(x+h)^2 +9}+ \sqrt{x^2 +9}}{\sqrt{(x+h)^2 +9} + \sqrt{x^2+9}} = \lim_{h \to 0} \frac{2xh + h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be obser... | $x = 0$ cannot be a root so suppose $x > 0$.
The equation is equivalently written as $nx^3+3/x = -4$.
But by AM GM, $nx^3+3/x \ge 4 \sqrt[4]{n} > -4$ so no positive roots.
Likewise, let $x = -r$ where $r$ is a positive number.
Then we have $nr^4-4r+3 = 0$ or $nr^3+3/r = 4$.
Again by AM GM, we have $nr^3+3/r \ge 4\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all complex numbers satisfying the equation $\bar{z}+1=iz^2+|z|^2$ Find all complex numbers z satisfying $$\bar{z}+1=iz^2+|z|^2$$
where $i=\sqrt{-1}$
I only know one way i.e. assuming $z=x+iy$ but that process is very cumbersome. I don't know how to proceed otherwise with a shorter approach.
| I do not find it so cumbersome to replace $z=x+iy$ if you write out the details (it looks like it might be a mess, and, without trying it out, it's hard to see where the problem simplifies). Substituting $z=x+iy$ into the equation gives
$$
(x+1)+(-y)i=(x^2+y^2-2xy)+(x^2-y^2)i=(x-y)^2+(x^2-y^2)i.
$$
Therefore
\begin{al... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show $y^2,z^2,u^2$ are roots of $(1)$ Euler's solution for the general quartic are as follows:
From the depressed quartic, $x^4+px^2+qx+r=0$, assume $x=y+z+u$ and it may be shown that $u^2,y^2,z^2$ are roots of the cubic$$t^3+\dfrac q2t^2+\dfrac {q^2-4s}{16}t-\dfrac {r^2}{64}=0\tag1$$
$$\vdots$$
However, how w... | COMMENT.-Euler himself used the following equation to apply his method:
$$x^4-8x^3+14x^2+4x-8=0\tag1$$
I reproduce this solution because it seems to me illustrative for who likes the subject and of certain historical interest.
First a synthesis of the Euler method to solve the quartic.
$$***$$
*
*$Ax^4+Bx^3+Cx^2+D... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | It just depends on context.
In some rare cases
$$
a\frac{c}{d}:=a+\frac{c}{d}
$$
which is the interpretation in your answer, but mostly
$$
a\frac{c}{d}:=a\cdot\frac{c}{d}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Find the pseudo-inverse of the matrix $A$ without computing singular values of $A$. Consider the following Least Square Minimization problem:$min_{x \in \mathbb C^3} |x_1 +x_3-3|^2+|x_2 -x_3|^2+|x_1+x_3-4|^2$
Find the pseudo-inverse of the matrix $A=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & -1 \\
1... | There are several methods to calculate $A^+$ without calculating singular values. Here are two of them.
Method 1.
Use the formula $A^+=\lim_{t\to0}A^T(AA^T+tI)^{-1}$. In your case, you should get something like
\begin{align*}
A^+&=\lim_{t\to0}A^T(AA^T+tI)^{-1}\\
&=\lim_{t\to0}\pmatrix{1&0&1\\ 0&1&0\\ 1&-1&1}\pmatrix{t+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be ... | HINT:
Let $a=p-q,b=p+q,c^2=p^2+q^2$
$\cos C=\cdots=\dfrac{c^2}{2ab}=\dfrac12+\dfrac{p^2+q^2}{2(p^2-q^2)}-\dfrac12\ge\dfrac12$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
| let $\sqrt{x^2+b^2}+x=y,$ then $\displaystyle \sqrt{x^2+b^2}-x = \frac{b^2}{y}$ and $\displaystyle 2x=y+\frac{b^2}{y}$
$\displaystyle f(x) = (2a-2x)(x+\sqrt{b^2+y^2}) = (2a-y-\frac{b^2}{y})\cdot y = (2ay-y^2-b^2) $
$f(y)= (a^2+b^2)-(y-b)^2\leq (a^2+b^2)$ equality hold when $y=b$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Diophantine system of equations I'm trying to know if there is an efficient way to find the smallest (i.e lexicographically) triplet $(a,b,c)$ of integers verifying $$a^2+b^2+c^2 = x$$ $$a^3+b^3+c^3=y$$ $$a^4+b^4 + c^4 = z$$ if $(x,y,z)$ is known.
We assume that a solution exists for that triplet $(x,y,z)$.
Originally... | If you have software to find real roots, the following result provides another way to find qualifying triples $a,b,c$ ...
Proposition: If $a,b,c$ satisfy $a^2 + b^2 + c^2 = x$, $a^3 + b^3 + c^3 = y$, $a^4 + b^4 + c^4 = z$, then $a,b,c$ are roots of the following 12'th degree polynomial:
$$
p(t) = \left(12\right)\left(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it? How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it?
My attempt: I tried to put numbers from left to right and count the ways but it was to hard an... | Suppose that the first three numbers satisfy the condition, but the whole seqence does not.
Since $1+2+3+4>8$ we must necessarily have $a_4>a_1+a_2+a_3$.
Case $1: a_4=7$, in this case we must have that $a_1,a_2,a_3$ are a permutation of $1,2,3$. So there are $3!\times 4!$ counterexamples of this form.
Case $2: a_4=8$, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alte... | What about AM-GM?
$$(1-x)^n(1+nx) = \text{GM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1} \color{red}{\leq} \text{AM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1}=1$$
and the inequality holds tight since $(1-x)\neq(1+nx)$. Done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots ... | since the definition of the $$\sqrt{x^2-1}$$ we get $$|x|>1$$ which is for $$x-\sqrt{x^2-1}$$ impossible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix $\alpha > 1$. Take $x_1 > \sqrt{\alpha}$, and define $$x_{n+1} = \frac{\alpha + x_n}{1+x_n} = x_... | Notice that
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n} = 1 + \frac{\alpha-1}{1+x_n}. $$
Since $\alpha-1>0$, we have that if $x_n<\sqrt{\alpha}$, then
$$ x_{n+1} > 1 + \frac{\alpha-1}{1+\sqrt{\alpha}} = \frac{1+\sqrt{\alpha}+\alpha-1}{1+\sqrt{\alpha}} = \frac{\sqrt{\alpha}+\alpha}{1+\sqrt{\alpha}} = \sqrt{\alpha}$$
and simil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Infinitely nested radical formulas for $\pi$ Has anyone come by the infinitely nested radical formula:
$$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty} 2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2... | I appreciate your work and finding. We may derive $\pi$ In infinite number of ways with the help of nested square roots
For explaining your derivation we have to think infinite nested radicals from the end.
Let us start from chord inside unit circle and value of lengths bisecting the chord
The length of chord is $2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to solve these general $n \times n$ determinants? How would you solve these two general determinants?
$$
\begin{vmatrix}
2 & 1 & 0 & \cdots & 0 & 0 \\
1 & 2 & 1 & \cdots & 0 & 0\\
0 & 1 & 2 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 &... | For the second one, in general, if $$\Delta_n=\begin{vmatrix}
a & b & b & \ldots & b\\
b & a & b & \ldots & b \\
b & b & a & \ldots & b \\
\vdots&&&&\vdots\\
b & b & b & \ldots & a
\end{vmatrix}.$$ First step $R_i\to R_i-R_n\;(i\ne n)$ implies $$\Delta_n=
\begin{vmatrix}
a & b & b & \ldots & b\\
b -a& a-b & 0 & \ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^... | We already have answers showing correct ways to solve the problem.
What I found interesting was the implied question:
where is the error in the steps taken in the body of the question,
which appeared to show that the limit is $1$?
The steps in question are
\begin{align}
\newcommand{\?}{\stackrel{?}{=}}
\lim_{x\to -\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all real solutions to the equation$4^x+6^{x^2}=5^x+5^{x^2}$ Find all real solutions to the equation:$$4^x+6^{x^2}=5^x+5^{x^2}$$
Obviously $0$ and $1$ satisfy this equation, but here $x$ is a real number, and I do not know how to do the next step, in fact, I guess this equation is not in addition to $0$ and $1$ oth... | For any $x \in \mathbb{R}$, apply MVT to $t^{x^2}$ on $[5,6]$ and $t^x$ on $[4,5]$,
we find a pair of numbers $\xi \in (5,6)$ and $\eta \in (4,5)$ (both dependent on $x$) such that
$$6^{x^2} - 5^{x^2} = x^2 \xi^{x^2-1}\quad\text{ and }\quad
5^x - 4^x = x\eta^{x-1}$$
This implies
$$
{\tt LHS}-{\tt RHS} = (4^{x} + 6^{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving $ {L}_{1} $ Regularized Least Squares Over Complex Domain I would like to solve the following Regularized Least Squares Problem (Very Similar to LASSO):
$$ \arg \min_{x} \frac{1}{2} {\left\| A x - b \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{1} $$
Where $ A \in {\mathbb{R}}^{m \times n} $ and $ b \in {\m... | Write $A = B + Ci$, $b=c+di$ and $x=y+zi$. The objective function is
$$
\begin{align*}f(x) &= ||(B+Ci)(y+zi)-c-di||_1^2 + \lambda ||y+zi||_1 \\
&= ||By-Cz-c+(Cy+Bz-d)i||_2^2 + \lambda ||y+zi||_1 \\
&= ||By-Cz-c||_2^2 + ||Cy+Bz-d||_2^2 + \lambda \sum_{j=1}^n \sqrt{y_j^2+z_j^2} \\
&= \left\Vert\begin{pmatrix}B \\ -C \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
| Let $f_a(x) = x^4+2ax^3+x^2+2ax+1$. If $x$ is a negative root, then we have
$$a = -\frac{x^4+x^2+1}{2x^3+2x}$$
That is, if we let
$$g(x) = \frac{x^4+x^2+1}{2x^3+2x}$$
then $a = g(-x)$. We have that $-x > 0$; using standard techniques, $g(-x)$ reaches a maxiximum (on $(-\infty, 0)$) of $3/4$ at $x=-1$.
This shows tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don'... | We are given
$$
\left\{ \begin{gathered}
bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\
\varepsilon = \pm \frac{c}
{d} \hfill \\
\end{gathered} \right.
$$
So we have the equation of a vertical parabola in $\varepsilon$, whose intercepts with the $x$ axis
must be symmetrical (vs.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to find the given limits I now that: $$\lim_{x\to 0}\frac{\sin x}{x}=1,$$ but I didnt now how to prove that:$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$
Please help me. Thanky very much for your help.
| First method:
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\left(\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\right)=\lim_{x\to 0}\frac{1-\cos^2 x}{x^2(1+\cos x)}$$
$$=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}=\lim_{x\to 0}\frac{\sin x\cdot \sin x}{x\cdot x (1+\cos x)}=\lim_{x\to 0}\left(\frac{\sin x}{x}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\df... | Let $A$ be the set of all natural numbers $n$ not containing $9$ in their decimal representation and let $B$ be the set of all natural numbers $n$ that contain only some of the digits $0,1,2,3,4$ in their decimal representation. It is obvious that $B\subset A$. So:
$$\displaystyle{\sum_{n\in B}}\frac{1}{n}\leq \display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi... | \begin{eqnarray}
\sin z &=& -\sin x-\sin y\\
\cos z &=& -\cos x-\cos y
\end{eqnarray}
so
\begin{eqnarray}
\sin^2z &=& \sin^2x+\sin^2y+2\sin x\sin y\\
\cos^2z &=& \cos^2x+\cos^2y+2\cos x\cos y
\end{eqnarray}
adding to these equations concludes
\begin{eqnarray}
1 &=& 1+1+2(\sin x\sin y+\cos x\cos y)\\
0 &=& 1+2\cos(x-y)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Generators of a quotient ring I have a polynomial ring $\mathcal{R} = \mathbb{Z}_3[x]/(x^2+x+2)$ and I am supposed to find a generator of the multiplicative group of $\mathcal{R}$.
Well, I know that the $\vert \mathcal{R} \vert = 9$, therefore the number of elements of the multiplicative group is $9-1 = 8$ and that th... | The ring is the splitting field of $x^9-x$; since
$$
x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^4+1)
$$
and $x^4+1=(x^2+x+2)(x^2+2x+2)$, the complete factorization is
$$
x^9-x=x(x^8-1)=x(x-1)(x+1)(x^2+1)(x^2+x+2)(x^2+2x+2)
$$
Note that if $\alpha$ is a root of $x^2+1$, then $\alpha^4=1$, so it is not good for the purpose. Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the law of cosines imply triangle exists? This is surely trivial, but I've never seen a triangle explicitly defined in this way before. Let $a,b,c>0$. Is it true that:
$$a,b,c\text{ form a triangle}\iff -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1.$$
My attempt:
($\implies$) Suppose $a,b,c$ form a triangle, then by the law... | Let $a,b,c>0$ and
$$ -1\leq\frac{b^2+c^2-a^2}{2bc}\leq 1. \tag1$$
Then your proof via the Law of Cosines requires the existence of
$\vartheta$ such that $0 \leq \vartheta \leq \pi$
and $\cos\vartheta = \frac{b^2+c^2-a^2}{2bc}.$
Such a value $\vartheta$ always exists under the conditions given
in Inequality $1$.
Then y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite integral of $4x(1+x)^3$ I'm trying to integrate by parts, and have done:$$\int^1_0{4x(1+x)^3}=4\int^1_0{x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\frac{(1+x)^4 }4\bigg]=4\bigg[\frac{(x-1)(1+x)^4}4\bigg]^1_0=
\\ \bigg[(x-1)(1+x)^4 \bigg]^1_0 = 1$$
Why is this wrong?
| Integration by parts should give, if we take $dv=(1+x)^3 dx$ so that $v=\frac{(1+x)^4}{4}$. And $u=x$ so that $du=dx$,
$$\int {4x(1+x)^3}=4\int {x(1+x)^3}= 4\bigg[\frac{x(1+x)^4}4-\int \frac{(1+x)^4 }4 \mathrm dx\bigg]$$
As $\int u \mathrm dv=vu-\int v \mathrm du$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solving Integration of trignometry In the following integral
$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$
My try: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.
| $\displaystyle I = \int\frac{1}{\sec x+\csc x}dx = \frac{1}{2}\int\frac{\sin 2x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\sin x+\cos x)^2-1}{\sin x+\cos x}dx$
$\displaystyle I = \frac{1}{2}\int (\sin x+\cos x)dx - \frac{1}{2}\int \frac{1}{\sin x+\cos x}dx$
$\displaystyle I =\frac{1}{2}(\sin x-\cos x)-\frac{1}{2\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Need help solving this Can I have help with this? I can't think of a way to get started...Thank you
a) If $k$ and $\alpha$ are constants and the following trigonometric
equation can be solved,
$$2(\cos\alpha+\sin\alpha)\cos\theta+4(\cos\alpha-\sin\alpha)\sin\theta=k$$
Prove that $k^2\le32$.
b) If $k=2\sqrt2$, ... | hint:
For part (a)
$$2(\cos (\alpha) +\sin(\alpha))\cos \theta + 4(\cos \alpha - \sin \alpha)\sin(\theta) = k$$
By harmonic addition theorem, http://mathworld.wolfram.com/HarmonicAdditionTheorem.html
I can express the $a \cos \theta + b \sin \theta$ in the form of $R cos(\theta + \delta)$ where $R^2=a^2+b^2$.
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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.