Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\... | Inequality is equivalent with $$\frac{(x-1)(x-6)}{(x+1)(x-4)}<0.$$ Function $f(x)=\frac{(x-1)(x-6)}{(x+1)(x-4)}$ has sign $+$ on intervals $(-\infty,-1)$, $(1,4)$, $(6,\infty)$ and sign $-$ on intervals $(-1,1)$ and $(4,6)$.
Therefore, only integer solutions that satisfy inequality are $0$ and $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Using substitution while using taylor expansion I am trying to prove to myself why can substitution be used while using taylor expansion. for example: the taylor expansion of $e^{(x-3)^2}$ around $a=3$ and order of $6$ can be done by first evaluating $e^x$ and than plugging in $(x-3)^2$ getting $P_{6}(x)=1+(x-3)^2+\fra... | This problem is related to series composition and what you did is almost perfectly correct.
Make a change of variable $(x-3)^2=y$ and use the fact that, around $y=0$ $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)\tag 1$$ Now, you just need to replace $y$ by $(x-3)^2$ to get $$e^{(x-3)^2}=1+(x-3)^2+\frac{1}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the integral $\int _{\ln 2} ^{\ln 3} \frac { e ^x} {e ^{4x}-1} \,d x$ How do I compute $$\int \limits _{\ln 2} ^{\ln 3} \frac {e ^x} {e ^{4x}-1} \, d x $$
? Thank you so much for your answer.
| $$\int_{\ln(2)}^{\ln(3)}\frac{e^x}{e^{4x}-1}\space\text{d}x=$$
Substitute $u=e^x$ and $\text{d}u=e^x\space\text{d}x$.
This gives a new lower bound $u=e^{\ln(2)}=2$ and upper bound $u=e^{\ln(3)}=3$:
$$\int_{2}^{3}\frac{1}{u^4-1}\space\text{d}u=$$
Use partial fractions:
$$\int_{2}^{3}\left[\frac{1}{4(u-1)}-\frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Anti-associativity of product of sum of squares $\newcommand{\P}{\mathbb{P}}$$\newcommand{\Z}{\mathbb{Z}}$ Let $\P$ be the set of prime numbers congruent to $1 \pmod 4$. I know that for every $p \in \P$ there's a unique couple $(a^2,b^2)\in \Z^2$ such that $p= a^2+b^2$ and $a < b \in \Z$. Moreover, if we take $p,q \in ... | I address a differently phrased question, but it should provide the answer to yours.
Suppose the positive integer $z$ is a product of primes $p_j$ all congruent to $1$ modulo $4$. Write $z = a^ 2 + b^2$, with $a, b$ positive, $a < b$. Then $$\tag{dec}z = (a + i b) (a - i b)$$ is the product of two conjugate Gaussian in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trying to show that any group of order four is either cyclic or isomorphic to V I know the question has already been asked. But I have trouble with the answer.
Having a non-cyclic group $\,G=\{1,a,b,c\}\,$, how can I show that $ab=c$?
In my attempt, I assume that $ab=1$, and then $c^2=1$. And I see no problem with th... | We surely have $a^2\ne a$. So $a^2=1$, $a^2=b$ or $a^2=c$. If $a^2\ne1$, it's not restrictive to assume $a^2=b$. So we can start building the Cayley table:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & & \\
b & b & & & \\
c & c & & &
\end{array}
$$
We can see that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$, if $\sin(\alpha+\beta) = 1$ and $\sin(\alpha-\beta) = \frac 12$ If $\sin(\alpha+\beta) = 1$, $\sin(\alpha-\beta) = \frac 12$ , where $\alpha$, $\beta$ belongs to $[0,\frac \pi2],$ then what is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$?
What should be my approach to the q... | Since
\begin{align}
0\leq \alpha \leq \frac{\pi}{2}, ~~~~~~~~~~~~~~0\leq \beta \leq \frac{\pi}{2},
\end{align}
we have
\begin{align}
0\leq \alpha+\beta \leq \pi,~~~~~~~~~~~~-\frac{\pi}{2}\leq \alpha-\beta \leq \frac{\pi}{2}.
\end{align}
Thus, from the equations given, we find that
\begin{align}
\sin(\alpha+\beta)&=1\R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How do you find one-sided limits *algebraically*? Find $$\lim_{x\to\ -0.5^-}\sqrt{\frac{x+2}{x+1}}$$
Sorry, I have no idea where to start. I know how to find regular limits algebraically, but not one-sided.
Thanks
| *
*$$\lim_{x \to -0.5} f(x) = \lim_{x \to -0.5^{-1}} f(x)$$
if $f$ is continuous at $x=-0.5$
Is $$\sqrt{\frac{x+2}{x+1}}$$ continuous at $x=-0.5$?
What we need to check is if $$\lim_{x \to -0.5} \sqrt{x} = \lim_{x \to -0.5^{-1}} \sqrt{x}$$
and if $$\lim_{x \to 0-.5} \frac{x+2}{x+1} = \lim_{x \to -0.5^{-1}} \frac{x+2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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solving trigonometry equation $20\cot\theta + 15\cot\theta\operatorname{cosec}\theta - 4\operatorname{cosec}\theta = 3(1 + \cot^2 \theta) ?$ how do you solve for $0 < \theta < 360$:
$$20\cot\theta + 15\cot\theta\operatorname{cosec}\theta - 4\operatorname{cosec}\theta = 3(1 + \cot^2 \theta) ?$$
I tried turning the $(1 +... | Let try to convert the equation to $\sin$ and $\cos$.
$$20\frac{\cos \theta}{\sin\theta} + 15 \frac{\cos \theta}{\sin\theta}\frac{1}{\sin\theta} - 4\frac{1}{\sin\theta} = 3\frac{1}{\sin^2\theta}$$
$$20\cos\theta\sin\theta + 15\cos\theta-4\sin\theta = 3$$
$$(4\sin\theta+3)(5\cos\theta-1) = 0$$
Now, you can solve the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize. Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,
First Solution :
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$
$=\lim\limits... | Setting $x^6=2y$ and using $\cos2y=1-2\sin^2y$ to find $$\lim_{y\to0^+}\dfrac{1-\cos2y}{(2y)^2}=\dfrac24\left(\lim_{y\to0^+}\dfrac{\sin y}y\right)^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$ I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.
Thanks!
| $$\begin{align}
\tan 70-\tan 20-2\tan 40 &= \left( \tan { 70-\tan { 40 } } \right) -\left( \tan { 40+\tan { 20 } } \right) \\
&=\frac { \sin { 70 } }{ \cos { 70 } } -\frac { \sin { 40 } }{ \cos { 40 } } -\left( \frac { \sin { 40 } }{ \cos { 40 } } +\frac { \sin { 20 } }{ \cos { 20 } } \right) \\
&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$
| It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^{12}]\frac{1}{(1-x^2)^5}
&=[x^{12}]\sum_{k=0}^\infty \binom{-5}{k}(-x^2)^{k}\tag{1}\\
&=[x^{12}]\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$
Prove that $$\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$$
I've tried multiple things that didn't work.
Maybe this would help
$$\sum_{k=0}^n \binom{3n-k}{2n}=\sum_{k=0}^n \binom{3n-(n-k)}{2n}=\sum_{k=0}^n \binom{2n+k}{2n}$$
| Your identity is a special case of the more general identity $$S(m,n) = \sum_{k=0}^n \binom{m+k}{m} = \binom{m+n+1}{m+1},$$ which you can prove by induction on $n$: note $S(m,0) = 1 = \binom{m+1}{m+1}$. Then observe $$\begin{align*} S(m,n+1) &= S(m,n) + \binom{m+n+1}{m} \\
&= \binom{m+n+1}{m+1} + \binom{m+n+1}{m} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
... | The fraction becomes
$$
\frac{A}{(x+1)} + \frac{B}{(x+1)^2}=
\frac{A(x+1)+B}{(x+1)^2}=
\frac{Ax+(A+B)}{(x+1)^2}
$$
so $A=0$ and $B=1$. Indeed
$$
\frac{1}{(x+1)^2}
$$
is already in “partial fractions” form. And
$$
\int\frac{1}{(x+1)^2}\,dx=-\frac{1}{x+1}+C
$$
It would be different if you started with
$$
\frac{x}{(x+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve:
$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$
I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?
| $$\int \frac { x+1 }{ \sqrt { 1-x^{ 2 } } } dx=-\left( \int \frac { -x }{ \sqrt { 1-x^{ 2 } } } dx-\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } } \right) =\\ =-\frac { 1 }{ 2 } \int \frac { d\left( 1-{ x }^{ 2 } \right) }{ \sqrt { 1-x^{ 2 } } } +\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } =-\sqrt { 1-x^{ 2 } } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 3
} |
Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do nex... | Hint: if we multiply by $\sin^{-6}\left(x\right)
$ we get $$I=\int\frac{\sin^{4}\left(x\right)}{\sin^{4}\left(x\right)+\cos^{4}\left(x\right)}dx=\int\frac{\csc^{2}\left(x\right)}{\csc^{2}\left(x\right)+\cot^{4}\left(x\right)\csc^{2}\left(x\right)}dx
$$ $$=\int\frac{\csc^{2}\left(x\right)}{\cot^{6}\left(x\right)+\cot^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$
Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$
I think I'm supposed to use the following chain of inequalities
$$|e^z -1|\leq e^{|z|}-1 \leq |z|e^{|z|}$$
But every time I try to solve it I get stuc... | Start with the series expansion
$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$
and use the triangle and reverse triangle inequalities to get the result.
By the triangle inequality and the condition $\lvert z\rvert \le 1$, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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If $5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $, then $x$ equals $-4$ For the equation
$$5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $$
$x$ is equal to $-4$, but I'm not sure why.
I've taken the right side of the equation ${ \left( \frac { 1 }{ 5 } \right) }^{ \frac { 1 }{ x } }$ and converted it to ... | Since $125 = 5^3$, we have $$5 \cdot 5^{\frac{3}{x}} = \frac{1}{5^{\frac{1}{x}}} \Rightarrow 5^{1 + \frac{3}{x}} = 5^{-\frac{1}{x}}$$
Equating powers gives $1 + \frac{3}{x} = -\frac{1}{x} \Rightarrow x = -4$.
Some explanations:
We have $a^b \cdot a^c = a^{b+c}$, this can be reasoned as such: $$a^b \cdot a^c = \und... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\f... | A tricky way to verify the monotonicity is by applying AM-GM inequality as follows:
\begin{align*}
a_n & = \left(1 + \frac{1}{n}\right)^n \\
& = 1\times \left(1 + \frac{1}{n}\right) \times \cdots \times \left(1 + \frac{1}{n}\right)\\
& \leq \left[\frac{1 + n\left(1 + \frac{1}{n}\right)}{n + 1}\right]^{n + 1} \\
& = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$
$\nabla f_x$ $=$ $2xy^{2}z^{2}$,
$\nabla f_y$ $=$ $2yx^{2}z^{2}$,
$\nabla f_z$ $=$ $2zx^{2}y^{2}$
$\nabla g_x$ $=$ $2x$,
$\nabla g_y$ $=$ $2y$,
$\nabla g_z$ $=$ $2z$
setting the sets equal to each other... | You didn't consider the case $x = y = \lambda = 0, z = 1$ which satisfies the nabla equations, and gives the minimum value for $f$ of $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{... | For the most general case,
$\sum\limits_{n=1}^\infty\dfrac{1}{n\binom{kn}{n}}$
$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+1)}{n\Gamma(kn+1)}$
$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n)\Gamma((k-1)n+1)}{\Gamma(kn+1)}$
$=\sum\limits_{n=0}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+k)}{\Gamma(kn+k+1)}$
$=~_3\Psi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.... | Hint $\,\ \sqrt a + \sqrt b\, = \sqrt{(\sqrt a + \sqrt b)^2} = \sqrt{a+b +2{\sqrt{ab}}}.\ $ Here $\ a+b=12,\ \color{}{\sqrt{ab}} = 4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Chemical reaction modeled by a differential equation I am badly stuck on the following question. Thus, I am asking for some help :)
Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as
$$A + B \xrightarrow{k} X$$
If $2 \, \rm{g}$ of $A$ and $... | Without checking if the two forms are both correct :
$$\begin{cases}
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct \qquad [1]\\
\ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c \qquad [2]
\end{cases}$$
The relationship between $a,b,c,k$ can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to express $\cos(20^\circ)$ with radicals of rational numbers? In showing that the trisection of an angle with ruler and compass is not possible in general one shows that $\cos(20^\circ)$ cannot be constructed (thus the angle $60^\circ$ cannot be trisected) by determining its minimal polynomial, which is $8x^3-6x-1... | $$
\cos (20^\circ) = \cos (\pi/9) = -\frac12 (-1)^{8/9} \left(1+(-1)^{2/9}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the real and imaginary parts of an equation
Find the real and imaginary parts of $\frac{1}{3z+2}$
So I have expanded it out to get $\frac{1}{3x+3iy+2}$
Thus giving $Re(\frac{1}{3z+2})=\frac{1}{3x+2}$ and $Im(\frac{1}{3z+2})=\frac{1}{3y}$
However in my answer book it says: $Re(\frac{1}{3z+2})=\frac{3x+2}{(3x+2)^... | you find it as the form $z=a+ib$ so
$$z=\frac { 1 }{ 3x+3iy+2 } =\frac { 3x+2-3iy }{ \left( 3x+2+3iy \right) \left( 3x+2-3iy \right) } =\frac { 3x+2 }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } +i\frac { -3y }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational... | Other way
$$x^2\left(x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}\right)=0$$
$x=0$ is not a solution of this equation , therefore
$$x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}=0$$
we have
$$\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0$$
set $t=x+\frac{1}{x}$, thus
$$t^2-4t+4=(t-2)^2=0$$
Hence
$$t=x+\frac{1}{x}=2$$
i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My atte... | This is more easily solved using pole-polar relationships, I think. The chord of contact the the two tangents is the polar of their intersection point, namely $$4(0\cdot x)-(4\cdot y)=36,$$ i.e., the horizontal line $y=-9$. Substitute this value into the equation of the hyperbola and solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality Challenge with $a+b+c=3$ I tried to prove this inequality, but I failed. It seemed that if one uses Cauchy-Schwarz inequality on the right hand side, it may expand RHS's value larger than LHS's. Problem is presented as follows:
If $a,b,c\ge0$ and $a+b+c=3$, prove that $$2(ab+bc+ca)-3abc\ge a\cdot \sqrt{\frac... | After using $\sqrt{\frac{a^2+b^2}{2}}\leq\frac{3a^2+2ab+3b^2}{4(a+b)}$,
which is just $(a-b)^4\geq0$, you'll get something obvious.
Indeed, it remains to prove that
$$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\sum\limits_{cyc}\frac{c(3a^2+2ab+3b^2)}{4(a+b)}$$ or
$$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\frac{3}{2}(ab+ac+bc)-abc\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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A Problem with the Generating Function of Fibonacci. So basically I want to find the closed form of $G_n = \sum_{k = 1}^n \binom{n+k - 1}{2k-1}$.
After checking for $n = 1,2,3,4$ the values are $1, 3, 8, 21$ respectively. I claim that it is $F_{2n}$ where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$. My idea was to... | Hint: There is just a small calculation error when transforming $\binom{n+k-1}{2k-1}$.
We obtain
\begin{align*}
\sum_{n=0}^\infty G_nx^n&=\sum_{n=0}^\infty\sum_{k=1}^n\binom{n+k-1}{2k-1}x^n\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\binom{n+k-1}{2k-1}x^n\tag{1}\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty\binom{n+2k-1}{2k-1}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:... | Clearly, $\tan\theta\tan2\theta\tan3\theta$ is undefined if $\theta\equiv\dfrac\pi2,\dfrac\pi4,\dfrac\pi6\pmod\pi$
Otherwise,
$$\tan3\theta=-\dfrac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=-\tan(\theta+2\theta)=\tan(-3\theta)$$
$$\implies3\theta=n\pi-3\theta\iff\theta=\dfrac{n\pi}6$$ where $n$ is any integer
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos... | Here is another take with minimum number of applications of L'Hospital's Rule. We have
\begin{align}
L &= \lim_{x \to 0}\dfrac{\cos x - 1 + \dfrac{x^{2}}{2}}{x^{4}}\notag\\
&= \lim_{t \to 0}\dfrac{\cos 2t - 1 + 2t^{2}}{16t^{4}}\text{ (putting }x = 2t)\notag\\
&= \frac{1}{16}\lim_{t \to 0}\dfrac{2t^{2} - 2\sin^{2}t}{t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathca... | An overkill is to exploit that for any sufficiently large $x>0$ we have that
$$ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \frac{1}{2}+\sqrt{x+\frac{1}{4}}$$
since both terms are a positive solution of $y^2=x+y$. On the other hand
$$ \sqrt{x+\sqrt{x}} \geq \frac{1}{2}+\sqrt{x-\frac{1}{4}} $$
is a trivial inequality b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0$ has a solution modulo $p$ I need to solve this question:
Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0 $ has a solution modulo $ p $.
My approach was: I checked for $p = 2$ and there is no solution.
Now if $p \neq 2 $ so the equatio... | $x^2+6x+15 \equiv 0\pmod{p}\Rightarrow(x+3)^2+6\equiv0\pmod{p}\Rightarrow (x+3)^2\equiv-6\pmod{p}$.
In other words, $y^2\equiv -6\pmod{p}$ or to rephrase the argument, $\left(\frac{-6}{p}\right)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
| It is easy to check that if $a,b$ are two positive real numbers, then $\frac{a^a b^b}{(a+b)^{a+b}}$ is the maximum value of the function $f(x)=x^a(1-x)^b$ over the interval $[0,1]$, attained at $x=\frac{a}{a+b}$. By Euler's beta function we have:
$$ \int_{0}^{1}f(x)\,dx = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
asymptotics of the Gamma function and remainder I have found the following asymptotic formula in a book:
$$\lim_{\vert y\vert\rightarrow\infty}\vert \Gamma(x+iy) \vert e^{\frac{\pi}{2}\vert y\vert}\vert y \vert^{\frac{1}{2}-x}= \sqrt{2\pi}. $$
I would like to know if there are even sharper estimates, in particular is i... | There is an exact solution for $x=\dfrac 12$ as shown in this answer :
$$\tag{1}\left|\Gamma\left(\frac 12+iy\right)\right|=\sqrt{\frac{\pi}{\cosh\left(\pi y\right)}}$$
More general asymptotic results were obtained here like your :
$$\tag{2}\left|\Gamma\left(x+iy\right)\right|=\sqrt{2\pi}\,e^{-\pi|y|/2}\,|y|^{x-1/2}(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 ... | $x\equiv 32 (\mod 81) \ and \ x\equiv 59 (\mod 64) \Rightarrow x=81a+32 \ and \ x=64b+59 \\ $
then $81a+32=64b+59 \\
81a-64b=27 \ ...I \\$
with the euclidean algorithm we get a solution
$ (a,b)=(-405,-513) \\
81(-405)-64(-513)=27 \ ...II \\ $
subtracting II from I we get $ 81(a+405)-64(b+513)=0 \Leftrightarrow 81(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}... | I hope this would be useful for you.
So your trying to evaluate $\displaystyle\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}$. Consider first $\displaystyle\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}$, after developing the product properly. Now you have to make sense of $$\lim_{k\to\infty}\sum_{n=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 0
} |
How many k-digit numbers are both divisible by 3 and include the digit 3? At first I thought I could simply count how many (k-1)-digit numbers are divisible by 3 and multiply by k, accounting for the different possible placements of the final 3. But It seems that method includes duplicates.
Any direction, such as Incl... | We get from first principles the generating function
$$f(z) = (z+z^2+A+z+z^2+1+z+z^2+1) \\ \times
\prod_{q=1}^{k-1} (1+z+z^2+A+z+z^2+1+z+z^2+1).$$
This is
$$f(z) = (A + 2 + 3z + 3z^2)
\prod_{q=1}^{k-1} (A + 3 + 3z + 3z^2).$$
We obtain for the desired count
$$\frac{1}{3}
\sum_{q=0}^2
\left. \left(f(z) - \left. f(z)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}... | Your first proposition should be corrected by noting that when you multiply numerator and denominator by the conjugate of $x^2+2ixy-y^2$, the denominator becomes $(x+iy)^2(x-iy)^2=(x^2+y^2)^2$ and not $x^2+y^2$. Moreover, there's no $i$ in the imaginary part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Analytic Geometry Question: Computing the equation of a circle given 2 points and center I want to derive the equation of a circle passing through the intersection of the two circles. My method gave me a nonsensical solution. I would love if someone could tell me how my method went wrong, although since my solution is ... | Less elegant than Mick's answer.
Considering $$x^2+y^2=100 \tag 1$$ $$(x-11)^2+(y-4)^2=9 \tag 2$$ Subtract $(1)$ from $(2)$ to get $$-22 x-8 y+228=0\implies y=\frac{1}{4} (114-11 x)\tag 3$$ Replace in $(1)$ to get, after simplification $$137 x^2-2508 x+11396=0\tag 4$$ One of the solutions corresponds to $$x=\frac{2}{13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generalizing Variant Proof of Basel Problem Recently I have been thinking a lot about variations of the Basel Problem, and methods to solve them. Here I found the following solution to the Basel Problem by Alfredo Z. (I include the entire answer due to its brevity)
Define the following series for $ x > 0 $
$$\frac{\si... | To start, the "factoring" step is known as the Weierstrass factorization theorem, which asserts that you can express some functions as products of their factors.
From here, take note that you had
$$\frac{\sin(x)}x=\dots\left(1+\frac{x^2}{2^2\pi^2}\right)\left(1+\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{1^2\pi^2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ if $x^2+y^2=1$. Show that $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$. It is given that $x^2+y^2=1$. $x,y$ are positive real numbers.
[From a Regional Mathematical Olympiad, 2013 in India]
| $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{1+x^2+1+y^2}{(1+x^2)(1+y^2)}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{3}{2+x^2y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{3+3xy+2+2x^2y^2}{2+2xy+x^2y^2+x^3y^3}\ge \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A quadratic equation Find all values of a for which the quadratic $$\cos^2x - (a^2 + a + 5) |\cos x| + (a^3 + 3a^2 + 2a + 6) = 0$$ has real solution(s)
A. $a=-3$, B. $a=-2$, C. $a=-1$, D. $a=0$
I have solved this question by putting the values of the given options :
Let $\cos x = t$
then it will look like $$t^2 - (a^2... | Doing the same as Robert Z in his answer, considering $$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=0$$ the roots of the quadratic are $$t_1=3+a\qquad , \qquad t_2=2+a^2$$ since, after simplification $$\Delta=\left(a^2-a-1\right)^2$$ Obviousle $t_2$ must be discarded and $t_1$ would be acceptable if $3+a \leq 1$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical Proof - Cube Roots of Perfect Cubes Using Vedic Mathematics
Please refer this site. A method is provided for finding cube
roots of perfect cubes.
As per the method explained, suppose we are finding cube root of
$157464$
First we write as $157,\quad 464$
Last digit of $464$ is $4.$ Hence RHS=$4$
$157-... | Lemma. Then function $x\mapsto x^3:\mathbb Z\to\mathbb Z$ induces a bijection $\mathbb Z/10\mathbb Z\to \mathbb Z/10\mathbb Z$.
Proof. It's enough to show that $x^3\equiv y^3\pmod{10}$ implies $x\equiv y\pmod{10}$.
If $x^3\equiv y^3\pmod{10}$, then $x\equiv x^3\equiv y^3\equiv y\pmod{2}
$ and $xy^2\equiv x^5y^2\equiv (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using a derivative to find fourier coefficients The previous question was:
Find the fourier coefficients of $f(x)=x^2+1$
To which I found $a_0=\frac{\pi^2}{3}+1$ and $a_n=\frac{2}{n^2}(-1)^n$ (I am unsure on $b_n$ I get $b_n=(\frac{-2}{n\pi}-\frac{2\pi}{n}+\frac{4}{n^3\pi})(-1)^n$ but this is more method than correct... | Firstly: $f(x)$ is an even function, so you expect all $b_n = 0$.
Now for $g(x)$. You are right to note that $g(x) = \frac{1}{2} f'(x)$, but why are you integrating? Futhermore: the last equation: LHS is a function of $x$ and RHS is just a constant.
To solve the problem write $f(x)$ as a sum of cosines (so in in it's F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Trigonometry proof based on if $\cos A+$.... If $$\cos A+\cos B+\cos C=\sin A.+\sin B+\sin C=0$$ then prove that: $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$
My Attempt;
Here,
$$\cos A+\cos B+\cos C=0$$
$$\cos A+\cos B=-\cos C$$
squaring on both sides,
$$\cos^2A+2\cos A.\cos B+\cos^2B=\cos^2C$$
Again,
$$\sin A+\sin B+\sin... | Maybe you could try a complex approach (which is simpler ;-))
Let $z,u,v$ be three complex numbers:
$z=\cos(A)+i\sin(A)$, $u=\cos(B)+i\sin(B)$, $v=\cos(C)+i\sin(C)$.
then by hypothesis $z+u+w=0$. Hence
$$0=(z+u+v)^3=z^3+u^3+v^3+3uv(u+v)+3uz(u+z)+3zv(z+v)+6zuv\\=
z^3+u^3+v^3+3uv(-z)+3uz(-v)+3zv(-u)+6zuv=z^3+u^3+v^3-3zu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show t... | Because $(a+2)^3+(b+2)^3+(c+2)^3\geq\frac{1}{9}(a+2+b+2+c+2)^3=81$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Is there a simple function that generates the series; $1,1,2,1,1,2,1,1,2...$ or $-1,-1,1,-1,-1,1...$ I'm thinking about this question in the sense that we often have a term $(-1)^n$ for an integer $n$, so that we get a sequence $1,-1,1,-1...$ but I'm trying to find an expression that only gives every 3rd term as positi... | Let $\omega \neq 1$ be a cubic root of unity. We have that $1 + \omega + \omega^2 = 0$, i.e. $\omega + \omega^2 = -1$. Also, $w^{-1} = \omega^2$ and $\omega^3 = 1$.
Put $a_n = \omega^n + \omega^{-n}$. We get the sequences
\begin{eqnarray*}
a_0 &=& \omega^{0} + \omega^{0} = 2 \\
a_1 &=& \omega^{1} + \omega^{-1} = \omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 18,
"answer_id": 0
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b =... | A slightly different (but longer) method:
Let roots of $x^2-x-1$ be $c$ and $d$
Using Vieta's formulas, we get
$$c+d=1\tag{1}$$
$$cd=-1\tag{2}$$
Using Factor theorem, we get
$$ac^{17}+bc^{16}=-1\tag{3}$$
$$ad^{17}+bd^{16}=-1\tag{4}$$
Multiplying $(3)$ with $d^{16}$ and using $(2)$, we get
$$ac+b=-d^{16}\tag{5}$$
Mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan ... | If $\cot3A=\cot3x,3x=n180^\circ+3A$ where $n$ is any integer
$x=n60^\circ+A$ where $n\equiv0,1,2\pmod3$
$$\cot3A=\cot3x=\dfrac1{\tan3x}=\dfrac{1-3\tan^2x}{3\tan x-\tan^3x}=\dfrac{\cot^3x-3\cot x}{3\cot^2x-1}$$
$$\iff\cot^3x-3\cot3A\cot^2x-3\cot x+\cot3A=0\ \ \ \ (1)$$
If $\cot x_1,\cot x_2,\cot x_3$ are roots of $(1),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way... | To find $\sum_{k=1}^{n}(k^4)$ you may follow this process.
Consider the identity, $(x+1)^5-x^5=5x^4+10x^3+10x^2+5x+1$.
Putting $x=1,2,3,...,(n-1),n$ successively, we get,
$2^5\space-\space 1^5=5\cdot1^4+10 \cdot1^3+10\cdot1^2+5\cdot1+1$
$3^5\space-\space 2^5=5\cdot2^4+10 \cdot2^3+10\cdot2^2+5\cdot2+1$
$\cdot\space\spac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.
Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.
OBS. $O_2$ is the zero matrix
Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $... | Let's rewrite the - correct - identity you obtained as follows:
\begin{align*}& A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2\\
\iff& t(A-B)=A^2 -B^2- (\det(A) - \det(B))I_2\end{align*}
Squaring both sides of your original identity gives
$$(A^2 -B^2)^2=t^2\underbrace{(A-B)^2}_{=0} + 2(\det(A) - \det(B))t(A-B) + (\det(A) - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show that $9|2^{2n+1}+2^{4n+1}-4$
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
1-Using induction
2-Don't use induction
I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
| It's true for all non-negative integers $n$.
Without induction:
$9\mid 2^{2n+1}+2^{4n+1}-4$ is equivalent to $$9\mid 2\cdot 2^{2n+1}+\left(2^{2n+1}\right)^2-8$$
$$=\left(2^{2n+1}+1\right)^2-9$$
I.e. $9=3^2\mid \left(2^{2n+1}+1\right)^2$, i.e. $3\mid 2^{2n+1}+1$, which is true,
because $2^{2n+1}\equiv (-1)^{2n+1}\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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A really interesting logarithm integral I need to prove that
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x =
\frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$
My approach
We know that
$$\int^1_0 \frac{\log x}{x^2-2\cos(\theta)x+1}\mathrm{d}x =- \frac{\mathrm{cl}_2(\theta)}{\sin(\theta)}$$
Let $\theta=\pi/3$
$$\int^1_0 \f... | Hint. One may write
$$
\int^1_0 \frac{\log x}{x^2-x+1}dx=\int^1_0 \frac{(1+x)\log x}{1+x^3}dx=\left.\partial_s\int^1_0 \frac{(1+x)\:x^s}{1+x^3}dx\right|_{s=0}
$$ then after a change of variable one may use
$$
\int^1_0 \frac{t^s}{1+t}dt=\frac12\psi\left(\frac{s}2+1\right)-\frac12\psi\left(\frac{s}2+\frac12\right), \quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| As $P(0)=1\cdot3\cdot5\cdot7\cdot9,$ and $9\nmid P(10),7\nmid P(8)$
The highest divisors of $P(n)$ must divide $3\cdot5$
$P(n)\equiv(n+1)n^2(n+2)^2\pmod3$
Clearly, $3\mid n(n+1)(n+2)$
$P(n)\equiv(n-2)(n-1)n(n+1)(n+2)\pmod5$
Clearly, $5\mid(n-2)(n-1)n(n+1)(n+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the ... | Since the expression is symmetric in $a$ and $b$, it is not restrictive to assume $b=a+1$, so
$$
a^2+b^2+a^2b^2=
a^2+(a+1)^2+a^2(a+1)^2=
a^4+2a^3+3a^2+2a+1
$$
If you don't see an easy factorization, note that for $a=0$ the statement is clear; for $a\ne0$ we can write
$$
a^4+2a^3+3a^2+2a+1
=
a^2\left(a^2+\frac{1}{a^2}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Show that $\sum^{6}_{i=1} a_{i}=\frac{15}{2}$ and $ \sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4} \implies \prod_{i=1}^{6} a_{i} \leq \frac{5}{2}$ Let $a_{i}$, $1 \leq i \leq 6,$ be real numbers such that
$\displaystyle\hspace{1.2 in}\sum^{6}_{i=1} a_{i}=\frac{15}{2}\;\;$ and $\;\;\displaystyle\sum^{6}_{i=1} a^{2}_{i}=\frac{4... | Here is a Calculus of Variations approach. Perhaps not terribly elegant, but it works.
$$
\sum_{j=1}^6a_j=\frac{15}2\implies\sum_{j=1}^6\delta a_j=0\tag{1}
$$
$$
\sum_{j=1}^6a_j^2=\frac{45}4\implies\sum_{j=1}^6a_j\delta a_j=0\tag{2}
$$
To maximize $\prod\limits_{j=1}^6a_j$, we want
$$
\sum_{j=1}^6\frac{\delta a_j}{a_j}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 0
} |
Show product of $(r^5-r+2)$ over the 5 roots of $x^5+x^3+1=0$ is $1$ Let $(a,b,c,d,e)$ be the five roots of $x^5+x^3+1=0$, and let $g(x) = x^5-x+2$.
Prove
$$
g(a)g(b)g(c)g(d)g(e) = 1
$$
preferably without pages of messy algebra.
| We'll do some computations. As noted in G. Sassatelli's comment, for roots of our polynomial,
$$x^5-x+2 - (x^5+x^3+1)=-x^3-x+1.$$
Use polynomial division to get
$$x^2+1 + x^2(x^3+x-1)=x^5+x^3+1=0$$
and
$$-(x^3+x-1)=(1+x^2)/x^2,$$
since none of the roots are zero.
Realize that $abcde=-1$, so our product is equivalent to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How do I prove thcd It makes me think of means... The RHS Is like a geometric mean but the "divided by 4" annoyes me. The LHS is probably an arithmetic mean... Can they be combined?
| By Cauchy-Schwarz,
$$\sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}}
=\sqrt{a^2 + b^2 + c^2 + d^2}\cdot \sqrt{4\cdot\frac{1}{4^2}}\geq \frac{a+b+c+d}{4}\\\geq \sqrt[3]{\frac{abc + abd + acd + bcd}{4}}$$
where in the last step we use
Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
| Hint: the left hand side equals $$(1+n)^3=(1+n)^2+n(1+n)^2$$
and $$\sum_{k=0}^nk=\frac{n(n+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge... | Hint: The given inequality is equivalent to
$$\sum_{cyc} \frac1{1-a} - \frac32\sum_{cyc}a^2 \ge 4 $$
which follows from noting that for $x \in (0, 1)$,
$$f(x) = \frac1{1-x}-\frac32x^2-\frac43 -\frac54\left(x-\frac13\right) = \frac{(1+2x)(1-3x)^2}{12(1-x)}\geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Showing $\frac{z}{1+z}+\frac{2z^2}{1+z^2}+...+\frac{2^{k}z^{2^k}}{1+z^{2k}}+...=\frac{z}{1-z}$
Prove that for $\left\lvert z \right\rvert<1$, $\dfrac{z}{1+z}+\dfrac{2z^2}{1+z^2}+...+\dfrac{2^{k}z^{2^k}}{1+z^{2k}}+...=\dfrac{z}{1-z}$. Also, justify any change in the order of summation.
This is a exercise from my textb... | Note that for each $k \ge 1$,
$$\frac{2^k z^{2^k}}{1+z^{2^k}} = \frac{2^k z^{2^k}(1 - z^{2^k})}{(1 + z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}(1+z^{2^k} - 2z^{2^k})}{(1+z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}}{1-z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}. $$
Further, since $\lvert z \rvert < 1$,
$$\lim_{k\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solution of $\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$
Find the solution of $$\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$$
I was able to bring it down to $$\frac{d(x^2-y^2)}{\sqrt{1+x^2-y^2}}=2\left(\frac{x.d(y/x)}{\sqrt{1-(y/x)^2}}\right)$$
Any he... | We can write the equation as follows:
$$ \frac{xdx-ydy}{x^2-y^2}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\frac{xdy-ydx}{x^2-y^2}$$
It's
$$ \frac{xdx-ydy}{x^2-y^2}=\frac{d\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}$$
and
$$\frac{xdy-ydx}{x^2-y^2}=\frac{1}{2}d\ln\frac{x+y}{x-y}$$
$\,\displaystyle \sqrt{x^2-y^2}\,$ and $\,\displaystyle \ln\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric Integral - Quotient Rule: $\int \frac{x\cos(x)}{(x+\cos(x))^2} dx$? $$\int \frac{x\cos(x)}{(x+\cos(x))^2} dx$$
I suspect that the integral can be solved by thinking of it as the derivative of a quotient with a denominator of $x + \cos(x)$. How else could this integral be evaluated?
| We have $$I=\int\frac{x\cos\left(x\right)}{\left(x+\cos\left(x\right)\right)^{2}}dx=\int\frac{1-\sin\left(x\right)}{\left(x+\cos\left(x\right)\right)^{2}}\frac{x\cos\left(x\right)}{1-\sin\left(x\right)}dx
$$ $$\stackrel{IBP}{=}-\frac{1}{x+\cos\left(x\right)}\frac{x\cos\left(x\right)}{1-\sin\left(x\right)}+\int\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy?
For small values of $x$, how good is the approximation $cos(x)\approx 1$? How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy?
My teacher told me, it would be easier to do the second part of the question first, so I need to find out ho... | If you haven't learned about Taylor series.
Will you accept that $\sin x \approx x$ when $x$ is small?
and $\sin x < x$ for all $x.$
You can do this geometrically, plot your unit circle and see that for small $x$, the distance traveled about the $x$ axis is very nearly equal to the length of the curve.
$\cos x = \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$ Although I can determine using a calculator that $\sqrt{5}-\sqrt{3}$ is larger than $\sqrt{15}-\sqrt{13}$, how would I go about proving that? My teacher gave us a hint which was to use the difference of two squares identity $(a^2-b^2) = (a-b)\cdot(a+... | Note that $$\sqrt { 5 } -\sqrt { 3 } =\frac { \left( \sqrt { 5 } -\sqrt { 3 } \right) \left( \sqrt { 5 } +\sqrt { 3 } \right) }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } ,\\ \quad \sqrt { 15 } -\sqrt { 13 } =\frac { \left( \sqrt { 15 } -\sqrt { 13 } \right) \left( \sqrt { 15 } +\sqrt { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
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What is the fewest number of squares required to cover a $11\times13\text{ cm}$ rectangle without overlap? I need help figuring out this math puzzle: I have a $11\times13\text{ cm}$ rectangle and I need help figuring out the least number of squares I need to cover the rectangle without overlap. I'm told the answer shou... | I can prove there is no 5-square solution.
The partitions of $11\times 13 = 143$ into sums of five squares can be enumerated:
$$ \matrix{1^2 &+ 1^2 &+ 2^2 &+ 4^2 &+ 11^2\cr
1^2 &+ 1^2 &+ 4^2 &+ 5^2 &+ 10^2\cr
1^2 &+ 2^2 &+ 5^2 &+ 7^2 &+ 8^2\cr
1^2 &+ 3^2 &+ 4^2 &+ 6^2 &+ 9^2\cr
2^2 &+ 4^2 &+ 5^2 &+ 7^2 &+ 7^2\cr
2^2 &+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$? Given an integer $n \ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \Pi_{k=1}^n k^k = n^n(n-1)^{(n-1)}\ldots2^21^1.$$
It's easy see that, $u(n) \le n^{n(n+1)/2}$, but this is not very interesting.
Update
We have $u(n) \le e^{\left... | Using Jensen's inequality:
Letting $A= \sum_{k=1}^n k = \frac{n (n+1)}{2}$ and $B= \sum_{k=1}^n k^2 = \frac{n (n+1)(2n+1)}{6}$
We have
$$
\begin{align}
\log(u(n))
&=\sum_{k=1}^n k \log(k) \\
&= A \sum_{k=1}^n \frac{k}{A} \log(k) \tag{1}\\
&\le A \log \left( \sum_{k=1}^n \frac{k}{A} k \right) =
A \log \left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
How to solve the following first order ODE I want to solve the following ODE: $$\left(x^2+x y(x)\right) y'(x)+y(x)^2+3 x y(x)=0.$$ However, I didn't find this fit any types in the textbook. For example, I expressed it as $$M(x,y)dx+N(x,y)dy=0$$ and checked whether $M_y=N_x$ which doesn't hold though. Does anyone know h... | $$\left(x^2+x y\right) y'+y^2+3 x y=0$$
Make substitution $y=xz$, where $z$ is function of $x$.
$$
\left(x^2+x^2z\right)\left(z+xz^{'}\right)+x^2z^2+3x^2z=0
$$
Factor out $x^2$ and solve separable ODE
$$
\left(1+z\right)\left(z+xz^{'}\right)+z^2+3z=0\to\\
x\frac{dz}{dx}=-z\frac{z+3}{z+1}-z=-z\frac{2z+4}{z+1}\to\\
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| The quadratic we get is $x^4-2ax^2-x+a^2-a$. We can factor it by trying to guess the coefficients (this will depend heavily on the equation and how good-written the problem is): Let's assume all coefficients will be integers. Let $p(x)=x^2-2ax^2+a^2-a$. There are two cases:
If we could factor $p(x)$ as a product of a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Picking certain number of balls without replacement and finding its probability distribution Suppose I am to choose three balls without replacement from a bag containing $5$ white and $4$ red balls. What will be the probability distribution of the red balls drawn ?.
According to my book, probability function will be
$$... | Simplest Approach
Without replacement, the simplest method is to compute how many ways there are to pick $k$ from the $4$ red balls and $3-k$ from the $5$ white balls. Then divide that by the total number of ways to pick $3$ from the $9$ balls in total:
$$
\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}
$$
Pick by Pic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| $$ax^3+8x^2+bx+6=(ax+c) (x^2-2x-3)=ax^3+(c-2a)x^2+(-2c-3a)x-3c$$
$$\begin{align}6&=-3c\Rightarrow c=-2\\8&=c-2a\Rightarrow 2a=c-8=-10\Rightarrow \color{red}{a=-5}\\b&=-2c-3a=4+15=\color{red}{19}\end{align}$$
Using this approach, you will also find out the third root is given by $ax+c=0\Rightarrow x=-\frac{2}{5}$
Also $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the stor... | We may prove it this way too. (Symbols are a bit different, apologies for that).
We want to show that
$$
\sum_{j=r}^n \binom{j-1}{r-1} (1-p)^{j-r}
= \sum_{j=r}^{n} \binom{n}{j} p^{j-r} (1-p)^{n-j}
$$
For $n=r$, LHS = 1 = RHS.
Assume the result is true for $n = r + m$, we prove it is true for $n = r + m + 1$.
LHS =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Deriving a recurrence relationship for derivatives of $\frac{\arctan(x)}{x}$ I was trying to derive a recurrence relantionship for computing the $k$-th derivative of the function (hoping no error during copy)
$$
f(x) = \frac{\arctan(x)}{x}
$$
Using maple I've seen the following derivatives
$$
\begin{array}{l}
f^{(0)}(x... | $n\in\mathbb{N}$
It's $$\frac{d^n}{dx^n }\arctan x=\frac{d^n}{dx^n}(xf(x))=n\frac{d^{n-1}}{dx^{n-1} }f(x)+ x\frac{d^n}{dx^n }f(x)$$ and for $x>0$ (and proofed by induction) one gets $$\frac{d^n}{dx^n }\arctan x=(-1)^{n-1}(n-1)!\frac{\sin(n \arctan\frac{1}{x})}{\sqrt{1+x^2}^n}$$
It follows that the recursion for the der... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
... | Here's the general method:
You have to find the roots of $\Delta=-15-8i$. Let $(x+iy)^2=\Delta$. This means
$$x^2-y^2+2ixy=-15-8i\tag{1}$$
This means $x^2-y^2=-15$, $\;xy=-4$. As it would be too complicated to eliminate one of the unknowns to determine the other, we'll obtain another equation from the moduli.
Observe t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of positive integer solutions to the equation: $ (x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77. $ Q: Find the number of positive integer solutions to the equation:
$$\begin{align}
(x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77.
\end{align}$$
The solution given is $\begin{pmatrix} 6\\2 \end{pmatrix}$ $\begin{pmatrix} 10\\3 \end{... | Hint:
You probably have specific examples of how to find the number of integer solutions to either $x_1+x_2+x_3=k$ or $y_1+y_2+y_3+y_4=k$, for any constant $k$. (If you don't: think about trying to distribute those $k$ $1$'s into boxes, such that each box gets at least one.)
To reduce there: note that $77=7\cdot11$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How do I go from $33/64 - 2x + x^2 \longrightarrow \bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$?
How do I go from $\frac{33}{64} - 2x + x^2 \longrightarrow\bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$ and then to $\big(x - 1-\... | $$
\begin{align}
&\frac{33}{64} - 2x + x^2 \\
&= x^2 - 2x + \frac{33}{64} \\
&= x^2 - 2x + 1 - 1 + \frac{33}{64} \\
&= (x-1)^2 - 31/64 \\
&= (x-1)^2 - 31/64 \\
&= (x-1)^2 - 31/64 \\
&= \left( (x-1) - \sqrt{31}/8 \right)\cdot\left( (x-1) + \sqrt{31}/8 \right) \\
&= \color{red}{\left( x-1 - \sqrt{31}/8 \right)\cdot\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solutions to ceiling equation system
Prove that there does not exist an $x$ with $1000 \leq x \leq 1990$ that can be expressed in the forms $$\dfrac{10000}{x} = \left \lceil\dfrac{10000}{x} \right \rceil -\dfrac{1}{m} \quad \text{and} \quad \dfrac{10000}{x-1} = \left \lceil\dfrac{10000}{x-1} \right \rceil-\dfrac{1}{k}... | Denote $\lceil\frac{10000}{x}\rceil=A$ and $\lceil \frac{10000}{x-1}\rceil=B$. We have $1000\le x\le 1990$.
A tedious calculation gives five values to be discarded because $A\ne B$ and five sets of candidates to be considered: $$\begin{cases}x=1000 \Rightarrow A=10\text{ and } B=11\\x=1012\Rightarrow A=9\text{ and } B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$
Q: H... | $$\newcommand{\PV}{\operatorname{PV}}
\begin{align}
R(x,y)
&=\frac12\,\PV\!\!\int_0^\infty\frac{\mathrm{d}z}{(z+y)\sqrt{z+x}}\tag{1}\\[6pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\frac{2\,\mathrm{d}z}{z^2-1}\tag{2}\\[3pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
| An interesting way may be the following: if we define
$$ a_n = \int_{0}^{\pi/2}\sin^n(x)\,dx \tag{1}$$
we clearly have that $\{a_n\}_{n\geq 1}$ is a decreasing sequence. On the other hand, integration by parts gives:
$$ a_{2n} = \frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad a_{2n+1}=\frac{1}{2n+1}\cdot\left(\binom{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sov... | Preliminary Result:
$\displaystyle \frac{2}{(x^2+1)(x^3+1)} \equiv \frac{x+1}{x^2+1}-\frac{x^2+x-1}{x^3+1}$
Proof: Obvious.
Consider the parametrised integral:
$\displaystyle f(\alpha) = \int \frac{x^\alpha \, \text{d}x}{(x^2+1)(x^3+1)} = \frac{1}{2} \int \frac{x^{\alpha+1}+x^\alpha}{x^2+1}-\frac{x^{\alpha+2}+x^{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
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Inductive proof for $\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$ I want to prove the following identity using induction (not double counting method). Although it is a specific version of Vandermonde's identity and its inductive proof is presented here, but I need a direct inductive proof on this, not the general f... | Suppose
$$
\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}\tag{1}
$$
then
$$
\begin{align}
\sum_{k=0}^{n+1}\binom{n+1}{k}^2
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]^2\tag{2a}\\
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}^2+\binom{n}{k-1}^2+2\binom{n}{k}\binom{n}{k-1}\right]\tag{2b}\\
&=\binom{2n}{n}\left(1+1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopita... | The following standard formula is well known $$\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = n = \lim_{t \to 0}\frac{(1 + t)^{n} - 1}{t}\tag{1}$$ and it appears that we can go very easily to the next step if $n$ is a positive integer and derive the formula $$\lim_{x \to 1}\frac{x^{n} - 1 - n(x - 1)}{(x - 1)^{2}} = \lim_{t \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 8
} |
A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers
A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers.
My Working
Let original number $n$ be $\ a_1+10a_2+10^2a_3+...+10^{k-1}a_k$.
The new number... | Assume that $n$ has at least two digits, then $n=10a+b$ where $a\geq 1$ and $b\in \{0,1,2,3,4,5,6,7,8,9\}$.
By assumption $a$ divides $n$, therefore $a$ divides also $b=n-10a$.
Hence if $b\not=0$ then $a\leq b\leq 9$, which implies that $n$ has only two digits.
It follows that the number has this property iff it ends... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Obtaining the bound without using only the trivial inequality
If $x,y,z$ are 3 real numbers such that $$x+y+z = 4$$ and $$x^2 + y^2 + z^2 = 6$$ show that each of $x,y,z$ lie in the closed interval $[\frac{2}{3},2]$
I obtained the upper bound, but I am unable to obtain the lower bound.
My approach
$x^2 + y^2 + (4-(x+... | $y+z=4-x$ and $(4-x)^2-2yz=6-x^2$, which gives $yz=x^2-4x+5$.
Thus, $(4-x)^2-4(x^2-4x+5)\geq0$, which is $(x-2)(3x-2)\leq0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$. Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$.
Since $p \mid m^2 + 1$, we have $m^2 \cong -1( \mod p)$ and h... | Let $p$ divide some numbers of the form $m^2 + 1$ and $n^2 + 2$. Then both $-1$ and $-2$ are quadratic residues modulo $p$. This means that $p \equiv 1 \pmod 4$ and that $p \equiv 1,7 \pmod 8$. This means that $p$ is of the form $8s + 1$.
Now from the Euler's Criterion we have that an coprime number $a$ of $p$ is a biq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_... | Hint: $x^2 - 1 = (x - 1)(x + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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By Induction, prove $1^n+2^n+3^n+...+n^n < (n+1)^n$ Here is the question:
Given that $$x^n + n( x^{n-1} ) \leq (x + 1) ^n \tag{A}$$
Prove: $$1^n + 2^n +3^n +...+n^n < (n+1)^n. \tag{B}$$
Here is my thinking.
Replace $n$ with $x$ in A can get $n^n + n(n^{n-1})$ which is $2n^n$
In inductive step, $1^{n+1} + 2^{n+1} + ..... | A rough sketch of the induction step for n:
Assume $1^n + \dots + n^n < (n+1)^n$. Our goal is to show that $1^{n+1} + \dots + (n+1)^{n+1} < (n+2)^{n+1}$. So,
\begin{alignat}{2}
(n+2)^{n+1} &\geq (n+1)^{n+1} + (n+1)(n+1)^n &\qquad \mbox{(by A)} \\
&> (n+1)^{n+1} + (n+1)(1^n + \dots + n^n) &\\
&\geq (n+1)^{n+1} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Confused with changes of bases. I have a problem to solve, and it makes me realise that I have a bad understanding of basis change and the linear applications that go along. I have a linear application who's matrix is expressed like this:
$A=\begin{bmatrix}
\frac{4a-b}{3}&\frac{2a-2b}{3} \\
\frac{-2a+2b}{3}& \frac{-... | HINT.-The determinant of $A$ is equal to $ab$ so if $ab\ne 0$ then the matrix is invertible. If you want to know the base $\{b_1,b_2\}$ where $b_1=\begin{pmatrix} x_1\\y_1\end{pmatrix}$ and $b_2=\begin{pmatrix} x_2\\y_2\end{pmatrix}$
you have to solve
$$A\begin{pmatrix} x_1\\y_1\end{pmatrix}=\begin{pmatrix} \frac{4a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \bino... | Here is an answer based upon a transformation of generating series.
We show
\begin{align*}
\sum_{j=0}^k\binom{k-j}{j}(-1)^j
=\frac{(-1)^{\lfloor k/3\rfloor}+(-1)^{\lfloor (k+1)/3\rfloor}}{2}
\qquad\qquad k\geq 0\tag{1}
\end{align*}
where $\lfloor x \rfloor$ denotes the floor function. We set as upper limit of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$. I need to describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$.
After testing out some $n \in \mathbb{N}$, I came to the conclusion that $3^{n} - 2^{n}$ is divisible by $5$ iff $n$ is even - i.e., if $n$ is of the for... | I think this is a little easier to see using modular arithmetic. Note that $5$ divides $3^n - 2^n$ means $3^n \equiv 2^n \pmod{5}$. Since $3 \cdot 2 = 6 \equiv 1 \pmod{5}$, then $3 = 2^{-1}$ in $\mathbb{Z}/5\mathbb{Z}$. Then
\begin{align*}
2^n \equiv 3^n \equiv (2^{-1})^n = 2^{-n} \pmod{5}
\end{align*}
and multiplyin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find a Sum for $\left(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\right)^2$ How can one find a formula for $\big(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\big)^2$ in the form
$$
\Big(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\Big)^2=\sum _{i=0}^N A_i \cos (B_i \,x),
$$
where $N$ depends only on $n,$ and the coeff... | We'll show that
\begin{align}\boxed{\text{ }
\\\quad\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2= \frac{n+1}{2}+\sum_{k=1}^{n-1} \Big(n-\frac{k-1}{2} \Big)\cos(kx)+\sum_{k=n}^{2n-2} \Big(n-\frac{k+1}{2} \Big)\cos(kx).\quad\\}
\end{align}
Note that this is in the form requested since the constant term is just the $\cos (0x)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Prove $4p-3$ is a square knowing that $n\mid p-1$ and $p\mid n^3-1$, $p$ prime I really need some help at this problem:
Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is a square.
So $p \mid (n-1)(n^2+n+1)$
What if $p \mid n-1$?
Treating t... | Note $p \mid n-1$ is impossible because $n \le p-1$, so we have $p \mid n^2+n+1$.
Since $n \mid p-1$, we can write $p = an+1$ for some integer $a \ge 1$. Since $p \mid n^2+n+1$, we can write $$n^2 + n + 1 = bp = b(an+1)$$ for some integer $b \ge 1$.
Reducing modulo $n$ gives $1 \equiv b \pmod{n}$, so write $b = rn+1$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove this trigonometry equation: $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
Prove that $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
I've tried writing $\sin 40^\circ$ as $\sin(40^\circ+10^\circ)$, then wrote $\sin(50^\circ+10^\circ)$ as $\sin 40^\circ... | You can use the general formula
$$ \sin\alpha \sin\beta = \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}.$$
There are also the similar formulas
$$ \cos\alpha \cos\beta = \frac{\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2}$$
and
$$ \sin\alpha \cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Right Triangles and a Circle In a test sometime ago, I had to deal with the following question:
Let $\triangle ABC$ be a right angled triangle. Let $\triangle DEF$ be a right triangle inscirbed in the incircle of $\triangle ABC$. Find the smallest possible ratio of $\frac{[ABC]}{[DEF]}$ where $[ABC]$ denotes the area... | If a right triangle $DEF$ is inscribed in a circle with radius $r$, its maximum area is given by $r^2$, since the longest side of such a triangle is also a diameter of the circumcircle of $DEF$. It follows that $\frac{[ABC]}{[DEF]}$ is lower bounded by $\frac{[ABC]}{r^2}$ where $r$ is the inradius of $ABC$. On the othe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability that the lot contains defective articles. I tried much in the problem but I didn't get my answer correct.
The question is---
A lot contains 20 articles.the probability that the lot contains exactly 2 defective articles is 0.4 and that the lot contains exactly 3 defective articles is 0.6.articles are drawn f... | If the lot contains two defective articles and the testing procedure ends at the $12^{th} $ testing, then the first defective article must be chosen among the first $11$ and the second must be the $12^{th} $. The probability that this occurs with the first defective article taken at the first test is
$$ \frac{2}{20} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Second partial derivative with $f(x,y)=x^3+5x^2y+y^3$ I have this problem:
I found $F_x=3x^2+10yx$ and $F_y=5x^2+3y^2$, then $D_uf=\frac{9}{5}x^2+6yx+4x^2$
$F_{x2}= \frac{58}{5}x + 6y, F_{y2}=6x+{24}{5}y$
$D_uf_2=\frac{174}{25}x+\frac{18}{5}y+\frac{32}{5}x+\frac{96}{25}y=>at(2,1)=>534/25$
The answer however is $\frac... | If $f(x,y)= x^3 + 5x^2y + y^3 $ and $u=\langle\frac35, \frac45\rangle $
$$\nabla f(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle$$
$$D_uf(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle \cdot \langle\frac{3}{5},\frac{4}{5}\rangle =\frac{1}{5}(29x^2 + 30xy +12y^2) $$
$$D^2_uf(x,y)=\frac{3}{5}\underbrace{\frac{1}{5}(58x+30... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A trigonometric equation - Finding the value of theta I have to find the value of $\theta$ in this equation "$\cos2\theta \sin\theta = 1$"
The problem is that I have to solve using algebra (making equations...etc) like this one, for example: \begin{align} 4\sin^2\theta - 1 = 0\\ \sin^2\theta = 0.25\\ \sin\theta = +0.5,... | Hint:
Use the identity-$\cos2\theta=\cos^2\theta-\sin^2\theta$
So,$$\cos2\theta\sin\theta=1$$
$$\implies(\cos^2\theta-\sin^2\theta)\sin\theta=1$$
Now,replace $\cos^2\theta$ by $1-\sin^2\theta$
So,$$(1-2\sin^2\theta)\sin\theta=1$$
$$\implies\sin\theta-2\sin^3\theta-1=0$$
Now,let,$\sin\theta=x$.So,this reduces to a cubic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Difficulty in the evaluation of double integral of a continuous function. Evaluate the double integral
$$\iint_D \frac{xy}{\sqrt{1-y^2}} dxdy,$$
where $D$ is the region of the first quadrant bounded by standard circle with unit radius $1$: that is $x^2+y^2=1$.
I have difficulty with its evaluation.
First of all I wa... | The fact that $D$ is a disc tells you that you should use polar coordinates. Therefore, let $x = r \cos t$ and $y = r \sin t$ with $r \in [0,1]$ and $t \in [0, \frac \pi 2]$, in order to transform your integral into
$$\iint \limits _{[0,1] \times [0, \frac \pi 2]} \frac {r \cos t r \sin t } {\sqrt {1 - r^2 \sin^2 t}} r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$ $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$
without D l hospital rule and series expansion.
i have s... | $b_n:=1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x$
$b_n-1=(b_{n-1}-1)\cdot \cos a_n x$ => $b_n=b_{n-1}\cdot \cos a_n x+1-\cos a_n x$
$\lim_{x\to 0}\frac{b_n}{x^2}=\lim_{x\to 0}(\frac{b_{n-1}}{x^2\cos a_n x}+\frac{1-\cos a_n x}{x^2\cos a_n x})=\lim_{x\to 0}\frac{b_{n-1}}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Real values of $k$ for which $x^2+(k+1)x+k^2$ has one root double the other
For what real values of $k$ does $x^2+(k+1)x+k^2$ have one root double the other?
For a start, I found the range of $k$ which endows this equation with real roots:
$$-\frac13\le k\le1$$
| Suppose the equation has one root double the other. Then it can be written as
$$(x-a)(x-2a)=x^2-3ax+2a^2$$
Comparing coefficients between this expression and $x^2+(k+1)x+k^2$ we have
$$k+1=-3a\qquad2a^2=k^2$$
From the first equation we have $k=-3a-1$; substituting this into the second equation yields
$$2a^2=(-3a-1)^2=9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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.