Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Does this matrix have a square root, and what is it? let $$M=\Biggl(\matrix{1&0&1\\0&1&0\\1&0&1}\Biggr)$$
does $M$ have a square root?
I believe that $M$ corresponds to a positive linear transformation, because it is a self-transpose, and has all non negative eigenvalues 2, 0, and 1 (according to numpy.linalg.eig). Bec... | Now, I see the problem. In programming languages there are two "squaring" operations on matrices. One squares each individual entry, the other is actual matrix multiplication.
You squared each individual entry of $M^{1/2} := \begin{bmatrix}\tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \tfrac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
Finding Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
what I tried:
let $S = 4x^2+4xy+4y^2+x-5$
$\dfrac{dS}{dx}=8x+4y+1$ and $\dfrac{dS}{dy}=4x+8y$
for center $\dfrac{dS}{dx}=0$ and $\dfrac{dS}{dy}=0$
getting center as $ x=-\dfrac{1}{6}$ and $y=\dfrac{1}{12}$
How do I ... | We have to turn axes through $\theta$ where
$$ \tan 2 \theta = \frac {h}{b-a}=\frac {2}{4-4}= \infty; \theta = \pi/4$$
in order to remove $xy$ term and have axes parallel to $x-,y-,$ axes
$$ x1=(x-y)/\sqrt2;\, y1=(x+y)/\sqrt2;$$
plug in and taking only quadratic terms as $(x,y)$ translations do not affect eccentricity ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Two distinct answers to same integral? I am wonder if any integral has 2 answers which might not be equivalent.
As far as I know the integral:
∫sec²x tanx dx
has 2 answers using substitution:
When tanx=t
sex²xdx=dt
=∫tdt
=t²/2 + C
=tan²x/2 + C
When secx=t
=secxtanxdx=dt
=∫tdt
=t²/2 + C
=sec²x/2 + C
If both answers abo... | The $C$ constant term in $ \frac{\tan^{2}x}{2} + C $ and $ \frac{\sec^{2}x}{2} + C $ need not be the same number, since these are solutions of an indefinite integral which is geometrically a family of curves - and they may refer to same or different curves depending on the value of their $C$s. It you are wondering how ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
interpretation on $1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}$ and $1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$ How to interpret the results
$$
1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}
\\
1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}
$$
I want to find a clear a... | Two different combinatorial interpretations of your first identity are given in the answers to this question. My answer (not the accepted one) was based on counting the number of ways you can place two white bishops on an $n\times n$ chessboard so that they guard each other. Calling this number $B_n$, and counting in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve binomial coefficient equation My book asks me to solve this equation:
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\begin{pmatrix} 7\\x \end{pmatrix}$$
The solution is $x=3$ and the formula $$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n\\k \end{pmatrix}=\begin{pmatrix} n+1\\... | We have
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{2!4!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$
$$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $h(t)=-16t^2+80t$ then simplify $\frac{h(a)-h(1)}{a-1}$ I am to simplify $\frac{h(a)-h(1)}{a-1}$ given $h(t)=-16t^2+80t$. The solution provided is
$\frac{-64+80a-16a^2}{-1+a}$ = $-16a+64$
I cannot see how this was arrived at. Here's as far as I got:
$\frac{h(a)-h(1)}{a-1}$
$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # subs... | Sometimes it's helpful (but not really necessary) to have positive leading coefficients on the terms of highest power.
$$\begin{align}
\frac{16(-a^2+5a-4)}{1-a} &= \frac{-16(a^2-5a+4)}{-(a-1)}\\
&= \frac{-1}{-1}\cdot\frac{16(a^2-5a+4)}{a-1}\\
&= \frac{16(a^2-5a+4)}{a-1}\\
\end{align}$$
In order to factor the polynomial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that the equation :$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=2$ has no solution in $\mathbb{N^{3}}$ Prove that the following equation :
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=2$
$x,y,z\in\mathbb{N}$
Has no solution in $\mathbb{N^{3}}$
I'm going to use the hint are given :
$\left(\frac{a+b+c}{3}\right)^{3}\geq abc$ ... | You used the inequality in the special case of $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$. You need to prove the result in general without this assumption, as follows:- $$\left(\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3}\right)^3 \ge \frac{x}{y}\frac{y}{z}\frac{z}{x}=1.$$ So
$$\frac{\frac{x}{y} + \frac{y}{z} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
r... | Don't forget to get a common denominator first, before you add or subtract fractions.
Starting from (4)
$$
\begin{align}
\Big(x-\frac{r}{1+r}\Big)^2-\Big(\frac{r}{1+r} \Big)^2+\frac{r}{r+1}+y^2&=\frac{1}{r+1}\tag 4\\
\Big(x-\frac{r}{1+r}\Big)^2+y^2&=\frac{1}{r+1} +\Big(\frac{r}{1+r} \Big)^2- \frac{r}{r+1}\\
&=\frac{r+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$
Find $$\lim_{(x,y) \to (0,0)} \frac{xy-\sin(x)\sin(y)}{x^2+y^2}$$
What I did was:
*
*Find the second order Taylor expansion for $xy-\sin(x)\sin(y)$ at $(0,0)$:
$P_2 (0,0)=f(0,0)+df_{(0,0)}+d^2f_{(0,0)}+r_2(0,0)$
*
*$f(0,0)=0$
*$df_{(0,0)}=\frac{... | We have that
$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\frac{xy}{x^2+y^2}\left(1-\frac{\sin(x)}x\frac{\sin(y)}y\right)\to 0$$
indeed $\frac{xy}{x^2+y^2}$ is bounded and $1-\frac{\sin(x)}x\frac{\sin(y)}y \to 0$
As an alternative by polar coordinates
$$\frac{xy-\sin(x)\sin(y)}{x^2+y^2}=\cos\theta\sin \theta-\frac{\sin(r\cos\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would ca... | A slightly faster variant of the same computation using the identity $$\tan^{-1} u \pm \tan^{-1} v = \tan^{-1} \frac{u \pm v}{1 \mp u v}$$ can be performed by observing that in the special case $u = v$ $$2\tan^{-1} u = \tan^{-1} \frac{2u}{1-u^2}.$$ Consequently, we iterate $g(u) = 2u/(1-u^2)$ twice for $u = 1/5$ to ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 1
} |
Normal functional equation: $ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that for any real numbers $ x $ and $ y $ we have
$$ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $$
What... | Note that for the rest of this proof $a$ and $b$ will be any number in $\mathbb{R}$.
First, consider the following inputs.
\begin{alignat}{3}
P(0,0) &\rightarrow f(f(0)) + f(0) = f(0) + 0 &&\rightarrow f(f(0)) = 0\\
P(0,1) &\rightarrow f(f(0)) + f(1) = f(0) + f(1) &&\rightarrow f(0) = f(f(0)) = 0\\
P(0,a) &\rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Challenging Sum: compute $\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)$ How to prove
$$\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)=\frac74\zeta(2)\zeta(3)-\frac{279}{16}\zeta(5)+\frac43\ln^3(2)\zeta(2)-7\ln^2(2)\zeta(3)\\+\frac{53}4\ln(2)\zeta(4)-\frac2{15}\ln^5(2)+16\operatorn... | A second solution in large steps by Cornel Ioan Valean
Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is
Let $n$ be a non-negative integer number. Then, we have
$$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Find $a$ and $b$ such that the improper integral converges Find $a$ and $b$ so that the integral $$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$ converges.
I know that $\int_1^\infty \frac{1}{x^p}$ converges if $p>1$ and I've tried integration by parts but neither has helped. Can someone give me a hint as to how I ... | We have that
$$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx+\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$
and
$$\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_0^1 \frac{y+1}{y^a(y^2+2y+2)^b}\, dy$$
which converges for $a<1$ since as $y \to 0^+$
$$ \frac{y+1}{y^a(y^2+2y+2)^b} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of $\frac{e^{-xy}}{1+x^2+y^2}$ in $x,y ≥ 0$ I am trying to figure out if $f(x,y)=\frac{e^{-xy}}{1+x^2+y^2}$ goes to zero, in the area x,y ≥ 0, when $x^2+y^2$ goes to infinity.
So I rewrote the function as $\frac{1}{e^{xy}}\frac{1}{1+x^2+y^2}$.
The function $\frac{1}{1+x^2+y^2} $ definitley goes to zero. But what ... | We have that for $x^2+y^2 \to \infty$ since
$$0< e^{-xy}\le 1$$
$$\frac{e^{-xy}}{1+x^2+y^2}\le \frac{1}{1+x^2+y^2}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sum of first n terms is prime.
$a_n$ is a non-decreasing sequence of positive integers. If an positive integer $k$ appears in $a_n$ exactly $k$ times and $S_n$ is the sum of the first n terms, find all $N$ such that $S_N$ is prime.
The sequence is
$$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots$$
If $n$ is a triangular numb... | Let $n=m(m+1)/2 + j$ with $m$ a positive integer and $0\leq j \leq m$.
Then $S_n=\frac{(2m+1)(m+1)m}{6} + (m+1)j$.
We write this as $\frac{(2m^2+m+6j)(m+1)}{6}$
Suppose $m+1$ is not a divisor of $6$, then the left side must be a divisor of $6$. This is clearly only possible if $m=1$, in which case $j=0$ gives $S_n=1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statem... | Hint :
\begin{eqnarray*}
31(4^n+5^n+6^n)= 4^{n+2}+5^{n+2}+6^{n+2} +15 \times 4^n+ 6 \times 5^n - 5 \times 6^n.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Evaluating $\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}\right)+\cos^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$
Evaluate
$$\sin^{-1}\left[\cot\left(\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}\right)+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}\right]$$
$$\sin^{-1}\left... | I think there are some errors in your expression. If you notice $5\pi /12>1$
Also $$\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}=\dfrac{\pi}{12}$$
So I don't think your expression is correct. But if you put the expression $$\sin^{-1}\sqrt{\dfrac{2-\sqrt{3}}{4}}+\cos^{-1}\left(\dfrac{\sqrt{12}}{4}\right)+\sec^{-1}\sqrt{2}$$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$
How to consider it?
| We have that
$$\frac{1}{\sqrt{n^2+n-nk}}\le \frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}\le \frac{(e-1)\frac{\log k}k+1}{\sqrt{n^2+n-nk}}$$
and since
$$\frac{1}{\sqrt{n^2+n-nk}}-\frac{1}{\sqrt{n^2-nk}}=$$
$$=\frac{n}{\sqrt{n^2+n-nk}\sqrt{n^2-nk}\left(\sqrt{n^2+n-nk}+\sqrt{n^2-nk}\right)}=$$
$$=\frac{\frac1{n^2}}{\sqrt{1+\frac1n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$ In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$
My attempt is as follows:-
$$\triangle=\dfrac{1}{2}bc\sin A$$
For having the maximum area,
$$A=90^{\ci... | Let $m_{1}=12,m_{2}=9$ and $m_{3}$ be the third median.
Let $$s=\frac{m_{1}+m_{2}+m_{3}}{2}=\frac{21+m_{3}}{2}$$
So $$s-m_{1}=\frac{m_{3}-3}{2};s-m_{2}=\frac{m_{3}+3}{2};s-m_{3}=\frac{21-m_{3}}{2}$$
Now, area of triangle the length of whose medians are $m_{1},m_{2},m_{3}$ is given by (http://mathworld.wolfram.com/Trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing ... | It' not a coincidence. Let $f(x)=x + g(x)$ where $g(g(x))=x$, that is, $g$ is an involution. We'll show that $f(x)=f(a-x)$, provided that $f(x)=a$.
Assume $f(x)=a$, that is $a-x=g(x)$. Now
$$f(a-x)=f(g(x))=g(x)+g(g(x))=g(x)+x=f(x)$$.
In our case define
$$g(x)=\frac{3x}{\sqrt{x^2 - 9}}.$$
Indeed, $g$ is an involutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x... | \begin{eqnarray}
\mathcal L &=& \lim_{x\to0}\frac{1-\cos x \sqrt{\cos 2x}}{x\sin x}=\\
&=& \lim_{x\to0}\frac{1-\cos^2 x \cos 2x}{x\sin x (1+\cos x\sqrt{\cos 2x})}=\\
&=& \lim_{x\to0}\frac{1-\cos^2x(2\cos^2x-1)}{2x\sin x}=\\
&=& \lim_{x\to0}\frac{1-2\cos^4x+\cos^2x}{2x\sin x}=\\
&=&\lim_{x\to0}\frac{2\left(\frac12 + \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Integral $\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$ I want to evaluate the integral $$\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$$
I tried using Feynman’s trick and introduced the parameter $e^{-\sqrt{1+t^2} x}$, then differentiating with respect to $x$ un... | Another solution for an interesting antiderivative.
$$I=\int\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt$$
First write
$$I=\frac12\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt+\frac12\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt$$
For the first integral, use ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Sequence: $u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$
Study the following sequence of numbers:
$$\forall n\in\mathbb{N}, u_n=\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}}$$
I tried to calculate $u_{n+1}-u_n$, but I couldn't simplify the expression.
Plotting the sequence shows arithmetic (or seems to be an) progression.... | We have that
$$\sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \ge n\cdot \frac{n}{\sqrt{2n^2}}=\frac{n}{\sqrt 2}$$
form which we conclude that $u_n \to \infty$, we can also obtain that
$$\frac{n}{\sqrt{2}}\le \sum_{k=n}^{2n}\frac{k}{\sqrt{n^2+k^2}} \le \frac{2n}{\sqrt{5}}$$
and by Riemann sum since
$$\lim_{n\to \infty }\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A problem about equilateral triangle In below equilateral triangle $ABC$, $AD=CE$, $DH=GH$. Prove that $BE=BG$
My thoughts: It looksl ike we need to prove that $ABGD$ are on the same circle but I am not sure how to use the condition $DH=GH$
| Let $A,B,C$ be vectors from $O(0,0)$ to $A,B,C$ resp, let $a=A-C,b=B-C$. And let $C$ be $(0,0)$ then.
Let's write all we have:
$D=(1-t)a, E=tb$,
$F$:$F=AE\cap BD$, $AE:ua+(1-u)tb$, $BD:vb+(1-v)(1-t)a$
$$\begin{cases}
u=(1-t)(1-v)\\\
(1-u)t=v
\end{cases}$$
$$u=(1-t)(1-t+ut)$$
$$u-t(1-t)u=(1-t)^2$$
$$u = \frac{(t - 1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to solve $2x^2-2\lfloor x \rfloor-1=0$ How do I solve $2x^2-2\lfloor x \rfloor-1=0$?
I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfl... | Suppose $x = n + \delta$ where $n$ is an integer and $0 \le \delta < 1$
Then
\begin{align}
2x^2-2\lfloor x \rfloor-1&=0 \\
x^2 &= \lfloor x \rfloor + \dfrac 12 \\
x &= \sqrt{\lfloor x \rfloor + \dfrac 12} \\
n+\delta &= \sqrt{n + \dfrac 12} \\
\delta &= \sqrt{n + \dfrac 12} - n \\
\end{align}
\begin{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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For $f(x)=x^2+a\cdot x + b$ prove $b\leq -\frac{1}{4}$ Equation $f(f(x))=0$, for $f(x)=x^2+a\cdot x + b$, have four real solutions and sum of two of them is $-1$. Prove that $b\leq-\frac{1}{4}$.
My progress so far:
Let $x_{1}$ and $x_{2}$ be real solutions for $f(x)=0$. We know that discriminant of this equation is pos... | I found out a solution:
Discriminant of equations $f(x)=x_{1}$ and $f(x)=x_{2}$ must be positive so:
$a^2-4b+4x_{1}\ge 0$ and $a^2-4b+4x_{2}\ge 0$, if we sum them we will get:
$2a^2-4a-8b\ge 0$ (by vieta's formula $x_{1}+x_{2}=-a$)
form this we have $b\le \frac{1}{4}(a-1)^2-\frac{1}{4}$ ... (*)
Now discussion by which ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find all solutions to equation in $\mathbb{R}$ Find all solutions to equation in $\mathbb{R}$
$$\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}} = \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}$$
Solution: For $1991\leq x<1996$,
$\frac{x-7}{1989}>\frac{x-1989}{7},... | You have a nice solution.
Anyway, consider that you are looking for the zero of function
$$f(x)=\left(\sqrt{\frac{x-7}{1989}} + \sqrt{\frac{x-6}{1990}} + \sqrt{\frac{x-5}{1991}}\right) -\left( \sqrt{\frac{x-1989}{7}} + \sqrt{\frac{x-1990}{6}} + \sqrt{\frac{x-1991}{5}}\right)$$
Perform a Taylor expansion at $x=1991$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we prove partial sums $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ for all $N$ Can we prove $\sum_{n=1}^{N} \frac{1}{n} \ge \frac{1}{2}\log_2(N)$ ?
I can figure out how to prove this where $N$ is a power of $2$ by adding the successive inequalities listed below (Cauchy condensation)
$$ \frac{1}{1} > 0 $$
$... | Here is one way.
$$
\sum_{n=1}^N \frac{1}{n} = \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{n}
\geq \sum_{n=1}^N \int_{n}^{n+1}\frac{dx}{x}
= \int_{1}^{N+1}\frac{dx}{x} = \Big[\ln x\Big]_1^{N+1}
= \ln (N+1)
$$
(I write $\ln$ for the natural logarithm.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Volume of ellipsoid in Cartesian co ordinates ( w/o changing to spherical or cylindrical systems) I tried triple integrating over the $$\int_{-a}^{a} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} {d}z {d}y {d}x... | For starters, I would rescale $x, y, z$ by $X = x/a, \ Y = y/b, \ Z = z/c$. Thus our triple integral becomes
$$ I = abc \int_{-1}^1 \int_{-\sqrt{1 - X^2}}^{\sqrt{1 - X^2}} \int_{-\sqrt{1 - X^2 - Y^2}}^{\sqrt{1 - X^2 - Y^2}} dZ dY dX. $$
Integrating with respect to $Z$
\begin{align*}
I & = 2abc \int_{-1}^1 \int_{-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the mistake in my solution to find roots of a parabola? The parabola is,
$\frac{1}{2}g{{t}_{t}}^{2}+u{t}_{t}-2H=0$
By the way, $u=\sqrt{\frac{3gH}{2}}$. I have found the discriminant as below,
$\Delta=b^2-4ac$
$\Delta=u^2-4(\frac{1}{2}g)(-2H)$
$\Delta=u^2+4gH$
$u^2=\frac{3gH}{2} \Rightarrow \Delta=\frac{3gH}{2}... | Both solutions produce $0$ in $f$.
\begin{align*}
f&\left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) \\
&= \frac{g}{2} \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right)^2 + u \left( \frac{-u + \sqrt{\frac{11gH}{2}}}{g} \right) - 2H \\
&= \frac{g}{2} \left( \frac{ \left( -u + \sqrt{\frac{11gH}{2}} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | Following the logic in your answer
$$
\lim_{n\to\infty}\overbrace{\left(\frac1n+\frac1n+\cdots+\frac1n\right)}^\text{$n$ terms}=0
$$
since each term tends to $0$. This is obviously false since the sum for each $n$ is $1$, so the limit is $1$.
We can say that a finite sum of terms which have a limit equals the finite su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 6
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Find all primes $p$ such that $x^2\equiv 10\pmod p$ has a solution
Find all primes $p$ such that $x^2\equiv 10$ mod $p$ has a solution
So I want $\left(\frac{10}{p}\right)=\left(\frac{5}{p}\right)\left(\frac{2}{p}\right)=1$
Then $\left(\frac{5}{p}\right)=\left(\frac{2}{p}\right)=1$
or $\left(\frac{5}{p}\right)=\left(... | Everything you do is correct (modulo noting that $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)$ by Quadratic Reciprocity, which is why you are checking for quadratic residues modulo $5$) and dealing with the primes $p=2$ and $p=5$.
If $p=2$ or $p=5$, then $x^2\equiv 10\pmod{p}$ clearly has solutions.
Otherwise, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proving systems of nonlinear modular equations have no solution I have reason to suspect this system of six nonlinear modular equations has no solution for $2 < x < y < z$ even integers.
$$
\left\{
\begin{aligned}
z(3y+2) \equiv y(3z+2) \equiv 0& \mod x\\
z(3x+2) \equiv x(3z+2) \equiv 0& \mod y\\
x(3y+2) \equiv y(3x+2... | Proof outline
*
*We first simplify the system into (almost) coprime modulus using only $3$ of the $6$ equations (ignoring the $\equiv 0$ part).
*This will allow us to derive 2 general classes of solutions.
*We will then use a fourth equation $x(3z+2)\equiv 0 \pmod y$ to show both are not feasible, concluding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\t... | The problem is not too difficult if you compose Taylor series
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\frac{5 x^7}{112}+O\left(x^9\right)$$
$$\sin(x)\sin ^{-1}(x)=x^2+\frac{x^6}{18}+\frac{x^8}{30}+O\left(x^{10}\right)$$ will give no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving a first order ordinary differential equation that is not linear and not separable Problem:
Solve the following differential equations.
$$ ( 3x - y + 1 ) dx - ( 3x - y) dy = 0 $$
Answer:
I am going to use the substitution $z = 3x - y + 1$.
\begin{align*}
\frac{dz}{dx} &=3 - \frac{dy}{dx} \\
dz &=3 \, dx - dy \\
... | $$( 3x - y + 1 ) dx - ( 3x - y) dy = 0$$
You made a sign mistake here: $$z{+(z-1)(3-z')}=0$$ it should be $$z\color{blue}{-(z-1)(3-z')}=0$$
$$z-3(z-1)+(z-1)z'=0$$
$$-2z+3+(z-1)z'=0$$
$$(z-1)z'=(2z-3)$$
$$\frac {(z-1)}{(2z-3)}dz=dx$$
Integrate:
$$2z+\ln|z-\frac 32|=4x+C$$
Substitute back $z=3x-y+1$:
$$2x-2y+\ln|3x-y-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $n$ is a natural number then how to show that $\frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{\sqrt{5}*2^n} $is also a natural number? This question is from Introduction to Analysis volume 1 by R. Courant, chapter 1, section 1.5, 9th question.
I tried using binomial theorem, root $5$ get an odd power in the numerator and ca... | We derive the formal power series $\sum_{n=0}^\infty F_nz^n$ with
\begin{align*}
F_n=\frac{1}{2^n\sqrt{5}}\left(\left(1+\sqrt{5}\right)^n-\left(1-\sqrt{5}\right)^n\right),\quad n\geq 0\tag{1}
\end{align*}
We do some calculations with its coefficients and conclude that $F_n$ is a natural number for each $n\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Hard inequality for positive numbers The problem is to prove that for $a,b,c>0$ we have
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+... | I will use the following inequality:
(Vasile Cirtoaje) If $x,y,z$ are positive real numbers, then:
$$(x+y+z)^3 \geq \frac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
Setting $x=\frac{a^2}{b^2}, y = \frac{b^2}{c^2}, z = \frac{c^2}{a^2}$, we find
$$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^3\geq \frac{27}{4}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Solve $\int x^2e^x\sin x$ Solve $\int x^2e^x\sin x$
My attempt is as follows:-
$$I_1=\int e^x\sin x$$
$$I_1=e^x\sin x-\int e^x\cos x$$
$$I_1=e^x\sin x-e^x\cos x-\int\sin (x)e^x$$
$$2I_1=e^x\left(\sin x-\cos x\right)$$
$$I_1=\dfrac{e^x\left(\sin x-\cos x\right)}{2}\tag{1}$$
$$I_2=\int e^x\cos x$$
$$I_2=e^x\cos x+\int e^... | Start with writing $\sin x=\color{red}{\Im}e^{ix}$
$$\int x^2e^x \sin x\ dx=\color{red}{\Im}\int x^2e^{(1+i)x}\ dx$$
$$=\color{red}{\Im}\left(\frac{x^2}{1+i}e^{(1+i)x}-\frac{2x}{(1+i)^2}e^{(1+i)x}+\frac{2}{(1+i)^3}e^{(1+i)x}\right)$$
$$=x^2 \color{red}{\Im} \frac{e^{(1+i)x}}{1+i}-2x \color{red}{\Im}\frac{e^{(1+i)x}}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Factoring $x^4-2x^3+2x^2+x+4$
I need to show that the polynomial is not irreducible and I am trying to factor the polynomial
$$x^4-2x^3+2x^2+x+4$$
I checked from a calculator that it has a factor but how do I get it by myself?
I tried grouping but It didnt work I got
$x^2(x^2-2x+2)+x+4$ And I dont know how should... | $$x^4-2x^3+2x^2+x+4$$
$$=(x^2+1)^2-2x^3+x+3$$
$$=(x^2+1)^2-2x(x^2+1)+3x+3$$
$$=(x^2+1)^2+x(x^2+1)-3x^3+3$$
$$=(x^2+1)^2+x(x^2+1)-3(x-1)(x^2+x+1)$$
$$=(x^2+x+1)(x^2+1-3x+3)$$
and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Problem (1) 1.A of Linear Algebra Done Right by Axler: Multiplicative Inverse My textbook, Linear Algebra Done Right, Third Edition, by Axler, has the following problem:
Suppose $a$ and $b$ are real numbers, not both $0$. Find real numbers $c$ and $d$ such that
$$\dfrac{1}{(a + bi)} = c + di.$$
This is effectively fi... | You have
$$\begin{equation}\begin{aligned}
\frac{1}{a + bi} & = \frac{a-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \frac{-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \left(\frac{-b}{a^2+b^2}\right)i
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Consider the solution of what $c$ and $d$ are in the original equation of
$$\fr... | {
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"url": "https://math.stackexchange.com/questions/3489660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\foral... | Observe that:
$$
a_{n+2} = \frac{a_{n+1} + 2}{3a_{n+1} + 2} = \frac{\frac{a_n + 2}{3a_n + 2} + 2}{3\frac{a_n + 2}{3a_n + 2} + 2} = \frac{a_n + 2 + 6a_n + 4}{3a_n + 6 + 6a_n + 4} = \frac{7a_n + 6}{9a_n + 10}
$$
Therefore:
$$
a_{n+2} - a_n = \frac{7a_n + 6}{9a_n + 10} - a_n = \frac{-9a_n^2 - 3a_n + 6}{9a_n + 10} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Error in different attempts to solve $\int \frac{dx}{(1+x^4)^{1/4}}$
$$
\int \frac{dx}{(1+x^4)^{1/4}}=\frac{1}{2}\bigg[\frac{1}{2}\log\Big|\frac{1+z}{1-z}\Big|-\tan^{-1}z\bigg]+C
$$
Find $z$
Method 1
Set $x^2=\tan t\implies 2x.dx=\sec^2t.dt\implies dx=\dfrac{\sec^2t.dt}{2\sqrt{\tan t}}$
$$
\int\frac{dx}{(1+x^4)^{1/... | Both solutions only differ by a constant and a mistake in sign because
$$\log\left|\frac{1+\frac{1}{z}}{1-\frac{1}{z}}\right| = \log\left|\frac{1+z}{1-z}\right|$$
and
$$\tan^{-1}\left(\frac{1}{z}\right) = \frac{\pi}{2} - \tan^{-1}(z)$$
which is consistent because there is a sign error in a method, I just can't immedi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Cauchy Schwarz inequality Taken from Titu Andreescu and Bogdan Enescu's Mathematical Olympiad Treasures on page 9 Problem 1.19, to prove,
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geqslant \frac{3}{2}$.
It is easy to see why $LHS$ may yield
$\geqslant\frac{(a+b+c)^2}{2(ab+bc+ca)}$ from Cauchy.
Yet how do... | Write $x:=b+c,\,y:=c+a,\,z:=a+b$ so $\frac{a}{b+c}=\frac{y+z-x}{2x}$. The desired result is equivalent to $\sum_\text{cyclic}\frac{y+z}{x}\ge6$, which is trivial because Cauchy-Schwarz implies $q+\frac1q\ge2$ (hint: use $\binom1q,\,\binom{q}{1}$) for $q\in\{\frac{x}{y},\,\frac{y}{z},\,\frac{z}{x}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$, $a^2-b^2+c^2-d^2=1749$. What are the possible values of $a^2+b^2+c^2+d^2$?
$a, b, c, d$ are all prime numbers. $a>3b>6c>12d$,
$a^2-b^2+c^2-d^2=1749$. What are the possible values of
$a^2+b^2+c^2+d^2$?
So far I've used difference of squares to get: $(a+b)(a-b)+(c+... | From Yahoo Answers:
The inequality can be rewritten as:
$$a ≥ 3b+1, b ≥ 2c+1, c ≥ 2d+1 \tag{i}$$
$$a^2−b^2+c^2−d^2 = 1749 \tag{ii}$$
Using $\text{(i)}$, the left side of $\text{(ii)}$ is always odd and $1749$ is also odd. Therefore $d^2$ must be even which forces $d = 2$.
With $d=2$, $\text{(ii)}$ becomes $1753 = a^2−... | {
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"url": "https://math.stackexchange.com/questions/3497826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Distributing balls into bins randomly Problem:
If $n$ balls are distributed at random into $r$ boxes (where $r \geq 3$), what is the probability that box $1$ at exactly $j$ balls for
$0 \leq j \leq n$ and box $2$ contains exactly $k$ balls for $0 \leq k \leq n$ ?
Answer:
Let $p$ be the probability that we see... | You have corrected the mistakes that were your initial solution. Your work is now correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In a ring where $(a-b)^2 = a - b$ for fixed $a,b$, then $(a-b)(a+b) = 1 \iff a^2 - b^2 = 1$. Let $(A,+,\cdot)$ be a ring where $(a-b)^2 = a - b$ for fixed $a,b$. Then $$(a-b)(a+b) = 1 \iff a^2 - b^2 = 1.$$
I was able to prove one implication:
Proof ($"\Rightarrow")$
We have that $(a-b)^2(a+b) = a-b = (a-b)(a+b) = 1$ th... | Idempotents are always a joy to work with. For the converse implication denote $a-b=e$ and notice that $e$ is an idempotent by hypothesis; substituting $a=b+e$ in the relation $a^2=b^2+1_A$ leads (after a few simple calculations and cancellations) to $e+be+eb=1_A$ and hence to
$$eb+be=1_A-e \tag{1}$$
Multiplying relat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Inequality Solution Correctness Let $a,b,c \in \Re $ and $ 0 < a < 1 , 0 < b < 1 , 0 < c < 1 $ & $ \sum_{cyc} a = 2$
Prove that
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
My solution
$$ \prod_{cyc}\frac{a}{1-a} \ge 8$$
or
$$ \prod_{cyc}a \ge 8\prod_{cyc}(1-a)$$
From $ \sum_{cyc} a = 2$ we can conclude $\prod_{cyc}a \le ... | You need to prove that:
$$\prod_{cyc}(1-a)\leq\frac{1}{8}abc.$$
You proved that $$\prod_{cyc}(1-a)\leq\frac{1}{27}.$$
Thus, it's enough to prove that $$\frac{1}{27}\leq\frac{1}{8}abc,$$ which is wrong, which says that your solution is wrong.
Since $$a+b-c=2-2c>0,$$ we see that $a$, $b$ and $c$ are sides-lengths of a tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find solution of system of recurrence relations by using generating function How can I use generating functions to find $\{a_n\}$ and $\{b_n\}$ if
$\left\{ \begin{array}{l}
a_n=a_{n-1}+b_{n-1};\\
b_n=n^2a_{n-1}+b_{n-1};\\
a_0=1, b_0=1.
\end{array} \right.$
Thank you in advance.
| Hint.
Taking
$\left\{ \begin{array}{l}
\sum_{n=1}\left(a_nx^n-a_{n-1}x^n-b_{n-1}x^n\right) = 0\\
\sum_{n=1}\left(b_nx^n-n^2a_{n-1}x^n-b_{n-1}x^n\right)=0\\
\end{array} \right.$
and calling
$$
\sum_{n=0}a_nx^n = A(x)\\
\sum_{n=0}b_nx^n = B(x)
$$
we have
$$
\cases{A(x)-a_0-xA(x)-xB(x) = 0\\
B(x) - b_0 -\sum_{n=1}n^2a_{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\gcd(a,b)=d$ then prove that $\gcd(\frac{a}{c},\frac{b}{c})=\frac{d}{c}$ My attempt:
$ax+by=d$(as gcd(a,b)=d).
Dividing by c
$\frac{a}{c}x+\frac{b}{c}y=\frac{d}{c}$
Let $k$ be the gcd of$ (\frac{a}{c},\frac{b}{c})$ so
$k|\frac{a}{c}$ and $ k|\frac{b}{c}$ then $k|\frac{a}{c} x + \frac{b}{c}y$ and so $k|\frac{d}{c}$... | Clearly $\frac {d}{c} \mid \frac {a}{c}$ and $\frac {b}{c}$ as $\gcd(a,b)=d$. Let $m$ be a common divisor of $\frac {a}{c}$ and $\frac{b}{c}$. Then $m$ divides the left side which implies that $m \mid \frac{d}{c}$ and $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving recursive function outputs $2^x \cdot {x \choose y}$ Consider following function
$f: \mathbb{N}\times\mathbb{N} \rightarrow \mathbb{N}$:
\begin{align*}
f(x ,y) =
\begin{cases}
0 & \text{if } x < y\\
2^x & \text{if } y = 0\\
2 \cdot (f(x - 1, y - 1) + f(x - 1, y)) & \text{else.}\\
\end{cases}
\end{alig... | short answer: Do induction on the values of $n = x -y$ starting with $n = 0$.
Prove the Base case and the induction step by induction on $y$ starting with $y=0$.
That is the Base case is to prove that the result is true for $x-y=0$ and showing this holds for $f(y,y)$ by doing base case $f(0,0)$ and doint induction on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Question on Integration solution I am looking through the solution of integrating: $$\int_{}^{} \frac{x^4+1}{x^3+1} dx$$
I understand it but for the following part:
My problem is limited to the first and penultimate column. I would very much appreciate it if someone could explain to me these, especially the penultimat... | An introduction to partial fractions.
If we have two fraction and we want to add them together, $\frac {a}{b} + \frac {c}{d} = \frac {ad + bc}{bd}$
We can do the same thing, with rational polynomials.
$ \frac {1}{x} - \frac {1}{x+1} = \frac{x+1}{(x)(x+1)} - \frac {x}{x(x+1)} = \frac {x+1-x}{x(x+1)} = \frac {1}{x(x+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
... | Using $e^x=1+x+O(x^2)$ along with $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2=O\left(n^3\right)$ , we assert that
$$\begin{align}
\sum_{k=1}^ne^{k/n^2}&=\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)+O\left(\frac{1}{n}\right)\\\\
&=n+\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)\\\\
&=n+\frac12+O\left(\frac1n\right)
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence... | $\begin{array}\\
\Delta^2_hf(x)
&= f(x-h)-2f(x)+f(x+h)\\
\text{so}\\
\Delta^2_1f(x)
&= f(x-1)-2f(x)+f(x+1)\\
&\approx (f(x)-f'(x)+f''(x)/2-f'''(x)/6+...)-2f(x)+(f(x)+f'(x)+f''(x)/2+f'''(x)/6+...)\\
&\approx f''(x)/2+O(f''''(x))\\
\end{array}
$
If
$f(x) = x^{1/2},\\
f'(x) = x^{-1/2}/2,\\
f''(x) = -x^{-3/2}/4,\\
f'''(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
... | Your error in algebra:
You have
$a^n = a^2 a^{n-1}
$
for $a = \frac12$ and
$\frac14$.
It should be
$a^n = a\cdot a^{n-1}
$.
$a^2 a^{n-1}
=a^{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| Starting from
$$\int \frac{x}{x^4+4}\,dx$$
Make the substitution $u=x^2$. Then, $\dfrac{du}{2}=x\,dx$ and the integral becomes
$$\frac{1}{2}\int \frac{1}{u^2+4}\,du$$
Now let $u=2\tan(\theta)$. Then, $du=2\sec^2(\theta)\,d\theta$ and the integral becomes
$$\frac{1}{4}\int \frac{\sec^2(\theta)}{1+\tan^2{(\theta)}}\,d\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove or disprove inequality between 1-norm, 2-norm and infinity norm. Prove or disprove:
$$
v\in R^n \Rightarrow ||v||_1||v||_{\infty}\leq \frac{1+\sqrt n}{2}||v||^2_2.
$$
Can someone help me with this? I have used several cases to verify it but cannot come up with proof.
| Here is a proof which also shows that the inequality is tight.
Instead of $|x_i|$, consider w.l.o.g. $x_i \ge 0$. Further, let $x_{\rm max} = x_n$.
Denote $\sqrt{\sum_{i=1}^{n-1} x_i^2} = ||\bar v||_2$.
First, by Cauchy-Schwarz
$$||v||_1 = \sum_{i=1}^{n-1} 1 \cdot x_i + x_n\le \sqrt{\sum_{i=1}^{n-1} 1} \sqrt{\sum_{i=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove th... | Another proof:
Since the inequality is symmetric in $a$ and $c$, WLOG, assume that $a \ge c$.
Due to homogeneity, assume that $c = 1$.
Let $a = 1 + s$ for $s \ge 0$.
We split into two cases:
1) $0 < b \le 1$: Let $b = \frac{1}{1+t}$ for $t \ge 0$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(2+t)(st+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^... | Both computations are incorrect. The first method amounts to writing
\begin{equation}
\begin{array}{l}
1 - \cos(x^2) = \frac{x^4}{2} + o(x^4)\cr
\arctan(x) = x + o(x)\cr
\sin(2x)= 2x + o(x)\cr
e^{x^2}= 1 + x^2 + o(x^2)
\end{array}
\end{equation}
but substituting these values in the fraction cannot tell us the limit bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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If $1−a\cos x−b\sin x−A\cos2x−B\sin2x\geq0$ for all real $x$, then $a^2+b^2\leq2$ and $A^2+B^2\leq1$
Four real constants $a$, $b$, $A$, $B$ are given, and
$$
f(x) = 1 − a\cos(x) − b\sin(x) − A\cos(2x) − B\sin(2x)
$$
Prove that if $f(x)\geq 0$ for all real $x$, then
$$
a^2+b^2\leq 2 \quad \mbox{and} \quad A^2 + B... | Observe
\begin{align}
a \cos()+b\sin()
&= \sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos(x)+\frac{b}{\sqrt{a^2+b^2}}\sin(x) \right)\\
&= \sqrt{a^2+b^2} \sin(x-\theta_0)
\end{align}
for some $\theta_0$ and likewise
\begin{align}
A \cos(2)+B\sin(2) = \sqrt{A^2+B^2} \sin(2x-\phi_0).
\end{align}
Then we see that
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Unifying the sum of two error functions. Let's propose you have the sum of two error functions:
$f(x) = \text{erf}(ax)+\text{erf(bx)}$
If you wanted to solve for x, you'd first unify them into a third function
$f(x) = \text{erf}(cx)$
Or something similar. How could you approach this problem?
| As already said in comments and answer, you will need some numerical method for solving for $x$ the nonlinear equation
$$y=\text{erf}(a x)+\text{erf}(b x)$$
I shall consider the case where $a$ and $b$ are positive (the other situations being more complex).
The first point is to notice that $y$ is bounded by $2\text{erf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find:$\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$
Evalute: $$\lim_{x\to
0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$
My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2... | Here is another way after your first step (removing the $\sin() $). Note that if $A, B$ tend to $1$ then $$A-B=B\cdot\frac{\exp(\log A - \log B) - 1}{\log A - \log B} \cdot(\log A - \log B) $$ and the first two factors above tend to $1$. Therefore $A-B$ can be replaced by $\log A - \log B$.
Thus the numerator in your c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find cube roots of 1 modulo power of two (if they exist)? It would be useful for efficiently implementing an algorithm if I could find a $c > 1$ where $c^3 \equiv 1 \pmod {2^{64}}$. It's plausible that such a $c$ exists because $2^{64} \equiv 1 \pmod 3$, so all non-zero values could be partitioned into groups of... | You can do this by induction. We begin with $1^3 \equiv 1 \pmod {2^k}$ where the start is $k=1.$ In paerticular, there are no others. Do we get any additional roots as $k$ increases? Just two choices.
This works: $1^3 \equiv 1 \pmod {2^{k+1}}$
Maybe:
$$ (1+2^k)^3 = 1 + 3 \cdot 2^k + 3 \cdot 2^{2k} + 2^{3k} \equiv 1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Generating function of binary string with minimum lengths for 0s and 1s a) Show that the generating function by length for binary strings where every block of 0s has length at least 2, each block of ones has length at least 3 is:
$$\frac{(1-x+x^3)(1-x+x^2)}{1-2x+x^2-x^5}$$
b) Give a recurrence relation and enough initi... | robjohn has given a generating function. Here is a recurrence based approach.
Suppose $b_n$ is the number of strings ending with $0$ and $c_n$ the number ending with $1$, and $a_n=b_n+c_n$ being what you want. It seems obvious that
*
*$a_n=b_n+c_n$ by definition
*$b_n=b_{n-1}+c_{n-2}$ by adding $0$ to an exist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving the trigonometric equation $483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$
I'm trying the value of $\alpha \in [0,\pi]$ that is solution to this trigonometric equation: $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
I... | Starting from @the_candyman's answer
$$\begin{cases}
\frac{483}{2}Y + \frac{483\sqrt{3}}{2}X + 16\sqrt{3}XY + 48X^2 - 4 = 0 \\
X^2 + Y^2 = 1
\end{cases}$$
eliminate $Y$ from the first equation
$$Y=\frac{-96 X^2-483 \sqrt{3} X+8}{32 \sqrt{3} X+483}$$ Plug it in the second to end with
$$12288 X^4+123648 \sqrt{3} X^3+928... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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value of expression having variables $p,q,r,x,y,z$
If $p,q,r,x,y,z$ are non zero real number such that
$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$
Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is
what try
$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$
$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2... | By C-S $$0=px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}\geq px+qy+rz+|px+qy+rz|$$ and since
$$px+qy+rz+|px+qy+rz|\geq0,$$ we obtain $$px+qy+rz+|px+qy+rz|=0,$$ which gives
$$px+qy+rz\leq0.$$
Also, the equality occurs for $$(x,y,z)||(p,q,r),$$ which says that there is $k<0$, for which
$$(p,q,r)=k(x,y,z).$$
Thus, $$\sum_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given that $x^2 + ax + b > 0$ and $x^2 + (a + np)x + (b + nq) > 0$, prove that $x^2 + (a + mp)x + (b + mq) > 0, m = \overline{1, n - 1}$.
Given that $$\large \left\{ \begin{align} x^2 + ax + b > 0\\ x^2 + (a + np)x + (b + nq) > 0 \end{align} \right., \forall x \in \mathbb R \ (p, q \in \mathbb R \setminus \{0\}, n \in... | U could always check out D.See if leading coefficient >0 and D is less than 0 the quadratic is >0 for all real x.
Given : $ a^2 - 4b <0$
Given : $ (a+np)^2 - 4(b+nq)= a^2-4b + n(np^2 +2(ap-2q))<0$
(as n decreases $np^2 +2(ap-2q)$ also decreases) $\Rightarrow$ the expression is lesser for lesser 'n' and hence less tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $2x^2-4xy+6y^2=9$, find the biggest and lowest value of $2x-y$ So, that's the problem. I tried factoring the quadratic equation to get something like $2x-y$ but it doesn't work. Dividing by $y^2$ won't work because the right side is $9$, not $0$. The last idea I have is to say that $2x-y=t => y=2x-t$ and rep... | By C-S:
$$3=\sqrt{2x^2-4xy+6y^2}=\frac{2}{3}\sqrt{\left(2(x-y)^2+4y^2\right)\left(2+\frac{1}{4}\right)}\geq$$
$$\geq\frac{2}{3}\sqrt{(2(x-y)+y)^2}=\frac{2}{3}|2x-y|,$$ which gives
$$-\frac{9}{2}\leq2x-y\leq\frac{9}{2}.$$
The equality occurs for $$\left(\sqrt2(x-y),y\right)||\left(\sqrt2,\frac{1}{2}\right),$$ which with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p {\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q. If ${\sqrt 3} - {\sqrt 2}, 4- {\sqrt 6}, p{\sqrt 3} - q {\sqrt 2}$ form a geometric progression, find the values of p and q.
So I take the second term $4-{\sqrt 6} =( {\sqrt 3} - {\sqrt 2}) (r... | Note
$$(4- {\sqrt 6})^2=({\sqrt 3} - {\sqrt 2})( p {\sqrt 3} - q {\sqrt 2})$$
or,
$$22-8\sqrt6 = 3p +2q -(p+q)\sqrt6$$
Therefore,
$$22= 3p +2q,\>\>\>\>\>p+q=8$$
Solve to obtain $p=6$ and $q=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \sin{x}\ +\ \frac{1}{2}\sin{2x}\ +\ \frac{1}{3}\sin{3x}\ +\ \frac{1}{4}\sin{4x}\ =\ \frac{2}{3}\left(\cos{x}+1\right)\left(\sin^5x\ +\ 4\right)$ SORRY IF MY TITTLE IS UNCLEAR WITH ONLY MATH FUNCTIONS. IT'S MORE THAN 150 CHARACTERS
*
*This is my math problem
$$ \sin {x}\ +\ \frac{1}{2}\sin {2x}\ +\ \frac{1}{3}\si... | An algorithmic approach for many of these problems can be:
You can call $t=\tan(\frac{x}{2})$. Then $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t}{1+t^2}$. Making this substitution your equation
$$6\sin(x)cos^2(x) − 2\cos(x)\sin(x) + 2\sin(x) − 2\sin^5(x) − 8 = 0$$
becomes a polynomial equation.
$$6(2t)(1-t)^2(1+t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Splitting field of $\alpha=(2+\sqrt{2})^{1/3}$ Given $\alpha=(2+\sqrt{2})^{1/3}$ The problem asks to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and find it's splitting field. I have the solution and I understood until the point where it shows that the minimal polynomial is $$x^6-4x^3+2=0$$
Now, sinc... | First, remember that if $\alpha$ is a (real or complex) root of $x^n-a=0$, $a\neq 0$, then the other roots are all of the form $\lambda^i\alpha$, where $\lambda$ is a primitive $n$th root of unity. That is, a complex number such that $\lambda^n=1$ but $\lambda^k\neq 1$ for $1\leq k\leq n-1$. This follows simply by noti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral with binomial to a power $\int\frac{1}{(x^4+1)^2}dx$ I have to solve the following integral:
$$\int\frac{1}{(x^4+1)^2}dx$$
I tried expanding it and then by partial fractions but I ended with a ton of terms and messed up. I also tried getting the roots of the binomial for the partial fractions but I got complex... | Use $\left(\frac x{x^4+1}\right)' = -\frac3{x^4+1} + \frac 4{(x^4+1)^2} $ to rewrite the integral as
$$I = \int \frac 1{(x^4+1)^2}dx=\frac x{4(x^4+1)}+\frac34\int\frac1{x^4+1} dx$$
where
\begin{align}\int\frac2{x^4+1} dx = &\int\frac{1+x^2}{x^4+1} dx + \int\frac{1-x^2}{x^4+1} dx\\
= &
\int\frac{d(x-\frac1{x})}{(x-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}$. While attempting to algebraically solve a trigonometry problem in (Question 3535106), I came across the interesting equation
$$
4\sin^{2}\frac{\pi}{9}-2\sqrt{3}\sin\frac{\pi}{9}+1=\frac{1}{4}\sec^{2}\frac{\pi}{9}
$$
which ar... | Evaluate
$$
\begin{aligned}
4\cos ^2\frac{π}{9}&\left( LHS-RHS \right) \\
= \ &16\sin ^2\frac{π}{9}\cos ^2\frac{π}{9}-8\sqrt{3}\sin \frac{π}{9}\cos ^2\frac{π}{9}+4\cos ^2\frac{π}{9}-1\\
=\ &4\sin ^2\frac{2π}{9}-4\sqrt{3}\sin \frac{2π}{9}\cos \frac{π}{9}+2\left( 1+\cos \frac{2π}{9} \right) -1 \\
= \ &2\left( 1-\cos \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{... | Like Steven Sadnicki said, you factored out the $(-1)^n$ term even though it's inside the sum, so the summands are of alternating sign. Here's a solution:
$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2 + n - 2 } = \sum_{n=2}^{\infty} \frac{(-1)^n}{(n-1)(n+2)} = \sum_{n=2}^{\infty} (-1)^n [ \frac{1}{3} \cdot ( \frac{1}{n-1} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove $a^3+b^3+3abc>c^3$ where a,b,c are triangle sides If $a,b,c$ are triangle sides prove
$a^3+b^3+3abc>c^3$
| Method 1. For any numbers $a,b,c$ let $F(c)=c^3-3abc-(a^3+b^3).$ Observe that $F(a+b)=0$ so $c-(a+b)$ is a factor of the polynomial $F(c).$ Using synthetic division we obtain $$F(c)=[c-(a+b)]\cdot [c^2+c(a+b)+(a^2-ab+b^2)].$$ Now if $a,b,c$ are positive and $c<a+b$ then $$c-(a+b)<0$$ while $$c^2+c(a+b)+(a^2-ab+b^2)>(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of $a$ and $b$ if the limits exist. Could someone solve this
$$f(x) =
\begin{cases}
x^3+a, & x<0 \\
a\sin\frac{\pi{x}}{3}+b, & 0\leq {x}<2 \\
3, & x=2 \\
\log_2x^{b+1}, & x>2\\
\end{cases}$$
If $\lim_{x\to0}$ and $\lim_{x\to2}$ both exist, find the value of a and b.
| We have $$\lim_{x\rightarrow 0^-}f(x)=a; \lim_{x\rightarrow 0^+}f(x)=b;$$ $$\lim_{x\rightarrow 2^-}f(x)=b+a\frac{\sqrt{3}}{2}; \lim_{x\rightarrow 2^+}f(x)=b+1$$
Because $\exists \lim_{x\rightarrow 0}f(x)$ and $\exists \lim_{x\rightarrow 2}f(x)$, then
$a=b$ and $b+1=b+a\frac{\sqrt{3}}{2}$ so $a=b=\frac{2\sqrt{3}}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a $3\times 3$ matrix $P$ such that $g_{j} = \sum_{i=1}^{3}P_{ij}f_{i}$ Let $V$ be the vector space over the complex numbers of all functions from $\textbf{R}$ into $\textbf{C}$, i.e., the space of all complex-valued functions on the real line. Let $f_{1}(x) = 1$, $f_{2}(x) = e^{ix}$ and $f_{3}(x) = e^{-ix}$.
(a) P... | (a) Your reasoning is not sound. Scalars $a,b,c$ are complex, so $a = -1 - i$, $b = 1$, $c = i$ satisfy your two equations also. A hint is that you are on the right track—just go one step further.
(b) I got the same $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can $6$ prizes be distributed among $4$ persons such that each one gets at least one prize? My understanding:
First select $4$ prizes and distribute it among the $4$ people in $^6C_4\times4!$ ways and then distribute the remaining $2$ prizes in two cases: when $2$ people have $2$ prizes: $\frac{4!}{2!}... | We can also count this using a generating function approach.
Since the prizes are not identical, we compute the exponential generating function. To remove the possibility of anyone getting no prizes we use $e^x-1$ in each factor. The number of ways to distribute $6$ prizes among $4$ people is
$$
\begin{align}
6!\left[x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$ Denote $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$, then how can we prove that
*
*$f(n,1)=\cdots=f(n,n-1)=0,$
*$f(n,n)=\frac{1}{2} (2 n)!,\ \ f(n,n+1)=\frac{1}{12} n (n+1) (2 n+1) (2 n)!, \cdots$
Moreover is there a general closed-for... | I would like to fill in the details for @MHZ. We seek to evaluate
$$F_{n,j} = \sum_{k=0}^n k^{2j} (-1)^{n-k} {2n\choose n-k}.$$
where $j\ge 1.$ With this in mind we introduce the function
$$F_n(z) = \frac{(2n)!}{2} z^{j-1} \prod_{q=1}^n \frac{1}{z-q^2}.$$
This has the property that the residue at $z=k^2$ where $1\le k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| $$x^2(x^2+6)=(x^2+3)^2-3^2$$
WLOG $x^2+3=3\csc t,t\to0^+$
$$\lim_{t\to0^+}3(\csc t-1)(3\csc t-3\cot t)$$
$$=\lim\dfrac{9(1-\sin t)(1-\cos t)}{\sin^2t}$$
$$=9\lim\dfrac{1-\sin t}{1+\cos t}=?$$ as $1-\cos t\ne0$ as $t\ne0$ as $\to0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ ... | Here's an alternative answer if you absolutely have to use L'Hopital's rule:
First rewrite the expression inside the limit as follows:
$$x^3\Big(\sin(\frac{1}{x+2})-2\sin(\frac{1}{x+1})+\sin(\frac{1}{x})\Big)=x^3\Big[(\sin(\frac{1}{x+2})-\frac{1}{x+2})-2(\sin(\frac{1}{x+1})-\frac{1}{x+1})+\sin(\frac{1}{x})\Big]+x^3\Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Closed form of integral $\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx$ Originally this integral was interesting for me as it represents the specific harmonic number $H_{-\frac{1}{5}}$ and also because Mathematica returned the wrong value $0$ for the integral. I posted the probem here https://mathematica.stackexch... | My working is quite long but I got the following in the end:
$$\begin{align}F(x) = \int\dfrac{1 - \frac1{\sqrt[5]x}}{1 - x}\,\mathrm dx =&\dfrac{-\left(5 + \sqrt 5\right)\ln\left(2\sqrt[5]{x^2} + \left(1 + \sqrt{5}\right)\sqrt[5]x + 2\right) - \left(5 - \sqrt5\right)\ln\left(2\sqrt[5]{x^2} + \left(1 - \sqrt{5}\right)\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Beginner Proof by Induction: Is this correct? I am asked to prove $\frac{1}{1.4} + \frac{1}{4.7} + ... + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}, n\geq 1$ by induction.
Can someone verify that I did this correctly. I am unsure about my Inductive Step.
Proof by induction
Inductive Hypothesis Let $P(k) = \sum\limits_{i=1... | I don't understand what you're doing in the last couple of lines for $P(k + 1)$, for example between the third & second last, you change the RHS value but leave the LHS the same.
Instead, I would normally go from one side of the equation, expand or otherwise manipulate it as need be so can use the assumption that $P(k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ Question: Solve the PDE $(3x+y-z)p+(x+y-z)q=2(z-y)$ by Lagrange's method of Solution.
Progress: Lagrange's auxiliary equations are.
$$\dfrac{dx}{3x+y-z}=\dfrac{dy}{x+y-z}=\dfrac{dz}{2(z-y)}$$ The three ratios are equal to $\dfrac{-dx+3dy+dz}{0}$. So we have $-dx+3dy+dz=0\imp... | We can find that,
$$\text{Each ratio} = \frac{dx-dy+dz}{2x-2y+2z} = \frac{dx+dy-dz}{4x+4y-4z}$$
or
$$\frac{d(x-y+z)}{x-y+z} = \frac{d(x+y-z)}{2(x+y-z)}$$
On integrating,
$$\ln(x-y+z) = \frac{1}{2}\ln(x+y-z) +\ln C_2 \implies C_2 = \frac{x-y+z}{\sqrt{x+y-z}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$ I have to find this integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}} dx$$
My attempt was to split the integral:
$$\int_{\frac{1}{4}}^4\frac{(x+1)\arctan x}{x\sqrt{x^2+1}}=\int_{\frac{1}{4}}^4\frac{x\arctan x+1}{x\sqrt{x^2+1}}+\int_{\... | The bounds suggest the subtitution $u=\frac{1}{x}\Rightarrow dx=-\frac{1}{u^2}du$
$$\int_{1/4}^4 \frac{(x+1)\arctan x}{x\sqrt{x^2+1}} \, dx=\int_{1/4}^4 \frac{(1+u)\arctan \frac{1}{u}}{u\sqrt{u^2+1}} \, du$$
Therefore, using $\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}$ for $x>0$, we get:
$$
\begin{aligned}
\int_{1/4}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $\lfloor k^4\rfloor$ where $k^3 - 5k^2 + 2 = 0$. Problem :
Find a value of $\lfloor k^4 \rfloor$ where $k$ is the biggest real root of equation $k^3 - 5k^2 + 2 = 0$.
Can't use calculator.
Let $f(x) = x^3 - 5x^2 + 2$. Then $f(4)f(5)<0$. so $k\in(4,5)$.
And I bounded more like : $$\frac{9}{2}<k<5.$$
But this is not... | If $a<b<c$ are roots of the equation, so by the Viete's theorem $a^4+b^4+c^4=585$ and since for another roots $x_i$ we have $|x_i|<0.7,$ we obtain the answer: $[c^4]=584.$
Indeed, $a+b+c=5,$ $ab+ac+bc=0$ and $abc=-2$.
Thus, $$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=$$
$$=(a+b+c)^4+4(a+b+c)abc=625-40=585.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Transformation of homogeneous coordinates to Euclidean coordinates? I understand that two vectors
$$
v_{1} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}
\quad
\text{,}
\quad
v_{2} = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix}
$$
in homogeneous coordinates are equivalent if
$$
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
=
... | $v$ is a point at infinity. There is no equivalent point in euclidean space, but you can think it as the point of intersection of the lines of a parallel bundle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $\frac{cb}{1-c} +\frac{ac}{1-a} + \frac{ba}{1-b}$ How do I minimize the expression
$$f(a,b,c)=\frac{cb}{1-c} +\frac{ac}{1-a} + \frac{ba}{1-b}$$
subject to the constraint $abc=(1-a)(1-b)(1-c)$ with $a$, $b$, $c \in (0,1)$. Conceptually, the Lagrange multiplier procedure can be utilized. But, the re... | Let $\frac{a}{1-a}=\frac{y}{x},$ $\frac{b}{1-b}=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $\frac{c}{1-c}=\frac{x}{z}$, $a=\frac{y}{x+y}$ and by Nesbitt we obtain:
$$\sum_{cyc}\frac{ab}{1-b}=\sum_{cyc}\left(\frac{y}{x+y}\cdot\frac{z}{y}\right)=\sum_{cyc}\frac{z}{x+y}\geq\frac{3}{2}.$$
The equality occurs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $(4-x^2)y''+2y=0$ by series I know that there are other ways to solve this, but I want to solve the ODE $(4-x^2)y''+2y=0$ by series close to $0$, that is, I'm looking for solutions of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}.$
Substituing in the ODE, I got:
$$(8a_{2}+2a_{0})+(24a_{3}+2a_{1})x+\sum_{n=2}^{\i... | We have a differential equation that doesn’t have a constant coefficient for the second derivative.
$p\left( x \right) = 4-{x^2} \hspace{0.25in}p\left( 0 \right) = 4 \ne 0$
So, assume that $x_0$ is an ordinary point for this differential equation. We first need the solution and its derivatives,
$y\left( x \right) = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the re... | Use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx.~~~(1)$$
$$I=\int_{0}^{\pi} \frac{dt}{3+2 \cos t}~~~~(2)$$
Using (1), we get
$$I=\int_{0}^{\pi} \frac{dt}{3-2 \cos t}~~~~(3)$$
Add (2) and (3), then
$$2I=\int_{0}^{\pi} \frac{6\,dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{\sec^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why after depressing a cubic does it have different roots? The cubic $(x-1)(x-2)(x-2)=0$ will have the roots 1 and 2. Expanding will give $x^3-5x^2+8x-4=0$ which is in the form $ax^3+bx^2+cx+d=0$. Depressing it by substituting $x = t - \frac{b}{3a} = t+\frac{5}{3}$ will give $t^3-\frac{1}{3}t+\frac{2}{27}=0$. This is i... | As you said in the first paragraph, you get the depressed cubic by substituting
$x = t - \frac{b}{3a} = t+\frac53$.
Conversely, that means $t = x - \frac53$ in your example.
So if $x = 1$ is a root of the original cubic, it had better be true that
$t = x - \frac53 = -\frac23$ is a root of the depressed cubic.
And if $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1-3^x+5^x-7^x = 0 \Leftrightarrow x=0$ Problem :
Prove that equation $$1-3^x+5^x-7^x = 0$$
has unique root $x=0$.
Since $\lim_{x\to-\infty}(1-3^x+5^x-7^x) = 1$ and $\lim_{x\to\infty}(1-3^x+5^x-7^x) = -\infty$,
I tried to show $x \to 1-3^x+5^x-7^x$ is decreasing function, but it wasn't easy to determine it... | Note that $$1-3^x+5^x-7^x = 0 \iff 1+5^x = 3^x+7^x$$
Since the exponential functions involved are increasing functions we have the following inequalities.
$$x>0\implies 3^x+7^x > 1+5^x$$
and $$x<0\implies 3^x+7^x < 1+5^x$$
Thus the equality happens if and only if $x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve indefinite integral $ \int \frac{x^4}{\sqrt{x^2+4x+5}}dx $ I need to solve the next problem:
$$
\int \frac{x^4}{\sqrt{x^2+4x+5}}dx
$$
I know the correct answer is
$$
(\frac{x^3}{4}-\frac{7x^2}{6}+\frac{95x}{24}-\frac{145}{12})*\sqrt{x^2+4x+5}\space+\space\frac{35}{8}\ln{(x+2+\sqrt{x^2+4x+5})}+C
$$
still, I canno... | Hint:
$$\int\frac{x^4\mathrm dx}{\sqrt{x^2+4x+5}}=\int\frac{x^4\mathrm dx}{\sqrt{(x+2)^2+1}}=\int(\tan\theta -2)^{4}\sec\theta\mathrm d\theta$$
Set $\tan\theta=x+2$, $\sec^2\theta\mathrm d\theta=\mathrm dx$. Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Greatest common divisor of $(x+1)^{4n+3} + x^{2n}$ and $x^3-1$. I have to find the greatest common divisor of
$$(x+1)^{4n+3} + x^{2n}$$
and
$$x^3-1$$
I know I can express the second polynomial as:
$$x^3-1 = (x-1)(x^2+x+1)$$
So I would have to check if the first polynomial is divisible by $(x^3-1)$, $(x^2+x+1)$ or $(x-... | Only $x^2+x+1$ can be an answer because $x-1$ is not valid.
Modulo $x^2+x+1$ we obtain:
$$(x+1)^{4n+3}+x^{2n}\equiv x^{2n}-x^{8n+6}=x^{2n}(1-(x^3)^{2n+2})\equiv0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the general solution to $\csc \theta + \sec \theta = 1$
Find the general solution to $$\csc\theta + \sec\theta =1$$
This is how I solved.
We have,
\begin{align}
\csc\theta + \sec\theta &=1\\
\frac1{\sin\theta} + \frac1{\cos\theta}& =1\\
\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} &=1\\
(\sin\theta + \cos... | Hint
Again avoid squaring as it immediately introduces
When do we get extraneous roots?
$\sin x\cos x=\sin x+\cos x=y,$
$y=\sqrt2\cos(x-45^\circ)\implies-\sqrt2\le y\le?$(say)
$$y^2=1+2\sin x\cos x$$
Let $y^2-1=2y\iff y^2-2y-1=0$
$y=1\pm\sqrt2$
$\implies y=1-\sqrt2$
$\implies\cos(x-45^\circ)=\dfrac1{\sqrt2}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Properties of three terms of a geometric series I’m [still!] working on the equation in this question, namely
$$(b^2+2)^2=(a^2+2c^2)(bc-a). \tag{$\star$}$$
where $a,b,c$ are integers. Evidently, $(\star)$ implies
$$\frac{b^2+2}{bc-a} = \frac{a^2+2c^2}{b^2+2}, \tag{1}$$
which is to say that $\{bc-a,b^2+2,a^2+2c^2\}$ a... | If I have understood the additional constraints to your original problem correctly, the problem is now to find all integral solutions to
$$(b^2+2)^2=(a^2+2c^2)(bc-a), \tag{$\star$}$$
such that
$$\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2},$$
and moreover $x:=\frac{b^2+2}{bc-a}=\frac{a^2+2c^2}{b^2+2}$
is an integral solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the area of the smaller right triangle?
The two diagonal lines are parallel, and the area of the region between them is $42$
What is $A,$ the area of the smaller right triangle?
So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x +... |
So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$
Continuing your equation, we get:
$\dfrac{xy + 3x + 4y + 12 - xy}{2} = 42$
$xy + 3x + 4y + 12 - xy = 2 \cdot 42$
$3x + 4y + 12 = 84$
$\boxed{3x + 4y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| $$x^2-x+e^x(x-2)=0 \implies e^{-x}=\frac{x-2}{x(1-x)}$$ and the solution cen write using the generalized Lambert function (have a look at equation $(4)$).
By inspection or graphing, the positive root is close to $x=2$. To have a better value, expand the lhs as a Taylor series
$$y=x^2-x+e^x(x-2)=2+\left(3+e^2\right) (x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators. My question is: Do we have an splitting equation where we can produce fractions with odd denominators?
To split an Egyptian fraction to Egyptian fractions, we can use the splitting equation belo... | Just a thought:
If you limit the problem to just the examples you gave; that is, can any given odd Egyptian fraction be split into three different odd Egyptian fractions?
The examples you gave have two main algebraic forms:
$$\frac{1}{n}=\frac{1}{b}+\frac{1}{an}+\frac{1}{abn}\tag{1}$$
and
$$\frac{1}{n}=\frac{1}{ab}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$a+ b+ c+ d+ e = 12$, and $a^4b^3c^3de=4\times 6^6$, $a,b,c,d,e$ are positive. Find value of $4a + 3b + 3c + d + e$ Given that $$a+ b+ c+ d+ e = 12\quad \&\quad a^4b^3c^3de=4\times 6^6$$ where $a,b,c,d,e$ are positive integers, find the value of $4a + 3b + 3c + d + e$
I tried by trial and error by putting values but it... | Since nobody wants to type an answer:
as suggested in the comment, use AM-GM inequality on the $12$ numbers $a/4, a/4, a/4, a/4, b/3, b/3, b/3, c/3, c/3, c/3, d, e$.
This gives:
$$\frac{a + b + c + d + e}{12} \geq \sqrt[12]{\frac{a^4b^3c^3de}{4\times 6^6}},$$ where equality holds if and only if all the $12$ numbers are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The parametric ratio $\frac{x}{y}$ with known $x+y$ and $x\cdot y$ $x$ and $y$ are in fact $\lambda_1$ and $\lambda_2$, the bigger and smaller eigenvalues of a parametric matrix $A'A$, and $t$ is a very small constant. I have that
$$
\begin{split}
x+y &= 1+ \frac{t^2}{2} \\
xy &= \frac{t^2}{4}
\end{split}
$$
How would... | So $x=\frac{1}{4}(t^2+2+\sqrt{t^4+4})$ and $y=\frac{1}{4}(t^2+2-\sqrt{t^4+4})$, $\dfrac{x}{y}=\dfrac{(t^2+2+\sqrt{t^4+4})^2}{4t^2}$ say $t>0$.
To prove that $$\dfrac{(t^2+2+\sqrt{t^4+4})^2}{4t}\ge 1\Leftrightarrow t^4+4+4t^2+t^4+4+2(t^2+2)(\sqrt{t^4+4})-4t\ge 0$$ but $t^2-t+1\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.